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2025-03-21T14:48:30.190662
| 2020-04-01T05:16:02 |
356248
|
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|
Stack Exchange
|
References for the connectivity of complements of smooth submanifolds
Suppose $M$ is a smooth $m$-manifold and $Z$ a codimension-$d$ closed submanifold. Neither $M$ nor $Z$ has a boundary and neither is assumed to be compact. I believe that the inclusion $M \setminus Z \to M$ is $d-1$-connected, in that one can construct an argument based on the argument of Proposition 4.1 of https://people.math.osu.edu/anderson.2804/eilenberg/appA.pdf
The idea is that to show $i_*: \pi_n(M\setminus Z, m_0) \to \pi_n(M, m_0)$ is surjective when $n \le d-1$, you can take any class in the target and find a representative $f: S^n \to M$ that is smooth and meets $Z$ transversely. Dimension counting then shows that $f(S^n) \cap Z = \emptyset$, so the class is in the image of $i_*$.
To show it is injective when $n < d-1$, you argue in a similar way, but this time with smooth maps $F: D^{n+1} \to M$ that restrict to $f: S^n \to M \setminus Z$ on the boundary, where $f$ represents a class in the kernel of $i_*$. Again, you can choose a $D^{n+1}$ that is smooth and transverse to $Z$, and again conclude by counting dimensions.
Does anyone know of a reference for this fact in the (published) literature?
This is Theorem 1.1.4 in the notes https://ivv5hpp.uni-muenster.de/u/jeber_02/skripten/bordism-skript.pdf . It seems hard to track down a published reference. Many authors would just say ``By a standard transversality argument...".
Thanks! I'm glad it appears somewhere online. The treatment of transversality in Guillemin & Polalck's Differential Topology (https://www.ams.org/books/chel/370/) is sufficiently strong to prove the result without too much extra work, since it handles manifolds with boundary. Other treatments I found restrict to manifolds without boundary for transversality, so aren't quite as convenient.
|
2025-03-21T14:48:30.190821
| 2020-04-01T06:32:21 |
356251
|
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"GH from MO",
"Vesselin Dimitrov",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356251"
}
|
Stack Exchange
|
Analytic equivalents for primes in arithmetic progressions
By way of context: it is known that the prime number theorem $\pi(x) \sim x/\log x$ is (nontrivially) equivalent to the statement that $\zeta(s)$ does not vanish on the line $\Re s=1$.
I would like to have it clarified what are the analogous equivalent statements for primes in arithmetic progressions. In particular, I would love to have specific references to papers or books in which the following equivalences are proved (or at least stated explicitly):
Dirichlet's theorem, that there are infinitely many primes in any admissible arithmetic progression (mod $q$). is equivalent to the nonvanishing of $L(1,\chi)$ for all Dirichlet characters $\chi\pmod q$. [Edit: as has been pointed out, this nonvanishing is more likely equivalent to the equality of the Dirichlet densities of primes among the reduced residue classes (mod $q$). What analytic statement could be equivalent to the mere infinitude of primes in all such classes?]
The prime number theorem in arithmetic progressions $\pi(x;q,a)\sim x/(\phi(q)\log x)$ is equivalent to the statement that $L(s,\chi)$ does not vanish on the line $\Re s=1$ for all Dirichlet characters $\chi\pmod q$.
Of course, if these statements themselves are incorrect, I would like to be corrected as well as being pointed to the literature.
I am confused. Dirichlet's theorem, the PNT in arithmetic progressions, and the nonvanishing of $L(s,\chi)$ for $\Re(s)=1$ are three true statements. True statements are always equivalent.
One sense in which such statements are equivalent is in that analogous statements are known to be equivalent in greater generality, for instance for L-functions in the Selberg class. You can deduce equivalence for PNT and for PNT in APs this way. As for Dirichlet's Theorem, I doubt it's equivalent to nonvanishing at 1. Instead, I would expect nonvanishing to be equivalent to the result about Dirichlet density of primes in APs.
Here are two equivalences.
Theorem 1. For each $m \geq 1$, the following are equivalent.
a) For all nontrivial Dirichlet characters $\chi \bmod m$, $L(1,\chi) \not= 0$.
b) For all $a \in (\mathbf Z/m\mathbf Z)^\times$, the set of primes $p \equiv a \bmod m$ has Dirichlet density $1/\varphi(m)$.
Proof of Theorem 1.
We will will compute the Dirichlet density of $\{p \equiv a \bmod m\}$ without assuming (a) and then see why (a) and (b) are equivalent. The trivial Dirichlet character modulo $m$ will be written as $\mathbf 1_m$.
For each Dirichlet character $\chi \bmod m$, $L(s,\chi)$ is analytic for ${\rm Re}(s) > 0$ except for $L(s,{\mathbf 1}_m)$ having a simple pole at $s = 1$. Set
$$
n(\chi) := {\rm ord}_{s=1}(L(s,\chi))
$$
so $n({\mathbf 1}_m) = -1$ and $n(\chi) \geq 0$ for all nontrivial $\chi$.
For ${\rm Re}(s) > 1$ and $(a,m) = 1$,
$$
\sum_{p \equiv a \bmod m} \frac{1}{p^s} =
\frac{1}{\varphi(m)} \sum_{p}\sum_{\chi} \frac{\chi(p)\overline{\chi}(a)}
{p^s} =
\frac{1}{\varphi(m)} \sum_{\chi}\overline{\chi}(a)\left(\sum_{p} \frac{\chi(p)}
{p^s}\right)
$$
where the sum on the right run over all primes $p$ and all Dirichlet characters $\chi \bmod m$.
For a Dirichlet character $\chi \bmod m$ and ${\rm Re}(s) > 1$,
$$
\log L(s,\chi) = \sum_p \frac{\chi(p)}{p^s} + \sum_{p,k\geq 2} \frac{\chi(p^k)}{kp^{ks}} = \sum_p \frac{\chi(p)}{p^s} + O(1),
$$
where the $O$-constant is $\sum_{p,k \geq 2} 1/(kp^k)$, so
$$
\sum_{p \equiv a \bmod m} \frac{1}{p^s} =
\frac{1}{\varphi(m)}\sum_{\chi \bmod m} \overline{\chi}(a)\log L(s,\chi) + O(1).
$$
Now let's bring in the order of vanishing $n(\chi)$ above. For $s$ near $1$, $L(s,\chi) = (s-1)^{n(\chi)}f_\chi(s)$ where $f_\chi(s)$ is an analytic function in a neighborhood of $s = 1$ and $f_\chi(1) \not= 0$. Therefore $f_\chi(s)$ has an analytic logarithm around $s = 1$ (well-defined up to adding an integer multiple of $2\pi i$), so for $s > 1$,
$\log L(s,\chi) = n(\chi)\log(s-1) + \ell_{f_\chi}(s)$,
where $\ell_{f_\chi}(s)$ is a suitable logarithm of $f_\chi(s)$. Thus
$$
\log L(s,\chi) = n(\chi)\log(s-1) + O_\chi(1)
$$
for $s$ near $1$ to the right, and plugging this into the above displayed formula,
\begin{align}
\sum_{p \equiv a \bmod m} \frac{1}{p^s} & = \frac{1}{\varphi(m)}\sum_{\chi \bmod m} \overline{\chi}(a)(n(\chi)\log(s-1)+ O_\chi(1)) + O(1) \nonumber \\
& =\frac{1}{\varphi(m)}\left(\sum_{\chi} \overline{\chi}(a)n(\chi)\right)\log(s-1) + O_m(1).
\end{align}
To compute a Dirichlet density,
we want to divide both sides by $\sum_p 1/p^s$ for
$s$ near $1$ to the right. For such $s$,
$$
\log \zeta(s) = \sum_p \frac{1}{p^s} + O(1) = -\log(s-1) + O(1).
$$
Therefore $\sum_p 1/p^s \sim -\log(s-1)$ as $s \to 1^+$, so
dividing through by $\sum_p 1/p^s$ and letting $s \to 1^+$ gives us
\begin{equation}
\lim_{s \to 1^+}
\frac{\sum_{p \equiv a \bmod m} 1/p^s}{\sum_p 1/p^s} =
\frac{1}{\varphi(m)}\left(-\sum_{\chi} \overline{\chi}(a)n(\chi)\right),
\end{equation}
which expresses the Dirichlet density of $\{p \equiv a \bmod m\}$ in terms of the
orders of vanishing $n(\chi)$ as $\chi$ runs over Dirichlet characters mod $m$.
If (a) is true then $n(\chi) = 0$ for all nontrivial $\chi$, so the right side of the above limit calculation
is $(1/\varphi(m))(-n({\mathbf 1}_m)) = 1/\varphi(m)$, which is (b).
Conversely, if (b) is true then
$$
\sum_{\chi} \overline{\chi}(a)n(\chi) = -1
$$
for all $a \in (\mathbf Z/m\mathbf Z)^\times$ by our limit calculation. Why does this imply $n(\chi) = 0$ for nontrivial $\chi$?
Using complex vectors indexed by all the Dirichlet characters mod $m$, let
${\mathbf n}_m = (n(\chi))_\chi$ and
${\mathbf v}_a = (\chi(a))_\chi$ for each $a \in (\mathbf Z/m\mathbf Z)^\times$. The space of all complex vectors $\mathbf z = (z_\chi)_\chi$ has
dimension $\varphi(m)$ and it has the Hermitian inner product
$\langle \mathbf z, \mathbf w\rangle = \frac{1}{\varphi(m)}\sum_{\chi} z_\chi\overline{w_\chi}$ for which the vectors ${\mathbf v}_a$ are an orthonormal basis by the orthogonality relations for Dirichlet characters mod $m$. The above displayed formula
says $\langle {\mathbf n}_m,{\mathbf v}_a\rangle = -1/\varphi(m)$ for all $a$ in $(\mathbf Z/m\mathbf Z)^\times$, so
$$
{\mathbf n}_m = \sum_{a} \langle {\mathbf n}_m,{\mathbf v}_a\rangle{\mathbf v}_a = -\frac{1}{\varphi(m)}\sum_{a}{\mathbf v}_a.
$$
For each nontrivial $\chi \bmod m$, the $\chi$-component of
$\sum_{a} {\mathbf v}_a$ is $\sum_a \chi(a)$, which is $0$. So the $\chi$-component of ${\mathbf n}_m$, which is $n(\chi)$, is 0. That is (a).
QED Theorem 1. (I only realized after copying and pasting this that I had already copy and pasted it earlier as an answer to the MO question here.)
Theorem 2. For each $m \geq 1$, the following are equivalent.
a) For all Dirichlet characters $\chi \bmod m$, $L(s,\chi) \not= 0$ when ${\rm Re}(s) = 1$.
b) $\sum_{n \leq x} \chi(n)\Lambda(n) = o(x)$
for nontrivial Dirichlet characters $\chi \bmod m$
and $\sum_{n \leq x} \chi_{{\mathbf 1}_m}(n)\Lambda(n) \sim x$,
c) For all $a \in (\mathbf Z/m\mathbf Z)^\times$, $|\{p \leq x : p \equiv a \bmod m\}| \sim (1/\varphi(m))x/\log x$.
Comparing the proof of Theorem 2 below to the sketch in the answer by<PHONE_NUMBER>, we will also be using a Tauberian theroem (to prove (b) implies (c)), but we will not need an explicit formula.
Proof of Theorem 2.
We will show (a) is equivalent to (b) and (b) is equivalent to (c).
First we show (a) implies (b).
Set $\psi_\chi(x) = \sum_{n \leq x} \chi(n)\Lambda(n)$ for all $\chi$, so
(b) says $\psi_\chi(x) = o(x)$ for nontrivial $\chi$ and $\psi_{{\mathbf 1}_m}(x) \sim x$.
For $\sigma > 1$, $-L'(s,\chi)/L(s,\chi) = \sum \chi(n)\Lambda(n)/n^s$, for all Dirichlet characters $\chi \bmod m$,
so $\psi_\chi(x)$ is a partial sum of coefficients of $-L'(s,\chi)/L(s,\chi)$.
Since $L(s,{\mathbf 1}_m) \not= 0$ on $\sigma = 1$ by (a), $-L'(s,{\mathbf 1}_m)/L(s,{\mathbf 1}_m)$ is holomorphic on $\sigma \geq 1$ except for a simple pole at $s = 1$ with residue 1 and it has nonnegative Dirichlet series coefficients with $\psi_{{\mathbf 1}_m}(x) = O(x)$. Therefore $\psi_{{\mathbf 1}_m}(x) \sim x$, which is part of (b), by Newman's Tauberian theorem. To get the rest of (b), namely $\psi_\chi(x) = o(x)$ for nontrivial $\chi$, we have $-L'(s,\chi)/L(s,\chi)$ being holomorphic on $\sigma \geq 1$ by (a) and its Dirichlet series coefficients satisfy $|\chi(n)\Lambda(n)| \leq {\mathbf 1}_m(n)\Lambda(n)$ for all $n$, so $\psi_\chi(x) = o(x)$ by a corollary of Newman's Tauberian theorem for $-L'(s,\chi)/L(s,\chi)$ using comparison Dirichlet series $-L'(s,{\mathbf 1}_m)/L(s,{\mathbf 1}_m)$ .
Thus (a) implies (b).
To show (b) implies (a), we will use the following fact. For a function $a(x)$ on $[1,\infty)$ that is bounded and Riemann integrable on $[1,T]$ for all $T \geq 1$, so $f(s) := \int_1^\infty (a(x)/x^s)dx/x$ is absolutely convergent on $\sigma > 1$, if $a(x) \to 0$ as $x \to \infty$ and $f$ extends to a meromorphic function on $\sigma = 1$ then $f$ in fact is holomorphic on $\sigma = 1$. (This is used to show the condition $\psi(x) \sim x$ implies $\zeta(s) \not= 0$ on $\sigma = 1$ by using $a(x) = \psi(x)/x - 1$.) Because of the integral representations
$$
-\frac{L'(s,\chi)}{sL(s,\chi)} = \int_1^\infty \frac{\psi_\chi(x)}{x} \frac{dx}{x^s}
$$
and
$$
-\frac{L'(s,{\mathbf 1}_m)}{sL(s,{\mathbf 1}_m)} - \frac{1}{s-1} = \int_1^\infty \left(\frac{\psi_{\mathbf 1}(x)}{x} -1\right)\frac{dx}{x^s},
$$
for ${\rm Re}(s) > 1$,
where $\chi$ is nontrivial in the first equation.
we can use the above fact when $a(x) = \psi_\chi(x)/x$ for nontrivial $\chi$ and
$a(x) = \psi_{{\mathbf 1}_m}(x)/x - 1$ to conclude that $L'(s,\chi)/L(s,\chi)$ is holomorphic on
$\sigma = 1$ for nontrivial $\chi$ and $L'(s,{\mathbf 1}_m)/L(s,{\mathbf 1}_m)$ is holomorphic on $\sigma = 1$ except for a simple pole at $s = 1$, so
$L(s,\chi)$ is nonvanishing on $\sigma = 1$ and
$L(s,{\mathbf 1}_m)$ is nonvanishing on $\sigma = 1$.
Thus (b) implies (a).
That (b) implies (c) follows from the above integral representation of $-L'(s,\chi)/L(s,\chi)$ for all nontrivial $\chi$ by a standard method to prove (c).
Our last step is showing (c) implies (b). Set $\pi(x;a \bmod m) = |\{p \leq x : p \equiv a \bmod m\}|$ when $(a,m) = 1$ and $\pi_\chi(x) = \sum_{p \leq x} \chi(p)$,
where $\chi$ is a Dirichlet character mod $m$.
Write $\chi$ as a linear combination of
delta-functions on $(\mathbf Z/m\mathbf Z)^\times$:
$\chi = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\delta_a$. Then
\begin{align*}
\pi_\chi(x) & = \sum_{p \leq x} \chi(p) \\
& = \sum_{p \leq x} \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\delta_a(p) \\
& = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\left(\sum_{p \leq x} \delta_a(p)\right) \\
& = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\pi(x; a \bmod m),
\end{align*}
so
$$
\frac{\pi_\chi(x)}{x/\log x} = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\frac{\pi(x;a \bmod m)}{x/\log x}.
$$
By (c), as $x \to \infty$ the right side tends to
$\sum_{a \in ({\mathbf Z}/m{\mathbf Z})^\times} \chi(a)/\varphi(m)$, which is 0 if $\chi$ is nontrivial. Therefore when $\chi$ is nontrivial we have $\pi_\chi(x) = o(x/\log x)$, which implies $\psi_\chi(x) = o(x)$ by
the same argument that $\pi(x) \sim x/\log x$ implies $\psi(x) \sim x$. To show (c) implies
$\psi_{{\mathbf 1}_m}(x) \sim x$, sum the relation in (c) over all $a$ in $(\mathbf Z/m\mathbf Z)^\times$ to
get $\pi(x) \sim x/\log x$, the Prime Number Theorem, which
is equivalent to
$\psi(x) \sim x$, so
$\psi_{{\mathbf 1}_m}(x) \sim x$ since $\psi_{{\mathbf 1}_m}(x) = \psi(x) + O_m(\log x)$.
QED Theorem 2.
Update (2022): The prime number theorem
$\pi(x) \sim x/\log x$ is equivalent to two properties involving the Moebius function: $\sum_{n \leq x} \mu(n) = o(x)$ and
$\sum \mu(n)/n = 0$. These each have analogues in Theorem 2. See here.
Regarding Dirichlet's Theorem, I think you need the quantitative version (in terms of Dirichlet density) for the equivalence to hold.
Suppose
$$\sum_{\substack{p\equiv a(q)\\ p<x}} \frac1p = \frac{\chi_0(a)}{\phi(q)} \log\log x+O(1).$$
Then summing by parts gives for $s=1+\delta$ that, uniformly in $\delta$,
\begin{align*}
\sum_p \chi(p)p^{-s}
&= \sum_{a(q)} \chi(a)\sum_{p\equiv a(q)} p^{-s} \\
&= \sum_{a(q)} \chi(a) \sum_n \bigg(\frac1{\phi(q)}\log\log n +O(1)\bigg)\big(n^{-\delta}-(n+1)^{-\delta}\big) \\
&= \left(\sum_{a(q)} \chi(a)\right) \left (\sum_n \frac{\log\log n}{\phi(q)}\big(n^{-\delta}-(n+1)^{-\delta}\big) \right) \\ &\qquad{}+ \sum_{a(q)} \chi(a) \sum_n O(1)\big(n^{-\delta}-(n+1)^{-\delta}\big) \\
&\ll \sum_{n\geq 1} \big(n^{-\delta}-(n+1)^{-\delta}\big) \ll 1
\end{align*}
And since $\log L(s,\chi) = \sum_p \chi(p)p^{-s} + O(1)$ we see that $L(1,\chi)\neq 0$.
On the other hand if you only told me there were infinitely many primes in each AP, but with different asymptotics for different $a$ mod $q$, I don't think the conclusion would follow.
I have no idea if this is written up anywhere but I'll now add it to the homework in my course notes.
In (2.27) of Iwaniec and Kowalski's book, it is explained how the divergence of $\sum 1/p$ over just the one arithmetic progression $1 \mod{N}$ yields at once the non-vanishing of all $L(1,\chi)$ to the conductor $N$.
For $\pi(x;q,a)\sim x/(\varphi(q)\log x)$: I don't know of a source that states such results for progressions to modulus $q$ except on p. 40 of Iwaniec and Kowalski, and only for $q=1$. Davenport does not seem to say anything on the matter. My next guess would be to check Montgomery and Vaughan or Chandrasekharan. One direction follows from the explicit formula, and the other direction follows from applying the Weiner-Ikehara tauberian theorem to each $-L'/L(s,\chi)$ and combining them via the orthogonality relations.
For the infinitude of primes in progressions, we have (for $s>1$)
$\displaystyle \sum_{p\equiv a\pmod{q}}p^{-s} = \frac{1}{\varphi(q)}\log\frac{1}{s-1}+\frac{1}{\varphi(q)}\sum_{\substack{\chi\pmod{q} \\ \chi\neq 1}}\bar{\chi}(a)\log L(s,\chi)+O_q(1)$
The boundedness of $\log L(s,\chi)$ for all $s>1$ is equivalent to the nonvanishing of $L(1,\chi)$ by means of Dirichlet's test for uniform convergence. Thus
$\displaystyle \sum_{\substack{p\equiv a\pmod{q}}}p^{-s}=\frac{1}{\varphi(q)}\log\frac{1}{s-1}+O_q(1)$ for all $s>1\iff L(1,\chi)\neq 0$ for all $\chi\pmod{q}$.
(See Chapter 1 of Davenport.) But this statement is stronger than merely the infinitude of primes in progressions.
|
2025-03-21T14:48:30.191590
| 2020-04-01T06:48:06 |
356252
|
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|
Stack Exchange
|
Algebras of the cone monad on Top?
Let us work in Top, the category of topological spaces - although the reader is welcome to replace this by their favorite convenient category of topological spaces.
If $X,Y$ are spaces, let $X\ast Y$ denote the join of spaces, $X\coprod X\times I\times Y\coprod Y/\sim$, where $\sim$ is the equivalence relation generated by $x\sim (x, 0,y)$ and $(x,1,y)\sim y$ for any $x,y$.
A good way to think of a point in the join is in the style of barycentric coordinates - if $X, Y$ are both nonempty, the join is homeomorphic to the set of formal linear combinations $t_0x+t_1y$, with $t_0,t_1\geq 0$ and $t_0+t_1= 1$.
It is not hard to see that $\ast$ is a monoidal bifunctor, with unit $\emptyset$. For the associativity axiom, note that $X\ast Y\ast Z$ (again assumed nonempty)can be naturally identified with the set of formal linear combinations $t_0x+t_1 y + t_2z$, $(t_0,t_1,t_2)$ a coordinate in the standard $2$-simplex; if $t_0'(t_0x + t_1y)+ t_1'z$ is a point in $(X\ast Y)\ast Z$, we send it to $t_0t_0'x + t_0't_1y + t_1'z$.
The singleton set $1$, i.e. the $0$-simplex, is a monoid with respect to this monoidal product. The unit maps $\emptyset\to 1$ and $1\ast 1\to 1$ need not be described, as they are unique. Associativity of multiplication is trivial, etc.
The $n$-fold join $1\ast\dots\ast 1$ can be naturally identified with the $(n-1)$-simplex. By the universal property of the augmented simplex category $\Delta_a$, the category of finite ordinals and nondecreasing maps between them, associated to the choice of monoidal product $\ast$ and and monoid $1$ there is a functor $\Delta_a\to \mathbf{Top}$ sending $[0]$ to $1$, sending the unit $\emptyset=[-1]$ to the unit $\emptyset$, and respecting the monoidal product, up to isomorphism. This functor is the standard embedding of the simplex category $\Delta_a$ into $\mathbf{Top}$ which sends $[n]$ to $\Delta^n$.
Because $1$ is a monoid, the functor $1\ast - : \mathbf{Top}\to \mathbf{Top}$ - commonly known as the cone functor - is a monad in a natural way. The multiplication can be described as follows: If $X$ is a given space, let $p_0, p_1$ denote two singleton sets; then the elements of $p_0\ast(p_1\ast X)$ are represented by formal linear expressions $t_0p_0 + t_1(t_0'p_1+t_1'x)$, where $t_0,t_1,t_0',t_1'$ are all $\geq 0$, and $t_0+t_1=1,t_0'+t_1'=1$. The multiplication identifies $p_0$ with $p_1$ and anything on the line between them, so $\mu : p_0 \ast (p_1\ast X) \to p_2 \ast X$ sends $t_0p_0 + t_1(t_0'p_1+t_1'x)$ to $(t_0 + t_1t_0')p_2 + t_1t_1'x$. If $h : 1\ast X\to X$ is an action, and representing $t_0p_0 + t_1x$ by the pair $(t_0,x)$ for concision, we can write that the associativity axiom of the monad asks that $h(t_0,h(t_0',x))= h(t_0+t_1t_0',x)$
My question is - what are the algebras of this monad? It is clear that an algebra of the monad is a contractible topological space, as the unit map $\eta : X\to C(X)$ sends the point $x$ to $0\cdot p_0 + 1\cdot x$. A left inverse to this map would necessarily, it is clear, contain the data of a contracting homotopy, as it would send the singleton $p_0$ to some distinguished point $x_0\in X$, and to every other $x$ it would associate a path from $x_0$ to $x$.
The associativity axiom is interesting, though. Translated from diagrammatic language into equational, if $h$ is an algebra map, viewed as a homotopy $h:I\times X\to X$, it should satisfy (for some distinguished choice of basepoint $x_0\in X$)
$$h(1,x) = x_0$$
$$h(0,x) = x$$
$$h(t,h(t',x)) = h(t+(1-t)t',x)$$
for all $t,t'$ in $[0,1]$.
Now, if $X$ is some convex polyhedron which is a subset of $\mathbb{R}^n$, such a contracting homotopy is easily found: set $h(t,x) = tx_0 + (1-t)x$. So motivated by this example, let us call such a contracting homotopy a "straight-line homotopy." So algebras of the cone monad are precisely the spaces admitting a contracting straight-line homotopy.
My question is: What spaces are these? Is there a nice characterization? Does every contractible space admit a contraction by a straight-line homotopy? (I have tried to show this directly by coming up with some clever elementary argument involving reparametrizing the rate of an arbitrary homotopy, in the style of standard lemmas in Chapter 0 of Hatcher, but so far it evades me.)
Could you write down an explicit multiplication $\mu$ for the cone monad? I think there's some choice there (up to homotopy?). In the last equation you're getting the $t + (1 - t) t'$ from a particular choice of $\mu$. I wonder if a different (but equivalent) $\mu$ would give you something more easily manageable (something involving $\max$ and $\min$).
Hi Prof. Bauer; let $p_0$ and $p_1$ denote two singleton points. Under the identification of $p_0\ast (p_1\ast X)\cong p_0\ast p_1 \ast X$ I mention above, a point in ${p_0}\ast (p_1 \ast X)$ is a formal linear combination $t_0p_0 + t_1p_1 + t_2 x$ for some $x$, where $(t_0,t_1,t_2)$ is a point in the standard $2$-simplex with its usual embedding in $\mathbb{R}^3$. The multiplication should send this to $(t_0+t_1)p_0 + t_2x$ in $p_0\ast X$. Here $t_0$ and $t_1$ are my $t, t'$ in the original post respectively.
I do not quite understand your comment about replacing $\mu$ with an equivalent one. If we work in Top, $\mu$ should be pinned down up to various commuting homeomorphisms; I had not considered the problem in the homotopy category, perhaps I just assumed that the problem would become trivial there.
I have edited the post to explain how I came up with the multiplication of the monad.
Things will work better if you reverse the direction of $I$ (i.e. conjugate by $t\mapsto 1-t$) and then introduce the notation $tx$ for $h(t,x)$. Then you have $1x=x$ and $s*(tx)=(st)x$ and $0x=x_0$, so we just have based spaces with continuous action of the monoid $I$ with the extra condition $0x=\text{basepoint}$.
|
2025-03-21T14:48:30.192086
| 2020-04-01T07:00:46 |
356253
|
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|
Stack Exchange
|
Non-density of continuous functions to interior in set of all continuous functions
Let $M$ be an $m$-dimensional manifold and $N$ be an $n$-dimensional manifold. Suppose also that the topology on $N$ can be described by a metric. Thus, the set $C(M,N)$ can be endowed with the topology of [uniform convergence on compacta][2].
Let $N'\subseteq N$ be a dense subset which is homeomorphic to $\mathbb{R}^n$. In this post's answer's comment it was remarked that $C(D,D-\{0\})$ is not dense in $C(D,D)$ where $D$ is the unit disc.
In general, when is $C(M,N')$ not dense in $C(M,N)$?
Actually, even the case of $N=S^1$, $N'=\mathbb{R}$ is an example of non-density. Note that if $N'$ is homeomorphic to $\mathbb{R}^n$, every map into $N'$ is null-homotopic. Null-homotopic maps are closed in $C(M,N)$, and so $C(M,N')$ is NOT dense as long as there is a non-null homotopic map from $M$ into $N$. So, if $N$ is not contractible, the identity map cannot be approximated. I am however not that well-versed in algebraic topology to prove that if $N$ is null-homotopic, $C(M,N')$ is dense.
Would you know a reference where I can find the fact that null-homotopic maps form a closed subspace?
Yeah, I thought that it is obvious when I was writing that comment, but now I cannot find neither a proof, nor a reference. I guess, time to ask yet another question.
1) About Erz' answer, and Annie's second question. It is a general property that for two differentiable manifolds $M$, $N$, such that $M$ is compact, and given a continuous map $f:M\to N$, any continuous map $g:M\to N$
close enough to $f$ is homotopic to $f$. This can be proved by many different ways, depending on taste. (One can assume that $N$ is also compact, since
we are only interested in a small neighborhood of the compact $f(M)$ in $N$):
a) Put a Riemannian metric on $N$, let $i>0$ be its injectivity radius: any two points $x,y\in N$ whose distance is less than $i$ are linked by a unique shortest geodesic, and this geodesic depends continuously on the pair $(x,y)$. Then, assuming that $d(f(x),g(x))<i$ for every $x\in M$, the geodesics from $f(x)$ to $g(x)$ provide the desired homotopy;
b) Embed $N$ in $R^n$ for $n$ large enough (Whitney); then $N$ has a tubular neighborhood $T$ in $R^n$ together with a continuous retraction $p:T\to N$; if $g$ is close enough to $f$, then the segments $[f(x),g(x)]$ in $R^n$ are contained in $T$ and
their projection under $p$ provide the homotopy;
c) Triangulate $N$ and use the straight lines in the simplices of the triangulation instead of the geodesics;
d) Use a good cover of $N$ by open subsets $U(i)$ such that every intersection
$U(i_1)\cap\dots\cap U(i_k)$ which is nonempty is contractible...
2) Erz' homotopical argument is excellent;
but even if $N$ is contractible, $C(M,N')$ is in general not dense in $C(M,N)$.
We should think to the complement $X$ of $N'$ in $N$.
For example, let $M$ be the compact interval $[-1,+1]$, $N$ be the complex plane $\mathbb{C}$ and $X$ be the nonpositive real halfline $\mathbb{R}_-$. Clearly, $\mathbb{C}\backslash X$ is diffeomorphic with the plane (being open and starred with respect to $1$.) Then, the embedding $[-1,+1]\to\mathbb{C}$: $t\mapsto-1+it$ can certainly not be approximated by continuous maps valued in $\mathbb{C}\backslash X$ (by the elementary "intermediate value theorem").
More generally, note that by Weierstrass' approximation theorem,
Annie's question is equivalent to: when is $C^\infty(M,N')$ not $C^0$-dense
in $C^\infty(M,N)$? The advantage of this differentiable setting is that
thanks to Thom's transversality theorem, one can answer in
all cases where $X$ is regular enough (technically, a finite or denumerable
union of smooth submanifolds $S_i\subset N$; e.g.
a polytope):
a) If one at least of the $S_i$'s is of codimension $\le m$ in $N$, then
there will be a smooth map $f:M \to N$ which intersects $S_i$ transversely in at least one point; and $f$ cannot be in the $C^0$-closure of $C(M,N')$.
b)
If on the contrary each $S_i$ is of codimension $>m$ in $N$, then by Thom's transversality theorem, $C^\infty(M,N')$ is $C^\infty$-dense in $C^\infty(M,N)$, and in particular $C^0$-dense. (As to Erz' second question: since $\dim(M)+\dim(S_i)<\dim(N)$, saying that $f:M\to N$
is transverse to $S_i$ amounts to say that $f(M)$ does not meet $S_i$).
I took liberty of latexifying your answer. I hope you don't mind. Could you please elaborate on your last point? I looked up Hirsh, and he says that the transversal maps are dense. But why are "avoiding" maps dense in the set of transversal ones?
Also, what if we apply this to $M=N$? $X$ then cannot contain a submanifold of any co-dimension, and so should be empty? And finally, is there a way to have similar reasoning without smoothness?
I also have a question/ confusion:
If $N=[0,1]^n,N'=(0,1)^n, X=\partial (0,1)^n$, and $M=\mathbb{R}^m$; with $m\geq 1$. Then unless $n\leq 1$, $C(M,N')$ is not dense in $C(M,N)$?
Doesn't that contradict @erz 's answer to this post:
https://mathoverflow.net/questions/354304/density-of-continuous-functions-to-interior-in-set-of-all-continuous-functions
?
I'm assuming that the argument somehow breakdown when looking at (smooth) manifolds with boundaries?
@AnnieTheKatsu From my understanding the argument goes like this: if $X$ contains a submanifold and a $f$ is transversal to this submanifold, then locally, you have the inclusion of $\mathbb{R}^m$ into $\mathbb{R}^m\oplus \mathbb{R}^k$, where $X$ contains $\mathbb{R}^k$. Surely, you cannot approximate such an inclusion avoiding $X$. However, this only works for submanifolds. The boundary of a manifold with boundary is not a submanifold in this sense.
Thanks, this makes more sense. However, I've never heard of a Weirestrass's theorem in such a setting (as far as I know Stone-Weierestrass requires function to map into $\mathbb{R}$)...
Of course, the transversality theorem does not hold when the submanifold is contained in the boundary. As for the other question, by "Weierstrass" I mean: every continuous map $f:M\to N$ between two manifolds can be approximated by a smooth one. This is proved by embedding $N$ into $R^d$ (Whitney), approximating $f$ by a smooth map $g:M\to R^d$ by the classical Weierstrass theorem, and reprojecting $g$ into $M$ by a local projection.
@GaelMeigniez Very interesting answer. I wonder one thing, if $M=N$ then if $F:M\rightarrow M$ is continuous injection then $C(M,Im(F))$ is only dense in $C(M,M)$ if $F$ is surjective? Is this a correct interpretation of your answer?
@ James_T: Yes, I think so.
|
2025-03-21T14:48:30.192521
| 2020-04-01T08:30:34 |
356257
|
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|
Stack Exchange
|
Random sparse and invertible matrices
Let $n\leq m$ and $0\leq k\leq (n\times m - \min\{n,m\})$ be in $\mathbb{N}$. Let $\mu$ be a probability measure dominated by the Lebesgue measure on $\mathbb{R}$ and generate a random $n\times m$ matrix $A$ as follows:
$$
A_{i,j} \sim \mu
$$
and set exactly $k$ elements of $A$ equal to $0$ where the indices of the zero entries are chosen uniformly at random.
What is the probability that $A$ has full rank?
For example, if $n=m$, $k=0$ and $\mu$ is Gaussian then $det(A)\neq 0$ with probability $1$. So $A$ has full rank... but what about in general?
From the bounds you have in the question, maybe you know this already? It’s just the probability that the pattern of non-zero entries contains a subset of the form $(1,j_1),\ldots,(n,j_n)$ with distinct j’s. For any fixed pattern satisfying this, the probability that the matrix obtained by filling the entries is full rank is 1 (just take a sub-determinant).
Actually, I didn't know this. Do you have a reference?
I indicated the proof. Consider the subdeterminant. Condition on all the entries except for the n entries I identified. Then with probability 1, the product of the n entries is not the exact number needed to make the subdeterminant vanish.
Yes but how to compute this probability?
I know how to compute it for small values of $k$ and for large values of $k$, but the point of my previous comment is that it reduces to a counting problem: out of the $\binom{mn}{k}$ arrangements, how many "contain" an "injection"? In particular for $k<m$, the probability is 1. If $k=mn-m$, then probability is $m(m-1)\ldots (m-n+1)/\binom{mn}{n}$.
|
2025-03-21T14:48:30.192668
| 2020-04-01T10:45:45 |
356258
|
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|
Stack Exchange
|
Resonances for Schrodinger operators with radial potentials
Let $V\in L^{\infty}(\mathbb{R}^3)$ be a radial, compactly supported potential, and consider the Schrodinger operator $H:=-\Delta + V$ on $L^2(\mathbb{R}^3)$. Let $\psi$ be a resonance for $H$, i.e. a function $\psi\in L^2(\mathbb{R}^3,\langle x\rangle^{-1-\varepsilon}dx)\setminus L^2(\mathbb{R}^3)$ which satisfies $(-\Delta + V)\psi=0$.
Is it true that $\psi$ is radial? If not, is it at least true that the orthogonal projection of $\psi$ into the space of radial functions is a resonance?
$\psi$ will factor into a radial function times an angular dependence, but it is not solely a function of the radial coordinate.
Ok thanks, but if I take the projection of $\psi$ into the space of radial function at least I get a radial function $\psi_r$ that solves $(-\Delta+V)\psi_r=0$. The point is to understand wheter we actually have $\psi_r\not\in L^2$.
you require $\psi\in L^2$, doesn't this imply $\psi_r\in L^2$?
No, I require $\psi\not\in L^2$.
Expand your resonance in spherical harmonics: $\psi = \sum_{\ell=0}^\infty \sum_m \psi_{\ell m}(r) Y_{\ell m}(\theta,\phi)$. Then each coefficient satisfies the radial Schrödinger equation
$$ -\frac{1}{r^2} \frac{\partial}{\partial r} r^2 \frac{\partial \psi_{\ell m}}{\partial r} + \frac{\ell(\ell+1)}{r^2}\psi_{\ell m} + V(r) \psi_{\ell m} = 0 . $$
Since $V(r)$ is compactly supported, for sufficiently large $r$, you must have $\psi_{\ell m} = A r^\ell + B r^{-\ell-1}$. Your asymptotic condition forces $A=0$ for all $\ell$. On the other hand, the asymptotics of the remaining term mean that, for $\ell \ge 1$, $\psi_{\ell m}(r) Y_{\ell m}(\theta,\phi)$ will be in $L^2(\mathbb{R}^3)$. So the component of $\psi$ orthogonal to radial functions is necessarily in $L^2(\mathbb{R}^3)$.
Only the $\ell=0$ case escapes $L^2(\mathbb{R}^3)$. And of course, when $\psi_{\ell=0}$ is non-vanishing, it is $O(r^{-1})$ at infinity and hence a resonance by your definition. So the answer to your second question is Yes, but if there are any solutions at higher $\ell$ (they will be normalizable eigenfunctions), then mixing them with an $\ell=0$ resonance will give you non-radial resonances.
|
2025-03-21T14:48:30.192842
| 2020-04-01T10:57:22 |
356260
|
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|
Stack Exchange
|
Differences between GAP and MAGMA
GAP and MAGMA are computer algebra systems. What are the objective differences between the two?
Which capabilities are not shared?
How do they compare on facilities for working with character tables?
How do they compare on facilities for working with group generators?
I agree. This is really not the right forum to discuss a topic that is affected by people's varying views on the morality of proprietary software, etc. It would be more legitimate to ask a more specific question like how do they compare on facilities for working with character tables.
GAP is better because it has QPA :)
I edited the title and question in response to the comment by @YCor . Is that why the upvotes were reset to zero?
@DerekHolt, I only mentioned open source/proprietary because I wanted to exclude that category of answers. Feel free to comment on facilities for working with character tables.
@Philip no, up/downvotes are not reset after any kind of revision. The score is just the number (upvotes minus downvotes).
Could anyone who voted to close, explain why this is an opinion based question? Do you believe the systems objectively have no differences? Do you believe that differences are merely a matter of opinion?
Of course the systems have some differences. I could try and summarise them, but if I did that then there would almost certainly be widespread discussion and disagreement, so if you substitute "to a large extent" for "merely" in your final question, then the answer is yes. That is why this question is not suitable for this forum. I use both systems extensively and they constitute an essential part of my research. I have used them in a wide variety of applications and I have often found that one performed better than the other for various types of calculations.
I addressed this question briefly in the introduction to the book "Handbook of Computational Group Theory" by Bettina Eick, myself, and Eamonn O'Brien, so I can refer you there, but that book was published in 2005, and both systems have undergone widespread development since then.
@YCor Is the title now acceptable?
Yes (I've removed my comment). The suitability of the question itself was discussed anyway.
I welcome someone's revising the question to make it more appropriate. I am mystified why questions asking for "the best book on X" are considered in scope for this forum, but for some reason it does not matter whether this question or title is objective if there is any chance of disagreement among answers. Best regards.
|
2025-03-21T14:48:30.193049
| 2020-04-01T11:07:59 |
356262
|
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|
Stack Exchange
|
Question on the center of universal enveloping algebra
Let $G$ be a unitary group $U(n)$ and $G’$ is a subgroup $U(m)$. (i.e. $m\le n$)
Let $\mathfrak{g},\mathfrak{g}'$ be the complexified Lie algebra of $G,G’$ and $\mathfrak{z},\mathfrak{z}’$ be the center their universal enveloping algebra.
Let $\mathfrak{t}’$ and $\mathfrak{t}$ be their maximal toral subalgebra of $\mathfrak{g}'$ and $\mathfrak{g}$ such that $\mathfrak{t}’ \subset \mathfrak{t}$.
Since $\mathfrak{t}’ \subset \mathfrak{t}$, I guess that there is a natural canonical surjection $\mathfrak{z}, \to \mathfrak{z},’$ which is compatible with domain restriction morphism $C^{\infty}((U(n)) \to C^{\infty}(U(m))$.(Here, we regards elements of $\mathfrak{z}$ and $\mathfrak{z}'$ as differential operators.)
More precisely, I am asking the exitence of canonical surjection $p:\mathfrak{z}, \to \mathfrak{z}’,$ such that for $\phi \in C^{\infty}((U(n))$ and $X \in \mathfrak{z}’$, $X \cdot (\phi|_{U(m)})=(Y \cdot \phi)|_{U(m)}$ where $Y$ is an arbitrary element in $p^{-1}(X)$.
Such $p$ does exists? If so how can we construct it? Some expert has alluded to me to use Harish-Chandra isomorphism. But I have no idea how to use it in this situation.
Thank you in advance!
For a particular case ZUgl(n) to ZUgl(N). Take a look at section 10. "Coherence property of quantum immanants and shifted Schur polynomials" , of the https://arxiv.org/abs/q-alg/9605042 there is kind of "averaging" map discussed there .
I assume $\mathfrak{t}, \mathfrak{t}'$ are the relevant maximal tori, which implies that you are choosing tori. Does this mean that your "canonical" surjection is allowed to depend on the choice of maximal tori?
@Alexander, Thank you! It seems to holds for unitary groups. Do you think that it does hold for other classical groups also?
@user44191, Yes! I mean canonical map as the natural map once we fix relevant maximal tori.
Does https://mathoverflow.net/questions/107389/about-the-map-s-mathfrakg-g-rightarrow-s-mathfrakh-h-for-h answer your question ?
@Alexander, Thank you very much! Though it is the dual version of what I am seeking for, it is similar. I am finding some sufficient condition such map be surjective! In the meanwhile, may I ask you some stupid question? I am wondering whether Lie algebra of complex Lie group equal to complexification of a Lie algebra.
For example, let $G=GL_N(\mathbb{R})$ and $G_{\mathbb{C}}= GL_N(\mathbb{C})$ and $g_{\mathbb{R}}$ and $g_{\mathbb{C}}$ be their corresponding real and complex Lie algebra.
Then $g_{\mathbb{C}}= g_{\mathbb{R}} \otimes_{\mathbb{R}} \mathbb{C}$?
Thank you very much always!
You are welcome. Yes, g_C = g_R \times C
@Alexander, Hello! Do you think that such coherence property also holds for reductive groups not only for classical groups? https://mathoverflow.net/questions/391427/the-coherence-property-of-center-of-universal-enveloping-algebra-for-reductive-l
|
2025-03-21T14:48:30.193254
| 2020-04-01T11:57:43 |
356264
|
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|
Stack Exchange
|
quadratic residues and cubic polynomials
I'm really not sure about this, but I've heard somewhere that for any prime $p$,
$|\sum_{x=0}^{p-1} (\frac{ax^3 +bx^2 +cx +d}{p} ) |\le \sqrt{2p}$ holds.
Does anyone know a proof for this inequality or a similar result?
It should be $2\sqrt{p}$. This is Hasse's theorem on elliptic curves.
At least for $a\neq 0$... For $a=c=d=0$, $b=1$, it is obviously false!
Wow thanks!! That was what I was looking for
A nice self-contained proof is given in Voloch's answer to https://mathoverflow.net/questions/87916.
|
2025-03-21T14:48:30.193322
| 2020-04-01T12:01:26 |
356265
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356265"
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|
Stack Exchange
|
Small radical of polynomial at integers
Let $f(x)$ be squarefree polynomial with coprime integer coefficients.
Assume $\deg f > 1$.
For natural $N$
define $G(f,N) = \#\{X : 1 \le X \le N, |X|^{\deg(f)-1 } \ge rad(f(X))\}$
where $rad(A)$ is the radical of $A$, the product of the prime factors of $A$.
How large can $G(f,N)$ be for some $f$? Can we get $G(f,N) \sim \log^\alpha{(N)}$ for $\alpha > 1$.
For $f=x(x-1)$ and $X=p^m$ for small $p$ and large $m$ we get
$G(f,N) \sim \log{(N)}$.
Another approach to get a near miss is solve $F(X) \mod p^m =0$,
then the radical is at most $\frac{p f(X)}{p^m}$.
We think abc implies that $G(f,N)$ can't be as large as
$n^\alpha,\alpha>0$ for all $f$,can we show this unconditionally?
Experimentally $G(x^4+1,10^5)= 2, G(x(x-1)(x-2)(x-3)(x-4),10^5)=4887$
|
2025-03-21T14:48:30.193404
| 2020-04-01T12:52:28 |
356268
|
{
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356268"
}
|
Stack Exchange
|
A comparison between a category and its group objects
Let $\mathbf{C} $ be a complete, cocomplete, and cartesian closed category. Denote by $\mathit{Grp}(\mathbf{C}) $ the category of group objects in $\mathbf{C} $.
1) Which of these features does $\mathbf{C} $ share with $\mathit{Grp}(\mathbf{C}) $?
2) Does the forgetful functor $\mathit{Grp}(\mathbf{C})\hookrightarrow\mathbf{C} $ have a left adjoint?
3) Is there any reference for a comparison between a category and its group objects?
The category $ Grp(C) $ may not be a cartesian closed category. For example, consider the category $ Set $. In this case, $ Grp $ is the category of its group objects, which is not cartesian closed.
In fact, the only case I can think of where $Grp(C)$ is cartesian closed would be the trivial case where $C = 1$ is the terminal category. Certainly if $C$ is locally presentable, then $Grp(C)$ will be locally presentable too (and in particular complete and cocomplete) and a free group functor will exist. Note also that $Grp(C) \to C$ is an equivalence iff $C$ is additive, so that $C = Grp(C) = Grp(Grp(C)) = Ab(C)$ is the category of abelian group objects in itself.
@Tim: These facts are very useful for me. It would be great if you mention some references as well.
@Emptymath This is probably a bit too specific of a question to have an explicit reference. Tim's points mostly follow from very general facts about locally presentable categories, for which Adamek and Rosicky's book with the same title is the canonical reference. The point about $C$ additive is essentially the Eckmann-Hilton argument.
I am not very familiar with locally presentable categories, although they are interesting for me. I particularly need a reference for the statement that if $ C $ is locally presentable, then $ Grp(C) $ is locally presentable.
$Grp(C)$ is pointed, so it can't be cartesian closed.
|
2025-03-21T14:48:30.193560
| 2020-04-01T13:12:25 |
356270
|
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|
Stack Exchange
|
Does the expression $x^4 +y^4$ take on all values in $\mathbb{Z}/p\mathbb{Z}$?
As the title asks: does there exist $N$ such that, for any prime $p$ larger than $N$, the expression $x^4 +y^4$ takes on all values in $\mathbb{Z}/p\mathbb{Z}$?
I have been thinking about this problem for days, but I failed to solve it. Does anyone know whether this is true, or does anyone know any partial results about it?
Partial results:
If $p=4k+3$, it easily works
if $p=4k+1$, if $g$ is a primitive root modulo $p$, and $A_i = \left\{ g^k : k \equiv i \pmod{4} \right\}$, then at least three of $A_i (i=0,1,2,3)$ must be expressed.
I'm not sure whether you're using $\mathbb{Z}_p$ to denote the ring of $p$-adics or for $\mathbf{Z}/p\mathbf{Z}$.
@YCor I'm not sure that matters, in the light of Hensel's lemma...by which I suspect that if the statement holds for $\mathbb{Z}/p\mathbb{Z}$, then it holds for $\mathbb{Z}_p$.
@FrankaWaaldijk sure but it would matter in the way to formulate an answer.
Ok, I get you now :-)
Actually it is quite important whether $\mathbb{Z}_p$ means the $p$-adic integers or the finite field. In the latter case the answer is yes, but not in the former case. The problem is that $x^4 + y^4 = p$ has a solution in the $p$-adics if and only if $p$ is split in the $8$th cyclotomic field (i.e. $p \equiv 1 \bmod 8$). The easiest way I know how to prove it over a finite field is via the Hasse-Weil theorem, but probably there is an elementary approach.
@Daniel ah i forgot that case, very silly of me, thank you for correcting me.
So are you clarifying that $\mathbb{Z}_p$ does indeed denote the $p$-adic integers?
Ah, I'm sorry. What I intended was $\mathbb{Z}/ p\mathbb{Z}$ , not p-adic integers
@Daniel could you explain why the answer is yes in that case please?
See this question.
emtom has found the right reference, but there is a more explicit result in that book (Ireland and Rosen, A Classical Introduction to Modern Number Theory). In fact, Theorem 5 of Chapter 8 (on page 103) directly implies that the number $N = N_{p,\alpha}$ of solutions to $x^4+y^4=\alpha$ in $\mathbb{F}_p$ satisfies the inequality
$$
\left| N - p \right| \leq M_0 + M_1 p^{1/2}
$$
for some $M_0$ and $M_1$ that are described explicitly in the statement of the theorem (and from that description it easily follows they can be bounded in a way that is independent of $p$ or $\alpha$).
It then automatically follows that for sufficiently large $p$, we will have $N_{p,\alpha}>0$ for all $\alpha$. In other words, for sufficiently large $p$, the expression $x^4+y^4$ assumes all values of $\mathbb{F}_p$ as $x$ and $y$ run through $\mathbb{F}_p$.
Of course, this is no better (and possibly slightly worse) than what you get from the Hasse-Weil bound which Dan Loughran referred to, but at least this reference has the virtue of providing a completely elementary proof.
Expanding on a comment, the curve $X^4+Y^4=aZ^4$ (for $a\ne0$) has genus $3$. So the Hasse-Weil bound says
$$ N_p(a) := \#\bigl\{ [X,Y,Z]\in\mathbb P^2(\mathbb F_p) : X^4+Y^4=aZ^4 \bigr\} $$
satisfies
$$ \bigl| N_p(a) - p - 1 \bigr| \le 2g\sqrt{p} = 6\sqrt{p}. $$
Thus
$$ N_p(a) \ge p + 1 - 6\sqrt{p}. $$
There are at most $4$ points with $Z=0$, so
$$
\#\bigl\{ (X,Y) \in\mathbb A^2(\mathbb F_p) : X^4+Y^4=a \bigr\} \ge p-3-6\sqrt{p}.
$$
So you'll always have a solution provided $p\ge3+6\sqrt{p}$, which means that there is always a solution provided $p\ge43$.
By adding a numerical computation, it seems to hold for all $p \geq 31$. For $p=29$, we get e.g. that there are no $x,y \in \mathbb{F}_{29}$ such that $x^4+y^4=4$.
@RP That makes sense, I took the worst possible case that there are 4 $\mathbb F_p$ rational points at infinity, but for many values of $p$, there will be less. It ultimately depends on whether $-1$ is a square root, and whether it is a fourth root.
Not exactly an answer, but in exercise $18$ in page $106$ of Ireland and Rosen's A Classical Introduction to Modern Number theory it states: Let $p\equiv 1\mod 4$ and let $p=A^2+B^2$ where we fix $A$ by requiring that $A\equiv 1\mod 4$. Then $N=\#\{(x,y)\in\mathbb{F}_p^2\mid x^4+y^4=1\}$ satisfies $N=p-3-6A$ if $p\equiv1\mod 8$ and $N=p+1+2A$ if $p\equiv5\mod 8$.
|
2025-03-21T14:48:30.193983
| 2020-04-01T13:53:51 |
356272
|
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|
Stack Exchange
|
Generalising injective modules
Free modules over a ring generalise to projective modules over a ring, which generalise to flat modules, which generalise to torsion free modules:
$$
\textrm{free} \to
\textrm{projective}
\to
\textrm{flat}
\to
\textrm{torsion free}.
$$
Does there exist an analogous chain of properties for injective modules.
There is a notion of a cofree module and there's also a notion of a divisible module over any type of ring, in the sense that appears in T.Y. Lam's Lectures on modules and rings around p 70. The last one always seemed like a fair dual to flatness. I've forgotten a dual for torsion-free modules, if I knew one. So at least Cofree -> injective -> divisible.
|
2025-03-21T14:48:30.194067
| 2020-04-01T14:23:12 |
356275
|
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|
Stack Exchange
|
Solving a delay-differential equation related to epidemiology
For some inexplicable reason, I have recently been interested in epidemiology. One of the classical and simplistic models in epidemiology is the SIR model given by the following system of first-order autonomous nonlinear ordinary differential equations in the real variables $s,i,r$ satisfying $s,i,r\geq 0$ and $s+i+r=1$:
$$
\begin{align}
s' &= -\beta i s\\
i' &= \beta i s - \gamma i\\
r' &= \gamma i
\end{align}
\tag{$*$}
$$
(where prime denotes derivative w.r.t. time, $s,i,r$ represent the proportion of “susceptible”, “infected” and “recovered” individuals, $\beta$ is the kinetic constant of infectiousness and $\gamma$ that of recovery; the “replication number” here is $\kappa := \beta/\gamma$). It is easy to see that for every initial value $(s_0,i_0,r_0)$ at $t=0$ the system has a unique $\mathscr{C}^\infty$ (even real-analytic) solution (say $r_0=0$ to simplify, which we can enforce by dividing by $s_0+i_0$); it does not appear to be solvable in exact form in function of time, but since $s'/r' = -\kappa\cdot s$ and $i'/r' = \kappa\cdot s - 1$ (where $\kappa := \beta/\gamma$) it is possible to express $s$ and $i$ as functions of $r$, viz., $s = s_0\,\exp(-\kappa\cdot r)$ and of course $i = 1-s-r$. This makes it possible to express, e.g., the values of $s,i,r$ at peak epidemic (when $i'=0$): namely, $s = 1/\kappa$ and $r = \log(\kappa)/\kappa$); or when $t\to+\infty$: namely, $s = -W(-\kappa\,\exp(-\kappa))/\kappa$ where $W$ is (the appropriate branch of) Lambert's transcendental W function; this is upon taking $i_0$ infinitesimal and $r_0=0$.
Now one of the many ways in which this SIR model is simplistic is that it assumes that recovery follows an exponential process (hence the $-\gamma i$ term in $i'$) with characteristic time $1/\gamma$. A slightly more realistic hypothesis is recovery in constant time $T$. This leads to the following delay-difference equation:
$$
\begin{align}
s' &= -\beta i s\\
i' &= \beta i s - \beta (i s)_T\\
r' &= \beta (i s)_T
\end{align}
\tag{$\dagger$}
$$
where $(i s)_T = i_T\cdot s_T$ and $f_T(t) = f(t-T)$. When I speak of a “$\mathscr{C}^\infty$ solution” of ($\dagger$) on $[0;+\infty[$ I mean one where the functions $s,i,r$ are $\mathscr{C}^\infty$ and satisfy the equations whenever they make sense (so the second and third are only imposed for $t\geq T$), although one could also look for solutions on $\mathbb{R}$.
I am interested in how ($\dagger$) behaves with respect to ($*$). Specifically,
Does ($\dagger$) admit a $\mathscr{C}^\infty$ (or better, real-analytic) solution on $[0;+\infty[$ for every initial value $(s_0,i_0,r_0)$ at $t=0$? Or maybe even on $\mathbb{R}$? Is this solution unique? (Note that we could try to specify a solution by giving initial values on $[0;T[$ and working in pieces, but this does not answer my question as it would not, in general, glue at multiples of $T$ to give a $\mathscr{C}^\infty$ solution.)
Can we express this solution in closed form or, at least, express $s$ and $i$ in function of $r$?
Qualitatively, how do the (at least continuous!) solutions of ($\dagger$) differ from those of ($*$), and specifically:
How do the behaviors compare when for $t\to 0$?
How do the values compare around peak epidemic $i'=0$?
How do the limits when $t\to+\infty$ compare?
I stumbled upon the SEIR model which differentiates between infected and exposed people - there is only a German Wikipedia page (https://de.m.wikipedia.org/wiki/SEIR-Modell). The model does not have delay, though.
Nice question. Regarding Equation $(*)$, how does the parameters, particularly $\beta$ and $\gamma$ control the "speed" (such as half life) of $I$ approaching the asymptotic state for time infinity? Do you have a reference on this issue?
+1. There is an exact closed form solution to the SIR model https://arxiv.org/pdf/1403.2160.pdf.
I was unable to make your model work (numerically, I haven’t tried your solutions yet). It seems that a regular SIR with a delay in the terms that have gamma works best with actual data ... that is, instead of gamma i, you would have gamma i_T in both the second and third equation of (*). I don’t know how to solve it analytically though. It would be nice. I will keep trying to make your model work ...
By the way, it might be worth pointing out that there a variety of models called SIR models and that the equations which you describe are specifically known as the Kermack-McKendrick equations.
After trying many Ansätze in the form of series, I stumbled upon the fact (which I find rather remarkable) that ($\dagger$) admits an exact solution in closed form. To express this, let me first introduce the following notations:
$\kappa := \beta T$ (reproduction number), which I assume $>1$;
$\Gamma := -W(-\kappa \exp(-\kappa))/\kappa$ the solution in $]0,1[$ to $\Gamma = \exp(-\kappa(1-\Gamma))$, which will be the limit of $s$ when $t\to+\infty$ (both in ($*$) and in ($\dagger$), starting with $i$ and $r$ infinitesimal);
$X := \exp(\beta(1-\Gamma) t)$, a change of variable on the time parameter, in which $s,i,r$ will be expressed.
The solution is then given by:
$$
\begin{align}
s &= \frac{(1-\Gamma)^2+\Gamma c X}{(1-\Gamma)^2+c X}\\
i &= \frac{(1-\Gamma)^4 c X}{((1-\Gamma)^2+c X)((1-\Gamma)^2+\Gamma c X)}\\
r &= \frac{\Gamma (1-\Gamma) c X}{(1-\Gamma)^2+\Gamma c X}
\end{align}
$$
where c is an arbitrary positive real parameter which merely serves to translate the solution.
From this we can deduce the following about the behavior at start, peak and end of the epidemic, and how it compares to ($*$) (for which I let $\kappa := \beta/\gamma$ and $\Gamma$ defined by the same formula):
Initially, $i$ grows like $c\,\exp(\beta(1-\Gamma) t)$ (and $r$ like $\frac{\Gamma}{1-\Gamma}$ times this). This is in contrast to ($*$) where $i$ initially grows like $c\,\exp((\beta-\gamma) t)$ (and $r$ like $\frac{\gamma}{\beta-\gamma}$ times this), i.e., for a given reproduction number $\kappa$, solutions of ($\dagger$) grow faster than those of ($*$).
Peak epidemic happens for $X = \frac{(1-\Gamma)^2}{c\sqrt{\Gamma}}$, at which point we have $s = \sqrt{\Gamma}$ and $i = (1-\sqrt{\Gamma})^2$ and $r = \sqrt{\Gamma}(1-\sqrt{\Gamma})$. This is in contrast to ($*$) where we have $s = \frac{1}{\kappa}$ and $i = \frac{\kappa-\log\kappa-1}{\kappa}$ and $r = \frac{\log\kappa}{\kappa}$; so, for a given reproduction number $\kappa$, solutions of ($\dagger$) have a peak epidemic with fewer uninfected ($s$) than those of ($*$).
The limit when $t\to+\infty$ is the same in ($\dagger$) as in ($*$), namely $s \to \Gamma$.
|
2025-03-21T14:48:30.194471
| 2020-04-01T14:25:43 |
356276
|
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"François Brunault",
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|
Stack Exchange
|
Beilinson regulator: a road map
I'm approaching to the Beilinson Conjecture and after studying some properties of the Deligne-Beilinson cohomology, I want to understand the regulator maps. But I don't know anything about K-theory and I would like to understand the necessary background. Is there any reference or some reasonable "chain" of references to follow to understand the construction of regulator? Secondly, I would like to clearly understand the relation between this regulator and L-functions. Can someone help me? Thanks you!
You could begin with examples. For K_2 of curves, see Dokchitser--de Jeu--Zagier, Numerical verification of Beilinson's conjecture for K_2 of hyperelliptic curves.
Thanks I will see this paper! What about the background in K-theory? What do you suggest to know about it in general for a depth knowledge of this subject (Beilinson conjectures, regulators etc...)? Thanks!
Beilinson's conjecture was originally formulated using K-theory, but a more modern approach is to use motivic cohomology (which is the same as long as we work with non-singular varieties and with rational coefficients). So it is not necessary that you learn K-theory (but the motivic approach uses heavy machinery like A^1-homotopy theory). You can find good references on Beilinson's conjectures here: https://mathoverflow.net/questions/126699/status-of-beilinson-conjectures/126714#126714 Personally I like Ramakrishnan's survey "Regulators, algebraic cycles, and values of L-functions" very much.
|
2025-03-21T14:48:30.194593
| 2020-04-01T15:04:22 |
356279
|
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"Joseph O'Rourke",
"LSpice",
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"sort": "votes",
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|
Stack Exchange
|
On 'fair bisectors' of planar convex regions
Definitions (https://www.ias.ac.in/article/fulltext/pmsc/122/03/0459-0467):
Given a planar convex region $C$ (could be smooth or polygonal), an area bisector of $C$ is any line that partitions $C$ into 2 pieces of equal area. A 'fair bisector' is a line that partitions $C$ into 2 pieces of equal area and equal perimeter.
Thru every point on the boundary of $C$, an area bisector can be drawn (for a description of their properties, please see 'Mathematical Omnibus' by Fuchs and Tabachnikov, Lecture 11). But it can be seen that a convex planar region can have just a single fair bisector (eg. for a thin isosceles triangle, the only fair bisector is the bisector of its apex angle) or a finite number of them (in which case, their number is necessarily odd as can be seen from simple continuity arguments; see reference at the top) or infinitely many.
Observations: For regions with a center of symmetry such as a circular disk or ellipse or regular polygon with even number of sides, all fair bisectors are concurrent. But, numerically, we see that for a general convex region $C$ with finitely many fair bisectors, the fair bisectors are not necessarily concurrent but usually very close to being so. Clearly,for a general $C$ with exactly 3 fair bisectors, they determine a small triangular region deep in the interior of $C$. For $C$s with more fair bisectors, their many possible intersections will divide the interior of $C$ into many regions. Let us refer to the union of those regions which do not share the outer boundary of $C$ as the 'core' of $C$.
The core must lie deep inside $C$.
Questions:
For which convex shape of $C$ is the area of the 'core' of $C$ the largest as a fraction of the area of $C$? Intuitively, a relatively large core is a measure of the asymmetry of $C$. Can one say (say) that such a shape is always one with exactly 3 fair bisectors?
Generalizing a bit, what about lines that break off the same fraction $t$ of the area and outer boundary length of $C$? For a circular disk, it appears that only for $t=1/2$, we have such lines (any diameter). Are there $C$'s for which such lines exist for several (maybe even arbitrarily many) different values of $t$?
Guess: All centrally symmetric convex regions (rectangles, ellipses,...) appear to give such only one single partitioning line that divides both area and outer perimeter in same ration - only for $t=1/2$. But general convex regions with no symmetry might give infinitely many such lines - one such partitioning line for each orientation - and a different value $t$ for each orientation. And the set of these lines might even have interesting envelopes.
These questions have obvious higher dimensional analogs.
This may be useful: Goldberg, Michael. "On area-bisectors of plane convex sets." The American Mathematical Monthly 70, no. 5 (1963): 529-531. He proves that there are convex regions with a point through which $n$ area-bisectors pass, $n \ge 4$. It is known that every convex region has a point through which at least $3$ area-bisectors pass.
Thanks Prof. O'Rourke. The reference does give a nice intuitive property of area bisectors. The problem with 'fair' bisectors seems that they could be just a few in number so many of the structures such as envelopes which have nice properties cannot be defined in general. Still one wonders how the intersections of fair bisectors could have structures which reflect the overall nature of the convex region.
Name of reference: Nandakumar and Ramana Rao - Fair partitions of polygons: An elementary introduction.
This is not an answer, and not even that helpful, but
I wanted to see the central pattern formed by the collection of
perimeter bisectors.
Prof. O'Rourke, our experiments too indicate such patters for the envelope for perimeter bisectors - the envelope has cusps and each segment of the envelope seems to be a parabola segment (those of area bisector envelopes are hyperbola segments). For lines that cut out some other fraction t of the outer boundary length, the envelope appears to have discontinuities so on the whole they may not be as nice as envelopes that cut out a specified fraction of the area.
An answer to the question 2 (written with K Sheshadri)
The guess made above that needs to be proved: For a general convex polygonal region with no symmetries, for every orientation, we have a unique line with that orientation that separates out the same fraction of both area and outer boundary length. The value of this common fraction of area and perimeter separated out will vary continuously with orientation.
Proof :
Consider the convex polygonal region $C$ and a given orientation (direction). Draw both tangents to $C$ in that orientation. We assume both these tangents to touch $C$ at a single vertex (coincidences of tangents with entire edges of $C$ can be dealt with by small perturbations). Let these parallel tangents be distance $D$ apart. By sliding a line coincident with one of the tangents perpendicular to itself until it coincides with the other tangent to $C$, we get a continuous range of cutting lines. Let these cutting lines be parametrized by $d$, the perpendicular distance from the tangent from which we began sliding the cutting line.
For each value of $d$, we have a line that cuts $C$. Plot against $d$, the fraction of area (call this fraction Af) of the full $C$ that the piece separated from $C$ has and also the fraction of perimeter (call this fraction Pf) for the same piece. Obviously, as $d$ goes from 0 to $D$ both Af and Pf go from 0 to 1.
Now, we observe that the plot of Af against d has a quadratic behavior at both ends. Its plot will be continuous and made of several parabolic segments - beginning with an upward parabolic piece (where, as $d$ starts from 0, Af also starts from 0) and ending with a downward parabolic piece (when Af tends to 1 as $d$ approaches $D$). Moreover, due to convexity of $C$, the curve of Af rises monotonically.
On the other hand, Pf has a linear behaviour throughout including at ends. This graph is a continuous polyline and also rises monotonically.
From the above observations, as $d$ is increased from 0, the Af curve (quadratic) begins lower than Pf (linear) curve and as $d$ tends to $D$, Af approaches 1 from above the Pf curve. This plus the monotonically rising nature of both graphs plus their start values both being 0 and end values both being 1 guarantee that they have to necessarily intersect at some intermediate value of $d$; at these intersections, obviously, Af = Pf. It appears that convexity of $C$ also guarantees there will be only one such intersection.
Thus we have, for every orientation, a value of $d$ for which Af and Pf have same value - as claimed. If $C$ is centrally symmetric (circle, ellipse, rectangle, regular polygon with even number of sides...), the only such value of $d$ is $D$/2 and the common value of the fractions is 1/2 for all orientations. This will not be the case for asymmetric convex polygonal $C$'s - we have different common Af and Pf values for different orientations. This fraction should change continuously with orientation.
We guess that the envelope etc. of the cutting lines with common Af and Pf for each orientation might have interesting properties.
|
2025-03-21T14:48:30.195035
| 2020-04-01T15:51:29 |
356283
|
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|
Stack Exchange
|
Tightness of Hilbert-space-valued arrays
Let $\mathcal{H}$ be a separable Hilbert space. Assume we have some triangular array $W_{n,j}, j=1, \ldots ,n $ of $\mathcal{H}$-valued random elements with $\mathbb{E} \Vert W_{n,j} \Vert_{\mathcal{H}}^2 < \infty$ that is strictly stationary (which means that $W_{n,1},\ldots , W_{n,n}$ is strictly stationary for every $ n$). After Remark 3.3 in "Central Limit and Functional Central Limit Theorems for Hilbert-Valued Dependent Heterogeneous Arrays with Applications" (Chen and White, 1998) the sequence $(\sum_{j=1}^n W_{n,j})_n $ is tight if we have
\begin{align*}
\lim_{n \rightarrow \infty} \mathbb{E} \bigg\Vert \sum_{j=1}^n W_{n,j} \bigg\Vert_{\mathcal{H}}^2 < \infty. \tag{1}
\end{align*}
I would like to understand why. It suffices to show (see Lemma 3.2 in the same paper)
\begin{align}
\lim_{k \rightarrow \infty} \limsup_{n \rightarrow \infty} \mathbb{E} \bigg\Vert \sum_{j=1}^n \sum_{l=k}^{\infty} \langle W_{n,j} , e_l\rangle_{\mathcal{H}} e_l \bigg\Vert_{\mathcal{H}} ^2 =0. \tag{2}
\end{align}
for some complete orthonormal basis $(e_l)_l$. So far I calculated using straightforward calculations that
\begin{align}
\mathbb{E} \bigg\Vert \sum_{j=1}^n \sum_{l=k}^{\infty} \langle W_{n,j} , e_l\rangle_{\mathcal{H}} e_l \bigg\Vert_{\mathcal{H}} ^2 = \sum_{i,j=1}^n \mathbb{E} \sum_{l=k}^{\infty} \langle W_{n,j},e_l\rangle_{\mathcal{H}}\langle W_{n,i},e_l\rangle_{\mathcal{H}}.
\end{align}
Obviously the term within the expectation above is bounded by
\begin{align*}
\sum_{l=1}^{\infty} \vert \langle W_{n,j},e_l\rangle_{\mathcal{H}}\langle W_{n,i},e_l\rangle_{\mathcal{H}} \vert,
\end{align*}
which is independent of $k$ and the respective expectations exist because
\begin{align}
\mathbb{E} \sum_{l=1}^{\infty} \vert \langle W_{n,j},e_l\rangle_{\mathcal{H}}\langle W_{n,i},e_l\rangle_{\mathcal{H}} \vert &= \sum_{l=1}^{\infty} \mathbb{E} \vert \langle W_{n,j},e_l\rangle_{\mathcal{H}}\langle W_{n,i},e_l\rangle_{\mathcal{H}} \vert \\
&\leq \sum_{l=1}^{\infty} \mathbb{E} \langle W_{n,1},e_l\rangle_{\mathcal{H}}^2 \\ &= \mathbb{E} \sum_{l=1}^{\infty} \langle W_{n,1},e_l\rangle_{\mathcal{H}}^2 < \infty
\end{align}
where we used monotone convergence, Cauchy-Schwarz, the strict stationarity and Parseval's identity. Therefore by dominated convergence
\begin{align}
\lim_{k \rightarrow \infty} \mathbb{E} \bigg\Vert \sum_{j=1}^n \sum_{l=k}^{\infty} \langle W_{n,j} , e_l\rangle_{\mathcal{H}} e_l \bigg\Vert_{\mathcal{H}} ^2 =0
\end{align}
because we have by Parseval's identity and Cauchy-Schwarz
\begin{align}
\sum_{l=1}^{\infty} \langle W_{n,j},e_l\rangle_{\mathcal{H}}\langle W_{n,i},e_l\rangle_{\mathcal{H}} = \langle W_{n,j},W_{n,i} \rangle_{\mathcal{H}} \leq \Vert W_{n,j} \Vert_{\mathcal{H}} \Vert W_{n,i} \Vert_{\mathcal{H}} < \infty
\end{align}
elementwise on the underlying probability space. Finally the claim would follow if we are able to interchange $\lim_{k \rightarrow \infty}$ and $\limsup_{n \rightarrow \infty}$ in Eq. (2) (probably using cond. (1)), but I do not know how I could verify this. I would really appreciate it if someone could help me with that.
|
2025-03-21T14:48:30.195200
| 2020-04-01T15:56:09 |
356284
|
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"D. Dona",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356284"
}
|
Stack Exchange
|
How to make algebraic dependence explicit
Here is a possibly standard question, from someone who is in no way an expert. The scenario is taken from Shafarevich, Basic Algebraic Geometry 1, ch. 1, §6.3, proof of Thm. 1.25(ii).
Let us have a regular map $f:V\rightarrow W$ ($V,W$ open in $\mathbb{A}^{N},\mathbb{A}^{M}$ respectively, and $f(V)$ dense in $W$); passing to the pullback $f^{*}$, this translates to an inclusion of the function field $k(W)$ into $k(V)$. We write $k[v_{1},\ldots,v_{N}]$ for $k[V]$, and $k[w_{1},\ldots,w_{M}]$ for $k[W]$ (the $v_{i}$ and $w_{j}$ are like the coordinates of $\mathbb{A}^{N},\mathbb{A}^{M}$ modulo the relations given by the definition of $V,W$, if my imagination serves me right). If $\dim(V)=n>m=\dim(W)$, then $k(V)$ has transcendence degree $n-m$ over $k(W)$, which means that wlog we have $v_{1},\ldots,v_{n-m}$ algebraically independent over $k(W)$ and that in principle we also have $F_{i}(v_{i};v_{1},\ldots,v_{n-m};w_{1},\ldots,w_{M})=0$ for some $F_{i}$ for all $n-m< i\leq N$: here each $F_{i}$ is a polynomial in one variable (i.e. the $v_{i}$) over $k(W)[v_{1},\ldots,v_{n-m}]$.
My question is: how can one explicitly obtain the $F_{i}$ (from $f$ and the polynomials defining $V,W$)? Or at least, can we find bounds for their degrees, including the degrees of their coefficients as functions of the $w_{j}$? If standard, references are welcome!
I think you can compute $F_i$ using elimination theory and grobner bases, e.g., with the software Macaulay2.
@wnx Thanks, that's exactly the input I needed! Just to answer my own question: one can look at the undergraduate book by Cox, Little and O'Shea (§3.1) and use the elimination theorem with the ordering $w_{1}\prec\ldots\prec w_{M}\prec v_{1}\prec\ldots\prec v_{n-m}\prec v_{i}\prec v_{n-m+1}\prec\ldots\prec v_{N}$ (add one $z$ to homogenize and multiply away the denominators, if necessary) to get inside a Gröbner basis an $F_{i}$ as the one we already know exists by algebraic dependence. Then Dubé (1990) gives degree bounds for reduced Gröbner basis elements wrt $N,M,\deg(V),\deg(W),\deg(f)$.
|
2025-03-21T14:48:30.195354
| 2020-04-01T16:19:03 |
356287
|
{
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"authors": [
"Joseph O'Rourke",
"Neil Strickland",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/356287"
}
|
Stack Exchange
|
How do I find the portion of a cell/voxel lying within a defined surface?
We have a 3-dimensional grid of voxels (or cells), with individual voxels being of volume $dx\,dy\,dz$ where $dx=dy=dz=1$.
A cone-like surface is defined by some function, $z = f(x, y)$, which in this case is specifically (where $\epsilon$, $w_0$ and $r_0$ are constants):
$$z=\left(\frac{x^2+y^2}{w_0^2r_0^{-2\epsilon}}\right)^{\frac{1}{2\epsilon}}$$
Over the entire grid, how does one compute the fraction of each voxel's volume lying within the volume enclosed by that cone-like surface?
Obviously values should range from $0\rightarrow1$ and, in this particular case, voxels positioned at large values of $x$ and $y$ and small values of $z$ have none of their volume within the cone, while those lying at large values of $z$, and small values of $x$ and $y$ are completely enclosed by the cone-like surface.
I am guessing that you have a particular application in mind. It would be helpful to explain that, and in particular, what level of accuracy you care about. The issues here are probably more about efficient computation than any real mathematical complexity, so this site is probably not the best one for your question. You could try math.stackexchange.com or scicomp.stackexchange.com.
A voxel is a cube, so you are asking for the intersection of a cube (aligned with the axis) and your function $f(x,y)$. Since a cube is defined by six planes, you need to compute the intersection of a plane with $f$, in particular, $f(x,c)$ for $y=c$ with $c$ a constant. The intersection curve is pretty simple, but in the end you will need to numerically approximate.
|
2025-03-21T14:48:30.195744
| 2020-04-01T16:42:37 |
356289
|
{
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"Jon Pridham",
"Roberto Pagaria",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/356289"
}
|
Stack Exchange
|
Simplicial differential graded algebra and a filtration
Let $A$ be a simplicial differential algebra, i.e. for each $n \in \mathbb{N}$ a differential graded algebra $(A_n,d_n)$ and for each weakly increasing map $f \colon [n] \to [m]$ a morphism $f_* \colon A_n \to A_m$ such that $f_*$ and $d$ commute.
Consider the increasing filtration $F_{\bullet} A_n$ defined by $F_k A_n = \sum_{f\colon [k] \to [n]} \operatorname{Im} f_*$ where the sum is taken over all weakly increasing maps.
I claim that $\operatorname{gr}_F^\bullet H^\bullet (A_n,d_n) \cong H^\bullet(\operatorname{gr}_F^\bullet A_n,d_n)$ are isomorphic as algebras. Is it true?
It is equivalent to prove that the differential is strict, i.e. $\operatorname{Im} d \cap F_kA_n = d(F_kA_n)$. For $k=n-1$ it follows from Lemma 8.3.7 of "Weibel - An introduction to homolgical algebra" by noticing that $F_{n-1}A_n = D_n(A)$ is the degenerate subcomplex.
I also wonder if the direct sum decomposition generalize to $A_n= \cap_{f\colon [n]\to [k]} \ker f_* \oplus F_kA_n$.
You don't seem to be using the algebra structure anywhere. Is this just a question about simplicial cochain complexes? If so, I'd use the Dold-Kan correspondence to write $A=K(NA)$ as the denormalisation of the chain cochain complex $NA$. That gives a natural expression for $A_n$ as a direct sum of various copies of $N_kA$ for $k\le n$.
I'd like to have $\operatorname{gr}_F^\bullet H^\bullet (A_n,d_n) \cong H^\bullet(\operatorname{gr}_F^\bullet A_n,d_n)$ as algebra, not only as vector space
But presumably you've already got a map in one direction, so you can disregard algebra structure when proving quasi-isomorphism. Strictness of the differential isn't an algebra statement.
|
2025-03-21T14:48:30.195874
| 2020-04-01T16:47:57 |
356291
|
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"Jochen Glueck",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/356291"
}
|
Stack Exchange
|
Equivalence of antiderivative in L1 sense and in the usual sense
We say that$\ f$ is differentiable w.r.t to $L_1$ if there exists a$\ g$ such that:
$$
\lim_{h\to 0}\left\Vert\frac{f(x+h)-f(x)}{h} - g(x)\right\Vert_1 = 0
$$
where $\Vert \cdot \Vert_1$ is the $L_1$ norm. Since $f$ is in $L_1$, the corresponding$\ g$ must be in$\ L_1$ too, and so by Lebesgue, it has an antiderivative $G$ which is differentiable a.e, with $G'(x)=g(x)$.
My question is: does $f=G$ a.e?
Here is my line of thought: if $G$ is in $L_1$, it can be shown that
$$
\hat{g}{(t)} = 2\pi it\hat{G}{(t)} = 2\pi it\hat{f}{(t)},
$$
which then implies that $f=G$ a.e. and so, in order to show that $f=G$ a.e, it is enough to show that$\ G$ is in$\ L_1$, and that's where i got stuck.
I guess everything's happening on $\mathbb R$?
This seems loosely related to Proposition 9.3 in "Haïm Brezis: Functional Analysis, Sobolev Spaces and Partial Differential Equations (2011)".
Most antiderivative of $g$ are not in $L^1(\mathbb{R})$, in your case only one antiderivative $G_0$ will be in $L^1(\mathbb{R})$, the one actually equal to $f$. All the other antiderivatives $G$ are equal to $G_0 + c$, with $c \neq 0$, which is not in $L^1(\mathbb{R})$.
To proof that there exist one antiderivative $G_0$ in $L^1(\mathbb{R})$.
You start by noticing that your $L^1(\mathbb{R})$ differentiability imply differentiability in the distributional sens so, for $\phi \in \mathcal{C}_{comp}^\infty(\mathbb{R})$, we have
$$
\langle f',\phi \rangle = \langle g,\phi \rangle. \qquad (1)
$$
Fix $G$ an antiderivative of $g$, you have $G' = g$ in the distributional sens.
The equation $(1)$ become
$$
\langle f',\phi \rangle = \langle G',\phi \rangle \implies \langle (f-G)',\phi \rangle = 0
$$
and than imply $f-G = c$ a constant.
Choosing $G_0 = G + c$, we have $G_0 = f \in L^1(\mathbb{R})$.
You can further show that there exists a constant $c_0$ such that
$$
G_0(x) = c_0 + \int_0^x g(y)\, dy.
$$
|
2025-03-21T14:48:30.196130
| 2020-04-01T17:36:51 |
356295
|
{
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"Wojowu",
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}
|
Stack Exchange
|
How balanced can abc triples be?
I was looking at the $241$ known "good" abc triples (i.e. with quality $\geqslant1.4$), wondering how frequently $a$ and $b$ would have more or less the same order of magnitude. The outcome is not very surprising. With rank numbers (rk) corresponding to the quality ranking, there are only 15 such triples with $1<b/a<10$, which are the following:
rk quality size merit b/a
95 1.4316 13.28 12.18 1.1951826
240 1.4003 16.79 14.68 1.5557841
66 1.4420 15.51 15.53 1.5715695
105 1.4290 10.44 8.74 1.6514252
151 1.4158 23.92 24.63 1.6673082
43 1.4526 9.43 8.28 2.8166179
173 1.4121 29.38 31.48 3.3264647
206 1.4061 8.95 6.49 3.6854690
225 1.4022 10.67 8.12 4.6701482
160 1.4145 8.81 6.56 4.8162473
226 1.4020 13.49 11.09 7.4162550
108 1.4284 11.77 10.25 7.7411486
9 1.5270 9.78 11.02 8.7781887
199 1.4071 16.28 14.49 9.3202338
72 1.4403 16.98 17.38 9.4437408
Let's define the balance of an abc triple as the proportion $a/b$. (The convention being of course always $a<b$.) As this criterion has nothing to do with prime factors, I am not sure that looking at balances would give any new insights into the abc conjecture. But nevertheless, I would be curious about their distribution, and of course there might be unexpected patterns.
Generally speaking, they tend to be very small (i.e. $a\ll b$), so:
Conjecture: For $0<\varepsilon<1$, there are only finitely many abc triples with balance $a/b>\varepsilon$.
I would not expect this to be weaker (in the sense of implied by) or stronger than the abc conjecture itself. But maybe those among the millions of known abc triples with a "not too bad" balance, say $>.8$, could exhibit some patterns?
Note that this conjecture may be completely wrong. There might be plenty of abc triples of quality just above $1$ with a high balance, though intuitively I would doubt that. At this point, I'm wondering if somebody with access to the corresponding computer power could look (or has already looked) for the most balanced abc triples and/or provide some statistics about the general distribution of balances. But I am aware that this is not supposed to be the goal of a question to be asked here, so I have a slightly related and more feasible question:
Is there a known infinite sequence of abc triples $(a_n,b_n,c_n)_{n\in\mathbb N}$ such that $a_{n+1}>a_n$?
I am only aware of such sequences with $a_n\equiv1$, i.e. the worst possible balance at all.
The list of high quality triples you looked at is not (known to be) complete and therefore suffers from a search bias. However the list of good triples is complete up to $10^{18}$ (see here: http://www.math.leidenuniv.nl/~desmit/abc/abctriples_below_1018.gz) and you perhaps could test or refine your conjecture there.
I assume by abc triple you mean good abc triple.
It is known that a single good abc triple gives rise to
infinite sequence of good abc triples.
Let $a,b,c=a+b$ be good abc triple.
Then $A=4ab,B=(b-a)^2,C=(A+B)=(a+b)^2=c^2$ is good abc triple
too, and it is twice bigger than the original, but in
general of lower quality. The radical of $AB(A+B)$ is
at most $(b-a)$ times the radical of $ab(a+b)$.
This construction is used in Bart de Smit high merit triples:
https://www.math.leidenuniv.nl/~desmit/abc/?set=3
check for common factors of the high merit triples.
Maybe we need to swap A,B and clear a common factor of four.
I suspect the conjecture is false.
Treating $A,B$ as polynomials in $a,b$,
$\deg(A(a,b)) = \deg (B(a,b))$ which shows they
are not necessarily very unbalanced as in your
construction with $a=1$. Likely a single balanced
solution will give infinite many balanced ones.
According to my google search yesterday, "abc triple" is often taken to mean a triple satisfying $rad(abc)<c$.
@Wojowu Could be. I have seen explicit distinction.
@Wojowu I suspect your conjecture is false by my construction when you work with symbolic $a,b$ and check the degrees of the triples.
What do you mean with my conjecture? (are you confusing me with the OP?) And what do you mean with symbolic $a,b$ and their degrees?
@Wojowu I confused you with the OP, sorry.
I mean it indeed in the sense of @Wojowu. The threshold of 1.4 for a "good" one being kind of arbitrary....
@Wolfgang my answer works for quality > 1
Well, if the balance is $a/b=\varepsilon$, then in this construction, $A/B=\frac{(1-\varepsilon)^2}{4}<\frac14$ so indeed, that looks like my conjecture should be false for $\varepsilon <\frac14$. I would consider balances $>\frac14$ much more interesting anyway.
@Wolfgang Don't you need epsilon in the denominator of the inequality?
Oops yes that was too quick. I'll think again and check with the numerical triples...
|
2025-03-21T14:48:30.196447
| 2020-04-01T17:41:46 |
356296
|
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"authors": [
"Federico Barbacovi",
"Jim Bryan",
"Piotr Achinger",
"Puzzled",
"Will Sawin",
"abx",
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"sort": "votes",
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|
Stack Exchange
|
Formal neighbourhood of a closed subscheme
Let $X$ be a variety and $Y \subset X$ a closed subvariety.
Edit: Assume they are both smooth.
Denote $N_{Y / X}$ the normal bundle of $Y$ in $X$. The formal neighbourhood of $Y$ in $X$ is the formal scheme $X_Y = \text{Spec}\left( \lim_{n} \mathcal{O}_X \left/ \mathcal{I}_Y^n \right. \right)$.
The question is the following: how distant is the formal neighbourhood of $Y$ in $X$ from the formal neighbourhood of $Y$ in the total space of $N_{Y / X}$?
The question should be related to the splitting of the tangent sequence
$$
0 \rightarrow T_Y \rightarrow T_X \vert_Y \rightarrow N_{Y / X} \rightarrow 0,
$$
and thus to the vanishing of $\text{Ext}^1_Y(N_{Y/X}, T_Y)$, but I can't figure out this relation.
Thanks.
You are probably assuming that $Y$ and $X$ are smooth, right?
A good example to keep in mind is where $Y$ is a non-singular fiber of a (non iso-trivial) elliptic fibration $X\to \mathbb{P}^1$. In this case $N_{Y/X}$ is trivial, but the formal neighborhood of $Y$ in $X$ is non-trivial. In this case the Ext group and the extension you write are both non-trivial, but in general I think you need to know about higher order obstructions to fully answer this question.
@abx Yes, I am assuming that they are smooth.
@Jim Bryan Thanks, I will have a look a that example. However, at a quick read, it seems it doesn't contradict the fact that the vanishing of that extension group (or of the splitting of the exact sequence), could imply the triviality of the formal neighbourhood. Am I right?
Note an additional complication in characteristic $p$: take an elliptic fibration as in Jim Bryan's comment and pull it back by the Frobenius on the base. Now Kodaira-Spencer is identically zero and the tangent sequence splits. However, the formal neighborhood remains nontrivial.
If you look at a family of elliptic curves like $y^2 = x^3 - x-t^n$ as a surface $X$ mapping to a curve $C$ with parameter $t$, and $Y$ the fiber over $0$, then the neighborhood modulo $\mathcal I_Y^n = (t)^n= (t^n)$ is trivial, hence equal to the $n$'th formal neighborhood of $Y$ in the total space of $N_{Y/X}$, but the neighborhood modulo $\mathcal I_Y^{n+1}$ is not. This shows that no finite list of obstruction classes will describe the formal neighborhood.
If the $n$th' order formal neighborhood is isomorphic to the $n$th order formal neighborhood in the normal bundle, then the ring of functions on it is isomorphic to $$\mathcal O_Y \oplus N_{Y/X}^\vee \oplus \operatorname{Sym}^{2} N_{Y/X}^\vee \oplus \dots \oplus \operatorname{Sym}^{n-1} N_{Y/X}^\vee $$ and the $n+1$st order neighborhood is locally the sum of that with $\operatorname{Sym}^n N_{Y/X}^\vee$ and then automorphisms respecting the algebra structure and the filtration but not the direct sum structure are given by $\operatorname{Sym}^n N_{Y/X}^\vee$-values derivations on that sum. A derivation is determined by the Leibnitz rule and its value on the first two factors, so the local automorphisms are given by $$\mathcal T_Y \otimes \operatorname{Sym}^n N_{Y/X}^\vee + N_{Y/X} \otimes \operatorname{Sym}^n N_{Y/X}^\vee$$ and so you obstruction classes in $H^1(Y, \mathcal T_Y \otimes \operatorname{Sym}^n N_{Y/X}^\vee )$ and $H^1(Y, N_{Y/X} \otimes \operatorname{Sym}^n N_{Y/X}^\vee)$, I think.
The first obstruction controls when the formal neighborhood can be viewed as a fibration over $Y$ and the second obstruction controls when you can identify the fibers with the normal bundle.
Dear Will, thanks for your answer and sorry for the late reply. I am going through the details of what you've written, but I get stuck pretty soon. I have two questions: are you considering $y^2 = x^3 - x -t^n$ inside $\mathbb{C}^3$? Moreover, you say that the $n$th neighbourhood of $Y$ in the total space of its normal bundle is trivial, but I don't see why. If you consider the family in $\mathbb{C}^3$, the normal bundle of $Y$ in $X$ is trivial, and the $n$th neighbourhood of $Y$ in $N_{Y/X}$ should not be trivial.
Finally, in the second line after the direct sum I think there is a typo ("the filtration but not the given by...") that I can't make sense of. Thanks.
@Federico I want to consider it as a family of closed elliptic curves, say in $\mathbb P^2 \times \mathbb A^1$ where the $\mathbb P^2$ has $x,y$ coordinates and the $\mathbb A^1$ has $t$ as a coordinate. As for the typo, I added in the missing words.
@Will Sawin: are you sure about the second cohomology group? According to Hartshorne "Algebraic Geometry" Exercise 4.10 of Chapter 3 (which deals with $n =1$) first order extensions of $Y$ are in $1$-to-$1$ correspondence with elements of $H^1(Y,N_{Y/X}^{\vee}\otimes T_Y)$. In general to have that the formal neighborhood of $Y$ in $X$ is isomorphic to the formal neighborhood of $Y$ in $N_{Y/X}$ isn't it enough to have that $H^1(Y,\text{Sym}^nN_{Y/X}^{\vee}\otimes T_Y) = 0$ for all $n > 0$?
@Arty I am not sure that obstructions in the second cohomology group are nontrivial. If you can prove that only the first obstruction is needed, that would be great! But I don't think what happens for $n=1$ is necessarily a great guide to what happens for all $n$.
|
2025-03-21T14:48:30.196782
| 2020-04-01T18:02:43 |
356298
|
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|
Stack Exchange
|
Real representations of SO(n) and U(n)
I would like to get some references where I can find the theory of the real representations of $\mathbf{SO}(n)$ and $\mathbf{U}(n)$.
In particular, I would like to know for which dimensions there exist irreducible representations of these groups and for which dimension, how many non-equivalent representations are?.
See Math Overflow: https://mathoverflow.net/questions/227869/is-there-a-formula-for-the-frobenius-schur-indicator-of-a-rep-of-a-lie-group
Take a look at John Frank Adams' Lectures on Lie groups. W. A. Benjamin, Inc., New York-Amsterdam 1969 xii+182 pp.. chapter 6.
Read about Frobenius-Schur anywhere. In a nutshell a complex irreducible $V$ of complex dimension $n$ can give 1 or 2 real irreducibles, whose real dimensions are $n$ or $2n$. This can be easily determined by computing the FS-indicator or dimensions of invariants in both $S^2V$ and $\Lambda^2 V$. The latter can be done in Lie for each particular representation.
Minor correction: $n$ or $2n$, not $n$ or $n/2$.
The book by Broecker and tom Dieck (Representations of Compact groups, Springer Graduate Texts in Math) has a very useful section about real, complex and quaternionic representations and how to pass between them.
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2025-03-21T14:48:30.196898
| 2020-04-01T18:13:10 |
356299
|
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|
Stack Exchange
|
When is the internal covering number of a metric space monotonic?
Given a radius $r > 0$, the internal covering number of a subset $T$ of a metric space $(X, d)$ is denoted $N_r(T)$ and is defined to be the smallest number of balls of radius $r$ (under $d$) with centres in $T$ such that $T$ is contained in the union of the balls.
Given another subset of $X$, $U$, which is a superset of $T$, it is not necessarily true that $N_r(T) \leq N_r(U)$.
My question:
Are there well known examples of sets for which $T \subseteq U$ but $N_r(T) > N_r(U)$?
Are there necessary/sufficient conditions on $X$ or $d$ such that the internal covering number is monotonic, i.e. $T \subseteq U \implies N_r(T) \leq N_r(U)$?
In case it is relevant, my application is to cases in which $X$ is generated by $T$ under some (infinite) set of transformations (e.g. a Lie group).
Do you mean balls of a particular radius? Otherwise the internal covering number of a nonempty bounded set is $1$ and the internal covering number of an unbounded set is $\infty$.
apologies. yes, I do. Typically you would specify a radius and ask how many balls of that radius are needed to cover a set. I'll update the question.
An easy example: if $U$ is the Euclidean $n$-dimensional punctured disc of radius $r$, and $T$ the punctured disk, then $N_r(T) = n + 1$ while $N_r(U) = 1$. For general $X$, this idea implies that for any $x \in X, r \in \mathbb{R}_{>0}$ there is some $x_r$ such that $d(x, y) < r \implies d(x_r, y) < r$ (or $\leq r$, depending on whether you mean the open or closed ball).
@user44191, a very clean example, nice -- punctured contained if the full (unpunctured) disk. (Your first word "punctured" was a typo).
Ultrametric spaces satisfy 2. (https://en.wikipedia.org/wiki/Ultrametric_space)
@AntonPetrunin Are ultrametric spaces the only spaces that satisfy 2?
Yes it seems that only ultrametric spaces satisfy condition 2 --- it sufficient to check 2-point sets in 3-point sets.
The most obvious example:
Suppose $U$ is a closed ball of radius $1$ in $\mathbb R^d$, and $T$ is the corresponding sphere. Then if $r = 1$, $N_r(U) = 1$ but $N_r(T) > 1$.
EDIT:
Let $X$ be any metric space such that there are three points $a,b, c$ with $d(a,b) \le d(a,c) < d(b,c)$. Then take $d(a,c) \le r < d(b,c)$, $T = \{b,c\}$ and $U = \{a,b,c\}$. We have $N_r(U) = 1$ but $N_r(T) = 2$.
Asking that this example does not exist is a rather severe restriction on the metric space!
I think that that restriction is equivalent to the space being ultrametric, as said in the comments to the question. Examples include p-adic spaces.
|
2025-03-21T14:48:30.197099
| 2020-04-01T18:40:48 |
356301
|
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|
Stack Exchange
|
Smooth subvarieties through a singular point
Suppose that $X$ is a singular variety of dimension $n$, with singular locus of dimension $a$. Does there always exist a smooth subvariety $V$ of given dimension $m$ (where $a < m < n$) which contains $Sing(X)$? Of course it is sometimes possible: for example the ruling of a singular quadric cone passes through the singular point. But I am not sure what the general situation is.
What is the obstruction to finding a $V$? If there always is a $V$, can you do better, for example by requiring the tangent space to $V$ at a given singular point to contain a particular tangent direction?
What if $a = n-1$ and $\operatorname{Sing}(X)$ is itself singular?
If $a = n-1$ then there is no $m$ with $a < m < n$. But anyway, assume $a = 0$ if you want; I'm curious about any results with this general flavor. Given a specific $X$ and $m$, how can I tell which way things go?
Oh, $m$ is given? That is not clear from the current formulation of the question. Please edit the question to clarify your exact assumptions.
I am pretty sure there are counterexamples with $n=2$, $a=0$ and $m=1$.
Do you know what the obstruction is in that case? Given a normal surface singularity, what do I need to compute to know whether there exists a smooth curve passing through it?
For a normal surface singularity, whether or not the exceptional divisor in some resolution has a reduced component is probably the relevant condition.
One obstruction is the following: if $X$ is factorial with $p \in X$ a singular point, then any codimension-1 subvariety of $X$ containing $p$ must be singular at $p$.
If $X$ is factorial, then every codimension $1$ subvariety passing through a singular point is singular itself, so this tells you that on such an $X$ the answer is negative for $m=n-1$.
ps: Uh, I've just noticed that the same condition was mentioned by @Pop, except that mathoverflow hides comments after a certain number, so you won't see it unless you click on "Show more"...
@SándorKovács: nice to receive your imprimatur. :)
A simplest example of a normal factorial surface singularity is $E_8$ with equation $x^2 + y^3 + z^5 = 0$. By comments of Pop and Sándor Kovács, there will be no smooth curves on this surface passing through the origin.
|
2025-03-21T14:48:30.197302
| 2020-04-01T20:06:27 |
356306
|
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|
Stack Exchange
|
Is the spectrum of a "self adjoint" operator real on $\ell^p$?
There might be an obvious answer to the question, but it doesn't come to mind.
Suppose we have an infinite matrix $A=(a_{ij})$, which defines a bounded linear operator on $\ell^p$, i.e. for all sequences $(x_i)\in \ell^p, p>1$
$$ \sum_{i=1}^\infty\big|\sum_{j=1}^\infty a_{ij}x_j\Big|^p \leq C \Vert x \Vert_{p}^p.$$
For some positive constant $C$.
Furthermore, assume that $a_{ij}=\overline{a_{ji}}$. Is it true that the spectrum of $A$ on $\ell^p $ is real ?
Do you know whether the point spectrum is always real?
Yes, Jochen, because the hypothesis implies that $A$ also is bounded on $\ell_q$, $1/p+1/q=1$, hence all eigenvectors are in $\ell_2$. Basically the same argument gives a positive answer to the OP's question (I think).
@BillJohnson Your argument doesn't seem to use the fact $p>1$; but there are known examples of "self-adjoint" (i.e. conjugate symmetric) convolution operators on $l^1$(free group) whose spectrum has non-empty interior
Comment for the OP: I suspect the answer is negative for suitable convolution operators on $\ell^p({\bf F}_2)$ but right now I don't remember if the details for the $p=1$ case work for $1<p<2$.
Actually, it may be enough to work on ${\bf FS}_2$, the free monoid generated by 2 elements: for $p=1$ the construction in the proof of Theorem 5.1 of Jenkins's 1970 paper https://projecteuclid.org/euclid.pjm/1102977529 but as I said, I have not tried to see if the same ideas wor for $1<p<2$.
There are self-adjoint operators in $L^2$, generating (analytic) semigroups in $L^p$ for all $p$, such that the spectrum is not real for $p \neq 2$. One can take $L=r^2D_{rr}+2rD_r$ in the half-line. The spectrum is a parabola which degenerates into the negative half-line when $p=2$. The resolvents and the semigroups have similar bad properties, by the spectral mapping theorem. Maybe a discrete version can be contructed using them.
I think I remember there is a paper by Gohberg and Krupnik in which they construct an operator on $\ell_p$ which is self-adjoint for $p=2$ and the spectrum is an ellipsoid por $p\neq 2$ which increases with $p>2$. Unfortunately, I have no access to my office.
@M. Gonzalez Very nice If you can recover that example (or give a more precise reference). I am also locked at home but I have online access to the library.
I guess the only correct part of my comment above is that the point spectrum is real when $p<2$ (because the operator is bounded on $\ell_2$ and $\ell_p \subset \ell_2$).
@Giorgio Metafune It is something I used in my Thesis, around 1983. I have tried to find the reference, but I failed.
It seems that I have found a counter example myself.
For the Hilbert matrix
$$ H_\lambda:= \big( \frac{1}{1-\lambda+k+n} \big)_{k,n\geq 0}, \lambda < 1 $$
Rosenblum in "On the Hilbert Matrix I, Proceedings of the AMS" proves that the pointspectrum considered as an operator on $\ell^p, p>2$ contains the set
$$ \{ \pi \sec(\pi u ) : | \Re ( u )| < 1/2-1/p \}. $$
If one could provide a more elementary counterexample I would be interested in looking at.
However, I think it is true for $1 < p \le 2$ because, for $q \ge p$, if we denote $\sigma^q(A)$ the spectrum of the unbounded operator $A$ on $\ell^q$ with domain ${x \in \ell^q \mid Ax \in \ell^q}$ then we can show that the spectrum is increasing with $q$ meaning that $\sigma^q(A) \subset \sigma^r(A)$ for $p \le q \le r$. If I am not mistaken, for $1 < p \le 2$, the operator $A$ induce a self-adjoint operator on $\ell^2$ so $\sigma^p(A) \subset \sigma^2(A) \subset \mathbb{R}$.
If you can provide the missing details please go ahead and post it as an answer, I will accept it.
|
2025-03-21T14:48:30.197583
| 2020-04-01T20:57:10 |
356310
|
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|
Stack Exchange
|
Covering number of $l_2$ Ball in $\mathbb{R}^d$
What is the covering number $N(\epsilon, B_2, ||\cdot||_2)$ of a ball $B_2$ in $\mathbb{R}^d$ of radius $r$ under the $l_2$ norm?
In what space exactly?
@NateEldredge The space is $\mathbb{R}^d$ for some dimension $d$. And the ball is the $l_2$ ball.
Okay. The focus on $l^2$ and the [tag:banach-spaces] tag made it sound like you had infinite-dimensional spaces in mind.
The $\epsilon$-covering number of the euclidean unit ball in $\mathbb R^d$ scales like $(1/\epsilon)^d$. More formally,
Lemma. If $\epsilon < 1$, then $(1/\epsilon)^d \le N(\epsilon, B_2) \le (3/\epsilon)^d$. Else $N(\epsilon,B_2) = 1$.
Proof. See Theorem 4.2 and Example 14.1 of this manuscript http://www.stat.yale.edu/~yw562/teaching/598/lec14.pdf. $\quad\quad\Box$
Now, use this to get (an estimate of) the covering number of $rB_2$, for any $r \ge 0$.
|
2025-03-21T14:48:30.197678
| 2020-04-01T21:06:46 |
356312
|
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|
Stack Exchange
|
Sufficient conditions on $ a_i,b_i$ for $a_1\phi(n)+b_1, \cdots, a_k\phi(n)+b_k$ to be simultaneously prime infinitely often?
I am really interested in sufficient conditions on $a_i, b_i$ guaranteeing that the linear forms $a_1\phi(n)+b_1,\dots, a_k\phi(n)+b_k$ become simultaneously prime for infinitely many positive integers $n$. Here, $\phi(n)$ is the Euler's totient function and $a_i, b_i$ are arbitrary positive integers.
Background. In the 1904 paper of L. E. Dickson titled A new extension of Dirichlet’s theorem on prime numbers, Messenger of Math. 33 (1904), 155–161,
the problem of if the forms $a_1 n+b_1 ,\cdots ,a_kn+b_k$ become simultaneously prime infinitely many times is discussed (See Dickson's Conjecture.) In my question here $n$ is replaced with $\phi(n)$.
Note: The motivation of this question is the estimation of primes using Euler's toitient function and also to know if there is any connection between Lucas-Lehmer Primality test and the titled affine-linear form for $d=1$ and $t=1$ because the latter depend on relationship between $n$ and $\phi(n)$
I find even the problem of constructing an infinite family of pairs $(a,b)$ with the property that each linear form $a\phi(n)+b$ becomes prime for infinitely many values of $n$ hard. The only example of such a form I have is $\phi(n)+1$ where $a=b=1$; its value is prime once $n$ is. But for instance, if you want to use the same approach to argue that the pairs $\phi(n)+1,\phi(n)-1$ or $\phi(n)+1,2\phi(n)+3$ become simultaneously prime infinitely often, you arrive at unsolved conjectures such as the infinitude of twin primes or Germain primes.
@KhashF this question was upvoted and now it downvoted however it received good answer
|
2025-03-21T14:48:30.197822
| 2020-04-01T22:04:39 |
356315
|
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|
Stack Exchange
|
Solve optimal control problem whose associated system is nonlinear
Solve the optimal control problem of the LQR kind
$$
\min_u \int_0^{+\infty} x_1^2+x_2^2+\gamma(u_1^2+u_2^2) \, dt \quad\text{such that}\quad \begin{cases}\dot x_1=\alpha(x_2-x_1)+u_1,& x_1(0)=1,\\\dot x_2=\beta(x_1-x_2)+u_2,& x_2(0)=-1\end{cases}
$$
where $\alpha>\beta>0$ and $\gamma>0$.
I notice that all $x_i$ and $u_i$ in the integrand are squared and that there are no subtractions, hence the integrand has a minimum value in $0$ reached when $x_i=u_i=0$. Moreover, since $x_i$ and $u_i$ are squared, I guess the integrand is a paraboloid, hence a convex function. Could this information be helpful in finding a solution?
Since I'm not able to find a trivial solution via heuristic arguments, I tried using some methods (based on Riccati equation).
First method
Introducing $Q=B=I_2,\ \ R=\gamma I_2,\ \ A=\begin{pmatrix}-\alpha & \alpha\\\beta & -\beta\end{pmatrix}$, the problem can be rewritten as
$$
\min_u \int_0^\infty (x^T Q x + u^TRu) \, dt \quad\text{such that}\quad \begin{cases}\dot x = Ax+Bu \\ x(0) = \begin{pmatrix}1\\-1\end{pmatrix}\end{cases}
$$
with the following associated Riccati equation
\begin{align*}\tag1
0 &= Q+A^TS+SA-SBR^{-1}BS,\qquad S=\begin{pmatrix}s_1 & s_2 \\ s_2 & s_3\end{pmatrix} \\ &= I+A^TS+SA-\frac1\gamma S^2
\end{align*}
Both $Q$ and $R$ are symmetric and positive definite, so the optimal control is given by
$$
u = -R^{-1}B^TSx = -\frac1\gamma Sx,
$$
plugging it in the expression for $\dot x$ and integrating we can find $x$. But first we have to find $S$. Equation $(1)$ is equivalent to the following nonlinear system
$$
\begin{cases}
-\dfrac{s_1^2}{\gamma}-2\alpha s_1 -\dfrac{s_2^2}{\gamma}+2\beta s_2+1=0\\
\alpha s_1 - \alpha s_2 - \beta s_2 + \beta s_3 - \dfrac{s_1s_2}{\gamma}-\dfrac{s_2s_3}{\gamma}=0\\
-\dfrac{s_2^2}{\gamma}+2\alpha s_2 - \dfrac{s_3^2}{\gamma} - 2\beta s_3+1 = 0
\end{cases}
$$
which apparently doesn't have a trivial solution. Since I don't know how to solve it by hand, I tried running the following Matlab code
syms x y z a b g
eqn1 = 0 == -x^2/g-2*a*x-y^2/g+2*b*y+1;
eqn2 = 0 == a*x-a*y-b*y+b*z-x*y/g-y*z/g;
eqn3 = 0 == -y^2/g+2*a*y-z^2/g-2*b*z+1;
sol = solve([eqn1, eqn2, eqn3], [x, y, z]);
but the provided solutions are huge in length, hence they are not handy.
Second method
Alternatively, $S$ could be found by determining the eigenvectors of the associated Hamiltonian matrix
$$
H = \begin{pmatrix}A & -BR^{-1}B^T \\ -Q & -A^T \end{pmatrix} = \begin{pmatrix}A & -\gamma^{-1}I_2 \\ -I_2 & -A^T \end{pmatrix}
$$
whose special feature is that if $λ$ is an eigenvalue, then also $−λ$, $\bar λ$ and $−\bar λ$ are eigenvalues. Moreover, denoting $\begin{pmatrix}V_1 \\ V_2\end{pmatrix}\in\mathbb C^{4\times2}$ the matrix whose two columns are the eigenvectors corresponding to the two eigenvalues of $H$ having negative real part (more info here), we have
$$
S = V_2 V_1^{-1}
$$
Apparently there is no explicit decomposition of $H$ (a decomposition is found here, however without explicit formulas for the eigenvalues and eigenvectors). The Matlab function icare finds $S$ using this method, but only if the parameters are known
syms a b g
B=eye(2); Q=eye(2); R=g*eye(2); A=[-a a;b -b];
icare(A,B,Q,R)
Error using icare
Conversion to logical from sym is not possible.
Neither method seems to provide a handy expression for the solution. Do you how to solve the problem using the two methods above or other methods?
Why do you believe a handy expression exists? In other words, why so you ecpect a simpler expression than the result of your method 1.
@PiyushGrover I guess a handy expression exists since the integrand is convex. What expression are you talking about? I did not write any expression for $x$
What does the problem being convex have to do with "handy expression" ? Convex just means you get a global minima. The expression I was referring to is the output of matlab from method 1.
Please do not double-post. Solve a 2-dimensional optimal control problem via Riccati nonlinear equation
@FedericoPoloni should I close this one, and update the old one adding the text written here?
@soundwave Frankly I don't see the difference between the two questions. Personally I would delete this question, and leave the other as it is (no need to add anything). Also, all evidence up to now points to "there is no nice closed-form expression", so unless you have a good answer to Piyush's comments I don't think there is more to add.
|
2025-03-21T14:48:30.198224
| 2020-04-01T22:48:08 |
356319
|
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|
Stack Exchange
|
Pros and cons of using integer programming alone or combined integer and global optimization?
First, I am not sure if this is the right question to ask in this forum. But I have been looking for answers for a long time and I have been also asking my university's "engineering" professors but I don't seem to get a mathematical answer.
I am trying solve a complex optimization problem involving network and short path optimization and at the same time solve for non-linear pressure, flows and diameter. I can apply one of the following methods:
The first method is to use a Mixed Integer Linear Programming method
(MILP) only and linearize the pressure, flows and diameter using
piece-wise linear approximation. The linearization is done since the
MILP only takes linear equations and look for an optimal solution.
The second method is to use a combined local optimization and global
optimization method. The local optimization would use MILP first to
find solution to sub-problems where continuous optimization would be
costly to use (e.g. allocation of production ). Then, I would use a
global optimization method using derivative-free genetic algorithm
to perturb the system (e.g. the network path) and
find a better global solution. Every time the system is perturbed,
the local MILP optimization is repeated.
I am after any recommendations, heads-up and things to watch out for if I implement one of these two methods. Is the second method mathematically acceptable compared to the first method?
What exactly is the problem you are trying to solve? Is this a flow optimization problem? A routing problem to find where to send the follow to a certain location? I am still trying to understand what your objective function would be.
|
2025-03-21T14:48:30.198371
| 2020-04-01T23:03:55 |
356320
|
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"Rohil Prasad",
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"url": "https://mathoverflow.net/questions/356320"
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|
Stack Exchange
|
When is this differential form harmonic?
Let $(M^3, g)$ be a (closed) Riemannian manifold and let $u: M \to S$ be a harmonic function, where $S$ is a closed orientable surface. If $\omega$ is a $2$-form on $S$, what are sufficient conditions on $\omega$ in order $u^* \omega$ to be a harmonic $2$-form on $M$?
The concrete case I am analyzing is the following: we have $u_1, u_2 : M \to \mathbb{S}^1$ harmonic functions, $u = (u_1, u_2):M \to \mathbb{S}^1 \times \mathbb{S}^1$ and $\omega = d\theta_1 \wedge d\theta_2$, where $d\theta_i$ denotes the volume form on each $\mathbb{S}^1$.
I’d be glad to have a solution in either case.
I think this is probably false most of the time in the concrete case that you're describing. Just like a product of two harmonic forms is not necessarily harmonic, there's no good a priori reason that a wedge product of harmonic 1-forms (which is $u^*\omega$ in your case) is harmonic.
In your particular case, it seems clear to sufficiency two conditions: the first is that $\nabla u^*(\omega) = 0,$ i.e, $u^*(\omega)$ is parallel.
The last is quite technical so let me introduce some concepts before:
Recall the Weitzenbock formula:
$$\Delta = \nabla^*\nabla + Q^p.$$
Since $\Sigma$ is closed we can intrdocute the $L^2$ inner product of $\Omega^p(\Sigma)$ in the following way
$$(\omega,\beta) := \int_{\Sigma}\omega\wedge\star\beta = \int_{\Sigma} g(\omega,\beta)d\mu_g.$$
Therefore, $\Delta$ is self-adjoint on this inner product and $\nabla^*$ is the dual of $\nabla$ on this inner product.
In the particular case of $n=3$, $Q^p$ is nothing but the Ricci tensor on $2$-forms, i.e, if $R$ denotes the Riemann tensor, then
$$R(X,Y)\omega := \nabla_X\nabla_Y\omega - \nabla_Y\nabla_X\omega -\nabla_{[X,Y]}\omega,$$
$$\langle R(X,Y)\omega,\omega\rangle $$
the Ricci tensor corresponds to
$$(\mathrm{Ric}~\omega)(X_1,X_2) = \sum_{s=1}^3\left(R(X_s,X_1)\omega(X_s,X_1) + R(X_s,X_2)\omega(X_s,X_2\right)),$$
where $\{X_1,X_2,X_3\}$ is an orthonormal pair.
Hence the Weitzenbock formula reduces to (for $p=2$)
$$\Delta = \nabla^*\nabla + \mathrm{Ric}.$$
Therefore, for any parallel $2$-form $\beta$ one has
$$(\Delta\beta,\beta) =\int_{\Sigma} \langle \Delta\beta,\beta\rangle =\int_{\Sigma} \langle \nabla^*\nabla\beta,\beta\rangle + \int_{\Sigma}\langle\mathrm{Ric}~\beta,\beta\rangle = \int_{\Sigma}|\nabla\beta|^2 +\int_{\Sigma}\langle\mathrm{Ric}~\beta,\beta\rangle = \int_{\Sigma}\langle\mathrm{Ric}~\beta,\beta\rangle.$$
Since
$$(\Delta\beta,\beta) = (d\delta\beta,\beta) + (\delta d\beta) = |\delta\beta|^2 + |d\beta|^2,$$
onde thus concludes that $u^*(\omega)$ is closed and co-closed, hence, harmonic, provided if
$$\int_{\Sigma}\langle\mathrm{Ric}~u^*(\omega),u^*(\omega)\rangle\leq 0,$$
and this is our second condition.
Note that in general, parallel forms are not harmonic, but in this particular that if $g$ is Ricci flat, we are done, since
$$\Delta u^*(d\theta_i) = 0,$$
and for $p=1$, $Q^p$ is also the Ricci tensor, it follows that
$$0 = \int_{\Sigma}\langle \mathrm{Ric}(u^*(d\theta_i)),u^*(d\theta_i)\rangle +\int_{\Sigma} |\nabla u^*(\theta_i))|^2 = \int_{\Sigma} |\nabla u^*(\theta_i))|^2,$$
and hence, $u^*(\theta_i)$ is parallel, hence, all of our conditions would be satisfied.
|
2025-03-21T14:48:30.198584
| 2020-04-02T02:42:49 |
356324
|
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|
Stack Exchange
|
The definition of Langlands' L-function $L(s,\pi,r)$ in the case of $\operatorname{GL}_1$
Let $G$ be a split reductive group over a $p$-adic local field $k$. For $\pi$ an unramified representation of $G(k)$, and $r$ a finite dimensional representation of the L-group $^LG$, Langlands showed how to attach an L-function $L(s,\pi,r)$.
I have computed this L-function in the case of $\operatorname{GL}_1$ and am getting something that looks "wrong" to me. I was hoping to state it here and perhaps someone can explain where I am making a mistake.
Definition of $L(s,\pi,r)$:
Let me first say in my own words how to define $L(s,\pi,r)$. Let $\mathscr H_G = C_c^{\infty}(G(\mathcal O_k) \backslash G(k) / G(\mathcal O_k))$ be the spherical Hecke algebra of $G$, and let $\mathscr H_T = C_c^{\infty}(T(k)/T(\mathcal O_k))$ be the one of $T$. The Satake transform
$$S: \mathscr H_G \rightarrow \mathscr H_T$$
$$S(f)(t) = \delta_B(t)^{1/2} \int\limits_U f(tu) du,$$
for $B = TU$ a Borel subgroup of $G$, defines an isomorphism of $\mathscr H_G$ with a subalgebra $\mathscr H_T^W$ of Weyl group invariant functions on $\mathscr H_T$. If $\pi$ is an unramified representation of $G(k)$, then there is a nonzero vector $v$ in the space of $\pi$, and a character $\xi$ of $\mathscr H_G$, such that
$$\xi(f)v = \int\limits_{G(k)} f(g)\pi(g)vdg \tag{$f \in \mathscr H_G$}.$$
There is an unramified character $\chi$ of $T(k)$, unique up to Weyl group equivalence, such that
$$\xi(f) = \int\limits_{T(k)} S(f)(t) \chi(t)dt.$$
The Harish-Chandra map $H_T: T(k) \rightarrow \operatorname{Hom}_{\mathbb Z}(X(T),\mathbb Z) = X(^LT)$ defined by $H_T(t)(\eta) = -\operatorname{ord}_k\eta(t)$ is surjective with kernel $T(\mathcal O_k)$, and induces an isomorphism
$$T(k)/T(\mathcal O_k) \rightarrow X(^LT)$$
and therefore we get bijections
$$\operatorname{Hom}_{\textrm{grp}}(T(k)/T(\mathcal O_k), \mathbb C^{\ast}) \rightarrow \operatorname{Hom}_{\textrm{grp}}(X(^LT),\mathbb C^{\ast}) \rightarrow ^LT$$
through which $\chi$ corresponds to an element $A_{\pi} \in \space ^LT \subset \space ^LG$. Then Langlands defines
$$L(s,\pi,r) = \operatorname{det}(1 - q^{-s} r(A_{\pi}))^{-1}.$$
Computation in the case $\operatorname{GL}_1$:
Let $T = \operatorname{GL}_1$, so that $^LT = \mathbb C^{\ast}$. Let $r: \space ^LT \rightarrow \operatorname{GL}_1(\mathbb C) = \mathbb C^{\ast}$ be the identity map.
If $\chi$ is an unramified representation (unramified character) of $T(k)$, then by the way the Harish-Chandra map $H_T$ is defined, we actually get $A_{\pi} = \chi(\varpi)^{-1}$, where $\varpi$ is a uniformizer. Then Langlands' L-function is
$$L(s,\chi,r) = (1 - q^{-s} \chi(\varpi)^{-1})^{-1}.$$
This doesn't seem right. Taking $r = \textrm{identity}$ should yield the "standard L-function" which should be
$$L(s,\chi,r) = L(s,\chi) = (1 - q^{-s}\chi(\varpi))^{-1}.$$
In fact, if you identity $k^{\ast}$ with the abelianized Weil group via the local Artin map (the one that sends a uniformizer to a geometric Frobenius),then $(1 - q^{-s}\chi(\varpi))^{-1}$ is the Artin L-function attached to the one-dimensional Artin representation corresponding to $\chi$.
Have I made a mistake in computing these L-functions? If not, how should one reconcile these L-functions?
Of course, one very easy way to force things to match up is to use a different definition of $H_T$:
$$H_T(t)(\eta) = \textrm{ord}_k \eta(t)$$
and I have seen some authors define $H_T$ this way. Another thing one could do is to define standard L-functions differently:
$$L(s,\pi) = L(s,\pi, \tilde{r})$$
for $\pi$ a representation of $\operatorname{GL}_n(k)$ and $r$ the identity map on $^L\operatorname{GL}_n = \operatorname{GL}_n(\mathbb C)$.
|
2025-03-21T14:48:30.198794
| 2020-04-02T03:08:59 |
356325
|
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"url": "https://mathoverflow.net/questions/356325"
}
|
Stack Exchange
|
Poisson equation on exterior of a ball
Let $ B_1^c$ denote the compliment of the unit ball centered at the origin in $ R^N$ where $N \ge 3$. I am interested in $ -\Delta u(x)=f(x)$ in $ B_1^c$ with $ u=0$ on $ \partial B_1^c$. In particular let $ \sigma:=N-2+\epsilon$ where $\epsilon>0$ is small. I am interested in solving the prior pde with an estimate of the form
$$ \sup_{x \in B_1^c}|x|^{\sigma+1} | \nabla u(x)| \le C \sup_{B_1^c} |x|^{\sigma+2} |f(x)|$$ where $C$ is independent of $f,u$. Note that all solutions $u$ cannot solve this since we can add $ c (|x|^{2-N}-1)$ to any solution. So my question is: i seem to be able to prove the existence of a solution $u$ with the above desired property (using spherical harmonics and a blow up argument). But this result seems a bit suspect to me. Any comments on whether this is trivally false or possibly true would be greatly appreciated.
thanks
using an abstract argument i seem to be able to prove this for general exterior domains (here the domain doesn't include the origin).
|
2025-03-21T14:48:30.198899
| 2020-04-02T03:42:24 |
356327
|
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"Federico Poloni",
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"landau"
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|
Stack Exchange
|
Diagonalizable stochastic matrix that satisfies an equation
Given an arbitrary discrete probability distribution $a = (a_1, ..., a_n)$ and another arbitrary discrete probability distribution $b = (b_1, ..., b_n)$, what is the easiest known way to find a diagonalizable $n \times n$ stochastic matrix $M$ such that $M \cdot a = b$?
Surely the easiest known way is to put $b$ in every column?
I would love that to work. How do we know the matrix is diagonalizable?
@TobiasFritz I would suggest you to post it as an answer, so that you can grab the bounty and this question can be marked as solved.
Take $M$ to be the stochastic matrix which has $b$ in every column. Then $M$ maps every probability vector to $b$, and in particular $Ma = b$.
Concerning diagonalizability, choose any basis which contains $b$ as first basis vector and such that all other basis vectors $c$ satisfy $\sum_i c_i = 0$. In such a basis, $M = \mathrm{diag}(1,0,\ldots,0)$.
|
2025-03-21T14:48:30.199005
| 2020-04-02T04:20:40 |
356329
|
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"Alex Ravsky",
"RobPratt",
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|
Stack Exchange
|
Transforming an optimization problem to maxmin formulation
Given $N=mn$ real numbers $a_i$, we seek to partition them into $n$ subsets $S_j$ ($1\le j\le n$), each containing $m$ numbers, so as to maximize $\prod_{j=1}^n \sum_{a_i\in S_j} a_i$. My questions are: (1) Can this problem be cast to a known problem? (2) Given its NP-hardness, how to design approximation algorithms with constant approximation factor?
After thinking over the problem, the transformation is impossible. Now the problem is how to solve the original problem, i.e., maximize $\prod_{j=1}^n\sum_{a_i\in S_i} a_i$
Looks like an NP-hard problem to me...
@dohmatob Thank you. Any idea on approximation algorithms?
You can approximate the problem by minimizing the range via integer linear programming. Let binary decision variable $x_{i,j}$ indicate whether number $i\in N$ is assigned to group $j\in\{1,\dots,n\}$. The problem is to minimize $u-\ell$ subject to:
\begin{align}
\sum_j x_{i,j} &=1 &&\text{for all $i$}\\
\sum_i x_{i,j} &= m &&\text{for all $j$}\\
\ell \le \sum_i a_i x_{i,j} &\le u &&\text{for all $j$}
\end{align}
To obtain a formulation for the min-max or max-min problem, omit the parts involving $\ell$ or $u$, respectively.
Thank you Rob for the comment. However, the ILP formulation is still relatively intractable. I want to have structural insight on the problem. Let me pose the following two questions: (1) is the problem NP-hard, if yes, how to prove this? (2) how to design approximation algorithms?
How big are $n$ and $m$?
In the general case we do not pose any condition on $n$ or $m$. However, I am interested in the case of large $n$ and small $m$ (e.g., $4$ or $5$).
How large is large? :) I would guess that even for fixed $m$, the problem is NP-hard, but that doesn't mean that a commercial ILP solver can't solve it quickly in practice. Do you have sample data in mind?
The size is beyond any commercial ILP solver. Therefore we seek a poly-time approximation algorithm with hopefully constant approximation factor. I have revised the problem formulation.
Without the constraint that each subset contains $m$ numbers, the Longest Processing Time (LPT) heuristic provides a 4/3-approximation for the min-max problem.
Thank you Rob for the comment. I am aware of that line of research on number partitioning. However, the problem is not exactly the same here,which is to maximize the product-sum. Have you got an idea if we can prove any approximation ratio using algorithms in that area?
This answer is partial.
We assume that all $a_i$ are non-negative. Put $A=\tfrac 1m \sum a_i$ and $A_j=\sum_{i\in S_j} a_i$. We want to maximize $\Pi=\prod A_j$. But by AM-GM inequality, $P^{1/m}\le \tfrac 1m \sum A_j=A$ and the equality holds iff each $A_j$ equals $A$.
Even when $m=2$ and all $a_i$ are positive integers the problem to check whether the equality can hold is a variant of the partition problem, and at the first reference is stated that it is $NP$-hard (but without a citation). So this should be a known problem and there can be already developed heuristic and approximation algorithms for it.
I guess that a heuristics can be based on some balancing idea, which looks promising when there are no big gaps between $a_i$’s.
For instance, I can propose the following algorithm to find an initial feasible partition. Assign $m$ piles, order $a_i$ in a non-increasing order, and then split this sequence of $a_i$’s into bags $B_1,\dots, B_n$ each containing $m$ consecutive $a_i$’s. An each step we pick a next bag and distribute its $a_i$’s into the piles, one number to each pile, trying to make the vector of the sums of elements of the piles more balanced (a measure of a balancedness of a vector $x=(x_1,\dots,x_m)$ can be a norm (for instance, $\ell_2$ or $\ell_\infty$) of a vector $x-\left(\frac 1m \sum x_i\right) (1,1,\dots,1)$). I guess this should be done as follows: put the biggest element of the bag to the pile with the smallest sum, the second biggest element of the bag to the second smallest sum, and so forth.
The obtained feasible partition (or even a random one) can be further iterative balanced by local search. Namely, given sets $S_1,\dots S_m$ and a small constant $b\ge 1$ (maybe even $b=1$ will provide a good approximation) we check all subsets $C_i$ of size $b$ in each of $S_j$ (so there are ${n\choose b}^m$ possibilities to consider in total). For each of the possibilities we consider a union $C$ of $C_i$ and try to redistribute $C$ between $S_j$ trying to make the sequence of their sums more balanced. In particular, when $b=1$ and $m=2$, we look for indices $i_1\in S_1$ and $i_2\in S_2$ such that when we swap $i_1$ and $i_2$ between $S_1$ and $S_2$, the difference $|A_1-A_2|$ will decrease.
Thank you Alex for the valuable comments. For the case $m=2$, I can show that the problem is already NP-hard by relating it to the number partitioning problem. Your idea actually coincides with the heuristics there, my question is on how to prove any approximation results using such idea.
@lchen I guess we can obtain approximation results if there are some restrictions on $a_i$’s, providing the proposed heuristics gives the values of $A_i$’s close to $A$.
Thank you Alex. Do we really need that. My feeling is that the heuristic gives quite good approximations, but fail to derive any bound.
|
2025-03-21T14:48:30.199346
| 2020-04-02T05:36:13 |
356330
|
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|
Stack Exchange
|
Independence of variables in curvilinear coordinate systems
Let $U$ be a connected open subset of $\Bbb{R}^n$, and let $(\xi_1,\dots,\xi_n)$ be a curvilinear smooth ($C^\infty$) coordinate system on $U$. Suppose $1\leq k<n$. A smooth function $f:U\rightarrow\Bbb{R}$ for which
$$
\frac{\partial f}{\partial\xi_{k+1}},\dots,\frac{\partial f}{\partial\xi_n}\equiv 0\quad (\star)
$$
does not necessarily admit a global representation of the form $g(\xi_1,\dots,\xi_k)$ on the whole $U$ even if the coordinate system $(\xi_1,\dots,\xi_n)$ coincides with the Euclidean system $(x_1,\dots,x_n)$. A typical example is
$$
f(x,y)=
\begin{cases}
0\quad\quad\,\,\,\, \text{if } x\leq 0\\
{\rm{e}}^{\frac{-1}{x}}\quad\,\,\,\, \text{if } x>0, y>0\\
-{\rm{e}}^{\frac{-1}{x}}\quad \text{if } x>0, y<0\\
\end{cases}
$$
which is a smooth function on $U:=\Bbb{R}^2-[0,\infty)$ with $\frac{\partial f}{\partial y}\equiv 0$, but cannot be written as $f(x,y)=g(x)$ for any $g:\Bbb{R}\rightarrow\Bbb{R}$.
Two observations regarding this example:
The function $f$ above is not analytic ($C^\omega$).
Defining a function $g$ of a single variable by setting $g(x_0)$ to be the value of $f$ on the vertical line $x=x_0$ does not work because there are vertical lines in $\Bbb{R}^2$ whose intersections with the domain $U$ of $f$ is disconnected, and thus $f$ may attain more than one value on $\{x=x_0\}\cap U$.
Going back to the original setting, it is not hard to show that given a curvilinear coordinate system $(\xi_1,\dots,\xi_n)$ on $U\subseteq\Bbb{R}^n$ and a smooth function $f:U\rightarrow\Bbb{R}$ satisfying $(\star)$, it may be written as $g(\xi_1,\dots,\xi_k)$ on the entirety of $U$ if either 1) $f$ is analytic or 2) any coordinate surface $\{\xi_{1}=a_1,\dots,\xi_k=a_k\}$ in $U$ is connected.
My Question:
Are there conditions on the function $f$ or on the coordinate system that guarantee the existence of a global representation of the form $g(\xi_1,\dots,\xi_k)$? Also is there a term for curvilinear coordinate systems $(\xi_1,\dots,\xi_n)$ for which all coordinate surfaces (i.e. the level sets of functions of the form $(\xi_{i_1},\dots,\xi_{i_m})$) are connected?
|
2025-03-21T14:48:30.199494
| 2020-04-02T06:21:05 |
356332
|
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"ARA",
"Andrej Bauer",
"Martin Brandenburg",
"fosco",
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|
Stack Exchange
|
Explicit description of exponentials of étalé spaces
It is well known that the category $\mathit{Sh}(X)$ of sheaves of sets on a topological space $ X $ is a topos.
On the other hand, there exists a natural equivalence of categories between $\mathit{Sh}(X)$ and the category $\mathit{Et}(X)$ of étalé spaces over $ X $, so that the category of etale spaces over $ X $ is a topos, too.
The constructions in $\mathit{Sh}(X)$ are translatable to $\mathit{Et}(X)$, and vice versa.
For example, products of sheaves correspond to fiber products of étalé spaces.
The description of the exponentials in $\mathit{Sh}(X)$ is given, e.g., in [Mac Lane and Moerdijk, Sheaves in Geometry and Logic]. The question that arises is
What is the explicit construction of exponentials in $Et(X)$?
Have you tried to compute this by yourself? All the ingridients are explictly available: the functors passing between sheaves and etale spaces, as well as the exponentials in sheaves. Where did you get stuck?
It is basically a nonlinear variant of the well-known description of inner homs in the category of (geometric) vector bundles over X.
Let $[E,E’]$ be the internal hom of two étale spaces $\pi : E \to X$ and $\pi’: E’ \to X$. I will write $\mathrm{Hom}_X(A,B)$ for the set of morphisms of étale spaces $A$ and $B$ over $X$, and $A \times_X B$ for the product of $A$ and $B$ as étale spaces over $X$.
To compute the sections of $[E,E’]$, you can use
$$\mathrm{Hom}_X(U,[E,E’]) \simeq \mathrm{Hom}_X(U \times_X E, E’) \simeq \mathrm{Hom}_X(U \times_X E, U \times E’).$$
The first isomorphism uses the universal property of the exponential, the second isomorphism uses that the image of $U \times E$ is always contained in $(\pi’)^{-1}(U) =U \times E’$, so there is a unique factorization through $U \times E’$.
The elements of the total space correspond to stalks, so the fiber of $[E,E’]$ above $x \in X$ is given by the set
$$[E,E’]_x \simeq \varinjlim_i \mathrm{Hom}_X(U_i \times_X E, U_i \times_X E’)$$
where $(U_i)_{i \in I}$ is a projective system of open neighborhoods of $x$ such that every open neighborhood of $x$ contains some $U_i$ (for example, if $X$ is a metric space, you can take open balls of shrinking radius).
A basis of open sets on $[E,E’] = \bigsqcup_x [E,E’]_x$ is given by the sets $A_{(U,s)} = \{ s_x : x \in U \}$, for $U$ in some basis of open sets of $X$, $s \in \mathrm{Hom}_X(U \times_X E, U \times_X E’)$, and $s_x$ the element of $[E,E’]_x$ corresponding to $s$. This topology is often not easy to visualize. The spaces $E$ and $E’$ are already typically non-Hausdorff, and even if they are Hausdorff there is no guarantee that $[E,E’]$ will be as well. But maybe it helps to keep in mind that the subspace topology on the fibers is always the discrete topology (as for any étale space).
For example, take $X= \mathbb{R}$, $E = \mathbb{R}-\{0\}$ and $E’ = \mathbb{R}\sqcup\mathbb{R}$ with $\pi$ and $\pi’$ the natural projections.
We first look at $x \neq 0$. For a small open interval $U$ containing $x\neq 0$, there are exactly two maps $E \times_X U \to E’ \times_X U$ (because $E \times_X U \simeq U$ and $E’ \times_X U \simeq U \sqcup U$).
However, if $U$ is a small open interval containing $x=0$, then $E \times_X U = U -\{0\}$ has two components, and because of this there are four maps to $E’ \times_X U \simeq (U-\{0\}) \sqcup (U - \{0\})$.
So in this example, $[E,E’]$ has four points above $x=0$ and two points above each $x \neq 0$. Further, $[E,E’]$ has four global sections corresponding to the four continuous maps $E \to E’$ over $X$. This gives four open sets $A_{(X,s)}$ in $[E,E’]$, with each of these homeomorphic to $\mathbb{R}$. But these four lines are glued together to a single line on $\{ x \in \mathbb{R} : x > 0\}$ iff the corresponding sections agree on $\{ x \in \mathbb{R} : x > 0\}$, and similarly over $\{x \in \mathbb{R} : x < 0 \}$.
Here is a picture:
The gluing of different lines is supposed to be “instant” (just as in the case of the line with two origins) and as a result the space is non-Hausdorff.
Depending on what you mean by "explicit", the following answer will be satisfying or not:
Let $F : {\sf Et}(X) \leftrightarrows {\sf Sh}(X) : G$ be the equivalence in subject. Let $[-,-]$ be the exponential of sheaves, i.e. on presheaves, i.e. the sheaf $U\mapsto {\sf Sh}(X)(yU \times A,B)$ (if I remember well, it is enough that $B$ is a sheaf for this to be a sheaf). Then, the exponential $\langle -,-\rangle$ on étale spaces is the functor $(E,E')\mapsto G[FE, FE']$.
Indeed, let's prove that ${\sf Et}(X)(E\times E', E'')\cong {\sf Et}(X)(E, \langle E',E''\rangle)$:
$$
\begin{align*}
{\sf Et}(X)(E\times E', E'') & \cong {\sf Sh}(X)(F(E\times E'), FE'') \\
& \cong {\sf Sh}(X)(FE\times FE', FE'')\\
& \cong {\sf Sh}(X)(FE, [FE', FE''])\\
& \cong {\sf Et}(X)(E, G[FE', FE'']) \\
& = {\sf Et}(X)(E, \langle E', E''\rangle).
\end{align*}
$$
I have used the following facts: 1. $F$ preserves products (and for that matter, all limits: it is an equivalence). 2. ${\sf Sh}(X)$ is cartesian closed.
By explicit I mean an object in $Et(X)$ that represents the exponential of $E,E'\in Et(X)$.
So what you don't like in my answer? $F,G$ and $[-,-]$ are explicitly determined(in McLM, or in other books on sheaf theory). $F$ sends an étale space to its sheaf of sections. $G$ is more difficult to write down properly, but you can do it with a little effort. ${\sf Et}(X)$ is cocomplete, as well as $G$, so it is enough to define it on representables in ${\sf Sh}(X)$ and then take a Kan extension.
In your answer, what I like to know is just the description of the etale space $ G[FE, FE'] $ with respect to $E$ and $E'$. At least, what is the total space of $ G[FE, FE'] $?
The total space of $G[FE,FE']$ is the disjoint union of all stalks $[FE, FE']_x$ at points of $X$. Properly topologized, this is étale over $X$.
@Fosco: I think the OP is asking precisely for the description you indicated in the comment (stalks). It would probably be helpful if you took thre trouble of amending your answer. which right now looks a bit like you're trying to play a joke on the OP.
Actually, $[FE,FE']_x$ is not $[E_x,E'_x]$, which makes the whole question rather non-trivial.
|
2025-03-21T14:48:30.200019
| 2020-04-02T06:40:17 |
356334
|
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|
Stack Exchange
|
The profinite topology on the Mordell Weil group
In this lecture of Serre on his open image theorem, around 6 minutes, Serre mentions the following theorem of Tate:
Let $A/k$ be an abelian variety over a number field and consider the Mordell-Weil group $A(k)$. It is a finitely generated group and we have the inclusion $A(k) \to \prod_v A(k_v)$ where $v$ ranges over the non archimedean valuations of $k.$
Consider the subgroups of $A(k)$ defined by finitely many congruence conditions under this map. Are these exactly the finite index groups? One direction is clear so really: Is every finite index subgroup of $A(k)$ determined by finitely many congruence conditions?
Serre mentions that Tate showed that this is equivalent to the following condition:
Let $\ell$ be a prime and consider $V_\ell(A) = \mathbb Q_\ell\otimes_{\mathbb Z_\ell}T_\ell(A)$ with the Galois action $G_k$. Then, if $H^1(G_k,V_\ell(A)) = 0$, then every finite index subgroup is cut out by congruence conditions.
How does one prove this? Answer in comments
I am a little confused because of the following argument:
By the standard Kummer pairing and taking an inverse limit, we have an inclusion:
$$0 \to A(k)\otimes\mathbb Z_\ell \to H^1(G_k,T_\ell).$$
To require $H^1(G_k,\mathbb V_\ell) = 0$ is to require $H^1(G_k,T_\ell)$ to be $\ell^{\infty}$ torsion but I don't think this is true for $A(k)\otimes\mathbb Z_\ell$?
Answer to above confusion: Tate/Cassels/Serre don't prove or need that the entire Galois cohomology is zero, they show that a particular Selmer group is $0$ and there is no contradiction.
Milne, Congruence subgroups of abelian varieties, Bull Sc Math 96 (1972) 333-338
Following that lead, see also Serre, Sur les groupes de congruence des variétés abéliennes.
|
2025-03-21T14:48:30.200171
| 2020-04-02T08:31:42 |
356339
|
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|
Stack Exchange
|
How to find the coordinates of these points?
Does anyone know any way or any algorithm that can exactly and/or numerically find the coordinates
of $n_{k}+2$ equally spaced points on the $(n_{k}-1)$-dimensional
unit sphere $S^{n_{k}-1}$ for some sequence of integers $1<n_{1}<n_{2}\cdots<n_{k}<\cdots$? Many thanks.
What does "equally spaced" mean?
Depends on what you mean by equally spaced.
If you want that all points have pairwise identical distances, then this is not possible. This is only possible for at most $n+1$ points, but not more.
If you want to maximize the minimal distance of $n+2$ points in the $(n-1)$-sphere, you have to pick some of the vertices of the $n$-dimensional crosspolytope, that is $(\pm1,...,\pm 1)$.
Here is a reference to that:
Conway, John Horton, and Neil James Alexander Sloane. Sphere packings, lattices and groups.
In Chapter 1, subsection 2.6 they introduce the function $A(n,\theta)$ to denote the maximal number of points on an $(n-1)$-dimensional sphere (in $\Bbb R^n$) so that any two points have an angle of at least $\theta$ between them.
They state
$A(n, \pi/2)=2n$ (for the vertices of the crosspolytope), and
$A(n,\theta)\le n+1$ for $\theta>\pi/2$ (for the vertices of a simplex).
Now, let's say you have an arrangement of $n+2$ points that maximizes the minimal distance between any two points.
Let $\theta$ be this (angular) distance.
By 2., we know that $\theta \le\pi/2$, and by 1. we know that $\pi/2$ can be achieved (and we actually have space to add more points).
Thank you for your answer, but why does one have to pick some of the vertices of the n-dimensional crosspolytope? Thanks.
Since you are saying "this is only possible for at most n+1 points, but not more", how about 6 and 20 points on $S^2$? Thanks.
@nancyD I will edit the answer later, to add some further information (right now, I can't). For now: for 6 and 20 points the pairwise distance is not identical, e.g. for 6 points there are antipodal pairs of vertices, and non-antipodal ones.
@nancyD I updates the answer with a reference.
$6$ and $20$ points on $S^2$ would seem to be irrelevant, nancy, since your question specifically asks for the number of points to be greater than the dimension by exactly three.
@M.Winter Thank you for your update with the reference. I am trying to read the reference and understand your answer. Thanks a lot.
|
2025-03-21T14:48:30.200362
| 2020-04-02T09:14:22 |
356342
|
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|
Stack Exchange
|
Convex Julia sets
Consider the classical Julia set $J_f$ associated with $f(z)=z^2+c$.
Since $J_c$ is completely invariant,
we know that $f^{-1}(J_f) \subseteq J_f$.
Now, let $H_f$ be the convex hull of $J_f$.
Is it true that $f^{-1}(H_f) \subseteq H_f$?
I have done some basic computer experiments, and it seem to hold for $c \in [0,1]^2 \subset \mathbb{C}$. Moreover, I suspect that the natural generalization of the statement above might hold for all polynomial maps. However, I have examples with rational maps where the statement is not true.
As an example, consider $f(z)=z^3-iz + 0.2 + 0.4i$.
The blue points is the Julia set $J_f$ associated with $f$. The shaded region is the convex hull $H_f$ of the Julia set.
Taking a uniform square grid $G$ on $H_f$, and plotting
the points $f^{-1}(G)$ gives the black dots.
As we can see, it is reasonable to guess that $f^{-1}(H_f)\subset H_f$.
This is indeed true for certain values of $c$ e.g. $c<-2$ in which case $J_{f_c}$ is a Cantor subset of the real axis and $H_{f_c}$ is a compact interval. But I can't see how this could be true when the filled Julia set has non-empty interior (special cases with smooth Julia sets such as $c=0$ aside)? Have you verified it for the basilica Julia set (c=-1)?
@KhashF Indeed, the pictures I can make definitely supports the conjecture that it works for c=-1 also.
Edited: The previously found sufficient condition is indeed necessary, but even better, it is satisfied by all polynomials of degree at least two. Thus the conjecture is true:
Theorem: Let $p$ be a complex polynomial of degree $d \geq 2$ and let $H_p={\rm conv}J_p$, the convex hull of the Julia set $J_p$ of $p$. Then $p^{-1}(H_p) \subset H_p$.
To prove this, I use several lemmata. First, there is the following version of Gauss-Lucas theorem, due to W. P. Thurston, which can be found in the preprint
A. Ch\'eritat, Y. Gao, Y. Ou, L. Tan: \emph{A refinement of the Gauss-Lucas theorem (after W.
P. Thurston)}, 2015, preprint, hal-01157602
Lemma 1: Let $p$ be any polynomial of degree at least two. Denote by $\mathcal{C}$ the convex hull of the critical points of $p$. Then $p: E \to \mathbb{C}$ is surjective for any closed half-plane $E$ intersecting $\mathcal{C}$.
From this we have the following:
Lemma 2: Let $p$ be any polynomial of degree at least two. Then all zeros of $p'$ belong to $H_p={\rm conv}J_p$, the convex hull of the Julia set $J_p$ of $p$.
Proof: Suppose there is an $x_0 \not \in H_p$ such that $p'(x_0)=0$. By the hyperplane separation theorem , there exists a closed half-plane $E$ such that $x_0 \in E$ and $E \cap H_p = \emptyset$. In particular, $E \cap J_p = \emptyset$. By Lemma 1, $p: E \to \mathbb{C}$ is surjective. Take a $z_0 \in J_p$. Then on one hand $p^{-1}(z_0) \subset J_p$, while on the other hand $p^{-1}(z_0) \cap E \neq \emptyset$, a contradiction.
The next lemma is a modification of Exercise 2.1.15 in
Lars Hörmander: Notions of Convexity
Publisher Springer Science \& Business Media, 2007 (Modern Birkhäuser Classics)
ISBN<PHONE_NUMBER>,<PHONE_NUMBER>854
Lemma 3:
Let $p(z)=\sum_{j=0}^d a_jz^j$ be a polynomial in $z \in \mathbb{C}$ of degree $d$. Let $B$ be a closed convex subset of $\mathbb{C}$ containing all zeros of $p'$. Then the set $C_B$ of all $w \in \mathbb{C}$ such that all the zeros of $p(\cdot)-w$ are contained in $B$ is a convex set.
Proof of Lemma: Note that by continuity of roots $C_B$ is closed when $B$ is. Let $w_1,w_2 \in C_B$ and $n_1,n_2 \in \mathbb{N}$ and consider the polynomial (in one complex variable $z$) $P(z):=(p(z)-w_1)^{n_1}(p(z)-w_2)^{n_2}$. Then all zeros of $P$ lie in $B$ (by definition of $C_B$), so the convex hull of zeros of $P$ is contained in $B$. By Gauss-Lucas theorem (standard version), all zeros of $P'$ are contained in $B$. The zeros of $P'$ are respectively all the zeros of $p(z)-w_1$, all the zeros of $p(z)-w_2$ (if $n_1, n_2 >1$), all the zeros of $p'$ and all the zeros of $p(\cdot)-\left (\frac{n_1}{n_1+n_2}w_1 + \frac{n_2}{n_1+n_2}w_2\right)$. By definition of $C_B$, $\frac{n_1}{n_1+n_2}w_1 + \frac{n_2}{n_1+n_2}w_2 \in C_B$. Varying $n_1,n_2$ and using the property that $C_B$ is closed, we get that $tw_1+(1-t)w_2 \in B$ for all $0 \leq t \leq 1$.
Proof of Theorem: Applying the Lemma 3 to $B=H_p={\rm conv}J_p$, we get that the set $C_p=\{w \in \mathbb{C}: \{z : p(z)-w=0\} \subset H_p\}$ is convex. Furthermore, for $w \in J_p$ all solutions of $p(z)-w$ are in $J_p \subset H_p$, so $J_p \subset C_p$. Hence $H_p \subset C_p$, which implies that $p^{-1}(H_p) \subset H_p$.
For the quadratic family $f_c(z)=z^2+c, \ c \in \mathbb{C}$ it is straightforward (without appealing to Lemma 2) to check that the critical point $0$ is the center of symmetry of the Julia set $J_c$, so it is a convex combination of two points in $J_c$.
|
2025-03-21T14:48:30.200698
| 2020-04-02T11:58:19 |
356351
|
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|
Stack Exchange
|
Siegel's bad character
Let $K$ be an imaginary quadratic field with discriminant $d_K$. Suppose that $d_K=gt$, where either $g,t$ are discriminants or have the value $g=1,t=d$. Let $f$ be an additional discriminant of a quadratic field, or $f=1$ and at the same time $g\neq1$. Denote by $G$ the positive discriminant from $fg,ft$, and $T$ the negative one. For a prime $\mathfrak p$ of $K$ such that $(\mathfrak p,f)=1$ let
$$\chi(\mathfrak p)=\begin{cases}\left(\frac{G}{N(\mathfrak p)}\right) \text{ if }\left(\frac{G}{N(\mathfrak p)}\right)\neq 0,\\ \left(\frac{T}{N(\mathfrak p)}\right) \text{ if } \left(\frac{T}{N(\mathfrak p)}\right)\neq 0.\end{cases}$$
Siegel writes that
"Für beliebige zu $f$ teilerfremde Ideale a wird
$\chi(\mathfrak a)$ auf multiplikative Art gebildet und erweist sich dann als eigentlicher
Charakter der Gruppe der Ringklassen mit dem Führer $\lvert f\rvert$."
What exactly he means by "Gruppe der Ringklassen mit dem Führer $f$"? Is it the class group of the order with conductor $f$? What is more systematic or modern way of viewing these characters (and the "Ringklassen")?
In other words, how to compute the weights in the following formula
$$\varepsilon_G^{h_Gh_T}=\prod_{C\in \text{Cl}_f}h(\tau_C)^{-\chi(C)},$$
where
$$h(\tau)=y^{1/2}\lvert \eta(\tau)^2\rvert, \\ \tau=x+iy,\\ \tau_C =\text{CM point representing the class $C$},\\\varepsilon_G = \text{the fundamental unit in $\mathbb Q(G^{1/2})$}.$$
See also my previous questions. According to these, in each class of $\mathcal O_f$ is an ideal of norm equal to $1$. But then is the character above trivial?
Am I the only one who expected a biographical discussion of Carl Ludwig Siegel?
"Gruppe der Ringklassen mit dem Führer $f$" means "ring class group of conductor $f$". The modern point of you is that you divide the idele class group by the relevant open subgroup, and you consider the characters of the quotient group. There are many books on the subject, e.g. Weil: Basic number theory, Neukirch: Algebraic number theory etc.
@GHfromMO Where can I find definition of the ring class group of conductor $f$?
@user131781 Relax!
I am quite relaxed, thank you very much.
The ring class group is associated to an order $\mathcal{O}$ of a number field. It equals the multiplicative group of invertible fractional ideals in $\mathcal{O}$ modulo the subgroup of invertible principal fractional ideals in $\mathcal{O}$. In an imaginary quadratic number field, there is a unique order of conductor $f$, and this gives rise to the ring class group of conductor $f$.
@GHfromMO How then do you explain the following puzzling phenomenon: in this question I found that each class contains a proper fractional ideal with norm equal to $1$. But it follows from this that the value of the character is always $1$. This does not make sense to me.
I added a remark to your other post. Note also that the ring class group there has $f+1$ elements, so it has $f+1$ characters. Siegel elaborates on the quadratic characters (also known as genus characters). Nontrivial quadratic characters exist if and only if $f+1$ is even (i.e. $f$ is odd).
The ring class group is a special kind of a ray class group, introduced connection with the theory of complex multiplication in complex quadratic number fields. Given an integer $f > 1$, consider the group $D_f$ of all ideals coprime to $f$. The group $P_f$ of principal ideals is the group generated by ideals $(\alpha) \in D_f$ with $\alpha \equiv z \bmod f$ for some integer $z$. The ring class group modulo $f$ is simply the quotient $D_f/P_f$, and the ring class field defined modulo $f$ is obviously a subextension of the ray class field modulo $f$. In Siegel's definition of the quadratic character, ${\mathfrak p}$ is an ideal in the maximal order, not in the order with conductor $f$. It is true, however, that the ring class group modulo $f$ is isomorphic to the ideal class group of the order with conductor $f$, but the isomorphism is not as natural as one might expect. You should definitely look into the books by Cox and Cohn that I have already recommended.
Computing the corresponding CM-points is usually done via quadratic forms: to $Q(x,y) = Ax^2 + Bxy + Cy^2$ you associate the root of the equation $Q(z,1) = 0$ in the upper half plane.
|
2025-03-21T14:48:30.201345
| 2020-04-02T12:47:08 |
356353
|
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|
Stack Exchange
|
Factoring cyclotomic polynomials over quadratic subfield
The quadratic subfield of $\mathbb{Q}(\zeta_p)$ is given by $\mathbb{Q}(\sqrt{p^*})$, where $p^*$ is the choice of $\pm p$ which is $1$ mod $4$. By some elementary Galois theory, the cyclotomic polynomial $\Phi_p = \frac{x^p-1}{x-1}$ factors into two irreducible polynomials of degree $\frac{p-1}{2}$ over this quadratic subfield, which one can describe as
$$
P_{QR}(x) = \prod_{k\ \text{QR}} (x-\zeta_p^k)
$$
$$
P_{QNR}(x) = \prod_{k\ \text{QNR}} (x-\zeta_p^k)
$$
where $k$ ranges over the (nonzero) quadratic residues and quadratic nonresidues modulo $p$ respectively.
Is there an explicit description for the coefficients of these polynomials in terms of $\sqrt{p^*}$? The first coefficient, the one in front of $x^{\frac{p-3}{2}}$, is $\frac{1\mp\sqrt{p^*}}{2}$ by the statement of Gauss sums.
Some trivial observations. We have
$$P_{QR}(1/x) x^{(p-1)/2} = \prod_{QR} (1 - x \zeta^k),$$
$$P_{QNR}(1/x) x^{(p-1)/2} = \prod_{QNR} (1 - x \zeta^k),$$
which are easier to work with. On the other hand,
if $p \ge 5$, the product of $\zeta^k_p$ over quadratic residues is one,
and the product over non-residues is also one. Hence we can write
$$\begin{aligned}
P_{QR}(1/x) x^{(p-1)/2} = & \ \prod_{QR} ( 1 - x \zeta^k) = \prod_{QR} ( \zeta^{-k} - x) \\
= & \
(-1)^{(p-1)/2} \prod_{QR} (x - \zeta^{-k}) \\
= & \ \begin{cases} P_{QR}(x), & p \equiv 1 \mod 4 \\
- P_{QNR}(x), & p \equiv 3 \mod 4 \end{cases} \end{aligned}$$
because $(-1/p) = (-1)^{(p-1)/2}$. The same swapping occurs for $P_{QNR}$.
Let's consider the highest powers of $P_{QR}(x)$ and $P_{QNR}(x)$,
or equivalently the lowest powers of $P_{QR}(1/x) x^{(p-1)/2}$ and
$P_{QNR}(1/x) x^{(p-1)/2}$. We have
$$\frac{P_{QR}(1/x) x^{(p-1)/2}}{P_{QNR}(1/x) x^{(p-1)/2}}
= \prod ( 1 - x \zeta^k)^{ \left( \frac{k}{p}\right)}.$$
As mentioned there is the Gauss sum:
$$\sum \left( \frac{k}{p}\right) \zeta^{k} = \sqrt{p^*},$$
and similarly
$$\sum \left( \frac{k}{p}\right) \zeta^{nk} = \left( \frac{n}{p} \right) \sqrt{p^*},$$
applying $[n] \in (\mathbb{Z}/p\mathbb{Z})^{\times} = \mathrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ to both sides; when $p|n$ we interpret the RHS as being zero, and this is still correct, although we won't actually care about terms this deep into the power series above.
We deduce that
$$- \log \left( \frac{P_{QR}(1/x) x^{(p-1)/2}}{P_{QNR}(1/x) x^{(p-1)/2}} \right) = \sum_{n=1}^{\infty} \frac{x^n}{n} \sum \left( \frac{k}{p}\right) \zeta^{nk}
= \sqrt{p^*} \sum_{n=1}^{\infty} \left( \frac{n}{p}\right) \frac{x^n}{n}.$$
Naturally
$$- \log \left( P_{QR}(1/x) x^{(p-1)/2} P_{QNR}(1/x) x^{(p-1)/2} \right)
= \log \left( \frac{1 - x^p}{1 - x} \right) = \log(1-x) + O(x^p) = - \sum \frac{x^n}{n},$$
and so (for example)
$$\log(P_{QR}(1/x) x^{(p-1)/2}) = \frac{1}{2} \sum \frac{x^n}{n} \left(1 - \left( \frac{n}{p}\right) \sqrt{p^*}\right) + O(x^p),$$
You can now formally expand this out to get the first few terms.
For example, the first non-zero term is
$$\frac{1 - \sqrt{p^*}}{2},$$
and the second is
$$ \frac{3 + p^* - 2 \sqrt{p^*}\left(1 + \left( \frac{2}{p}\right) \right) }{8}$$
For example, if $p \equiv 3,5 \mod 8$ so $(2/p) = -1$, this is
$$\frac{3 + p^*}{8}.$$
Note the conditions on $p$ ensure that this is an algebraic integer, as it has to be.
As you keep going, you get more and more terms involving the quadratic residues $(n/p)$ for small $n$,
and it becomes messier and messier, and dependent on $p$ modulo higher integers.
The third term, for example, is
$$\frac{15 - 9 \sqrt{p^*} + 3 p^* - p^* \sqrt{p^*} + 6(p^* - \sqrt{p^*})(2/p) - 8 \sqrt{p^*} (3/p)}{48}.$$
Also, one can internet-search on "Aurefeuillian factorization".
Thanks, this is great!!
|
2025-03-21T14:48:30.201597
| 2020-04-02T12:49:17 |
356354
|
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|
Stack Exchange
|
On dense embedding of Banach spaces
Disclaimer: When I came up with this question yesterday, I suspected it to be trivial (trivially true or trivially false). Then it kept me awake several hours tonight... (I still hope, though, this is just due to my ignorance.)
Question. Let $E,F$ be Banach space and suppose that $E$ embeds densely and continuously into $F$ (so we consider $E$ as a subspace of $F$ from now on). Assume that there exists a constant $M \in (0,\infty)$ with the following property:
For each $f \in F$ we can find a sequence $(e_n)$ in $E$ that converges to $f$ with respect to $\|\cdot\|_F$ and that satisfies $\|e_n\|_E \le M \|f\|_F$.
Does it follow that $E = F$?
Remark. I first thought the answer should be yes due to some application of the open mapping theorem: clearly, it suffices to show that $\|\cdot\|_E$ and $\|\cdot\|_F$ are equivalent on $E$, and by the open mapping theorem this is true iff $\|\cdot\|_F$ is complete on $E$; but I wasn't able to prove that latter property.
Am I overlooking some simple argument, or a simple counterexample?
Edit. It is probably worthwhile to note the following fact:
As observed by Nate Eldredge in a (now deleted) comment, it is easy to see that the answer is "yes" if $E$ is reflexive: Given $f \in F$ and $(e_n) \subseteq E$ as above, we can choose a subsequence of $(e_n)$ that converges weakly (in $E$) to a vector $e \in E$. For each $f' \in F'$ this implies that $\langle f', f\rangle = \langle f', e\rangle$, so $f = e \in E$.
Great question! What you need is Sandy Grabiner's approximation lemma:
Lemma: Let $E$ and $F$ be Banach spaces, let $T \in B(E,F)$, and let $M > 0$ and $0 < r < 1$. Suppose that for each $y \in [F]_1$ there exists $x_0 \in [E]_M$ with $\|y - Tx_0\| \leq r$. Then for each $y \in F$ there exists $x \in E$ with $\|x\| \leq \frac{M\|y\|}{1-r}$ and $Tx = y$.
I think this immediately implies that $F = E$ in your question. You can probably prove the lemma yourself very easily (hint: geometric series), but for the sake of completeness a reference is: Theorem 3.35 of my book Measure Theory and Functional Analysis.
It's a great lemma and deserves to be better known. For instance, both the open mapping theorem and Tietze's extension theorem follow easily from it.
Ha! Thanks a lot, you just saved my day! I thought I'd loose my mind...
@NateEldredge: I didn't notice that, fixed.
@JochenGlueck: you are welcome!
Aargh, switched them back.
Probably it was when I was a graduate student when I first saw what is called the "little open mapping theorem". It says that if $T$ is in $L(X,Y)$ and the closure of $T\mathring{B}_X(1)$ contains $\mathring{B}_Y(r)$, then $T\mathring{B}_X(1)$ contains $\mathring{B}_Y(r)$. The first step is of course the proof of what you call Sandy's approximation lemma. The open mapping theorem is an immediate consequence of the LOMT. A non linear version of this was useful when I was working on Lipschitz quotient mappings $20+$ years ago. After proving the LOMT in class, an easy HW problem is...
"Every separable Banach space is isometrically isomorphic to a quotient of $\ell_1$."
There is a version for complete metric groups of this lemma in Tougeron's book Ideaux de Fonctions Differentiables which I can't check now because of Corona.
After Nik Weaver answered the question and Bill Johnson pointed out in a comment that what one needs is actually a part of the usual proof of the open mapping theorem, I thought about it once again, and I think it is worthwhile to explicitly remark that the following more general assertion is true:
Theorem. Let $E$, $F$ be Banach space and suppose that $E$ embeds densely and continuously into $F$ (so we may consider $E$ as a subspace of $F$). Suppose that for each $f \in F$ there exists a sequence $(e_n)$ in $E$ which converges to $f$ with respect to $\|\cdot\|_F$ and which is bounded with respect to $\|\cdot\|_E$. Then $E = F$.
Discussion. The point here is that, in constrast to the statement in the original question, we do not assume a priori that the bound $\sup_n \|e_n\|_E$ is uniform with respect to $\|f\|_F$.
Proof. Actually, this is almost exactly the proof of the open mapping theorem, but with a (very) small additional perturbation argument. The details are as follows:
Let $j$ denote the embedding of $E$ into $F$. Let $B_E(e,r)$ denote the open ball in $E$ with radius $r$ and center $e$, and likewise for $F$. It follows from the assumption that
$$
\bigcup_{n \in \mathbb{N}} \overline{ j B_E(0,n)}^F = F.
$$
Due to Baire's theorem we can thus find an integer $n \in \mathbb{N}$ such that $\overline{ j B_E(0,n)}^F$ has non-empty interior in $F$, i.e. it contains a ball $B_F(f,\varepsilon)$.
Since $j(E)$ is dense in $F$, we can choose $e \in E$ such that $j(e)$ is closer than $\varepsilon / 2$ to $f$ (with respect to $\|\cdot\|_F$). (This is the "perturbation argument" mentioned above - well, "argument" is quite an exaggeration...) Hence,
$$
\overline{ j B_E(0,n)}^F \supseteq B_F(f,\varepsilon) \supseteq B_F(j(e), \varepsilon/2).
$$
Now, we proceed again as in the proof of the open mapping theorem: We can write each $x \in B_F(0, \varepsilon/2)$ as $x = (x+j(e)) - j(e)$, and the vector in brackets is in $B_F(j(e), \varepsilon/2)$ and can thus be approximated (wrt $\|\cdot\|_F$) by a sequence $(j(e_k))$ with $\|e_k\|_E < n$. Since $j(e_k - e)$ approximates $x$ (wrt $\|\cdot\|_F$), we obtain $x \in \overline{ j B_E(0,n + \|e\|)}^F$.
We proved that the $F$-closure of $j B_E(0,n + \|e\|)$ contains $B_F(0, \varepsilon/2)$, so it follows from the small open mapping theorem (see Bill Johnson's comment to Nik Weaver's answer) that $j B_E(0,n + \|e\|)$ itself contains an open ball in $F$ centred at $0$. Hence, $j(E) = F$.
Edit 2023-05-02. I found a reference where the theorem above along with
a different proof is explicitly stated: Theorem 5 in "Evgeniĭ A. Lifshits: Ideally convex sets (1970)" (link to zbMATH)
|
2025-03-21T14:48:30.202008
| 2020-04-02T13:35:42 |
356356
|
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|
Stack Exchange
|
Is there a name for a tree with all leaf vertices identified with each other?
Is there a name for those graphs that can be formed by taking a tree and identifying all the vertices of degree 1 (leaves) with each other?
Or, if I understand correctly, an equivalent definition may be: Those graphs that can be formed by taking a tree, and then adding a vertex that shares an edge with all vertices of degree 1.
Are you asking about finite graphs? In the finite case all these graphs have connectivity $\kappa(G)=2$, but ne converse is not true. The second definition is not the same as the first because you cannot obtain the graphs you would get in the first one when you start from a tree with two or three vertices.
In a paper of mine with Alan Sokal (https://arxiv.org/abs/1307.1721), graphs of this type were important and because we could not find a name for them at the time, we used the term "leaf-joined tree" to refer to a graph formed from a particular tree (in our case a complete k-ary tree of certain depth) by identifying all the vertices.
Thanks @m-winter for pointing out that the definitions are not equivalent. I am interested in whether either class of graphs has a known name. (Yes, finite case.)
|
2025-03-21T14:48:30.202133
| 2020-04-02T13:38:45 |
356357
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356357"
}
|
Stack Exchange
|
Spitzer's condition, a slowly varying function and its behavior
Let $S$ denote a random walk that satisfies Spitzer's condition $$ \frac{1}{n} \sum _{k=1}^n P (S_k > 0 ) \to \rho$$ for some $\rho \in (0,1)$. From the book Regular Variation (Bingham, Goldie, Teugels), the function $$ l(n) = \exp \left( \sum_{k=1}^\infty \frac{(1-\frac{1}{n})^k} {k} \Big( P(S_k > 0) - \rho \Big) \right) $$ is slowly varying. I am interested in learning more about the behavior of $l(n)$ for $n$ large. In particular, I am asking myself whether the following statement can be made:
$$ l(n) = o(n^\epsilon) \ \ \forall \epsilon > 0.$$
I am thankful for any hints, references and ideas.
Edit (sharing my progress/thoughts): By monotone convergence, $$l(n) \to \exp \left( \sum_{k=1}^\infty \frac{1} {k} \Big( P(S_k > 0) - \rho \Big) \right) $$ as $n \to \infty$. If the right hand side is finite, the claim that $l(n) = o (n^\epsilon)$ for all $\epsilon > 0$ is immediate. So, only the case where the sum on the right hand side diverges, has to be considered. Note that Spitzer's condition is equivalent to $P(S_n > 0 ) \to \rho$ as $n \to \infty$.
All slowly varying functions are $o(n^\epsilon)$ for every $\epsilon > 0$, are they not?
@MateuszKwaśnicki Oh, I did not know. Would you have a reference for that?
Proposition 1.3.6(v) in B-G-T. ("...proofs may be left to the reader.")
|
2025-03-21T14:48:30.202253
| 2020-04-02T13:41:29 |
356358
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356358"
}
|
Stack Exchange
|
About the orientation of index formula on orbifold
Let $X$ be a closed oriented orbifold with singularity $\Sigma X$. The singularity is defined as
$$ \Sigma X=\{(x)|~x\in X,~G_x\neq1\}, $$
where $G_x$ is the isotropy group.
For $u\in K_v(TX)$, as stated in Kawasaki's paper https://projecteuclid.org/euclid.nmj/1118786571 we have
$$ Ind(u)=(-1)^{\dim(X)}<ch(u)\hat A^2(X),[TX]>+\sum^n_{i=1}\frac{(-1)^{\dim(\Sigma_i)}}{m_i}<ch^\Sigma(u)\hat A^{2,\Sigma}(\Sigma_iX),[T\Sigma_i X]>,$$
where $\{\Sigma_iX\}$ denotes all the components of the singularity set.
Q
I do not follow how to choose the orientations of the singularity set $\Sigma X$.
It seems that if we reverse the orientation of the singularity, then the formula changes?
|
2025-03-21T14:48:30.202448
| 2020-04-02T13:53:21 |
356360
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356360"
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|
Stack Exchange
|
Why is the first conjugate time continuous on the unit tangent bundle?
Let (M,g) be a complete, connected Riemmanian manifold and SM the unit tangent bundle. Define the map
$con:SM\to (0,\infty]$ such that for $v\in SM, con(v)$ is the first positive time such that $\gamma_v(con(v))$ is conjugate to $\gamma_v(0)$ along $\gamma_v$.Set $con(v)=\infty$ if no such time exists. (Here $\gamma_v$ is the geodesic with $\dot\gamma_v(0)=v$)
Does anybody have a reference for the continuity of con?
This, I believe, follows from the fact that, for a parameterized system of ODEs, the solutions to a specified initial value problem depend continuously on the parameter. The idea is to apply this to the Jacobi equation.
Thank you but in how far is this a initial value problem? Isn't it a boundary problem?
This is Proposition 1.2 in the article The Lipschitz Continuity of the Distance Function to the Cut Locus by J. Itoh and M. Tanaka. As the title suggests, they even show the Lipschitz continuity for the $k^\text{th}$ focal time.
|
2025-03-21T14:48:30.202565
| 2020-04-02T14:03:44 |
356362
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356362"
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|
Stack Exchange
|
Picard group modulo codimension 2
Let $X$ be a normal (possibly singular) projective surface over $\mathbb{C}$. Consider the set $M_X$ of all coherent sheaves $F$ on $X$ such that there exists a finite subset $Y\subset X$ such that $F$ restricted to $X\setminus Y$ is a line bundle. $M_X$ becomes a monoid via the tensor product. Now let $G_X$ be the set of equivalence classes of $M_X$ where two sheaves $F_1,F_2\in M_X$ are equivalent if there is a finite subset $Y\subset X$ such that $F_1$ and $F_2$ are isomorphic on $X\setminus Y$. This equivalence relation is compatible with the tensor product and so $G_X$ is a group.
In general, one has the group homomorphism $\textrm{Pic}(X)\to G_X$ that sends a line bundle to its equivalence class. If $X$ is smooth, then this is an isomorphism and thus $G_X$ is just the usual Picard group.
My hope is that in general we can understand $G$ in terms of a desingularisation $f: X'\to X$. Because $X$ is normal, it has only finitely many singularities. Away from these singularities $f$ is an isomorphism and the group $G_{X'}$ is just the Picard group of $X'$. So my hope is that we can identify $G_X$ with $\textrm{Pic}(X')$. Is something like that true?
I am confused. Why is $Pic(X) \to G_X$ an isomorphism? Doesn't $G_X$ parametrize coherent sheaves? What is the inverse of $[F]$ in $G_X$?
Not any coherent sheaves, but such $F$ that are a line bundle on $U=X\setminus Y$ for a finite set $Y$. Then the inverse of $[F]$ is any extension of the inverse of $F|_U$ to $X$.
Ok, thank you. I misread the definition, sorry.
The group $G_X$ can be identified with the group of rank 1 reflexive sheaves on $X$ ($F$ is reflexive if the canonical morphism $F \to F^{\vee\vee}$ is an isomorphism) by taking a sheaf $F$ to the reflexive sheaf $F^{\vee\vee}$. The monoidal structure on the set of all reflexive sheaves is given by
$$
(F,G) \mapsto (F \otimes G)^{\vee\vee}.
$$
Furthermore, the group $G_X$ can be identified with the class group $\operatorname{Cl}(X)$ of Weil divisors on $X$.
The relation of $\operatorname{Pic}(X')$ to $\operatorname{Cl}(X)$ is given by the following exact sequence
$$
0 \to \bigoplus \mathbb{Z}[E_i] \to \operatorname{Pic}(X') \to \operatorname{Cl}(X) \to 0,
$$
where $E_i$ are the components of the exceptional divisor of $X' \to X$.
Thanks! So the situation is even better than I have expected. Am I right in assuming that the reflexive sheaf corresponding to a Weil divisor is just the subsheaf of the quotient field of all functions with the prescribed pole and zero loci?
There are two nice constructions. 1) Restrict the divisor to the smooth locus of the surface, take the corresponding line bundle, and then push it forward to the surface. 2) If the divisor is effective, take the dual of its ideal sheaf; and in general extend this association by linearity.
|
2025-03-21T14:48:30.202794
| 2020-04-02T14:45:43 |
356369
|
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"Ben Barber",
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|
Stack Exchange
|
Voronoi diagram on (weighted) graphs
Suppose I have a graph $G$ (possibly with weights on edges), and I have a subset $S$ of $k$ vertices $s_1, \dotsc, s_k$. I want to solve the post office problem: that is, I want to partition the vertices of $G$ into subsets $D_1, \dotsc, D_k,$ so that $s_i$ is the closest vertex of $S$ to every vertex in $D_i.$ I assume this has been studied - what is the most efficient algorithm?
Add a new root node $r$ connected with $s_1,\ldots,s_k$ with edges of weight $1$, say. Then find a minimum weight spanning tree $T$ with root $r$ using Dijkstra's algorithm, for example. Given any node $n$ the path from $n$ to $r$ in $T$ will lead to the nearest post office at the next to last step.
Dijkstra on an augmented graph is the right call, but its output is not a minimum weight spanning tree but a shortest path tree.
@BenBarber Indeed. For some reason I hadn't heard of the shortest path tree. If you want to make this an answer, I am happy to accept.
@BenBarber Thanks for the correction! A minimum weight spanning tree would also work, but Dijkstra's algorithm is quick and to the point.
@FrançoisG.Dorais, I'm not sure that's right: if everywhere is 2 from the root but 1 from each other then a minimum spanning overestimates the distance to almost everywhere.
@IgorRivin neither had I!
Once you decide that the elements of $S$ should look for the vertices in their part rather than the other way round the naive approach of exploring edges one at a time in increasing distance from $S$ is the best you can do (up to administrative overhead) because you might have to examine all of the edges anyway. This is essentially Dijkstra's algorithm.
|
2025-03-21T14:48:30.202952
| 2020-04-02T16:06:26 |
356374
|
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|
Stack Exchange
|
A Bessel-like integral
I encounter the following integral when trying to find the inverse Fourier transform of the characteristic function of a certain sum of random variables. Here, $0\le\lambda\le1$, $p\ge0$, $q\ge0$ are real, and $n$ is an integer. I want to compute the following integral:
$$\int_0^{2 \pi} e^{p \cos (\lambda \tau) + q \cos ((1 - \lambda) \tau)}
\cos (n \tau) \frac{d \tau}{2 \pi}$$
This is a generalization of a Bessel integral, in that for $q = 0$ and $\lambda=1$, I know that:
$$\int_0^{2 \pi} e^{p \cos (\tau)}\cos (n \tau) \frac{d \tau}{2 \pi}
=I_n(p)$$
where $I_n(p)$ is the modified Bessel function of the first kind.
This is not a complete answer, but it's a bit long to fit in a comment, so I'm posting it here in case it's useful to someone.
Let's denote:
$$
I(p, q, \lambda; n)
:=
\int_0^{2 \pi} e^{p \cos (\lambda \tau) + q \cos ((1 - \lambda) \tau)}
\cos (n \tau) \frac{\mathrm d \tau}{2 \pi}
$$
There's a symmetry in this integral of the form:
$$
I(p, q, \lambda; n)
=
I(q, p, 1 - \lambda; n)
$$
Also note that the integrand is even, i.e. $f(-\tau) = f(\tau)$.
Now, as mentioned in the OP, we have:
$$
I(p, 0, 1; n)
=
I(0, p, 0; n)
=
I_n(p)
$$
Actually, a slight generalization for $q \neq 0$ gives us:
$$
I(p, q, 1; n)
=
I(q, p, 0; n)
=
e^q I_n (p)
$$
Due to the symmetry property above, taking $\lambda = 1/2$ we actually obtain a rather simple result:
\begin{align*}
I(p, q, \frac{1}{2}; n)
=&
I(q, p, \frac{1}{2}; n)
=
\frac{1}{2 \pi}
\int_0^{2\pi}
\mathrm d\tau\,
e^{(p + q)\cos(\tau / 2)}
\cos (n\, \tau)\\
=&
\frac{1}{\pi}
\int_0^{\pi}
\mathrm dy\,
e^{(p + q)\cos y}
\cos (2\, n\, y)
=
I_{2n}(p + q)
\end{align*}
A special case (probably of limited interest though) is $q = 0$, and $\lambda = 1 / m$, where $m \in \mathbb N$.
Of course, if $m = 1$ or $m = 2$, we obtain the previous results, namely $I_n(p)$ and $I_{2n}(p)$, so we need to take $m \geq 3$ to obtain anything new.
The integral can then be converted into the following form:
$$
I(p, 0, 1/m; n)
=
\frac{m}{2 \pi}
\int_0^{2 \pi / m}
\mathrm dx\,
e^{p \cos x}
\cos (m\, n\, x)
$$
Unfortunately, a general analytic solution doesn't seem likely.
One value which seems to be solvable though is $m = 4$ (originally found by experimenting in Mathematica); in this case, we write the integral as:
$$
I(p, 0, 1/4; n)
=
\frac{2}{\pi}
\int_0^{\pi/2}
\mathrm dx\,
e^{p \cos x}
\cos (4\, n\, x)
$$
We can then use the $n$ angle expansion, so our integral turns into:
$$
I(p, 0, 1/4; n)
=
\frac{2}{\pi}
\sum\limits_{k\;\mathrm{even}}
(-1)^\frac{k}{2}
\begin{pmatrix}4 n\\ k\end{pmatrix}
\int_0^{\pi/2}
\mathrm dx\,
e^{p \cos x}\,
\cos^{4n - k} x\,
\sin^k x
$$
Let's focus on the integral in the above; using the fact that $4 n - k := 2 \ell$ is even, we can rewrite $\cos^{2\ell} x = (1 - \sin^2 x)^\ell$, expand this using the binomial theorem, and additionally expand $e^u = \cosh u + \sinh u$, so that our final result is a linear combination of integrals of the following form:
$$
\mathcal{I}(\alpha, p)
=
\int_0^{\pi/2}
\mathrm dx\,
\sinh(p \cos x)\,
\sin^\alpha x\\
\mathcal{J}(\alpha, p)
=
\int_0^\pi
\mathrm dx\,
\cosh(p \cos x)\,
\sin^\alpha x
$$
for some integer values of $\alpha$.
Note that the second integral has shifted limits, $[0,\pi]$, instead of $[0,\pi/2]$; this is because the integrand is actually symmetric around $\pi/2$, so we can write $\int_0^{\pi/2}(\cdots) = \frac{1}{2} \int_0^\pi (\cdots)$ (the easiest way to see this is to shift the origin to $\pi/2$, i.e. use the substitution $x \rightarrow x - \pi / 2$; then it's simple to demonstrate that $f(-x) = f(x)$, where $f$ denotes the integrand).
These kinds of integrals are known in the literature, see for instance Gradshteyn and Ryzhik, 7th ed., formulas 3.997.1 and 3.997.2:
$$
\mathcal{I}(\alpha, p)
=
\frac{\sqrt \pi}{2}
\left(\frac{2}{p}\right)^\alpha
\Gamma\left(\frac{\alpha + 1}{2}\right)
\mathbf{L}_\frac{\alpha}{2}(p)\\
\mathcal{J}(\alpha, p)
=
\sqrt \pi
\left(\frac{2}{p}\right)^\alpha
\Gamma\left(\frac{\alpha + 1}{2}\right)
I_\frac{\alpha}{2}(p)
$$
where $I_\frac{\alpha}{2}$ is the modified Bessel function of the first kind, and $\mathbf{L}_\frac{\alpha}{2}$ is the modified Struve function, see for instance DLMF, chapter 11.
The final result can be shown to be:
\begin{align*}
I(p, 0, 1/4; n)
=&
\frac{2}{\pi}
\sum\limits_{k\;\mathrm{even}}
\sum\limits_{\ell=0}^{2n - k/2}
\begin{pmatrix} 4n \\ k \end{pmatrix}
\begin{pmatrix} 2n - k / 2 \\ \ell \end{pmatrix}
(-1)^{2 n - \ell}
\left[
\mathcal{I}(4n - 2\ell, p)
+
\frac{1}{2}
\mathcal{J}(4n - 2\ell, p)
\right]
\end{align*}
Evidently, using the symmetry property, this automatically gives us $I(0,p,3/4;n)$ as well.
However, I haven't found anything which works for general $m$; while the procedure itself can more or less be repeated for any even $m$, the integrals $\mathcal{I}$ and $\mathcal{J}$ have different integration limits, which do not have any obvious representations in terms of special functions like the above.
|
2025-03-21T14:48:30.203255
| 2020-04-02T16:32:23 |
356380
|
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Stack Exchange
|
A question about subobjects of the unit in a rigid abelian tensor category
I have a question about Proposition 1.17 in Deligne and Milne, Tannakian Categories (see here), in the last 4 lines of the proof.
I don't know how it follows from $U\otimes U\simeq U$ that $T=\ker(U\to U^\vee\otimes U)=0$ (or the largest subobject $T$ of $U$ such that $T\otimes U=0$ is $0$). Even if admitting this, I don't know why $\mathbb{1}=U\oplus U^\perp$ (or the SES at the beginning splits).
Could anyone give some details about these points? Thanks!
I think they key point is that you have a coevaluation map $\eta\colon 1\to U\otimes U^\vee$, an isomorphism $\delta\colon U\to U\otimes U$ and an evaluation map $\epsilon\colon U\otimes U^\vee\to 1$, so you have $(1\otimes\epsilon)\circ(\delta\otimes 1)\circ\eta\colon 1\to U$. This will split the inclusion $U\to 1$.
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2025-03-21T14:48:30.203348
| 2020-04-02T17:23:59 |
356383
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"Taras Banakh",
"Wlod AA",
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|
Stack Exchange
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A topologically transitive dynamical system without dense orbits
By a dynamical system I understand a pair $(K,G)$ consisting a compact Hausdorff space and a subgroup $G$ of the homeomorphism group of $K$.
We say that a dynamical system $(K,G)$
$\bullet$ is topologically transitive if for every non-empty open set $U\subseteq K$ its orbit $GU=\{g(x):g\in G,\;x\in U\}$ is dense in $K$;
$\bullet$ has dense orbit if for some point $x\in K$ its orbit $Gx$ is dense in $K$.
It is easy to see that a dynamical system is topologically transitive if it has a dense orbit. If the space $K$ is metrizable and nonempty, then the converse also is true.
On the other hand, under $\mathrm{non}(\mathcal M)<\mathfrak c$, there exists a subgroup $G\subset S_\omega$ of cardinality $|G|\le\mathrm{non}(\mathcal M)<\mathfrak c$ that induces a topologically transitive action of the Stone-Cech remainder $\omega^*=\beta\omega\setminus\omega$. The dynamical system $(\omega^*,G)$ does not have dense orbits since the space $\omega^*$ has density $\mathfrak c>|G|$. I am interested if such an example can be constructed in ZFC.
Problem. Is there topologically transitive dynamical system without dense orbits?
Does $S_\omega$ have a dense orbit on $\omega^*$? (Its action is obviously topologically transitive.) I thought I knew the answer to be negative, but I don't see right now.
By the way the action on the empty set is trivially topologically transitive without dense orbits, so the assertion for $K$ metrizable should assume $K$ non-empty.
@YCor The action of $S_\omega$ on $\omega^*$ is minimal: all orbits are dense.
The answer is yes:
there is a topologically transitive dynamical system without dense orbits.
Indeed,
let X be a topological space that is not separable. Let $\ K=X^{\Bbb Z},\ $ and let $\ G\ $ be the group of homeomorphism of $\ K,\ $ induced by shifts $\ s_n\ (n\in\Bbb Z)\ $ of $\ \Bbb Z:\ $
$$ \forall_{n\in\Bbb Z}\forall_{x\in\Bbb Z}\quad
s_n(x):= x+n $$
Let $\ p:=(p_n)\in K\ $ be arbitrary, and let
$$ P:=\{p_n:n\in \Bbb Z\} $$
Then there exists non-empty open $\ U\ $ in $\ X,\ $ disjoint with set $\ P.\ $ Then non-empty open in $\ K\ $
set $\ W,$
$$ W\ :=\ \pi_0^{-1}(U)\ $$
is disjoint with the orbit $\ p.\ $
On the other hand, let $\ \emptyset\ne G\subseteq K, $
where $\ G\ $ is open in $\ K.\ $ Then there exists
non-empty $\ H\ $ and integer $ a\ge 0\ $ such that
$\ H\ $ is an open subset of $\ X^{(-a)..a}\ $ (Perl notation "s..t") and
$$ \emptyset\ \ne\ \pi_{(-a)..a}^{-1}(H)\ \subseteq G $$
Obviously, the orbit of $\ \pi_{(-a)..a}^{-1}(H)\ $, hence of $\ G,\ $ is dense in $\ K.$
Great!
Thank you for the great answer. And what about the case of subgroups $G\subseteq S_\omega$ acting on $\omega^*$?
@TarasBanakh, you simply overlooked this and I just jumped in. Is $\ \omega^*\ $ simply an infinite countable set (in this context)? And $\ S_{\omega}\ $ the group of all permutations (self-bijections)? For any group which contains the one in the answer, the situation is the same (so simple and nice :) ).
By $\omega^*$ I denote the remainder of the Stone-Cech compactification of $\omega$ and this is the principal space I am interested in. In this case the group $\mathbb Z$ (and any other countable group) does not work, unfortunately.
"[...] the remainder of the Stone-Cech [...]" -- no wonder that your mind was far away from any cartesian products. :)
@TarasBanakh at this point if you're specifically interested in $G\subset S_\omega$ acting on $\omega^*$ it would be worth a separate question.
Well, the argument shows that if $G$ acts on the discrete set $I$ with no finite orbit and $X$ is an arbitrary topological space then the action of $G$ on $X^I$ is topologically transitive (because by BH Neumann's lemma for every finite subset $F$ of $I$ there exists $g$ such that $gF\cap F=\emptyset$). Next, there are many ways to ensure that there is no dense orbit, e.g., that $G$ is countable while $X^I$ is not separable (which holds if $I$ is nonempty and $X$ non-separable, or if $X$ is non-indiscrete and $I$ has cardinal $>c$).
@YCor Thank you for the suggestion. I will then accept the answer of Wlod AA and write this specific question separately.
After all, possibly the Tikhonov embedding into cube may lead to an example, it's to be seen. That would make the Cartesian product a hero again :).
Taras, thank you (and to YCor :) ) for accepting.
You are welcome! Thank you for the involvement.
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2025-03-21T14:48:30.203671
| 2020-04-02T17:52:47 |
356386
|
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"Denis Nardin",
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|
Stack Exchange
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Geometric interpretation of nonconnective, non-coconnective chain complexes / spectra?
Let's stipulate that
Connective -- i.e. nonnegatively-(homologically)-graded -- chain complexes have a very natural geometric interpretation: by the Dold-Kan theorem, they are a way of thinking about simplicial abelian groups.
Dually, coconnective -- i.e. nonpositively-graded -- chain complexes have a very natural geometric interpretation as complexes of functions on spaces.
The motivations for considering the category of all (unbounded) chain complexes, then, are quite good -- this category provides a home for both the connective and coconnective chain complexes, and has excellent formal properties like stability and a good duality theory.
However, these motivations are quite formal in nature -- they don't provide a geometric interpretation in line with (1) or (2) above. For the most part, these motivations operate "one category level higher", discussing properties of the category of chain complexes. I'm specifically looking for something which gives a geometric, natural way to think about an individual chain complex.
Question 1: What is a good geometric interpretation of nonconnective, noncoconnective chain complexes?
Notes:
A similar discussion would more generally stipulate that grouplike $E_\infty$-spaces have a natural geometric interpretation, and ask for a similarly "geometric" interpretation of more general spectra. I'd be equally happy with a discussion in this setting.
Similarly, I'd be happy with a discussion in the context of complexes of sheaves of various flavors.
Guess: Here's a guess of a picture which might be appropriate, based on my understanding of Tyler and Adeel's answers to this question, and inspired by Sanath's comment below.
A general $E_\infty$ ring spectrum $A$ can be thought of as the global sections $A = \Gamma(X, \mathcal O_X)$ of a spectral scheme $X$ (note that the structure sheaf $\mathcal O_X$ takes values $\Gamma(U,\mathcal O_X)$ in connective $E_\infty$ ring spectra when $U$ is affine open, but its global sections can be nonconnective).
Thus the negative-dimensional homotopy groups of $A$ can be thought of as measuring the cohomology of $X$, i.e. the global structure of how $A$ is glued together from affines. The positive-dimensional homotopy groups of $A$ can be thought of as some sort of nilpotent thickening / infinitesimal structure of $X$ -- or perhaps it's better to think of them as encoding the "stacky" part of the structure of $X$.
A spectrum $M$ is a module over an $E_\infty$-ring spectrum $A = \Gamma(X, \mathcal O_X)$, and so we should think of $M = \Gamma(X,\mathcal F)$ as the global sections of a quasicoherent sheaf on $X$ (where the values $\Gamma(U,\mathcal F)$ of $\mathcal F$ on affine opens $U$ are connective, but its global sections need not be).
Thus the negative-dimensional homotopy groups of $M$ can be thought of as measuring the cohomology of $\mathcal F$, and the positive-dimensional homotopy groups of $M$ can be thought of as the infinitesimal (or rather "stacky") structure of $\mathcal F$.
Question 2: Is this a good picture to have in mind when trying to think about nonconnective, noncoconnective chain complexes / spectra geometrically?
I think I feel a bit more confident in thinking about $E_\infty$ ring spectra this way than I do about thinking about general spectra (or module spectra) in this way.
Question 3: For example, do we have $KU = \Gamma(X, \mathcal O_X)$ for a natural (or even canonical) spectral scheme $X$?
Are "reduced excisive functors" an acceptable answer?
@DenisNardin Hmmm... Maybe? I guess my issue is that a functor is typically an object that lives a category level up. I suppose we've learned to think of sheaves as first-class geometric objects -- so maybe if you could convince me that reduced excisive functors are as "geometric" as sheaves are?
Somewhat stupid tongue-in-cheek response: if you were to allow a slight broadening of your question to asking about spectra which have unbounded (positive and negative) homotopy, then you get examples from the myriad of periodic cohomology theories. This actually isn't as facetious as it might seem to be at first sight. A broadening of the perspective you mentioned on nonpositively graded chain complexes is as the derived global sections of the structure sheaf on a (derived connective) scheme: cohomology is concentrated in negative homological degree.
One could then interpret your question as asking for natural sources of derived/spectral schemes whose derived ring of functions is concentrated in unbounded degrees. Highly structured periodic cohomology theories provide the most natural example of such objects (e.g., BZ/2 giving rise to KO). Not sure if this counts as "geometric" to you.
@skd Sure, but again there's the issue that cohomology theories live "one category level up", as functors defined on the category of spaces. I'm hoping for something which "lives at category level zero" in some sense.
Are you happy with the answer that the unbounded chain complexes are combinatorial spectrum objects in Abelian groups? Also, there's this paper of Paul Lessard where he identifies spectra with locally finite (∞,Z)-categories: https://arxiv.org/abs/1812.00122 . Somehow the identification between simplicial abelian groups and bounded below complexes is kind of too strict to be a homotopy-invariant statement. It's an equivalence of presentations of homotopy theories rather than an equivalence of homotopy theories.
@HarryGindi Well, I'm finding nonconnective spectra to be just as mysterious as nonconnective chain complexes. So if you could convince me that spectrum objects -- "combinatorial" or otherwise -- are "geometric" in the super-imprecise sense that I'm failing to make clear, then sure. I don't know what an $(\infty,\mathbb Z)$-category is -- I think I get $\infty$-groupoids via the homotopy hypothesis -- if there's a "geometric picture" for what it means to pass to $\mathbb Z$, that would be cool.
@skd You know what -- if I had a good "geometric" way to think about the fact that a sheaf of connective $E_\infty$ ring spectra can have nonconnective global sections, I think that would satisfy me. Actually, what's a good example of this? Eg is $KU$ the global sections of a natural sheaf of connective ring spectra on something? The sense I got here is that nonconnective $E_\infty$ rings inevitably take one away from geometric intuition. Should I resign myself to this?
@TimCampion My intuition is that affine should mean "even periodic" rather than "connective" (this is not true in SAG at the moment).
Question 3 can be interpreted in at least two ways. The first is as you said in your comment; namely, whether KU is the global sections of a sheaf of connective E_oo-rings on a scheme. I do not know the answer to this, but I would not believe any such statement to be true. The second is whether there is some derived scheme whose global sections is KU. The answer to this question is yes: there is a sheaf of even-periodic E_oo-rings on Spec(Z) whose global sections is KU. In general, it seems that the realm of spectral algebraic geometry is divided into two kingdoms.
The first is the world of algebraic geometry modeled on connective E_oo-rings as the affines; this is what Lurie's books study. The other is the world of algebraic geometry modeled on even-periodic E_oo-rings as the affines (as Denis mentioned); this is motivated by the chromatic picture, and Lurie's approach to this in Elliptic-N is by developing the connective theory and then localizing. The output of taking global sections on a spectral scheme in this latter world will only yield periodic things, so ...
... perhaps this is acceptable as a motivator for "unbounded" algebraic geometry if you believe that even-periodic E_oo-rings are sufficiently geometrically motivated.
As for question 2: I agree with the interpretation of positive homotopy/homology groups as nilpotent fuzz, and negative homotopy groups as cohomology (concentrated in positive cohomological degree). One comment: while one will get examples of spectra/chain complexes with homotopy concentrated in positive and negative dimensions, most natural algebro-geometric objects have finite cohomological dimension, so the resulting spectra will be bounded below.
So, in this setting, the picture is somewhat close to behaving like "usual" connective algebraic geometry (but, as pointed out in the link you mentioned in the question, is still quite different): examples of such spectra arise naturally from algebro-geometric considerations. However, the sort of E_oo-ring spectra (or, more generally, spectral schemes) which arise from algebro-topological (I'm thinking "chromatic") considerations are unbounded. This is getting dangerously close to philosophy, but my intuition is that the "source" of these unbounded (periodic) E_oo-rings is just different.
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2025-03-21T14:48:30.204201
| 2020-04-02T18:26:03 |
356391
|
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|
Stack Exchange
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Hasse invariant of abelian varieties with complex multiplication
Is there a good way to compute Hasse invariants of elliptic curves or higher dimensional Abelian varieties with complex multiplication?
For example, if $E$ is an elliptic curve with CM by an imaginary quadratic field $K$, what are the possibilities for the valuation of the Hasse invariant of $E$ at $p$ depending on if $p$ splits/is inert/ramifies in $K/\mathbb{Q}$? What if $A$ is an Abelian variety of dimension $d$ with CM by a CM field $K$ with $[K:\mathbb{Q}] = 2d$, depending on what $p$ does in $K/K^+$ and in $K^+/\mathbb{Q}$?
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2025-03-21T14:48:30.204276
| 2020-04-02T18:56:34 |
356394
|
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"authors": [
"Aleksei Kulikov",
"Brian Hopkins",
"Carlo Beenakker",
"Johnny Cage",
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"url": "https://mathoverflow.net/questions/356394"
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Stack Exchange
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Meinardus theorem at use: problems with conditions
I am working on an enumerative problem related to knot theory, and I have found the following generating function
$$F(z)=\prod_{n\geq 1} \frac{1}{(1-z^{2n+1})^2}.$$
I am interested on getting asymptotic estimates for this GF. I am not an expert on partition function theory, but it seems that trying to apply Meinardus Theorem on this context should be a good idea (in this case, the sequence of coefficients to be studied is $a_n=2$ for $n=2k+1$ and $\geq 3,$ and $a_n=0$ in the rest of the cases.
Once trying to apply Meinardus Theorem (as it is stated in the reference book of George Andrews: The theory of partitions) I have seriuos troubles with condition III, namely that
$$f(t,y)=Re(g(e^{-t-2\pi i y}))-g(e^{-t})\leq -C t^{-\varepsilon}$$
for $g(z)=\sum_{n\geq 1} a_n z^n$, $|y|\leq 1/2$, $t$ small enough and convenient choices of positive $C$ and $\varepsilon$. Essentially, the problem I am finding is that when taking $y=1/2$ and $t$ small (or in a neighbourhood), the value I get for $f(t,y)$ is $-2t+o(t)$, which cannot fit with the condition $\leq -C t^{-\varepsilon}$.
So: which are the alternatives that one can use to get the assymptotics in such a situation (without having to make the computation of the contour integral computation?)
I was thinking also to try to get results exploiting the asymptotics of the partition function, but I do not see a direct way.
This is a postscript to Carlo Beenakker's answer, a combinatorial explanation for the relationship between $a_n$ and $c_n$.
Your $F(z)=\sum_n c_nz^n$ is the generating function for the number of partitions of $n$ into odd parts 3 or greater where there are two kinds of each part. Carlo's $G(z) = \sum_n a_nz^n$ allows two kinds of 1's also, matching one of the descriptions given for A022567.
Rewrite the equation in Carlo's comment as $a_n = c_n + 2a_{n-1} - a_{n-2}$. To see this, condition the partitions counted by $a_n$ by whether they contain (either kind of) 1 as a part. Those with no 1's are given by your $c_n$. To build these partitions of $n$ with 1's from smaller partitions, take the partitions of $n-1$ counted by $a_{n-1}$ and include a $1_1$. Repeat and include a $1_2$ to complete the $2a_{n-1}$ term. Some partitions of $n$ arise twice, though---precisely those that include both a $1_1$ and a $1_2$; the number of those is $a_{n-2}$ by adding $1_1$ and $1_2$ to each partition of $n-2$. So $a_n = c_n + (2a_{n-1}-a_{n-2})$.
Glad this was helpful. Curious what kind of knot theory enumeration brought this up.
Let me first consider the ratio
$$G(z)=\frac{F(z)}{(1-z)^2}=(1-z)^{-2}\prod_{n\geq 1} \frac{1}{(1-z^{2n+1})^2}=\frac{1}{\left(\left(z;z^2\right){}_{\infty }\right){}^2},$$
with $(a;q)_\infty$ the q-Pochhammer symbol.
The series expansion
$$G(z)=\sum_{n=0}^\infty a_n z^n$$
has coefficients $a_n$ listed in OEIS:A022567.
The large-$n$ asymptotics has been derived (using a variation of the Meinardus method) by V. Kotěšovec in A method of finding the asymptotics of q-series based on the convolution of generating functions, see page 8:
$$a_n=\tfrac{1}{4} 6^{-1/4}n^{-3/4}\,e^{\sqrt{2\pi^2 n/3}}\bigl(1+{\cal O}(n^{-1/2})\bigr).$$
Higher order terms are listed in the OESIS entry, the factor $1+{\cal O}(n^{-1/2})$ expands further into $1+q_1 n^{-1/2}+q_2 n^{-1}+{\cal O}(n^{-3/2})$ with
$$q_1=\frac{\pi}{12\sqrt 6}-\frac{\sqrt{27/2}}{8\pi},\;\;q_2=\frac{\pi^2}{1728}-\frac{45}{256\pi^2}-\frac{5}{64}.$$
The coefficients in the series expansion of $F(z)=\sum_{n=0}^\infty c_n z^n$ follow by equating $c_n=a_n-2a_{n-1}+a_{n-2}$ and expanding $n^{3/4}e^{-\sqrt{2\pi^2 n/3}}c_n$ in powers of $1/\sqrt n$. After some algebra I find
$$c_n= \frac{\pi^2}{6n}a_n\left(1+{\cal O}(n^{-1/2})\right).$$
The terms $q_1$, $q_2$ do not contribute to this order.
great, many thanks! what do you mean by "smaller by a factor $\Pi^2/6n$"? This means that $[z^n]F(z)=\Pi^2/6n a_n$? if yes, how you reach (in a fast way, I assume) such result?
if $F(z)=\sum_n c_nz^n$, then $c_n=a_n-2a_{n-1}+a_{n-2}$; inserting the asymptotics for $a_n$ gives $c_n=(\pi^2/6)n^{-1}a_n[1+{\cal O}(n^{-1/2})]$.
While I do not doubt the final result, it seems to me that the derivation in your comment doesn’t work: if you just plug in the bound from your post then the error term will be bigger than the main term. It can be fixed by a more precise expansion, though (up to $o(1/n)$), and I would be very surprised if it would not follow from the same method.
@AlekseiKulikov --- thanks, I have expanded the derivation in the answer box, which I think is now complete (the higher order terms in the expansion of $a_n$ do not seem to contribute to $c_n$ to order $1/\sqrt n$).
|
2025-03-21T14:48:30.204715
| 2020-04-02T19:13:38 |
356396
|
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|
Stack Exchange
|
Relation between compact vertical cohomology and local cohomology groups
I'm reading the books by Bott & Tu and Milnor & Stasheef simultaneously. The following is my doubt:
The Thom isomorphism in Bott & Tu is obtained as $H_{cv}^{*+n}(E)\rightarrow H^*(M)$, where $\pi\colon E\to M$ is $n$ plane bundle over the manifold of dimension $m$ manifold $M$ and the isomorphism is given by the integration along fiber map $\pi_*$. By the projection formula, the Thom isomorphism $\mathscr{T}$, inverse to $\pi_*$, is then explicitly given by $\omega\mapsto \pi^*(\omega)\wedge\Phi$, where $\Phi=\mathscr{T}(1)$ is the Thom class of $E$.
In Milnor & Stasheef, the isomorphism is $\mathscr{T}\colon H^*(M)\to H^{*+n}(E,E_0)$, where $E_0$ is the complement of zero section and the map factors through $H^*(E)$ i.e. $H^*(M)\xrightarrow{\pi^*} H^*(E)\xrightarrow{\smile\text{fundamental class}} H^{*+n}(E,E_0)$.
I know that $H_{c}^{*+n}(E)=\varinjlim H_{c}^{*+n}(E, E-K)$ over the directed set of compact subsets of $E$. My question is, how can I relate $H^{*+n}(E,E_0)$ and $H_{cv}^{*+n}(E)$.
Any hint would be helpful. Thanks.
Compactly supported homology can be seen to be equal to the reduced homology of the one point compactification, essentially by definition of the topology on the one point compactification.
A similar argument shows that vertical compactly supported homology is the reduced homology of the space obtained by compactifying the fibers and then identifying each of the points at infinity to a single point.
Excision then relates this to the relative homology $H^* (E, E_0)$. $H^* (E, E_0) \cong H^*(D(E),S(E)) \cong \bar{H}^*(D(E)/S(V))$. Here $D(E),S(E)$ are the disk and sphere bundles. This quotient can be seen to be homeomorphic to the space obtained by compactifying the fibers and then identifying the points at infinity. This space is called the Thom space of E. This then yields the isomorphism between vertical compactly supported homology and the relative homology of the total space and the total space with deleted zero section
On the off chance you see this, would you be willing to expand your first paragraph? It's not clear to me (but I'm not much good at topology!)
@AlekosRobotis First off, let me mention everything I wrote should be in cohomology not homology. Now, suppose that our space $X$ behaves nice enough so that there is a well behaved contractible neighborhood of the point at infinity in the compactification (maybe this is always true). Excision type results will imply that the inclusion of cochains that are zero valued in some neighborhood of infinity into cochains that are zero valued at infinity is an isomorphism on cohomology. But such a complex is isomorphic to the complex of compactly supported cochains, just by forgetting $\infty$.
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2025-03-21T14:48:30.204919
| 2020-04-02T19:36:42 |
356401
|
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"Taras Banakh",
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"url": "https://mathoverflow.net/questions/356401"
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Stack Exchange
|
A permutation group inducing a topologically transitive action without dense orbits on $\omega^*$
Let $G$ be a subgroup of the permutation group $S_\omega$ of the countable infinite set $\omega$. Each bijection $g\in G$ admits a unique extension to a homeomorphism $\bar g$ of the Stone-Cech compactification $\beta\omega$ of $\omega$. The homeomorphism $\bar g$ induces a homeomorphism of the remainder $\omega^*=\beta\omega\setminus\omega$ of the Stone-Cech compactification. So, we obtain a continuous action of the group $G$ on the compact Hausdorff space $\omega^*$. I am interested in properties of the obtained dynamical system $(\omega^*,G)$. Namely, I would like to know the answer to the following
Problem. Is there a subgroup $G\subseteq S_\omega$ such that the dynamical system $(\omega^*,G)$ is topologically transitive (=each nonempty open set has dense orbit) but does not have a dense orbit.
An example of such subgroup $G$ exists under the assumption $\mathrm{non}(\mathcal M)<\mathfrak c$. So, the question actually ask about the situation in ZFC.
Remark. If a group $G\subseteq S_\omega$ induces a topologically transitive action on $\omega^*$, then $G$ has large cardinality, namely, $|G|\ge\mathsf \Sigma\ge\max\{\mathfrak b,\mathfrak s,\mathrm{cov}(\mathcal M)\}$. More information on the cardinal $\mathsf \Sigma$ can be found in this preprint.
Identify $\omega$ with a dense countable order $Q$. Let $G$ be the group of piecewise monotone permutations of $Q$ (i.e., cut $Q$ into finitely many convex pieces, and rearrange them to finitely many convex pieces through order preserving or reversing partial isomorphisms). I think $G$ acts topologically transitively on $\beta^*Q$. But I don't see if there's a dense orbit.
Note: For $G$ acting on $\omega$, it's clear that if for every $I,J\subset\omega$ with $J,\omega-I$ infinite there exists $g\in G$ such that $gI\subset^* J$, then $G$ acts minimally on $\beta^\omega$. Here $\subset^$ means inclusion modulo finite subset. In particular $S_\omega$ acts minimally on $\beta^*\omega$ (as you told me in answer to a comment of mine in your previous question; I'm just writing the easy argument to help the reader.
You are right the group of piecewise monotone functions acts topologically transitively on $Q^*$ (because each sequence contains a monotone subsequence). Concerning dense orbit, let me think a bit.
Yes. I needed piecewise, so as to handle the case of a bounded above increasing sequence, vs an unbounded above increasing sequence.
The action of your group on $Q^*$ has a dense orbit if there exists a Ramsey ultrafilter. More precisely, the orbit of the Ramsey ultrafilter will be dense. It remains to understand what happens if Ramsey ultrafilters do not exist. But anyway, this is not a ZFC-example.
The action of the group $G$ on $Q^*$ will have many dense orbits: just take any ultrafilter living on a monotone sequence; its orbit will be dense.
That's true, it doesn't work.
It turns out that this problem is independent of ZFC because of the following simple
Theorem. Under $\mathfrak t=\mathfrak c$, every topologically transitive continuous action of a group $G$ on $\omega^*$ has a dense orbit.
Proof. Let $(A_\alpha)_{\alpha\in\mathfrak c}$ be an enumeration of all infinite subsets of $\omega$. By transfinite induction we shall construct a transfinite sequence of infinite subsets $(U_\alpha)_{\alpha\in\mathfrak c}$ of $\omega$ and a transfinite sequence $(g_\alpha)_{\alpha\in\mathfrak c}$ of elements of the group $G$ such that for every $\alpha\in\mathfrak c$ the following conditions are satisfied:
(a) $U_\alpha\subseteq^* U_\beta$ for all $\beta<\alpha$;
(b) $g_\alpha(U_\alpha)\subseteq^* A_\alpha$.
To start the inductive construction, put $U_0=A_0$ and $g_0$ be the identity of the group $G$. Assume that for some ordinal $\alpha\in\mathfrak c$, a transfinite sequence $(U_\beta)_{\beta<\alpha}$ satisfying the condition (a) has been constructed. By the definition of the tower number $\mathfrak t$ and the equality $\mathfrak t=\mathfrak c>\alpha$, there exists an infinite subset $V_\alpha\subseteq\omega$ such that $V_\alpha\subseteq^* U_\beta$ for all $\beta<\alpha$. The infinite sets $V_\alpha$ and $A_\alpha$ and determine clopen sets $\overline V_\alpha=\{p\in\omega^*:V_\alpha\in p\}$ and $\bar A_\alpha=\{p\in\omega^*:A_\alpha\in p\}$ in the space $\omega^*=\beta\omega\setminus\omega$. Since the action of the group $G$ on $\omega^*$ is topologically transitive, there exist $g_\alpha$ and an infinite subset $U_\alpha\subset V_\alpha$ such that $g_\alpha(\overline U_\alpha)\subseteq \bar A_\alpha$, which implies $g_\alpha (U_\alpha)\subseteq^* A_\alpha$. This completes the inductive step.
Adter completing the inductive construction, extend the family $\{U_\alpha\}_{\alpha\in\mathfrak c}$ to a free ultrafilter $\mathcal U$ and observe that its orbit intersetcs each clopen set $\bar A_\alpha$, $\alpha\in\mathfrak c$ and hence is dense in $\omega^*$. $\qquad\square$
Interesting. A follow-up question would be the same question (esp. in ZFC+CH) for $G\subset\mathrm{Homeo}(\beta^*\omega)$.
What is $\beta^\omega$? If $\beta^\omega=\omega^*$, then the theorem answer this question under $\mathfrak t=\mathfrak c$ and hence under CH.
Yes I mean $\beta^X=\beta X-X$ (I'm boycotting the notation $X^$ which is incomprehensible without context, and I like to retain the letter $\beta$ to denote the Stone-Cech remainder). Ah indeed I didn't notice your answer doesn't assume $G\subset S_\omega$, so is much more general than what you initially asked.
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2025-03-21T14:48:30.205253
| 2020-04-02T20:21:06 |
356404
|
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"Hussain Rashed",
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|
Stack Exchange
|
The set of all functions which vanish at infinity is a subset of the set of all functions which have vanishing variation
Let $X$ be a coarse space, we define the following:
$D_b(X)$ is the set of all bounded functions $f:X\rightarrow \mathbb{C}$
$f\in $$D_b(X)$ is said to vanish at infinity if for each $\varepsilon$>0 there is a bounded set $K\subseteq X$ such that $\left \| f|_{X\setminus K} \right \|<\varepsilon$, i.e. the set \begin{Bmatrix}
x\in X:\left | f(x) \right |\geq \varepsilon
\end{Bmatrix} is bounded in $X$. The collection of all functions that vanish at infinity is denoted by $D_0(X)$.
Let $f \in D_b(X)$, and $E\subseteq X\times X$ is a controlled set (an entourage), the $E$-variation of f is the function $\operatorname{Var}_E(f)$ defined by
$$ \operatorname{Var}_E(f)(x
)=\sup\left \{ \left | f(x)-f(y) \right |:(x,y) \in E \right \} $$
$f \in D_b(X)$ is said to have vanishing variation if $\operatorname{Var}_E(f)\in D_0(X)$, for all controlled sets $E\subseteq X\times X$.The collection of all functions that have vanishing variation is denoted by $D_h(X)$.
I need to verify that $D_0(X)\subseteq D_h(X)$. To this end, let $f \in D_0(X)$, $E$ be a controlled set, and $\varepsilon >0$, then there exists a bounded subset $K\subseteq X$ such that $\left \| f|_{X|K} \right \|<\frac{\varepsilon }{2}$. We aim to show that $\operatorname{Var}_E(f)(x)<\varepsilon$ for all $x\in X|L$, for some $L$ bounded subset. Indeed taking $L=K$, if $x\in X|K$ with $(x,y)\in E$, we consider two cases:
Case 1: $y\in X|K$, then $\left | f(x)-f(y) \right |<\varepsilon$.
Case 2:$y\in K$, in this case I cannot get that $\left | f(x)-f(y) \right |<\varepsilon$. Can we find a bounded subset $L$ of $X$ such that $ \operatorname{Var}_E(f)(x)<\varepsilon$ outside $L$?
Difference between $\operatorname{Var}(f)|{X-K}$ and $\operatorname{Var}(f|{X-K})$.
$\operatorname{Var}_E(f)(x)$ where $x \in X-K$ is the largest distance between $f(x)$ and $f(y)$ where $(x,y)\in E$, while $\operatorname{Var}E(f)|{X-K}(x)$ is the largest distance between $f(x)$ and $f(y)$ where $(x,y)\in E \cap (X-K)\times (X-K)$. How would that help?
Try an example: $X = E = \mathbb{R}$, $f(x) = 1/(1+x^2)$, so $\operatorname{Var}(f)(x) = 1 - 1/(1+x^2)$ for all $x$. Now whatever $K$ is, if $x \notin K$, then $\operatorname{Var}(f)|_{\mathbb{R}-K}(x)$ is still $1-1/(1+x^2)$. The problem is that you shouldn't just take all values $y$. You need to restrict to $y \in X-K$ as well. ... Can I ask you where this question came from?
I am reading this paper https://arxiv.org/pdf/1711.06836.pdf, and it stated on page 16 that $D_0(X)$ is a subset of $D_h(X)$, and that is not clear to me.
At this point it may be a question for the authors.
For $f(x) = 1/(1+x^2)$, $\operatorname{Var}(f)(x) = \max{1/(1+x^2), 1-1/(1+x^2)}$ for all $x$, and the maximum is $1-1/(1+x^2)$ for $|x| \geq 1$. The point is the same though.
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2025-03-21T14:48:30.205433
| 2020-04-02T20:27:33 |
356405
|
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|
Stack Exchange
|
Pro-trivial cosimplicial tower of spaces
Let $\{X^\bullet_s\}$ be a cosimplicial tower of spaces. In other words, for each fixed "tower degree" n, we have a (let's assume Reedy fibrant) cosimplicial space $X^\bullet_n$, and for each fixed cosimplicial degree $k$, we have a tower of spaces $\{X^k_s\}_s$. Suppose that for each $\ast$, the tower of (let's assume abelian) groups $\{\pi_\ast X^k_s\}_s$ is pro-trivial. Are there reasonable conditions under which we can conclude that the same is true after taking totalizations? That is, are there conditions under which $\{\pi_\ast \mathsf{Tot}X^\bullet_s\}_s$ is pro-trivial?
It seems like strong convergence of each homotopy spectral sequence should be enough, but I haven't been able to prove this. I'm wondering if anyone has any suggestions for suitable conditions.
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2025-03-21T14:48:30.205514
| 2020-04-02T20:42:28 |
356406
|
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|
Stack Exchange
|
Why are Serre functors always exact?
Let $k$ be a field and $\mathcal{T}$ be a $k$-linear triangulated category with finite dimensional spaces of morphisms. Bondal and Kapranov proved that every Serre functor on $\mathcal{T}$ is exact (Proposition 3.3 here). A different proof was given by Huybrechts in his Fourier-Mukai book (Proposition 1.46). The proof by Bondal and Kapranov is a bit technical and relies on the notion of a mattress. For now, I decided not to read it thoroughly. Concerning the one given by Huybrechts, I could not convince myself that it is complete. Could someone please shed a light on this, possibly providing a simple proof or explaining how the desired linear form in Huybrechts' proof is constructed?
The Serre duality isomorphism $Hom(C_0,S(C)) \cong Hom(C,C_0)^\vee$ represents $\xi$ as a linear form. Is that what you were asking?
@Sasha No, I did not understand how to define the linear form for a general map $u : C \to C_0$. In Huybrechts' proof, it is clear how to do this if $u$ factors through $\varphi$ or $h$, but I do not know how to deal with the general case
(Notations as in Huybrechts' book.) Consider the maps $\mathrm{Hom}(C,S(B))\rightarrow\mathrm{Hom}(C,C_0)$ and $\mathrm{Hom}(A[1],C_0)\rightarrow\mathrm{Hom}(C,C_0)$, and let $I$ and $J$ be their images. Huybrechts tells you how to define linear forms $\Xi_I:I\rightarrow k$ ("condition $\mathrm{i}')$") and $\Xi_J:J\rightarrow k$ ("condition $\mathrm{ii}')$"), and shows that $\Xi_I|_{I\cap J}=\Xi_J|_{I\cap J}$. The latter condition says exactly that there is a linear form on $I+J$ restricting to $\Xi_I$ on $I$ and $\Xi_J$ on $J$; extending this form to all of $\mathrm{Hom}(C,C_0)$ gives the desired form $\Xi$ Serre dual to $\xi$.
Thanks, that works! Sometimes we forget we are dealing with vector spaces and can extend maps!
|
2025-03-21T14:48:30.205651
| 2020-04-02T20:57:58 |
356407
|
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"sort": "votes",
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|
Stack Exchange
|
Strong chains in $[\omega_2]^{\omega_2}$ mod finite of length $\omega_3$
Probing a bit the difference between $[\omega_1]^{\omega_1}$ and $[\omega_2]^{\omega_2}$ modulo the finite sets:
Question
Can there exist a family $\langle X_\alpha:\alpha<\omega_3\rangle$ of
sets in $[\omega_2]^{\omega_2}$ such that for $\alpha<\beta<\omega_3$,
$|X_\beta\setminus X_\alpha|=\omega_2$, and
$ X_\alpha\setminus X_\beta$ is finite?
The existence of such a family (a strong chain of length $\omega_3$ in $[\omega_2]^{\omega_2}$ mod finite) implies the existence of a strongly almost disjoint family in $[\omega_2]^{\omega_2}$ of size $\omega_3$: let $A_\alpha:=X_{\alpha+1}\setminus X_\alpha$, and we get a collection of size $\omega_3$ in $[\omega_2]^{\omega_2}$ with pairwise finite intersection. The existence of such a family is consistent, as shown by Baumgartner [1].
Koszmider [2] showed that you can have long chains (of length $\omega_2$) in $[\omega_1]^{\omega_1}$ modulo the finite ideal, so this question is an attempt to lift his result to $[\omega_2]^{\omega_2}$. (I do not even know the answer for $[\omega_2]^{\omega_2}$ if ``finite'' is replaced by countable, but obviously a consistency result there should be easier to obtain, I just don't know if it's been done.)
There are differences between $\omega_1$ and $\omega_2$ that indicate this question may be interesting. Koszmider [3] showed that one can have a sequence $\langle f_\alpha:\alpha<\omega_2\rangle$ that is strictly increasing mod finite, while Shelah [4] has shown in ZFC that this phenomenon CANNOT happen at $\omega_2$: there is no sequence $\langle f_\alpha:\alpha<\omega_3\rangle$ of functions in $^{\omega_2}\omega_2$ that is strictly increasing mod finite.
[1] Baumgartner, James E., Almost-disjoint sets, the dense set problem and the partition calculus, Ann. Math. Logic 9, 401-439 (1976). ZBL0339.04003.
[2] Koszmider, Piotr, On the existence of strong chains in $(\wp(\omega_1)/ \text{Fin})$, J. Symb. Log. 63, No. 3, 1055-1060 (1998). ZBL0936.03043.
[3] Koszmider, Piotr, On strong chains of uncountable functions, Isr. J. Math. 118, 289-315 (2000). ZBL0961.03039.
[4] Shelah, Saharon, On long increasing chains modulo flat ideals, Math. Log. Q. 56, No. 4, 397-399 (2010). ZBL1200.03031.
|
2025-03-21T14:48:30.205814
| 2020-04-02T21:16:44 |
356408
|
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|
Stack Exchange
|
Finding k items in a binary tree
Let us be given a binary tree of height $n$ (and $2^n$ leaves) among which we search $k$ items, where $k < < 2^n$. Suppose we have a test that shows if in the children and childrens-children generations of a node at least one item is hidden (sorry for my prose speak). The question is: how many tests do we need in average (or expectation under uniform law) to find all $k$ items? In computer science speak it would be the expected run-time of the following algorithm. Of course, finding one item needs $n$ tests (and so $k \times n$ is a trivial upper bound to find $k$ items). Now, running through the tree we cut out useless subtrees on each "hunt" for one item, we expect less and less tests for the following ones to be needed, and so I expect much less than $k \times n$ tests needed in average.
I guess this question is well-known, but I lack keywords in combinatorics / computer science to find it. Thank you!
We can associate leaf elements of the binary tree with binary codes of length $n$. The code we assign to a leaf is the sequence of left and right traversals that you must make to reach the element in the binary tree. Assign a 0 for each left traversal and a 1 for each right traversal. For instance, if you must traverse the tree by going to a left child, then another left child, then a right child, the code for this node in the tree would be "001".
Let $x_1, \ldots, x_k$ be the binary codes associated with the $k$ elements that you are searching for in the binary tree. (note that these binary representations will have length $n$, the depth of the tree).
In the worst case it takes time $\Omega(k + k(n-\log k))$ to find all $k$ elements that you are searching for
In particular, for $k \ll 2^n$ this is $\Omega(k\cdot n)$.
proof:
We construct worst case elements as follows:
We chose the elements such that the first $\log k$ digits of the code of $x_i$ is the binary representation of $i$.
Note that the prefix consisting of the first $k$ digits of the binary codes for each $x_i$ differ, so the $x_i$ are ancestors in the tree of distinct nodes at depth $\log k$. Traversing this tree of depth $\log k$ to get to these "parent nodes" takes time $\Omega(k)$. From there, we must traverse each subtree starting at the parent node for each of the $k$ items. This takes time $\Omega(k(n - \log k)$ as we must traverse $n-\log k$ more levels / digits in the binary representations.
In answer to your other question about average running time, I believe that the answer is also $\Omega(n\cdot k)$ average running time.
The probability that two randomly chosen leaves in your tree share a prefix of length $\ell$ is $1/2^\ell$.
Thus the expectation of the number of items that share a prefix of length $\ell$ is approximately (it is approximate because you can't actually chose the same element twice)
${k \choose 2} \cdot 2^{-\ell}$.
Setting $\ell = 2\log k$ we get that the expected number of elements that share a prefix of length $2\log k$ is less than $1$.
Hence you should still expect to need to spend time $\Omega(k\cdot n)$ on the problem.
|
2025-03-21T14:48:30.206045
| 2020-04-02T21:19:58 |
356409
|
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|
Stack Exchange
|
Is there a characterization of the class of first-order formulas that are closed in every compact Hausdorff structure?
Fix a relational language $\mathcal{L}$. (I don't think relational really matters that much but I don't want to worry about it.) A topological $\mathcal{L}$-structure is an $\mathcal{L}$-structure $M$ together with a topology $\tau$ on $M$ such that for every relation symbol $R \in \mathcal{L}$ (including equality), $R^{M}$ is closed as a subset of $M^n$ (where $n$ is the arity of $R$). Note that since the diagonal is closed in $M^2$, a topological $\mathcal{L}$-structure is automatically Hausdorff. For any topological property $P$ (such as compact or second countable), a topological $\mathcal{L}$-structure is $P$ if the underlying topological space is $P$.
For any $\mathcal{L}$-structure $M$ and any $\mathcal{L}$-formula $\varphi(\bar{x})$ with a designated tuple of variables $\bar{x}$ such that the free variables of $\varphi$ are contained in $\bar{x}$, let $\varphi(M)$ be the subset of $M^n$ of tuples $\bar{a}$ such that $M \models \varphi(\bar{a})$, where $n = |\bar{x}|$.
Let $T$ be the set of $\mathcal{L}$-sentences true in every compact $\mathcal{L}$-structure. $T$ is certainly non-trivial (for non-trivial $\mathcal{L}$). For instance, if $\mathcal{L}$ contains a binary relational symbol $\leq$, then $T$ contains a sentence that says that if $\leq$ is a linear order then there is a greatest element.
For lack of a better term, call an $\mathcal{L}$-formula $\varphi(\bar{x})$ closed if for every compact $\mathcal{L}$-structure $M$, $\varphi(M)$ is a closed subset of $M^n$. Broadly my question is about syntactic characterization of closed formulas, but before we can even talk about that there's a deeper issue.
Question 1. For a fixed language $\mathcal{L}$, are $T$ and the collection of closed $\mathcal{L}$-formulas set theoretically absolute? What if we change the definition of $T$ and closed formula to restrict attention to second countable compact $\mathcal{L}$-structures?
If they aren't then I'm not very hopeful for a syntactic characterization. There are a few easy facts about the collection of closed $\mathcal{L}$-formulas:
Any $\mathcal{L}$-sentence is closed.
Any positive atomic formula is closed.
Any positive Boolean combination of closed formulas is closed.
Any variable substitution of a closed formula is closed (i.e. if $\varphi(x,y)$ is closed, then $\varphi(x,x)$ is closed).
If $\varphi$ is closed, then $\exists x \varphi(x)$ is closed (this uses the fact that the structures are compact, so the projection maps are closed).
If $\varphi$ is closed, then $\forall x \varphi(x)$ is closed (this uses the fact that the projection maps are open).
Another subtler operation under which closed formulas are closed is special relative universal quantification.
If $\varphi$ is closed and $\psi$ is arbitrary, but has $x$ as its only free variable, then $\forall x (\psi \rightarrow \varphi)$ is closed. (EDIT: I realized you don't even need $\psi$ to be closed for this.)
This last operation is part of the definition of generalized positive formulas, which are relevant to positive set theory. A hopeful guess as to a syntactic characterization is the following.
Question 2. Is every closed $\mathcal{L}$-formula logically equivalent over $T$ to one of the form $\bigvee_{i<n} \varphi_i \wedge \psi_i$, where each $\varphi_i$ is a sentence and each $\psi_i$ is a generalized positive formula? Again, what if we restrict attention to second countable structures?
Where the generalized positive formulas are the smallest class of formulas containing $\bot$, $\top$, and the positive atomic formulas and closed under positive Boolean combination, existential quantification, and the formation of formulas of the form $\forall x (\psi \rightarrow \varphi)$, where $\psi$ is arbitrary and has at most $x$ free and $\varphi$ is generalized positive.
Yes that is what I mean.
How about $\forall x,\forall y,(\psi(x,y)\to\varphi)$ with $\psi$ arbitrary?
More generally, I think you need to consider formulas $\phi(\vec x,\vec p)$ such that for every fixed $\vec a\in M$, $\phi(M,\vec a)$ is closed. (Let’s call them parametric closed formulas.) These include all atomic formulas, all formulas in which only $\vec p$ are free, and are closed under monotone Boolean operations, $\exists x_i$, and $\forall p_i$. (You can simulate $\forall x_i$ with $\forall p_i$, and I think you can also simulate variable substitution by quantifiers, so these are not necessary.) Then closed formulas are the parametric closed formulas with no free $\vec p$.
This allows not just $\forall p_1,\forall p_2,(\psi(p_1,p_2)\to\dots)$ as above, but also more complicated things like $\forall p_0,\exists x_1,\forall p_1,\dots,\exists x_n,\forall p_n,(\psi(\vec p)\to\dots)$, which you’d have hard time to introduce without separating the $\vec x$ and $\vec p$ variables.
I believe the formulas in my previous comments should give counterexamples to Question 2, btw.
In your second comment are you saying you think that this is a characterization of parametric closed formulas?
No, only that those are the obvious closure conditions. (However, since they are fairly simple, I do have a hope that this might be a complete characterization up to provable equivalence.)
Okay. Should I modify my question 2 to include your observation?
This is up to you.
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2025-03-21T14:48:30.206491
| 2020-04-02T21:28:05 |
356410
|
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|
Stack Exchange
|
Quotients of the irrationals
Everyone knows that there is a closed equivalence relation $\sim$ on the Cantor set $C$ such that each non-trivial equivalence class has exactly $2$ points and $[0,1]\simeq C/\sim$. Thus a closed quotient of a zero-dimensional space may not be zero-dimensional, even if the equivalence classes are compact. Note that in this case, if we specify that $C/\sim$ must be totally separated, then $C/\sim$ is automatically zero-dimensional, because every totally separated compact metric space is zero-dimensional. Totally separated means that for every two points $x$ and $y$ in the space, there is a clopen set containing $x$ and missing $y$. Zero-dimensional means that the space has a basis of clopen sets.
This question is about similar quotients of the irrationals $\mathbb P$.
Question. Let $\sim$ be a closed equivalence relation on $\mathbb P$ such that $\mathbb P/\sim$ is Polish and every equivalence class is compact. If $\mathbb P/\sim$ is totally separated, then is $\mathbb P/\sim$ necessarily zero-dimensional?
Note that every Polish space is a closed quotient of $\mathbb P$; shown here. So the condition that the equivalence classes are compact is critical. I believe my question is equivalent to: Is every totally disconnected Polish perfect image of $\mathbb P$ zero-dimensional? A continuous mapping is perfect if it is closed has compact point preimages.
"Everyone knows": is it an exercise? the obvious relation I can think of has countably many equivalence classes of cardinal 2 and others have cardinal 1.
@YCor that's what I meant by non-trivial equivalence classes have size 2
Is there any nonempty separable metrizable space that's not homeomorphic to a quotient of $\mathbf{P}$?
If you take the projection $\mathbf{R}\to\mathbf{R}/\sqrt{2}\mathbf{Z}$ and restrict it to $\mathbf{P}\to\mathbf{R}/\sqrt{Z}\mathbf{Z}$, it's surjective. Is it a quotient map? (Restricting to $\mathbf{P}\cap [0,3]$ which is homeomorphic to $\mathbf{P}$, we can assume fibers are finite of cardinal $\le 3$.)
@YCor So just define $p\sim p'$ if $p$ and $p'$ differ by an integer multiple of $\sqrt{2}$. This is a closed equivalence relation on $\mathbb P$, and the quotient is connected?
@YCor There are separable metrizable spaces (e.g. the non-analytic spaces) which are not quotients of $\mathbb P$. But maybe every Polish space is a quotient of $\mathbb P$?
Yes, it seems to work with the circle.
"Polish spaces" are metric complete? Then they are absolute $G_\delta$-spaces hence analytic and images of irrationals.
(I missed "non-trivial" in the formulation of the OP question too.
Does a closed (equivalence) relation $\ W\subseteq X^2\ $ mean that $\ W\ $ is closed in $\ X^2\ $ (or simply, in the case of equivalence, that each class is closed in $X$)?
@WlodAA Engelking's definition (which I use) is that the natural mapping $X\to X/\sim$ is closed. This guarantees $X/\sim$ is metrizable if $X$ is metrizable. I think he gives an example showing this definition is stronger than yours.
I am justified in not knowing -- I am the only Polish topologist who at the time was not gifted their own copy from Engelking of his classic. He told me that he wanted to give me the new edition but... then crazy political times came and I emigrated and he had more urgent matters to attend to. In this and only this way I am a unique Polish topologist.
(Indeed, the difference between the two definitions is common and can be dramatic).
@WlodAA Your comment about Polish spaces is correct. Engelking proved that every Polish space is a continuous closed image of $\mathbb P$, and hence a closed quotient of $\mathbb P$. I've simplified my question now.
The answer to this question is negative and follows from the characterization:
Theorem. A topological space $X$ is an image of the space of irrationals $\mathbb P$ under a perfect map $f:\mathbb P\to X$ if and only if $X$ is Polish and nowhere locally compact.
Proof. To prove the "only if" part, assume that a topological space $X$ is the image of $\mathbb P$ under a perfect map $f:\mathbb P\to X$. By Theorem 3.7.20 of Engelking's "General Topology" (denoted later by [EGT]), the space $X$ is regular, by Theorem 3.7.19 in [EGT], $X$ is second-countable and by the Uryssohn Metrization Theorem, $X$ is metrizable and separable. By Theorem 3.9.10 of [EGT], the space $X$ is Cech-complete and being separable and metrizable is Polish. Assuming that $X$ contains a compact subset $K$ with non-empty interior, we can apply Theorem 3.7.2 of [EGT] and conclude that the preimage $f^{-1}(K)$ is compact and by the continuity of $f$, it has non-empty interior in $\mathbb P$. On the other hand, it is well-known that $\mathbb P$ contains no compact sets with nonempty interior. This contradiction shows that $X$ is nowhere locally compact.
To prove the "if" part, assume that $X$ is a nowhere locally compact Polish space. Let $\bar X$ be any metric compactification of $X$. By Theorem 4.18 in Kechris' "Classical Descriptive Set Theory" (denoted later by [CDST]), $\bar X$ is the image of the Cantor cube under a continuous map $\tilde g:2^\omega\to\bar X$. Using the Kuratowski-Zorn Lemma, choose a minimal closed subset $Z\subseteq 2^\omega$ such that $f(Z)=\bar X$. Let $g=\tilde g{\restriction}Z$. The minimality of $Z$ and closedness of the map $g:Z\to\bar X$ ensure that for any non-empty open set $U\subset Z$ the image $g(U)$ has nonempty interior in $\bar X$.
Being Polish, the space $X$ is a $G_\delta$-set in $\bar X$ and its preimage $P=g^{-1}(X)$ is a $G_\delta$-set in the zero-dimensional compact space $Z$. We claim that $P$ is nowhere locally compact. Assuming that $P$ contains a compact subset $K$ with nonempty interior, we can use the minimality of $Z$ and conclude that $g(K)$ is a compact set with nonempty interior in $X$, which contradicts the nowhere local compactness of $X$. This contradiction shows that the Polish zero-dimensional space $P$ is nowhere locally compact. By the Aleksandrov-Urysohn characterization of $\mathbb P$ (see Theorem 7.7 in [CDST]), the space $P$ is homeomorphic to the space of irraionals $\mathbb P$. It remains to apply Proposition 3.7.6 of [EGT] to see that the restriction $f=g\restriction P:P\to X$ is perfect.
Nice! Such a clean result!
|
2025-03-21T14:48:30.206919
| 2020-04-02T22:03:02 |
356415
|
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|
Stack Exchange
|
Real manifolds and affine schemes
I noticed the following strange (to me) fact. If $M$ is a real manifold (smooth or not) and $R = C(X, \mathbb{R})$ is the ring of real functions (smooth functions in the smooth case) then the affine scheme $X = \mathrm{Spec}(R)$ has a natural map $M \to X$ which is a homeomorphism on real points i.e. $M \to X(\mathbb{R})$ is a homeomorphism. Even better, in the smooth case, we can identify $M$ with the ringed space $(X(\mathbb{R}), \mathcal{O}_X)$. Even more better, the functor $M \mapsto \mathrm{Spec}(C(M, \mathbb{R}))$ from the category of smooth manifolds to the category of affine schemes is fully faithful.
Add to this the Serre-Swan theorem which states that there is an equivalence between the category of vector bundles on $M$ and the category of finite projective $R$-modules i.e. vector bundles on $X$.
These facts seem to imply that smooth manifolds may be thought of "as" affine schemes. This observation leads me to ask the following questions:
(1) Do you know of any fruitful consequences or applications of looking at manifolds in this light?
(2) Is there anywhere this identification fundamentally fails?
(3) Is there an algebraic classification for what these rings look like? In particular, if $A$ is an $\mathbb{R}$-algebra then when is $X(\mathbb{R})$ a topological manifold and when can $X(\mathbb{R})$ be given a smooth structure compatible with $\mathcal{O}_{X}$?
(4) What do the "extra" points of $X$ look like? Is there a use for these extra points in manifold theory, the way that generic points have become important in algebraic geometry?
For question (4), I believe that maximal ideals of $R$ should correspond to ultrafilters on $M$ identifying the closed points of $X$ with the Stone-Cech compactification of $M$. What about the other prime ideals?
Many thanks.
The main issue is that commutative algebra doesn't tell you very interesting things about the spectrum of the kinds of commutative algebras over $\mathbb{R}$ that arise as algebras of functions on manifolds. But, this is a nice question, which hopefully someone more qualified than me will answer.
https://arxiv.org/abs/1104.4951
(1): Serre-Swan makes thinking about vector bundles in the context of K-theory much more natural, at least for me. (I can't summarise that relation more effectively than most references you can search for can.)
The book Nestruev, Smooth Manifolds and Observables is precisely about this algebraic perspective on manifolds. There you also find a characterization of the commutative algebras that are isomorphic to rings of smooth functions on manifolds.
(1) This is a highly productive way of looking at smooth manifolds.
It is responsible for synthetic differential geometry and derived smooth manifolds.
Both of these subjects heavily rely on this identification.
Synthetic differential geometry looks at spectra of finite-dimensional
real algebras, and interprets these as geometric spaces of infinitesimal shape.
This allows one to make infinitesimal arguments of the type used by
Élie Cartan and Sophus Lie perfectly rigorous.
(Text)books have been written about this approach:
Anders Kock: Synthetic differential geometry;
Anders Kock: Synthetic geometry of manifolds;
René Lavendhomme: Basic concepts of synthetic differential geometry;
Ieke Moerdijk, Gonzalo Reyes: Models for smooth infinitesimal analysis.
Derived smooth manifolds start by enlarging the category of real algebras
of smooth functions to a bigger, cocomplete category
(such as all real algebras, or, better, C^∞-rings, see below).
One then takes simplicial objects in this category of algebras,
i.e., simplicial real algebras.
This category admits a good theory of homotopy colimits,
which are given by derived tensor products and other constructions
from homological algebra.
The opposite category of simplicial real algebras
should be thought of as a category of geometric spaces of some kind.
This category has an excellent theory of homotopy limits,
in particular, one compute correct (i.e., derived) intersections
of nontransversal submanifolds in it, and get expected answers.
For example, this can be used to define canonical representative
of characteristic classes, one could take the Euler class to be
the derived zero locus of the zero section, for example.
Other (potential) applications include the Fukaya ∞-category,
which in presence of nontransversal intersections can be turned
into a category in a more straightforward way.
Dominic Joyce wrote a book about this:
Dominic Joyce: Algebraic geometry over C^∞-rings.
Other sources:
David Spivak: Derived smooth manifolds.
Dennis Borisov, Justin Noel: Simplicial approach to derived differential manifolds.
David Carchedi, Dmitry Roytenberg: On theories of superalgebras of differentiable functions.
David Carchedi, Dmitry Roytenberg: Homological algebra for superalgebras of differentiable functions.
David Carchedi, Pelle Steffens: On the universal property of derived smooth manifolds.
(2) Yes, for example, Kähler differentials are not smooth differential 1-forms,
even though derivations are smooth vector fields.
This is (one of) the reasons for passing to C^∞-rings instead.
In the category of C^∞-rings, C^∞-Kähler differentials are precisely
smooth differential 1-forms.
(3) Not really, this question was discussed here before,
here is one of the discussions: Algebraic description of compact smooth manifolds?
Dmitri, while sometimes useful, and maybe even occasionally essential (especially to build moduli spaces), do you really mean highly productive? I mean, any student of algebraic geometry would be ignored today if they didn't think of schemes in this fashion, but your average student studying most aspects of smooth manifolds probably doesn't even know that the theory you describe exists, ditto for their advisors. I'm not trying to be negative, and I think this perspective is important. I hope I don't give the opposite impression.
@AndySanders: "Highly productive" is not the same thing as "well-known". My meaning is quite literal, that this approach produced a lot of good research. I did not say that the majority of those who work with smooth manifolds are familiar with it.
I guess we have different definitions of highly, but it's not worth arguing semantics. Regardless, thank you for this detailed and helpful post. I agree with your statement that it has produced good research.
Thank you @DmitriPavlov for the detailed answer. I clearly have a lot of reading here to do. I do have a question about your remark regarding the Kahler differentials.
if we define $\Omega_A$ in the standard way for A an $\mathbb{R}$-algebra then we get $Der(A, A) = Hom_A(\Omega_A, A)$ and for $A$ the ring of smooth functions on a manifold, $Der(A, A)$ is exact the tangent fields. So $Hom_A(\Omega_A, A)$ give the correct thing but $\Omega_A$ does not?
Hmm ... on $\mathbb{R}$ we cannot write $d(e^x)$ in terms of $d(x)$ even though any A-linear map $\phi : \Omega \to A$ is defined by $\phi(dx)$ since we can always write by Hadamard lemma $f = x \cdot g + c$ for $c$ constant then $d(f) = g \cdot dx + x \cdot d(g)$ and $h = \phi(d(f)) = g \phi(d(x)) + x \phi(d(g))$ so at $x = 0$ we have $h(0) = g(0) \phi(d(f))$ and we can play the same game around any point to determine $\phi(d(f))$ in terms of $d(x)$.
@BenC: The question about Kähler differentials for smooth functions is considered in great detail here: https://mathoverflow.net/questions/6074/kahler-differentials-and-ordinary-differentials.
|
2025-03-21T14:48:30.207672
| 2020-04-02T22:09:50 |
356416
|
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|
Stack Exchange
|
Algebras from a basis of a Frobenius algebra
Let $A$ be a commutative Frobenius algebra over a field $K$ (we can assume that $A$ is local).
We can assume $A=K[x_1,...,x_r]/I$ for an ideal $I$ with $J^n \subseteq I \subseteq J^2$ where $J=<x_i>$, the ideal of $K[x_1,...,x_r]$.
Let $B=\{v_i \}$ be a monomial vector space basis (meaning $v_i$ is a homogeneous polynomial in $x_i$) of $A$ containing the unit of $A$.
Let $M_i:=v_i A$ and $M:= \bigoplus_{}^{}{M_i}$ and $C:=\underline{End_A}(M)$ the stable endomorphism ring of $M$.
Question: Is $C$ independent of the choosen basis $B$ up to isomorphism?
You can choose a basis consisting of invertible elements and then get $C=0$. On the other hand you can choose another basis containing a non-invertible element and get $C\ne0$.
@VictorOstrik Thanks, I forgot to add that the basis should consist of monomials.
|
2025-03-21T14:48:30.207768
| 2020-04-02T23:33:28 |
356417
|
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|
Stack Exchange
|
Unique products of integers in fixed-size intervals
Let $d$ be fixed positive integer. For an integer $n$ consider a function $f_n: 2^{[d]} \to \mathbb{N}$ defined by $f_n(S) = \prod_{i \in S}(n + i)$. I'm interested in the smallest value $n_0(d)$ such that for any $n \geq n_0(d)$ the function $f_n$ is injective. In other words, $n_0(d)$ is the threshold on $n$ so that any subset of $[n + 1, n + d]$ should be reconstructible by the product of its elements.
Q: what are good bounds on $n_0(d)$? Both lower and upper bounds are interesting.
One easy upper bound is $n_0(d) = O(d!)$, since for $n > c \cdot d!$ we can look at digits of $f_n(S)$ in $n$-ary system to obtain the coefficients of the polynomial $\prod_{i \in S}(x + i)$. If we use balanced $(n + d / 2)$-ary system, we can lower the bound to $n_0(d) = O(((d / 2)!)^2)$. It appears $n_0(d) = O(d^c)$ or even $O(d)$ should be possible.
One lower bound would be $n_0(d) \geq d/2 + O(1)$, provided by a set containing $k, k + 1, 2k, 2k + 2$ for $k \sim d / 2$. Later edit: comments have examples that show $n_0(d) = \Omega(d^3)$, it's interesting if the method can be generalized.
Any product that has a prime factor q greater than d will single out that multiple of q in the interval. So non unique products will be d-friable numbers, and there aren't that many in an interval of length d. Indeed, I think you can replace d! by lcm(1..d) in your upper bound by adapting an argument of Langevin. This relates to Grimm maps and jumping primes. Gerhard "Check Out MathOverflow Question 243490" Paseman, 2020.04.02.
An example of $k^2; (k+1)^2, (k+2)^2$ and $k(k+1),(k+1)(k+2),k(k+2)$ shows that $n\geq d^2/16+O(1)$.
@IlyaBogdanov Nice! We can also have $(k^3 - 16k)(k^3 - 7k - 6)(k^3 - 7k + 6) = (k^3 - 4k)(k^3 - 13k - 12)(k^3 - 13k + 12) = (k - 4)(k - 3)(k - 2)(k - 1)k(k + 1)(k + 2)(k + 3)(k + 4)$ for $n_0(d) = \Omega(d^3)$ for large enough $d$. I wonder if this can be generalized to show superpolynomiality...
Yes, I already came up with the same example. in order to proceed further, it suffices to find an $k\times k$ weak magic square (all row and column sums are equal) sich that the $i$th powers of the entries also form a weak magic square, for $i$ up to $k(1-\alpha)-1$. This provides $\Omega(d^{1/\alpha})$. In your example, $\alpha=1/3$.
One can beef up the lower bound a little bit. As an example, take a prime quadruplet such as 101,103,107,109. One has 11021*11009=10807*11227. If d is a little over 400 (421 in this case), $n_0$ can't be much smaller than (d/4)^2. One can try tight constellations of (more than 4) primes to improve asymptotically the exponent of the lower bound.
As observed in a comment, a non unique product must have all numbers involved lacking prime factors greater than d. A nice argument of Langevin maps a unique prime divisor to each member of [n+1,n+d] by using the largest prime power map: this works to provide an objective map when n is at least lcm(1..d), and can work for smaller values of n. Likely these prime powers can be used as tags to determine the presence or absence of a member forming a product of numbers in this interval. If so, this would give a tighter upper bound than d factorial on $n_0$.
Gerhard "Prime Powers Powering A Proposal" Paseman, 2020.04.02.
Indeed, one can replace primes in the quadruplet with arbitrary integers in a tight constellation, but it helps in thinking about this to choose a coprime set. This convinces me that a polynomial upper bound in d is not feasible for $n_0$. Gerhard "Use Bigger And Bigger Counterexamples" Paseman, 2020.04.02.
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2025-03-21T14:48:30.208020
| 2020-04-03T00:22:45 |
356418
|
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|
Stack Exchange
|
Does Cantor Bernstein hold in a Closed Symmetric Monoidal Category?
In a closed symmetric monoidal category with $[I,X] \cong X$ for all $X$ is it true that having monomorphisms $m :A \rightarrow B$ and $m: B \rightarrow A$ is enough to imply $A \cong B$ ?
I tried to figure this out, and expected to end up using the Yoneda Embedding, problem is, my proof seems to hold for all categories, which is clearly wrong. Here is my attempt!
Given $m: A \rightarrow B$, using the covariant hom functor $Hom(X,-):C \rightarrow SET$ there is a morphism $Hom(X,m): Hom(X,A) \rightarrow Hom(X,B)$ Defined by $Hom(X,m)(h) \equiv m \circ h$.
Since $m$ is a monomorphism $Hom(X,m)(h_1) = Hom(X,m)(h_2) \iff m \circ h_1 = m \circ h_2 \iff h_1 = h_2$. Since there are both ways monomorphisms, there are both ways injections, so there is a bijection between $Hom(X,m)(h_1)$ and $Hom(X,m)(h_2)$. Any injection between the two must then be a bijection.
Now $M_i \equiv Hom(X_{i},m)$ form the components of a natural transformation $M: Hom(-,A) \rightarrow Hom(-,B)$, since for any $h \in Hom(X_1,A)$ and $f: X_2 \rightarrow X_1$ then $Hom(f,B)(Hom(X_1,m)(h)) = m \circ h \circ f = Hom(X_2,m)(Hom(f,A)(h))$.
So $M$ is a natural transformation with every component a bijection, a natural isomorphism. So then $Hom(-,A) \cong Hom(-,B)$ and by Yoneda $A \cong B$.
If you have two monomorphisms X -> Y and Y -> X in any category, then Hom(X, Z) = Hom(Y, Z). The issue is that this cannot necessarily be made natural in Z, so X need not be isomorphic to Y. I haven't read what you wrote, but I'm guessing some overlooking of naturality is where the problem shows up.
It's certainly very far from true. Just take for example the category of compactly generated spaces, and consider the spaces $[0, 1]$ and $(0, 1)$. For another simple example, take the category of abelian groups, and the coproduct of $\mathbb{Z}$ and countably many copies of $\mathbb{Q}$.
Your proof first shows, using Cantor-Bernstein in Set, that there is a bijection between hom(X,A) and hom(X,B). However, after that you claim that "any injection between the two must then be a bijection" - this is false if the hom-sets are infinite.
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2025-03-21T14:48:30.208198
| 2020-04-03T00:54:09 |
356419
|
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|
Stack Exchange
|
Rigid Brunnian links for $n \geq 4$
Brunnian links consist of $n$ linked un-knot components such that the cutting of any component leaves all components unconnected. The most famous example is the three-component Borromean rings (or links). In 1954 Milnor classified all Brunnian links up to link homotopy.
In $\mathbb{R}^3$, it is simple indeed to see that one can construct Borromean rings from identical rigid components. (The simplest case involves three identical rigid ovals---same intrinsic shape, same intrinsic size.) I have sketched six classes of rigid (un-knot) components that can be linked to form Borromean rings, and I'm sure there are an infinite number of such.
Question
Is it possible to create identical rigid components interlocked to form Brunnian links for $n \geq 4$? Is there an upper limit for $n$ for which this can be done?
To clarify, by rigid Brunnian link, do you require that after removing any component, the remaining unlink may be split into unknots by rigid motion of each component (say so that each component can be moved arbitrary far from the others)? This would be the natural interpretation of your terminology.
I hadn't really thought about that, but let's assume "yes."
It looks to me like you have answered your own question. If there is anything that remains, could you perhaps edit your question to highlight your remaining concerns?
@RyanBudney: I guess I have no other concerns so I'll accept my own answer. I do wonder if this result is publishable, as an example of what one might call "Rigid knot theory" or "Homology-free knot theory." It explicitly breaks one of the fundamental tenets of classical knot theory, but it seems to me to be a sub-discipline worthy of explicit naming. It would include knot theory for Medieval chain mail and some puzzles, for instance. I haven't seen this sub-discipline explicitly identified... have you?
These kinds of examples, and other examples like it come up in several peoples' work, but they came up in slightly different contexts. My paper on splicing mentions some of these properties -- that you can find links where all the components have the same shape comes up naturally in some instances of hyperbolic links -- basically you get the result you are looking for, for free, if the link is hyperbolic with a large symmetry group. You can also describe the space of symmetry positions. Kanenobu has a paper on Brunnian properties of links that also touches on this topic.
I think this link projection shows that one can create such rigid, identical components for arbitrary $n$.
It’s not clear to me that after removing one component, these can be rigidly unlinked. Do you have an idea of why this should be possible?
@IanAgol: I think that if the radius of the "loop" terminations are made extremely large, then the links can be rigidly removed (after one has been removed). I'll keep working on it. I've been looking for references on what could be called "rigid knot theory" or "homology-free knot theory," without success. Any pointers?
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2025-03-21T14:48:30.208425
| 2020-04-03T02:07:45 |
356424
|
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|
Stack Exchange
|
Expansion of polytabloids in the standard basis
would like to know the most efficient way to write a polytabloid in terms of standard ones.
I know the Garnir elements, but using them to do calculations is hard. I also read about "quadratic elements" in Fulton's Young Tableaux, but I hope a better construction exists.
The thing I would like to have is, in a perfect world, an algorithm that has a random polytabloid and a standard one as input, and the coefficient of the standard polytabloid in the expansion of the random polytabloid as an output. I don't think such algorithm exists, but it can help figure out what kind of algorithm I am looking for.
Are you asking for a straightening rule? A recent preprint claims to give an explicit one: Reuven Hodges, A closed non-iterative formula for straightening fillings of Young diagrams, arXiv:1710.05214v1.
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2025-03-21T14:48:30.208510
| 2020-04-03T03:01:06 |
356427
|
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|
Stack Exchange
|
The weak restriction of the Jacquet module
Let $P= MN$ be a parabolic subgroup of a reductive p-adic group $G$, and $(\pi, V)$ is an irreducible, admissible representation of $G$. The Jacquet module is the representation $(\pi_N, V_N)$, where $\pi_N$ is the restriction of $\pi$ on the $M$-invariant subspace $V_N= V/V(N)$, and $V(N)$ is the linear span of $v−\pi(n)v; ~ v\in V,n\in N.$
If $\pi$ is tempered, the weak (= tempered) Jacquet module is $(\pi^w_N, V^w_N)$, where $V^w_N$ is the maximal tempered quotient of $V_N$. Using the character if $\Theta$ is the character of $\pi$, then $\Theta^w_M$ is the character of $\pi^w_N$(following: Harish-Chandra it is the weak constant term of $\Theta$ along $P$)
Now acoording to J.Arthur, if $\sigma$ is a discrite series of $M$ there is a bijection between the set of irreducible constituents of the representation induced by $\sigma$ denoted $Ind^G_M(
\sigma)$ and the set of irreducible representations of the Knapp-Stein R-group $R_\sigma$. If in addition $L$ is a Levi subgroup of $G$, with split component $A_L$ and $\mathcal{a}_L$ is the Lie algebras of $A_L$, containing $M$ and satisfies the Compatibility Condition
of Arthur, "CCA" ( that: $\mathcal{a}_L \cap\overline{\mathcal{a}^{+}_\sigma}$ contains an open subset of $\mathcal{a}_L$; where $\mathcal{a}^{+}_\sigma$
denotes
the positive chamber corresponding to the reduced roots for which the Plancherel measure of $\sigma$ vanishes), then $R^L_\sigma$ is the Knapp-Stein R-group of $Ind^L_M(\sigma)$ and so as at $G$ we obtain another bijection...
My question is: If $L$ is as above but it does not satisfies "CCA", Is $\Theta^w_L = 0$ ??
Thanks in advance
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2025-03-21T14:48:30.208768
| 2020-04-03T03:55:17 |
356429
|
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|
Stack Exchange
|
A smooth map on a Banach manifold whose pointwise rank is finite but its rank is not globally bounded
Is there a connected Banach manifold $M$ and a smooth map $f:M \to M$ such that the rank of $Df_x$ is finite for every $x\in M$ but this rank is not uniformly bounded
I presume you want to include the assumption that $M$ be connected, or else it should be possible to give examples where the rank depends on which connected component you are in.
@GabeK Yes of course thank you for your comment. I add the word "connected"
Let $\mathbb H$ and $(e_n)_{n\geq0}$ be your favourite second countable Hilbert space and orthonormal Hilbert basis. Mine is the space $\ell^2(\mathbb N)$ of square integrable sequences, with $e_n$ the indicator function of the singleton $\lbrace n\rbrace$. I write, for any $x\in\mathbb H$, $x^n$ for the $n$th coordinate $\langle e_n,x\rangle$ of $x$.
Choose a smooth function $\phi:\mathbb R\to\mathbb R$ such that $\phi$ is identically zero over $(-\infty,1/2)$ and $\phi'(1)\neq0$. Then
$$f:x\mapsto\sum_{n\geq0} \phi(x^n)e_n$$
is a function as you describe. Indeed,
if $|x^n|<1/3$ for all $n\geq N$, then in a neighbourhood of $x$ the condition still holds with $1/2$ instead of $1/3$, and the function $f$ has values in the finite dimensional space generated by $(e_0,\cdots,e_{N-1})$ hence its differential has finite rank;
since $x$ can have $|x^n|\geq1/3$ for only a finite number of $n$, $f$ is well-defined and smooth;
the differential of $f$ at $x=e_0+\cdots+e_N$ is
$$Df_x:u\mapsto\sum_{0\leq n\leq N} \phi'(1)u^ne_n,$$
which has rank $n$.
Just for fun, here is a stronger topological result. Say that $f$ has finite rank if $Df_x$ has finite rank for all $x$, and bounded rank if the rank of $Df_x$ is bounded uniformly in $x$. We say that $x$ is a nice point for $f$ if there exists a neighbourhood of $x$ over which $f$ has bounded rank, otherwise we say it is bad.
Theorem
Let $E$ be a subset of $\mathbb H$. Then there exists $f:\mathbb H\to\mathbb H$ of finite rank whose set of nice points is exactly $E$ if and only if $E$ is open dense.
This makes it possible to have many bad points, for instance there exists an $f$ with finite rank such that for all $\varepsilon>0$, there exists $x$ in the unit ball such that $B(x,\varepsilon)$ contains infinitely many bad points.
The direct implication is a consequence of the Baire category theorem. Indeed, the set $E_n$ of points $x$ such that $Df_x$ has rank at most $n$ is closed, and the set $E=\bigcup_{n\geq0}\operatorname{int}E_n$ of points where the rank is locally bounded is clearly open. To show that $E$ is dense, fix a non-empty open set $U$. Since $f$ has finite rank, $\bigcup_{n\geq0}(E_n\cap U)=U$, hence one of the $E_n\cap U$ must have non-empty interior, which is the same as saying that $\operatorname{int}E_n\cap U$ is non-empty.
In the other direction, this answer shows that in a Hilbert space, any closed set is the vanishing set of some smooth real-valued map with all derivatives uniformly bounded ($\forall k,\exists M,\forall x,\|D^k\rho_x\|\leq M$). Let $d:\mathbb H\to\mathbb R$ be the distance to $E^\complement$; note that it is continuous. Let $E_0$, resp. $E_n$ for $n>0$, be the inverse image by $d$ of $(1,+\infty)$, resp. $(\frac1{n+1},\frac2n)$. Obviously, $E_n$ is open for all $n$ and $E$ is the union of the $E_n$. Moreover, since $E$ is open dense, $x\in E$ if and only if there exists a neighbourhood of $x$ that intersects finitely many of the $\overline{E_n}$.
Let $\rho_n:\mathbb H\to\mathbb R$ be a smooth function with zero set $E_n$ and all derivatives uniformly bounded. Then
$$ x\mapsto \rho_n(x)\sum_{k\leq i\leq \ell}x^ie_i $$
has rank zero in the complement of $\overline{E_n}$, and at least $k-\ell$ on $E_n$ (because the derivative of $\rho_n$ may only “kill” at most one direction in the image of the projector). Moreover, all its derivatives are uniformly bounded. It means that there exists a sequence $(\varepsilon_n)_{n\geq0}$ of positive numbers such that all the derivatives of the series
$$ f:x\mapsto\sum_{n\geq0}\varepsilon_n\rho_n(x)\left(x^{n^2+1}e_{n^2+1}+\cdots+x^{(n+1)^2}e_{(n+1)^2}\right) $$
converge uniformly over $\mathbb H$, so that $f$ is well-defined and smooth (see the reference above for more details).
The rank of $Df_x$ is at least $2n$ but finite over $E_n$, and the rank of $Df_x$ is zero (hence finite) at all points $x$ of $E^\complement$. Moreover, $x$ is nice for $f$ if and only if one of its neighbourhoods intersects only finitely many $\overline{E_n}$, if and only if $x\in E$.
|
2025-03-21T14:48:30.209103
| 2020-04-03T04:43:40 |
356431
|
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|
Stack Exchange
|
Are morphisms from affine schemes to Artin stacks affine morphisms?
It is explained in this MO discussion that if one has a morphism $f:X\rightarrow Y$ of schemes such that $X$ is an affine scheme, then $f$ need not be an affine morphism. However, if $Y$ is separated, then $f$ is indeed affine.
Question: what happens is if $Y$ is an Artin stack? (But $X$ is still an affine scheme). What are some natural conditions on $Y$ to ensure that $f$ is an affine map?
The example I have in mind is when $Y$ is the stack of rank $n$ vector bundles.
The obvious natural condition is that the diagonal $\Delta_Y:Y\to Y\times Y$ is affine. This holds for stacks of vector bundles: it just means that if $E$ and $F$ are rank $n$ vector bundles on $X$, then $\underline{\mathrm{Isom}(}E,F)$ is (representable and) affine over $X$; it is in fact a $\mathrm{GL(E)}$-torsor.
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2025-03-21T14:48:30.209207
| 2020-04-03T05:36:40 |
356435
|
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"url": "https://mathoverflow.net/questions/356435"
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|
Stack Exchange
|
Projective/injective object in functor category
Let $\mathcal{C}$ denote the functor category $Fun(\textbf{Man} , \textbf{Ab})$, where $\textbf{Man}$ and $\textbf{Ab}$ denote the category of smooth manifolds and abelian groups respectively. I want to know what are the projective objects or injective objects in this category $\mathcal{C}$? More precisely, I am interested in resolving objects of $\mathcal{C}$ by either projective or injective objects. So for that I need to know what are these objects? Any kind of partial help or comments would be great.
The approach which I thought of is the following: If we can find a category $\mathcal{D}$ with two functors from $ F : \mathcal{C} \rightarrow \mathcal{D}$ and from $ G : \mathcal{D} \rightarrow \mathcal{C}$ such that both are in adjunction. Then under some conditions using the unit or co-unit map, we can resolve objects.
For any essentially small category $\mathcal{T}$, a generating class of projective objects of the functor category $Fun(\mathcal{T}, \mathbf{Ab})$ is given by the linearisation of reprensentable functors $\mathcal{T}(t,-)$ (thanks to the Yoneda lemma). So you get natural projective resolutions of functors by the multi-object bar construction. You can dualise it to get injective resolutions. I am afraid that it is not easy to say something more precise and useful in practice in your specific setting.
You might find something useful in the following paper (data from MathSciNet): MR0204497 (34 #4336) 18.20
Watts, Charles E.
A homology theory for small categories. 1966
Proc. Conf. Categorical Algebra (La Jolla, Calif., 1965) pp. 331–335 Springer, New York
One way to construct injectives on a presheaf category $[\mathscr C^{\operatorname{op}},\mathbf{Ab}]$ is to consider the forgetful functor
$$i^* \colon \big[\mathscr C^{\operatorname{op}},\mathbf{Ab}\big] \to \big[\mathscr C^{\operatorname{disc,op}},\mathbf{Ab}\big]$$
induced by the inclusion $i \colon \mathscr C^{\operatorname{disc}} \to \mathscr C$ (where $\mathscr C^{\operatorname{disc}}$ is the subcategory with only identity morphisms). If $\mathscr C$ is small, then $i^*$ has left and right adjoints $i_!$ and $i_*$ given by
\begin{align*}
\big(i_! \mathscr F\big)(c) = \bigoplus_{c' \to c} \mathscr F(c'),\\
\big(i_* \mathscr F\big)(c) = \prod_{c \to c'} \mathscr F(c').
\end{align*}
In particular, $i^*$ is an exact left adjoint to $i_*$, so $i_*$ takes injectives to injectives [Stacks, Tag 015N]. But in $[\mathscr C^{\operatorname{disc,op}},\mathbf{Ab}]$ injectives are computed pointwise, so this gives a recipe to construct injectives in $[\mathscr C^{\operatorname{op}},\mathbf{Ab}]$.
See for example [Stacks, Tag 01DJ] for a brief discussion, or [SGA IV$_1$, Exp. I, Prop. 5.1] for a more general discussion of adjoints (but without the mention of injectives).
In general the colimit for $i_!$ is taken over the opposite of the comma category $(i \downarrow c)$ (whose objects are $(i(c') \to c)$), which in this case is just a discrete category since $\mathscr C^{\operatorname{disc}}$ is, so we get a direct sum; similarly for $i_*$.
References.
[SGA IV$_1$] M. Artin, A. Grothendieck, J.-L. Verdier, Séminaire de géométrie algébrique du Bois-Marie 1963–1964. Théorie de topos et cohomologie étale des schémas (SGA 4), 1: Théorie des topos. Lecture Notes in Mathematics 269. Springer-Verlag (1972). ZBL0234.00007.
[Stacks] A.J de Jong et al, The stacks project.
For projective objects, see here: Necessary conditions for cofibrancy in global projective model structure on simplicial presheaves.
As explained there for presheaves of sets (and the same argument works
for abelian groups), projective presheaves are precisely coproducts of retracts of representables.
As for injective presheaves, the general consensus is that there is no
general criterion to characterize them other than by their lifting properties.
This question has been discussed before, see, for example, here:
What are the fibrant objects in the injective model structure?
As for the last paragraph, this approach is known as the bar construction.
It's seems very helpful at least projective case. Thanks.
|
2025-03-21T14:48:30.209519
| 2020-04-03T07:37:59 |
356439
|
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Stack Exchange
|
Subsupermanifolds defined using ideal, transversal example
I am currently learning about algebraic viewpoint on closed embedded subsupermanifolds. In particular, I am struggling with something which should be ''easy to see''. Namely, I refer to the lemma just above Proposition 3.2.9 in "Introduction to the Theory of Supermanifolds" by D. A. Leites (http://iopscience.iop.org/0036-0279/35/1/R01).
Remark: I know there is an answer below. It, however, does not work for supermanifolds (I am not able to generalize).
The setting: Suppose we have a map $\psi: L \rightarrow M$ of supermanifolds, such that it is transversal to a given closed embedded subsupermanifold $N \subseteq M$.
Now, to every closed embedded subsupermanifold, one may assign a unique ideal $\mathcal{J}_{N} \leq C^{\infty}(M)$ in the superalgebra of its global functions, defined by $\mathcal{J}_{N} = \{ f \in C^{\infty}(M) \; | \; j^{*}(f) = 0 \}$, where $j: N \rightarrow M$ is the embedding. It can be then shown that $j^{\ast}: C^{\infty}(M) \rightarrow C^{\infty}(N)$ is a superalgebra epimorphism and $\mathcal{J}_{N}$ is its kernel, whence $C^{\infty}(N) \cong C^{\infty}(M) / \mathcal{J}_{N}$
Next, by definition of the map of supermanifolds, we have a superalgebra morphism $\psi^{\ast}: C^{\infty}(M) \rightarrow C^{\infty}(L)$. We can thus consider a subset $\psi^{\ast}(\mathcal{J}_{N}) \subseteq C^{\infty}(L)$. As $\psi^{\ast}$ is usually not surjective, this is in general not an ideal. We can, however, consider the ideal $\mathcal{I} = \langle \psi^{\ast}(\mathcal{J}_{N}) \rangle \leq C^{\infty}(L)$ generated by this subset.
THE ACTUAL QUESTION: $\mathcal{I}$ is supposed to have the following property: Let $\{ f_{\mu} \}_{\mu \in J}$ be any collection of functions in $\mathcal{I}$, such that $\{ supp(f_{\mu}) \}_{\mu \in J}$ is locally finite. Then also their sum $\sum_{\mu \in J} f_{\mu}$ must be function in $\mathcal{I}$.
I am stuck at this very point. According to D.A. Leites, this should be easy to see.
For the sake of completeness, let me recall some definitions:
A collection $\{C_{\mu} \}_{\mu \in J}$ of subsets of any topological space is locally finite, if for every compact subset $K$, $C_{\mu} \cap K \neq \emptyset$ only for finitely many $\mu \in J$. On (ordinary) manifolds, this is equivalent to every point $m$ having a neighborhood $U_{m}$, such that $C_{\mu} \cap U_{m} \neq \emptyset$ only for finitely many $\mu \in J$.
For a function $f$ on a supermanifold $M$, its support $supp(f)$ is a set of points $m$ of the underlying manifold $|M|$, where the germ $[f]_{m}$ of $f$ does not vanish.
Some comments: (i) The ideal $\mathcal{J}_{N}$ has this very property. For each $U \subseteq M$ open, let $j^{\ast}_{U}: C^{\infty}(U) \rightarrow C^{\infty}(U \cap N)$ be the superalgebra morphism induced by the pullback by $j$. For each point $m \in M$, we can pick a precompact neighborhood $U_{m}$. If $f_{\mu} \in \mathcal{J}_{N}$ for every $\mu \in J$, we obtain
$(j^{\ast}( \sum_{\mu \in J} f_{\mu} ))|_{U_{m} \cap N} = j^{\ast}_{U_{m}}( (\sum_{\mu \in J} f_{\mu} )|_{U_{m}}) = \sum_{\mu \in J} j^{\ast}_{U_{m}}( f_{\mu}|_{U_{m}}) = \sum_{\mu \in J} (j^{\ast}(f_{\mu}))|_{U_{m} \cap N} = 0,$
where it was important that after restriction to $U_{m}$, the sum is finite. But $m$ was arbitrary and $\{ U_{m} \cap N \}_{m \in M}$ forms an open cover of $N$, which proves that $j^{\ast}( \sum_{\mu \in J} f_{\mu} ) = 0$, that is $\sum_{\mu \in J} f_{\mu} \in \mathcal{J}_{N}$.
(ii) I don't know whether the transversality of $\psi$ to $N$ is somewhat important at this point.
(iii) D. A. Leites only assumes that the indexing set $J$ is countable. This is not very important though, as for general $J$, one can always find a countable subset $J'$, such that $\sum_{\mu \in J} f_{\mu} = \sum_{\mu' \in J'} f_{\mu'}$.
I think Proposition 3.2.6 of the same paper discusses this.
Update: I added a previous version (Version 2) which does not use dimension theory and tubular neighborhoods and it might be possible to generalize it to supermanifolds. However, it requires that $N$ can be covered by finitely many coordinate balls.
VERSION 1: FOR SMOOTH MANIFOLDS AND ANY N AND $\psi$ (uses that $N$ can be covered by $\mathrm{dim}(N)+1$ charts and that there is a tubular neighborhood of $N$)
It holds
$$
\mathcal{I} = \Bigl\{ \sum_{i=1}^k (g_i \circ \psi) h_i \ \Bigl|\ k\in\mathbb{N}, g_i\in C^\infty(M), h_i\in C^\infty(L): g_i(N) = 0 \Bigr\}.
$$
Given a locally finite collection of smooth functions $(f_i\in \mathcal{I} \mid i\in I)$, we will show that
$$ \sum_{i\in I} f_i \in \mathcal{I}. $$
Lemma 1: We can assume that $f_i = (g_i \circ \psi) h_i$ for all $i\in\mathcal{I}$.
Proof: Pick a partition of unity $(\eta_j \mid j\in \mathcal{J})$ on $L$ such that $\eta_j$ has compact support for every $j\in\mathcal{J}$.
Let $\mathcal{A}:= \mathcal{I}\times\mathcal{J}$, and define
$$ f_\alpha:= \eta_j f_i $$
for all $\alpha=(i,j)\in \mathcal{A}$.
The system $(f_{\alpha}\mid \alpha\in\mathcal{A})$ is locally finite, and it holds
$$
\sum_{\alpha\in\mathcal{A}} f_{\alpha} = \sum_{j\in \mathcal{J}} \eta_j \sum_{i\in\mathcal{I}} f_i = \sum_{i\in\mathcal{I}} f_i.
$$
For every $i\in\mathcal{I}$, there is an $m_i\in \mathbb{N}$ such that $f_i = \sum_{l=1}^{m_i} (g_{il}\circ\psi)h_{il}$.
It follows that for every $\alpha = (i,j)\in\mathcal{A}$, it holds
$$
f_\alpha = \eta_j f_i = \sum_{l=1}^{m_i} (g_{il}\circ\psi)(\eta_j h_{il}) = \sum_{i=1}^{m_i} (g_{il}\circ\psi)h_{\alpha l} \in \mathcal{I},
$$
where we defined
$$ h_{\alpha l} := \begin{cases} 0 & \text{if }f_\alpha=0, \\
\eta_j h_{il} & \text{otherwise.}
\end{cases} $$
Let $x\in L$.
There is an open neighborhood $U$ of $x$ and a finite subset $\mathcal{J}_0\subset \mathcal{J}$ such that $\mathrm{supp}(\eta_j) \cap U = 0$ for all $j\in \mathcal{J}\backslash\mathcal{J}_0$.
Because $\sum_{j\in \mathcal{J}}\eta_j = 1$, it holds $U\subset \bigcup_{j\in \mathcal{J}_0} \{ x\in M \mid \eta_j(x)\neq 0\}$.
Because $\bigcup_{j\in \mathcal{J}_0} \mathrm{supp}(\eta_j)$ is compact, there exists a finite subset $\mathcal{I}_0\subset \mathcal{I}$ such that
$$
\mathrm{supp}(f_i)\cap \bigcup_{j\in \mathcal{J}_0} \mathrm{supp}(\eta_j) = \emptyset
$$ for all $i\in \mathcal{I}\backslash\mathcal{I}_0$.
Suppose that $\mathrm{supp}(h_{\alpha l})\cap U \neq 0$ for some $\alpha = (i,j)$ and $l\in \{1,\dotsc,m_i\}$.
Because $\mathrm{supp}(h_{\alpha l})\subset\mathrm{supp}(\eta_j)$, it must hold $j\in \mathcal{J}_0$.
By definition of $h_{\alpha l}$, it holds either $h_{\alpha l} = 0$ if $f_\alpha = 0$, which is equivalent to $\{x\in M \mid f_i(x)\neq 0\}\cap\{x\in M\mid \eta_j(x)\neq 0\}=\emptyset$ which is equivalent to $\mathrm{supp}(f_i)\cap\{x\in M\mid \eta_j(x)\neq 0\} = \emptyset$, or $h_{\alpha_l} = \eta_j h_{il}$.
The second option can possibly occur only for $i\in\mathcal{I}_0$.
This shows that the collection
$$ ((g_{il}\circ\psi)h_{\alpha l} \mid \alpha=(i,j)\in\mathcal{A}, l\in\{1,\dotsc,m_i\}) $$
is locally finite.
Its sum equals $\sum_{\alpha\in\mathcal{A}} f_\alpha$ and hence $\sum_{i\in \mathcal{I}}f_i$ by construction. QED
By Lemma 1, we can assume that $f_i = (g_i\circ\psi)h_i$ for $g_i\in C^\infty(M)$ with $g_i(N)=0$ and $h_i\in C^\infty(L)$ without lost of generality.
Denote $k:=\dim(N)$ and $n:=\dim(M)$.
Pick a tubular neighborhood $\mathcal{N}(N)$ of $N$ in $M$.
The $k$-dimensional manifold $N$ can be always covered by $k+1$ (not necessarily connected) charts $U_1$, $\dotsc$, $U_{k+1}$.
Every chart $U_j$ on $N$ induces a submanifold chart $V_j = \mathcal{N}(U_j)$ on $M$. Let $V_0\subset M$ be an open subset disjoint from $N$ such that $M = \cup_{j=0}^{k+1} V_j$.
Let $\lambda_0$, $\dotsc$, $\lambda_{k+1}$ be a subordinate partition of unity. Let $\mu$ be a bump function which equals $1$ on $\mathrm{supp}(\lambda_0)$ and vanishes on $N$.
Let $(x_j,y_j)\in \mathbb{R}^n$ be coordinates on $\mathcal{N}(U_j)$ such that $x_j = (x_j^1,\dotsc,x_j^k)$ gives coordinates on the base and $y_j = (y_j^1,\dotsc,y_j^{n-k})$ on fibers.
An important feature of $\mathcal{N}(U_j)$ is that it contains the vertical line $\gamma(t) = (x_j,0) + t((x_j,y_j)-(x_j,0))$ connecting $(x_j,0)$ and $(x_j,y_j)$.
The Fundamental theorem of calculus in the form
$$ f(\gamma(1))-f(\gamma(0)) = \int_{0}^1 (\nabla f)(\gamma(t))\cdot\gamma'(t) \mathrm{d}t $$
then asserts that the following holds for all $i\in I$ and $j\in \{1,\dotsc,k+1\}$ on the entire $\mathcal{N}(U_j)$:
$$ (\lambda_j g_i)(x_j,y_j) - \underbrace{(\lambda_j g_i)(x_j,0)}_{=0} = \sum_{a=1}^{n-k} y^a_j \underbrace{\int_{0}^1 \frac{\partial(\lambda_j g_i)}{\partial y^a_j}(x_j,ty_j) \mathrm{d}t}_{\displaystyle=:u_{i a}^j}. $$
Let $\tilde{y}^a_j$ and $\tilde{u}_{ia}^j$ be the smooth functions on $M$ obtained from $y^a_j$ and $u_{ia}^j$, respectively, by multiplication with a bump function which is $1$ on $\mathrm{supp} \lambda_j$ and $0$ on a neighborhood of the closure of the complement of $\mathcal{N}(U_j)$.
For all $i\in I$ and $j\in \{1,\dotsc,k+1\}$, we have the following relations on $M$:
$$ \lambda_0 g_i = \mu \lambda_0 g_i\quad\text{and}\quad\lambda_j g_i = \sum_{a=1}^{n-k} \tilde{y}^a_j \tilde{u}_{ia}^j. $$
Using this, we compute
\begin{align*}
\sum_{i\in I} (g_i \circ \psi) h_i
&= \sum_{i\in I} \sum_{j=0}^{k+1} (\lambda_j g_i \circ \psi) h_i \\
& = \sum_{i\in I} (\lambda_0 g_i \circ \psi) h_i + \sum_{i\in I} \sum_{j=1}^{k+1} \sum_{a=1}^{n-k} (\tilde{y}^a_j \circ \psi)(\tilde{u}_{ia}^j \circ \psi)h_i \\
& = (\mu\circ\psi)\sum_{i\in I}(\lambda_0 g_i \circ \psi) h_i+ \sum_{j=1}^{k+1} \sum_{a=1}^{n-k} (\tilde{y}^a_j\circ \psi) \sum_{i\in I} (\tilde{u}_{ia}^j \circ \psi)h_i\\
& = (G_0 \circ \psi) H_0 + \sum_{j=1}^{k+1} \sum_{a=1}^{n-k} (G_{ja}\circ\psi)H_{ja},
\end{align*}
where we denoted
$$ G_0:= \mu,\quad G_{ja}:=\tilde{y}^a_j,\quad H_0:=\sum_{i\in I}(\lambda_0 g_i \circ \psi) h_i,\quad H_{ja}:=\sum_{i\in I} (\tilde{u}_{ia}^j\circ \psi)h_i. $$
It holds $G_0$, $G_{ja}\in C^\infty(M)$, $G_0(N)=G_{ja}(N) = 0$, $H_0$, $H_{ja}\in C^\infty(L)$, and it follows that $\sum_{i\in I} (g_i \circ \psi) h_i \in \mathcal{I}$.
VERSION 2: PROOF WHEN $N$ CAN BE COVERED BY FINITELY MANY COMPATIBLE COORDINATE BALLS (not using dimension theory and tubular neighborhood)
Write $\mathbb{R}^n = \mathbb{R}^k\times \mathbb{R}^{n-k}$ with coordinates $(x,y)$. Let $f: \mathbb{R}^n\rightarrow \mathbb{R}$ be a smooth function vanishing at $\{(x,y) \mid x = 0\}$. Then the fundamental theorem of calculus asserts that the following holds for all $(x,y)\in \mathbb{R}^n$: $$ f(x,y) = \sum_{j=1}^{k} x^j \int_{0}^1 \frac{\partial f}{\partial x^j}(tx,y) dt. $$ Let $U_\alpha$ $(\alpha\in\mathcal{A})$ be a cover of $N$ by coordinate balls, and let $\lambda_\alpha$ $(\alpha\in\mathcal{A})$ be a subordinate partition of unity. Suppose that we are given $\sum_{i\in I} (g_i \circ \psi) h_i$ as above and we want to show that it lies in $\mathcal{I}$. We can even assume that the support lies in an arbitrary small neighborhood of $\psi^{-1}(N)$. Using the analytical fact above, there are smooth functions $x_{\alpha}^j$ vanishing on $N$ and smooth functions $u^{\alpha}_{ij}$ for all $i\in I$, $\alpha\in\mathcal{A}$ and $j\in\{1,\dotsc,k:=\mathrm{codim} N\}$ such that $$ \lambda_\alpha g_i = \sum_{j=1}^k x_\alpha^j u_{ij}^\alpha. $$ We compute \begin{align*} \sum_{i\in I} (g_i \circ \psi) h_i = \sum_{j=1}^k \sum_{\alpha\in\mathcal{A}} (\lambda_\alpha x^j_\alpha\circ\psi) \sum_{i\in I} (u_{ij}^\alpha \circ \psi) h_i. \end{align*} If $\mathcal{A}$ is finite, then we are done.
Your answer related to the original formulation of this question (I have edited it significantly). In a summary - you give an answer for the case where all the manifolds are ordinary manifolds. However, in supermanifolds, one does not have a full apparautus of tubular neighborhoods, or the fact that k-dimensional manifold can be covered by k+1 charts.
Version 2 can be significantly simplified. By the modification of Lemma 1 above, you can assume that $f_{i} = (g_{i} \circ \psi) h_{i}$, where ${ supp(g_{i}) }_{i \in I}$ is locally finite.
It then suffices to cover $N$ by finitely many precompact sets ${U_{\alpha} }{\alpha=1}^{m}$ (e.g. coordinate balls). Let $U{0} := M - N$ and let ${ \lambda_{\alpha} }{\alpha = 0}^{m}$ be the corresponding partition of unity. Note that $\lambda{0} \in \mathcal{J}{N}$. Now, as $U{\alpha}$ are compact for $\alpha > 0$, there is a finite subset $I_{\alpha} \subseteq I$ such that
such that $\lambda_{\alpha} \cdot g_{i} \neq 0$ only for $i \in I_{\alpha}$. Also note that $\lambda_{\alpha} \cdot g_{i} \in \mathcal{J}_{N}$. Then
$\sum_{i \in I} (g_{i} \circ \psi) h_{i} = \sum_{i \in I} ((\sum_{\alpha=0}^{m}\lambda_{\alpha}) \cdot g_{i} \circ \psi) h_{i} = (\sum_{i \in I} (g_{i} \circ \psi) h_{i}) \cdot (\lambda_{0} \circ \psi)$
$+ \sum_{\alpha=1}^{m} \sum_{i \in I_{\alpha}} (\lambda_{\alpha} \cdot g_{i} \circ \psi) h_{i}$.
The first summand is in $\mathcal{I}$ as $\lambda_{0} \in \mathcal{J}_{N}$ and the rest is a finite sum of elements in $\mathcal{I}$.
|
2025-03-21T14:48:30.210251
| 2020-04-03T08:07:30 |
356441
|
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|
Stack Exchange
|
Infinitesimal categories and left duality
I have been reading Kassel's Quantum groups and there is something I can not understand.
In Section 4 of chapter $XX$, he introduces the notion of a Infinitesimal symmetric category, that is
a strict tensor category $(\mathcal{S}, \bigotimes, I)$ in which all the Hom sets are vector spaces together with, for every object $V,W$ of $\mathcal{S}$, a symmetry $\sigma_{V,W}: V \otimes W \rightarrow W \otimes V $ and an infinitesimal braiding $t_{V,W} :V \otimes W \rightarrow V \otimes W$ having the following properties:
$$\sigma_{V,W} \circ t_{V,W} = t_{W,V} \circ \sigma_{V,W}$$
and
$$ t_{U,V\otimes W} = t_{U,V}\otimes \mathrm{id}_W + (\sigma_{U,V}\otimes \mathrm{id}_W)^{-1} \circ(\mathrm{id}_v \otimes t_{U,W}) \circ (\sigma_{U,V}\otimes \mathrm{id}_W).$$
He then claims that if the infinitesimal symmetric category category $\mathcal{S}$ also has a left duality $V \mapsto V^*$ with structure maps $b_V^0 : I \rightarrow V \otimes V^*$ and $d_V^0 : V^* \otimes V \rightarrow I$, then we have have
$$t_{V,W} = \frac12 (C_{V\otimes W}- C_V \otimes \mathrm{id}_W- \mathrm{id}_V \otimes C_W)$$
where the $(C_V : V \rightarrow V)_V$ is a natural family of endomorphisms of $\mathcal{S}$ given by :
$$ C_v = - \big(\mathrm{id}_V \otimes(d_v^0 \circ t_{V^*.V})\big) \circ (b_V^0 \otimes \mathrm{id}_V).$$
I simply do not see at all why this is true, so I would glad welcome so kind of explanation. I have tried something using the fact that $(\mathrm{id}_V \otimes d_V^0) \circ (b_V^0 \otimes \mathrm{id}_V) = \mathrm{id}_V$ but this has not been very successful.
Did you try drawing it ? This is fairly easy to prove using graphical calculus. The following example might help your intuition: let $g$ be a Lie algebra and $t\in S^2(g)^g$ where the superscript means $g$ invariant. Then $t$ induces an infinitesimal braided structure on $g$-mod. Now write $t=\sum e_i \otimes e^i$ where $e_i$ is a basis and $e^i$ the dual basis w.r.t the pairing induced by $t$. Then C is nothing but the associated Casimir element, i.e. $C=\frac12 \sum e_ie^i$ regarded as an element in $U(g)$ (ie you replace the tensor product by the product in $U(g)$).
Thank you for answer. I totally understand it when the category $\mathcal{S}$ is the module category for some Hopf algebra. My question is maybe to be understand in the most general setting possible, when we do not know much about $\mathcal{S}$.
Apologies for my terrible handwriting, but here is an image with a graphical proof. The first equality is one of the defining property of $t$, represented as a chord diagram, ie a dotted line between two strand (the other defining property is the obvious other version of that). For me it matter which side of a strand the dotted line is attached to, and switching from one to the other at a given vertex change the sign of the expression. Then there is the definition of $c$ (which I realize is a bit different but equivalent to yours, the end dotted line can be moved to the rightmost strand at the cost of a sign), and a computation of $c_{U\otimes V}$ which should give what you want.
|
2025-03-21T14:48:30.210481
| 2020-04-03T08:34:35 |
356444
|
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|
Stack Exchange
|
Conceptual definition of derived functors allowing for quick proof of comparison theorems for sheaf cohomology
There are several approaches of increasing sophistication and simplicity to defining derived functors. I know of universal $\delta$-functors and Kan extensions along localizations. More definitions are discussed in this paper by Hinich.
I am looking for a definition of derived functors that answers all of the following criteria.
It's reasonably conceptual (i.e defined by universal properties) and does not require too many constructions (so nothing like injective resolutions).
It admits a reasonably quick path to proving the classical comparison theorems for sheaf cohomology.
No mention of triangulated categories or their structure.
I haven't been able to find a reference that gives a good definition of derived functors and also uses it to prove the comparison theorem, so if you know one, a reference would be great.
I asked about the relation between the $\delta$-functor approach and Kan extension along localizations here, but ideally, I'd like to avoid $\delta$-functors entirely. Is there such a path?
Does your desire to avoid triangulated categories extend to stable ∞-categories? If so I cannot see how you could avoid $\delta$-functors.
I don't know what stable $\infty$-categories are but I have nothing against them. My fear is that they don't admit a "reasonably quick path" to concrete comparison theorems, but I'd be happy to discover otherwise. I should emphasize that I want to be able to write down all the details.
See my answer here for a quick rundown of stable ∞-categories: https://mathoverflow.net/questions/344219/replacing-triangulated-categories-with-something-better/344221#344221 . Admittedly you need to develop some theory before working with them. And what do you mean by "write down all details"? Starting from ZFC set theory (I assume not)? What kind of things are you willing to take for granted?
Thanks for the link. I'm happy to take basic topology and category theory for granted, but I don't want to assume something like the Quillen equivalence between simplicial sets and topological spaces. Sorry I can't be more specific - I just have no idea how the $\infty$-category approach looks like.
@DenisNardin: If the OP already consider injective resolutions to have "too many constructions", then suggesting stable ∞-categories is a bit weird. Recall that Lurie's definition of the derived ∞-category (Definition <IP_ADDRESS> in Higher Algebra) uses projective resolutions, for example. So you suggested approach definitely violates OP's requirement of not using resolutions.
@DmitriPavlov Yeah, I decided that it was not a suitable approach, or I'd have written an answer (although there's no need of injective resolutions to define the derived ∞-category of an abelian category, and it's arguably not even the most natural approach). That said, I don't think you can go much lower tech than $\delta$-functors...
Theoretically, some time in the future, homotopy type theory may be able to provide enough tools to state and prove comparison theorems without referring to injective resolutions, triangulated categories, or similiar tools. But right now, in the current state of the field, there is hardly anything simpler than injective resolutions, model categories, triangulated categories, stable ∞-categories, etc., all of which inevitably pass through some form of injective or projective resolutions.
If you care mostly about sheaf cohomology (as opposed to the general theory of derived functors, which I'd argue is essentially the same thing as the general theory of injective resolutions etc or of derived categories), then Cech cohomology is pretty concrete and conceptually simple. (The fact that these have long exact sequences is then a theorem rather than a definition).
|
2025-03-21T14:48:30.210842
| 2020-04-03T09:49:16 |
356449
|
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"Jeremy Rickard",
"Mare",
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"Todd Trimble",
"YCor",
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|
Stack Exchange
|
Classification of Frobenius algebras of small dimensions
Despite (commutative) Frobenius algebras over a field $K$ being a very popular class of algebraic objects, it seems no attempt of classification (up to $K$-algebra isomorphism) for them has been attempted yet. While in nearly every book on group theory, you can find a classification of groups of small order, no such classification of Frobenius algebras of small dimension seems to exist in the literature (despite there being at least 3 books on Frobenius algebras).
We can assume that they are local and of the form $K[x_1,...,x_r]/I$, where being Frobenius means here (without loss of generality, assuming they are local and split) that $J^n \subseteq I \subseteq J^2$ for $J=<x_1,...,x_r>$ and that there is a unique 1-dimensional ideal in $K[x_1,...,x_r]/I$, or equivalently a "unique" non-zero polynomial of highest degree survives in $K[x_1,...,x_r]/I$.
In particular it seems to be unknown whether there are only finitely many Frobenius algebras of a given vector space dimension and whether their classification is independet of the given field.
This motivates the following question:
Question: Can we give a classification of all Frobenius algebras (up to isomorphism of $K$-algebras) of vector space dimension $d$ for small $d$ (lets say $1 \leq d \leq 10$)? In particular is there a computer algebra system that can do this in case the field is finite, lets say with two or three (to have signs) elements?
Here a start:
$d=1$: We only have $K$.
$d=2$: We only have $K[x]/(x^2)$.
$d=3$: We only have $K[x]/(x^3)$.
$d=4$: We have $K[x]/(x^4)$ and $K[x,y]/(x^2,y^2)$. (there is also $K[x,y]/(xy,x^2-y^2)$ but should be isomorphic to $K[x,y]/(x^2,y^2)$?)
$d=5$: We have $K[x]/(x^5)$ and $K[x,y]/(xy,x^2-y^3)$ and $K[x,y,z]/(yz,xz,xy,y^2-z^2,x^2-z^2)$, did I forgot one?
$d=6$: We have $K[x]/(x^6)$, $K[x,y]/(x^2,y^3)$, $K[x,y]/(xy,x^3-y^3)$ and $K[x,y,z]/(yz,xz,xy,y^2-z^3,x^2-z^3)$ but here Im not sure whether there are more (probably there are).
I'm confused by the definition of Frobenius algebra: on Wikipedia it is defined as an algebra endowed with some bilinear form, but then they say that some given algebra "is a Frobenius algebra". Is Frobenius algebra both used for "algebra endowed with a bilinear form with some properties", and for "an algebra that possesses at least one such bilinear from"?
@YCor I guess the most common definition is that it is one with at least one such bilinear form (at least that is the definition used in all books I know about the topic). An alternative equivalent definition is that it is a finite dimensional algebra $A$ such that $A \cong D(A)$ as right modules.
Thanks. It's a pity there are not two distinct words (like "metric" vs "metrizable") since both notions are useful.
Huh -- I've always heard it used in the sense of an algebra with certain extra structure (which can take various forms, for example coalgebra structure suitably compatible with the algebra structure). In other mathematics, there is the coinage "Frobenious", an adjectival form that suggests having a certain property -- maybe it would be good to adapt that here. :-)
@Todd The term has been used for an associative algebra with a certain property since the 1930s. Maybe the people who stole the name should think of a new one for the algebra with extra structure. :-)
I've sometimes seen "Frobenius algebra" for the structure and "algebra with the Frobenius property" for the property.
Or, following the HoTT book, "Frobenius algebra" and "merely Frobenius algebra." The issue is whether there "exists" an A-mod iso $A \cong A^*$ or there "merely exists" such an iso.
|
2025-03-21T14:48:30.211084
| 2020-04-03T10:22:27 |
356451
|
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|
Stack Exchange
|
Support of closed random walk on $\mathbb Z$
I am researching closed random walks on graphs and have the following problem that I haven't been able to find a reference for.
Consider a random walk on $\mathbb Z$ starting at 0 and at each step it moves $-1$ or $+1$ each with probability $1/2$. If the walk has length $2n$ it is well-known that the support (or how many elements of $\mathbb Z$ that are covered by the walk) is $\Theta(\sqrt n)$ with high probability.
Suppose now that our walk is closed, i.e. we condition on that the walk starts and ends at 0. Is it still the case that the walk has support $\Theta(\sqrt n)$ with high probability?
I would be happy if I could just show that the support is at least $\Omega(n^{\varepsilon})$ for some constant $\varepsilon>0$.
This process is a bridge, and in one dimension it is just a question of the difference between the max and min, and I would be surprised if a pretty good treatment were not out there.Certainly, the limiting process, Brownian bridge, is well understood.
One way to do this is as follows. We have to show that
$$P(M_n\ge x|S_n=0)\to1$$
(as $n\to\infty$) if $x=o(\sqrt n)$, where $S_n$ is the position of the walk at time $n$ and $M_n:=\max_{0\le k\le n}S_k$. By the reflection principle (see e.g. Theorem 0.8) and the de Moivre--Laplace theorem , for natural $x$ such that $x=o(\sqrt n)$,
$$P(M_n\ge x,S_n=0)=P(S_n=2x)\sim P(S_n=0),$$
whence
$$P(M_n\ge x|S_n=0)=\frac{P(M_n\ge x,S_n=0)}{P(S_n=0)}\to1,$$
as desired.
I found an answer after discussion with @AndersAamand. The result follows by applying Stirling's approximation to observe that the probability that a regular random walk of length $n$ ends at $k$ (which must have the same parity as $n$) is independent of $k$ up to constant factors as long as $k=\Theta(\sqrt n)$.
First, we prove the above statement. For simplicity, we shall do so for walks of length $2n$. Let $\omega=\omega_1\omega_2\dots\omega_{2n}$ be a random walk on $\mathbb Z$ of length $2n$ starting at $0$. Then by Stirling's approximation
$$P(\omega_{2n}=2k) = 2^{-2n}\binom{2n}{n+k} \approx 2^{-2n}\sqrt{\frac{2n}{2\pi(n-k)(n+k)}}\cdot \frac{(2n)^{2n}}{(n+k)^{n+k}(n-k)^{n-k}} = \Theta(1/\sqrt n)$$
whenever $k=O(\sqrt n)$. Where the constants in the two $O$-notations depend on each other.
Second, let $T$ denote the number of closed walks of length $2n$ on $\mathbb Z$. For each $k\in \mathbb Z$ let $a_k$ denote the number of walks of length $n$ from 0 to $k$. Then $T=\sum_{k\in Z}a_k^2$. Note that by the above, $a_k=\Theta(a_j)$ for every $k, j=O(\sqrt n)$. Hence, for a closed random walk $\omega=\omega_1\dots\omega_{2n}$ of length $2n$ and some $k=O(\sqrt(n))$,
$$P(|\omega_n|\leq t) = \frac{\sum_{j=-t}^t a_j^2}{\sum_{j\in \mathbb Z}a_j^2}\leq \frac{\sum_{j=-t}^t a_j^2}{\sum_{j=-\sqrt n}^{\sqrt n}a_j^2} = O\left(\frac t{\sqrt n}\right).$$
And since this holds for the midpoint of the walk, it must hold for the support as well.
|
2025-03-21T14:48:30.211284
| 2020-04-03T11:28:25 |
356453
|
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"Dylan Wilson",
"Marc Hoyois",
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|
Stack Exchange
|
Fibrant objects in $\mathbb{S}$-local model structure on $Top_*$
Let $\mathbb{S}$ be the sphere spectrum. We can localize the category of based spaces, $Top_*$ at a homology theory, and hence at $\mathbb{S}$.
Equipping $Top_*$ with the Quillen model structure (weak homotopy equivalences, and Serre fibrations), and Bousfield localizing at $\mathbb{S}$ gives a model structure with the weak equivalences the isomorphisms on stable homotopy groups, i.e. $\pi_*^\mathrm{st}$-isomorphisms.
Is there a good characterization of the fibrant ($\mathbb{S}$-local) objects in this model structure?
In particular, are "nice" spaces fibrant in this model structure, e.g. nilpotent spaces, simply connected spaces, etc?
Isn't this the same as localizing at $H\mathbb{Z}$? Anyway: It looks like Eilenberg-MacLane spaces for abelian groups are local (since cohomology is a stable invariant), and local objects are closed under homotopy limits, so every nilpotent space is local. Since this is the same as HZ-localization, you can probably find lots more information in Bousfield's original paper inventing "Bousfield localization" for spaces ;)
@DylanWilson I believe this is precisely the same thing. Do you have a reference for this fact? I’ve tried showing it but got nowhere quick
Elaborating on Dylan's comment: the $\mathbb{S}$-local model structure is actually the same as the $H\mathbb{Z}$-local one, since homology detects equivalences between bounded below spectra. $H\mathbb{Z}$-local spaces are called semi-$\mathbb{Z}$-complete in Bousfield and Kan (VII 2.2), but not much is said about them. Any $\mathbb{Z}$-complete space is $H\mathbb{Z}$-local, since by definition it is a homotopy limit of spaces of the form $\Omega^\infty(H\mathbb{Z}\wedge X)$. Any nilpotent space is $\mathbb{Z}$-complete, hence $H\mathbb{Z}$-local.
I suppose one way to show this is to show that an isomorphism on stable homotopy groups is an isomorphism on integral homology? Then both localized model structures are the same.
@User1236262625 yes, this is true (for bounded below spectra), as Marc said- it's the Hurewicz theorem.
|
2025-03-21T14:48:30.211447
| 2020-04-03T12:20:56 |
356455
|
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|
Stack Exchange
|
Markov semigroups and resolvents, difference of continuity
Let $(E,d)$ be a locally compact separable metric space. We have a Markov process $X=(\{X_t\}_{t \ge 0},\{P_x\}_{x \in E})$ on $E$. For bounded measurable function $f$ on $E$, we define
\begin{align*}
P_tf(x)&=E_{x}[f(X_t)],\quad t>0,\\
R_{\alpha}f(x)&=\int_{0}^{\infty}\exp(-\alpha t)P_tf(x)\,dt,\quad \alpha>0.
\end{align*}
We assume that for any bounded measurable function $f\colon E \to \mathbb{R}$ and $\alpha>0$, the function $R_{\alpha}f$ is $\lambda$-Hölder continuous function on $(E,d)$. That is, it holds that
\begin{align*}
|R_{\alpha}f(x)-R_{\alpha}f(y)|\le C\|f\|_{\infty}d(x,y)^{\lambda},\quad x,y \in E
\end{align*}
Here, $\lambda>0$ and $C>0$ are a positive constant independent of $x,y$, and $f$.
We denote by $\|f\|_{\infty}$ the sup-norm of $f$.
My question
We assume that the Markov process $X$ is symmetric with respect to a $\sigma$-finite measure $m$.
Under the above conditions, the semigroup $P_tf(x)$ is also Hölder continuous in $x$?
If the semigroup is ultracontractive in the sense that $P_t(L^1(E,m)) \subset L^{\infty}(E,m)$, for any $t>0$ and $f \in L^2(E,m)$, we can find a bounded measurable function $h\colon E \to \mathbb{R}$ such that $P_tf=R_{1}h$. Therefore, then, the semigroup $P_tf(x)$ is also Hölder continuous in $x$.
Without the ultracontractivity, can we prove the Hölder continuity of $P_tf$?
What is the dependence of the constant $C$ on the function $f$?
@JohnDawkins Thank you for your comment. We assume that $C=C'|f|_{\infty}$, where $C'$ is a constant independent of $f$.
|
2025-03-21T14:48:30.211577
| 2020-04-03T13:41:18 |
356458
|
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|
Stack Exchange
|
Olympiad inequality as a generalizing result due at the origin to Vasile Cirtaoje
Claim:
let $a,b,c>0$ and $p\geq 1$ then we have :
$$\left(\frac{a^3}{13a^2+5b^2}\right)^p+\left(\frac{b^3}{13b^2+5c^2}\right)^p+\left(\frac{c^3}{13c^2+5a^2}\right)^p\geq 3\left(\frac{a+b+c}{54}\right)^p$$
The case $p=1$ have been proved by user RiverLi and I offer a partial proof too in this case (see here).
The motivation:
In my partial proof (and a large part is due to user Alex Ravsky here) I use the well-know inequality called Karamata's inequality and Buffalo's way.
The fact is : the use of this inequality allow us to a generalisation because I apply the Karamata's inequality with the exponential function so it works for $e^x$ then it works also for $e^{xp}$.
Now and have a look to Alex's proof the only values for wich the inequality is valid is $p\geq 1$ .It's explained by another inequality called Jensen's inequality.
At the origin the inequality is due to Vasil Cirtoaje and strenghened by user Michael Rozenberg who found the coefficient $13$ and $5$.
Some details of the proof :
Let $a,b,c>0$ such that $\frac{a^3}{13a^2+5b^2}\geq \frac{b^3}{13b^2+5c^2}\geq \frac{c^3}{13c^2+5a^2}$ and $a\geq b\geq c$ and $13a^2+5b^2\geq 13b^2+5c^2\geq 13c^2+5a^2$ then $\exists n>1$ such that :$$\frac{\frac{a^3}{13a^2+5b^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\geq \frac{a+b+c}{54}$$
$$\Big(\frac{\frac{a^3}{13a^2+5b^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\Big(\frac{\frac{b^3}{13b^2+5c^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\geq \frac{(a+b+c)^2}{54^2}$$
$$\Big(\frac{\frac{a^3}{13a^2+5b^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\Big(\frac{\frac{b^3}{13b^2+5c^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\Big(\frac{\frac{c^3}{13c^2+5a^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\geq \frac{(a+b+c)^3}{54^3}$$
Remains to apply Karamata's inequality to get the desired result .By Karamata's inequality I mean this special case :
If $a_1\geq a_2\geq a_3\geq\cdots\geq a_n$ and $b_1\geq b_2\geq b_3\geq\cdots\geq b_n$ are two sequences of positive real numbers then we have $\frac{a_1}{n}+\frac{(n-1)b_1}{n}\geq \frac{a_2}{n}+\frac{(n-1)b_2}{n} \geq\cdots\geq \frac{a_n}{n}+\frac{(n-1)b_n}{n}$and $b_1\geq b_2\geq b_3\geq\cdots\geq b_n$ satisfying the following conditions(call the conditions $C$):$$\frac{a_1}{n}+\frac{(n-1)b_1}{n}\geq b_1,(\frac{a_1}{n}+\frac{(n-1)b_1}{n})(\frac{a_2}{n}+\frac{(n-1)b_2}{n})\geq b_1b_2,\cdots,(\frac{a_1}{n}+\frac{(n-1)b_1}{n})(\frac{a_2}{n}+\frac{(n-1)b_2}{n})\cdots (\frac{a_n}{n}+\frac{(n-1)b_n}{n})\geq b_1b_2\cdots b_n,$$ Then we have :
$$a_1+a_2+a_3+\cdots+a_n\geq b_1+b_2+b_3+\cdots+b_n$$
To get the power $p$ we use Jensen's inequality because we have :
Let $u,v>0$ and $p$ a real number such that $p\geq 1$ and $n$ a natural number large enought:
$$u^p\frac{1}{n}+v^p\frac{n-1}{n}\geq \left(u\frac{1}{n}+v\frac{n-1}{n}\right)^p$$
To go further :
We have also a stronger statement :
let $a,b,c>0$ and $p\geq 1$ then we have :
$$\left(\frac{a^3}{13a^2+5b^2}\right)^p+\left(\frac{b^3}{13b^2+5c^2}\right)^p+\left(\frac{c^3}{13c^2+5a^2}\right)^p\geq \frac{a^p+b^p+c^p}{18^p}\geq 3\left(\frac{a+b+c}{54}\right)^p$$
To show it we come back at the end proof where I use Jensen's inequality .The inequality in $u,v$ can be strenghened using strong convexity and a modulus $m$ .I have not try but it seems promising .see here Theorem 2.
Using the idea above we got :Let $a,b,c>0 $ and $m=\min(\frac{a^{3}}{13a^{2}+5b^{2}},\frac{b^{3}}{13b^{2}+5c^{2}},\frac{c^{3}}{13c^{2}+5a^{2}},\frac{a+b+c}{54})$ and $p\geq 2$ a real number we have :
$$\left(\frac{a^{3}}{13a^{2}+5b^{2}}\right)^{p}+\left(\frac{b^{3}}{13b^{2}+5c^{2}}\right)^{p}+\left(\frac{c^{3}}{13c^{2}+5a^{2}}\right)^{p}-\left(3\left(\frac{\left(a+b+c\right)}{54}\right)^{p}+p\left(p-1\right)\left(m\right)^{\left(p-2\right)}\cdot0.5\cdot\left(\sum_{cyc}\left(\frac{\left(a+b+c\right)}{54}-\left(\frac{a^{3}}{13a^{2}+5b^{2}}\right)\right)^{2}\right)\right)\geq 0$$
Ps:The sign at the end is not a superior strict but superior or equal .
Question: How to show the claim properly?
The case $p>1$ easily follows from the case $p=1$. Indeed, let $M_p(x,y,z) := \left(\frac{x^p+y^p+z^p}3\right)^{1/p}$. Then for $p>1$ we have
$$M_p\big( \frac{a^3}{13a^2+5b^2}, \frac{b^3}{13b^2+5c^2}, \frac{c^3}{13c^2+5a^2}\big) \stackrel{(1)}{\geq} M_1\big( \frac{a^3}{13a^2+5b^2}, \frac{b^3}{13b^2+5c^2}, \frac{c^3}{13c^2+5a^2}\big) \stackrel{(2)}{\geq} \frac{a+b+c}{54},$$
where (1) the power mean inequality and (2) is the case $p=1$.
Ok thanks ! What do you think about the last inequality have you a hint (my proof of it is partial )?
|
2025-03-21T14:48:30.211831
| 2020-04-03T13:55:46 |
356460
|
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|
Stack Exchange
|
Generations until fixation: A nontrivial generalization of a dice convergence problem
In spite of its "recreational" aspect, this question appears to me to be research-level and (I hope) clearly formulated and tagged.
Edit 4/4/20: You can find a related question with the same motivator as posed by Greg Egan:
Background:
Here is a relatively straightforward question from a recent 538 Riddler:
Last week, you started with a fair 6-sided die and rolled it six times, recording the results of each roll. You then wrote these numbers on the six faces of another, unlabeled fair die. For example, if your six rolls had been 3, 5, 3, 6, 1 and 2, then your second die wouldn’t have had a 4 on it; instead, it would have two 3s.
Next, you rolled this second die six times. You took those six numbers and wrote them on the faces of yet another fair die, and you continued this process of generating a new die from the previous one.
Eventually, you’d have a die with the same number on all six faces. What was the average number of rolls it would take to reach this state?
The link above contains an exact solution, but there is also a follow-up with some work done, yet which lacks a closed form solution.
My Question:
What is the precise answer for the version of this question in which you start with a fair $N$-sided die?
The 538 piece above includes a link to work from Angela Zhou, who also tweets:
The limiting version of this has been studied in mathematical biology; a reference is The Average Number of Generations Until Fixation of a Mutant Gene by Kimura and Ohta.
Unfortunately, I don't think the above reference gives an answer any more precise than the given answer - it just comes up with the same proven estimate (not the conjectured one) - so I don't think I have anything that would constitute an answer, just another place to look to see if someone has answered.
On reexamination, the paper has a better estimate than I remembered, though it still only has limiting behavior (and I'm not entirely confident in that limit). Still, I will post an answer once I translate it from biology-speak to dice-speak.
Down-voter: Please include some sort of comment about why you have voted in such a way. This question appears to me to be research-level and (I hope) clearly formulated and tagged. Recommendations for improvements will be most welcome; thanks.
@BenjaminDickman a naive Bayes classifier would most likely, based on the title and the first few lines, detect this question as off-topic, and I admit I did so as well (but I then read the comments). I'd suggest to start the post with a few lines that indicate that this not an exercise.
@YCor If you have an idea about how better to start the post (or title it, tag it, etc) then I would be happy to see this edited over! Please feel free to modify at will if/when you have the bandwidth. [I've changed it a bit already in a first attempt...]
Done: I basically copied your comment.
(I think) this is the simplest (discrete) case of Kingman's coalescence. See .e.g https://arxiv.org/abs/0809.4233 and the explanations and refernces there.
@esg I don't see how to transform this into an answer. Do you? I would love to see it!
View the results of the $N$ throws of the first round as value table $(f_1(1),\ldots,f_1(N))$ of a random mapping $f_1$.
Imagine that you write the pairs $(i,f_1(i))$ on the faces of the unlabeled die (instead of just the values $f_1(i)$, as you do).
Then the results of N throws of the second round may be viewed as producing the value table $\big((f_2(1),f_1(f_2(1)),\ldots,(f_2(N),f_1(f_2(N))\big)$,
(where $f_2$ is a random mapping, and independent of $f_1$),
etc,
and you are interested in the "coalescence time" $T$ of this process, i.e. the smallest $t$ for which the composition $f_1\circ\ldots\circ f_t$ of independent (uniform)
$N$ to $N$ mappings becomes constant.
@esg A posted answer would be much better than comments, if you find the time. Thanks.
A similar idea was considered by Motoo Kimura and Tomoko Ohta in The Average Number of Generations Until Fixation of a Mutant Gene. They used a diffusion model, which should correspond to looking at the behavior as $N \rightarrow \infty$. The relevant line should be:
NOW, in a population consisting of N individuals, if we assume that each mutant gene is represented only once at the moment of its occurrence, p = 1/ (2N), and from formula (14), the average number of generations until fixation of a neutral mutation becomes
$\bar{t}_1(\frac{1}{2N}) = -8 N N_e (1 - \frac{1}{2N}) \log_e(1 - \frac{1}{2N})$
Our $N$ is Kimura and Ohta's $2N$; if I understand correctly, their $N_e$ should equal $N$ in our situation ($N_e$ denotes an "effective" population size where individuals act differently; here, that isn't true). Correspondingly, our limiting behavior should be:
$-2 N^2 (1 - \frac{1}{N}) \ln(1 - \frac{1}{N}) = 2 N (N - 1) \sum_{i = 1}^\infty \frac{1}{i N^i} = 2 \left(\sum_{i = 1}^\infty \frac{1}{i N^{i - 2}} - \sum_{i = 1}^\infty\frac{1}{i N^{i - 1}}\right)$
$= 2N + 2\sum_{i = 0}^\infty \frac{1}{N^i} (-\frac{1}{i + 1} + \frac{1}{i + 2}) = 2N - 2\sum_{i = 0}^\infty \frac{1}{(i + 1)(i + 2) N^i}$
$ = 2N - 1 - \frac{1}{3N} - O(\frac{1}{N^2})$
This is subject to the diffusion model being accurate to this degree of approximation, which I have no specific support for. I wouldn't be surprised to see the logarithmic term in Angela Zhou's answer show up when considering the effect of the first "round".
Tracing back the references a bit, I found this paper, which may be more obviously relevant (as well as already being written for mathematicians directly).
I can propose a simple heuristic but it doesn't go as far as Angela Zhou conjecture.
For each face the probability that it appears $k$ times in $N$ toss is $\frac{N!}{k!(N-k)!}\frac{(N-1)^{n-k}}{N^N}$. For $N$ large this can be approximate by the Poisson distribution $\frac{1}{k!}e^{-1}$.
As repeating the process, for any fixed number the number of face of that number should behave rougthly like a Galton Watson process as each face are almost independent.
So my claim is that this problem should behave for $N$ large as $N$ iid Galton Watson process with poisson distribution:
$Z_{n+1}:=\sum_{i=1}^{Z_n}P_{i,n}$ with $P_{i,n}$ iid Poisson variable. Considering the generating function
$$\mathbb{E}(s^{Z_1})=\sum_{k}\frac{s^k}{k!}e^{-1}=e^{s-1}$$we get that $\mathbb{E}(s^{Z_n})=u_n(s)$ where $u_0(s)=s$ and for all $n$ $u_{n+1}(s)=e^{u_n(s)-1}$.
In particular as $\mathbb{P}(Z_n=0)=u_n(0)$, the proportion of numbers that have disappeared at time $n$ is about $Nu_n(0)$, which are for $n=1:$ $Ne^{-1},$ $n=2: Ne^{e^{-1}-1},$ $n=3: Ne^{e^{e^{-1}-1}-1},\cdots$).
As $u_n(0)\rightarrow 1$, we also have that $$u_{n+1}(0)-1=u_n(0)-1+\frac{1}{2}(u_n(0)-1)^2+\mathcal{O}((u_n(0)-1)^3)$$ which allow one to get the behavious $(u_n(0)-1)\sim-\frac{2}{n}$ (This assuming $N(u_n(0)-1)=1$ gives the $2N$ expected rolls).
This is the simplest (discrete) case of Kingman's coalescence. See e.g. https://arxiv.org/abs/0809.4233 and the explanations and references there.
The relation to the process considered there can be seen as follows:
view the results of the $N$ throws of the first round as value table $(f_1(1),\ldots,f_1(N))$ of a random mapping $f_1$.
Imagine that you write the pairs $(i,f_1(i))$ on the faces of the unlabeled die (instead of just the values $f_1(i)$, as you do).
Then the results of the $N$ throws of the second round may be viewed as producing the value table $\big((f_2(1),f_1(f_2(1)),\ldots,(f_2(N),f_1(f_2(N))\big)$,
(where $f_2$ is a random mapping, and independent of $f_1$),
and so on, and you are interested in the "coalescence time" $T$ of this process, i.e. the smallest $t$ for which the composition $f_1\circ\ldots\circ f_t$ of independent (uniform)
$N$ to $N$ mappings becomes constant.
|
2025-03-21T14:48:30.212306
| 2020-04-03T14:04:24 |
356461
|
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|
Stack Exchange
|
Completeness hypothesis in the positive mass theorem
I am trying to understand and further formalize Witten's proof of the positive mass theorem. Dan Lee, in his book "Geometric relativity" did a wonderful job with formalizing and carrying out the details of Parker and Taubes' work, which was already a formalization of Witten's work.
The statement of the theorem in his book is more or less the following:
Theorem:
Let $(N,g)$ be a complete asymptotically euclidean spin $n$-manifold with nonnegative scalar curvature and $n \geq 3$.
Suppose further that $N$ has a well defined ADM mass.
Then the ADM mass of each end is nonnegative.
Moreover, if the mass of any end is zero, then $(N,g)$ is globally isometric to euclidean space.
I do not particularly like the completeness hypothesis as in most cases of interest in physics, the manifold is not complete.
Therefore I am wondering why the completeness hypothesis is necessary.
The only place I can find in the proof in his book where the completeness hypothesis is used explicitly is for positive mass rigidity, that is to say to prove that if the mass of any end is zero, then $(N,g)$ is globally isometric to euclidean space.
The completeness hypothesis is almost never stated in other surveys.
Parker and Lee, in their survey on the Yamabe problem state the theorem as follows:
Theorem:
Let $(N,g)$ be an asymptotically flat Riemannian manifold of dimension $n \geq 3$ such that the ADM mass is well defined, and with nonnegative scalar curvature.
Then its mass $m(g)$ is nonnegative, with $m(g) = 0$ if and only if $(N, g)$ is isometric to $\mathbb{R}^n$ with its Euclidean metric.
The positive mass rigidity part in this theorem is plainly false, as $\mathbb{R}^n \setminus \{0\}$ satisfies all hypotheses but is not isometric to $\mathbb{R}^n$, so for this part the completeness is necessary.
However, without completeness it is possible to prove that the manifold has to be flat.
Hence I think the following theorem is also true:
Theorem:
Let $(N,g)$ be an asymptotically euclidean spin $n$-manifold with nonnegative scalar curvature and $n \geq 3$.
Suppose further that $N$ has a well defined ADM mass.
Then the ADM mass of each end is nonnegative.
Moreover, if the mass of any end is zero, then $(N,g)$ is flat.
Can someone confirm this?
Notice that our definition of an asymptotically flat Riemannian manifold is that $N$ is the union of a compact set $N_0$ together with a set $N_\infty$ that is diffeomorphic to the complement of closed ball in $\mathbb R^n$, with a metric that approaches the Euclidean metric at a certain rate. This implies that $N$ must be complete. It's not too hard to show that every Cauchy sequence converges, but this comment is too small to contain the proof. :-)
Completeness is necessary. Otherwise you can just take a maximal spatial slice of the negative-mass Schwarzschild solution (i.e. a constant $t$ slice in Boyer-Lindquist coordinates) and it has vanishing scalar curvature with $m < 0$. There are however no complete maximal spatial sections of the negative-mass Schwarzschild solution.
For Lee and Parker, the implicit assumption of completeness is harmless: for applications to the Yamabe problem your asymptotically flat manifold comes from stereographic projection of a compact (closed) manifold about a point, so is always complete.
In the Witten proof included in the Appendix to Lee and Parker, completeness is used in the integration by parts argument equation (A.5). Here $N_R$ is assumed to be a compact manifold with boundary $S_R$. If your original manifold were not complete, but asymptotically flat, then for sufficiently large $R$ there must be another component of the boundary $N_R$ corresponding to the ends of the terminating geodesics.
Finally, if you make your definitions carefully, then incomplete manifolds are also ruled out of the Lee and Parker statement.
Recall that their definition of Asymptotically Flat manifold starts with a Riemannian manifold $N$ which is decomposed into some non-compact ends each of which are asymptotically flat in the usual sense, plus a compact portion.
If you start with an incomplete Riemannian manifold (here we are careful in having "manifolds with boundary" being not "manifolds"), and take away the asymptotically flat ends, you will necessarily still be left with something non-compact due to the incompleteness. (Incompleteness will always leave the manifold [which by definition cannot include a boundary] non-compact [Cauchy sequence alone the incomplete geodesic], and finite-length means it cannot be coming from an AF end.) So "complete" can be regarded as implicit.
(I would also like to dispute the statement that "in most cases of interest in physics, the manifold is not complete." In physics, the space-time manifold may not be complete, but the spatial sections generally are.
Thank you very much for your answer. When I said that in most cases of interest in physics, the manifold is not complete I was thinking primarily to a maximal spatial slice of the positive mass Schwarzschild solution. As your counterexample satisfies also the hypothesis for Schoen and Yau's theorem I guess that in such case the positive mass theorem cannot simply be used?
A maximal spatial slice of the maximally extended positive mass Schwarzschild solution passes through the bifurcate sphere on the event horizon and continues to the "other" domain of outer communications. This is a complete Riemannian manifold with two asymptotically flat ends. It shrinks to a throat at the bifurcate sphere, and gives the usual popular-science illustration of what a blackhole/wormhole looks like.
|
2025-03-21T14:48:30.212778
| 2020-04-03T14:05:46 |
356462
|
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|
Stack Exchange
|
Existence of large first return times
Let $(X,T,\mu)$ be a measure preserving system, with $\mu$ a probability measure. Let $E \subset X$ of positive measure and $\tau_E$ be the first return time to $E$. Then the Kac Lemma asserts that $$\int_E \tau_E d \mu = 1,$$
that is the average return time to $E$ is $\mu(E)^{-1}$. My question is when is it known that there is some $x \in E$ so that $\tau_E(x) \geq C \mu(E)^{-1}$, where $C$ is some large (fixed) constant? Any pointers to relevant literature would be much appreciated!
Hi. Can you be a bit more precise about "when"? Are you in a fixed dynamical system and are asking which $E$ have some $x$ with $\tau_E(x) \ge C\mu(E)^{-1}$. Or are you asking about properties of a dynamical system that guarantee the existence of some $E$? Note that for any dynamical system, $E=X$ won't have $||\tau_E||_\infty \ge C\mu(E)^{-1}$.
@mathworker21 Sure. I am interested in theorems of the form: suppose a dynamical system (or measure preserving system) and a set E satisfy a certain property. Then there is an x in E with a large return time. Homomorphisms of the torus are of particular interest to me.
Perhaps it helps having a look at https://link.springer.com/article/10.1007/s002200100427. The notion of "long return time" seems to be quite close to what you want.
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2025-03-21T14:48:30.213153
| 2020-04-03T14:07:04 |
356463
|
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"authors": [
"Mateusz Kwaśnicki",
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"https://mathoverflow.net/users/50406",
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"url": "https://mathoverflow.net/questions/356463"
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|
Stack Exchange
|
Hardy-Littlewood-Sobolev for "componentwise product" of Riesz kernels
Let $d\in\mathbb N$ and $0<\alpha<d$. Define the Riesz kernel $K_\alpha(x):=|x|^{\alpha-d}$, and the associated convolution operator
$$K_\alpha f(x):=\int\frac{f(y)}{|x-y|^{d-\alpha}}~dy.$$
The classical Hardy-Littlewood-Sobolev inequality states that we have the following boundedness property: for every $q>1$,
$$\|K_\alpha f\|_q\leq C\|f\|_p$$
for some universal constant $C>0$ and $1/p=1/q+\alpha/d$.
Now let $0<\alpha_1,\ldots,\alpha_d<1$, and suppose instead that we consider the convolution operator given by the kernel
$$K_{(\alpha_1,\ldots,\alpha_d)}(x)=\prod_{i=1}^d|x_i|^{\alpha_i-1},\qquad x=(x_1,\ldots,x_d),$$
which is what I call a "componentwise product of Riesz Kernels."
Question. Given $q>1$, does there exist some constant $C>0$ and a $p>1$ such that $$\|K_{(\alpha_1,\ldots,\alpha_d)} f\|_q\leq C\|f\|_p?\tag{1}$$
By a simple scaling argument, we can see that if (1) holds, then it must be for $1/p=1/q+\sum_i\alpha_i/d$. Indeed, letting $f_c(x):=f(cx)$, then $\|f_c\|_p=c^{-d/p}\|f\|_p$
and $\|K_{(\alpha_1,\ldots,\alpha_d)}f_c\|_q=c^{-\sum_i\alpha_i-d/q}\|K_{(\alpha_1,\ldots,\alpha_d)}f\|_q$.
I would write $K_{(\alpha_1,\ldots,\alpha_d)}$ as a composition of operators $K_j$, $j = 1, \ldots, d$, defined by $K_j f(x) = \int_{-\infty}^\infty f(x - y e_j) |y|^{\alpha_j-1} dy$, where $e_j = (0, \ldots, 0, 1, 0, \ldots, 0)$ has $1$ at $j$-th coordinate. Then it seems sufficient to apply the usual H-L-S inequality to each $K_j$. Have you tried that?
@MateuszKwaśnicki Thanks for pointing out this very simple reduction; I did not think of that. If each $K_j$ does indeed satisfies a HLS inequality of the form $|K_jf|_q\leq C|f|_p$ with $1/p=1/q+\alpha_j/d$, then we have the result!
I realise I was too optimistic in my comment: no such ineqaulity holds in general. Indeed, consider $f(x) = \prod_j f_j(x_j)$. Then
$$ K_{(\alpha_1,\ldots,\alpha_d)} f(x) = \prod_j K_{\alpha_j} f_j(x_j) $$
and so
$$ \|f\|_p = \prod_j \|f_j\|_p , \qquad \|K_{(\alpha_1,\ldots,\alpha_d)} f\|_q = \prod_j \|K_{\alpha_j} f_j\|_q $$
So by the usual Hardy–Littlewood—Sobolev inequality (the "only if" part), in order to get the bound we are interested in, we would need
$$ \frac{1}{p} = \frac{1}{q} + \frac{\alpha_j}{1} , $$
that is,
$$ \alpha_1 = \alpha_2 = \ldots = \alpha_n \qquad \text{and} \qquad \frac{1}{p} = \frac{1}{q} + \frac{\alpha}{d},$$
with $\alpha = \alpha_1 + \alpha_2 + \ldots + \alpha_d$. In this case, the "if part" of the Hardy–Littlewood—Sobolev inequality implies that indeed the desired inequality holds.
Thanks for pointing out yet another simple observation. The result then only holds when all $\alpha_i$ are equal...
|
2025-03-21T14:48:30.213346
| 2020-04-03T15:23:37 |
356470
|
{
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"authors": [
"Noam D. Elkies",
"Stanley Yao Xiao",
"https://mathoverflow.net/users/10898",
"https://mathoverflow.net/users/14830"
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"url": "https://mathoverflow.net/questions/356470"
}
|
Stack Exchange
|
Upper bounds on the length of the shortest vector in lattices associated to polynomial congruences
We consider a lattice $\Lambda \subset \mathbb{Z}^2$, and put $\lambda_1(\Lambda), \lambda_2(\Lambda)$ for successive minima of the lattice $\Lambda$.
By a well-known theorem of Minkowski, one has that
$$\displaystyle \lambda_1(\Lambda) \leq 2(\det(\Lambda)/\pi)^{1/2}.$$
For me it is quite annoying that the constant on the right hand side is greater than one, so I am wondering if it is possible to obtain an upper bound with a constant that is less than one, at least in some special cases.
One particular case where this is possible is the following. Consider the congruence
$$\displaystyle x^2 + 1 \equiv 0 \pmod{d}.$$
The solutions to this congruence lie on a finite union of lattices, defined by
$$\displaystyle u \equiv \omega v \pmod{d}$$
for some $\omega \in \mathbb{Z}$. Note that $\det(\Lambda_\omega) = d$. For such a lattice $\Lambda_\omega$, we must have $\lambda_1(\Lambda_\omega) \leq d$, since in each such lattice it is possible to find a vector $(r,s)$ such that $r^2 + s^2 = d$.
Is it possible to improve the bound provided by Minkowski's theorem for $\lambda_1(\Lambda)$ for general $\Lambda \subset \mathbb{Z}^2$? What about lattices that arise from polynomial congruences in general?
You're asking in effect for the Hermite constant for lattices in the plane.
It's well known (and I'm sure you know it well but momentarily forgot)
that the largest possible constant occurs for a triangular lattice.
If we scale this lattice to have $\lambda_1 = 1$ then the determinant is
$\sqrt{3}/2$, so the best we can do is $\lambda_1 \leq (4/3)^{1/4} \Delta^{1/2}$
where the constant is still a bit above $1$. For a square lattice such as
you get from $x^2+1 \equiv 0$, the constant is exactly $1$. I don't know
where the factor $1/\sqrt2$ comes from.
@NoamD.Elkies yes I made a mistake in the question; I will fix the error
Well $x^2+x+1 \equiv 0$ always gives a triangular lattice.
|
2025-03-21T14:48:30.213510
| 2020-04-03T16:09:29 |
356473
|
{
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"authors": [
"Dylan Wilson",
"Mike Shulman",
"fosco",
"https://mathoverflow.net/users/49",
"https://mathoverflow.net/users/6936",
"https://mathoverflow.net/users/7952"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/356473"
}
|
Stack Exchange
|
Category in a coend depends on the integration variable of an outer coend
Let $\mathsf{Finj}$ be a skeleton of the category of finite sets and injections. Let $n$ denote its typical object, and as usual $n/\mathsf{Finj}$ the coslice category at $n \in \mathsf{Finj}$. So, an object in $n/FInj$ is a pair $(m, f : n\to m)$. Finally, let $A$ be a set and $A^\bullet : \mathsf{Finj}° \to \mathsf{Set}$ be the functor $n\mapsto A^n$.
Now. I've been told that the following functor is the object part of a monad on $[\mathsf{Finj},\mathsf{Set}]$:
$$
TX : n \mapsto \left(\int^{(m,f) \in n/\mathsf{Finj}} A^m \times Xm\right)^{A^n}
$$
The mate of the initial cowedge yields a candidate unit map. No problem. Multiplication seems quite a bit more painful to find, because if I try to evaluate $TTX$ on $n$, I obtain the scary
$$
\begin{align*}
TTXn &= \left( \int^{m\in n /\mathsf{Finj}} A^m \times \left( \int^{p\in m /\mathsf{Finj}} A^p \times Xp \right)^{A^m} \right)^{A^n}\\
& \to \left(\int^{m \in n/\mathsf{Finj}} \int^{p\in m /\mathsf{Finj}} A^p \times Xp \right)^{A^n}\\
& {\color{red} \to} \left(\int^{p\in m /\mathsf{Finj}} A^p \times Xp \right)^{A^n} \\
& = TXn
\end{align*}
$$ In the first step I used the evaluation map of the cartesian closed structure of $\mathsf{Set}$, but I have no idea why the red map should exist.
The problem is that in the integral
$$
\int^{m \in n/\mathsf{Finj}} \int^{p\in m /\mathsf{Finj}} A^p \times Xp
$$ the innermost integral depends on the integration variable of the outer one. I must admit my coend-fu is too weak to find how this can be further reduced, and this situation directly leads to the following vague but general question:
Let $\mathcal I$ be a category, and let $-/\mathcal I : \mathcal I° \to \sf Cat$ be the coslice functor $i\mapsto i/\mathcal I$. How can one evaluate the following?
$$
\int^{i\in k/\mathcal I} \int^{X\in i/\mathcal I} T(X,X)
$$
Here's a guess: maybe(?) if $X \to Y$ is bicartesian, then $\mathsf{Tw}(X) \to \mathsf{Tw}(Y)$ is cocartesian (or cartesian, depending on the convention you used to define Tw...) in which case one could compute colimits over $\mathsf{Tw}(X)$ fiberwise. (Applying this to $A\times B \to B$ gives the usual Fubini)
I'll be glad to read an answer, even if it does not solve the problem, where you explain what you mean precisely by "compute colimits fiberwise": I'm not familiar with the practice.
In general, if $E \to B$ is a map, then I can compute the colimit over $E$ by left Kan extending to B and then computing the colimit. The value of the left Kan extension at b is the colimit over $E \downarrow b$. When $E \to B$ is cocartesian, the natural map $E_b \to E \downarrow b$ is final, so you may compute the colimit over $E_b$ instead.
Where by "is cocartesian" I presume you mean "is an opfibration"? I've seen people call these "cocartesian fibrations" but never simply "cocartesian".
@MikeShulman Yes, thank you- I was being a bit informal but I see that it might be confusing: I mean "cocartesian fibration" (and "bicartesian fibration" and "cartesian fibration" etc.) in my comments.
|
2025-03-21T14:48:30.213743
| 2020-04-03T16:40:03 |
356475
|
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"Connor Malin",
"Dylan Wilson",
"Tyrone",
"Wlod AA",
"YCor",
"erz",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/356475"
}
|
Stack Exchange
|
Is limit of null-homotopic maps null-homotopic?
The question is motivated by my failed comment to this one.
Let $M$ and $N$ be path connected locally compact, locally contractible metric spaces (you may assume that they are manifolds).
Let $\varphi_{n}:M\to N$ be null-homotopic and convergent to $\varphi:M\to N$ in the compact open topology.
Does it follow that $\varphi$ is null-homotopic?
Note that homotopy between maps is a path in $C(M,N)$ (for nice $M$, $N$), and so what my question asks is whether the path component of a constant map is closed in $C(M,N)$.
I am waaay out of my depth here, but perhaps there is a continuous or positive lower semi-continuous functional on $C(M,N)$ akin to the topological degree such that null-homotopic maps would be the zero-set of that functional?
Maybe try: Take $\phi_0=*$ the constant map and write homotopies $\phi_i \sim \phi_{i+1}$, and run them during time $[1/(i+2), 1/(i+1)]$ this defines a map $M\times (0,1] \to N$ which you can complete(?) to $M\times[0,1] \to N$ placing $\phi$ at time $0$. (I don't anything about topology, so maybe there's some point set horror that makes this not work).
@DylanWilson yeah on principle your argument is like this: you have a convergent sequence in the same path component, then take a long path that runs through all these points, and voila. However (again this is just why this argument doesn't work on principle, i hope it does in this situation), if your space is the topologist's sine curve this long path is jut too long to converge.
If $M,N$ are manifolds then they have CW homotopy type. Thus if $M$ is compact, then $C(M,N)$ (compact-open topology) has CW homotopy type (this is Milnor). In particular it is locally contractible, so locally path connected, so the path components coincide with the connected components and are clopen.
@Tyrone Local contractibility is not a homotopy invariant. For example, a comb space is homotopy equivalent to a point, but is not locally connected.
@ConnorMalin you're right, locally contractible is not enough, but the statement still stands: it is enough that the $C(M,N)$ is homotopy equivalent to a space which admits an ambiently contractible open covering - this property is homotopy invariant.
@Tyrone Ah I see, a continuous map induces a continuous map between topologized path components and a homotopy equivalence induces a bijection on path components, so in fact the homotopy equivalence induces a homeomorphism on the topologized path components.
@ConnorMalin, to clarify, I was half assuming that 'manifold' means 'smooth manifold', in which case $M,N$ are CW complexes, so the local contractibility statement applies. If $M,N$ are only $C^0$ and of dimensions not equal to $4$, then they are homeomorphic to CW complexes, so the statement applies again. In the case that it is $4$-dimensional, then it is homotopy equivalent to a CW complex (it is not known in general if it is homeomorphic). If it is compact then it is homotopy equivalent to a finite CW complex. Thus even in this case $C(M,N)$ is homotopy equivalent to a CW complex.
In this case, as you point out, the local contractiblity statement I made is not necessarily true, but a CW complex admits a (numerable) ambiently contractible open covering (I have in mind that it is a Dold space). Such a covering guarantees that its path components are open. Since such a covering may be pulled back along a homotopy equivalence, this guarantees that anything homotopy equivalent to a CW complex has open path components (although it need not be locally contractible - for instance the comb space).
I'll provide a general theorem (then one has to apply it to specific circumstances). There is a micro-dictionary/Notation at the bottom of this note.
B-assumption: Space $\ N\times N\times[0;1]\ $ is normal.
Every metric space $\ N\ $ satisfies B-assumption.
Notation Let $\ \mathcal W_N\ $ be the set of all closed
neighborhoods of diagonal $\ \Delta_N\ :=\ \{(y\ y):\ y\in N\}\ $
in $\ N\times N.$
Family $\ \mathcal W_N\ $ is a basis of all neighborhoods of the diagonal $\ \Delta_N.$
A-assumption: Space $\ N\ $ is an ANR, meaning that
for every normal space $\ X\ $ and closed subset $\ A\ $ of $\ X,\ $
and for every continuous function $\ f:A\to N\ $ there exists
a neighborhood $\ U\ $ of $\ A\ $ and continuous $\ F:U\to N\ $ such
that $\ F|A=f.$
Thus, $\ N^2\ $ is an ANR too.
Definition: Sequence $\ f_n:M\to N\ $ is d-convergent to
$\ f:M\to N\ \Leftarrow:\Rightarrow $
$$ \forall_{V\in\mathcal W_N}\exists_{m\in\Bbb N}
\forall_{n\ge m}\quad (f_n\triangle f)(M)\,
\subseteq V $$
Only continuous functions are meant:
THEOREM Let sequence $\ f_n:M\to N\ $ be d-convergent
to $\ f:M\to N.\ $ Then there exists $\ m\in\Bbb N\ $ such that
$\ f_n\ $ and $\ f\ $ are homotopic for every $\ n\ge m.$
PROOF Diagonal $\ \Delta_N\ $ is an ANR because it is
homeomorphic to $\ N.\ $ Also, $\ \Delta_N\ $ is closed in $\ N^2\ $ since $\ N\ $ is Hausdorff. Thus, there exists $\ U\in\mathcal W_N\ $
and a retraction $\ \rho:U\to\Delta_N\ $ (it is an extension of the
identity map on $\ \Delta_N.)$
Consider the function $\ g\ $ from a closed subset of $\ N^2\times[0;1]\ $ into $\ N^2\ $ given as follows:
$\ \forall_{y\in N^2}\quad g(y\ 0)\ :=\ y; $
$\ \forall_{y\in\Delta_N}\forall_{t\in[0;1]}
\quad g(y\ t)\ := y; $
$\ \forall_{y\in U}\qquad g(y\ 1)\ :=\ \rho(y). $
The arguments of $\ g\ $ belong to the union of three closed subsets
of $\ N\times[0;1],\ $ where the three parts of the definition of
$\ g\ $ coincide on the overlaps hence $\ g\ $ is well defined. This
$\ g\ $ admits an extension $\ G_0\ $ over a closed neighborhood of its closed $3$-part domain. This neighborhood includes $\ V\times[0;1],\ $ where $\ V\subseteq U\ $ is a closed neighborhood of $\ \Delta_N,\ $
because $\ [0;1]\ $ is compact.
Now, by (very elementary and great) Borsuk's homotopy extension
theorem, there is homotopy
$$ H:N^2\times[0;1]\to N^2 $$
such that:
$\ \forall_{y\in N^2}\qquad H(y\ 0)\ :=\ y; $
$\ \forall_{y\in V}\forall_{t\in[0;1]}
\quad H(y\ t)\ := G_0(y\ t); $
Let $\ m\in\Bbb N\ $ and $\ n\ge m\ $ be as in
Definition. Let homotopies
$\ h_n\ h:M\times[0;1]\to N\ $ be given as
$$ h_n\ :=\ \pi'\circ H\circ
((f_n\triangle f)\times\Bbb I );$$
$$ h\ :=\ \pi''\circ H\circ
((f_n\triangle f)\times\Bbb I );$$
where $\ \pi'\ \pi'':N^2\to N\ $ are the canonical projections,
and $\ \Bbb I:[0;1]\to[0;1]\ $ is the identity map.
We see that:
$$ \forall_{x\in M}\quad h_n(x\ 0)\ =\ f_n(x); $$
$$ \forall_{x\in M}\quad h(x\ 0)\ =\ f(x); $$
$$ \forall_{x\in M}\quad h_n(x\ 1)\ =\ h(x\ 1). $$
Define $\ \gamma_n:M\to Y\ $ by $\ \gamma_n(x):=h_n(x\ 1)=h(x\ 1).\ $
We see that $\ f_n\ $ is homotopic to $\ \gamma_n\ $ is homotopic
to $\ f.\,\ $ Remember (observe) that $\ H\ $ in the expressions for
$\ h_n(x\ 1)\ $ and $\ h(x\ 1)\ $ is equal to $\ G_0\ $ (we have
$\ (f_n(x)\ f(x))\in V).\ $ End of PROOF
NOTATION
For functions $\ f:P\to Q\ $ and $\ g:R\to S,\ $ the cartesian product $\ f\times g:P\times Q\to R\times S\ $ is given by
$$ \forall_{(p\ r)\in P\times R}\quad
(f\times g)(p\ r)\ :=\ (f(p)\ g(r)\,) $$
Let $\ P=R\ $ and $\ \Delta_P:=\{(p\ p): p\in P\}.\ $ Then
$\ f\triangle g: P\to Q\times S\ $ is given as follows:
$$ f\triangle g\ := (f\times g)\circ \delta_P $$
where $\ \delta_P:P\to P\times P\ $, and
$\ \forall_{p\in P}\ \delta_P(p):=(p\ p).$
As one can see, the argument space M (the source) is ARBITRARY.
Could you relate your assumption on $N$ to that of the OP (in particular local contractibility)? Could you state whether you claim you're answering the OP's question?
I assume that $\ N\ $ is an ANR.
Can you relate ANR and local contractibility? It maybe obvious to you but not to me.
True, I left some (perhaps even subtle) general consideration untouched. However "ANR assumption" is still very general. The classical fundamental paper by Karol Borsuk showed that in the case of finite-dimensional compact metric spaces, local contractability implies ANR.
I remember that mathematicians like Dugundji and Hu have studied the general scope of the ANR concept.
Thank you for your answer, but I have a hard time following it. What is $(f\Delta f_n)$? Is it the cartesian product of $f$ and $f_n$? If it is, is the $d$-convergence the same as uniform convergence in the case when $N$ is metric? When you apply Borsuk's theorem, what is $Y$? Is it supposed to be $N$? Also, which homotopy are you extending? I guess, you assume that $\Delta_N$ is a deformation retract in $U$, right? does it follows from being ANR?
@erz, I'll add a tiny dictionary. And I'll fix the typo you've spotted, thank you. (It should be N not Y).
For compact spaces, d-convergence is uniform.
@erz, for non-compact metric spaces d-convergence, implies uniform convergence but the converse is false.
Thank you! But why is $\rho$ homotopic to the identity on $U$?
@erz, at a moment of weakness I've simplified my proof too much. I'll fix it, no sweat. Thank you for your high-quality reading.
@erz, mathematically -- no sweat, but editing-wise -- torture. I hope I fixed everything (let's keep fingers crossed). #### The proof works in a smaller neighborhood $\ V\ $ hence the Definition refers now to $\ V\ $ The Definition is (logically) the same but the Proof refers to $\ V\ $ hence the convenient trivial modification of the definition.
I think now I understand. Thank you! I corrected $Y$ to be $N$ in 2 more places. However, this solves a slightly different problem from what I asked: the convergence of functions is different, and I cannot see how to reduce my problem to this one. I'll wait a bit, and if nobody comes up with a compact-open version, I'll accept your answer.
@erz, thank you very much for everything, including fixing $\ Y\to N$ translation (the standard notation $\ f:X\to Y,\ $ with $\ Y\in \text{ANR},\ $ got imprinted in my brain, I guess). I hope that my answer provides a good environment for your topic (even if only partial -- it's always partial :) ).
Please see the answer to Annie's question.
Non-density of continuous functions to interior in set of all continuous functions
|
2025-03-21T14:48:30.214406
| 2020-04-03T16:43:01 |
356476
|
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"Artemy",
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|
Stack Exchange
|
When is rank-1 perturbation to a positive operator still positive?
Let $A : \mathcal{H} \to \mathcal{H}$ and $B : \mathcal{H} \to \mathcal{H}$ be trace-class (hence compact) Hermitian operators on a separable Hilbert space. Assume that $A$ is strictly positive and that $B$ is positive and rank-1. I'm interested in conditions when $A - \epsilon B \ge 0$ for some strictly positive $\epsilon \in \mathbb{R}$ (as usual, $>$ and $\ge$ for operators refers to being positive (semi)definite).
If $A$ is finite dimensional, then $A - \epsilon B \ge 0$ for some $\epsilon > 0$ always. This is because the smallest eigenvalue of $A - \epsilon B$ obeys $\lambda_\min >0$ for $\epsilon=0$ and varies continuously with $\epsilon$.
If $A$ is infinite dimensional and $B=\vert \phi\rangle\langle\phi\vert$ for some eigenvector $\vert \phi\rangle$ of $A$ (with corresponding eigenvalue $\lambda>0$), then it is clear that $A - \epsilon B\ge 0$ for $\epsilon\le \lambda$.
What about when $A$ is infinite dimensional and $B$ does not have the form of $\vert \phi\rangle\langle\phi\vert$?
It is not always true in finite dimensions, if $A$ can be positive semidefinite. You'd need all of its eigenvalues to be strictly positive. In infinite dimensions you need $\phi$ to belong to the spectral projection $\chi_{[\epsilon,\infty)}(A)$, assuming $|\phi|=1$.
Sorry, I should have been more explicit. $A$ is strictly positive.
In "On Majorization, Factorization, and Range Inclusion of Operators on Hilbert Space (1966)", R. G. Douglas proved the following result (Theorem 1 in the paper):
Theorem. Let $C$ and $D$ be bounded linear operators on a real or complex Hilbert space $\mathcal{H}$; then the following are equivalent:
(i) $C\mathcal{H} \subseteq D \mathcal{H}$.
(ii) There exists a number $\lambda \in [0,\infty)$ such that $CC^* \le \lambda^2 DD^*$.
(iii) There exists a bounded linear operator $E$ on $\mathcal{H}$ such that $C = DE$.
Now, if you choose $C$ in the theorem as the positive square root $\sqrt{B}$ of $B$ and $D$ in the theorem as the positive square root $\sqrt{A}$ of $A$, you can characterize the property you are interested in by means of a range condition.
More precisely:
Corollary. Write your rank-$1$ operator $B$ as $B = \alpha \vert \phi\rangle\langle\phi\vert$ for a number $\alpha > 0$ and a vector $\vert \phi\rangle \in \mathcal{H}$ of norm $1$ (not necessarily an eigenvector of $A$). Then the following are equivalent:
(i) There exists $\varepsilon > 0$ such that $A \ge \varepsilon B$.
(ii) $\vert \phi\rangle$ is an element of the range of $\sqrt{A}$.
|
2025-03-21T14:48:30.214597
| 2020-04-03T18:33:16 |
356485
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356485"
}
|
Stack Exchange
|
Friedrich's extension of the generator of a continuous time markov chaoin
Consider the infinitesimal generator $G$ of a Markov chain with state space $\mathbb{Z}$ such that it is symmetric with respect to a measure $\mu$ on $\mathbb{Z}$. Then, the operator $(G,C_c(\mathbb{Z}))$, ($(C_c(\mathbb{Z})$ is the set of all functions with finite support) is symmetric on $L^2(\mathbb{Z},\mu)$. Is the operator $(G,C_c(\mathbb{Z}))$ essentially self-adjoint on $L^2(\mathbb{Z},\mu)$?
Could you specifiy what you mean by $C_c(\mathbb{Z})$? The functions with finite support?
Yes, it is the set of functions with finite support. I have edited it now.
Thanks for your response! How do you know that $C_c(\mathbb{Z})$ is contained in the domain of $G$?
I think it should be straight forward. Since the functions of the form $1_n$ (indicator functions) are in the domain, so is their span.
Well yes, but my question is: why are all of these indicator functions in the domain? I'm not really convinced that this follows from your assumptions. (But I might be overlooking something simple or well-known.)
I just found this paper by Ornstein; in the middle of the first page, he claims that there are examples where not all the indicator functions $1_n$ are in the domain of $G$: "if $i=j$ there are examples where it [it = the derivative of the transition probability $P_{ij}(t)$ at $t=0$] is infinite".
@JochenGlueck: That likely depends on your favourite definition of a Markov chain. As I understand, if we assume that the paths are càdlàg in the discrete topology on $\mathbb{Z}$, then the indicator functions are in the domain of the generator.
@MateuszKwaśnicki: I see, thank you very much! I sometimes tend to see things through my "semigroup filter" - so when I hear "Markov process with countable state space", I just think of a positive $C_0$-semigroup on $\ell^1$ which is norm-preserving on the positive cone.
Not necessarily, I think. Here is an example that comes to my mind, but I did not check all details carefully.
Map $\mathbb{Z}$ to $A = \{2^{-n} : n = \in \mathbb{Z}\}$, and consider a symmetric, martingale continuous-time Markov chain $X_t$ on $A$, with counting measure as the reference measure. Thus, the transition rates are (up to a constant factor) $$q_{2^{-n},2^{-n+1}} = q_{2^{-n+1},2^{-n}} = 2^n,$$ and zero otherwise. This process can be seen as the trace left on $A$ by a Wiener process $W_t$. More precisely, $X_t$ changes its state at moments of time given by $W_t$ hitting a point in $A$, namely:
$$T_{n+1} = \inf \{t > T_n : W_t \in A \setminus \{W_{T_n}\}\} ,$$
and we set $X_t = W_{T_n}$ for $t \in [T_n, T_{n+1})$.
Now depending on what we do with the Wiener process $W_t$ at zero (say: kill or reflect), we will get different traces $X_t$. In other words: imposing different boundary conditions "at zero" for $X_t$, we get different Markov chains.
Of course, this can be mapped back to $\mathbb{Z}$ rather than $A$.
|
2025-03-21T14:48:30.214948
| 2020-04-03T20:54:29 |
356492
|
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|
Stack Exchange
|
How to find the associated conservation law from a given symmetry
It is a very well-known fact that any conservation law associated with some given PDE has an associated invariance (by Noether's Theorem). However, it is completely mysterious for me how to compute/derive these conservation laws just by knowing the invariances of the equation. For example, the one-dimensional nonlinear wave equation $$
u_{tt}-u_{xx}+f(u)=0, \qquad (t,x)\in\mathbb{R}\times\mathbb{R},
$$
is invariant under space translations. On the other hand, it is "well-known" that associated to this space translation invariance is the momentum conservation of the equation, that is, $$
P(u,v)(t):=\int_{\mathbb{R}} u_x(t,x)v(t,x)dx=\int_{\mathbb{R}}u_{0,x}(x)v_0(x)dx=P(u,v)(0).
$$
Nevertheless, I have no idea how to derive this conservation law (generally speaking) just by knowing that the equation is invariant under space translations. What about time-translations for example, what is its associated conservation law? Please, don't misunderstand me, I do know how to explicitly derive the momentum conservation directly from the equation, what I would like to know is how to derive it from the space-translations invariance. Any hint suggested reading or answer is very welcome!
The equation you wrote is variational. So you can use the general formalism there to make the computation starting from an appropriate Lagrangian. (Here the action is $$-u_t^2 + u_x^2 + F(u)$$ where $F$ is an antiderivative of the scalar function $f$.) Following https://en.wikipedia.org/wiki/Noether%27s_theorem#Field_theory_version you can get a conserved current $j$. Integrating $j^0$ along ${t = const}$ gives you the time-independent conserved quantity.
Conceptually, Noether's theorem is saying that a 1-parameter family of diffeomorphisms of the space F of fields induces a Hamiltonian flow on T^*F, and the Hamiltonian (Noether charge) is then conserved under that flow. I wrote a blog post about this a couple of years ago: http://jde27.uk/blog/noether.html - in particular the conservation of field momentum associated with space translation is worked out. Admittedly, I find it easier to think about going from the conserved quantity to the symmetry (from Hamiltonian to its flow) but that's just because differentiating is easier than integrating.
Time-translation invariance implies the conservation of energy.
I don't know if it'll help to answer your specific question, but I hope you won't mind my taking the excuse to share Baez's exposition Noether's theorem in a nutshell, which I've always found second to none on this subject.
This theorem is known to hold for variational problems in very broad generality, but I don't think there is a theorem of this type for nonvariational problems in any generality.
@BenMcKay Every conserved current is equivalent to one s.t. its conservation condition takes the form $\mathrm{d} j[\phi] = \xi \cdot E[\phi]$ (no derivatives of the equations $E[\phi]=0$). Applying the Helmhotz operator to both sizes gives $H(\xi \cdot E[\phi])=0$, implying differential conditions on $\xi$. On appropriate equivalence classes, $j \leftrightarrow \xi$ is a bijection. When $E[\phi]=0$ is variational, the condition on $\xi$ is the same as being a symmetry of the Lagrangian.
I should have written "...applying the Euler-Lagrange operator to both sides...". The Euler-Lagrange operator starts off a sequence, in the same way that the exterior derivative starts off the de Rham sequence, where the Helmholtz operator is next in the sequence.
To be blunt, the answer to your question is Noether's theorem (often precised as Noether's first theorem). So, essentially you already knew the answer to your own question.
However, the other answers are missing a degree of pragmatism. The calculation of the conserved current, once you know the Lagrangian and the symmetry is straightforward and mechanical. Namely, suppose you have a Lagrangian density $L[\phi] = L(x,\phi(x),\partial \phi(x), \partial^2\phi(x), \ldots)$, which depends your dynamical field $\phi(x)$. The variational principle will be $S(\phi) = \int L[\phi] \, \mathrm{d}x$, where $\mathrm{d}x$ is the coordinate volume form.1 An infinitesimal local field transformation $\phi^a \mapsto \phi^a + \delta_{\xi}\phi^a$ is allowed to be coordinate and field dependent, $\delta_\xi \phi^a = \xi^a[\phi] = \xi^a(x,\phi(x), \partial \phi(x), \partial^2 \phi(x), \ldots)$, and commutes with coordinate derivatives, namely $\delta_\xi \partial^n \phi^a = \partial^n (\delta_\xi \phi^a) = \partial^n \xi^a[\phi]$ for any $n\ge 0$. The example of time translation $\xi^a[\phi] = \frac{\partial}{\partial t} \phi^a$ is illustrative.
Such a local field transformation is a symmetry of the Lagrangian when its variation vanishes modulo a total divergence, $\delta_\xi L[\phi] = \partial_i J_0^i[\phi]$. The next step is a bit unintuitive, but it makes the calculation of the conserved current mechanical. Consider now the variation $\delta_{\varepsilon \xi}$, where $\varepsilon = \varepsilon(x)$ is an arbitrary function of the coordinates $x^i$. Using integration by parts, we can put the variation of the Lagrangian into the form
$$ \tag{$*$}
\delta_{\varepsilon \xi} L[\phi]
= \varepsilon\partial_i J^i_0[\phi] + (\partial_i\varepsilon) J^i_1[\phi] + \partial_i(-)^i .
$$
The leading term has to agree with $\delta_\xi L[\phi]$ when we set $\varepsilon \equiv 1$. The desired conserved current corresponding to $\xi$ is
$$ J_\xi^i[\phi] = J_0^i[\phi] - J_1^i[\phi] . $$
You can get the current in one step if you use integration by parts to directly put the variation of the Lagrangian into the form $\delta_{\varepsilon \xi} L[\phi] = -J_\xi^i[\phi] (\partial_i \varepsilon) + \partial_i(-)^i$, which is a formula that can be found in some physics textbooks on QFT.
The proof of Noether's theorem in this form is also straightforward (and a reshuffling of the standard proof). It only relies on the usual lemma that any density $N[\varepsilon, \ldots]$ that linearly depends on an arbitrary function $\varepsilon = \varepsilon(x)$ (and possibly any other fields) has a unique representative modulo total divergence terms, namely $N[\varepsilon, \ldots] = \varepsilon N_0 + \partial_i(-)^i$, with $N_0$ unique. The Euler-Lagrange equations $E_a[\phi]=0$ are defined by the identity $\delta_\xi = \xi^a E_a[\phi] + \partial_i(-)^i$ for arbitrary $\xi$. So, when $\xi$ is a symmetry, using $(*)$ and one more integration by parts, we find the identity
$$
\delta_{\varepsilon \xi} L[\phi]
= \varepsilon \xi^a E_a[\phi] + \partial_i(-)^i
= \varepsilon \partial_i J^i_\xi[\phi] + \partial_i(-)^i ,
$$
which implies that $\partial_i J^i_\xi[\phi] = \xi^a E_a[\phi]$, which vanishes when $E_a[\phi] = 0$. In other words, $J^i_\xi[\phi]$ is a conserved current.
1 If you change the independent coordinates $x^i$, the Lagrangian will change by the appropriate Jacobian. Working with differential forms allows you to keep everything more manifestly invariant.
You can find an overview of methods to obtain conservation laws from a wave equation in On the structure of conservation laws of (3+1)-dimensional wave equation. Noether's method requires that the PDE follows from a variational principle for a Lagrangian (as pointed out by Willie Wong). A direct algorithmic method to obtain conservation laws from a PDE without variational structure is described in the cited paper.
It's been a while since I've thought about this stuff so take this with a grain of salt. Also, I'm only familiar with this in the context of a finite dimensional phase space, whereas the phase space is infinite-dimensional in the example you give and I'm not sure what extra subtleties that introduces. But the finite-dimensional case might still provide some useful intuition.
Those two caveats aside, I hope the following is of some use:
We can define a canonical symplectic form on phase space in terms of the Lagrangian. The symplectic form gives us a bijective correspondence between tangent vector fields and 1-forms (it works the same as with Riemannian manifolds, the key is just that we have a perfect pairing on tangent spaces). We also have a Poisson bracket operation {A, B} between scalar fields A and B. {A, B} is the Lie derivative of B along the tangent vector field corresponding to the exterior derivative of A (obtained using the correspondence between 1-forms and tangent vector fields provided by the symplectic form).
A continuous symmetry is a flow that preserves the Hamiltonian H and the symplectic form. To a continuous symmetry, we can associate a unique tangent vector field that generates it, which corresponds to a 1-form (which happens to be closed). We can then integrate that 1-form to get a scalar field, which I will call S. We have that {S, H} = 0, this basically says that the vector field corresponding to S generates a symmetry. But the Poisson bracket is anticommutative, so {H, S} = 0, implying that S is a conserved quantity (since the vector field corresponding to H generates the time-evolution flow). Therefore, continuous symmetries correspond to conserved quantities.
|
2025-03-21T14:48:30.215513
| 2020-04-03T20:55:58 |
356493
|
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|
Stack Exchange
|
Which $p$-adic valuations of Weil numbers (that is, eigenvalues of Frobenius) are possible?
Let $C$ be a smooth projective curve over a finite field $\mathbb F_q$, $q$ is a power of the characteristic $p$. It is well-known that if $\alpha$ is an eigenvalue of Frobenius acting on $H^1_{et}(C,\mathbb Q_{\ell})$ (that is, on $\ell$-adic etale cohomology, where $\ell$ is a prime distinct from $p$) then $\alpha$ is an algebraic integer and $\alpha \overline \alpha=q$ (i.e., $\alpha$ is a Weil number).
My question is: what can one say about the maximal rational $v$ such that $\alpha\cdot q^{-v}$ is an algebraic integer (certainly, this can be re-formulated in terms of valuations of $\alpha$ at the primes lying above $p$; hence a rational maximum $v$ exists indeed). Can $v$ be distinct from $0$ and $1/2$?
Actually, I am interested about the generalization of this question to all cohomology spaces of arbitrary smooth projective varieties over $\mathbb F_q$; yet it appears that Honda's "Isogeny classes of abelian varieties over finite fields" (along with the Rieman Hypothesis over finite fields and other properties of etale cohomology) reduce this general question to the case of curves.
Every value $c \in \big[0,\tfrac{1}{2}\big] \cap \mathbf Q$ can occur as the smallest slope of an abelian variety over $\mathbf F_q$; see the corollary below.
What Honda actually proves [Hon68] (see [Mil94, Prop. 2.6] for a motivic reinterpretation) is:
Theorem (Honda). Let $q$ be a power of a prime $p$. Then the map
\begin{align*}
\frac{\{\text{simple abelian varieties } A \text{ over } \mathbf F_q\}}{\{\mathbf F_q\text{-isogeny}\}} &\to \frac{\{q\text{-Weil numbers}\}}{\text{conjugacy}}\\
A &\mapsto \operatorname{Frob}_A
\end{align*}
is a bijection.
In other words, every $q$-Weil number (of weight $1$) is realised inside some abelian variety; in particular inside some smooth projective curve (cut by smooth hyperplanes).
So the only question is: what are the $q$-Weil numbers? This is a purely number theoretic question. If you only want to know the valuation¹ of $\alpha$ and all its conjugates, it is not so hard to come up with examples.
Lemma. Let $a, b \in \mathbf Z$ be coprime with $0\leq a \leq \tfrac{b}{2}$, and let $\alpha$ be a root of
$$f(x) = x^{2b} + q^ax^b + q^b.$$
Then $\alpha$ is a $q$-Weil number with slopes
$$\big\{\underbrace{\tfrac{a}{b},\ldots,\tfrac{a}{b}}_b,\underbrace{\tfrac{b-a}{b},\ldots,\tfrac{b-a}{b}}_b\big\}.$$
Proof. If $g(x) = x^2 + q^ax + q^b$, then $\beta = \alpha^b$ is a root of $g$. Note that $g$ is irreducible over $\mathbf Q$ (even over $\mathbf R$) since
$$\Delta = q^{2a} - 4q^b < 0.$$
Hence $\beta\bar\beta = q^b$ (the constant term of $g$), so $\alpha\bar\alpha = q$. Clearly the Newton polygon of $f$ has slopes $\tfrac{a}{b}$ and $\tfrac{b-a}{a}$, both with multiplicity $b$. $\square$
In fact it's not hard to see that $f$ is irreducible when $q = p$, using Newton polygons and irreducibility of $g$, but we don't need this.
Corollary. For every $c \in \big[0,\tfrac{1}{2}\big] \cap \mathbf Q$, there exists a simple abelian variety $A$ over $\mathbf F_q$ such that the smallest slope of $H^1_{\operatorname{\acute et}}(A_{\bar{\mathbf F}_q},\mathbf Q_\ell)$ is $c$.
Proof. Write $c = \frac{a}{b}$ with $a$ and $b$ coprime, take $\alpha$ as in the lemma, and apply Honda's theorem to get a simple abelian variety over $\mathbf F_q$ with slopes the conjugates of $\alpha$. $\square$
Remark. The question which precise set of slopes can occur on smooth projective curves is a very difficult one, and this is an active area of study. For example, it is not known if for every $(g,p)$ there exists a curve $C$ of genus $g$ in characteristic $p$ such that all slopes are $\tfrac{1}{2}$ (i.e. $C$ is supersingular). If you only care about one eigenvalue and its conjugates, you can reduce to the answer above.
¹ We choose a prime above $p$ (equivalently, pick an embedding $\bar{\mathbf Q} \hookrightarrow \bar{\mathbf Q}_p$) and normalise the valuation so that $v(q) = 1$. Then studying the valuations of $\alpha$ at primes above $p$ has become studying the valuation of the conjugates of $\alpha$, which are called the slopes of $\alpha$.
References.
[Hon68] T. Honda, Isogeny classes of abelian varieties over finite fields. J. Math. Soc. Japan 20, p. 83-95 (1968). ZBL0203.53302.
[Mil94] J. S. Milne, Motives over finite fields. Motives (Seattle, WA). Proc. Symp. Pure Math. 55.1, p. 401-459 (1994). ZBL0811.14018.
Thank you very much! I was actually interested in the effectivity of motives (and their tensor products). It appears that in characteristic $0$ only "integral effectivity" is possible, whereas the effectivity of motives in characteristic $p$ can be "fractional".
@MikhailBondarko: ah, interesting. I think I agree with you, and I had never thought of that!
|
2025-03-21T14:48:30.215839
| 2020-04-03T21:05:06 |
356494
|
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|
Stack Exchange
|
Spherical objects and K-theory
My question goes as follows: given a ring $R$ (with $1\neq 0$). Define $\mathbf{Perf}_{R}$ the the category of Prefect complexes over $R$. I want to prove that the Waldhausen $K$-theory of the category of perfect complexes is the same as the $K$-theory of $\mathbf{Proj}_{R}$ the category of finitely projective $R$-modules, i.e. $$ K(\mathbf{Perf}_{R})\sim K(\mathbf{Proj}_{R})$$ The proof that I am looking for is the one using the notion of $n$-spherical objects developed by Wladhausen in his famous article "Algebraic K-theory of spaces " (section 1.7 if I remember correctly).
Lets me try to adapt the Waldhausen theorem in the algebraic context. Let say we have a ring $R$. And Let $\mathbf{C}$ be an essentially small full subcategory of the category chain complexes over $R$. Such that $\mathbf{C}$ is a Waldausen category where weak equivalences are quasi-isomorphisms and cofibrations are those coming from the (projective) model structure on the category of chain complexes over $R$.
Definition $B$ is an $n$-spherical object in $\mathbf{C}$, if the homology $H_\ast (B)$ is concentrated in degree $n$. Lets denote by $\mathbf{C}^{n}$ the category of $n$-spherical objects in $\mathbf{C}$. If I an not wrong $\mathbf{C}^{n}$ is a waldhausen category where weak equivalences are quasi-iso and cofibrations are ordinary cofibrations such that the cofiber is also an object in $\mathbf{C}^{n}$.
Now the Wladhausen theorem says that $hocolim_{n}K(\mathbf{C}^{n})\sim K(\mathbf{C})$.
Since I'm not sure If my understanding of the theorem is corrected, I wanted to ask experts for some clarification. In case my interpretation of the theorem is completely wrong, I would be happy to get some help. Thank you!
I believe the theorem you are trying to prove is due to Brinkmann who was an early student of Waldhausen.
What you are suggesting as the proof is only part of the story.
In addition to section 1.7, you will also need to apply section 1.8, which is about split cofibrations.
There is a difference between strictly spherical objects (i.e., complexes of projectives which are concentrated in a single degree) and spherical objects in your sense (in the sense of homology). You need to show that the K-theory of these coincide.
Fair enough! So in the general version $K(\mathbf{C}^n)\sim K(\mathbf{C}{strict}^n)$ where $\mathbf{C}{strict}^n$ is the category of strict $n$-spherical objects in $\mathbf{C}$?
Under suitable assumptions yes. The assumption is that every cofibration sequence must split, I believe.
|
2025-03-21T14:48:30.216028
| 2020-04-03T21:31:43 |
356499
|
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"url": "https://mathoverflow.net/questions/356499"
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|
Stack Exchange
|
Elementary but intractable minimum
Let $q$ be a power of a prime $p$ and $f\in\mathbb N^*$. For $n\in\mathbb N$, denote by $\overline n$ the residue of $n$ modulo $q^f$ in $\{0,\cdots, q^f-1\}$. Let $m,\alpha$ be elements of $\mathbb N^*$.
Can one show that the quantity
$$\sum_{i=mq^f}^{(m+\alpha)q^f-1}q^{l_i+\overline{i-l_i}}$$
admits one and only one minimum (that one will determine) when $(l_i)$ describes the set of sequences of distincts integers in $\{0,\cdots,\alpha q^m-1\}$?
Thanks in advance.
|
2025-03-21T14:48:30.216102
| 2020-04-03T21:54:17 |
356502
|
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"LSpice",
"Lior Silberman",
"Monty",
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"https://mathoverflow.net/users/29422",
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"paul garrett"
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|
Stack Exchange
|
Left translation of automorphic form satisfies $K$-finiteness?
Does a left translation of an automorphic form satisfy left $K$-finiteness?
Let $F$ be a number field and $G$ is an algebraic group. Let $\phi$ is an automorphic form on $G$. Let $K$ be a maximal compact subgroup of $G(A)$.
Then $\phi$ has $K$-finiteness.
For arbitrary $x$, let $\phi_x(g)=\phi(xg)$. Then I am wondering whether $\phi_x$ has also $K$-finiteness.
Decompose $K=K_{\infty}K^{\infty}$. I checked $K^{\infty}$-finiteness of $\phi_x$ but cannot check $K_{\infty}$-finiteness.
Does this hold? If so, how can we prove it?
Though "left translation" appears in the title, it does not appear in the question, nor do you give a good context. Please amplify?
@paul, oh I am very sorry. I edited my question. Would you see it again? Thank you!
I think, but am not sure, that the usual term is "is $K$-finite", not "has $K$-finiteness". Anyway, doesn't it just mean that the set of $K$-translates lies in a finite-dimensional space? Then testing the $K$-finiteness of $\phi_x$ is just testing the $x^{-1}K x$-finiteness of $\phi$, which we have already established.
@LSpice, since $\phi_x$ is the left translate of $\phi$, I think it is not of the $x^{-1}Kx$-finiteness.
Is $\phi$ $K$-finite on the left? Or right?
@Subhajit, Usually K-finite means about right K translation.
Whether you want my answer or Paul Garrett's depends on whether the full group is acting from the same side as the compact subgroup, or from the opposite side.
Presumable (in an automorphic forms context with contemporary left-right conventions) you mean that $\varphi$ is right $K$-finite. This could apply to any (complex-valued) function $\varphi$ on a topological group $G$, with compact subgroup $K$. It means that the space of functions obtained by right translation $R_k$ by $k\in K$ is finite-dimensional. These are functions $R_k\varphi(g)=\varphi(gk)$.
Left translation $T_x$ by $x$ stabilizes that collection of functions, because left and right translation commute: $T_xR_k\varphi=R_kT_x\varphi$.
I am very sorry for late reply and always thank you very much! I am learning much from your answers.
Dear Prof. @Paul, if you are fine, would you take a look my another question related to Harish-Chandra isomorphism? Thank you very much! https://mathoverflow.net/questions/357661/is-there-a-functorial-relation-of-center-of-universal-enveloping-algebra
The answer is yes at the finite places, but no at the infinite place.
Because each finite-dimensional complex representation of a totally disconnected group factors through a finite quotient, a vector is $K_\textrm{f}$-finite iff its stabilizer in $G(\mathbb{A}_\textrm{f})$ is open in that group, and this condition is obviously invariant under right-translation by $G(\mathbb{A}_\textrm{f})$.
On the other hand, at infinite places this is false: translating a $K_\infty$-finite vector by an element of $G(F_\infty)$ will generically give a non-$K_\infty$-finite vector. For example consider the action of $\mathrm{SL}_2(\mathbb{R})$ on $\left\{ f\colon\mathbb{R}^2\setminus{0}\to \mathbb{C} \vert f(rx) = r^{\frac12+it}f(x) \right\}$. A vector is $\mathrm{SO}(2)$-finite iff its restriction to the circle is a trigonometric polynomial. But it is easy to check that for essentially any $g\in\mathrm{SL}_2(\mathbb{R})\setminus\mathrm{SO}(2)$, if $f(x)$ is a non-constant trigonometric polynomial then $f(xg)$ isn't.
This is one of the motivation for passing to $(\mathfrak{g},K_\infty)$-modules: since $\mathfrak{g}$ is a finite-dimensional representation of $K_\infty$, its action on smooth vectors preserves $K$-finiteness.
I am sorry for late reply but thank you very much for your kind reply. Since I didn’ t know the connection of modular form and automorphic form, I studied modular form for a while to understand your answer. But I couldn’t understand your last comment. Why the finite dimension of $\mathfrak{g}$ guarantee the $\mathfrak{g}$-action preserves the $K$-finiteness?
Let $v$ be a $K$-finite smooth vector in a representation $V$ of a Lie group $G$. Then $v$ lies in a $K$-invariant subspace $U$. If $X\in\frak{g}$ then $Xv$ lies in the image of a map $\frak{g}\otimes U\to V$, whose image is a f.d. rep. of $K$ since its domain is the tensor product of two such representations.
|
2025-03-21T14:48:30.216393
| 2020-04-03T22:15:23 |
356503
|
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|
Stack Exchange
|
Relative completeness of a relative cocompletion of a subcategory
I'm going to use the language from Lack and Rosicky's Notions of Lawvere theory, but I won't be touching on actual enriched category theory.
Suppose I have a category $\mathbb{C}$ with a class of limits and colimits, $\Phi, \Psi$ respectively, that commute with each other. Now suppose I have a subcategory $\mathbb{C}'$ that is closed to $\Phi$-limits, are there any conditions under which the $\Psi$-cocompletion of $\mathbb{C}'$ in $\mathbb{C}$ will be $\Phi$-complete?
|
2025-03-21T14:48:30.216475
| 2020-04-03T23:16:55 |
356505
|
{
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"Gael Meigniez",
"J. Doee",
"abx",
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|
Stack Exchange
|
vector field on the torus
Is the following statement true?
Let $T$ be diffeomorphic to the solid torus. Let $v$ be a vector field such that $v$ and $curl(v)$ are both tangent to $\partial T$ everywhere and $|v|$ is constant on $\partial T$. Then $v$ is zero vector field.
What is $curl(v)$? What is $|v|$?
the metric is euclidean and $curl(v)= \nabla \times u$
the question does not seem to be well written. Since you made no hypothesis at all for $v$ in the interior of $T$, but only on the boundary, you cannot conclude in the interior either. For example, any smooth vector field identically zero on a small neighborhood of $\partial T$ but not identically zero in $T$ is a counterexample.
The answer is no.
The vector field $\nabla\times \vec{v}$ is tangent to $\partial T$ iff for any region $\Sigma\subset\partial T$ bounded by the Jordan curve $\partial\Sigma$ the flux integral $\iint_{\Sigma}(\nabla\times \vec{v}).\vec{n}\,{\rm{d}}S$ is zero; this is because the tangency to $\partial T$ means orthogonality to the unit normal vector $\vec{n}$ at any point of $\partial T$. Stokes' Theorem now implies that $\oint_{\,\partial\Sigma}\vec{v}.{\rm{d}}\vec{r}$ is zero. Thus the line integral of $\vec{v}$ along any simple closed curve that encloses a region in $\partial T$ (i.e. is a boundary) must be zero. This provides a reformulation of the condition on $\nabla\times \vec{v}$: Denoting the differential $1$-form corresponding to $\vec{v}$ by $\omega$, the $2$-form ${\rm{d}}\omega|_{\partial T}=d\left(\omega|_{\partial T}\right)$ vanishes.
Now I construct a counter-example: Write the solid torus $T$ as $\overline{\Bbb{D}}\times\Bbb{S}^1$ where $\overline{\Bbb{D}}$ with $\Bbb{S}^1$ and are the unit closed disk and the unit circle in $\Bbb{R}^2$. Equip $T$ with the Euclidean norm induced from $\Bbb{R}^2\times\Bbb{R}^2$. The vector field
$-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}$ on $\overline{\Bbb{D}}$ could be thought of as a vector field on $\overline{\Bbb{D}}\times\Bbb{S}^1$ which is independent of the third coordinate. I claim that it works as $\vec{v}$. On
$\partial T=\Bbb{S}^1\times\Bbb{S}^1$ the vector field $\vec{v}$ is given by
$$
-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}|_{\Bbb{S}^1}=\frac{\partial }{\partial \theta}
$$
which is tangent to the first component and is of length one. It remains to check the tangency condition imposed on $\nabla\times \vec{v}$. The $1$-form $\omega$ corresponding to $\vec{v}$ is the pullback of the $1$-form $-y{\rm{d}}x+x{\rm{d}}y$ on $\overline{\Bbb{D}}$ via the projection
$p_1:T=\overline{\Bbb{D}}\times\Bbb{S}^1\rightarrow\overline{\Bbb{D}}$. Restricted to the boundary
$\partial T=\partial\overline{\Bbb{D}}\times\Bbb{S}^1=\Bbb{S}^1\times\Bbb{S}^1$, $\omega|_{\partial T}$ is the pullback of the closed $1$-form ${\rm{d}}\theta$ via
$p_1:\partial T=\Bbb{S}^1\times\Bbb{S}^1\rightarrow\Bbb{S}^1$. Consequently, $\omega|_{\partial T}$ is closed as well.
Thank-you very much for your answer. If I understand correctly, your counterexample is the unit vector field in the poloidal direction. What if one furthermore requires that the flow of the vector field generate $T$, namely that evolving a crosssection of $T$ by the flow of $v$ sweeps out all of $T$? An example of such a field would be the toroidal vector field, $e_\phi$ whose curl is in the $\hat{z}$ direction and is not tangent to the torus except on two circles.
|
2025-03-21T14:48:30.216728
| 2020-04-03T23:16:56 |
356506
|
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Guest",
"KhashF",
"Nathaniel Johnston",
"https://mathoverflow.net/users/11236",
"https://mathoverflow.net/users/128556",
"https://mathoverflow.net/users/149682"
],
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"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:627776",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/356506"
}
|
Stack Exchange
|
Commuting matrices of complex functions
If $A(z) :=[A_{ij}(z)] $ and $B(z) :=[B_{ij}(z)] $ are two invertible $n\times n$ matrices of entire complex valued functions entries $A_{ij}(z)$, and $B_{ij}(z) $ with
(1). $AA^{\#}=A^{\#}A$ where $A^{\#}(z)=\left(\overline{A(\bar{z})} \right)^{T}$.
(2). $BB^{\#}=B^{\#}B$
(3).$ AB=BA$
Is it true that
$$B(A A^{\#}) =(AA^{\#}) B$$.
Thanks in advance.
Yes, this is true, and it has nothing to do with entire functions. It's just the fact that if $A$ and $B$ are normal matrices (your conditions (1) and (2)) that commute (condition (3)) then they are simultaneously orthonormally diagonalizable, so each of $A$, $B$, $A^{#}$, and $B^{#}$ commute with each other.
@NathanielJohnston I thought the same first. But notice that the $A^#(z_0)$ is the Hermitian adjoint of $A(\bar{z_0})$ not $A(z_0)$.
Let $z_0\in\Bbb{R}$ be arbitrary. The matrices $A_0:=A(z_0)$ and $B_0:=B(z_0)$ are normal; in view of $A_0^*=A^\#(\bar{z_0})=A^\#(z_0)$ and $B_0^*=B^\#(\bar{z_0})=B^\#(z_0)$ they commute with their Hermitian adjoints. They also commute with each other. So $A_0$ and $B_0$ could be simultaneously diagonalized by a unitary matrix. That matrix also diagonalizes $A_0^*$ and $B_0^*$. So the four matrices $A_0=A(z_0),B_0=B(z_0),A_0^*=A^\#(z_0),B_0^*=B^\#(z_0)$ could be simultaneously diagonalized which implies that they all commute with each other. So $B(AA^\#)=(AA^\#)B$ at any point of the real line. Because the entries are entire functions, they coincide throughout the whole complex plane.
Added: As mentioned by @Guest and @MarkSapir, $A^\#(z)$ (respectively
$B^\#(z)$) is the Hermitian conjugate of $A(\bar{z})$ (resp. $B(\bar{z})$) rather than that of $A(z)$ (resp. $B(z)$). Indeed, the Hermitian conjugates $A^*(z):=(\overline{A(z)})^{\rm{T}}$ and $B^*(z):=(\overline{B(z)})^{\rm{T}}$ do not vary holomorphically with $z$; they are anti-holomorphic. On the contrary, $A^\#(z)$ and $B^\#(z)$ are holomorphic because in their definitions the complex conjugation is applied both in the domain and the range. To circumvent this difficulty, one notices that $A^\#(z)$ (respectively
$B^\#(z)$) is indeed the Hermitian conjugate of $A(z)$ (resp. $B(z)$) if $z$ is real. So by a linear algebra argument, one can establish $B(z)\left(A(z)A^\#(z)\right)=\left(A(z)A^\#(z)\right)B(z)$ for $z\in\Bbb{R}$. This persists on the rest of $\Bbb{C}$ because the entries of these matrices are entire functions (holomorphic on the whole $\Bbb{C}$).
$A^{}$ is not the same as $A^{#} $. $A^{#} (z) =A^{} (\bar{ z}) $.
@Guest You are right, I edit my answer shortly.
@MarkSapir Can you point out what's the mistake? I think $A(z)$ and $B(z)$ are normal once $z$ is real. I am only using the normality over the real line.
@MarkSapir I think "entire" means analytic on $\Bbb{C}$: https://en.wikipedia.org/wiki/Entire_function
@KhashF Do you mean that if $A(z) $ (with entire entries) is a normal matrix on the real line it will be normal on the whole complex plane?
@Guest I am saying if $z$ is real the matrices are normal and you have the desired equality. The equality persists throughout $\Bbb{C}$ due to Identity Theorem (https://en.wikipedia.org/wiki/Identity_theorem).
@MarkSapir I don't know what being "normal on the real line" means. Are you talking about a single matrix or a matrix-valued function?
@MarkSapir $A^#(z)=(\overline{A(\bar{z})})^{\rm{T}}$ is holomorphic since you are conjugating twice (Schwarz Reflection); but $(A(z))^*=(\overline{A(z)})^{\rm{T}}$ is an anti-holomorphic function of $z$.
@Mark Yes. I hope so.
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