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2025-03-21T14:48:31.455314
| 2020-07-09T12:06:21 |
365225
|
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"authors": [
"Mushu Nrek",
"Nik Weaver",
"S.Surace",
"https://mathoverflow.net/users/160649",
"https://mathoverflow.net/users/23141",
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}
|
Stack Exchange
|
Integrate Radon-Nikodým derivatives against Lebesgue measure
I am struggling for quite some time, because of a problem involving Radon-Nikodým derivatives. I will try to describe the main features and perhaps somebody has an idea how to solve it.
I consider two sequences of measures on some compact set of $\mathbb{R}^2$: $\Xi_n$ and $\Lambda_n$ which are absolutely continuous in the sense $\Xi_n \ll \Lambda_n \ll \mathrm{Leb}$. I know that $\Lambda_n$ converges to some measure $\Lambda$ which is not absolutely continuous wrt. to Lebesgue measure anymore. (I suspect that it has a non-zero absolutely continuous part, but I am not quite sure how to prove this.) Furthermore, $\Xi_n$ also converges to some measure $\Xi$ satisfying $\Xi \ll \Lambda$.
The weird thing is that I need to integrate $\dfrac{d\Xi}{d\Lambda}$ against Lebsegue measure. In the prelimit case, there is no problem, because $\Xi_n$ and $\Lambda_n$ both have densities $\xi_n > 0$ and $\lambda_n > 0$ with respect to $\mathrm{Leb}$, so that simply
$$
\dfrac{d\Xi_n}{d\Lambda_n} = \dfrac{\xi_n}{\lambda_n} \quad \mathrm{Leb}-\text{a.e.}\quad\text{(because $\lambda_n > 0$)}
$$
Finally, $0\leq \dfrac{\xi_n}{\lambda_n}\leq 1$ so that the measure given by $d\kappa_n(x) := \dfrac{\xi_n(x)}{\lambda_n(x)}\cdot d\mathrm{Leb}(x)$ so that $(\kappa_n)$ is tight. For the moment, suppose that there is only one limiting point, i.e. $\kappa_n \to \kappa$, where $\kappa$ is some finite measure. I am wondering if one could conclude that
$$
d\kappa(x) = \dfrac{d\Xi}{d\Lambda}(x)\cdot d\mathrm{Leb}.
$$
In particular that would imply that $\Xi$ and $\Lambda$ have a non trivial absolutely continuous part. (Note that we may assume that $\mathrm{Leb}\ll\Lambda$ so that the Radon-Nikodým derivative is defined $\mathrm{Leb}$-a.e.)
I am not deep into measure theory, so I am really struggling with this problem. Especially because I find it strange to integrate the Radon-Nikodým derivative of $\Xi$ wrt. to $\Lambda$ against Lebesgue measure. I would be glad for any ideas or references that might go into this direction.
To give a larger context: I am working with a sequence $\Lambda_n(t)$ of measure-valued stochastic process that converges to a white noise process $\Lambda(t)$. And now I am looking at a related process $\Xi_n(t)$ and want to understand its limiting behaviour. So I would also be very glad about any ideas or references wrt. absolutely continuous parts of random fields and random measures wrt. to Lebesgue measure.
Edit: The convergence takes place at least in the distributional sense. I think I can also get it for Lipschitz continuous functions so that it is weak convergence.
Edit2: Initially, the integral
$$
\int \dfrac{\xi_n}{\lambda_n}(x) \phi_n(x) dx
$$
for some sequence of continuous (or smooth if you like) functions converging pointwise to some continuous (or smooth) limit phi appears in the problem. Since I could not figure out what will happen, I translated it towards the Radon-Nikodym derivative, hoping that since I require only weak convergence, the measure theoretic setting would be more helpful.
I'm curious: what do you need to integrate $d\Xi/d\Lambda$ for? Either I missed it or you left it out on purpose. Note that $d\Xi/d\Lambda$ is only defined a.s. up to $\Lambda$-nullsets, so you may obtain different results when integrating against Lebesgue measure depending on which version you choose.
In what sense does $\Lambda_n$ "converge" to $\Lambda$?
@S.Surace The integral with respect to Lebesgue measure appears in the infinitesimal generator of the process I am considering. Since $\Lambda$ can be decomposed into an absolutely continuous part and a singular part wrt. Lebesgue measure, I was thinking that a) for the absolutely continuous part, there is no problem and b) the singular part only changes things on a Lebesgue-nullset, so that different versions of this derivative shouldn't yield different results...
@NikWeaver I edited my post. It is convergence in the distributional sense or weak convergence of measures.
You still have the problem that $d\Xi/d\Lambda$ is defined only up to $\Lambda$-nullsets, giving potentially very different results when integrated against Lebesgue measure (some versions are not even integrable). For example, take $\Xi_n$ and $\Lambda_n$ each to be a weighted sum of a Gaussian centered at $0$ with variance $1/n$, and a uniform distribution on $[1,2]$. Then $d\Xi/d\Lambda$ may take arbitrary values on $(-\infty,0)\cup(0,1)\cup(2,\infty)$, which is a $\Lambda$ but not a Lebesgue nullset. Maybe try to multiply by indicator of $\text{supp}(\Lambda)$.
@S.Surace This is true in general, but in this case, Lebesgue measure is absolutely continuous wrt $\Lambda$. And that means that the derivative is defined Lebesgue-a.e. It is similarly to the non limiting case, where $\lambda > 0$.
I'm confused. The question says that the limit of $\Lambda_n$ is not absolutely continuous wrt. Lebesgue measure. But the issue is not this. The problem arises even in the absolutely continuous case when $\lambda(x)=0$ for $x$ in some set of positive Lebesgue measure. There is no problem for $d\Xi/d\Lambda$ as that is defined only up to $\Lambda$-nullsets, but since you want to integrate this wrt. Lebesgue measure, you need to fix a version which is Lebesgue integrable, e.g. by setting it to zero on $\Lambda$-nullsets. This concerns your conjecture about $\kappa$ (2nd display equation).
But only you can decide which one is the correct version.
Exactly, but even if $\Lambda$ is not absolutely continuous wrt. Lebesgue, it is possible that Lebesgue is absolutely continuous wrt. $\Lambda$. And in this case there is no problem. See, in the prelimiting case, Lebesgue is also absolutely continuous wrt. $\Lambda_n$, because $\lambda_n > 0$, so that the density $1/\lambda_n(x)$ is well-defined. In the limit, I do not have any representation anymore, but I still have that Lebesgue is absolutely continuous wrt. $\Lambda$. I tried to clarify this is my edit. @S.Surace
There is no reason for the limit measure $\kappa$ to be related in any way to the limit measures $\Xi$ and $\Lambda$ (and, in particular, to their Radon-Nikodym derivative).
More precisely, if your sequences $\Xi_n,\Lambda_n$ on a compact $X\subset\mathbb R^2$ are such that $\Lambda=\lim\Lambda_n$ is singular with respect to $\text{Leb}$, then for any prescribed measure $\kappa$ on $X$ there are sequences
$$
\Xi'_n\ll\Lambda'_n\ll \text{Leb}
$$ with
$$
\|\Xi_n-\Xi'_n\|, \|\Lambda_n-\Lambda'_n\|\to 0
$$
(so that, in particular, $\Xi'_n\to\Xi, \Lambda'_n\to\Lambda$), and such that the measures
$$
\kappa'_n = \frac{d\Xi'_n}{d\Lambda'_n}\,\text{Leb}
$$
weakly converge to $\kappa$. In fact, the presence of an ambient Euclidean space is completely irrelevant here, and instead of the Lebesgue measure one can talk about any reference measure on $X$.
The idea of the construction is very simple (I skip the details). Since $\lim\Lambda_n$ is singular with respect to the measure $\text{Leb}$, there are subsets $X_n\subset X$ with $\text{Leb}(X_n)\to\text{Leb}(X)$, whereas $\Xi(X_n),\Lambda_n(X_n)\to 0$. Fix a sequence $\epsilon_n\to 0$. Then the restrictions of $\Xi'_n$ and $\Lambda'_n$ to $X\setminus X_n$ are the multiples of $\Xi_n|_{X\setminus X_n}$ and $\Lambda_n|_{X\setminus X_n}$, respectively, chosen in such a way that
$\Xi'_n(X\setminus X_n)=\Lambda'_n(X\setminus X_n)=1-\epsilon_n$, whereas on $X_n$ the measures $\Xi'_n$ and $\Lambda'_n$ can be defined in such a way that the Lebesgue measure multiplied by their ratio converges to $\kappa$.
Thank you very much! I see that in general, there is no way of getting a positive result. But in my case, I know (strongly suspect) that Lebesgue measure is absolutely continuous wrt $\Lambda$. This means that $\Lambda$ and Lebesgue are not completely singular, and in particular, I don't see how in that case, we can find the sets $X_n$ you mention. It would be great if you could elaborate on that.
|
2025-03-21T14:48:31.455912
| 2020-07-09T12:33:56 |
365227
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/365227"
}
|
Stack Exchange
|
When is the Kochen-Stone inequality an equality?
The Kochen-Stone theorem says that if $A_n$ is sequence of events with $\sum_{i=1}^{\infty} P(A_i) = \infty$, then:
$$
P(A_n \mbox{ i.o.}) \ge \limsup_{n \to \infty} \frac{(\sum_{i=1}^nP(A_i))^2}{\sum_{i, j= 1}^nP(A_i \cap A_j)}
$$
I am interested in cases where the $A_n$ are not mutually independent, but where the inequality is actually an equality. Any hints or references to results or examples of this kind would be much appreciated.
In case it's of interest to others, I can now show that if $A_n$ is any increasing sequence of events then the Kochen-Stone inequality is an equality. This was enough for my application (I am looking at issues of computability and wanted to arrange for equality when the probabilities $P(A_n)$ form a Specker sequence).
To see why the Kochen-Stone inequality is an equality for increasing $A_n$, first note that in A Simple Proof of Two Generalized Borel-Cantelli Lemmas (Springer Lecture Notes in Mathematics vol. 1874), Yan showed that the diagonal terms in the sums in the Kochen-Stone inequality are negligible:
\begin{equation} \label{yan-eq}
\limsup_{n \to \infty} \frac{(\sum_{k=1}^n P[A_k])^2}{\sum_{i,k=1}^n P[A_i A_k]} = \limsup_{n \to \infty} \frac{\sum_{1 \le i < k \le n} P[A_i]P[A_k]}{\sum_{1 \le i < k \le n}P[A_iA_k]} \tag{1}
\end{equation}
Now assume $A_1 \subseteq A_2 \subseteq A_3 \subseteq \ldots$ and assume, for simplicity, that $P(A_1) > 0$. Put $q_n = P(A_n)$. Then $P(A_n \mbox{ i.o.}) = q$ where $q = \lim_{n \to \infty}q_n$. Noting that $P(A_iA_k) = P(A_i)$ when $i < k$, we can write the fraction on the right-hand side of equation (1) as $w_n = \frac{u_n}{v_n}$ where:
\begin{align}
u_n &= q_1\sum_{k= 2}^n q_k + q_2\sum_{k=3}^n q_k + \ldots + q_{n-1}q_n\\
v_n &= (n-1)q_1 + (n-2)q_2 + \ldots + q_{n-1}\\
\end{align}
and then the Kochen-Stone inequality is $q \ge \limsup_{n \to \infty} w_n$. I claim that, in fact, $w_n \to q$ as $n \to \infty$, so that $q = \lim_{n \to \infty} w_n = \limsup_{n \to \infty} w_n$. To see this, note that:
\begin{align}
\sum_{k=i}^nq_k &= q\left(n - i + 1 - \frac{1}{q}\sum_{k=i}^n (q - q_k) \right)
\end{align}
Let us write $\sigma_i^j$ for $\sum_{k=i}^j(q - q_k)$. From the above, multiplying by $q_{i-1}$ and summing for $i$ from $2$ to $n$, we have:
\begin{equation}
u_n = qv_n - \sum_{i=2}^{n} q_{i-1}\sigma_i^n
\end{equation}
Define the sequence $r_n$ by:
\begin{align}
r_n &= \frac{\sum_{i=2}^{n} q_{i-1}\sigma_i^n}{v_n} \tag{2}
% %
% &= \frac{q_1\left(\sum_{k=2}^{n}(q - q_k)\right) + q_2\left(\sum_{k=3}^{n}(q - q_k)\right) + \ldots + q_{n-1}(q - q_{n-1})}
% {q_1 (n-1) + q_2 (n-2) + \ldots + q_n}
\end{align}
I claim that $r_n \to 0$ as $n \to \infty$ so that $w_n = u_n/v_n = q - r_n \to q$, which is what we want to prove. To see this, given $\epsilon > 0$, put $\epsilon_0 = \frac{q_1}{4}\epsilon$. Then there is an $N$ such that for all $n > N$, $q - q_n < \epsilon_0$. Then for $n > N$, we have:
\begin{equation}
r_n = \frac{C}{v_n} + \frac{s_n}{v_n}
\end{equation}
where $C$ and $s_n$ are obtained by grouping the monomials $q_iq_j$ in the numerator of the fraction on the right-hand side of equation (2) so that $C$ comprises the $q_iq_j$ for which $i$ and $j$ are most $N$ and $s_n$ contains all the rest. Observing that $s_n$ contains at most $(n-1)(n - N)$ monomials all of which evaluate to at most $\epsilon_0$ and that $v_n > q_1((n-1) + (n - 2) + \ldots + 1)= q_1\frac{1}{2}(n-1)n$, we have:
\begin{equation}
r _n \le \frac{C}{v_n} + \frac{1}{q_1}\cdot\frac{(n-1)(n-N)\epsilon_0}{\frac{1}{2}(n-1)n} \to \frac{2}{q_1}\epsilon_0 = \frac{\epsilon}{2}\mbox{ as $n \to \infty$}
\end{equation}
So there is an $M > N$ such that for all $n > M$, $|r_n - \frac{\epsilon}{2}| < \frac{\epsilon}{2}$, but then $r_n < \epsilon$ and we are done.
|
2025-03-21T14:48:31.456134
| 2020-07-09T14:34:18 |
365233
|
{
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],
"authors": [
"Tyrone",
"https://mathoverflow.net/users/105652",
"https://mathoverflow.net/users/54788",
"nikola karabatic"
],
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"url": "https://mathoverflow.net/questions/365233"
}
|
Stack Exchange
|
Does the suspension spectrum functor preserve weak equivalences?
Let $\Sigma^{\infty}$ denote the suspension spectrum functor from pointed topological spaces (=CGWH spaces) to orthogonal spectra. As usual, a weak equivalence of spaces is a continuous map inducing a bijection on $\pi_0$ and an iso on all $\pi_i$, $i\ge 1$, for all choices of basepoint. For orthogonal spectra, I refer to the usual notion of stable equivalence, see e.g. Mandell-May-Shipley-Schwede.
Question: If $f\colon X\longrightarrow Y$ is a weak equivalence, is $\Sigma^\infty f\colon \Sigma^\infty X\longrightarrow \Sigma^\infty Y$ a stable equivalence?
Remarks: 1. This is true if $X$ and $Y$ are well-pointed, see this MO discussion.
In general, if $X$ and $Y$ are not well-pointed, then the reduced suspension does not preserve weak equivalences. For example, take the set $\{0\} \cup\{\frac 1n\}$ which receives a weak equivalence from a countable discrete set. The reduced suspension is a map from a wedge of spheres to the Hawaiian earrings which is not a weak equivalence.
Doesn't your counterexample in $2$ answer this in the negative if $X,Y$ are not well-pointed?
@Tyrone: it is not clear to me whether the map becomes a weak equivalence after suspending further
Enough of the homology of the Hawaiian earing is known and is not the homology of the wedge. In fact $\pi_n$ of the $n$-dimensional Hawaiian earing is known (and it is not even free abelian if I recall).
Hmm you are right about the homology, but I think this argument uses the suspension iso for reduced suspension and reduced homology implicitely and this does NOT hold for not-well-pointed spaces (since the reduced cone is not acyclic, at least I don't see why it should be).
Or did you intend to argue in another way?
What do you mean by acyclic here? If you don't want to use homology, then just use the computation of $\pi_n$ of the $n$-dimensional Hawaiian Earing (which the the $(n-1)$-fold reduced suspension of the Hawaiian Earing).
This sounds good, didn't know the computation about the n-dimensional HE
@Tyrone if you write this as an answer I will accept it :)
|
2025-03-21T14:48:31.456285
| 2020-07-09T14:53:10 |
365234
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Iosif Pinelis",
"Martin Sleziak",
"darij grinberg",
"https://mathoverflow.net/users/2530",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/8250"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/365234"
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|
Stack Exchange
|
Bijective proof of a combinatorial identity: $\sum\limits_{k=0}^n\binom nk^2 \binom k{n-m}=\binom nm \binom{n+m}m$
Identity
\begin{equation}
\sum_{k=0}^n\binom nk^2 \binom k{n-m}=\binom nm \binom{n+m}m \tag{1}
\end{equation}
was used in an answer here. As shown in that answer, (1) easily reduces to
\begin{align}
\sum_{k=0}^n\binom nk \binom m{n-k}=\binom{n+m}n, \tag{2}
\end{align}
which latter admits an obvious bijective proof.
Question: Is there a bijective proof of the original identity (1), without the reduction to (2)?
A possibly relevant reference is to automated search for bijective proofs.
First apply the trinomial revision formula (and symmetry of binomial coefficients) to get $\dbinom{n}{k}\dbinom{k}{n-m} = \dbinom{n}{m} \dbinom{m}{k-n+m}$. This lets you factor out $\dbinom{n}{m}$, and the sum turns into an easy Vandermonde convolution. All the steps can be made bijective, since they just use trinomial revision (in the case when everything is a nonnegative integer), symmetry of binomial coefficients and Vandermonde convolution. Thus you get a bijective proof by pasting together these bijections.
I have tried to put this identity into some search engines, specifically Approach0 and SearchOnMath. There is this post, where the answer gives a combinatorial proof: Some binomial coefficient identity
Proving $\sum_{m=0}^n\binom{n}{m}^2 \binom{m}{n-k}=\binom{n}{k}\binom{n+k}{k}$ The answers there point out that this is a special case of $\sum_{k\ge0}\binom ak\binom bk\binom kc=\binom ac\binom{a+b-c}a$.
@darijgrinberg : Thank you for your comment, which actually answers the question. Your identity is the same I used in the linked answer to reduce (1) to (2). However, I did not realize that it is an instance of the trinomial revision (and I did not even know that term, "trinomial revision").
@MartinSleziak : Thank you for your comments, one providing another bijective proof and the other providing a generalization of the identity. Also, I think you previously suggested this valuable resource, approach0, and I actually bookmarked it, but then forgot about it.
|
2025-03-21T14:48:31.456439
| 2020-07-09T14:54:33 |
365235
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/365235"
}
|
Stack Exchange
|
Subalgebras of a polynomial ring carved out by (families of) coefficient equalities
Let $\mathbf{k}$ be a field, and let $P=\mathbf{k}\left[ x_{1},x_{2}
,\ldots,x_{n}\right] $ be a polynomial ring over $\mathbf{k}$ in $n$
variables $x_{1},x_{2},\ldots,x_{n}$. Alternatively, $P$ can be a power series
ring, or $n$ can be $\infty$.
A "monomial" shall mean a monomial in $x_{1},x_{2},\ldots,x_{n}$. For each
monomial $\mathfrak{m}$ and each $f\in P$, we let $\left[ \mathfrak{m}
\right] \left( f\right) $ denote the coefficient of $\mathfrak{m}$ in $f$.
Fix a scalar $c_{\mathfrak{m}}\in\mathbf{k}$ for each monomial $\mathfrak{m}$.
Let $S$ be a set of pairs of monomials. Let $M$ be the subset
\begin{align}
\left\{ f\in P\ \mid\ c_{\mathfrak{m}}\cdot\left[ \mathfrak{m}\right]
\left( f\right) =c_{\mathfrak{n}}\cdot\left[ \mathfrak{n}\right] \left(
f\right) \text{ for all }\left( \mathfrak{m},\mathfrak{n}\right) \in
S\right\}
\end{align}
of $P$. It is clear that $M$ is a $\mathbf{k}$-vector subspace of $P$.
Question. What are examples of choices of $c_{\mathfrak{m}}$ and $S$ for
which $M$ is a $\mathbf{k}$-subalgebra of $P$ ?
Here are three examples where this is satisfied:
Example 1. Let $c_{\mathfrak{m}}=1$ for each monomial $\mathfrak{m}$. Let
$S$ be the set of all pairs $\left( \mathfrak{m},\mathfrak{n}\right) $ of
monomials such that $\mathfrak{n}$ is obtained from $\mathfrak{m}$ by a
permutation of the variables. Then, $M$ is the ring of symmetric polynomials
in $x_{1},x_{2},\ldots,x_{n}$ over $\mathbf{k}$; this is a $\mathbf{k}
$-subalgebra of $P$.
Example 2. Let $c_{\mathfrak{m}}=1$ for each monomial $\mathfrak{m}$. Let
$S$ be the set of all pairs $\left( \mathfrak{m},\mathfrak{n}\right) $ of
monomials such that $\mathfrak{m}$ and $\mathfrak{n}$ are pack-equivalent
(i.e., we can write $\mathfrak{m}$ and $\mathfrak{n}$ in the forms
$\mathfrak{m}=x_{i_{1}}^{a_{1}}x_{i_{2}}^{a_{2}}\cdots x_{i_{k}}^{a_{k}}$ and
$\mathfrak{n}=x_{j_{1}}^{a_{1}}x_{j_{2}}^{a_{2}}\cdots x_{j_{k}}^{a_{k}}$,
where $a_{1},a_{2},\ldots,a_{k}$ are positive integers and $i_{1},i_{2}
,\ldots,i_{k},j_{1},j_{2},\ldots,j_{k}$ are elements of $\left\{
1,2,\ldots,n\right\} $ satisfying $i_{1}<i_{2}<\cdots<i_{k}$ and $j_{1}
<j_{2}<\cdots<j_{k}$). Then, $M$ is the ring of quasisymmetric polynomials in
$x_{1},x_{2},\ldots,x_{n}$ over $\mathbf{k}$; this is a $\mathbf{k}
$-subalgebra of $P$.
Example 3. Let $c_{\mathfrak{m}}=2^{\ell\left( \mathfrak{m}\right) }$
for each monomial $\mathfrak{m}$, where $\ell\left( \mathfrak{m}\right) $
denotes the number of distinct variables that appear in $\mathfrak{m}$. (For
example, $\ell\left( x_{3}^{5}x_{4}x_{7}^{2}x_{8}\right) =4$.) Let $S_{1}$
be the set $S$ from Example 2. Let $S_{2}$ be the set of all pairs $\left(
\mathfrak{m},\mathfrak{n}\right) $ of monomials, where $\mathfrak{m}
=x_{i}^{2}\mathfrak{p}$ and $\mathfrak{n}=x_{i}x_{i+1}\mathfrak{p}$ for some
$i\in\left\{ 1,2,\ldots,n-1\right\} $ and some monomial $\mathfrak{p}$ that
contains neither of the variables $x_{i}$ and $x_{i+1}$. (For example,
$\left( x_{1}^{3}\underline{x_{2}^{2}}x_{5}^{4},x_{1}^{3}\underline{x_{2}
x_{3}}x_{5}^{4}\right) \in S_{2}$, where the underlining merely marks the
$x_{i}^{2}$ and $x_{i}x_{i+1}$ parts of the monomials.) Let $S=S_{1}\cup
S_{2}$. Then, $M$ is a $\mathbf{k}$-subalgebra of $P$. (This is more or less
the $\mathbf{k}$-subalgebra $\operatorname*{span}\left( \overline
{K}_{n,\Lambda}\right) _{n\in\mathbb{N};\ \Lambda\subseteq\left[ n\right]
}$ constructed in Theorem 4.4 (a) of my The eta-basis of
QSym,
except that I am now working in finitely many variables.)
Example 1 is trivial to check; Examples 2 and 3 are easiest to verify by
finding bases of the respective subspaces $M$ and showing that products of two
elements of these bases expand as linear combinations of these bases. As an
offshoot of a project I proposed, a student is looking for some general
criteria that ensure that $M$ is a $\mathbf{k}$-subalgebra of $P$; the
question above is meant to gather a few more examples to experiment upon.
|
2025-03-21T14:48:31.456677
| 2020-07-09T15:01:30 |
365236
|
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"authors": [
"Benoît Kloeckner",
"Mircea",
"https://mathoverflow.net/users/4961",
"https://mathoverflow.net/users/5628"
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"url": "https://mathoverflow.net/questions/365236"
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|
Stack Exchange
|
How are these "Voronoi-dual" configurations called?
If $\mathscr P\subset \mathbb R^d$ is a discrete point configuration, take the Voronoi diagram of $\mathscr P$ and call $\mathscr P'$ the vertices of that diagram.
I would like to know if configurations such that $\mathscr P = \mathscr P''$ have an established name, and if people studied them.
Some first facts and questions that may make the above property interesting:
If $\mathscr P$ has the property, then $\mathscr P'$ has it.
If $\mathscr P$ has the property, then if we join the vertices of $\mathscr P$ to their nearest neighbors in $\mathscr P'$, we form a bipartite graph with edges of equal length. Not all such graphs give $\mathscr P$ satisfying the property, but does there exist a simple extra convexity condition that can be added to get it?
It seems that $\mathscr P$ that have the property are unbounded, is that true? I could not prove or find a counterexample so far.
Do there exist $\mathscr P$ that have the property and that have accumulation points?
Examples: (1) All simple lattices I tried as $\mathscr P$ have the property. Are there lattices that do not have the property? (2) Vertices of rhombic tilings of $\mathbb R^2$ have the property. In particular this includes Penrose tilings.
For $d=1$, you can describe $\mathcal{P}$ by an increasing sequence $(x_n)$, and you have explicit linear equations giving you a sequence $(x_n'')$ defining $\mathcal{P}''$. Then the condition is that there is a $k$ such that $(x_n'')=(x_{n+k})$. For each $k$, you get a linear induction formula, so that $(x_n)$ should be a linear combination of products of polynomials and exponentials. They thus cannot be bounded. This leaves open the possibility of a accumulation point, but I have not worked out the computations ($k=0,1,2$ do not work if I did not messed up).
@BenoîtKloeckner thanks! in one dimension are there examples not congruent to the integers? (cf. point 2 in the list of facts I mention)
Ah, you're right pointing this out! Point 2 shows that when $d=1$ all example must be congruent to lattices: consecutive points must be at distance twice the constant length of edges of the bipartite graph. This make my comment moot.
|
2025-03-21T14:48:31.456858
| 2020-07-09T15:38:50 |
365240
|
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}
|
Stack Exchange
|
Modular and primitive ideals of $C_{0}(X,A)$
Let $A$ be $C^{\ast}$- Algebra and $X$ be a locally compact Hausdorff space and $C_{0}(X,A)$ be the set of all continuous functions from $X$ to $A$ vanishing at infinity. Define $f^{\ast}(t)={f(t)}^{\ast}$ (for $t\in X$). It is well known that ideals of $C_{0}(X,A)$ are of the form $\{ f\in C_{0}(X,A): f(x) \in I_x \; \forall x\in X \}$ where $I_x$ is closed ideal in $A$ for all $x$.
What’s known about the modular and primitive ideals of $C_{0}(X,A)$.
Any references or ideas?
The primitive ideals $P$ of $C_0(X,A)$ are of the form $P=\{f\in C_0(X,A): f(x)\in Q\}$ where $x\in X$ and $Q\in {\rm Prim}(A)$.
The modular ideals are more difficult to describe. They have to have the form $I=\{ f\in C_0(X,A): f(x_i)\in I_i\}$
where $\{x_i\}$ is a compact subset of $X$ and $I_i$ is a modular ideal in $A$ for each $i$. But probably not every such ideal will be modular.
The example of the C$^*$-algebra of continuous functions $f$ from $[0,1]$ to the $2\times 2$ matrices for which $f(1)={\rm diag}(\lambda (f), 0)$ shows the sort of problem that can crop up: every primitive quotient is modular but the ideal $\{0\}$ is not modular. One can probably build such an example into a $C_0(X(A)$.
Dear Douglas, Thank you! Do you have any reference for the results you mentioned?
Not really, I am afraid. The description of the primitive ideals is standard. Modular ideals are rather seldom referred to in the world of C*-algebras.
|
2025-03-21T14:48:31.456995
| 2020-07-09T17:45:15 |
365248
|
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|
Stack Exchange
|
Are cohomology functors sheaves?
Question is the following:
Is the functor $H^n_{dR}:\text{Man}\rightarrow \text{Set}$ a sheaf with respect to open cover topology on $\text{Man}$?
More generally, are cohomology functors sheaves in general (in any reasonably non trivial Grothendieck topology)?
I am also interested in cohomology functors that arise in Algebriac geometry/topology.
Is there a way of sheafification in this setup?
I have nothing much to support this question, this is completely out of curiosity.
Edit : I am also interested in answers/references related to the comment of Piotr Achinger; that reads
"in what way is cohomology a sheaf" leads one to notions like $\infty$-topoi etc.
No, in fact the cohomology presheaves $U\mapsto H^n_{\rm dR}(U)$ sheafify to zero for $n>0$.
No, as almost any example demonstrates (a sphere, for instance, covered by two contractible open sets); if cohomology were a sheaf the Mayer-Vietoris sequence would split as short exact sequences of $H^k$ for each $k$. $C^*(M)$ forms a sheaf of complexes, but the gluing property does not survive passing to cohomology.
Oh, no. I did not think about that @MikeMiller. That is straight forward that it is not expected to be sheaf.. this is a stupid question. I should not have asked it here (not in this form) :D
@PiotrAchinger Ok. It is not immediate for me, I will think about it. Thanks :)
The question is not stupid - trying to make sense of "in what way is cohomology a sheaf" leads one to notions like $\infty$-topoi etc.
@PiotrAchinger that is interesting. can you please suggest some (possibly short) reference that introduce $\infty$-topoi when trying to understand in what sense cohomology is a sheaf?
This upvote downvote game is funny :D 3 upvotes, 2 downvotes.
I don't know a particularly good introduction because, like many of the core concepts that motivate infinity categories, this idea is folklore which only (relatively) recently accquired a more precise accepted meaning. But the point is that there is an infinity category of chain complexes $Ch$, and the functor $U \mapsto C^*(U)$ becomes a sheaf in the infinity categorical sense. (In turn $\infty$-categories of sheaves of spaces (instead of chain complexes) are the paradigmatic example of $\infty$-topoi).
@PhilTosteson Ok. I am afraid I might lost the path if I google for infinity topoi and look for relation with Piotr Achinger's comment. Can you suggest some starting point towards this?
Is the functor H^n_dR:Man→Set a sheaf with respect to open cover topology on Man?
As already pointed out in the comments, the answer is no for n>0, yes for n=0.
"in what way is cohomology a sheaf" leads one to notions like ∞-topoi etc.
In the context of this question,
the assignment M↦Ω(M) yields a contravariant
functor from smooth manifolds to cochain complexes,
and this functor satisfies the homotopy descent condition.
This was first proved by Weil in Sur les théorèmes de de Rham (DigiZeitschriften, DOI, EuDML).
An accessible exposition is given by Bott and Tu in §8 of Differential Forms in
Algebraic Topology.
The link http://www.springerlink.com/index/10.1007/BF02564296 is broken..
I will read the part you mentioned from Bott and Tu. I am hearing the word “homotopic descent condition” for the first time. I can guess what it is, but I will read about it and ask if I have any further questions..
@PraphullaKoushik Try https://eudml.org/doc/139040 for Weil's paper, though EuDML is down for maintenance at the moment. Here's an unofficial link: https://www.maths.ed.ac.uk/~v1ranick/papers/weil.pdf
@DavidRoberts ok
@PraphullaKoushik: I have no links to SpringerLink in my post.
@PraphullaKoushik: The homotopy (not homotopic) descent condition is also known as the ∞-sheaf condition, and it is also used to define stacks in groupoids as presheaves of groupoids that satisfy the homotopy descent condition.
Previously you have mentioned http://www.springerlink.com/index/10.1007/BF02564296 as link for Weil’s paper, that is what I was mentioning. Yes, I misread homotopy as homotopic.. Thanks for the edit..
@PraphullaKoushik you should compare the actual URL Dmitri posted (eg by looking at the mouseover url, copying the link, looking at the answer source markdown etc) compared to where it lands after redirection. That said, both EuDML and Sprinklink are both still not working for me, so I hope the ahem other link is useful.
@DavidRoberts: There is also a perfectly official version from DigiZeitschriften that I just added. Theoretically, DOI entries should be updated with new URLs, but I guess this does not always happen in practice.
Thanks for the edit
|
2025-03-21T14:48:31.457330
| 2020-07-09T18:24:13 |
365251
|
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"Alexandre Eremenko",
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|
Stack Exchange
|
Did Euler ever use anything similar to Cauchy's inequality?
This could be asked more provocatively, indeed how it arose, as "how did Euler do so much mathematics without using and/or knowing Cauchy's inequality?", something that came up in the context when reading a comment about how ubiquitous said inequality/variants is in various fields today.
But I think the title question is at least fact-based, and of historical interest.
Q. Did Euler ever use anything similar to Cauchy's inequality?
Particularly I might have expected something like the integral version (thus with Schwarz's name) to appear somewhere in calculus of variations. Alternatively, something beyond simply a geometric-arithmetic mean might have occurred in number theory arguments. (I figure, as a first go, asking mathematicians is better than historians, especially as if there is "something similar" I'd like to hear it.)
For clarification, it is the Cauchy inequality that appears in Note II here, and not his bounds on derivatives (or Taylor series coefficients).
Which Cauchy inequality?
I'm far from an expert on math history, but I would venture that part of the answer is that Euler and his contemporaries were much less focused on proving estimates than modern mathematicians are. Off hand I can think of many more equalities and formulas attributed to Euler and his contemporaries than inequalities. And Cauchy's inequality isn't particularly helpful in proving the former. That said, I would be surprised if there isn't "something similar," as you say, somewhere in his prodigious works.
That was exactly my feeling too - estimates just weren't vogue back then (partially because Euler didn't do them!)
There are lots of inequalities in Note II. Do you mean Theorem 16, the one that says $a \cdot b \le |a| |b|$?
following up on @MarkLewko 's comment: one inequality attributed to Euler is a triangle inequality
|
2025-03-21T14:48:31.457490
| 2020-07-09T18:40:04 |
365254
|
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"url": "https://mathoverflow.net/questions/365254"
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|
Stack Exchange
|
Injective resolution of representable sheaves
Let $S$ be a scheme and consider the abelian category $\mathcal{A}$ of sheaves of abelian groups on the fppf site over $S$. It is known that $\mathcal{A}$ has enough injectives. My question is: given a representable sheaf , say $\mathbb{G}_m$, I'm wondering if we can explicitly construct the first few terms of an injective resolution that are representable. In other words, can we have an exact sequence of abelian sheaves $1 \to \mathbb{G}_m \to \mathcal{I}_0 \to \mathcal{I}_1 \to \mathcal{I}_2$ where $\mathcal{I}_i$ are representable for $i=0,1,2$?
|
2025-03-21T14:48:31.457665
| 2020-07-09T20:28:00 |
365255
|
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"Maxime Ramzi",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365255"
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|
Stack Exchange
|
Surprising examples of functors which preserve cofiltered limits but not all limits?
Question: What are some "surprising" examples of functors (resp. $\infty$-functors) $F$ which preserve cofiltered limits?
I'm not quite sure what "surprising" means, but I think that
"Surprising" should probably mean that $F$ doesn't preserve all connected limits (resp. all contractible limits) -- though one of my main examples below seems to violate this, so take it with a grain of salt.
"Surprising" should also mean that $F$ is not just obtained as $G^{op}$ where $G$ is a functor which preserves filtered colimits for some familiar reason.
I've learned of two interesting examples from Joyal's paper on combinatorial species (Theorem 1 in the appendix). Namely,
In $Set$ or $Spaces$, arbitrary coproducts commute with cofiltered limits. I believe this generalizes to any topos or $\infty$-topos, but I'm not quite sure.
In $Set$ or $Spaces$, quotients by a finite group action commute with cofiltered limits. Actually, I think this generalizes to an arbitrary action of a group object in $Spaces$, and to all contractible limits, so maybe it's not actually "surprising" according to (1) above. Nonetheless, I'm still surprised by it since it's a colimit construction, so I'll count it. I'm not quite sure if this generalizes to an arbitrary $\infty$-topos.
For the generalization of your second example, you might be interested in https://ncatlab.org/nlab/show/commutativity+of+limits+and+colimits#coproducts_commute_with_connected_limits
|
2025-03-21T14:48:31.457779
| 2020-07-06T17:06:05 |
364974
|
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|
Stack Exchange
|
Gaussian bounds for discrete (graph) Dirichlet heat kernel
(this is an attempt to refine a previous question; I was told that it would be better to create a new question than edit the previous one, I hope this is the correct ettiquete.)
Let $\Omega$ be a finite, connected subset of $\mathbb{Z}^n$, $W_t$ a simple discrete-time random walk on $\mathbb{Z}^n$ started at $x$, and $T_\Omega$ the first time at which $W_t$ leaves $\Omega$; consider
$$
P^D_\Omega(x,y;t) := \mathbb{P}[W_t=y \text{ and } T_\Omega>t],
$$
the discrete or graph heat kernel on $\Omega$ with Dirichlet boundary conditions.
By analogy with some known results on infinite graphs and continuous regions with boundaries one might expect a bound (at least for times smaller than the square of the diameter of $\Omega$) of the form
$$
P_\Omega^D(x,y;t)
\le
C_\Omega
\frac{\phi_\Omega(x,t) \phi_\Omega(y,t) }{t^{n/2}}
e^{- c |x-y|^2/t},
$$
with $\phi_\Omega:\mathbb{Z}^n \times \mathbb{N} \to [0,1]$ vanishing outside $\Omega$ and under certain circumstances (not too close to reentrant corners?) with $\phi_\Omega(x,t) \le d(x, \Omega^{\mathsf{c}}) /\sqrt{t}$.
Is this correct/known, in particular for $\Omega$ which is not an interval/rectangle/rectangular solid?
You can get a reasonable estimate by splitting time in three pieces, say [0,t/3], [t/3,2t/3], [2t/3,t] and estimating from above by the product of
--- the probability starting at x of staying in the domain up to time t/3
--- the probability starting at y of the the reversed walk staying in the domain up to time t/3
--- the conditional probability that S(t/3) = S(2t/3) given this which can be estimated as the supremum over all possibilities for the positions at t/3, 2t/3
If |x-y|^2 is much larger than t you might replace t/3 by the minimum of t/3 and the first time the walk traverses a distance of |x-y|/4 so that the walks in the first two steps are still distance |x-y|/2 away from each other.
|
2025-03-21T14:48:31.457954
| 2020-07-06T17:11:19 |
364975
|
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|
Stack Exchange
|
Is there a reversible fully polynomial-time approximation scheme for polygonal billiards?
Let $P \subset \mathbb{R}^2$ be a polygon with rational coordinates, and consider discrete billiards inside $P$, where a ball (of zero radius) moves by steps of fixed length on each step, in a rational direction, bouncing off walls. This system is reversible, and you can approximate it efficiently (by fixed or floating point arithmetic). Can you approximate it reversibly and efficiently?
You could extend this in many directions, by generalizing $P$ (higher dimensions, curves, obstacles inside), or you could have multiple bouncing bodies, friction, etc. I'll concentrate on polygonal billiards but a general method is of interest.
First, let's agree on a specific flavor of billiards. Fix once and for all some (non-intersecting) closed polygon $P \subset \mathbb{R}^2$ with rational coordinates (we just need that we can algorithmically quickly figure out approximations to them). Fix also a number $R$, and let $D$ be the closed disc of radius $R$. Now, billiards is the (abstract) dynamical ($\mathbb{N}$-)system $s : P \times D \to P \times D$ where $s(p, d)$ is obtained by moving $p$ forward by offset $d$, with usual (zero radius) billiard collisions. (If you hit a corner, we just canonically pick an infinitesimal side that's hit.)
I'll write some formulas for completeness. We define $s(p, d) = s'(p, d, d)$ where $s' : P \times D \times D \to P \times D$ is defined like so: Write $[u, v] = \{\alpha u + (1-\alpha) v \;|\; \alpha \in [0,1]\}$ for the line segment between $u$ and $v$, and $[u, v)$ is the half-open segment. If $[p, p + d'] \subset P$, define $s(p, d, d') = (p + d', d)$. Otherwise, there is a maximal $\alpha \in [0,1)$ such that $[p, p + \alpha d'] \subset P$, write $q = p + \alpha d'$. Suppose the line segment containing $q$ is $[u, v)$ where $u, v$ are vertices of $P$ in counterclockwise order. Let $v_0 = (v-u)/|v-u|$ be the tangent, and $v_1 = e^{\pi/2 i} v_0$ the normal, write $d = d_0 v_0 + d_1 v_1$, $d' = d_0' v_0 + d_1' v_1$ and define
$$ s'(p, d, d') = s'(q, d_0 v_0 - d_1 v_1, d_0' v_0 - d_1' v_1). $$
A couple of points.
For these formulas to well-define $s$, you need that there are finitely many bounces, i.e. the induction reaches the base case. This is not obvious syntactically, but if you've played billiards before it's easy to see there is actually a bound on the number of bounces independently of $(p, d) \in P \times D$. One way to see this: The main issue are corners. When the ball gets near a corner, think of the sides as mirrors and unfold the corner sort sort of like the Riemann surface of $\log$, and observe that the billiard path becomes a straight line; you see that the maximal number of bounces before leaving the vicinity of the corner is bounded by $\pi/\alpha + 1$ where $\alpha$ is the angle.
The map $s$ is bijective on $P \times D$. Namely, $s^{-1}(p, d) = (q, -d')$ where $s(p, -d) = (q, d')$.
If you like, you can do everything with rationals, since $s(p, d) \in \mathbb{Q}^4$ whenever $p \in P \cap \mathbb{Q}^2$ and $d \in D \cap \mathbb{Q}^2$, and $P$ has rational coordinates. By the formula for the inverse, this restriction of $s$ is also bijective.
Second, let's agree on what an approximation scheme for such a system is. If $M$ is a metric space and $f : M \to M$ is a function, an $\epsilon$-approximation scheme for $(M, f)$ is a tuple $(S, g, \imath, o)$ where $g : S \to S$, $\imath : M \to S$, $o : S \to M$, such that $\imath \circ o = \text{id}_S$, and $o \circ i \approx_{\epsilon} \text{id}$ and $o \circ g \circ \imath \approx_{\epsilon} f$ where $\approx_{\epsilon}$ for two functions means they are at Hausdorff distance less than $\epsilon$ as relations. You can think of the conditions on $\imath \circ o$ and $o \circ \imath$ as saying that $S \subset M$ is an $\epsilon$-net, and the condition on $o \circ g \circ \imath$ says that every point $x$ has a close by point of $S$ that's moved by $g$ roughly as $f$ does, and vice versa. An approximation scheme is finite if $S$ is.
On $P \times D$, we use the Euclidean metric. The following should be straightforward to prove, just use fixed-point arithmetic for the calculation. For the direction that every $x \in M$ is tracked by some $y \in S$, observe that discontinuities are rare and take a small enough $\epsilon > 0$. I haven't written an argument down; if you can refute this claim that's an acceptable answer.
Claim. For all $\epsilon$, there exist a finite $\epsilon$-approximation scheme for $(P \times D, s)$.
For the rest of this post I'll assume that's true. Now, while $(P \times D, s)$ is reversible, it is not clear that the arithmetic magically always rounds so there are no collisions, and so the naive fixed-point approximation might not be bijective. Let us say a finite $\epsilon$-approximation scheme is reversible if $g$ is bijective on $S$. My gut says that the following is easy to obtain from Hall's theorem, though I'm not sure how to proceed to get a clean proof:
Claim-ish conjecture. For all $\epsilon$, there exists a finite reversible $\epsilon$-approximation scheme for $(P \times D, s)$.
The idea for the proof would be that you fix a discrete set of directions and a discrete set of positions. If you have a single wall, you can match up the directions to roughly their mirrored versions, and if you have a point cloud hit the wall, some geometry of numbers shows that the point cloud's $\epsilon$-blow up contains more points of $S$ than the cardinality of the point cloud, which is what Hall's theorem needs. For multiple collisions interacting there are some issues and I didn't do the calculations. If you can refute the conjecture, that's an acceptable answer. If you have a clean proof of the conjecture, that's very interesting (but not an acceptable answer). For the rest of the post I'll assume this conjecture is true.
The third and final ingredient is the fully polynomial-time approximation scheme. (I'm borrowing the term from complexity theory since I feel it's close in spirit.) A fully polynomial-time approximation scheme or FPTAS for a system $(M, f)$ is a four-tuple $(S, g, \imath, o) = (S_n, g_n, \imath_n, o_n)_{n \in \mathbb{Z}_+}$. The tuple $(S_n, g_n, \imath_n, o_n)$ must be a $1/n$-approximation scheme for $(M, f)$. Additionally, $S_n \subset \Sigma^{p(n)}$ for some polynomial $p(n)$ and finite alphabet $\Sigma$, and it should be polynomial-time computable uniformly in $n$. Similarly, $g_n : S_n \to S_n$ is a polynomial-time computable function (so it's a function in the class FP) uniformly in $n$. We similarly define a strong FPTAS, with the difference that $(S_n, g_n, \imath_n, o_n)$ is a $1/2^n$-approximation scheme for all $n$. Such a scheme is reversible if all the $(S_n, g_n, \imath_n, o_n)$ are.
Claim. There exists a strong FPTAS for $(P \times D, s)$.
I haven't written a careful proof of this, but its proof should be the same as for the claim above: the proof is completely algorithmic and the number of bounces is bounded by a constant (independent of $\epsilon$) so the calculations go through in polynomial time. (If you can refute this claim that's an acceptable answer.)
Now, assuming the above claims are true, we can state the question I'm interested in.
Question. Does there exist a reversible FPTAS for $(P \times D, s)$? A reversible strong FPTAS?
For the first question, you could imagine just listing a concrete discretization, and realizing Hall's theorem (if it works the way I think), but I don't know if there is a relevant "polynomial-time Hall's theorem". For the latter, you'd presumably need some real understanding of what's going on.
You are right, no need for that, what's important is that it's quick to compute finite geometric approximations to the polygon, and rational coordinates implies that, but does not change the general problem otherwise (to my knowledge).
This paper and book by Beck are about this topic:
https://doi.org/10.1515/9783110317930.17
https://doi.org/10.1142/9913
I could not access the paper. I read the intro to the book and did not see anything even faintly related to my problem. Is there a deeper connection than just that the book deals with billiards, hidden in the mathematics, and if so, could you elaborate in an answer?
I'll note also that while I concentrated on billiards because I like to ask a very precise question, an answer with some sort of bijective approximation schemes for natural dynamical systems is of more interest to me (as an answer to this question) than general information about billiards.
|
2025-03-21T14:48:31.458769
| 2020-07-06T18:02:13 |
364978
|
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
Is there a bijective proof of an identity enumerating independent sets in cycles?
Let $C_m$ be the cycle with $m$ vertices, defined so that $C_1$ has a self-loop on its unique vertex. Let $p_m$ be the generating function enumerating the number of ways to choose $k$ vertices in $C_m$ so that no two are adjacent. Thus the coefficient of $z^k$ in $p_m(z)$ is the number of independent sets in $C_m$ of size $k$.
For instance, $p_1(z) = 1$, $p_2(z) = 1+2z$, $p_3(z) = 1+3z$, $p_4(z) = 1+4z + 2z^2$, $p_5(z) = 1 + 5z+5z^2$ and $p_6(z) = 1 + 6z + 9z^2 + 2z^3$. Set $p_0 = 2$.
It is not hard to show by algebraic arguments (related to the theory of Chebyshev polynomials) that if $\ell, m \in \mathbb{N}_0$ with $\ell \ge m$ then
$$p_\ell p_m = p_{\ell+m} + (-1)^m z^{m} p_{\ell-m}.$$
In particular, $p_m^2 = p_{2m} + 2(-1)^m z^{m}$, and so if $k < m$ then the coefficients of $z^k$ in $p_m^2$ and $p_{2m}$ are equal. I would like a bijective proof of this, or ideally, of the more general identity above.
Is there a bijective proof that if $k < m$ then the number of independent sets of size $k$ in the disjoint union $C_m \sqcup C_m$ is equal to the number of independent sets of size $k$ in $C_{2m}$?
But how do you prove that identity? A priori the $p_m$ are just some polynomials, which of their properties do you use in the proof?
In fact it seems that $p_m(x)=2(-x)^{\frac m2}T_m(\frac1{2\sqrt{-x}})$, but how to prove this?
I think this is in I. Gutman, F. Harary, Generalizations of the matching polynomial, Utilitas Mathematica 24 (1983) 97-106.
The result also appears as Problem 11898, American Math. Monthly 123 (March 2016), but the published solution (in a later issue that I don't have a reference for) is not bijective.
The solution appears in The American Math. Monthly 125 (January 2018).
Here is a link to R. Tauraso's solution: mat.uniroma2.it/~tauraso/AMM/AMM11898.pdf. Incidentally Richard is very modestly not mentioning that he proposed the problem.
Mark Wildon's link doesn't work for me but this one does.
It seems that I’ve seen this question here before, but I am not sure whether it had a bijective answer. Anyway, here you are.
Enumerate the vertices in two copies of $C_m$ as $1,2,\dots,m$ and $1’,2’,\dots,m’$, respectively. Take any independent set of size $k<m$ in the union of these cycles (regard it as marking some vertices). Choose the smallest $i$ such that both $i$ and $i’$ are not in the set. Arranging the vertices in the order
$$
1,2,\dots,i,(i+1)’,(i+2)’,\dots, m’, 1’,
\dots,i’,i+1,i+2,\dots,m
$$
you get a $C_{2m}$ with an independent set being marked.
The inverse map is to take $k$ marked vertices in $C_{2m}$, choose the smallest $i$ such that both $i$ and $i+m$ are not marked, cut $C_{2m}$ after them, and glue into two copies of $C_m$.
The same argument works for an arbitrary number of copies of $C_m$ (and still $k<m$).
|
2025-03-21T14:48:31.459027
| 2020-07-06T19:23:16 |
364982
|
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"Connor Malin",
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|
Stack Exchange
|
Smooth map homotopic to Lie group homomorphism
Let $G$ and $H$ be connected Lie groups. A Lie group homomorphism $\rho:G\to H$ is a smooth map of manifolds which is also a group homomorphism.
Question: Can we find a smooth (or real-analytic) map $f:G\to H$ which is not homotopic to any Lie group homomorphism?
For example, if $G=H=S^1$, it seems the answer is no. For simplicity, we may begin with the same question but assuming some extra conditions, such like (i) $G,H$ are torus, (ii) $G,H$ are compact, etc.
If you construct two Lie groups such that $H \simeq G$, but $BH \not\simeq BG$ then any homotopy equivalence $H \simeq G$ cannot be homotopic to a homomorphism, since applying $B$ to it would deloop it to a weak equivalence.
A non connected example of this is given by $H= \mathbb{Z}/2 \times \mathbb{Z}/2$ and $G= \mathbb{Z}/4$.
@ConnorMalin Don't all such examples need to be disconnected?
@NajibIdrissi $SO(4)$ and $S^3\times SO(3)$. It is true however that if $G,G'$ are connected simple lie groups, then $G,G'$ are isomorphic if and only if they are homotopy equivalent.
If $G$ is a compact simply-connected simple Lie group, then any nontrivial homomorphism $G\to G$ is an automorphism (it is injective because $G$ is simple, and any immersion of closed connected manifolds of the same dimension is covering map), and in particular it has degree $\pm 1$. For example, if $f: S^3\to S^3$ is a map of degree $d$ with $|d|>1$, then $f$ is not homotopic to a homomorphism.
On the other hand, by obstruction theory any self-map of an $n$-torus is homotopic to a map induced by multiplication by an $n\times n$ matrix with integer entries, which is a homomorphism.
In fact, compact connected simple Lie groups have (see https://en.wikipedia.org/wiki/List_of_simple_Lie_groups) finite fundamental group and hence their self-coverings are diffeomorphisms. Thus in the above answer one can drop the assumption ``$G$ is simply-connected''.
Thank you for the great answer! Could you explain more about the self-map of torus? I'm not familiar with obstruction theory. Or, any reference is also good enough.
@Hang If $X$, $K$ are connected CW complexes and $K$ has contractible universal cover, then the set of homotopy classes $[X,K]$ of maps from $X$ to $K$ is bijective to the set of conjugacy classes of the induced $\pi_1$-homomorphism (e.g. Spanier, "Algebraic Topology", Ch.11, Section1, Theorem 11). Thus $[T^n, T^n]$ is bijective to the set of endomorphisms of $\mathbb Z^n$. Those are $n\times n$ matrices over $\mathbb Z$. Any such matrix (as a map of $\mathbb R^n$) descends to a self-map of an $n$-torus.
As Igor shows, every endomorphism of a simple Lie group $G$ has degree $\in\{0,\pm 1\}$.
On the other hand, every compact Lie group admits self maps of other degrees. Namely, the $k$-th power map $g\mapsto g^k$ has degree $k^r$, where $r$ is the rank of the group. So, each $k$ with $|k|\geq 2$ gives an example of a smooth map which is not homotopy equivalent to a homomorphism.
One way to compute the degree of the $k$-th power map is as follows. First, we can find an element $g\in G$ which lies in a unique maximal torus $T^r$ and which is also a regular value of the $k$-th power map. The uniqueness of the maximal torus implies that all $k$-th roots of $g$ lie in $T^r$, so this reduces the degree calculation to $T^r$, where it is obvious.
I’m not sure if this is what you’re looking for, but the map $\mathbb{Z}/2\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$ given by sending $0\mapsto 1\,, 1\mapsto 0$ isn't homotopic to a homomorphism. More generally if the codomain is disconnected the answer to your question seems to be positive.
Thanks! This is a good point, but it could be more interesting to assume the connectedness.
|
2025-03-21T14:48:31.459311
| 2020-07-06T19:33:29 |
364984
|
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|
Stack Exchange
|
Iterative method for solving certain systems of linear equations
I've noticed that a method for calculating the stationary distribution of a finite-state rational-transition-probabilities Markov chain introduced by Arthur Engel many years ago generalizes to give an inefficient but charming method for solving certain systems of linear equations with rational coefficients. E.g., consider the linear system regarding three people sharing money described by Leonardo of Pisa in his Liber Abaci, as recounted by David Mumford in the April 1, 2020 installment of his blog (http://www.dam.brown.edu/people/mumford/blog/2020/Ridiculous.html); we can rewrite it as system of equations $x = \frac{5}{6} x + \frac{7}{18} y + \frac{4}{9} z$, $y = \frac{1}{6} x + \frac{5}{9} y + \frac{5}{18} z$, $z = \frac{1}{18} y + \frac{5}{18} z$, characterizing $(x,y,z)$ as a fixed point. It's easy to verify that (33,13,1) is a solution, as is any multiple; the question is how one can find such a solution.
Given a triple $t=(x,y,z)$ of positive integers, define a new triple $t'=(x',y',z')$ where $x'$ is $\frac{5}{6} x + \frac{7}{18} y + \frac{4}{9} z$ rounded up to the next multiple of 18, $y'$ is $\frac{1}{6} x + \frac{5}{9} y + \frac{5}{18} z$ rounded up to the next multiple of 18, and $z'$ is $\frac{1}{18} y + \frac{5}{18} z$ rounded up to the next multiple of 18. (More generally, you always round each variable up to the next multiple of the common denominator of the fractions associated with the right hand side of the associated equation.)
I claim that for any initial triple of positive integers, the process $t \rightarrow t’ \rightarrow t’' \rightarrow \cdots$ converges to a positive integer solution to the original underdetermined system.
This method is not efficient; if you start from (18,18,18), you don't arrive at (594,234,18) until 34 steps later. Still, I find it interesting that this iterative method always terminates. Has anyone characterized the sorts of problems for which this sort of method works?
I'd also be curious to know whether the approach can be modified so that one is guaranteed to find the minimal positive integer solution, or can be modified to converge more quickly.
|
2025-03-21T14:48:31.459470
| 2020-07-06T20:54:13 |
364990
|
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|
Stack Exchange
|
Conditions for a function to vanish almost nowhere on its support?
Let $f:\mathbb{R}^d\rightarrow\mathbb{R}$ be a continuous function and $\mathrm{supp}(f) := \mathrm{cl}\{x\in\mathbb{R}^d\mid f(x)\neq 0\}$ its support.
Under which conditions is it true that $f≠0$ (Lebesgue-) almost everywhere on $\mathrm{supp}(f)$?
(I.e., under which conditions on $f$ is the set $\{x\in\mathrm{supp}(f)\mid f(x)=0\}$ necessarily a Lebesgue nullset?)
If the support is compact, this happens iff the support is Jordan measurable.
@LiviuNicolaescu : Thank you. (You don't happen to know how compactness of supp can be traded off against regularity conditions on f?)
|
2025-03-21T14:48:31.459546
| 2020-07-06T20:54:54 |
364991
|
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|
Stack Exchange
|
Example of a projective variety over a field of characteristic zero which is uniruled but not ruled
A variety $ Z $ over a field $ k $ of characteristic zero is ruled if there is a variety $ M $ and a dominant, birational map $ \phi: M \times \mathbb{P}^{1}_{k} \dashrightarrow Z $. A variety $ Z $ over a field $ k $ of characteristic zero is uniruled if there is a variety $ M $ and a dominant, rational map $ \phi: M \times \mathbb{P}^{1}_{k} \dashrightarrow Z $. Does anyone know of a projective variety $ Z $ over a field $ k $ of characteristic zero which is uniruled, but not ruled?
I don't know much about this, but I would guess a cubic threefold.
I think you just need to know that there exists a threefold $X$ which is unirational but not rational (e.g. the cubic threefold). If $X$ is birational to $S\times \mathbb{P}^1$, there is a dominant rational map $X -\!-\!\!\!> S$, thus $S$ is unirational, hence rational by Castelnuovo's theorem. Therefore $X$ is rational.
Some conic bundles that are not birationally trivial do the job. For explicit examples, see pages 143-148 of
K. Matsuki: Introduction to the Mori program, Universitext. New York, NY: Springer (ISBN 0-387-98465-8/hbk). xxiii, 478 p. (2002). ZBL0988.14007,
The case of a cubic threefold $W_3 \subset \mathbb{P}^4$, cited by Roy Smith in his comment, belongs to this family of counterexamples. In fact, the blow-up $X=\mathrm{Bl}_L(W_3)$ of $W_3$ along a line $L \subset W_3$ is a conic bundle over $\mathbb{P}^2$. By Clemens-Griffths we know that $W_3$ is not rational, so $X$ is not rational, and this implies that its conic bundle $X \to \mathbb{P}^2$ is not birationally trivial.
|
2025-03-21T14:48:31.459657
| 2020-07-06T21:05:22 |
364992
|
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|
Stack Exchange
|
Two closed lagrangians embedded in $\mathbb{R}^{2n}$ can be shifted and perturbed to intersect at two points
In Polterovich paper "The surgery of Lagrange manifolds". It is stated in page 204 - in proof of Theorem 7a - the following,
If $L_1$ and $L_2$ are two Lagrangian submaifolds of $\mathbb{C}^{n}$, then after suitable linear shift and small perturbation we can assume $L_1$ and $L_2$ have exactly two intersection points.
I will be glad if some one help me see how such small perturbation is made so that we get exactly two points of intersections.
|
2025-03-21T14:48:31.459723
| 2020-07-06T21:44:25 |
364996
|
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|
Stack Exchange
|
Non-conformity colorings
Motivation. Recently I attended a little party (adhering to physical distancing and in accordance to other COVID19-related laws). An attendee told me that he chose drink $Y$ since at least half of his acquaintances at the party had drink $X$. I was sober enough to put this into a graph-theoretic property - and that's what this question is about.
Formal version. If $G=(V,E)$ is a simple, undirected graph and $v\in V$, let $N(v) = \{w\in V:\{v,w\}\in E\}$. Given a cardinal $\kappa > 0$ we say a map $c:V\to \kappa$ is a non-conformity coloring if for all $v\in V$ with $N(v)\neq \varnothing$ we have
$$|N(v)\cap c^{-1}(\{c(v)\})| < |N(v)\setminus c^{-1}(\{c(v)\})|.$$ (Note that this is the formal version of "$v$ has another drink than more than half of $v$'s neighbors" if we view $c$ as "drink assignment".) The non-conformity chromatic number $\chi_{nc}(G)$ is the smallest cardinal $\kappa$ such that there is a non-conformity coloring $c:V\to \kappa$.
If $K_3$ denotes the complete graph on $3$ vertices, it is easy to see that $\chi_{nc}(K_3) = 3$.
Question. Given a cardinal $\kappa > 3$, is there a graph $G$, finite or infinite, such that $\chi_{nc}(G) = \kappa$?
This is called unfriendly (or friendly) coloring, which has been studied a lot.
If $\kappa$ is an infinite cardinal, there is a graph $G$ of order $\kappa$ with $\chi_{nc}(G)=\kappa$. If $\kappa$ is regular, $G$ can be a complete graph; if $G$ is singular, $G$ can be a disjoint union of complete graphs.
If $G$ is a finite graph, then $\chi_{nc}(G)\le3$; just take a $3$-coloring of $G$ which minimizes the number of edges joining two vertices of the same color. The usual sort of compactness argument shows that $\chi_{nc}(G)\le3$ holds if $G$ is locally finite. Thus, if $\kappa\gt3$, there is no locally finite graph $G$ with $\chi_{nc}(G)=\kappa$.
For $n\lt\aleph_0$ the complete $n$-partite graph $G=K_{\underbrace{\aleph_0,\dots,\aleph_0}_n}$ is a countable graph with $\chi_{nc}(G)=n$.
|
2025-03-21T14:48:31.459871
| 2020-07-06T23:34:36 |
365003
|
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|
Stack Exchange
|
Do sparse graphs contain a single regular pair?
An easy corollary of the Szemerédi Regularity Lemma is that dense graphs contain linear sized $\varepsilon$-regular bipartite subgraphs whose density is similar to that of the parent graph. As noted by Tim Gowers in (Is there a weak strong regularity lemma?) there are easier ways of seeing this, with better bounds.
I'm wondering if a meaningful statement with the above flavor holds for sparse graphs, of density $\Omega(n^{-1/t})$. That is, graphs which are dense enough to necessarily contain $K_{t,t}$ subgraphs.
What I'm looking for exactly is a subgraph $(A,B)$ that satisfies $|A|=|B|=k$, $e(A,B)=\Omega(k^{2-1/t})$, and further, $(A,B)$ is $\varepsilon$-regular, in the sense that any subgraph $(A', B')$ with $\varepsilon k$ vertices on each side satisfies $e(A,B)=\Omega(k^{2-1/t})$. Note that this is a lot weaker than the usual notion of $\varepsilon$-regularity in that we allow that the density of a subgraph to be off by a constant factor from the density of the the parent graph, all that we insist on is that they are of the same order of magnitude.
I'm okay with aiming for a sub-linear sized regular pair (i.e. take $k=o(n)$) as the graph itself could be almost entirely filled with isolated vertices, except for a small clique. I would expect one could take $k$ polynomial in $n$, but I'm interested in any range where $k$ grows with $n$.
I'm also okay with assuming that the graph is $K_{10t, 10t}$-free (say), although I cannot tell if there is an easy construction that shows the necessity of such an assumption.
From what I can tell, the sparse versions of Regularity Lemma do not say anything immediately meaningful here, as they do not forbid all edges of the sparse graph being between non-regular pairs.
How large should $k$ be? Otherwise one can just take an edge.
It's of course only sensible if $k$ grows with $n$. I would expect that one can take $k$ to be polynomial in $n$, the exponent being a constant depending on $t$.
Theorem 1.1 here answers (a very close approximation of) my question. https://www.combinatorics.org/ojs/index.php/eljc/article/view/v9i1r1
In the regime where $1/t \ll \varepsilon$, in a graph with density $\Omega(n^{-1/t})$, we may find a subgraph $(A,B)$ with $k$ vertices on each part, where $k\geq n^{1-\gamma}$, and $(A,B)$ is $\varepsilon$-regular, and has density $\Omega(n^{-1/t})$.
The first caveat is that we need $1/t \ll \varepsilon$ to apply this theorem so that $\gamma$ is small. The second is that the density of $(A,B)$ is not $\Omega(k^{-1/t})$, but $\Omega(n^{-1/t})$, hence polynomially smaller than what I asked for. Still, this is a lot of useful information.
|
2025-03-21T14:48:31.460168
| 2020-07-07T05:09:46 |
365015
|
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|
Stack Exchange
|
Is there any relation between $h_{d_1},h_{d_2}$ and $h_{d_1d_2}$?
$h_{d_1}, h_{d_2}$ and $h_{d_1d_2}$ are class number of $\Bbb Q(\sqrt{d_1}),\Bbb Q(\sqrt{d_2}),\ and \ \Bbb Q(\sqrt{d_1d_2})$ respectively.
There is a relation between the class numbers of those quadratic fields and the class number of the biquadratic field $\mathbf Q(\sqrt{d_1},\sqrt{d_2})$. The key term to look for is "Brauer relation", which has a wider scope than just this example. A reference for this topic has already been mentioned in an earlier MO question here. See also equations (2.1) and (2.2) of Bosma and de Smit's "Class number relations from a computational point of view" (J. Symbolic Computation ${\mathbf{31}}$ (2001), 97-112).
|
2025-03-21T14:48:31.460260
| 2020-07-07T07:38:34 |
365019
|
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|
Stack Exchange
|
Is new $n$-conjecture as follows correct?
Given a positive integer $P>1$, let its prime factorization be written as$$P=p_1^{a_1}p_2^{a_2}p_3^{a_3}\cdots p_k^{a_k}.$$
Define the functions $h(P)$ by $h(1)=1$ and $h(P)=\min(a_1, a_2,\ldots,a_k).$
Is new the $n$-conjecture, formulated as follows, correct?
Conjecture: if ${P_1,P_2,...,P_n}$ are positive integer and pairwise coprime, then,
$$\min\{h(P_1), h(P_2),...,h(P_n), h(P_1+P_2+...+P_n)\} \leq n+1.$$
I proposed the case $n=2$ two years ago here (Is the conjecture A+B=C following correct?). Now I reformulate that question as follows:
Let ${P_1,P_2}$ are coprime, then:
$$\min\{h(P_1), h(P_2), h(P_1+P_2)\} \leq 3$$
If the conjecture is right. Then there are no positive integer $x, y, z$, $n, m, k>3$ satisfy the equation $x^n+y^m=z^k$
But one has to formulate it in a way to exclude cases like $2^n+2^n=2^{n+1}$
I want mean $x,y$ are positive integer and pairwise coprime
Here's a near miss for $n=3$: $2^{14}+7^5+13^6=2^5\cdot3^5\cdot5^4$.
Thanks to dear @ThomasBrowning
Draw attention because the nice problem was check with $P_1+P_2< 10^{18}$
Such attempted generalizations of ABC to four or more variables
often fail to specializations of the identity
$$
(x^2+xy-y^2)^3 + (x^2-xy-y^2)^3 = 2 (x^6 - y^6).
\label{1}\tag{*}
$$
One can use elliptic curves to make both $x^2 + xy - y^2$ and $x^2 - xy - y^2$
"powerful" (of the form $A^2 B^3$), which makes each of the four terms
$(x^2+xy-y^2)^3$, $(x^2-xy-y^2)^3$, $2x^6$, $2y^6$ have $h=6$
but for a stray factor of $2$ which should not matter in the context of
the ABC conjecture. For example, the pairwise prime numbers
$a,b,c,d$ below satisfy $2a^6 + b^6 + 61^9 c^6 = 2d^6$.
Here $d$ is even but $a$ is odd, so $2a^6$ has a "stray factor of $2$",
and the expansion to $a^6 + a^6 + b^6 + 61^9 c^6 = 2d^6$ loses
pairwise coprimality; so either way we don't quite get a counterexample.
Still, this suggests that generalizations of ABC to four or more variables
can run afoul of identities such as \eqref{1}.
(It is "well known" that the Mason-Stothers theorem forbids the disproof
of ABC itself by such an identity.)
a =<PHONE_NUMBER>691921314835532892967014277786302791344455107816139963763145069687359424810667270039489345929029301393007247303344511065237
b =<PHONE_NUMBER>365458069945118311017282675402206671149976403498624129498574702167905733126870212117037684063261425637225699359421949547271
c =<PHONE_NUMBER>6852061412855232703108032231130723932854745057900981539571749281974534306702514113168069346943754838515856358759614674721
d =<PHONE_NUMBER>123658957500070687744472849236478810555581279857582626862200039312130562436302012081022720213179152015505627679021327325170
Prof. Elkies, I hyperlinked the formula: however, if you do not like it, please feel free to revert.
Is the case n=2 true? Dea Professor @NoamDElkies
The "n=2 case" (no solution of A+B=C in coprime 4th-powerful numbers)
is plausible -- and it would be reduced to a finite computation
if we knew an effective ABC conjecture in the form N >> C^r with any r > 3/4
-- but it seems very difficult and I don't think a proof is known.
@NoamD.Elkies Thank you so much, I hope you enjoy this issue. Or introduce it with your student
|
2025-03-21T14:48:31.460479
| 2020-07-07T08:00:43 |
365020
|
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|
Stack Exchange
|
A limit calculation
I wonder if the limit below $$\lim_{x\rightarrow +\infty} e^{-x}\sum_{j=0}^{\infty}\frac{x^{j+a}}{\Gamma(j+a+1)}$$
equals 1, for real constant $a>0$, and how shall we get this result?
Could you cite any reference where you got this limit ? And am not sure if this gives 1 because if $\sum_{j=0}^{\infty}\frac{x^{j+a}}{\Gamma(j+a+1)}$ convergent to any finit number this yield to your limit to be 0 , You have one chance if this sum diverge
@zeraouliarafik the sum should be convergent to a function related to x. You may see that each term of this series is positive and goes to ∞ while x→∞. So the limit is possible to have a non-zero result.
If we take $$\chi(x)=\sum_{j=0}^{\infty} \frac{x^{j+a}}{\Gamma(j+a+1)}$$ then we get, $$\chi'(x)-\chi(x)=\frac{x^{a-1}}{\Gamma(a)}$$
Hence, the solution gives $$\psi(x)=\chi(x)e^{-x}=\frac{1}{\Gamma(a)}{\int e^{-x} x^{a-1} dx}+c$$
This means $$\lim \limits_{x \to \infty} \psi(x) -\psi(0)=1$$
As, $$\color{grey}{{\int_{0}^{\infty} e^{-x}x^{a-1} dx=\Gamma(a)}}$$
and $$\color{cadetblue}{\psi(0)=0} $$
So, $$\lim \limits_{x \to \infty} \psi(x)=e^{-x}\sum_{j=0}^{\infty} \frac{x^{j+a}}{\Gamma(j+a+1)} =1$$
Alternatively: $\chi(x) = e^x \bigl(1 - \Gamma(a, x)/\Gamma(a))$ where $\Gamma(a,x) = \int_x^\infty t^{a-1} e^{-t} \mathrm{d} t$ is the incomplete $\Gamma$ function. So the limit is $1 - \lim_{x \rightarrow \infty} \Gamma(a,x)/\Gamma(a)$ which is clearly $1$.
|
2025-03-21T14:48:31.460611
| 2020-07-07T09:36:21 |
365024
|
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|
Stack Exchange
|
Is the number of solutions of $\phi(x)=n!$ bounded? If yes, what is its bound?
Pillai showed in 1929 that the function $A(n)$ giving the number solutions of the equation $\phi(x)=n$ is unbounded in (S. Pillai, On some functions connected with $\varphi(n)$, Bull. Amer. Math. Soc. 35 (1929), 832–836). I'm interested to know about bounds on solutions of $\phi(x)=n!$ which is assigned A055506 in OEIS, where it is claimed that if $\phi(x) = n!$, then $x$ must be a product of primes $p$ such that $p - 1$ divides $n!$. It is unclear to me if this allows me to prove that there are finitely many solutions of the equation $\phi(x) = n!$. Probably an equivalent question is to ask: is the number of solutions of $\phi(x)=n!$ bounded? If yes, what is its bound ?
Related question:
https://math.stackexchange.com/q/3747571/156150
For any fixed $n$, the number of solutions to $\phi(x)=n!$ is bounded, since $\phi(x)$ tends to infinity. However, this number is probably not bounded independently of $n$.
Quotes from A055506 and A055487 respectively: "All solutions to $\phi(x) = n!$ are in the interval $[n!,(n+1)!]$", and "According to Tattersall, in 1950 H. Gupta showed that $\phi(x) = n!$ is always solvable". The first gives a bound for fixed $n$, while the latter says that there exists a solution for any $n$. (I do not know arguments backing up either claim.)
Actually Gupta proves that the number of solutions goes to infinity with $n$, see Hansraj Gupta,
The American Mathematical Monthly,
Vol. 57, No. 5 (May, 1950), pp. 326-329.
Thanks for the reference , I will check it
UPD. Bound simplified.
Here is a constructive bound for the number of solutions to $\phi(x)=m$.
Let $\varphi(a) = m$. If $p^k\mid a$ for some $k\geq 1$, then $p^{k-1}(p-1)\mid m$, and thus $k\leq 1+\frac{\log(m)}{\log(p)}\leq 1+\frac{\log(m)}{\log(2)}$.
Then the number of such $a$ is bounded by
$$\prod_{d\mid m} (2+\frac{\log(m)}{\log(2)}) = (2+\frac{\log(m)}{\log(2)})^{\tau(m)}.$$
For $m>40$, we have $2+\frac{\log(m)}{\log(2)}\leq 2\log(m)$, and thus we generously bound the number of solutions by
$$(2\log(m))^m.$$
I don't understand the upper bound on E. The integral from 2 to m is lower bounded by m/log m. Even if you don't do the integration but estimate the sum over divisors, I am not seeing your power of log as an upper bound. Gerhard "Am I Adding This Wrong?" Paseman, 2020.07.07.
@GerhardPaseman: Nice catch! That was the price of blindly copying bounds from Wikipedia (it seems that $\exp(x/2)$ factor is lost in the bounds for $\mathrm{Ei}(x)$ there). Anyway, this approach was somewhat overengineered, and I've replaced it with a simpler one.
Indeed. However you can replace the exponent with 2sqrt(m) , and if you look at divisor pairs, you can make the base smaller also. Gerhard "Likes To Improve Through Simplification" Paseman, 2020.07.07.
I think you also have to worry about one divisor of m corresponding to two different primes p. Gerhard "Maybe Leave Exponent At M" Paseman, 2020.07.07.
@GerhardPaseman: One divisor cannot correspond to different $p$, since $d$ in my answer essentially stands for $p-1$ (and different divisors give different $p$'s). I did not attempt to make the bound as small as possible -- that's a separate question.
But phi (27)=phi(19). How does your enumeration account for cases where 27 "is replaced by " 19 as factors of a potential solution a? Gerhard "One Divisor Unequals One Prime" Paseman, 2020.07.07.
@GerhardPaseman: In your example for $m=18$, $27=3^3$ corresponds to $d=2$ (and exponent 3 from its exponent range), $19$ corresponds to $d=18$. I'm not sure what you mean under "replaced by".
I am completely misreading your formula. I factor m into components each of the form (p^{k-1})(p-1) and call that component d, and think that each term counts the number of possible p that give rise to the same d. Until I figure this out, assume we are talking at cross purposes for now. Gerhard "Needs Some More Head Scratching" Paseman, 2020.07.07.
The bound is certainly overkill. First, it is well known that $n/\varphi(n)=O(\log\log n)$, whence $\varphi(n)\le m\implies n=O(m\log\log m)$. Even better, Wikipedia cites a result that $|{n:\varphi(n)\le m}|=O(m)$. In fact, the expression given there implies $|{n:\varphi(n)=m}|=O\bigl(m(\log m)^{-k}\bigr)$ for all constant $k$.
@EmilJeřábek: Of course, it's an overkill. My goal was to give a simple self-contained argument for bounding the number of solutions.
@GerhardPaseman: The product here is taken over all $d=p-1\mid m$ for candidate primes $p$ (and possibly some composites), where the product terms are the upper bound for the number of possible exponents. I believe you think more of the bound ala A045778(m), which needs to account for existence of multiple primepower preimages of the same number. My approach is free from this issue.
@MaxAlekseyev It’s easy to give a simple self-contained argument for bounds much closer to optimal. See my answer.
We have
$$\frac n{\varphi(n)}=\prod_{p\mid n}\bigl(1-p^{-1}\bigr)^{-1}
\le2\prod_{\substack{p\mid n\\p\ge3}}\frac32
=2\prod_{\substack{p\mid n\\p\ge3}}3^{\log_3(3/2)}
\le2\prod_{\substack{p\mid n\\p\ge3}}p^{\log_3(3/2)}
\le2n^{\log_3(3/2)}$$
(where $p$ runs over primes), hence
$$\varphi(n)\le m\implies n\le(2m)^{(1-\log_3(3/2))^{-1}}=(2m)^{\log_23}.$$
Using a larger cut-off $k$ in place of $3$, the same argument gives
$$\varphi(n)\le m\implies n\le(c_km)^{\log_{k-1}k},$$
where
$$c_k=\prod_{p<k}\bigl(1-p^{-1}\bigr)^{-1}.$$
Notice that $\log_{k-1}k\approx1+\frac1{k\log k}$ for large $k$.
I will not go into details, but it is easy to prove by well-known elementary arguments that $c_k=O(\log k)$, hence if we choose $k\approx\log m$, we obtain
$$\varphi(n)\le m\implies n\le c\,m\log\log m$$
for some constant $c$.
The nice idea you have used is the compositional inverse of Euler totiont function
Um, the argument I present above can be adapted to show that phi(n)=m implies n is at most (k+1)m, where k is the exponent of 2 that divides m. When m has many prime factors, this can be changed to ( Clog k) m using work of Mertens or simpler approximations. Gerhard "Not Quite Gilding The Lily" Paseman, 2020.07.09.
@GerhardPaseman I have no idea what all this talk about lilies is about, but did I just see you include parts of my answer in your answer without as much as a thank you?
I think this could blow up into a major misunderstanding if we aren't careful. It is not clear to me what part for which you want credit. However, the edit I made to my post has an argument similar to the one in my original post. If you need recognition, I need to know more precisely for what. I thought the product over primes had too generous a bound in the first part of your post. Gerhard "Not In It For Points" Paseman, 2020.07.09.
If you want credit for writing n/\phi(n), I am willing to recognize you as a recent source of inspiration. Gerhard "Will That Help This Situation?" Paseman, 2020.07.09.
@Gerhard The argument in my answer is extremely simple, hence it is no doubt “similar” to any other argument estimating $\varphi(n)$, such as the one in your answer. Nevertheless, the first paragraph you added basically copies what I already wrote, yielding the same $O(m\log\log m)$ bound, and I find this rather uncool. And yes, one of the points of my answer is that even though the bound on the product looks generous, it leads to the same asymptotic estimate as if it were bounded in a more complicated way, which is why I did the simple way.
I hope to avoid a misunderstanding. If by first paragraph of my answer you mean of my post, that appeared a day before your post. (I leave the ramifications of this unstated.) If you instead mean first paragraph of the edit, indeed some of it was inspired by the first paragraph of my answer and by the first line of your answer, where n/\phi(n) appears. However your simple argument giving a fractional exponent bound does not appear as simple to me as the one that gives a logarithmic (in m) bound on the multiplier. Gerhard "Please Add Clarity To Comment" Paseman, 2020.07.10.
Here is a simpler bound, based on the comment of R. van Dobben de Bruyn.
Let a solution of the equation be broken into two parts, c and d, where c is the n-smooth part of the solution, and is coprime to d, which of necessity is square free and has all prime factors bigger than n. (I leave the case n=1 to the reader.) Then c is at most n! (2/1)(3/2)(5/4)...((n+1)/n)which for n greater than three is strictly less than (n+1)n!. So c is less than (n+1)! .
Turning to d, each prime divisor of d contributes at least one power of 2 when subject to Euler's phi, so d has fewer than n prime divisors. So d is less than e times n!.
Since the shrinkage under phi of the product is at most e(n+1), the original solution must be less than 3(n+1)!. This is also a weak upper bound on the total number of solutions, but can probably be improved to show that the number and location of solutions generally is less than (n+1)!, leaving the case of small n to the reader where all creation (counterexamples, arrghh spellcheck!) must lie.
Edit 2020.07.09. GRP:
The argument above for bounding $n$ given $m=\phi(n)$ is made even simpler, as $ n/\phi(n)$ is a product of
$k$ many terms of the form $p/(p-1)$ where the $p$ are distinct primes. This bounded above by $(2/1)(3/2 )(5/3)...$, which for all $k$ is less than $k+1$ and for large $k$ grows like $\log k$. Since $k$ is bounded by a function smaller than $\log m$, we can get an upper bound on $n$ that looks like $Cm\log\log m$, likely for $C$ less than 4. Even when $k$ is large, $n$ can't have many more distinct primes than powers of 2 dividing $m$.
Towards the original question, note that there are easy solutions of totient value being a factorial, and that some of them can be extended by replacing certain powers of small primes by a prime $q$ such that $q$ is bigger than the base of the factorial and such that $q-1$ equals the powers of the small primes and $q$ is not already a prime factor of the solution being modified.
Thus it seems very likely that the number of solutions is not bounded as the size of the factorial grows.
End Edit 2020.07.09. GRP.
Gerhard "Leaving Hard Work To Others" Paseman, 2020.07.07.
If we let k be the exponent of the exact power of two dividing m, we get that any solution to the general equation must (for k not too small) be less than 3(k+1)m, giving a weak but non exponential (in m) upper bound on the number of solutions. Gerhard "Really Likes This Re-engineering Stuff" Paseman, 2020.07.07.
Toward a lower bound, which is the main thrust of the question. Note first that for P the product of the primes at most n one has $\phi((n!)P/\phi(P))=n!$ and one has an alternate representation when n+1 is an odd prime. There also is a representation involving a prime q bigger than n whenever $\phi(q)$ divides $n!/(P\phi(P))$. Proving the existence of $q$ for every sufficiently large n should show the number of solutions grows without bound as $n$ grows. Gerhard "Is Almost Convinced Of Unboundedness" Paseman, 2020.07.07.
|
2025-03-21T14:48:31.461296
| 2020-07-07T10:34:08 |
365029
|
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|
Stack Exchange
|
Extension of closed-form solvability of polynomial equations of x and exp to rational expressions?
Is my conjecture below true?
It's a new conjecture. It's an extension of Lin's theorem in [Lin 1983] which is cited in [Chow 1999] - see both references at the bottom of this page.
It seems that Ferng-Ching Lin was a student of Stephen Hoel Schanuel: https://www.genealogy.math.ndsu.nodak.edu/id.php?id=192925
The proof is very simple because the preconditions of my conjecture contain the preconditions of Lin's theorem except irreducibility. The irreducibility follows directly from very simple arguments from elementary algebra.
$\mathbb{L}$ are the elementary numbers, or Liouvillian numbers:
http://mathworld.wolfram.com/LiouvillianNumber.html, https://en.wikipedia.org/wiki/Elementary_number, http://mathworld.wolfram.com/ElementaryNumber.html, [Chow 1999].
$\ $
Theorem [Lin 1983]:
If Schanuel's conjecture is true and $\tilde{P}(X,Y)\in\overline{\mathbb{Q}}[X,Y]$ is an irreducible polynomial involving both $X$ and $Y$ and $\tilde{P}\left(\alpha,e^{\alpha}\right)=0$ for some non-zero $\alpha$ in $\mathbb{C}$, then $\alpha$ is not in $\mathbb{L}$.
$\ $
Conjecture:
Let
$f$ a non-constant elementary function,
$n\in\mathbb{N}_{\ge 1}$,
$P(x,y)\in\overline{\mathbb{Q}}[x,y]\setminus(\overline{\mathbb{Q}}[x]\cup\overline{\mathbb{Q}}[y])$ irreducible over $\mathbb{C}$,
$Q(x,y)\in\overline{\mathbb{Q}}[x,y]$ not zero,
$p(x)\in\overline{\mathbb{Q}}[x]$,
$q(x)\in\overline{\mathbb{Q}}[x]$ not zero,
$R(x,y)=\frac{P(x,y)}{Q(x,y)\cdot q(x)^n}$,
$r(x)=\frac{p(x)}{q(x)}$ not constant
so that
$P(x,y)$ and $Q(x,y)$ coprime over $\mathbb{C}$,
$p(x)$ and $q(x)$ coprime over $\mathbb{C}$.
Suppose Schanuel's conjecture is true.
If a $\tilde{P}(x,y)\in\overline{\mathbb{Q}}[x,y]\setminus(\overline{\mathbb{Q}}[x]\ \cup\ \overline{\mathbb{Q}}[y])$ of $degree_x=n$ with $\frac{P(x,y)}{{q(x)}^n}=\tilde{P}(r(x),y)$ exists and $R(f(z_0),e^{r(f(z_0))})=0$ for $z_0\in\mathbb{C}$ and $r(f(z_0))\neq 0$, then $z_0$ is not an elementary number.
$\ $
Proof draft:
We will prove our conjecture using Lin's theorem. Lin's theorem makes statements about equations of the form $\tilde{P}(\alpha,e^\alpha)=0$, wherein $\alpha\in\mathbb{C}$ and $\tilde{P}(x,y)$ is irreducible over $\mathbb{C}$. We consider the equation $R(f(z_0),e^{r(f(z_0))})=0$ from our conjecture and will show that under the preconditions of our conjecture there exists an equation equivalent to it in the form from Lin's theorem.
Setting $f(z_0)=x$ and $e^{r(f(z_0))}=y$ gives the equation $R(x, y)=0$. We consider it in the form $\frac{P(x, y)}{Q(x, y)\cdot q(x)^n}=0$ from the preconditions of our conjecture. Because $P(x,y)$ and $Q(x,y)$ are coprime over $\mathbb{C}$ by precondition, we can multiply both sides of the equation by $Q(x,y)$ and get the equation $\frac{P(x,y)}{q(x)^n}=0$ equivalent to it. By precondition, $\frac{P(x,y)}{{q(x)}^n}=\tilde{P}(r(x),y)$ holds. So our equation becomes $\tilde{P}(r(x),y)=0$. Backsubstitution of $x$ and $y$ gives $\tilde{P}(r(f(z_0)),e^r(f(z_0)))=0$.
We have $\forall z_0\in\mathbb{C}\colon r(f(z_0))\in\mathbb{C}$. We can therefore set $r(f(z_0))=\alpha$, which gives the form $\tilde{P}(\alpha,e^\alpha)=0$ from Lin's theorem. To fulfill all preconditions from Lin's theorem, we still have to show that $\tilde{P}(x,y)$ is irreducible over $\mathbb {C}$.
If $\tilde{P}(x,y)$ is reducible over $\mathbb{C}$, then
$n_1\in\mathbb{N}_{\ge 1}$,
$n_2\in\mathbb{N}_0$,
$a_0,...,a_{n_1}\in\mathbb{C}[y]$,
$b_0,...,b_{n_2}\in\mathbb{C}[y]$ not all constant
exist so that
$a_{n_1},b_{n_2}\neq 0$,
$\tilde{P}(x,y)=(a_0+a_1x^1+...+a_{n_1}x^{n_1}) \cdot(b_0+b_1x^1+...+b_{n_2}x^{n_2})$.
The highest degree regarding $x$ of the polynomial product on the right-hand side of the equation is $n_1+n_2$. This must be equal to the highest degree regarding $x$ of the polynomial $\tilde{P}(x,y)$ on the left-hand side of the equation. This was $n$ according to the preconditions of our conjecture. We have therefore $n_1+n_2=n$.
By substitution of $x$ by $r(x)$, we get
$$\tilde{P}(r(x),y))=(a_0+a_1r(x)^1+...+a_{n_1}r(x)^{n_1})\cdot(b_0+b_1r(x)^1+...+b_{n_2}r(x)^{n_2}).$$
Because, according to the preconditions of our conjecture, $r(x)=\frac{p(x)}{q(x)}$, and $p(x)$ and $q(x)$ are coprime over $\mathbb{C}$, we have
$$\tilde{P}(r(x),y))=(a_0+a_1\left(\frac{p(x)}{q(x)}\right)^1+...+a_{n_1}\left(\frac{p(x)}{q(x)}\right)^{n_1})\cdot(b_0+b_1\left(\frac{p(x)}{q(x)}\right)^1+...+b_{n_2}\left(\frac{p(x)}{q(x)}\right)^{n_2})$$
$$=(a_0+a_1\frac{p(x)^1}{q(x)^1}+...+a_{n_1}\frac{p(x)^{n_1}}{q(x)^{n_1}})\cdot(b_0+b_1\frac{p(x)^1}{q(x)^1}+...+b_{n_2}\frac{p(x)^{n_2}}{q(x)^{n_2}})$$
$$=(a_0+\frac{a_1p(x)^1}{q(x)^1}+...+\frac{a_{n_1}p(x)^{n_1}}{q(x)^{n_1}})\cdot(b_0+\frac{b_1p(x)^1}{q(x)^1}+...+\frac{b_{n_2}p(x)^{n_2}}{q(x)^{n_2}})$$
$$=\frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+...+a_{n_1}p(x)^{n_1})}{q(x)^{n_1}}\cdot\frac{(b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+...+b_{n_2}p(x)^{n_2})}{q(x)^{n_2}}$$
$$=\frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+...+a_{n_1}p(x)^{n_1})\cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+...+b_{n_2}p(x)^{n_2})}{q(x)^{n_1+n_2}}.$$
Because, as stated above, $n_1+n_2=n$, we have
$$\tilde{P}(r(x),y))=\frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+...+a_{n_1}p(x)^{n_1})\cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+...+b_{n_2}p(x)^{n_2})}{q(x)^n}.$$
According to the preconditions of our conjecture, $\tilde{P}(r(x),y))=\frac{P(x,y)}{q(x)^n}$. We have therefore
$$\frac{P(x,y)}{q(x)^n}=\frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+...+a_{n_1}p(x)^{n_1})\cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+...+b_{n_2}p(x)^{n_2})}{q(x)^n}.$$
Multiplying by $q(x)^n$ yields
$$P(x,y)=(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+...+a_{n_1}p(x)^{n_1})\cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+...+b_{n_2}p(x)^{n_2}).$$
If $\tilde{P}(x,y)$ is reducible over $\mathbb{C}$, then $P(x,y)$ is accordingly reducible over $\mathbb{C}$. Because, according to the preconditions of our conjecture, $P(x,y)$ is irreduible over $\mathbb{C}$, $\tilde{P}(x,y)$ is irreducible over $\mathbb{C}$.
Thereby all preconditions from Lin's theorem are fulfilled and we can apply it to our equation. According to Lin's theorem, $\alpha$ is not in $\mathbb{L}$. Because $\alpha=r(f(z_0))$, $r(f(z_0))\notin\mathbb{L}$. Because $r$ and $f$ are elementary functions, $r(f(z))\in\mathbb{L}$ holds for all $z\colon z\in dom(r\circ f)\ \land\ z\in\mathbb{L}$. But because, as just stated, $r(f(z_0))\notin\mathbb{L}$, $z_0\notin\mathbb{L}$, so $z_0$ is not an elementary number.
q.e.d.
$\ $
[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50
[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448
Here is a recent article using another method, Khovanskii's method:
[Belov-Kanel/Malistov/Zaytsev 2020] Belov-Kanel, A.; Malistov, A.; Zaytsev, R.: Solvability of equations in elementary functions. Journal of Knot Theory and Its Ramifications 29 (2020) (2) 204-205
Related question from same user: https://mathoverflow.net/questions/333027/why-are-there-elementary-equations-that-are-not-solvable-in-closed-form
Please don't create such a tag as "equations", it's far too broad and imprecise to be useful.
|
2025-03-21T14:48:31.461597
| 2020-07-07T10:51:18 |
365031
|
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|
Stack Exchange
|
The space of all constructions of $(\infty, 1)$-categories
In the paper How to glue derived categories the following is written about $(\infty, 1)$-categories:
This idea was discussed quite a lot in the early 1990-ies, and probably before that,
but people were reluctant to try to turn it into actual mathematics. Here is a rough
summary explaining why:
(i) It was felt that it certainly can be done, and in several ways.
(ii) All of the constructions would be very complicated and ad hoc. There is no
preferred construction.
(iii) This would not matter: all of the constructions will be equivalent, and moreover,
“the space of all possible constructions is contractible”.
(iv) Moreover, one could even turn the last statement into a theorem.
(v) However, to do so, one needs to make sense of “the space of all constructions”.
This is certainly possible, and there are several ways to do it, just as in (i). But
then one also has (ii), (iii), (iv), (v), and the infinite loop.
Have there been efforts to make precise the points (iv) and (v)? There is a paper of Toën titled "Vers une axiomatisation de la théorie des catégories supérieures" but I don't know anything besides that.
There is the work of Barwick--Schommer-Pries: https://arxiv.org/abs/1112.0040
@OP, have you read the paragraph that follows the text you have quoted? There are a lot of pointers. Anyway you might also be interested in Bergner's The homotopy theory of $(\infty, 1)$-categories which is kind of a survey of the whole thing.
The fact that $\infty$-groupoid of automorphisms of the category of $\infty$-category is understood (and very simple: it is the discrete group corresponding to taking opposite categories at various level) is I think the best to this question: it shows that up to confusion between a category and its opposite there is always a unique way to identifies two given models.
|
2025-03-21T14:48:31.461894
| 2020-07-07T12:27:06 |
365035
|
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|
Stack Exchange
|
How many “linkage shapes” in the plane have $n$ “joints”?
A linkage shape in the plane with $n$ joints means a choice of $n$ points (a.k.a., vertices) in the plane joined by struts (a.k.a., edges). The joints are flexible pivots for the struts connecting them. The rules for a linkage shape are: (1) the struts are connected end to end to form a closed path (a.k.a., cycle); (2) no joint can be straightened; (3) no two struts can intersect (except at their joints); (4) struts can be expanded or contracted in length (but never length 0). Two linkage shapes with $n$ joints are considered equivalent if either one can be obtained by continuous motion from the other by (a) changing angles at joints, or by (b) changing strut lengths, or by (c) translation or rotation (not reflection), all the while during the motion adhering to rules (1)-(4). I believe there are exactly 8 linkage shape equivalence classes with $6$ struts. In general, how many equivalence classes of linkage shapes with $n$ joints are there? (At a hardware store one may get a "folding ruler" for experiments with linkage shapes.)
In the literature, your linkage shapes in one equivalence class are called non-crossing morphs of one another. For example: Gotsman, Craig, and Vitaly Surazhsky. "Guaranteed intersection-free polygon morphing." Computers & Graphics 25, no. 1 (2001): 67-75. PDF dowload.
What are your counts to $n=3,4,5$? Clearly there is just one polygon triangle. $n=4,5$?
Many thanks for the reference. Yes, for n=3 there is 1; for n=4 there are 2; for n=5 there are 4; for n=6 there are 8. These were calculated by pencil and paper drawings. For n=7 and 8 there are computer simulations which currently lead me to believe there are 10 and 16 respectively, but those are not certain.
$1,2,4,8,10,16$ is not yet enough to pinpoint in OEIS...
Yeah, I tried that. I am revamping my simulations to get a longer (and possibly provably more reliable) sequence of counts.
|
2025-03-21T14:48:31.462054
| 2020-07-07T12:35:53 |
365036
|
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|
Stack Exchange
|
Computing whether a set of polynomials cuts out a projective variety
I have a set of multivariate polynomials over $\mathbb{Q}$, and I want to compute whether they cut out a projective variety, i.e. whether the radical of the ideal $I$ that they generate is homogeneous. My first idea is to compute $\sqrt{I}$ and then check whether the homogeneous components of each generator of $\sqrt{I}$ are in $\sqrt{I}$. I have verified in Macaulay2 that $I$ itself is not homogeneous. It appears that Macaulay2 will not compute $\sqrt{I}$ in a reasonable amount of time, so I am posting here to ask for other ideas (or an efficient implementation).
Here are the nuts and bolts of my ideal $I$. Let's arrange 16 variables into a $4 \times 4$ matrix
$$X=
\begin{bmatrix} A && B \\ C && D\end{bmatrix},
$$
where $A,B,C,D$ are $2 \times 2$ matrices of indeterminates. The ideal $I$ is generated by $96$ polynomials. The first $48$ polynomials are the $3 \times 3$ minors of $X$, the $3 \times 3$ minors of
$$
\begin{bmatrix} A^T && B^T \\ C^T && D^T\end{bmatrix}.
$$
and the $3 \times 3$ minors of
$$
\begin{bmatrix} \vec{A} && \vec{B} && \vec{C} && \vec{D}\end{bmatrix},
$$
where $\vec{A}$, etc. is the $2 \times 2$ matrix $A$ viewed as a $4 \times 1$ column vector. Note that this indeed gives $3*{4 \choose 3}^2=48$ polynomials (which generate a homogeneous radical ideal). The other 48 polynomials are the same, with $X$ replaced by
$$
\begin{bmatrix} 0&&1&&1&&0\\1&&0&&0&&0\\1&&0&&0&&0\\0&&0&&0&&0\end{bmatrix}+
\begin{bmatrix} A && B \\ C && D\end{bmatrix}.
$$
Note: This has been Cross-posted from Math.SE.
One might be able to guess one point say $X_0$ (in the affine space) satisfying the equations. Then one can check if $\lambda X_0$ satisfy the equations for every $\lambda$ in the field. (Since the equations are inhomogeneous, perhaps there is a $\lambda$ such that $\lambda X_0$ is not a solution.)
Your first set of $48$ polynomials depends on $X$ and two other matrices, so it isn't clear from your current description how to change these other two matrices to get the other $48$ polynomials. Do you mean to say replace A by A + the corresponding block, etc?
@tim Yes, you understand correctly.
|
2025-03-21T14:48:31.462219
| 2020-07-07T13:14:58 |
365037
|
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|
Stack Exchange
|
$G_{\mathbb Q}$ and primes of $\overline{\mathbb{Z}}$
I know that if $K/\mathbb Q$ is a finite Galois extension (i.e. a Galois number field), then for any prime $(p)\subseteq \mathbb Z$, the Galois group $G=\operatorname{Gal}(K/\mathbb Q)$ acts transitively on the finite set of primes of $\mathcal O_K$ lying over $p$. I was wondering if something similar is true for the infinite case $K=\overline{\mathbb{Q}}$, but the proof I know for the finite case does not generalize to the infinite case at all, because it asks us to use the CRT on the set of primes (whose finiteness is established by the unique factorization in the Dedekind domain $\mathcal O_K$). The Going-Up Lemma tells us that if $(p)\subseteq \mathbb Z$ is a prime, then there is at least one prime $\mathfrak p\subseteq \overline{\mathbb{Z}}$ lying over it. However, since $\overline{\mathbb{Z}}$ is no longer a Dedekind domain, this proof does not apply. Do we know if there are only finitely many in this case too, or can there be infinitely many such primes? Do we know if the absolute Galois group $G_\mathbb{Q}$ acts transitively on this set of primes too?
Not only there can be, there necessarily will be. In fact I believe the set of primes lying over any prine $p\in\mathbb Z$ will be uncountable, and I think $G_{\mathbb Q}$ does act transitively on it. To prove it, note that for any two such primes $\mathbb p_1,\mathbb p_2$, the set of automorphisms taking $K\cap\mathbb p_1$ to $K\cap\mathbb p_2$ form an inverse system as we go over all Galois extensions $K/\mathbb Q$, and the inverse limit is hence nonempty.
@Wojowu, can you point me to a source for your first assertion?
There are infinitely many prime ideals $\mathfrak p$ in $\overline{\mathbf Z}$ that lie over $p$ since you can find an arbitrarily large (finite) number of prime ideals lying over $p$ in suitable number fields. I'll show how to do this using composites of quadratic fields. For each $n \geq 1$ there are $n$ primes $q_1, \ldots, q_n$ such that $p$ splits completely in $\mathbf Q(\sqrt{q_i})$ by using $q_i \equiv 1 \bmod 4p$ for all $i$. When a prime splits completely in number fields $K_1, \ldots, K_n$, it splits completely in the composite field $K_1\cdots K_n$, and the composite of the fields $\mathbf Q(\sqrt{q_i})$ has degree $2^n$ over $\mathbf Q$ because those quadratic fields are linearly disjoint, either since $q_1, \ldots, q_n$ are multiplicatively independent mod squares (Kummer theory proof), or since the discriminants of those quadratic fields are $q_1, \ldots, q_n$ and these are pairwise relatively prime (not Kummer theory proof). Thus $p$ has $2^n$ distinct primes lying over it in the number field $\mathbf Q(\sqrt{q_1}, \ldots, \sqrt{q_n})$. For each prime over $p$ in $\mathbf Q(\sqrt{q_1},\ldots,\sqrt{q_n})$, pick a prime ideal lying over it in $\overline{\mathbf Z}$ and that gives you $2^n$ different primes over $p$ in $\overline{\mathbf Z}$.
The transitivity of the action of $G_\mathbf Q$ on the primes over $p$ in $\overline{\mathbf Z}$ is proved in the appendix on infinite Galois theory in Washington's book on cyclotomic fields. The proof uses transitivity of the Galois action on number fields plus compactness of $G_\mathbf Q$, as mentioned in Wojowu's comment above.
Thank you for the excellent answer! That was very helpful!
|
2025-03-21T14:48:31.462433
| 2020-07-07T13:40:38 |
365038
|
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|
Stack Exchange
|
Who was Bickart?
The term "Bickart points" is often used for the foci of the Steiner circumellipse of a triangle. Who was Bickart, and what was the first publication to use the term?
As you have doubtless discovered by Googling it, the main information available on the internet, doubtless all copied from one another, is that these foci are "also called" the Bickart points. Wow, you'd think someone before would have felt a little guilty about just slinging that around with no citation …. One author seems to mention finding out the name from Norman Wildberger. Did you ask him?
Sur l’hypocycloïde à trois rebroussements (1913) gives an early reference to L. Bickart (Revue de Mathématiques spéciales, 1908).
This reference says M. Bickart, but that "M." stands for "monsieur", the initial is "L."
Here is one paper by L. Bickart on this geometric topic.
I cannot quite recover the circumellipse construction from this paper, but the constructions seem sufficiently similar to support the conclusion that this is the right Bickart.
|
2025-03-21T14:48:31.462542
| 2020-07-07T14:37:57 |
365043
|
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|
Stack Exchange
|
Convexity at a point and Jensen inequality
I am looking for a reference for the following claim:
Let $\phi:\mathbb (a,b) \to \mathbb R$ be a continuous function, and let $c \in (a,b)$ be fixed.
Suppose that "$\phi$ is convex at $c$". i.e. for any $x_1,x_2>0, \alpha \in [0,1]$ satisfying $\alpha x_1 + (1- \alpha)x_2 =c$, we have
$$
\phi(c)=\phi\left(\alpha x_1 + (1- \alpha)x_2 \right) \leq \alpha \phi(x_1) + (1-\alpha)\phi(x_2) .
$$
Then $\phi$ satisfies Jensen ineqaulity "at $c$".
Finite form:
Given $\lambda_i \in [0,1],x_i\in(0,\infty),i=1,\dots,k$ such that $\sum_{i=1}^k \lambda_i=1,\sum_{i=1}^k \lambda_ix_i=c$, we have $$\phi(\sum_{i=1}^k \lambda_ix_i) \le \sum_{i=1}^k \lambda_i \phi(x_i).$$
A more general probabilistic (measure-theoretic) form:
Given a random variable $X \in (a,b)$ with expectation $E(X)=c$, we have
$$
\phi(c)=\phi(E(X)) \le E(\phi(X)).
$$
In addition, if $\phi$ is strictly convex at $c$, then equality holds if and only if $X$ is constant a.e..
Both of these forms of Jensen inequality follow from the existence of a supporting line to the graph of $\phi$ at $c$.
The proof of the latter fact is not hard, but I couldn't find a source in the literature that presents this "localized" form of Jensen inequality, under the sole assumption of "convexity at a point". (In fact, I couldn't even find the term "convex at a point" anywhere...).
I find it impossible to believe that this doesn't show up in existing literature. Any help would be welcomed.
Comment:
Convexity at $c$ does not imply that the one-sided derivatives exist, so the standard proof for the existence of a supporting line (subgradient) does not apply here. (When the function is convex on an interval, every number between the two-sided derivatives form a subgradient).
For any real numbers $u,v,c$ such that $u\le c\le v$, let $\mu_{c;u,v}$ denote the unique probability distribution on the set $\{u,v\}$ with mean $c$.
Your generalization of Jensen's inequality follows immediately from the well-known fact that any probability distribution $\mu$ on $\mathbb R$ with a given mean $c\in\mathbb R$ is a mixture of probability distributions of the form $\mu_{c;u,v}$. See e.g. formula (2.13).
Details: Indeed, that formula implies
$$Ef(X)=\int_{S_c} Ef(X_{u,v})\,\nu_X(du\times dv)$$
for some probability measure $\nu_X$ (depending on the distribution of $X$) on the set $S_c:=\{(u,v)\in\mathbb R^2\colon u\le c\le v\}$ and all functions $f\colon\mathbb R\to\mathbb R$ such the function $\mathbb R\ni x\mapsto f(x)-kx$ is bounded from below for some some real $k$.
Now, if $f$ is convex at $c$, then $Ef(X_{u,v})\ge f(EX_{u,v})=f(c)$ for all $(u,v)\in S_c$, and hence
$Ef(X)\ge f(c)$.
Answer to your additional question concerning the strict convexity at $c$:
Moreover, if $f$ is strictly convex at $c$, then $Ef(X_{u,v})> f(EX_{u,v})=f(c)$ for all $(u,v)$ in the set, say $S_c^\circ$, of all $(u,v)\in S_c$ such that $u<c<v$. Hence,
$Ef(X)>f(c)$ unless $\nu(S_c^\circ)=0$.
On the other hand, the condition $(u,v)\in S_c\setminus S_c^\circ$ implies that $Eg(X_{u,v})=g(c)$ for all functions $g\colon\mathbb R\to\mathbb R$. So, the condition $Ef(X)=f(c)$ implies $\nu_X(S_c^\circ)=0$, which in turn implies
that $Eg(X)=\int_{S_c\setminus S_c^\circ} g(c)\,\nu_X(du\times dv)=g(c)$ for all (say) nonnegative $g\colon\mathbb R\to\mathbb R$, which means that
$P(X=c)=1$.
Thank you for this answer. Can you please elaborate on how proposition 3.18 implies the required form of Jensen inequality? I am having a bit of trouble deducing it. BTW, this approach seems to be a bit of an overkill, isn't it? (it seems less elementary than just proving the existence of a supporting line. I hoped there would be a more elementary reference, say some book on convex analysis, rather than a paper from 2009. I am quite sure this version of Jensen was known long before that...).
Anyway, I am interested in understanding better this more general approach, so if you could elaborate more on how it implies Jensen, it would be great.
Oh, and last question: Does the approach you have described imply that if $\phi$ is strictly convex at $c$, and there is an equality in Jensen, then the random variable is constant a.e.?
@AsafShachar : I have added the details you requested, and even an (affirmative answer) to your additional question concerning the strict convexity at $c$.
@Iosif: somewhat tangentially: is it true that any probability measure on $\mathbb{R}^2$ with mean $c$ can be obtained from the mixture of probability measures with mean $c$ that are supported on 3 points? (And the obvious follow up with $\mathbb{R}^n$ and $n+1$ points.)
@WillieWong : This is so. Indeed, consider first the probability measures with a compact support. For any compact $K\subset\mathbb R^n$, let $P_{K,c}$ be the set of all probability measures with support $\subseteq K$ and mean $c$. Then $P_{K,c}$ is convex and compact and hence, by the Choquet–Bishop–de Leeuw theorem, any point in $P_{K,c}$ is a mixture of extreme points of $P_{K,c}$, which latter are probability measures with support of cardinality $\le n+1$. It remains to note that any probability measure on $\mathbb R^n$ is a mixture of probability measures with a compact support.
Thanks! This argument is so amazingly simple.
|
2025-03-21T14:48:31.462865
| 2020-07-07T15:20:01 |
365046
|
{
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|
Stack Exchange
|
A nontrivial principal bundle which satisfies Leray-Hirsch theorem
What is an example of a nontrivial principal bundle whose fibre space $G$, total space $P$ and base space $M$ are compact connected manifolds (the fiber $G$ is a compact Lie group) such that $$H^*(P,\mathbb{Q})=H^*(G,\mathbb{Q})\otimes H^*(M,\mathbb{Q})$$
Hatcher Corollary 4D.3 should provide examples if I am not mistaken.
The examples pointed out by Ulrich Pennig are the bundles $U(n-1) \to U(n) \to S^{2n-1}$, $SU(n-1) \to SU(n) \to S^{2n-1}$, and $Sp(n-1) \to Sp(n) \to S^{4n-1}$ so one doesn't have to look far to find such examples. Imagine what Lie group theory would be like if all these bundles were trivial, so all these groups would be products of odd-dimensional spheres!
@AllenHatcher Thank you for adding the details. I was in a rush and did not have time to fill them in.
Let $P$ be any $SU(2)$-bundle on $X$ with vanishing second Chern class $c_2(P)$. The hypotheses of the Leray-Hirsch theorem are satisfied if there is a class in $H^3(P)$ which restricts to the generator of $H^3(SU(2))$. This happens if and only if in the Leray spectral sequence, the map $d_3: H^0(X,H^3(SU(2))) \to H^4(X,H^0(SU(2))$ vanishes (since $H^i(SU(2))$ is concentrated in degrees $0$ and $3$, so the only nontrivial differentials are on the third page). This map is exactly the top Chern class.
There are nontrivial rank 2 complex bundles with vanishing Chern classes. The vanishing of $c_1$ implies that the $U(2)$ structure may be reduced to $SU(2)$, at which point the above argument shows that the hypotheses of Leray-Hirsch are satisfied.
An example of a non-trivial $SU(2)$-bundle with top Chern class zero is $SU(2) \to SU(3) \to S^5$, which is one of the examples Allen Hatcher mentions in the comments above.
Suppose $M = S^n$ is a sphere with $n$ odd and at least $5$. Pick your favorite Lie group $G$ for which $\pi_{n-1}(G)$ is non-trivial. (Many examples may be found here. For example, for any $n > 3$, $G= SU(\frac{1}{2}(n-1))$ works.) Since principal $G$-bundles over $M$ are classified by $[M,BG]$ which is in bijection with $[S^{n-1},G]$, there is a non-trivial principal $G$-bundle.
Suppose $P\rightarrow M$ is any such non-trivial bundle. Then $H^\ast(P;\mathbb{Q})\cong H^\ast(G;\mathbb{Q})\otimes H^\ast(M;\mathbb{Q})$. One way to see this is to note that Borel showed the universal bundle $EG\rightarrow BG$ is totally transgressive: differentials originating on the fiber are trivial, except possibly when they land in the base. Thus, the same must be true in the bundle $P\rightarrow M$. But since the rational cohomology ring of $G$ is generated in odd degrees and $H^\ast(M;\mathbb{Q})$ is concentrated in odd degrees, all the differentials must vanish.
There is also a low-dimensional example. Consider the canonical map $\mathbb RP^3 \to \mathbb RP^{\infty} \to \mathbb CP^{\infty}$, classifying the unique non-trivial principal $S^1$-bundle with base $\mathbb RP^3$. Its rational Serre spectral sequences collapses. Of course, its integral Serre spectral sequence does not collapse. The total space of this bundle is the 4-manifold $E = S^1 \times_{\mathbb Z/2} S^3$ whose fundamental group is $\mathbb Z$, not $\mathbb Z \times \mathbb Z/2\mathbb Z$.
|
2025-03-21T14:48:31.463085
| 2020-07-07T16:02:45 |
365049
|
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"Francesco Polizzi",
"R. van Dobben de Bruyn",
"https://mathoverflow.net/users/7460",
"https://mathoverflow.net/users/82179"
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|
Stack Exchange
|
Irreducibility of a polynomial when the sum of its coefficients is prime
I came up with the following proposition, but don't know how to prove it.
I used Maple to see that it holds when $ a + b + c + d <300 $.
Let $a,b,c$ and $d$ be non-negative integers such that $d\geq1$ and $a+b+c\geq1$.
If $a+b+c+d$ is a prime number other than $2$, the polynomial $ax^3+bx^2+cx+d$ is irreducible over $Z[x]$.
The sum of the coefficients of $$(\alpha x^2 + \beta x + \gamma)(ux +v)$$ is $$(\alpha+\beta+ \gamma)(u+v).$$
@FrancescoPolizzi: do we know anything about positivity of the factors?
@R.vanDobbendeBruyn: in fact no, and I am not claiming that this is a proof of anything. For a counterexample (if exists) we must have either $\alpha+\beta+\gamma$ or $u+v$ invertible in $\mathbb{Z}$, and the other quantity an odd prime number. Moreover, the four quantities $$\alpha u, \quad \alpha v + \beta u, \quad \beta v + \gamma u, \quad \gamma v$$
must be all non-negative.
Francesco Polizzi's idea is enough to solve the problem:
Lemma. Let $f = ax^3 + bx^2 + cx + d \in \mathbf Z_{\geq 0}[x]$ nonconstant with $d > 0$, such that $f(1)$ is a prime $p > 2$. Then $f$ is irreducible in $\mathbf Z[x]$.
Proof. Suppose $f = gh$ for $g,h \in \mathbf Z[x]$; we must show that $g \in \{\pm 1\}$ or $h \in \{\pm 1\}$.
First assume $\deg g > 0$ and $\deg h > 0$. Since $\deg f \leq 3$, without loss of generality we may assume $\deg g = 1$ and $m = \deg h \in \{1,2\}$; say
\begin{align*}
g = ux + v, & & h = \alpha x^2 + \beta x + \gamma ,
\end{align*}
with $u > 0$ and $\alpha > 0$ or $\alpha = 0$ and $\beta > 0$. If $v \leq 0$, then $f$ has a nonnegative real root, which is impossible because $f$ has non-negative coefficients and $f(0) > 0$. Thus, $v > 0$, and therefore $g(1) \geq 2$, forcing $g(1) = p$ and $h(1) = 1$. By symmetry, this rules out the case $\deg h = 1$. Thus, we may assume $\alpha > 0$. The formulas
\begin{align*}
a = \alpha u, & & b = \beta u + \alpha v, & & c = \gamma u + \beta v, & & d = \gamma v
\end{align*}
give
\begin{align*}
\alpha > 0, & & \alpha v \geq -\beta u, & & \gamma u \geq -\beta v, & & \gamma > 0.\label{1}\tag{1}
\end{align*}
We will show that $h(1) \geq 2$, contradicting the earlier assertion that $h(1) = 1$. This is clear if $\beta \geq 0$, since we already have $\alpha > 0$ and $\gamma > 0$. If $\beta < 0$, then \eqref{1} shows
$$\frac{\gamma}{-\beta} \geq \frac{v}{u} \geq \frac{-\beta}{\alpha} > 0,\label{2}\tag{2}$$
hence
\begin{align*}
h(1) = \alpha + \beta + \gamma &= \left(1 - \left(\frac{-\beta}{\alpha}\right) + \left(\frac{\gamma}{-\beta}\right)\left(\frac{-\beta}{\alpha}\right)\right)\alpha\\
&\geq \left(1 - \left(\frac{-\beta}{\alpha}\right) + \left(\frac{-\beta}{\alpha}\right)^2 \right)\alpha.
\end{align*}
The function $1-x+x^2$ attains a minimum of $\tfrac{3}{4}$ at $x = \tfrac{1}{2}$, and is bigger than $1$ on $(1,\infty)$. Thus, we only have to consider $x = \tfrac{-\beta}{\alpha} \in (0,1]$. For $x < 1$, we must have $\alpha \geq 2$, so $(1-x+x^2)\alpha \geq \tfrac{3}{2} > 1$. Finally, for $x = 1$, i.e. $\beta = -\alpha$, \eqref{2} shows that $v \geq u$. Since $p = v+u$ is odd, we must have $v > u$, so \eqref{2} gives $\gamma > -\beta$, so $h(1) = \gamma \geq 2$.
Thus, we conclude that $\deg g = 0$ or $\deg h = 0$; say $g = n$ for $n \in \mathbf Z_{>0}$ without loss of generality. If $n > 1$, then $n = p$, which is impossible since $f(0) = d \in \{1, \ldots, p-1\}$ is not divisible by $p$. So we conclude that $g = 1$. $\square$
Remark. The argument above relies crucially on the assumption $\deg f \leq 3$. More importantly, for higher degree polynomials the same result is false without a lower bound on $p$ that grows at least linearly in $\deg f$:
Example. Let $p$ be an odd prime. Then the polynomial $f = x^{2p-2} + x^{2p-4} + \ldots + x^2 + 1$ has $f(1) = p$, but it factors as
$$\Big(x^{p-1}+x^{p-2}+\ldots+x+1\Big)\Big(x^{p-1}-x^{p-2}+\ldots-x+1\Big).$$
Indeed, the above reads
$$\zeta_p(x^2) = \zeta_p(x)\zeta_{2p}(x) = \zeta_p(x)\zeta_p(-x),$$
which is true because both sides are monic and have the same roots in $\mathbf C$ (namely the $2p^\text{th}$ roots of unity except $\pm 1$, all with multiplicity $1$).
|
2025-03-21T14:48:31.463335
| 2020-07-07T16:13:26 |
365050
|
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|
Stack Exchange
|
Asymptotics of a quotient related to a simple random walk
Let $\lambda_0 < \lambda_1$ and $\lambda_0 \lambda_1 > 1$ (i.e. at least $\lambda_1 > 1$). Further, let $S_n$ denote a simple random walk with increment distribution $$ P(X = 0)= P(X= 1) = 1/2.$$
Note that, using the SLLN, one can easily prove that (as $n \to \infty$) $$ \lambda_0^{S_n} \lambda_1^{n-S_n} \to \infty \ \ a.s.$$
Now, my goal is to find a deterministic sequence $(a_n)_n$ such that $$ E \left[ \frac{\lambda_0^{S_n} \lambda_1^{n-S_n}} {\left(\sum_{k=0}^{n-1} \lambda_0^{S_k} \lambda_1^{k-S_k}\right)^2}\right] \Big{/} a_n \longrightarrow c \tag{1} $$
for some constant $c > 0$ as $n \to \infty$. To put it shortly, I am interested in the asymptotic behavior of the stated expectation.
My (not very educated) guess for such a sequence is $$ a_n = E \left[ \frac{1}{\lambda_0^{S_n} \lambda_1^{n-S_n}}\right].$$ The reason for this guess is that in a deterministic case (for any $\lambda > 0$), we have as $n \to \infty$ that $$ \frac{\lambda^n}{\sum_{k=0}^{n-1} \lambda^k} \to c> 0.$$
I am looking for ideas on how to handle the expectation in (1) and deduce such a "normalizing" sequence. Handling the expression in (1) seems quite complicated... Any ideas are much appreciated!
I would use a "Girsanov like" trick to modify the expectation: For a general random walk with $\mathbb{P}(X=1)=p$ that we denote the associate law $\mathbb{E}_p$.For $\alpha\in \mathbb{R}$ we have
$$\mathbb{E}_p(f(\boldsymbol{S}))=\sum_\boldsymbol{S} f(\boldsymbol{S})p^{S_n}(1-p)^{n-S_n}=\sum_{\boldsymbol{S}} f(\boldsymbol{S})\lambda_0^{\alpha S_n}\lambda_1^{\alpha(n-S_n)}(\frac{p}{\lambda_0^\alpha})^{S_n}(\frac{1-p}{\lambda_1^\alpha})^{n-S_n} = (\frac{p}{\lambda_0^\alpha}+\frac{1-p}{\lambda_1^\alpha})^n\mathbb{E}_{\tilde{p}}[f(\boldsymbol{S})\lambda_0^{\alpha S_n}\lambda_1^{\alpha(n-S_n)}]$$ with $\tilde{p}=\frac{p}{\lambda_0^\alpha}(\frac{p}{\lambda_0^\alpha}+\frac{1-p}{\lambda_1^\alpha})^{-1}$. This is to be understood as follow: one can change the law of the random walk $p\rightarrow \tilde{p}$ (adding a drift) if we add the weight $\lambda_0^{\alpha S_n}\lambda_1^{\alpha(n-S_n)}$. In our case $\tilde{p}=\frac{1}{\lambda_0^\alpha}(\frac{1}{\lambda_0^\alpha}+\frac{1}{\lambda_1^\alpha})^{-1}$.
Fist Case: If for $\alpha=1$, $$\tilde{p}\log(\lambda_0)+(1-\tilde{p})\log(\lambda_1)> 0$$
then choosing $\alpha=1$
$$\mathbb{E}_{1/2}(\frac{\lambda_0^{ S_n}\lambda_1^{(n-S_n)}}{(\sum_k \lambda_0^{ S_k}\lambda_1^{(k-S_k)})^2})= (\frac{1}{2\lambda_0}+\frac{1}{2\lambda_1})^n\mathbb{E}_{\tilde{p}}[\frac{(\lambda_0^{ S_n}\lambda_1^{(n-S_n)})^2}{(\sum_k \lambda_0^{ S_k}\lambda_1^{(k-S_k)})^2}]= (\frac{1}{2\lambda_0}+\frac{1}{2\lambda_1})^n\mathbb{E}_{\tilde{p}}[\frac{1}{(\sum_k \lambda_0^{S_k-S_n}\lambda_1^{(k-n+S_n-S_k)})^2}] $$ Because of the assumption on $\tilde{p}$, with probability that goes to 1 $\lambda_0^{S_k-S_n}\lambda_1^{(k-n+S_n-S_k)}\rightarrow 0$ exponentially fast as $n-k\rightarrow \infty$. And then we should obtain $$0<c\leq \mathbb{E}_{\tilde{p}}[\frac{1}{(\sum_k \lambda_0^{S_k-S_n}\lambda_1^{(k-n+S_n-S_k)})^2}]\leq 1 $$
Second Case :If the assumption on $\alpha=1$ fails then we choose $\alpha\leq 1$ such that $$\tilde{p}\log(\lambda_0)+(1-\tilde{p})\log(\lambda_1)=0$$ For the corresponding law with paramatter $\tilde{p}$, $S_k\log(\lambda_0)+(k-S_k)\log(\lambda_1)$ should behave for large $k$ as a brownian motion $\sigma B_k$. We could then guess
$$\mathbb{E}_{1/2}(\frac{\lambda_0^{ S_n}\lambda_1^{(n-S_n)}}{(\sum_k \lambda_0^{ S_k}\lambda_1^{(k-S_k)})^2})= (\frac{1}{2\lambda_0^\alpha}+\frac{1}{2\lambda_1^\alpha})^n\mathbb{E}_{\tilde{p}}[\frac{(\lambda_0^{ S_n}\lambda_1^{(n-S_n)})^{1+\alpha}}{(\sum_k \lambda_0^{ S_k}\lambda_1^{(k-S_k)})^2}] \approx (\frac{1}{2\lambda_0^\alpha}+\frac{1}{2\lambda_1^\alpha})^n\mathbb{E}[\frac{e^{(1+\alpha)\sigma B_n}}{(\max_k e^{\sigma B_k})^2}]\\= (\frac{1}{2\lambda_0^\alpha}+\frac{1}{2\lambda_1^\alpha})^n\mathbb{E}[e^{(1+\alpha)\sigma B_n-2\max_k \sigma B_k}] $$
where we make the assumption that the sum is dominated by the largest term. Hopefully this last espectation with the brownian motion can be estimated.
Thank you! I am slightly confused by the notation when you are presenting the "Girsanov like" trick. What does $\mathbf{S}$ denote and how come that you are taking the expectation but after the first equality there are still $S_n$ in the sum?
$\boldsymbol{S}=(S_1,S_2,\cdots, S_n)$ and $\mathbb{P}(\boldsymbol{S}=(S_1,S_2,\cdots, S_n))=p^{S_n}(1-p)^{n-S_n}1_{\forall i, |S_i-S_{i+1}|=1}$.
|
2025-03-21T14:48:31.463574
| 2020-07-07T16:15:30 |
365051
|
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"url": "https://mathoverflow.net/questions/365051"
}
|
Stack Exchange
|
Triangle-free graph with chromatic number >= 4 and one vertex of a color
I was wondering about the existence of an example of a graph in the title - ie a triangle-free graph which is not 3-colorable such that one of the colors only colors one vertex. The Groetzsch graph needs at least 2 of each color as far as I can tell. I would prefer an example which is 4-colorable.
Any ideas on how to find such a graph other than a brute force computer search? My understanding is that although the Groetzsch graph is the smallest triangle-free graph which is not 3-colorable it does not necessarily imply that a larger graph cannot have the property I'm looking for.
Just to clarify: are you asking for a triangle-free graph which is not 3-colorable, but in which there is a vertex s.t. when coloring this vertex with a 4th color the remaining graph is 3-colorable?
yes, and the graph should be triangle-free
In that case I think the Groetzsch graph should work. If you color the unique vertex of degree 5 with the fourth color, the rest is 3-colorable. In fact, a smallest triangle-free graph that is not 3-colorable must necessarily work: if you would need to color two vertices with the fourth color, you could just remove one of them to get a smaller graph which is still triangle-free and still not 3-colorable, contradicting the minimality.
Ah, it does indeed work! Thank you
|
2025-03-21T14:48:31.463812
| 2020-07-07T16:29:39 |
365052
|
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"Adam P. Goucher",
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"bof",
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|
Stack Exchange
|
Intuition about ordinal fixed points $\alpha = \aleph_\alpha$
I wanted to ask for your intuition about ordinal fixed points $\alpha = \aleph_\alpha$, where $\aleph_\alpha$ stands for the $\alpha$-th Aleph number in the Aleph sequence of cardinalities.
For background why I am asking this. I was surprised when I first learned $|\mathbb{Q}| = |\mathbb{N}|$ and $|{\cal P}(A)|>|A|$ for all sets $A$, but gained an intuition over time. It took me a bit longer to understand the convergence proof for Goodstein sequences, because initially I did not understand why a strictly decreasing sequence of ordinals is zero after finitely many steps; I had the wrong intuition about well-ordering (I thought "going downstairs" would be symmetric to "going upstairs").
Now, I am still unable to find the right intuition for ordinal fixed points, especially for the Aleph sequence. I am aware of the fixed-point lemma for normal functions of ordinals from Veblen. But I have not really gained an intuition from knowing the proof. In a sense, I can understand the proof formally only, but not "morally".
In my intuition (which might be wrong), I am starting from
$0\mapsto\aleph_0\\
1\mapsto\aleph_1\\
2\mapsto\aleph_2$
and so on. The difference in size between the ordinal index $\alpha$ and the Aleph number $\aleph_\alpha$ gets enormously big in a very fast way. My intuition is, the index $\alpha$ can never catch up with $\aleph_\alpha$, even when $\alpha$ is a limit ordinal. In my mental picture, a limit ordinal $\alpha$ can be an extremely "high jump", but it can never really catch up with all the extremely high jumps of the Aleph sequence $(\aleph_\alpha)$ which happen in every single step.
Please could you help me find the right intuition, or maybe point me to the error in my current mental picture? I might be overlooking something obvious!
The first aleph fixed point is the limit of $0, \aleph_0, \aleph_{\aleph_0}, \aleph_{\aleph_{\aleph_0}}, \dots$. Each ordinal $x$ below this limit lies in a 'bucket' between 2 consecutive terms of this sequence, and the function $x \mapsto \aleph_x$ only pushes that ordinal into the next bucket.
An easier analogy is the first fixed point of $x \mapsto \omega^x$, namely $\varepsilon_0$. If you can convince yourself that $\varepsilon_0$ is a thing, then what's wrong with the first aleph fixed point?
@AdamP.Goucher Thanks a lot Adam for your comment! Your statement that $x\mapsto \aleph_x$ just pushes it to the next bucket is great. I never looked at it this way. I would like to upvote that, would you be willing to post this as an answer?
Does your intuition tell you that, if a sequence $a_0,a_1,a_2,\dots$ increases very very rapidly, then $a_0,a_1,a_2,\dots$ can never catch up with $a_1,a_2,a_3,\dots$?
@bof Thanks a lot - you make a valid point. Learning from all the answers and comments here, getting a better understanding of the idea of limit ordinal
To the original poster and all the repliers, how do you know that your fixed point in aleph isn't a proper class?
@Louis_R The answer to your question is replacement. In ZC (= set theory without replacement) it is indeed consistent that there are no aleph-fixed-points, but ZFC proves that aleph-fixed-points exist.
Your intuition is finitary, and therefore wrong. Compare, for example, the two sequences:
$\alpha_n=n$, and
$\beta_n=2^n$.
It is easy to see that $\alpha_n<\beta_n$ for all $n$. We even know from elementary calculus that the rate of change between them is growing very fast as well, so there is no possible way for $\alpha_n$ to be equal to $\beta_n$. Game over, then, I suppose. We can all stop reading this and go get a beer.
But wait a minute, you might say, what about their limit? What is $\sup\alpha_n$ and how does it compare to $\sup\beta_n$? Well, both are $\omega$.
See, limit steps are a great opportunity for the slow and steady to catch up with the rapid. As long as the two sequences are continuous and increasing, they will catch up with one another at some limit points.
And so indeed, setting $\alpha_0=\omega$ and $\alpha_{n+1}=\omega_{\alpha_n}$, will have you with $\alpha=\sup\alpha_n=\omega_\alpha$. And therefore $\alpha=\aleph_\alpha$. The key point here is that the sequence is increasing incredibly fast. Enough to catch up at the limit point. And of course we may replace $\alpha_0$ with any ordinal as our starting point.
What does it mean for a sequence to be continuous?
$\sup\alpha_i=\alpha_{\sup i}$.
Ah, so it's an ordinal-indexed sequence? I thought the $n$ was meant to suggest a natural-indexed sequence.
Well, natural numbers are ordinals...
@AsafKaragila Thanks a lot Asaf for a great answer! Really helpful for me. So it is really the "jump" at the limit step which can be so incredibly large that $\alpha$ catches up with $\aleph_\alpha$. I think you are right in saying my mental model was more finitary!
@ClausDollinger: It's only natural. I used to see freshman year students who would always fall back to their intuition that "set" meant a finite set of natural numbers.
We can all stop reading this and go get a socially distanced beer.
@Carl-FredrikNybergBrodda: You don't even drink.
I know this great bar that can accommodate as many of us as we like, while maintaining distance. It's at the Hilbert Hotel...
|
2025-03-21T14:48:31.464432
| 2020-07-07T16:43:44 |
365053
|
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"Simon Henry",
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|
Stack Exchange
|
Algebras for products or limits of monads
If a category $C$ has limits of a certain type, then the category of monads on $C$ has the same type of limits, and these limits are computed "levelwise" (i.e. are preserved by the forgetful functor from the category of monads on $C$ to the category of endofunctor on $C$).
I wonder if algebras for such limits have been studied somewhere ?
If I focus on product of monads for simplicity there seems to be some interesting to say:
Given $M$ and $N$ two monads on $C$ the monad $M \times N$ on $C$ can be seen as induced by the adjunction:
$$(M , N) : C \leftrightarrows M\text{-Alg} \times N\text{-Alg} : U \times U'$$
Where $U \times U'$ is the functor sending a pairs of algebra $(X,Y)$ to $U(X) \times U'(Y)$ where $U$ and $U'$ denotes the forgetful functors, and $(M,N)$ is the functor sending $X$ to the pair $(M(X),N(X))$.
This adjunction is unfortunately not always monadic, but it is often not very far from it.
If I look at the simplest example: $C= Set$, $M=N=Id$, then I can prove by hand that a non-empty $Id \times Id$-Algebra is the same as a pair of non-empty sets, so that the category of $Id \times Id$-Algebra identifies with a the full subcategory of (Set $ \times $ Set) of pairs of sets such that one is empty if and only if the other is also empty.
I haven't check it in full details, but it seems there are many examples where the $(M \times N)-Algebras$ are really pairs of algebras. For example if $M$ and $N$ are monads on Set whose algebras are always non-empty.
Has this been studied somewhere in more detail ? Is there some assumptions under which we can describe algebra for some limits of monads in terms of algebras for the individual monads ?
By the formal theory of monads, the 2-functor $Alg: Mnd \to Cat$ is right 2-adjoint to the 2-functor which sends a category to the identity monad on it (here $Mnd$ is the 2-category of monads in $Cat$). So perhaps one should ask which limits in $Mnd(C)$ are also limits in the larger 2-category $Mnd$. I'm a bit confused, though, because there's some weirdness about which direction the coherence 2-cell points in the definition of a 1-morphism of monads, which might not line up properly for $Mnd(C)$ to actually be a sub-2-category of $Mnd$.
@TimCampion That's interesting. There are several way to construct a category Mnd of monads, which one are you referring to in your comment ? is it equivalent to the full subcategory of $\mathbf{Cat}^\to$ of monadic right adjoint functors ? I have the impression that the underlying idea of your comment actually say something about colimits of monads instead of limits (due to a hidden opposite category in the convention), for which it is indeed easy to describe the category of algebras...
For finitary commutative monads on $\mathbf{Set}$, this has been studied in Faro–Kelly's On the canonical algebraic structure of a category. I have reworded Proposition 11 ibid. below in terms of finitary monads rather than algebraic theories.
Proposition (Faro–Kelly).
For an finitary monad $\mathscr T$, write $(\mathscr T\text{-Alg})'$ for the full subcategory of $\mathscr T\text{-Alg}$ given by the non-empty algebras.
Let $\mathscr T$ and $\mathscr S$ be non-degenerate commutative finitary monads on $\mathbf{Set}$; so that $\mathscr T \times \mathscr S$ is another such. If both $\mathscr T$ and $\mathscr S$ have nullary operations; we have
$$(\mathscr T \times \mathscr S)\text{-Alg}\cong \mathscr T\text{-Alg}\times \mathscr S\text{-Alg}$$
Otherwise $(\mathscr T \times \mathscr S)\text{-Alg}$ is obtained from $((\mathscr T \times \mathscr S)\text{-Alg})'$ by freely adding an initial object; and $((\mathscr T \times \mathscr S)\text{-Alg})'$ is itself isomorphic to $(\mathscr T\text{-Alg})' \times (S\text{-Alg})'$.
The proof appears to generalise at least to arbitrary commutative monads with rank on $\mathbf{Set}$.
|
2025-03-21T14:48:31.464706
| 2020-07-07T16:53:22 |
365054
|
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|
Stack Exchange
|
Is it possible to compute the basis of this module?
Let $A$ be a polynomial algebra in $n$ variables over field $\mathbb{F}$ of characteristic zero which is algebraically closed. Assume that $a_1,\ldots, a_n, b_1,\ldots, b_n\in A$ are such that $a_1b_1+\ldots+ a_nb_n=1$. Consider the free left $A$-module $_{A}M$ with basis $r_1,\ldots, r_n$.
From Suslin-Quillen theorem we know that $_{A}M/(A(a_1r_1+\ldots+a_nr_n))$ is free $A$-module of rank $n-1$.
$\bf{Question.}$ Is there any formula of a basis of this module in terms of $a_i, b_i$?
No, it is what is known as a stably free module, but in general not free. The standard example is the tangent bundle of the real sphere. Let $A=\mathbb{R}[x,y,z]/(x^2+y^2+z^2-1)$ and take $a_1=x, a_2=y, a_3=z$.
Is there any such examples if $\mathbb{F}$ is algebraically closed. And $A$ is polynomial algebra?
If $A$ is the polynomial ring, then no such exists, so called Quillen-Suslin theorem, answering a question of Serre. If $A$ is not a polynomial ring, there are many such, even for algebras over algebraically closed fields.
I edited the question.
As I said, this is true, and the characteristic of the field is irrelevant.
I agree so I think I can delete the question. But is there any algorithm to construct its basis if given polynomials $a_i$?
"Algebraically closed" is also irrelevant.
@solver6 please don't delete your question
https://doi.org/10.1016/0021-8693(92)90189-S
So You can post it as answer. This article will be useful for me. I am solving some problem where polynomials $a_i$ are given only by some properties. And I think that there doesn't exist any formula to the basis of this free module and I should use existence of basis only as an abstract result.
|
2025-03-21T14:48:31.464860
| 2020-07-07T16:59:59 |
365057
|
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|
Stack Exchange
|
Question on countably homogeneous structures
A homogeneous structure is a countable first order structure $M$ over a relational language such that any isomorphism between finite substructures of $M$ can be extended to an automorphism of $M$.
Lachlan proved that if $M$ is any stable countably homogeneous structure over a finite relational language $\mathcal{L}$, then $M$ is a union of a chain $\{M_n : n ∈ \Bbb{N}\}$ of finite homogeneous $\mathcal{L}$-structures, and each sentence $σ ∈ Th(M)$ holds in all but finitely many of $M_n$.
I think that Lachlan proved this theorem based upon graph theory because he mainly works on model based graph theory. I wonder which of Lachlan's papers (or papers/books by others) contains the proof of this theorem.
Let's first observe that any stable countable homogeneous structure in a finite relational language is $\aleph_0$-categorical and $\aleph_0$-stable. This is explained in A survey of homogeneous structures by Macpherson: $\aleph_0$-categoricity is Corollary 3.1.3 on p. 17, and $\aleph_0$-stability is in the paragraph just before Example 3.3.2 on p. 22.
Then the result in your question follows directly from Corollary 7.4 in $\aleph_0$-categorical, $\aleph_0$-stable structures, by Cherlin, Harrington, and Lachlan.
I'm sure that the result also follows from the work in the pair of papers On countable structures which are homogeneous for a finite relational language by Lachlan and Stable finitely homogeneous structures by Cherlin and Lachlan. But after taking a quick look at those papers, I wasn't able to find a clear statement of the result in the form you quoted it.
By the way, it's completely wrong to think that Lachlan is "mainly a graph theorist". He is a first-rate logician, and while he did a lot of work at the intersection of model theory and combinatorics, he has broad interests in pure model theory and computability theory. The work on stable homogeneous structures does not use graph theory (though the results do apply to certain theories of graphs). Instead, this work uses a lot of group theory, including the classification of finite simple groups.
|
2025-03-21T14:48:31.465012
| 2020-07-07T17:09:15 |
365060
|
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|
Stack Exchange
|
Ekedahl sieve for composite moduli
The following version of a theorem of Ekedahl, known as the Ekedahl sieve, can be found in this paper by Bhargava and Shankar for example, is stated as follows: let $B$ be a compact subset of $\mathbb{R}^n$ having finite measure, and let $Y$ be any closed subscheme of $\mathbb{A}^n(\mathbb{Z})$ having co-dimension $k \geq 1$. Let $r, M$ be positive real numbers. Then
$$\displaystyle \# \{\mathbf{a} \in rB \cap \mathbb{Z}^n : \mathbf{a} \pmod{p} \in Y(\mathbb{F}_p) \text{ for some prime } p > M \} $$
$$ = O \left(
\frac{r^n}{M^{k-1} \log M} + r^{n-k+1} \right)$$
with the implied constant depending only on $Y$ and $M$.
If we replace the set on the left hand side with the set
$$\displaystyle S_M(r) = \{\mathbf{a} \in rB \cap \mathbb{Z}^n : \mathbf{a} \pmod{p} \in Y(\mathbb{F}_p) \text{ for all } p | q \text{ for some square-free } q > M \}$$
can we obtain an upper bound of comparable strength to the original theorem?
Minor correction: In the linked paper by Bhargava and Shankar, the implied constant should depend only on $Y$ and $B$.
|
2025-03-21T14:48:31.465111
| 2020-07-07T17:29:25 |
365061
|
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|
Stack Exchange
|
An extension of Lomonosov Theorem
Let $H$ be a complex infinite dimensional separable Hilbert space. There are various extensions of the following well known result:
Theorem (Lomonosov): Every nonscalar $T \in B(H)$ which commutes with a nonzero compact operator $K$ has a nontrivial hyperinvariant subspace.
It has been shown that there exist operators $T$ which do not commute with any nonzero compact $K$. This led to the following two generalisations; the first one can be found, for instance, in the book Kubrusly, C. S. Hilbert space operators. Birkhauser, Boston, 2003 (Problem and Solution 12.4), while the second one is obtained in Lauric, V. (1997). Operators $\alpha$-Commuting with a Compact Operator. Proceedings of the American Mathematical Society, 125(8), 2379-2384.
Theorem: Let $T \in B(H)$ be nonscalar. If there exists a nonzero compact $K$ such that $\operatorname{rank} (TK-KT) \leq 1$, then $T$ has a nontrivial hyperinvariant subspace.
Theorem: Let $T \in B(H)$ be nonscalar. If there exists a nonzero compact $K$ such that $TK= \alpha KT$ for some $\alpha \in \mathbb{C}$, then $T$ has a nontrivial hyperinvariant subspace.
I was wondering if the following natural generalisation is true: if there exists a nonzero compact $K$ such that $\operatorname{rank}(TK - \alpha KT) \leq 1$ for some $\alpha \in \mathbb{C}$, then there exists a nontrivial hyperinvariant subspace.
Looking again at Solution 12.4 in Kubrusly's book, I have noticed that the proof can be used even to prove the statement above, with some small changes. I will briefly sketch such small modifications: instead of considering the operator $TS-ST$, we will consider $C:=TS-\alpha ST$ ($\alpha \neq 0$). The claim:
If $\operatorname{rank}(C) =1$, then $\mathcal{R}(C) \subseteq \mathcal{R}(S)$.
(where $\mathcal{R}(\cdot)$ denotes the range) still holds, and it can be proved as in the cited book. The remaining part of the proof of (a) does not need any modification. The next step in Kubrusly's book (part (b)) is still applicable, with $C:=TS-\alpha ST$ and not as $TS-ST$ (as before). This time, the case $LC=0$ leads to a contradiction because of Lauric extension of Lomonosov Theorem (see the reference above). The case with $\operatorname{rank}(LC)=1$ is proved via part (a). We obtain again a contradiction and thus we conclude the proof of the above statement. I refer to Kubrusly, C. S. Hilbert space operators. Birkhauser, Boston, 2003 (Problem and Solution 12.4) for all the missing details and notations.
|
2025-03-21T14:48:31.465282
| 2020-07-07T17:50:15 |
365062
|
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|
Stack Exchange
|
Prove the optimal solution to maximizing nuclear norm with constraints is attained at corner points of feasible region
The nuclear norm (trace norm) of a matrix $X \in \Bbb R^{m \times n}$ is defined as
$$\|X\|_* := \sum_{i=1}^{\min(m,n)} \sigma_i(X)$$
where $\sigma_i(X)$ are the singular values of $X$.
The optimization problem I met is as follows,
$$
\max_X \|X\|_*
$$
where $X\in \Bbb R^{m\times n}$ needs meet the constraints: $X_{ij}\ge 0$ and $\sum_{j=0}^n X_{ij}=1$. That is to say, each row of $X$ is a probability distribution.
Question: I want to prove that the optimal solution $X^*$ is only attained at corner points of the feasible region, i.e., the row of $X^*$ is from the set $\{e_1,...,e_n\}$, where $e_i$ is a standard orthogonal basis vector of the space of $\Bbb R^n$.
What I have done is constructing a new optimization problem $F(X_k)$ as follows, and prove $F(X_k)$ is strictly convex. Then the problem can be solved by proof by contradiction. However, I can't prove it. I also posted a question on Mathematics StackExchange, but there hasn't been an answer until now.
$$\max_{X_k}F(X_k)=\|X\|_*$$
where $X_k$ is the $k$-th row of $X$, and the constraints for $X$ are the same as abovementioned.
The nuclear norm $\|\cdot\|_*$ is a norm and hence a convex function. On the other hand, the set
$$S:=\{X\in\mathbb R^{m\times n}\colon
X_{ij}\ge0\text{ and }\sum_{j=1}^n X_{ij}=1\ \forall i,j\}$$
is convex and compact. So, the maximum of the nuclear norm $\|\cdot\|_*$ on the set $S$ is attained at an extreme point of $S$, which is clearly a matrix $X\in S$ such that one entry of each row of $X$ is $1$ and the other entries are $0$. (You have to allow the non-strict inequalities $X_{ij}\ge0$ for the maximum to be attained.)
I'm sorry but there is ambiguity in my question. I mean the optimal solution in only attained at extreme points. I have updated the question accordingly. For example, $F(X)=1$ is a convex function, but the optimal solution is attained at any feasible solution.
@Jack Why is Iosif Pinelis' answer not correct?
@MarkL.Stone My question is to prove that the optimal solution is only attained at the extreme point of $S$ and can't be other points. That is to say, being an extreme point is a necessary condition for an optimal solution. I gave a counterexample based on losif Pinelis's answer. For example, $F(X)=1$ is a convex function, but the optimal solution is attained at any feasible solution. Thus, it's not enough to only depend on the convexity of nuclear norm.
Yes, the result quoted by @Iosif Pinelis is a well-known result that if there is a global optimum of a concave programming problem (i.e., minimizing a concave function (or, equivalently, maximizing a convex function) subject to convex constraints) with compact (convex) constraints, there is a global optimum at an extreme of the constraints. That does not rule out that there could be additional global optima located elsewhere..Why do you think that situation could be avoided for your problem?
@MarkL.Stone I plotted the objective values in the case of two dimensions. I found what I want to prove is true. And I believe it's also true for higher dimensions. I think the main difficulty is from the unusual definition of nuclear norm of a matrix. The definition seems disconnected from the original elements of the matrix because of the operation of SVD. So I can't find the proper tool to study the behavior of nuclear norm with respect to the change of elements of the matrix.
I think you should be able to differentiate the nuclear norm with respect to the matrix elements, using $|X|_* = \sqrt{trace(X^TX)}$
@MarkL.Stone : The difficulty is that the nuclear norm, given by the formula $|X|_*:=\text{trace}\sqrt{X^TX}$, is not strictly convex.
You mean it's convex, not strictly convex? Well, you're the one claiming there can't be a non-extreme global optimum. Neither @Iosif Pinelis nor I have made such a claim.
@MarkL.Stone Strict convexity can be used to prove my conclusion. However, my conclusion is possible to be true for convex functions which are not strictly convex.
@MarkL.Stone and Iosif , can I ask you two to please take a look at my question over here? https://math.stackexchange.com/questions/4757229/relationship-between-trace-norm-a-k-a-schatten-1-norm-of-a-matrix-and-the-vec
I discovered this question while hunting for an answer to my own and I'm hopeful one of you two might know how to answer my question
|
2025-03-21T14:48:31.465575
| 2020-07-07T18:15:56 |
365063
|
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|
Stack Exchange
|
Reference describing supersingular elliptic curves over algebraically closed field in characteristic 2
I'm looking for a reference for the fact that over an algebraically closed field of characteristic two, there is (essentially) only one supersingular elliptic curve.
This fact appears on Wikipedia, and there may be an exercise in a book of Silverman, but it would be great to cite a book or paper.
Thanks!
Why not cite the exercise in Silverman's book AEC? It is exercise 5.7. Use Theorem 3.1(iii).
You could get it by brute force - the supersingular j-invariants have to lie in $\mathbb{F}_{2^2}$, so you can just check each of them.
Another explicit approach: Let $E: y^2 + (a_1x + a_3)y = x^3 + a_2x^2 + a_4x + a_6$ be an elliptic curve over $\mathbb{F}_2[a_1,a_2,a_3,a_4,a_6]$. The $2$-division polynomial $\Psi_2(x)$ is $a_1^2x^2 + a_3^2$, and $j(E) = a_1^{12}/(a_1^3(\cdots) + a_3^4)$. Now $E$ is supersingular if and only if $\Psi_2$ has no roots (over the algebraic closure), which is precisely when $a_1 = 0$, and then this forces $j(E) = 0$.
This is stated and proven in Section 8.2 of
Max Deuring's Die Typen der Multiplikatorenringe elliptischer Funktionenkörper., Abh. Math. Semin. Hansische Univ. 14, 197-272 (1941). ZBL67.0107.01 (the case $p = 2$ starts at the bottom of Page 252).
|
2025-03-21T14:48:31.465686
| 2020-07-07T18:23:52 |
365064
|
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|
Stack Exchange
|
From fractional matching to integral matching in tripartite hypergraphs
Let $G = (X\cup Y, E)$ by a bipartite graph with $n = |X|\leq |Y|$.
A fractional matching is a function assigning a non-negative weight to every edge in $E$, such that the sum of weights near each vertex is at most $1$. A symmetric fractional matching is a matching in which the total weight near every vertex of $X$ is the same (denote it by $w_X$) and the total weight near every vertex of $Y$ is the same (denote it by $w_Y$; note that $w_Y\leq w_X$).
Suppose that $G$ admits a symmetric fractional matching in which $w_X = 1$. Then $G$ admits an $X$-perfect matching (a matching in which all vertices of $X$ are matched). Two ways to see this are:
It is well-known that, in bipartite graphs, the maximum matching size (denoted $\nu(G)$) equals the maximum fractional matching size (denoted $\nu^*(G)$). Here we have $\nu^*(G)=|X|\cdot w_X = n$, which implies $\nu(G)=n$.
For every subset $X_k \subseteq X$ of size $k$, the sum of weights near vertices of $X_k$ is $k\cdot w_X = k$; for each subset $Y_k \subseteq Y$ of size $k$, the sum of weights near vertices of $Y_k$ is $k\cdot w_Y\leq k$. Hence, each such $X_k$ must have at least $k$ neighbors in $Y$. Hall's marriage theorem implies that an $X$-perfect matching exists.
I am looking for conditions under which this theorem holds for tripartite hypergrpahs. Let $H = (X\cup Y \cup Z, E)$ by a 3-partite hypergraph with $n = |X|\leq |Y|\leq |Z|$.
Suppose that $H$ admits a fractional matching in which $w_X = 1$. The following example shows that $H$ need not have an $X$-perfect matching. Consider an hypergraph with the following edges:
$(x_1,y_1,z_1), (x_1,y_2,z_2), (x_2,y_2,z_1), (x_2,y_1,z_2)$
The fractional matching assigning a weight of $1/2$ to all edges is symmetric and has $w_X=w_Y=w_Z=1$. However, $H$ does not admit an $X$-perfect matching.
I would like to know whether this can be "fixed" by enlarging $Z$. In particular, I will be happy to prove the following theorem:
Wanted-Theorem 1. If $|X|=|Y|=n$ and $|Z|\geq 2n-1$, and $H$ admits a symmetric fractional matching with $w_X=w_Y=1$ (and $w_Z=n/(2n-1)$), then $H$ admits an $X$-perfect matching.
The theorem is true for $n=2$.
Proof. Denote the parts of the hypergraph by $X = \{x_1,x_2\}, Y = \{y_1,y_2\}, Z = \{z_1,z_2,z_3\}$. The sum near $x_1$ is 1/2, so the sum near either $(x_1,y_1)$ or $(x_1,y_2)$ or both must be at least 1/4; w.l.o.g. assume the sum near $(x_1,y_1)$ is at least 1/4. Similarly, the sum near $x_2$ is 1/2. The sum near $(x_2,y_1)$ is at most $1/4$, so the sum near $(x_2,y_2)$ must be at least 1/4. Now $H$ can be reduced into a bipartite graph: one side contains the two vertices $x_1 y_1$ and $x_2 y_2$, and the other size is $Z$. Each of $x_1 y_1$ and $x_2 y_2$ has at least one neighbor in $Z$ - since the weight near them is positive. Moreover, both vertices together must have at least two neighbors in $Z$ - since the total weight near them is 1/2 which is larger than 1/3. Therefore, by Hall's marriage theorem there is a matching of size $2$ in the reduced graph. It corresponds to a matching of size $2$ in the original hypergraph.
However, so far, my attempts to extend it to any $n$ failed:
I first looked for an extension of the fact $\nu(H)=\nu^*(H)$ from bipartite graphs to tripartite hypergraphs. Indeed I found one by Füredi (1981): he proves that, in an $r$-partite graph, $\nu(H) \geq \nu^*(H)/(r-1)$. In particular, in tripartite hypergraphs $\nu(H) \geq \nu^*(H)/2$. But this is insufficient for the wanted theorem. Füredi shows that the factor $r-1$ is tight, but his example uses a hypergraph in which $|X|=|Y|=|Z|$, so hope is not lost yet.
I also looked at various Hall-type theorems for hypergraphs, but I could not prove that the conditions of Wanted-Theorem 1 imply the sufficient conditions for any of them.
Any more ideas about how I can prove or disprove this? References to papers relating fractional matching to integral matching in hypergraphs are also welcome.
|
2025-03-21T14:48:31.466057
| 2020-07-07T19:11:50 |
365066
|
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|
Stack Exchange
|
Schlessinger criterion and finiteness of tangent space
Schlessinger's criterion allows us to study whether or not a functor $\mathcal{F}:C_{\Lambda}\rightarrow \text{Set}$ from the category of local Artin $\Lambda$-algebras to sets is representable. One of the necessary consditions is that the tangent space $t_{\mathcal{F}}:=\mathcal{F}(k[\epsilon]/(\epsilon^2))$ is a finite dimensional $k$-vector space, where $k$ is the residue field of $\Lambda$.
A generalization or maybe an analogue of this is the derived Schlessinger criterion, due to mainly Lurie if I'm not mistaken (see https://nms.kcl.ac.uk/ashwin.iyengar/lntsg2019/G8.pdf Thm 3.5). If I'm not mistaken, in this case we allow the tangent space to be possibly infinite, by which I mean that $H_0(\mathfrak{t}\mathcal{F})$ can be infinite dimensional.
So if I start with a "nice" functor $\mathcal{F}:C_{\Lambda}\rightarrow \text{Set}$, whose tangent space is not finite dimensional, can the induced functor
$$s\mathcal{F}:sC_{\Lambda}\rightarrow s\text{Set}$$
be pro-representable? In a more optimistic sense, if $\mathcal{F}$ does satisfy every condition of Schlessinger's criterion except the finiteness of $t_{\mathcal{F}}$, is $s\mathcal{F}$ pro-representable?
A nice functor on $C$ won't induce a nice functor on $sC$ - that's extra structure. Infinite-dimensional tangent spaces were successfully handled by Grothendieck before Schlessinger - see the FGA paper cited by Schlessinger. The derived Schlessinger criterion is really just Brown representability as formulated by Heller or Jardine.
I should perhaps add that http://arxiv.org/abs/0705.0344 appeared 3 years before Lurie's ICM address.
Dear Jon, I suppose I owe you an apology for missing your paper. I'm not certain to which Grothendieck paper you refer, could you maybe give me a more precise reference? Thanks a lot!
Technique de descent et theoremes de d'existence en geometrie algebrique, 11, Seminaire Bourbaki, Expose 195, 1959/1960
|
2025-03-21T14:48:31.466210
| 2020-07-07T20:48:01 |
365070
|
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|
Stack Exchange
|
Intuition behind stability and instability in model theory
In A survey of homogeneous structures by Macpherson (Discrete Mathematics, vol. 311, 2011), a stable or unstable theory is defined as (Definition 3.3.1):
A complete theory $T$ is unstable if there is a formula $\varphi(\overline{x}, \overline{y})$ (where $\ell(\overline{x}) = r$ and $\ell(\overline{y}) = s)$, some model $M$ of $T$, and $\overline{a_i} \in M^r$ and $\overline{b_i}\in M^s$ (for $i \in \Bbb{N}$) such that for all $i, j \in\Bbb{N}$,
$$M \vDash \varphi(\overline{a_i}, \overline{b_j})\iff i \leqslant j$$
The theory $T$ is stable otherwise.
I wonder what is the intuition behind this definition. More specifically, under what situation is it conceived? Why is it called stable theory? Does the name stable related to the notion of stability in geometry or physics?
What sort of intuition are you looking for? What this means? What are the consequences? Why anyone cares about this? The definition itself basically says that there is no infinite set linearly ordered by $\varphi$, which is why it is also called the order property. A stronger property, called the strict order property, says that (basically) $\varphi$ defines a poset with an infinite chain. Regarding the name "stable", I think the alternative definition given by Gabe is more closely related (at least according to my intuition, I am not sure about the historical origins).
The use of "stable" almost certainly has no connection to uses in geometry and physics. My impression is that the word arises in the sense that the cardinalities of type spaces are "stable" with respect to the sizes of parameter sets. I am going to add a little bit to my answer, and also give an example of a stable theory.
I am going to explain some motivation by relating this definition to stability to other definitions, and discussing some examples. For simplicity I am going to assume we are working in a countable language.
An alternate definition of stability, for a complete theory $T$ is as follows.
Definition. $T$ is stable if there is some infinite cardinal $\lambda$ such that $T$ is $\lambda$-stable, i.e., for any $M\models T$, and any $A\subseteq M$, if $|A|\leq\lambda$ then for any $n$, $|S_n^T(A)|\leq\lambda$.
(In the previous definition, $S_n^T(A)$ is the space of complete $n$-types with parameters from $A$. One usually drops $T$ from this notation when $T$ is fixed. Also, in checking stability of $T$, it is enough to restrict to $1$-types by an easy exercise.)
So the slogan is that stable theories have "few types", since in general $|S^T_n(A)|$ is only bounded by $2^{|A|+\aleph_0}$. Since types describe the behavior of elements of models (or rather the "potential" behavior in elementary extensions), then we have the idea that stable theories are nice because the number of possible behaviors is constrained.
Counting types this way is an important part of stability theory and Shelah's work in classifying first-order theories according to the spectrum function $I(T,\kappa)$, which counts the number of non-isomorphic models of $T$ of cardinality $\kappa$. Part of the idea is that in order for this function to be well-behaved, and for their to be hope of classifying the models of a theory, one needs a stability (in a strong way).
It is also worth mentioning here that there are strong restrictions on the set of $\lambda$ for which a theory $T$ can be $\lambda$-stable. For example if $T$ is stable in some $\lambda$ then it is stable in unboundedly many $\lambda$. Moreover, if $T$ is $\aleph_0$-stable then it is stable in all infinite $\lambda$. But this is getting off track a little so I won't say more.
To give some intuition for the connection between the two definitions of stability, consider the following example.
Example 1. Let $T$ be the theory a dense linear order without endpoints (this determines a complete theory). We can see in $T$ that there a lot of types. For example just consider $1$-types over $\mathbb{Q}$. For any irrational $\alpha\in\mathbb{Q}$, we have a type $p_{\alpha}$ which says $x<r$ for any rational $r>\alpha$, and $x>r$ for any rational $r<\alpha$. This type is finitely-satisfiable in $\mathbb{Q}$ (hence is a bona fide type, and complete by QE for this theory). It is clear that different irrationals give distinct types, and so we obtain $2^{\aleph_0}$ types over a countable parameter set. Conclusion: DLO is not $\aleph_0$-stable (by the second definition). This argument can be generalized to any $\lambda$ (this is Exercise 4.5.21 in Dave Marker's book Model Theory: An Introduction).
In the previous example, we kill stability and obtain a lot of types by using cuts in orderings. So the amazing result is that, in some sense, this is precisely the way to kill stability, but with respect to a more "local" notion of ordering as described by the definition you've given. Indeed, the following is part of Shelah's Unstable Formula Theorem, which is Theorem 2.2 in Chapter 2 of Classification Theory.
Unstable Formula Theorem (abridged).
Let $\phi(\bar{x},\bar{y})$ be a formula. The following are equivalent.
$\phi(\bar{x},\bar{y})$ is unstable in every $\lambda\geq\aleph_0$, i.e., for every $\lambda\geq\aleph_0$, there is a subset $A$ (of some model of $T$) such that $|A|\leq\lambda<|S_\phi(A)|$ (where $S_\phi(A)$ denotes the space of $\phi$-types with parameters from $A$).
$\phi(\bar{x},\bar{y})$ is unstable in some $\lambda\geq\aleph_0$.
$\phi(\bar{x},\bar{y})$ has the order property (i.e., satisfies the condition in your definition).
The proof of this theorem is quite beautiful, and draws from both model theory and combinatorics, going back to a combinatorial result of Erdos and Makkai from this paper. In order to fully establish the equivalence between the two notions of stability for a theory (rather than a formula), one needs to show that if $T$ is not $\lambda$-stable for any $\lambda$ (which means there are a lot of complete types) then this can be detected by a single formula (i.e. there are a lot of $\phi$-types for a fixed $\phi(\bar{x},\bar{y})$). This is Theorem 2.13 in the same chapter of Shelah's book.
Remark (on the word "stable"). Stable theories were first defined by Shelah in the 1969 paper Stable Theories. Much of the motivation in this paper comes from work of Morley on totally transcendental theories, which are a subclass of stable theories (in a countable language, totally transcendental is the same as $\aleph_0$-stable). In this paper, Shelah proves the first "spectrum theorem" for type-counting. In particular he proves that if $T$ is a complete theory (we are still in a countable language) then one of the following holds:
For any model $M$ and $A\subseteq M$, $|S_1(A)|\leq |A|+2^{\aleph_0}$.
For any model $M$ and $A\subseteq M$, $|S_1(A)|\leq |A|^{\aleph_0}$; and (1) fails.
For any infinite $\lambda$, there is a model $M$ and $A\subseteq M$ such that $|A|=\lambda<|S_1(A)|$.
Shelah calls the first two cases "stable", presumably because the the cardinalities of the type spaces are "stable" as functions of the cardinalities of the parameter sets (modulo some cardinal arithmetic). For example, in the first case, we can see that $T$ is $\lambda$-stable according to the above definition whenever $\lambda\geq 2^{\aleph_0}$. In the second case $T$ is $\lambda$-stable whenever $\lambda^{\aleph_0}=\lambda$. This result narrows the "stability spectrum" for a particular theory, and Shelah further refines this in his book.
Example 2 (an $\aleph_0$-stable theory).
Let $T$ be the theory of algebraically closed fields of characteristic $0$ (this determines a complete theory in the language of rings). This theory has quantifier elimination in the language of rings (this is known to algebraic geometers as the result of Chevalley saying that the projection of a constructible set is constructible). We can use this to deduce that $T$ is $\lambda$-stable for any infinite $\lambda$. Indeed, suppose $A$ is a subset of some model of $T$. We can assume for simplicity that $A$ is a model (note that $|acl(A)|=|A|+\aleph_0$). By quantifier elimination, a $1$-type over $A$ is determined by polynomial equations and inequations with variables in $A$. So one of two things can happen. Either the $1$-type contains a polynomial equation (and thus specifies an element in $A$) or the $1$-type says that no polynomial equation is satisfied (and thus describes a "transcendental" over $A$). So there are $|A|+1=|A|$ many $1$-types, i.e., $|S_1(A)|=|A|$.
Example 3 (a superstable theory).
Let $T$ be the theory of the additive group of integers. This theory does not have quantifier elimination in the language of groups. In order to obtain QE, one needs binary relation symbols $\equiv_n$ specifying congruence modulo $n$ for all $n\geq 2$ (note that each $\equiv_n$ is definable in the language of groups using existential quantifiers). We can use these congruence relations to see that $T$ is not $\aleph_0$-stable. In particular, for any (possibly infinite) set $P$ of primes, there is a $1$-type over $\emptyset$ which contains $x\equiv_n 0$ for all $n\in P$ and $x\not\equiv_n 0$ for all primes $n\not\in P$. So we get $2^{\aleph_0}$ types over $\emptyset$.
On the other hand, $T$ is $\lambda$-stable for any $\lambda\geq 2^{\aleph_0}$. This takes a little more work to write out, but one essentially uses QE to show that a $1$-type over a model $A$ either specifies an element of $A$, or specifies a new element plus divisibility conditions for primes. So we get $|A|+2^{\aleph_0}$ types. In general, a theory that is $\lambda$-stable for sufficiently large $\lambda$ (i.e. case (1) in the Remark above) is called superstable.
In both of the previous examples, a good thought experiment is whether it would be easy to check stability using the order property definition (i.e., show that no formula admits an order). Indeed, in my experience the order property definition is a good way to show that a theory is unstable, but in order to show a theory is stable one usually uses type-counting or more sophisticated methods (e.g., forking). On the other hand, going back to Example 1, while type-counting in dense linear orders wasn't all that hard, it is completely trivial to find a formula with the order property.
What I've said only scratches the surface of stability, and there is much more to say. Probably others will add. Indeed, if this website were called Model Theory MathOverflow, then your question could easily be a community wiki.
A relevant elaboration on this would be to get into the proof of Shelah's Unstable Formula Theorem, and see the combinatorial connection between the order property and coding "binary trees", which is a way to build lots of types.
If you are interested in homogeneous structures, then you might also be interested in the following (equivalent) formulation, using the notion of a half-graph:
Let $V=\{v_1,v_2,\ldots, v_n\}$ be the vertex set of a graph $G=(V,E)$. We say that $G$ is a half-graph if $v_i\mathrel{E}v_j$ iff $i\leq j$.
(Intuitively, this means just that the graph of $E$ as a binary relation consists of exactly the points above the diagonal, including the diagonal --- although for the following definitions it does not really matter what happens on the diagonal.)
Given a model $M$ and a formula $\varphi(\bar x,\bar y)$, where $\bar x$ has length $n$ and $\bar y$ has length $m$, you can consider the infinite graph $G_{\varphi,M}=(M^{n+m},E_\varphi)$, where for $(\bar a_1,\bar b_1), (\bar a_2,\bar b_2)\in M^{n+m}$ we have a $E_\varphi$-edge between them when $\varphi(\bar a_1,\bar b_2)$ holds.
Using this, we can define stability in the following way.
Fix a theory $T$.
A formula $\varphi(\bar x,\bar y)$ in $T$ is $k$-stable if for every (if $T$ is complete, then equivalently, some) model $M$ of $T$, the graph $G_{\varphi,M}$ does not contain a half-graph of size $k$.
A formula is stable if it is $k$-stable for some $k$.
$T$ is stable if every formula $\varphi(\bar x,\bar y)$ is stable.
For example, if $T$ is the theory of $M$, which is the Fraisse limit of a Fraisse class $\mathcal C$ in a finite language, which happens to be uniformly locally finite (e.g. it is relational), then you have quantifier elimination, which gives the following corollary.
A quantifier-free formula $\varphi(\bar x,\bar y)$ is $k$-stable if no $G_{\varphi,C}$ (with $C\in \mathcal C$) contains a half-graph of size $k$.
$T$ is stable iff every quantifier-free formula is $k$-stable.
In other words, stability just means that for each $\varphi$, there is some $k$ such that no graph in the class $(G_{\varphi,C})_{C\in \mathcal C}$ of finite graphs contains a half-graph of size $k$. So stability amounts to having some forbidden graphs. Many other model-theoretic tameness notions can be worded similarly.
Here's a topological way to understand stability of a formula. Depending on your background, you might find it intuitive.
Given a formula $\varphi(\bar x,\bar y)$, a model $M$ and a tuple $\bar b$ in $M$ of the same length as $\bar y$, you can define a formula $f_{\varphi,\bar b}\colon M\to \{0,1\}$ by putting $f_{\varphi,\bar b}(\bar a)=1$ if and only if $\varphi(\bar a,\bar b)$ holds.
Now, let us endow $M^{\lvert x\rvert}$ with the weakest topology which makes all these functions continuous. We may attempt to define stability in the following way (which is not yet complete).
"Definition" (I think this is not equivalent, although I don't have any counterexamples offhand): the formula $\varphi$ is stable if the set $\{f_{\varphi,\bar b}\mid \bar b\in M^{\lvert y\rvert}\}\subseteq C(M^{\lvert x\rvert})$ is relatively weakly compact.
Unfortunately, the topology on $M^{\lvert \bar x\rvert}$ itself may not be so nice (for starters, it will often not be Hausdorff). Instead, we consider the space $S_{\varphi}(M)$ --- in model theory, this is the space of $\varphi$-types over $M$. Topologically, it is the compactification of $M^{\lvert \bar x\rvert}$ induced by the family $(f_{\varphi,\bar b})_{\bar b\in M}$ (i.e. the closure of the image of $M^{\lvert \bar x\rvert}$ under the map $M\to \{0,1\}^{M^{\lvert y\rvert}}$, where $\bar a\mapsto (f_{\varphi,\bar b}(a))_{\bar b\in M^{\lvert y\rvert}}$). Note that the functions $f_{\varphi,\bar b}$ uniquely extend to continuous functions on $S_{\varphi}(M)$ (namely, the extensions are the projections).
Having this, we may define stability in the following way:
The formula $\varphi$ is stable when for every $M\models T$, the set $A_{\varphi,M}:=\{f_{\varphi,\bar b}\mid \bar b\in M^{\lvert y\rvert}\}\subseteq C(S_\varphi(M))$ is relatively weakly compact.
Hey Tomasz. Just a notational point. Do you want your $A_{f,M}$ to be $A_{\varphi,M}$ in your final equation? This makes sense since your collection of functions depends on $\varphi$. And maybe I will just add that if you just require $A_{f,M}$ (for a fixed $\varphi$) to be relatively weakly compact for a particular choice of $M$, you get the definition of ``stable in a model". The gates of stability theory seem to still be open here (i.e. If $\varphi$ is stable in $M$, then every type in $S_{\varphi}(M)$ and $S_{\varphi^*}(M)$ is definable, $S_{\varphi}(M)$ is scattered, etc).
Yes, that name is certainly more apt. If I recall correctly, the relative weak compactness of $A_{\varphi,M}$ (for a particular $M$) should be exactly equivalent to the nonexistence of an infinite sequence $(b_n)_n$ in $M$ such that for some infinite sequence $(a_n)_n$ in the monster model such that $\varphi$ codes a half-graph on $(a_nb_n)n$. In particular, for a fixed $M$, I don't expect this to dualize to $\varphi^*$. Regarding scatteredness or definability of types, I would have to think about it, but I would rather expect the types in $S{\varphi^*}(M)$ to be definable in this case.
The point is that relative weak compactness means that the pointwise limit of a convergent sequence of formulas is a formula, which should make a pointwise limit of isolated types definable.
I could be reading this wrong, but from 2.10 in https://arxiv.org/pdf/1410.3339.pdf, I believe we have that $A_{\varphi,M}$ is relatively weakly compact if and only if $A_{\varphi^*M}$ is relatively weakly compact. The scattering claim follows from 5.3.6 in Theory of Charges (by Bhaskara Rao) and Theorem 2.8 on this note on my webpage: https://www3.nd.edu/~kgannon1/Stability.pdf. In particular, no strongly continuous positive measures (which is what is shown in my note) implies scattered. But this discussion has deviated from the essence of OP's original question.
@Kyle: Sure, but as long as it stays in the comments, it's not too bad.
Would you mind defining what is “relatively weakly compact”?
@EmilJeřábek: The closure in weak topology is compact. In this case, this is equivalent to saying that whenever $(f_n)n$ and $(p_n)n$ are sequences in $A{\varphi,n}$ and $S\varphi(M)$, respectively, such that $\lim_n\lim_m f_n(p_m)$ and $\lim_m\lim_n f_n(p_m)$ exist, then the two limits are equal. Or that when $(f_n)_n$ is pointwise convergent, then the limit is continuous.
Oh I see, thank you.
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2025-03-21T14:48:31.467291
| 2020-07-07T23:39:33 |
365082
|
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|
Stack Exchange
|
Possible cardinalities of the remainders of compactifications of $\Bbb R$
With the usual topology on $\Bbb R$, a compactification $\mathrm{id}_{\Bbb R}:\Bbb R\to v\Bbb R$ can have a remainder $v\Bbb R \setminus \Bbb R$ of cardinality $1,2, 2^{\aleph_0}=\mathfrak c,$ or $2^{\mathfrak c}.$ The only possibilities less than $\mathfrak c$ are $1,2.$
Suppose $\mathfrak c^+<2^{\mathfrak c}.$ What possible cardinals between $\mathfrak c$ and $2^{\mathfrak c}$ can be the cardinals of such remainders?
Is there perhaps a Forcing argument that can answer or partly answer this?
This link deals with the question why the remainder cannot have cardinality $\aleph_0$: Existence of a compactification of $\mathbb{R}$ with $\aleph_0$ remainder. (This is already mentioned in your posts - still, I thought that link migt be useful, since some people might be curious about this.)
Every connected compact Hausdorff space of weight $\aleph_1$ is the remainder $v \mathbb R \setminus \mathbb R$ of some compactification of $\mathbb R$. In particular, $[0,1]^{\aleph_1}$ is the remainder of a compactification of $\mathbb R$, and therefore $\mathbb R$ has a compactification with remainder of cardinality $2^{\aleph_1}$.
Using forcing, one can show that it is consistent to have $\mathfrak{c} < 2^{\aleph_1} < 2^{\mathfrak{c}}$. (For example, Easton's Theorem immediately implies that we may get a model where $2^{\aleph_0} = \aleph_2$, $2^{\aleph_1} = \aleph_3$, and $2^{\aleph_2} = \aleph_4$, although Easton's Theorem is a bit overkill for this.) Thus it is consistent that $\mathbb R$ has a compactification with cardinality in $[\mathfrak{c}^+,2^{\mathfrak{c}})$.
The result about weight-$\aleph_1$ continua is proved by Dow and Hart in this paper. But the special case of $[0,1]^{\aleph_1}$ is actually much easier to prove, using the fact that $[0,1]^{\aleph_1}$ is separable. Let $\{d_1,d_2,d_3,\dots\}$ be a countable dense subset of $[0,1]^{\aleph_1}$. Map $\mathbb R$ into $[0,1] \times [0,1]^{\aleph_1}$ as follows. First map $\mathbb R$ onto the ray $[1,\infty)$, and then map $[1,\infty)$ into $[0,1] \times [0,1]^{\aleph_1}$ by linearly mapping each interval $[n,n+1]$ to the line segment connecting $(\frac{1}{n},d_n)$ to $(\frac{1}{n+1},d_{n+1})$ in $[0,1] \times [0,1]^{\aleph_1}$. This mapping embeds the ray $[1,\infty)$ in $[0,1] \times [0,1]^{\aleph_1}$, and its boundary in this embedding is precisely the set $\{0\} \times [0,1]^{\aleph_1} \approx [0,1]^{\aleph_1}$.
Edit: It is also possible to find a compactification $v \mathbb R$ of $\mathbb R$ such that $|v \mathbb R \setminus \mathbb R|$ has countable cofinality. In fact, I claim that the set
$$T = \{|v\mathbb R \setminus \mathbb R| \,:\, v\mathbb R \text{ is a compactification of } \mathbb R \}$$
includes all cardinals of the form $2^\kappa$, where $\aleph_0 \leq \kappa \leq \mathfrak{c}$, and all countable limits of such cardinals. So, for example, in a model of set theory where $2^{\aleph_n} = \aleph_{\omega+n+1}$ for all $n$ (which is consistent, by Easton's Theorem), there is a compactification $v \mathbb R$ of $\mathbb R$ such that $|v \mathbb R \setminus \mathbb R| = \aleph_{\omega+\omega}$.
Lemma: Suppose $X$ is a connected compact Hausdorff space, and $X$ has a dense subspace $D$ that is both separable and path connected. Then there is a compactification of $\mathbb R$ whose remainder is (homeomorphic to) $X$.
Proof: The main ideas are already present in the third paragraph above. Let $\{d_1,d_2,d_3,\dots\}$ be a countable dense subset of $D$. Map $\mathbb R$ into $[0,1] \times X$ as follows. First map $\mathbb R$ onto the ray $[1,\infty)$, and then map $[1,\infty)$ into $[0,1] \times X$ by linearly mapping each interval $[n,n+1]$ to some path connecting $(\frac{1}{n},d_n)$ to $(\frac{1}{n+1},d_{n+1})$ in $[0,1] \times D$. This mapping embeds the ray $[1,\infty)$ in $[0,1] \times X$, and its boundary in this embedding is precisely the set $\{0\} \times X \approx X$.
My claim above follows almost immediately from this lemma. Each of the spaces $[0,1]^\kappa$, where $\aleph_0 \leq \kappa \leq \mathfrak{c}$, is separable and path connected, and so $|[0,1]^\kappa| = 2^\kappa \in T$ by the lemma.
To get countable limits of such cardinals, fix some infinite cardinals $\kappa_1,\kappa_2,\kappa_3,\dots \leq \mathfrak{c}$. Let $Y$ be the space obtained by gluing the endpoints of an interval to some (any) point of $[0,1]^{\kappa_1}$ and some (any) point of $[0,1]^{\kappa_2}$, gluing the endpoints of another interval to $[0,1]^{\kappa_2}$ and $[0,1]^{\kappa_3}$, gluing the endpoints of another interval to $[0,1]^{\kappa_3}$ and $[0,1]^{\kappa_4}$, and so on. (In other words, $Y$ is obtained by stringing together the $[0,1]^{\kappa_n}$ like beads on a necklace.) Finally, let $X$ be the one point compactification of $Y$. Then $X$ satisfies the hypotheses of the lemma, and $|X| = \sup_n 2^{\kappa_n}$.
Interestingly, this method seems hopeless for getting a compactification of $\mathbb R$ with a remainder of size $\aleph_\omega$. I wonder if this is possible?
I originally asked this on MSE as https://math.stackexchange.com/questions/3674921/what-is-the-set-of-cardinals-of-the-remainders-of-compactifications-of-bbb-r . Asaf Karagila conjectured that $cf (|v\Bbb R$ \ $\Bbb R |)\ne \omega.$
@DanielWainfleet: Hmm, that's an interesting idea. I'll have to think about it.
@DanielWainfleet: OK, I think we can get cofinality $\omega$. The short version: By the argument in my last paragraph, we can get $[0,1]^\kappa$ for any $\kappa \leq \mathfrak{c}$ as a remainder, and by modifying the argument a bit, we can get a connected space that contains any countably many such spaces. I'll post some details this afternoon when I have time.
Neat! That is pretty cool.
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2025-03-21T14:48:31.467653
| 2020-07-08T01:27:22 |
365086
|
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|
Stack Exchange
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Quotient Jordan property
The Jordan property for finite subgroups of ${\rm GL}_n(\mathbb{C})$ says that there exists a constant $c(n)$ so that for any finite subgroup $G$ of ${\rm GL}_n(\mathbb{C})$ there is a normal abelian subgroup $A$ of $G$ of index at most $c(n)$, and we can embed $A$ in a maximal torus of ${\rm GL}_n(\mathbb{C})$.
I was wondering if there exists some sort of "quotient version". I am expecting some result of the form:
There exists a constant $c(n)$ satisfying the following: Let $G$ be a subgroup of ${\rm GL}_N(\mathbb{C})$ and $T_1$ be a subtorus of rank $N-n$ of a maximal torus $\mathbb{G}_m^N$ of ${\rm GL}_N(\mathbb{C})$. Assume that there exists an exact sequence $1\rightarrow T_1\rightarrow G\rightarrow G/T_1\rightarrow 1$, where $G/T_1$ is a finite group. Then, there exists a normal abelian subgroup $A\leqslant G/T_1$ of index at most $c(n)$ so that its pre-image on $G$ lies in the maximal torus (or some conjugate).
If $T_1$ is the trivial torus, this gives back the Jordan property. I am not an expert on representation theory. Some results in algebraic geometry I am working on point out to a statement like this, so I wanted to know if there is anything in the literature.
I don't understand what you mean by "suppose that there exists an exact sequence $1\to T_1\to G$". Do you assume that $T_1\subset G$? Or is $T_1\to G$ allowed to be an arbitrary group homomorphism, unrelated to the context? that would sound weird.
It's a short exact sequence, so $T_1$ is a subgroup of $G$. In trying to write down a counterexample I see what the OP might be getting at. If a subgroup $G$ of $X=\mathrm{GL}_N(\mathbb{C})$ contains a big subtorus $T_1$ of $X$, then $G/T_1$ is either 'small' (dependent on the codimension of the subtorus in the maximal torus) or it contains a torus.
I've not seen anything like this. Can I not take the product of the normalizer of $T_1$ with the Weyl group of the remaining part of the torus? Actually, I don't even need that small symmetric group.
Does $G$ need to act irreducibly? Surely.
OK, a clearer statement is "let $G$ be a subgroup of $\mathrm{GL}_N(\mathbb{C})$ with a normal subgroup of finite index $T_1$ that is a subtorus (...)".
I think it's known that $G=T_1F$ for some finite subgroup $F$ of $G$ (with possibly nontrivial intersection). One can apply Jordan to $F$, and also the fact that finite subgroups of $\mathrm{GL}_{N-n}$ have bounded order, to infer that there is an abelian subgroup $F'$ of $F$ centralizing $T_1$, of bounded index. So $F'T_1$ is abelian and has bounded index. Nevertheless here 'bounded' also depends on $N$, not only on $n$, and OP didn't quantify on $N$.
But if you have a bound on $N$ and $n$ then you can just apply standard Jordan. So I think this is either clear or false.
@Ycor Could you give me a reference for your first statement? And what you mean with possibly non-trivial intersection?
Also, I think the statement is false, it suffices to take $T_1$ acting on the first variables as
$(t_1x_1,\dots,t_{N-n}x_{N-n},x_{N-n+1},\dots,x_N)$, and then take $G_1$ to be an extension of $T$ adding elements of $S_{N-n}$ permuting the first $N-n$ variables. Since $S_{N-n}$ won't satisfy any Jordan property, then this violates what I wanted... but I think that under some mild conditions on $T_1$ this may be true.
Oops, I possibly learnt this fact here on MO, I don't remember a proper reference. Possibly nontrivial intersection: I mean we can't always require $F\cap T_1={1}$.
@user73577 What do you mean by "mild conditions on T_1" ? Do you really want a bound depending only on $n$ and not on $N$ ?
@user73577 your counterexample is just the one I gave in the second comment. Which is why you will need a condition on $G$ probably, not $T_1$. Irreducible is too strong, as that probably needs $T_1=0$ or $T_1$ is a maximal torus. But I don't know the origin of your ideas, so I don't know what you would want.
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2025-03-21T14:48:31.468069
| 2020-07-08T03:18:23 |
365090
|
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|
Stack Exchange
|
Arithmetic expansion of harmonic sum
Note: I have modified the initial question as follows:
Let $w_1, w_2, \ldots, w_d$ be positive weights, and $x_1, x_2, \ldots, x_d$ be positive variables. Now, let us consider the following harmonic sum $H$ of the $d$ variables w.r.t. the weights:
$H = \frac{1}{\frac{w_1}{x_1} + \frac{w_2}{x_2} + \cdots + \frac{w_d}{x_d}}$.
Then, I am wondering if $H$ can be approximately expressed with some arithmetic formalization as follows:
$H \geq \sum_{i=1}^{\infty} (\tau_i(w_1, \ldots, w_d)\cdot \theta_i(x_1, \ldots, x_d)),$
where each $\tau_i$ (resp. $\theta_i$) is any function independent from the variables $x_1, x_2, \ldots, x_d$ (resp. the weights $w_1, w_2, \ldots, w_d$).
The equality would hold when $w_k = x_k$ ($1\leq k \leq d$).
I could not handle this problem in my knowledge.
I should be pleased to have any comments or suggestions (on the mathematical field related to this problem).
What's wrong with $H\ge \frac 1{\sqrt{ \sum_i w_i^2}} \frac 1{\sqrt {\sum_i x_i^2}}$?
Thank you very much. That's very helpful for me. And now, I notice that I should have written the equality condition as $w_k = \frac{1}{x_k}$, not $w_k = x_k$. I am sorry for that. My concern is that though the Schwarz inequality holds, when we consider $w_k = \frac{1}{x_k}$ as the expected equality condition, $H$ does not reach the right term; $\frac{1}{\sqrt{\Sigma_i w_i^2}}\frac{1}{\sqrt{\Sigma_i x_i^2}}$.
It should be better to modify my question as follows. Suppose that $H$ is written as $\frac{1}{\frac{w_1}{x_1} + \cdots + \frac{w_d}{x_d}}$. Then, is it possible to express $H$ in an arithmetic form like $H \geq \Sigma_i (\tau_i(w_1, \ldots, w_d)\cdot \theta_i(x_1, \ldots, x_d))$ with equality $w_k = x_k (1\leq k \leq d)$, where $\tau_i$ (resp. $\theta_i$) is any function independent from the variables $x_1, \ldots, x_d$ (resp. the weights $w_1,\ldots, w_d$)?
I am sorry for the inconvenience to change the initial query (whose solution must be collect with yours). I was happy to have your helpful comment on the problem which troubles me.
|
2025-03-21T14:48:31.468223
| 2020-07-08T03:41:43 |
365092
|
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|
Stack Exchange
|
Functions with a Jacobian whose columns are orthogonal
I am interested in vector fields whose Jacobian has orthogonal columns; i.e. if $\mathbf{f}(\cdot):\mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is a function where $\mathbf{f}(\mathbf{x})=[f_1(\mathbf{x}), f_2(\mathbf{x}),~\dots, f_n(\mathbf{x})]^{\rm T}$, I am looking for all such functions that:
$\forall~\mathbf{x}\in\mathbb{R}^{n}\quad\&\quad 1\leq i,j\leq n:\quad \big(\nabla f_i(\mathbf{x})\big)^{\rm T}\nabla f_j(\mathbf{x}) = \begin{cases}0~&:i\neq j\\g_{i}(\mathbf{x})&:i=j \end{cases}$
A similar question has been asked here. As I understood, in Liouville's theorem for conformal maps all the diagonal elements of the Jacobian $\nabla\mathbf{f}(\mathbf{x})$ are the same. Here, however, I am looking for a generalized case where the diagonal elements are not necessarily the same. Do we have something similar to Liouville's theorem for this case?
Thanks.
Clearly, there are more examples here than in the case of Liouville's theorem: if $g$ is a conformal map and $f_i(x) = h_i(g_i(x))$ for an arbitrary collection of $h_i : \mathbb{R} \to \mathbb{R}$, then the gradients of $f_i$ are orthogonal.
Just one remark. Your definition does not imply (is not equivalent to) that the rows be orthogonal. Thus there would be a symmetric question concerning those fields whose Jacobian has orthogonal rows.
@DenisSerre: A good remark, but, in fact, the two problems are equivalent up to computing an inverse function. If one regards $x$ and $f$ as columns of height $n$, the Jacobian $J$ satisfies $df = J,dx$, and the stated condition is that $J^T,J$ be diagonal. If $K = J^{-1}$, then $dx = K,df$, and we see that $$K^T,K = (J^{-1})^T,(J^{-1})=( J,J^T)^{-1,},$$ hence $K^TK$ is diagonal if and only if $J,J^T$ is diagonal. Thus, $f:\mathbb{R}^n\to\mathbb{R}^n$ satisfies the column condition if and only if the inverse function $f^{-1}:\mathbb{R}^n\to\mathbb{R}^n$ satisfies the row condition.
You are asking about the subject of orthogonal (coordinate) systems. There is an extensive literature on this subject, in particular by Darboux when $n=3$, and if you search on "triply orthogonal systems", you will see references to Darboux, Eisenhart, etc. plus many more recent references. There are many classical examples.
The equations are underdetermined when $n=2$, determined when $n=3$, and overdetermined when $n>3$, but they are always involutive and 'integrable' (in the sense of integrable systems).
Thanks for your response. Actually I'm interested in the case where $n>3$ and the function is invertible. Does the over-determined means that they don't have any solution?
@oasis93: There are plenty of invertible solutions for all $n$. For example, if $f_j$ is an invertible function of $x_j$ only for all $j$, you have a solution. Of course, this is not all of the solutions. The involutivity property implies that there exist many solutions locally and describes how to construct them using PDE. Overdetermined means that you can't arbitrarily prescribe initial conditions for the PDE on a hypersurface, say, of the form $\mathbb{R}^{n-1}\subset\mathbb{R}^n$.
Can you please introduce some references for the general case of n>3? I tried terms such as "over-determined orthogonal coordinate systems" on Google, but it seems there's no exact result. When searching for "orthogonal coordinate systems" almost all the results are for the case of n=3, which is not of interest to me. I would appreciate if you could introduce a book or paper on the general case of n>3 that characterizes the solutions. Huge thanks.
@oasis93: Well, if you read French, you can start with Gaston Darboux' 1898 book, Leçons sur les Systèmes Orthogonaux et les Coordonnées Curvilignes, which contains considerable information about the $n$-dimensional case for $n>3$. It covers pretty much everything that was known about the subject before 1900. There is an informative and thorough review of this book (in English) by E. O. Lovett published by the AMS at https://www.ams.org/journals/bull/1899-05-04/S0002-9904-1899-00584-6/S0002-9904-1899-00584-6.pdf
|
2025-03-21T14:48:31.468510
| 2020-07-08T06:10:12 |
365095
|
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|
Stack Exchange
|
Is there any maximal 1-planar or 2-planar graph that is not 3-connected
A graph is $k$-planar if it can be drawn in the plane so that each edge is crossed at most $k$ times. A $k$-planar graph $G$ is maximal if $G+uv$ is not $k$-planar for any non-adjacent vertices $u,v\in V(G)$.
Is there any maximal $1$-planar graph that is not $3$-connected ?
Is there any maximal $2$-planar graph that is not $3$-connected ?
I believe those examples exist but didn't find any references mentioned this.
This is a non-3-connected 1-planar example...
The paper On Properties of Maximal 1-planar Graphs by Dávid Hudák, Tomáš Madaras and Yusuke Suzuki describes the construction of maximal 1-planar graphs with minimum degree 2 and with (relatively) small number of edges, namely, (8/3)*(n-2) (which is far less than the number of edges of a maximal n-vertex planar graph).
|
2025-03-21T14:48:31.468594
| 2020-07-08T08:00:00 |
365101
|
{
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"Dieter Kadelka",
"Erich Friedman",
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"url": "https://mathoverflow.net/questions/365101"
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|
Stack Exchange
|
Packing L's in Tans and L's in L's
I'm a young researcher and I'm pretty new in this field. I want to work on packing problem "L's in Tans" and "L's in L's" as presented on https://erich-friedman.github.io/packing . On link can be found some known optimal packings, but I can't find any papers related to this problem.
Can someone show me some papers related to this problems?
Thanks in advance!
Too many questions at once. Please be more specific. What is the connection of genetic algorithms and the link?
@DieterKadelka I edited question, hope now is more specific.
The packing center is now located at https://erich-friedman.github.io/packing/index.html
It might be hard to find literature on the specific families of packing problems you mentioned. However, they are special cases of a more general pattern of asking for the optimal packing of $N$ congruent copies of an object inside another object. You can find many papers tackling this type of problem, see what kind of computational methods they use, and try to apply them to your problems. For example, here is a paper studying the problem of packing Platonic solids into a sphere using a simulated-annealing-like Monte Carlo approach.
|
2025-03-21T14:48:31.468701
| 2020-07-08T08:50:47 |
365103
|
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|
Stack Exchange
|
$\|(A_n-z)^{-1} - (A-z)^{-1}\|\to 0\;\Rightarrow\; \|e^{-tA_n}-e^{-tA}\|\to 0$ for general $C_0$ semigroups?
In short, the question is whether norm-resolvent convergence implies operator-norm convergence of the assocoated semigroups. More specifically, assume the following:
The $A_n$ generate contraction semigroups for all $n$,
$A$ generates a contraction semigroup,
One has $\|(A_n+1)^{-1} - (A+1)^{-1}\|\to 0$ in operator norm.
Then, can one get anything better than strong convergence for the associated semigroups (preferably operator norm convergence for some fixed t>0)?
The answer is relatively simple for analytic semigroups, since in that case one has an integral formula that expresses the semigroup in terms of the resolvent. For general $C_0$ semigroups, it's much less obvious...
In general, no, at least not without some uniformity with respect to z. Note that $(tA-z)^{-1}$ is continuous in the operator norm for $t>0$ and any $z$ where it is defined. But $e^{-tA}$ need not be continuous in operator norm.
|
2025-03-21T14:48:31.468794
| 2020-07-08T09:40:11 |
365105
|
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|
Stack Exchange
|
Characterization of topological entropy?
Let $V$ be a smooth Anosov vector field on a compact $n$ dimensional manifold $X$. Let $D(X)$ denote the set of distance functions $d$ on $X$ that are equivalent to fixed Riemannian distance. For each $d\in D(X)$ consider the log Lipschitz constant of the time one flow
$$L_d (V) := \sup_{x\neq y} \log \frac{ d(e^Vx,e^Vy)}{d(x,y)}.$$
Then I think it is well known (eg. Thm 7.15 in Walters's ergodic theory book) that the topological entropy of $V$ satisfies $$h_{\mathrm{top}} (V) \leq n L_d (V).$$ But the question is whether the infimum of the right-hand side over all such equivalent distances $d\in D(X)$ equals the entropy
$$h_{\mathrm{top}} (V) = \inf_{d\in D(X)} \ nL_d (V) ? $$
|
2025-03-21T14:48:31.468875
| 2020-07-08T09:47:43 |
365106
|
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|
Stack Exchange
|
Which posets can occur from commutative Frobenius algebras?
Let $A$ be a commutative Frobenius algebra. We can assume that $A$ is local and $A=K[x_i]/(I)$ for some variables $x_1,...,x_n$ and an admissible ideal.
Then the non-zero monomials $u_i$(including 1) form a basis of $A$ and we can look at the poset of submodules of the form $X_i=u_i A$.
(in case this is needed, assume $I$ only contains field elements 0,-1 and 1. Im not sure whether it might cause problems with regard to uniqueness when other field elements might be allowed).
This poset is bounded. For example when $A=K[x_1,...,x_n]/(x_i^2)$ then the poset is the Boolean lattice, but not every bounded poset can occur, for example when the top of the poset covers a unique elements then the poset must be a chain (and $A=K[x]/(x^n)$).
Question: Is there a (homological) characterisation which posets can occur from commutative Frobenius algebras?
Can you please clarify what you mean by "$I$ only contains field elements $0$, $-1$, and $1$" ?
@ZachTeitler $I$ is generated by elements $r_i$ and in these relations you can only find field elements 0,1,-1. So for example $r_i=xyx$ is ok but $r_i=xyx-3*yxy$ not. But Im not sure whether this matters, probably not.
|
2025-03-21T14:48:31.468982
| 2020-07-08T09:47:47 |
365107
|
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|
Stack Exchange
|
Weighted Co/ends?
Recall: Limits
Recall that the limit of a functor $D\colon\mathcal{I}\to\mathcal{C}$ is, if it exists, the pair $(\mathrm{lim}(D),\pi)$ with
$\lim(D)$ an object of $\mathcal{C}$, and
$\pi\colon\Delta_{\lim(D)}\Rightarrow D$ a cone of $\lim(D)$ over $D$
such that the natural transformation
$$\pi_*\colon h_{\lim(D)}\Rightarrow\mathrm{Cones}_{(-)}(D),$$
is a natural isomorphism, where
$\mathrm{Cones}_{(-)}(D)\overset{\mathrm{def}}{=}\mathrm{Nat}(\Delta_{(-)},D)$, and
The component at $X\in\mathrm{Obj}(\mathcal{C})$ of $\pi_*$ is the map $(\pi_*)_X \colon \mathrm{Hom}_\mathcal{C}(X,\lim(D))\to \mathrm{Cones}_X(D)$ sending a morphism $f\colon X\to\lim(D)$ to the cone
$$\Delta_X\xrightarrow{\Delta_f}\Delta_{\lim(D)}\to D$$
of $X$ over $D$.
Recall: Ends
Now, the end of a functor $D\colon\mathcal{I}^\mathsf{op}\times\mathcal{I}\to\mathcal{C}$ is the representing object of the functor
$$\mathrm{Wedges}_{(-)}(D)\colon\mathcal{C}^\mathsf{op}\to\mathsf{Sets}$$
with
$$\mathrm{Wedges}_{(-)}(D)\overset{\mathrm{def}}{=}\mathrm{ExtNat}(\overline{\Delta_{(-)}},\overline{D}),$$
where
$\overline{D}\colon\mathsf{pt}\times\mathcal{I}^\mathsf{op}\times\mathcal{I}$ is the unique functor restricting to $D$ under the isomorphism $\mathsf{pt}\times\mathcal{I}^\mathsf{op}\times\mathcal{I}\cong\mathcal{I}^\mathsf{op}\times\mathcal{I}$ and similarly for $\overline{\Delta_{(-)}}$, and where
We are now working with extranatural transformations.
That is, the object $\int_{A\in\mathcal{C}}D^A_A$ of $\mathcal{C}$ such that
$$h_{\int_{A\in\mathcal{C}}D^A_A}\cong\mathrm{Wedges}_{(-)}(D).$$
Recall: Weighted Limits
We can generalise limits by replacing $\Delta_{(-)}$ with an arbitrary functor $W\colon\mathcal{C}\to\mathsf{Sets}$. This leads to the notion of the weighted limit of $D\colon\mathcal{I}\to\mathcal{C}$ with respect to the weight $W$. This is the object $\lim_W(D)$ of $\mathcal{C}$ for which we have a natural isomorphism
$$h_{\lim_W(D)}(-)\cong\mathrm{Nat}(W,\mathrm{Hom}_\mathcal{C}(-,D)).$$
Question: Weighted Ends
Just as with weighted limits, we may define the weighted end of a functor $D\colon\mathcal{I}^\mathsf{op}\times\mathcal{I}\to\mathcal{C}$ with respect to a weight $W\colon\mathcal{I}^\mathsf{op}\times\mathcal{I}\to\mathsf{Sets}$ as the object $\int_{A\in\mathcal{C}}^W D^A_A$ of $\mathcal{C}$ (if it exists) such that we have a natural isomorphism
$$h_{\int_{A\in\mathcal{C}}^W(D)}(-)\cong\mathrm{ExtNat}(\overline{W},\overline{\mathrm{Hom}_\mathcal{C}(-,D)}).$$
(Or rather that a certain natural transformation $W_*$ induced by $W$ is a natural isomorphism. Note that precomposing extranatural transformations with natural transformations gives back an extranatural transformation, so $\mathrm{ExtNat}(\cdots)$ is indeed a functor.)
Now (finally!) for the actual questions:
This notion seems to be very natural. Has it been considered somewhere in the literature?
Provided that $\mathcal{C}$ has cotensors, we may write any weighted limit on $\mathcal{C}$ as an end. Can we similarly express weighted ends in terms of ends or limits (possibly weighted)?
Are there any natural occuring examples of this notion?
Everything above can be categorified to the setting of bicategories (with pain). Is there anything remarkable about the resulting notion of a "weighted pseudo biend"?
Ends are themselves examples of (weighted) limits: the end of a functor F : I^op x I -> C can be defined as the limit of F weighted by Hom_I, or as the limit of F composed with Tw(I) -> I^op x I where Tw(I) is the twisted arrow category of I. So "weighted ends" can probably also be expressed as weighted limits...
This is a back-of-the-envelope conjecture, but it seems plausible to me that the "weighted coend" of $T : {\cal I}^\text{op}\times {\cal I}\to {\cal V}$ by $W : {\cal I}^\text{op}\times {\cal I}\to {\cal V}$ is no more than the usual coend of the composition $W\diamond T$, regarding both $W,T$ as endo-profunctors of $\cal I$.
Of course, this only works for coends, and $T$ must be valued on the base of enrichment. But hey, that's a start :-)
In this very special case it's very easy to see that the coend of $T$ weighted by $W$ is the same as the coend of $W$ weighted by $T$; thus this seems to behave like a tensor product of functors.
Googling information about "Coends as traces" (there's a wonderful talk by S. Willerton on the topic), and "shadows for bicategories" (see here, slide 37 of 46) might tell you something interesting.
|
2025-03-21T14:48:31.469244
| 2020-07-08T10:06:37 |
365109
|
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|
Stack Exchange
|
Hilbert scheme of real curves
Morally speaking, my question is whether every real smooth projective curve can be deformed in as many real directions as complex directions. Let me make the question precise.
Let $H$ be the Hilbert scheme of curves of degree $d$ and genus $g$ in $\mathbb{P}^n$. Then real points $p\in H(\mathbb{R})$ correspond to curves defined over $\mathbb{R}$. Now assume that $p\in H(\mathbb{R})$ is a point that corresponds to a smooth irreducible curve and let $U\subset H(\mathbb{R})$ be an open neighbourhood (in the euclidean topology) of $p$.
Question 1. Is it true that $U$ is Zariski dense in the irreducible component of $H$ containing $p$?
The answer is "yes" if $p$ is a smooth point of $H$ but we know that even if $p$ corresponds to a smooth curve, it is not necessarily a smooth point of $H$. However, for our purposes it is already sufficient if $p$ is a smooth point of $H_{\textrm red}$ (reduced scheme associated to $H$). So a positive answer to my question would follow from a positive answer to this question which can be phrased over an arbitrary ground field:
Question 2. If $p\in H$ corresponds to a smooth and irreducible curve, is $p$ a smooth point of $H_{\textrm red}$?
|
2025-03-21T14:48:31.469342
| 2020-07-11T12:42:16 |
365400
|
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|
Stack Exchange
|
Empty Weyl/Fredholm spectrum of an operator on an infinite dimensional Banach space
Let $X$ be a complex infinite dimensional Banach space, and let $T \in B(X)$ be nonscalar. The Fredholm spectrum of $T$ is defined by:
$$ \sigma_{\Phi} (T) := \lbrace \lambda \in \mathbb{C} : T- \lambda Id \not \in \Phi \rbrace $$
(an operator is Fredholm iff $\dim \ker T, \dim \ker T^* < \infty$, where $T^*$ denotes the adjoint of $T$; $\Phi$ is the set of Fredholm operators in $B(X)$). Moreover, the Weyl spectrum of $T$ is defined by:
$$ \sigma_{w}(T):= \lbrace \lambda \in \mathbb{C} : T- \lambda Id \not \in \Phi_0 \rbrace $$
where $\Phi_0$ is the set of Fredholm operators with index $0$, i.e. $\dim \ker T = \dim \ker T^* < \infty$. It can be seen that:
$$ \sigma_{\Phi}(T) \subseteq \sigma_{w}(T) \subseteq \sigma(T) $$
While it is known that $\sigma(T)$ is always nonempty, is it possible that $\sigma_{\Phi}(T) = \emptyset$? According to formula (8.50) in V. Rakočević, E. Malkowsky. Advanced Functional Analysis, 2019, we have (at least when $X$ is Hilbert):
$$ \sigma(T)=\sigma_{w}(T) \cup \sigma_p(T) $$
(where $\sigma_p(T)$ denotes the point spectrum, i.e. the eigenvalues of $T$), so if $T$ does not have any eigenvalue the Weyl spectrum is certainly nonempty. More generally, what about $\sigma_{\Phi}$? Is it always nonempty or can it be empty in some cases?
The Fredholm spectrum of $T$ coincides with the spectrum of the equivalence class of $T$ in the Calkin algebra of $X$, so it is indeed non-empty.
Thank you very much!
|
2025-03-21T14:48:31.469462
| 2020-07-11T12:48:24 |
365401
|
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"Tyrone",
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|
Stack Exchange
|
Is every open topological $d$-manifold homotopy equivalent to a CW-complex of dimension $\leq d-1$?
Let $M$ be a connected open topological $d$-manifold (without boundary).
Whitehead showed that if $M$ has a PL structure, there exists a subcomplex of dimension $\leq d-1$ onto which $M$ deformation retracts.
Can we still find a homotopy equivalent CW complex of dimension $\leq d-1$ when $M$ is not PL?
I think this is true. If $X$ is homotopy equivalent to a (connected) CW complex and $H^i(X;M)=0$ for all $\pi_1X$-modules $M$ and all $i>n$, then $X$ is a homotopy equivalent to a CW complex of dimension $\leq n$, provided $n\geq3$. The cases $n=1,2$ can be sorted out by hand (although I only know how to do $n=2$ under some basic finiteness assumptions). Bearing in mind that I am quoting 20+ year old literature, so maybe there is no problem at all.
@Tyrone Manifolds of dimension $\leq 3$ always have smooth, hence PL structures, so those are certainly okay.
It is explained in https://mathoverflow.net/questions/201944/topological-n-manifolds-have-the-homotopy-type-of-n-dimensional-cw-complexes that a topological $n$-manifold has homotopy type of an $n$-dimensional complex of dimension $\le n$. It remains to exclude dimension $n$ if the manifold is open.
The proof for $\le n$ should be generalizable to $<n$, i.e. one has to check that the universal covering of your open $n$-manifold has zero cohomology in dimensions $\ge n$ with any local coefficients.
A universal cover has no nontrivial local systems. I think you mean the cohomology of the actual manifold. It holds by Poincaré duality.
@archipelago: that's what I should have said, thanks.
$\DeclareMathOperator{\co}{H}
\DeclareMathOperator{\ch}{C}
\newcommand{\zz}{\mathbb{Z}}
\newcommand{\nn}{\mathbb{N}}
\newcommand{\A}{\mathcal{A}}
\newcommand{\B}{\mathcal{B}}
\DeclareMathOperator{\lf}{lf}$Let me put together an answer following the pointers in the comments. By Whitehead's result stated in the question and smoothability in lower dimensions we may assume $d \geq 4$. Write $\pi := \pi_1(M)$ for brevity.
As an ANR, $M$ has the homotopy type of a CW-complex, so by a result of Wall it suffices to show that $\co^{j}(M; \A) = 0$ whenever $j \geq d$ and $\A$ is a $\zz \pi$-module. Writing $w$ for the orientation $\zz \pi$-module, by Poincaré duality we have $$\co^j(M;\A) \cong \co^{\lf}_{d-j}(M; \A \otimes_{\zz} w)\,,$$
where $\co^{\lf}_{*}$ denotes the locally finite singular homology (sometimes called the Borel-Moore homology). Therefore the only nontrivial thing to check is the vanishing of the 0-th locally finite homology for every $\zz \pi$-module $\B$. Writing $p\colon \tilde{M} \rightarrow M$ for the universal cover, this amounts to showing that the first differential
$$\partial_1 \otimes_{\zz\pi} \B \colon \ch^{\lf,\pi}_1(\tilde{M}) \otimes_{\zz\pi} \B \rightarrow \ch^{\lf,\pi}_0(\tilde{M}) \otimes_{\zz\pi} \B$$
is surjective, for which $\partial_1$ being surjective before tensoring is enough.
Here $\ch^{\lf,\pi}_{*}(\tilde{M})$ is the locally $\pi$-finite (singular) chain complex of $\tilde{M}$, that is, the subcomplex of locally finite chains in $\tilde{M}$ that project to locally finite chains in $M$.
We can verify $\partial_1$ is surjective by elementary means: Fix a locally $\pi$-finite singular 0-chain $\tilde{\sigma}$; then $\sigma := p(\tilde{\sigma})$ in $M$ is necessarily supported on a discrete subset of $M$. Since $M$ is non-compact (and second-countable), we can find countably infinite discrete subsets
$$
\{\tilde{x}_n : n \in \nn\} \subseteq \tilde{M}
\\
\{x_n : n \in \nn\} \subseteq M
$$
such that $p(\tilde{x}_n) = x_n$ and the former contains the support of $\tilde{\sigma}$. Thus $\tilde{\sigma}$ is a formal sum of the form $$\tilde{\sigma} = \sum_{n \in \nn}a_n \tilde{x}_n$$
with $a_n \in \zz$. Now for each $n \in \nn$ we can find a path $\tilde{\gamma}_n : [0,1] \rightarrow \tilde{M}$ connecting $\tilde{x}_{n+1}$ to $\tilde{x}_{n}$ such that the formal sum $$\tilde{\tau} := \sum_{n \in \nn}b_n \tilde{\gamma}_n$$
with the coefficients $b_n := \sum_{j \leq n} a_j$, is a locally $\pi$-finite $1$-chain in $\tilde{M}$ (because its projection $\tau:= p(\tilde{\tau})$ is locally finite in $M$) with $\partial_1(\tilde{\tau}) = \sigma$.
|
2025-03-21T14:48:31.469737
| 2020-07-11T13:16:18 |
365403
|
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|
Stack Exchange
|
Lorentzian cobordism through the dominant energy condition
Is the answer to the following problem, or some close variant thereof, known? Briefly:
Given two initial data sets $I_1=(M,g_1,k_1)$ and $I_2=(M,g_2,k_2)$, is there a time-oriented spacetime satisfying the dominant energy condition which has both $I_1$ and $I_2$ as Cauchy surfaces? i.e. can two initial data sets be filled in by the dominant energy condition?
In more detail:
Let $M$ be a smooth manifold, $g_1$ and $g_2$ two Riemannian metrics and $k_1$ and $k_2$ two symmetric 2-tensors.
Does there necessarily exist a time-oriented Lorentzian manifold $(N,g')$ satisfying the dominant energy condition into which there are hypersurface embeddings $f_1$ and $f_2$ of $M$ such that:
$f_1^\ast g'=g_1$ and $f_2^\ast g'=g_2$
$k_1$ and $k_2$ are the second fundamental forms (relative to the time-oriented unit normal) of $f_1$ and $f_2$
every maximally extended timelike geodesic of $(N,g')$ intersects both $f_1(M)$ and $f_2(M)$ exactly once?
This seems to be somewhat analogous to a question studied in the Riemannian setting, see e.g. the article of Xue Hu and Yuguang Shi here https://www.emis.de/journals/SIGMA/Gromov.html
|
2025-03-21T14:48:31.470204
| 2020-07-11T13:29:58 |
365404
|
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|
Stack Exchange
|
Reference request : Global boundedness of weak solution for Neumann problem
I have some question on global boundedness of weak solution to Neumann problems.
Let $u\in W^{1,2}(\Omega)$ is a weak solution for Neumannn problem
$$ \mathrm{div} (A \nabla u )= \mathrm{div}\, F\quad \text{in } \Omega,\quad A\nabla u\cdot \nu = F \cdot \nu$$
i.e., $u$ satisfies
$$ \int_{\Omega} A\nabla u\cdot \nabla \phi\, dx =\int_{\Omega} F \cdot \nabla \phi\, dx $$
for all $\phi \in W^{1,2}(\Omega)$. Here $A(x)=(a^{ij}(x))$ is uniformly elliptic, i.e.,
$$ \Vert A \Vert_{L^\infty (\Omega)} \leq B, \quad a^{ij}(x) \xi_i \xi_j \geq \delta |\xi|^2$$
for all $x\in \Omega$ and $\xi \in \mathbb{R}^n$.
I want to show that if $F \in L^q(\Omega)$, $q>n$, then $u\in L^\infty(\Omega)$ and
$$ \Vert u \Vert_{L^\infty(\Omega)} \leq C \Vert F \Vert_{L^q(\Omega)} $$
for some constant $C>0$.
In the case of Dirichlet problem, we can prove similar conclusion via De Giorgi type argument due to Stampaccia and Moser type argument. It seems that the result is quite classical, but I cannot find exact reference on this result. I tried to prove it via Moser type argument, but I fail to conclude the desired result.
Thanks in advance.
Add : Here is my trial (not rigorous for the check whether I can use Moser's approach) Set $w=u^+$. Note that
$$ w^\beta \nabla w = \frac{1}{\beta+1} \nabla(w^{\beta+1})$$
Testing $\phi=w^{\beta}$, we obtain via Young's inequality that
$$ \frac{4}{(\beta+1)^2} \int_\Omega |\nabla w^{(\beta+1)/2}|^2 dx =\int_\Omega w^{\beta-1} |\nabla w|^2 dx \leq \int_\Omega w^{\beta-1} |F|^2 dx.$$
Using Sobolev's embedding theorem and Holder's inequality we get
\begin{align*}
\Vert w^{(\beta+1)/2} \Vert_{L^{2^*}(\Omega)} &\leq C\Vert w^{(\beta+1)/2} \Vert_{L^2(\Omega)}+C\Vert \nabla(w^{(\beta+1)/2}) \Vert_{L^2(\Omega)}\\
&\leq C\Vert w\Vert_{L^{(\beta+1)}}^{(\beta+1)/2} + C(\beta+1) \Vert{w^{\beta-1}}\Vert_{L^{p/(p-2)}(\Omega)}^{1/2} \Vert F \Vert_{L^p(\Omega)}^{1/2}
\end{align*}
I'm not sure to conclude the iteration due to the term $\Vert w\Vert_{L^{(\beta+1)}}^{(\beta+1)/2}$.
You are right, this is usually stated in the case of Dirichlet boundary condition. But what is the problem trying to repeat the usual proof? In the case where $\Omega$ is the whole space $R^n$ both De Giorgi's approach or Moser's work.
@GiorgioMetafune Thanks for your answer. I edit my question. In the case of Dirichlet boundary problem, our function is lie on $W^{1,2}_0$, but in the case of Neumann problem, the original function does not lie on that space. This makes me hard to use Sobolev embedding.
In the case of De Giorgi approach, it seems okay to obtain the desired result, but I'm not sure whether the desired assertion can be deduced via Moser's argument.
What kind of regularity do you assume for the boundary of $\Omega$? (I have serious doubts whether this is true for arbitary bounded domains.)
Take $|F|_p=1$ and iterate your inequalities until $\beta>p-1$. Then you can absorbe the first term on the RHS into the second and go on with the usual iteration. I did not check the details but it should work.
@JochenGlueck Of course, we need to assume the regularity of $\Omega$, at least Lipschitz domain to guarantee the Sobolev embedding theorem.
I think you have to assume something on $|u|_1$ or on $\int u$ otherwise, you have a solution up to a cosntant. This should fix the problem of lower order terms by bootstrap arguments.
|
2025-03-21T14:48:31.470426
| 2020-07-11T14:48:24 |
365407
|
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|
Stack Exchange
|
regularity of the solutions of Prandtl equation on the segment
Let $p(x)$ be a positive measurable function on $(-1,1)$. Consider the Prandtl equation
$$
u(x)-\frac{p(x)}\pi \int_{-1}^1 \frac{u'(t)}{t-x}dt=p(x)h_0(x),\quad u(1)=u(-1)=0.\quad\quad(\star)
$$
What is state of the art of the existsence and regularity theory for $(\star)$, what is a standard reference?
Recall that $(\star)$ may be rewritten as $u+p(x)\sqrt{-d^2/dx^2}u=p(x)h_0(x)$, where $-d^2/dx^2$ is Dirichlet Laplacian on $[-1,1]$, that relates this to the regularity theory for fractional Laplacians.
For the fractional Laplacian formulation: for the existence, one can refer, for example, to the 20-year-old paper by K. Bogdan and T. Byczkowski Potential theory of Schrödinger operator based on fractional Laplacian. Regularity might be treated in a much more recent paper The fractional Schrödinger equation with general nonnegative potentials. The weighted space approach by J.I. Díaz, D. Gómez-Castro and J.L. Vázquez. The latter is purely analytic and gives a lot of references.
With $q=1/p$, let me write your equation as
$$
\vert D\vert u+ q u= h_0, \quad u(\pm 1)=0.
$$
Multiplying the equation by $u$, we get
$$\Vert{u}\Vert_{H^{1/2}_0}^2\le
\Vert{u}\Vert_{H^{1/2}_0}^2+\underbrace{\langle qu, u\rangle_{L^2}}_{\ge 0}=\langle h_0, u\rangle_{L^2}\le \Vert{u}\Vert_{H^{1/2}_0}\Vert{h_0}\Vert_{H^{-1/2}},
$$
entailing
$
\Vert{u}\Vert_{H^{1/2}_0}\le \Vert{h_0}\Vert_{H^{-1/2}}.
$
Assuming $h_0\in H^{-1/2}$ will give $u$ in $H^{1/2}_0$ (completion of smooth compactly functions supported in $(-1,1)$ for the $H^{1/2}$ norm; here $H^{-1/2}$ stands for the dual space of $H^{1/2}_0$). With no more information on the regularity of $1/p$, it seems questionable to improve the regularity $H^{1/2}$.
|
2025-03-21T14:48:31.470556
| 2020-07-11T14:54:09 |
365408
|
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|
Stack Exchange
|
Heat Equation Boundary Value Problem - alternative expressions for solution
Let $B_t$ be a Brownian motion, with with density function $f(t,x)dx = P(B_t \in dx)$. Then $f$ solves the heat equation $\partial_t = \frac{1}{2} \partial_{xx}f(t,x)$. Let for a fixed $u > 0$, $\bar{B}_t = \sup_{s \leq t}B_s$, and let $$q(t,x)dx = P( \bar{B}_t > u, B_t \in dx).$$
Then $q$ still solves the heat equation with initial condition $q(0,x) = 0$ for $x < u$ and boundary condition $q(t,u) = p(t,u)$, $t \geq 0$. Using the classical theory of heat equation, one can write down an explicit expression for $q$:
$$
q(t,x) = \int_0^t f(s,u) \partial_x f(t-s,x-u)ds. \label{1}\tag{1}
$$
Show that $q$ can also be written as
$$
\begin{align}
q(t,x) &= \int_0^t f(t-s,x-u) \partial_u f(s,u)ds \label{2}\tag{2} \\
&= \int_0^t f(t-s,x-u) \frac{u}{s}f(s,u) ds. \label{3}\tag{3}
\end{align}
$$
I can think of a heuristic probabilistic argument involving path reversal of Brownian motion. Is there an analytic way to go from expression \eqref{1} to \eqref{2} and \eqref{3}?
I can see that the expression \eqref{2}/\eqref{3} satisfies the heat equation, but haven't been able to show that it satisfies the boundary condition. That may be sufficient.
@DanieleTampieri thanks for the edit :-)
|
2025-03-21T14:48:31.470671
| 2020-07-11T16:32:46 |
365411
|
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"Adam P. Goucher",
"Albert van der Horst",
"Joel David Hamkins",
"Noam D. Elkies",
"Will Sawin",
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|
Stack Exchange
|
Can a fixed finite-length straightedge and finite-size compass still construct all constructible points in the plane?
I am hoping that the brilliant MathOverflow geometers can help me out.
Question 1. Suppose that I have a fixed finite-length straightedge and fixed finite-size compass. Can I still construct all constructible points in the plane?
I know the answers in several variations.
If I have compasses of arbitrary size, then I don't need a straightedge at all. This is the Mohr–Moscheroni compass-only theorem.
If I have an infinite straightedge (or arbitrarily large), then I need only a single compass of any fixed size. This is the rusty compass theorem.
Indeed, the Poncelet–Steiner theorem shows that I need only an infinite straightedge and a single circle of known center and radius.
But what I don't know is the remaining case, where both the straightedge and compass are limited in size. The difficult case seems to be where you have two points very far apart and you want to construct the line joining them.
Will Sawin's comment answers the question I had asked above. But it doesn't seem to answer the relative version of the question, which is what I had had in mind:
Question 2. Suppose that I have a fixed finite-length straightedge and fixed finite-size compass. Can I still construct all constructible points in the plane, relative to a fixed finite set of points?
In other words, is the tool set of a finite straightedge and finite compass fully equivalent to the arbitrary size toolset we usually think about.
If $(a,b)$ is a constructible point, can't we construct $( \frac{a}{n}, \frac{b}{n})$ for $n$ sufficiently large by pretending my straightedge is infinite and then shrinking my construction? Then can I walk along that line in steps of size $\frac{a}{n}, \frac{b}{n}$ by using a special case of the rusty compass theorem construction?
Ah, I've realized that this answers the question I asked, but not the question I was thinking of, which was the relative constructibility question. Given points, can you still construct the same points relative to them? In other words, is the tool set fully equivalent? (Meanwhile, Will, please post your comment as an answer.)
For the difficult case you mention, suppose you have two points P and Q that are very far apart. Suppose you pick a line leaving $P$, approximately in the direction of $Q$, and count along it, one unit interval at a time, until you reach a point $R$ within one unit interval of $Q$, after $n$ unit intervals in total. Then to construct the line from $P$ to $Q$ we need to take $Q-R$, shrink my a factor of $n$, and then translate it to the point one unit interval off from $P$. These steps all seem like they can probably be done with a bounded straightedge but I didn't check.
@WillSawin That is what I was trying to do, but didn't see how to carry out the scaling part. If you could post an answer, I'd be grateful.
"approximately in the direction of" is a tall order if P,Q are far away
and you must get within one unit. A more robust solution is to
tile the plane with squares (or even parallelograms) and count
to determine the coordinates $(a,b)$, and thus also $(a/n,b/n)$,
within one unit.
(We can construct the tiling in square-spiral order
to make sure we get from P to Q in a finite number of steps.)
A bounded-length straightedge can emulate an arbitrarily large straightedge (even without requiring any compass), so the rusty compass theorem is sufficient.
Note that, in particular, it suffices to show that there exists an $\varepsilon > 0$ such that a straightedge of length $1$ is capable of joining two points separated by any distance $\leq 1 + \varepsilon$ (and therefore emulates a straightedge of length $1 + \varepsilon$, and therefore arbitrarily long straightedges by iterating this process).
We can use Pappus's theorem to establish this result for any $\varepsilon < 1$:
https://en.wikipedia.org/wiki/Pappus%27s_hexagon_theorem
In particular, given two points $A$ and $c$ (separated by a distance slightly greater than 1) which we wish to join, draw a line $g$ through $A$ and a line $h$ through $c$ which approach relatively close to each other. Then add arbitrary points $B, C$ on $g$ and $a, b$ on $h$ such that the four new points are within distance $1$ of each other and the two original points. We assume wlog $b$ is between $a, c$ and $B$ is between $A, C$.
Then we can construct $X$ by intersecting the short (length $< 1$) lines $Ab, Ba$, and construct $Z$ by intersecting the short lines $Bc, Cb$. Then $Y$ can be constructed by intersecting the short lines $XZ$ and $Ca$.
Now, $Y$ is positioned collinear with $A$ and $c$ and between them, so we can use it as a 'stepping stone' to draw a straight line between $A$ and $c$.
The result follows.
EDIT: I decided to do this with the edge of a coaster and two points slightly too far apart to be joined directly:
The construction means that you can henceforth proceed as though you have a length-$(1 + \varepsilon)$ straightedge instead of a length-$1$ straightedge. By the same argument, you can proceed as though you have a length-$(1 + \varepsilon)^2$ straightedge (by using the length-$(1 + \varepsilon)$ straightedge to draw each line in the diagram).
Note that this naive recursive approach means that, in particular, joining two points separated by a distance of $(1 + \varepsilon)^n$ would take $10^n$ applications of the original straightedge! (It's still polynomial in the length, though, so not too bad.)
Thanks for your excellent answer. This is great!
See also this 2008 article (in French) by Xavier Caruso http://xavier.toonywood.org/papers/publis/troppetit.pdf, which seems to use the same method.
@Goucher. Note that one can do an arbitrary line of lenght n by n applications of the ruler. So the arbitrary lines can solve the problem in one step, if you're lucky enough. That makes it seem like your argument is way to pessimistic. On the other hand, it seems that there is a carpenters way, but no mathematicians way to select the directions. Next thing you know the problem is NP complete. Groetjes Albert
@AlbertvanderHorst Doesn't Noam's observation (comment on the OP) enable you to pick the right directions? You can tile the whole region between the points before choosing the lines.
@joel-david-hamkins
Indeed I was too pessimistic. Instead of tiling one could proceed with 4 perpendicular lines shooting from each point. Prolong them in turn. After a certain number of steps O(L/l) you will find an intersection. Here L is the distance between the points and l is the length of the ruler.
|
2025-03-21T14:48:31.471089
| 2020-07-11T16:41:40 |
365412
|
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|
Stack Exchange
|
A series that is algebraic?
This question is a follow-up of question A series that is rational? . Let $k=\mathbb F_q(T)$. Can one prove (or disprove) that the series $\sum_{n\ge0}(1-TX^{q^n})Y^{q^n}\in k[[X,Y]]$ is algebraic over $k(X,Y)$? In the mentioned link it is proved that the series does not belong to $k(X,Y)$.
If the series would be $\sum_{n\ge0}(1-X^{q^n})Y^{q^n}$, it is easy to prove...
The series is $f-Tg$ where $f=\sum_{n\geq0}Y^{q^n}$ and $g=\sum_{n\geq0}(XY)^{q^n}$. Both $f$ and $g$ are algebraic since $f^q-f=-Y$ and $g^q-g=-XY$.
|
2025-03-21T14:48:31.471167
| 2020-07-11T16:49:54 |
365414
|
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|
Stack Exchange
|
Definition of the cotangent complexes of Artin stacks
I am studying the notion of the cotangent complexes of Artin stacks reading LMB's book and Olsson's paper. According to them, the cotangent complexes are defined as projective systems in their derived category.
However, in Lemma 2.1 and Proposition 2.2 of Zhang's paper(https://arxiv.org/pdf/1111.6294), the cotangent complexes of Artin stacks are elements of their derived category, where we consider their lisse-etale sites.
Is this correct ? (Did the author take the limit?)
I am confused...
Any comment welcome! Thank you!
Edit:I would appreciate if you could tell me any references to cotangent complexes of Artin stacks。
This is covered in Toën-Vezzosi Homotopical Algebraic Geometry II, at least. but I don't know of any books or papers that redevelop this in completely modern language (HAG II is a nice paper, but it is written in a language that people don't really use anymore). Jacob Lurie's book «Spectral Algebraic Geometry» specifically avoids dealing with Artin stacks (mentioned in the introduction). Anyway, the key thing to note about the cotangent complex is that it only makes sense as an object of the derived category. It is a fundamentally homotopical object (even back to Quillen).
There is an upgrade of Olsson's result in Laszlo-Olson: https://math.berkeley.edu/~molsson/article-Iweb.pdf
The relevant part is Example 2.2.5, which applies Theorem 2.2.3 to the case of a hypercovering of Artin stacks. This gives an equivalence of the derived category of quasicoherent sheaves on the stack with the one of the hypercover (rather than just between the categories of projective systems), which together with Olsson's original paper gives the cotangent complex as an object in the derived category of quasi-coherent sheaves (in the lisse-étale site).
Maybe I'm missing something, but I don't see how one passes from the pro-system which is the original definition of the cotangent complex to an element of the derived category of quasi-coherent sheaves on the hypercover. Is this also covered in Olsson's original paper somewhere, or is this something generally known?
In the original paper, he constructs a complex on the hypercover, and the pro-system is obtained by truncating this complex. This is only needed to use the descent result in that paper which doesn't work for the unbounded derived category. With the improved result, you simply omit passing to the pro-system.
|
2025-03-21T14:48:31.471350
| 2020-07-11T17:12:32 |
365416
|
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Stack Exchange
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Subgroup of $PGL(n(n-1)/2, \mathbb K)$ preserving the grassmannian $Gr(2, n)$
How can we determine the subgroup of $PGL(\wedge^2 \bar{\mathbb Q}^n)$ which preserves the grassmamnnian $Gr(2, n)$ embedded as a projective variety in $\wedge^2(\bar{\mathbb Q}^n)$ via the Plucker embedding. For $n = 4$, this variety is a quadric and it is a known fact that the transformation group of $Gr(2,4)$ is the orthogonal group of the corresponding quadratic form (defined on $\bar{\mathbb Q}^6$ by a single Plucker relation).
One may approach this question for $n = 5$ by considering the Plucker relations for $n = 5$ each of which yields a quadratic form and provided this is non-degenerate determine the corresponding orthogonal groups and finally take their intersection.
Is this a good approach. Is there an alternative method?
The automorphism group of $Gr(2,n)$ is $PGL(n)$. The question would be more appropriate on MSE.
This is of course true. Here we are actually seeking a subgroup of $PGL(\binom{n}{2}, \bar{\mathbb Q})$ that preserves the image of $Gr(2,n)$ in $\wedge^2 (\bar{\mathbb Q}^n)$. This is a non-trivial question (e.g. W. Nemitz Transformations Preserving the Grassmannian
William C. Nemitz,
Transactions of the American Mathematical Society Vol. 109, No. 3 (Dec., 1963), pp. 400-410)
It is the image of $\wedge^2:PGL(\bar{\mathbb{Q}}^n)\rightarrow PGL(\wedge^2\bar{\mathbb{Q}}^n)$.
Actually this is holds when n is not 4. For n = 4 Hodge-star operator is also preserving.
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2025-03-21T14:48:31.471465
| 2020-07-11T17:24:23 |
365420
|
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Search for a general formula from known iterative relation
$F$ is a mapping among $\{\theta_{n_1n_2}\}$, with $\eta_{1/2}$ being arbitrary constants involved.
$F: \theta_{n_1n_2} \rightarrow \theta_{n_1+1n_2}+\theta_{n_1n_2+1}+\eta_{1}n_1\theta_{n_{1}-1n_{2}}
+\eta_{2}n_2\theta_{n_{1}n_{2}-1}$
so what is $F^k:\theta_{n_1n_2} \rightarrow ?$ ($F$ acts $k$ times)
Is there a general formula?
Consider the generating function:
$$H(x,y) := \sum_{i,j} \theta_{i,j} \frac{x^i}{i!} \frac{y^j}{j!}.$$
Extending $F$ by linearity, define
$$F(H)(x,y):= \sum_{i,j} \left(\theta_{i+1,j}+\theta_{i,j+1}+\eta_{1}i\theta_{i-1,j}+\eta_{2}j\theta_{i,j-1}\right) \frac{x^i}{i!} \frac{y^j}{j!}$$
so that the question amounts to finding $n_1!\cdot n_2!\cdot [x^{n_1}y^{n_2}]\ F^{(k)}(H)(x,y)$.
The definition of $F$ implies that
\begin{split}
F(H)(x,y) &= \frac{\partial}{\partial x} H(x,y) + \frac{\partial}{\partial y} H(x,y) + (\eta_1 x + \eta_2 y)H(x,y) \\
&= e^{-\frac{\eta_1}2x^2-\frac{\eta_2}2y^2}\left(\frac{\partial}{\partial x}+\frac{\partial}{\partial y}\right) e^{\frac{\eta_1}2x^2+\frac{\eta_2}2y^2} H(x,y).
\end{split}
Then
$$F^{(k)}(H)(x,y) = e^{-\frac{\eta_1}2x^2-\frac{\eta_2}2y^2}\left(\frac{\partial}{\partial x}+\frac{\partial}{\partial y}\right)^k e^{\frac{\eta_1}2x^2+\frac{\eta_2}2y^2} H(x,y),$$
from where the coefficient of $x^{n_1}y^{n_2}$ can be extracted by standard means as follows.
$$
[x^{n_1}y^{n_2}]\ F^{(k)}(H)(x,y) =
\sum_{a=0}^{\lfloor n_1/2\rfloor} \frac{(-\eta_1/2)^a}{a!}
\sum_{b=0}^{\lfloor n_2/2\rfloor} \frac{(-\eta_2/2)^b}{b!}
\sum_{c=0}^k \binom{k}c (n_1-2a+c)_c (n_2-2b+k-c)_{k-c}
\sum_{d=0}^{\lfloor (n_1-2a+c)/2\rfloor} \frac{(\eta_1/2)^d}{d!}
\sum_{f=0}^{\lfloor (n_2-2b+k-c)/2\rfloor} \frac{(\eta_2/2)^f}{f!}
\frac{\theta_{n_1-2a+c-2f,n_2-2b+k-c-2f}}{(n_1-2a+c-2f)!(n_2-2b+k-c-2f)!}.
$$
Introducing $s:=a+d$ and $t:=b+f$, we rewrite the above as
$$k!\sum_{c=0}^k
\sum_{s=0}^{\lfloor (n_1+c)/2\rfloor} \frac{(\eta_1/2)^s}{s!}
\sum_{t=0}^{\lfloor (n_2+k-c)/2\rfloor} \frac{(\eta_2/2)^t}{t!}
C_{n_1,s,c}C_{n_2,t,k-c}
\frac{\theta_{n_1+c-2s,n_2+k-c-2t}}{(n_1+c-2s)!(n_2+k-c-2t)!},
$$
where
$$C_{n,s,c}:=\sum_{a=0}^s \binom{s}{a} \binom{n-2a+c}c (-1)^a = (-1)^n\sum_{a=0}^s \binom{s}{a} \binom{-c-1}{n-2a} (-1)^a$$
is the coefficient of $x^n$ in $(-1)^n (1-x)^s (1+x)^{s-c-1}$. Since $s\leq \lfloor (n_1+c)/2\rfloor$, it follows that $C_{n_1,s,c}=0$ if $s\geq c+1$; and similarly for $C_{n_2,t,k-c}$. Hence, we obtain a closed formula:
$$n_1!n_2!\ [x^{n_1}y^{n_2}]\ F^{(k)}(H)(x,y) =
n_1!n_2!k!\sum_{c=0}^k
\sum_{s=0}^{c} \frac{(\eta_1/2)^s}{s!}
\sum_{t=0}^{k-c} \frac{(\eta_2/2)^t}{t!}
C_{n_1,s,c}C_{n_2,t,k-c}
\frac{\theta_{n_1+c-2s,n_2+k-c-2t}}{(n_1+c-2s)!(n_2+k-c-2t)!}.
$$
|
2025-03-21T14:48:31.471607
| 2020-07-11T18:03:38 |
365422
|
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Can the Mandelbrot set be designed through inequalities?
Many years ago I found an inequality that directly controlled whether a point $\displaystyle c$ belongs or does not belong to the Mandelbrot set. Roughly, it was something like this: If $\displaystyle g(c) \leq 1/4$ then $\displaystyle c$ belongs to $\displaystyle M$ (ie to the Mandelbrot set), where $\displaystyle g(c)$ is a definite function. This inequality was efficient on any scale and the only limitation was the numerical accuracy of the computer. The same function $ \displaystyle g (x)$ was easy to adapt to the design of any set defined by the relation $\displaystyle f_c (z)=z^n + c$. The set $\displaystyle M$ contains discs and cardioids, some of which are distorted. The same function $\displaystyle g$ was functional in all these cases. These are included in the good news.
However, the $\displaystyle g$ function was not complete, as it often malfunctioned at seemingly random points $\displaystyle c$, but it was possible to fix the problem with additional interventions. I had started to improve the $\displaystyle g$ function by gradually removing these interventions, but then the computer's hard drive broke down. I had not printed the method, and since then I have not dealt with this issue. I've even forgotten the basics about the Mandelbrot set. However, I recently found in the margins of a book some of my illegible notes that were the starting point for the construction of the $\displaystyle g$ function. I am unable to verify the accuracy of these notes, but I remember that on the basis of these it was possible to check whether a point $\displaystyle c$ belongs to the large cardioid of $\displaystyle M$ with an initial form of the above inequality.
$\displaystyle r_m=\frac{\sqrt{3 \pm 2 \sqrt{3-8x_m}}}{4}$
$\displaystyle \left (= \sqrt{x_m^2 + y_m^2} \right)$
$\displaystyle y_m= \pm \frac{\sqrt{3-16x_m^2 \pm 2 \sqrt{3-8x_m}}}{4}$
I'm not happy that I can only provide this information, but the purpose is to motivate research.
Note : inequality can not be used to color the points that do not belong to $\displaystyle M$, so they must be colored through the standard method of repetition. This is due to the fact that a significant part of the necessary information is removed in return for the speed. Of course this is important because all the beauty of the Mandelbrot ensemble is in its atmosphere.
Secondary note: At many intersections of $\displaystyle M$ it is obvious that if we zoom in on this area an island will appear. It is possible to find out how large the zoom must be for nz the island to appear by checking a random point near the intersection, but i do not remember the relevant relationship.
You should state your question more clearly. Yes, Mandelbrot set can be described by an inequality, namely $M=\{ z:u(z)\leq 0\}$ where $u(c)=v_c(c),$ and
$$v_c(z)=\limsup_{n\to\infty}2^{-n}\log|(f^n_c)|,$$
This function is relatively easy to compute approximately for any $c$ since the convergence to the limit is fast. It does not follow that it is easy or possible to compute its zero
set. Indeed, it is known proved that $M$ is not semi-algebraic, and thus "not computable"
in any strictly defined sense,
L. Blum, M. Shub, S. Smale, On a theory of computation and complexity over the real numbers: NP-completeness, recursive functions and universal machines. Bull. Amer. Math. Soc. (N.S.) 21 (1989), no. 1, 1–46.
There is also the book: Complexity and Real Computation
By Lenore Blum, Felipe Cucker, Michael Shub, Steve Smale.
M. Braverman and M. Yampolsky,
Computability of Julia sets.
Algorithms and Computation in Mathematics, 23. Springer-Verlag, Berlin, 2009.
|
2025-03-21T14:48:31.471891
| 2020-07-11T19:11:38 |
365425
|
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"authors": [
"Emil Jeřábek",
"Fedor Pakhomov",
"Gerhard Paseman",
"Mirco A. Mannucci",
"Pace Nielsen",
"Ville Salo",
"https://mathoverflow.net/users/123634",
"https://mathoverflow.net/users/12705",
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"https://mathoverflow.net/users/3199",
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"https://mathoverflow.net/users/36385"
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The inconsistency of Graham Arithmetics plus $ \forall n, n < g_{64}$
As you all know, Ronald Graham just passed away. He is famous for many fabulous contributions to finite combinatorics, and much much more, but perhaps none of them is as popular as the infamous Graham number (see here): $g_{64}$.
This number is truly huge, though in more recent times it has been dwarfed by several other "finite inaccessible numbers" (allow me to call them that way), such as Friedman's TREE(3) .
Anyway, it is big enough to play the role of "infinite number" in what follows.
Now, start with Robinson's arithmetics $Q$ and "beef it up" so that it contains all the recursive definitions necessary to talk about $g_{64}$ (for instance, all equations to define upper arrow notation).
As a tribute to Graham, let us call this theory GRAHAM.
Define T as GRAHAM + $ \forall n, n < g_{64}$
Clearly, T is classically inconsistent. But: just like Parikh's feasible arithmetics, there are, as far as I see, no short proofs of its inconsistency without appealing to induction.
Now the question:
assuming a fixed proof system (say Gentzen), can we give a measure of
the shortest induction-free proof that T is not consistent?
NOTE: I have mentioned Parikh's aritmetics, because obviously T and Parikh system share some common features.
But in his case, he keeps some induction and augments arithmetics with an additional unary predicate expressing feasibility for which induction does not apply , whereas here there is no such predicate, but on the other hand there is no induction whatsoever.
As a practice try two simpler theories first, using the Paseman theory (all numbers are less than 7) and the Mannucci theory (less than 2401), and see how proof length grows. My guess is the lengths are super exponential and that it will be challenging to estimate this length for the Mannucci theory. Gerhard "Remember, Baby Steps At First" Paseman, 2020.07.11.
LOL. Good take. Yes, Gerhard, but in Mannucci's theory (as I mentioned several years ago on MO) even 5 is inaccessible, why bothering with such a huge number ? I can barely count beyond 2. Anyway, jokes aside, good guess, but I need an exact gauge (I have something in mind..)
@EmilJeřábek thanks! 1) would love if you (at your leisure ) expanded this comments into a full blown answer (looks like I have to do many nested cuts to make room for iterated tetartion.... ). 2) also, any useful ref you can think of is appreciated ( probably Pudlak has something along these lines)
So which theory should be the minimal one to have such definable tetartion-cut?
I will try to write it up, but I’m not sure when I will be able to get to it.
As much as it is honorific to put someone's name to a concept, I don't know that I'd like someone to tie my name to an inconsistent system that I never studied.
I hear you Pace, in fact you may have a point. What do you suggest? Calling it Mannucci's arithmetics is a bit pretentious, right? PS Actually GRAHAM is not inconsistent, only GRAHAM PLUS "graham number is infinite" is. So, yes, Graham perhaps would not be happy to have his name tagged to a subsystem of arithmetics based on his number, but at least this system is rock solid.
Sorry about the super naive question, but if you can define $g_{64}$, is there really not a short proof that just says "a number that is larger than itself does not exist, yet $g_{64}$ exists"?
@VilleSalo how would you prove the statement " "a number that is larger than itself does not exist"? Remember, "GRAHAM" has no induction. You can calculate in it, and you can use basic logic, but nothing else
Well it's a "local" property in the integers so my very vague knowledge of Robinson suggested it's an explicit axiom or corollary of them. But ok, it's not.
@VilleSalo Robinson’s arithmetic can indeed prove that $n\nless n$ for every standard numeral $n$, such as $n=g_{64}$, but since it cannot prove the universal statement $\forall x,(x\nless x)$, there is no uniform proof; you need separate proofs for each $n$, and their lengths will grow with $n$. The most straightforward proof uses $O(n)$ lines. The basic point of my answer is to find shorter proofs of these statements where possible.
At the request of the OP, I’m writing a lengthy nonanswer showing that there are short proofs of inconsistency of similar theories where the “big number” is given by a term in the usual language of arithmetic $L_{PA}=\{0,S,+,\cdot\}$, possibly expanded by the exponential function. The argument does not work for languages including faster growing functions such as tetration, let alone the Ackermann function needed to succinctly represent the Graham number.
Let $|t|$ denote the size (= number of symbols) of a syntactic object $t$ (a term, a formula, etc.).
Theorem 1: For any closed $L_{PA}$-term $t$, there is a proof of $t\nless t$ in $Q$ (and therefore a proof of inconsistency in $Q+\forall x\,x<t$) with $O(|t|)$ lines, each of size $O(|t|)$.
Proof: We will use the fact that there are $Q$-definable cuts that interpret fragments of arithmetic such as $I\Delta_0$, see [1,§V.5(c)]. Specifically, there exists a formula $I(x)$ such that $Q$ proves
$$\begin{align}
&I(0),\\
&\forall x\,\forall y\:\bigl(I(x)\land I(y)\to I(S(x))\land I(x+y)\land I(x\cdot y)\bigr),\\
&\forall x\,\forall y\:\bigl(I(x)\land y<x\to I(y)\bigr),\\
&\forall x\:\bigl(I(x)\to x\nless x\bigr).
\end{align}$$
Let us fix $I$ and a $Q$-proof of the above. Then we prove
$$I(t)$$
by (meta)induction on the complexity of a closed term $t$: if, say $t$ is $t_0+t_1$, we instantiate one of the formulas above to obtain
$$I(t_0)\land I(t_1)\to I(t_0+t_1),$$
and we conclude $I(t_0+t_1)$ using the induction hypothesis and modus ponens. This argument involves $O(1)$ proof lines for each subterm of $t$, where each line has size $O(|t|)$. QED
In fact, the same argument shows more: since every $\Pi_1$ sentence $\psi$ provable in $I\Delta_0+\exp$ is interpretable on a cut in $Q$ by [1,Thm. V.5.26], we can take the cut $I$ above to satisfy $\psi$, and obtain
Theorem 2: Let $\theta(x)$ be a fixed $\Delta_0$ formula such that $I\Delta_0+\exp\vdash\forall x\,\theta(x)$. Then given a closed $L_{PA}$ term $t$, we can construct a $Q$-proof of $\theta(t)$ with $O(|t|)$ lines, each of size $O(|t|)$.
In order to adapt the argument to exponentiation, we need more work, as there are no definable cuts in $Q$ closed under exponentiation. Let $Q(\exp)$ be the theory in language $L_{\exp}=L_{PA}\cup\{x^y\}$ axiomatized by $Q$ and
$$\begin{align}
x^0&=1,\\
x^{S(y)}&=x^y\cdot x.
\end{align}$$
Theorem 3: Let $\theta(x)$ be a fixed $\Delta_0$ formula such that $I\Delta_0+\exp\vdash\forall x\,\theta(x)$. Then given a closed $L_{\exp}$ term $t$, we can construct a $Q(\exp)$-proof of $\theta(t)$ with $O(|t|)$ lines, each of size $O(|t|)$.
In particular, we can construct a proof of inconsistency in $Q(\exp)+\forall x\,x<t$ with such parameters.
Proof: As above, we fix a definable cut $I_0(x)$ that, provably in $Q(\exp)$, is closed under $+$ and $\cdot$, and satisfies $\mathrm{PA}^-$ and $\forall x\,\bigl(I_0(x)\to\theta(x)\bigr)$. Moreover, we can make sure $Q(\exp)$ proves
$$\begin{align}
\forall x\,\forall y\,\forall z\:\bigl(I_0(x)\land I_0(y)\land I_0(z)\to x^{y+z}&=x^y\cdot x^z\bigr),\\
\forall x\,\forall y\,\forall z\:\bigl(I_0(x)\land I_0(y)\land I_0(z)\to\:\, x^{y\cdot z}&=(x^y)^z\bigr).
\end{align}$$
We now define a sequence of shorter and shorter cuts by
$$I_{k+1}(x)\iff I_k(x)\land\forall y\:\bigl(I_k(y)\to I_k(y^x)\bigr).$$
Using the properties of $I_0$, it is easy to construct by metainduction on $k$ $Q(\exp)$ proofs that $I_k$ is a cut closed under $+$ and $\cdot$, using $O(1)$ proof lines for each $I_k$, i.e., $O(k)$ lines in total to prove the properties for $I_0,\dots,I_k$. Each line has size $O(|I_k|)$.
As defined, $I_{k+1}$ involves two occurrences of $I_k$, hence $|I_k|=O(2^k)$. Pretend for the moment that we can rewrite the definition of $I_{k+1}$ so that it only refers to $I_k$ once. Then $|I_k|=O(k)$, hence the proof so far has $O(k)$ lines, each of size $O(k)$.
$\DeclareMathOperator\ed{ed}$For any closed term $t$, we define the exponentiation nesting depth $\ed(t)$ by
$$\begin{align}
\ed(0)&=0,\\
\ed(S(t))&=\ed(t),\\
\ed(t\circ u)&=\max\{\ed(t),\ed(u)\},\qquad\circ\in\{+,\cdot\},\\
\ed(t^u)&=\max\{\ed(t),1+\ed(u)\}.
\end{align}$$
Then we construct $Q(\exp)$ proofs of
$$I_{k-\ed(t)}(t)$$
by induction on the complexity of a closed term $t$ such that $\ed(t)\le k$, using the properties of $I_0,\dots,I_k$ above. We use $O(1)$ proof lines for each $t$ on top of the induction hypothesis, hence $O(|t|+k)$ lines in total, each of size $O(|t|+k)$. Choosing $k=\ed(t)\le|t|$, we obtain a proof of
$$I_0(t),$$
and therefore of $\theta(t)$, with $O(|t|)$ lines, each of size $O(|t|)$.
It remains to show how to present the definition of $I_k$ so that it has size only $O(k)$. The basic idea is to use the equivalences
$$\begin{align}
\psi(x)\lor\psi(y)&\iff\exists z\:\bigl((z=x\lor z=y)\land\psi(z)\bigr),\\
\psi(x)\land\psi(y)&\iff\forall z\:\bigl((z=x\lor z=y)\to\psi(z)\bigr),
\end{align}$$
however, the definition of $I_{k+1}$ involves both a positive and a negative occurrence of $I_k$, and these cannot be contracted directly. To fix this, we encompass both polarities in a single predicate
$$J_k(x,a)\iff(a=0\land I_k(x))\lor(a\ne0\land\neg I_k(x)).$$
In order to make the notation manageable, let me write
$$\def\?{\mathrel?}(\phi\?\psi_0:\psi_1)\iff\bigl((\phi\land\psi_0)\lor(\neg\phi\land\psi_1)\bigr).$$
We can express $J_{k+1}$ in terms of $J_k$ as
$$\begin{align}
J_{k+1}(x,a)&\iff\bigl[a=0\?\forall y\,(J_k(y,1)\lor J_k(y^x,0)):\exists z\,(J_k(z,0)\land J_k(z^x,1))\bigr]\\
&\iff\begin{aligned}[t]
\bigl[a=0&\?\forall y\,\exists u,v\:\bigl((v=0\?u=y^x:u=y)\land J_k(u,v)\bigr)\\
&\,:\exists z\,(J_k(z,0)\land J_k(z^x,1))\bigr]
\end{aligned}\\
&\iff\begin{aligned}[t]
\forall y\,\exists z,u,v\:\bigl[a=0&\?(v=0\?u=y^x:u=y)\land J_k(u,v)\\
&\,:J_k(z,0)\land J_k(z^x,1)\bigr]
\end{aligned}\\
&\iff\forall y\,\exists z,u,v\:\bigl[\bigl(a=0\to(v=0\?u=y^x:u=y)\bigr)\\\
&\qquad\qquad{}\land\forall u',v'\:\bigl[\bigl(a=0\?u'=u\land v'=v:(v'=0\?u'=z:u'=z^x)\bigr)\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\to J_k(u',v')\bigr]\bigr].
\end{align}$$
Notice that even though the last expression looks complicated, it contains only one occurrence of $J_k$ (even if we expand the abbreviations), hence we use it as the definition of $J_{k+1}$. This way, we define formulas $J_k$ of size $O(k)$, and we put $I_k(x)\iff J_k(x,0)$.
Let me point out that a general method how to eliminate such nested definitions of predicates is given by Avigad [2].
References:
[1] Petr Hájek, Pavel Pudlák: Metamathematics of first-order arithmetic, Springer, 1994, 2nd ed. 1998, 3rd ed. Cambridge Univ. Press 2017.
[2] Jeremy Avigad: Eliminating definitions and Skolem functions in first-order logic, ACM Transactions on Computational Logic 4 (2003), no. 3, pp. 402–415, doi: 10.1145/772062.772068.
Legendary Emil! Well you got my vote for the first installment, and of course waiting for the second one . THANK YOU
You're welcome. (The missing part is really a boring technicality.)
Emil, even though yours is a non-answer, it is a VERY GOOD non-answer, so obviously you get my green. I have not had the time to go through each line of your last addition, but I still do not see an obstruction to generalize the method: suppose I add to Q much more than the totality of EXP, I add in fact the totality of Tetartion, let us call it GRAHAM+ Why in that theory shouldn't I prove the existence of contradiction using definable cuts?
You are thinking about it the wrong way. You don’t get cuts closed under various functions by exploiting functions total in the theory, but by exploiting inner stuctural properties of the cuts. Functions of the ambient theory are basically irrelevant. I suggest you study §V.5(c) in Hájek–Pudlák in more detail to see how it works. But basically, recall that the very first step is that I need a cut satisfying $\forall x,x\nless x$ (this is where the contradiction ultimately derives from). Now, this is obtained by taking an instance of induction for a $\Delta_0$ formula $\theta(x)$ that ...
... implies the property I want, and consider the definable set $I_0={u:\theta(0)\land\forall x,(\theta(x)\to\theta(x+1))\to\theta(u)}$. Pure logic proves that this is an inductive set, i.e., it is closed under $S$, but no matter how much functions I have in the ambient theory, there is no reason for $I_0$ to be closed under anything else than $S$. At this point you lost all the extra functions, and you are never going to get them back. There is a general method where you take a cut closed under $S$ and define inside it a cut closed under $+$, then inside that a cut closed under $\cdot$,...
... and you can carry on to get closure under the $\omega_k$ functions, but that’s it. You can’t get cuts closed under faster growing functions such as exponentiation.
Ok thanks for the clarificaton
Let me add one more illustration. Let $T$ be an arbitrarily strong finitely axiomatized sequential theory, such as $I\Sigma_n$, or even NGB. Then there is a definable cut $I$ such that $T$ proves $I\Delta_0$ + cut-free consistency $CFCon(T)$ on $I$. Assuming $T$ is consistent, there is no definable cut $J$ inside $I$ that is $T$-provably closed under exponentiation (because then $T$ would interpret $I\Delta_0+\exp+CFCon(T)$, contradicting Gödel’s theorem, as $I\Delta_0+\exp$ proves the standard derivability conditions for cut-free provability). There is also no cut $J$ such that ...
... $T$ proves $J(x)\to\exists y,(I(y)\land y=2^0_x)$ (because by using cut elimination, $J$ would be an interpretation of $I\Delta_0+Con(T)$ in $T$, again contradicting Gödel’s theorem). And in fact, there is no definable cut $J$ such that, $T$-provably, $J(x)\to\exists y,(y=2^x_k\land I(y))$ for all constants $k$, as this would again contradict a version of Gödel’s theorem proved by Pudlák.
The length of the least proof of contradiction in $\mathsf{Graham}+\forall n (n<g_{64})$ should be inbetween $(\log_2^*(g_{64}))^{1/N}$ and $(\ln^*(g_{64}))^{N}$, where $\ln^*(x)=\min\{n\mid \log_2^n(x)<0\}$ and $N$ is some reasonable small integer (which could be figured out by a careful examination of the proof).
So let me first spell out what exactly I take as $\mathsf{Graham}$. Namely I'll assume that $\mathsf{Graham}$ is $\mathsf{Q}$ with additional function $x\uparrow^{y}z$ and axioms
$x\uparrow^z0=x$
$x\uparrow^0S(y)=x(x\uparrow^0y)$
$x\uparrow^{S(z)}S(y)=x\uparrow^z(x\uparrow^{S(z)}y)$
Note that here $x\uparrow^{y}z$ should corresponds to $x\underbrace{\uparrow\ldots\uparrow}\limits_{y+1\;\text{arrows}}z+1$ rather than $x\underbrace{\uparrow\ldots\uparrow}\limits_{y\;\text{arrows}}z$. This change of notations is due to the fact that in $\mathsf{Q}$ we start naturals with $0$ rather than $1$.
Numbers $g_{n}$ thus would be denoted by the terms
$g_1=\underline{3}\uparrow^{\underline{3}}\underline{2}$
$g_{n+1}=\underline{3}\uparrow^{g_n}\underline{2}$
Proof of the upper bound essentially follows from Emil's answer. Using axioms of $\mathsf{Graham}$ by polynomial in $\log_2^*(g_{64})$ proof we would show that $g_{64}=\underline{3}\uparrow^{\underline{1}} \underline{b}$, for appropriate $b<\log_2^*(g_{64})$. Next,since $\uparrow^0$ is exponentiation operation we could apply Emil's analysis to show by a polynomial in $\log_2^*(g_{64})$ proof in $\mathsf{Graham}$ that $\lnot\underline{3}\uparrow^{\underline{1}} \underline{b}<\underline{3}\uparrow^{\underline{1}} \underline{b}$. This gives an upper bound for the length of the proof of contradiction.
I'll sketch how to give the lower bound using fulfillment sequences see [1].
I'll assume that we use the variant of first-order language with $\forall,\exists,\land,\lor$ and negations that could be only used with atomic formulas; for non-atomic $\varphi$ we denote as $\lnot\varphi$ the formula that is obtained from $\varphi$ by replacing $\forall$ with $\exists$, $\land$ with $\lor$, non-negated atomic $\psi$, with $\lnot \psi$ and vice versa.
Let $\tau=\langle \tau_i\mid i\le n\rangle$ be a sequence of non-decreasings subsets of some model $\mathfrak{M}_{\tau}$ of the language $L(\mathsf{Graham})$. We call $\tau$ of this form fulfulment sequences. We define fulfilment relation $\tau, i\Vdash \varphi$, for $i\le n$ and $L(\mathsf{Graham})$ formulas $\varphi$ with parameters from $\tau_i$:
$\tau,i\Vdash \varphi\iff \mathfrak{M}\models \varphi$, if $\varphi$ is atomic or negation of atomic
$\tau,i\Vdash \varphi\land \psi \iff \tau,i\Vdash \varphi\text{ and }\tau,i\Vdash \psi$
$\tau,i\Vdash \varphi\lor \psi \iff \tau,i\Vdash \varphi\text{ or }\tau,i\Vdash \psi$
$\tau,i\Vdash \forall x\;\varphi(x) \iff \text{for all $i\le j\le n$ and $a\in \tau_j$ we have }\tau,j\Vdash \varphi(a)$
$\tau,i\Vdash \exists x\;\varphi(x) \iff \text{either $i=n$ or for some $a\in \tau_{i+1}$ we have }\tau,i+1\Vdash \varphi(a)$
Note that for $i\ge j$ we have $\tau, j\Vdash \varphi\Rightarrow \tau,i\Vdash \varphi$. And note that for a formula $\varphi$ of logical depth $l$ and $i\le \mathsf{len}(\tau)-l$ we couldn't have $\tau,i\Vdash \varphi$ and $\tau,i\Vdash \lnot \varphi$ at the same time.
Essential idea here is that a fulfillment sequence is a finite approximation for a first-order model.
Let us say that a fulfulment sequence $\tau$ supports set of formulas $\Gamma$ if the following holds:
For any $\varphi(\vec{x})$ that is a subformula of formula from $\Gamma$, any $i\le \mathsf{len}(\tau)$, and $\vec{a}\in \tau_i$ we have $i\Vdash \varphi(\vec{a})\lor\lnot\varphi(\vec{a})$.
We have $t(\vec{a})\in \tau_{i+1}$, for any term $t(\vec{x})$ appearing in $\Gamma$, any $i< \mathsf{len}(\tau)$, and $\vec{a}\in \tau_i$.
The following lemma connects finite deductions with fulfilement sequences:
Lemma 1. Suppose $\Gamma(x)$ is a sequent of logical depth $l$, $P$ is a deduction of $\Gamma$ in Tait calculus (with cuts) of the depth $k$, and $\tau$ is a fulfilment sequence that supports formulas appearing in $P$. Then for any $i$ between $k$ and $\mathsf{len}(\tau)-l$ and $\vec{a}\in\tau_i$ we have $\tau,i\Vdash \bigvee\Gamma(\vec{a})$
Proof. By induction on structure of $P$.
Suppose we have some finite sets of formulas $\Gamma\subseteq \Delta$ that are closed under subformulas. And suppose that we have some long enough fulfilment sequence $\tau$ that supports $\Gamma$. The key construction that we need is the construction of a shorter $\tau'$ such that $\tau'$ supports $\Delta$, $\mathfrak{M}_\tau=\mathfrak{M}_{\tau'}$, and for some function $f\colon \{0,..,\mathsf{len}(\tau')\}\to \{0,..,\mathsf{len}(\tau)\}$ we have
$\tau'_i\subseteq \tau_{f(i)}$,
$\tau',i\Vdash \varphi(\vec{a})\iff \tau,f(i)\Vdash \varphi(\vec{a})$, for any $\varphi(\vec{x})\in \Gamma$ and $\vec{a}\in \tau'_i$.
Let $s$ be the sum of lengthes of formulas from $\Delta$ and $k$ be the least number such that for any $\varphi\in \Delta$ and it's depth $k$ subformula $\psi$ we have $\psi\in \Gamma$. It would be always possible to construct $\tau'$ of the length $n$ as long as $\tau$ had the length $\ge P(n,s)\uparrow \uparrow k+1$, for some polynomial $P$. I'll skip technical details of construction of $\tau'$ from $\tau$.
Now our goal would be to construct long enough fulfulment sequence $\tau$ that would fulfill and support the set of all axioms of $T=\mathsf{Graham}+\forall x (x<g_{64})$. Let us construct finite sets of naturals $A_0=\{0\}$ and $$A_{i+1}=A_i\cup \{\max(0,a-1)\mid a\in A_i\}\cup \{\min(t(\vec{a}),g_{64})\mid \vec{a}\in \tau_i\text{ and }t(\vec{x}) \text{ occurs in axioms of $T$}\}.$$ Let $n$ be the last step so that $A_{n}\ne [0,g_{64}]$ and let $s$ be the least so that $[s,g_{64}]\subseteq A_n$. The model $\mathfrak{M}_\tau$ is the model with the domain $[0,s]$ that is obtained from $\mathbb{N}$ by collapsing all numbers $>s$ to $s$. We put $\mathsf{len}(\tau)=n$ and $\tau_i=(A_i\cap [0,s))\cup\{s\}$. It is easy to see that
$\tau$ supports the set of axioms of $T$
$\tau,i\Vdash \varphi$, for any $i\le \mathsf{len}(\tau)$ and axiom $\varphi$ of $T$
$\mathsf{len}(\tau)>\log_2(\log_2(g_{64}))$.
Finally assume for a contradiction that $P$ is a proof of the sequent $\lnot \mathsf{Graham},\exists x \lnot(x<g_{64})$ of the length $k$, where $k\le(\log_2^*(g_{64}))^{1/N}$. Then from $\tau$ constructed above we construct $\tau'$ so that
$\tau'$ supports the set of all formulas occuring in $P$
$\tau',i\Vdash \varphi$, for any $i\le \mathsf{len}(\tau)$ and axiom $\varphi$ of $T$
$\mathsf{len}(\tau')\ge k+l$, where $l$ is the logical depth of $\bigvee \lnot \mathsf{Graham}\lor \exists x \lnot(x<g_{64})$.
We get to a contradiction since from Lemma 1 should have that $\tau',k\Vdash \bigvee \lnot \mathsf{Graham}\lor \exists x \lnot(x<g_{64})$ but at the same time from 2. we have $\tau',k\Vdash \bigwedge \mathsf{Graham}\land \forall x (x<g_{64})$. And the latter is impossible since $k$ is too far away from the end of the sequence $\tau'$.
[1] J.E. Quinsey, "Some problems in logic: Applications of Kripke's Notion of Fulfilment", PhD thesis, St. Catherine's College, Oxford, 1980, https://arxiv.org/abs/1904.10540
Fedor, simply fantastic! Technically, you should have the green, but I have already given it to Emil. Truth is, you both deserve it. I am proud that there are folks like Emil and you here, not only knowledgeable, but willing to share. Follow up question: looks to me that the same methods would do for essentially any other beast as for instance TREE(n), mutatis mutandis (gotta add the recursive definitions, etc). Thoughts?
@MircoA.Mannucci Yes the construction haven't relied on on any special properties of $g_{64}$. If you would add to $\mathsf{Q}$ axioms for recursive definitions of some other functions and instead of $g_{64}$ would take some other very large number based on this functions this would still work.
|
2025-03-21T14:48:31.473269
| 2020-07-11T19:20:06 |
365427
|
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|
Stack Exchange
|
What is the Zhu algebra of a lattice vertex algebra?
Associated to a vertex algebra $V$ is an associative algebra $A(V)$, the Zhu algebra. Its defining property is approximately that the representations of $V$ and of $A(V)$ are the same.
In vertex operator algebras associated to affine and Virasoro algebras, Frenkel and Zhu prove for instance that the Zhu algebra of the affine vertex algebra $V_k(\mathfrak{g})$ is $U(\mathfrak{g})$.
Question: Is the Zhu algebra of the vertex algbera $V_k(L)$ associated to lattice $L$ known?
Writing $\mathfrak{h}=L\otimes_{\mathbf{Z}}\mathbf{C}$, there is a map $V_k(\mathfrak{h})\hookrightarrow V_k(L)$, which gives a map $U(\mathfrak{h})\to A(V_k(L))$.
Shouldn't it be the group algebra of $L^*/L$?
@TheoJohnson-Freyd I have no idea. If you find a reference, I'd be happy to accept that as an answer.
https://arxiv.org/abs/q-alg/9605032
I don't know a reference. I just (perhaps mistakenly) had assumed it. I will have to read the link Reimundo posted. @ReimundoHeluani I suggest you repost that link (perhaps with one-sentence summary) as an answer?
The truth of the matter is: I never think about, nor use, Zhu's algebra. I recognize that it is really important. But in RCFT, what you care about is the category Rep(V) of representations for V, not some particular algebra A whose module theory approximates Rep(V). I mean, we don't even have, in general, an equivalence Mod(A(V)) = Rep(V)!
Lattice VOAs were, afaik, one of the very first classes of VOAs whose representations were completely understood. In the positive definite case, L*/L is well-known to parameterize the irreps of V_L, and also of the group algebra of L*/L. Hence my (wrong) assumption.
sorry @TheoJohnson-Freyd for not being verbose, still at the clinic for the birth of my second baby. That paper computes the Zhu algebra in the even lattice case, it starts with an example in rank one to show it's not something trivial. There are newer results in the odd case that escape my memory, but a quick search through the papers referencing this one should find them.
@ReimundoHeluani Congratulations!
In RCFT, there is a 1-1 correspondence between Irr(V) <---> Irr(A(V)), W<---> lowest weight subspace of W (let us call it W_0). Then A(V) is equivalent to \bigoplus End(W_0), where the direct sum is over all (equivalence classes of) irreducibles of V.
I'm expanding Reimundo Heluani's link, which gives the answer when $L$ is an even positive definite lattice. Write $\mathfrak{h}=L\otimes_{\mathbf{Z}}\mathbf{C}$. Every $\alpha\in L$ gives two elements, $E_\alpha\in \mathbf{C}[L]$ and $\alpha\in\mathfrak{h}$.
The Zhu algebra is
$$A(V_L)\ =\ U(\mathfrak{h})\otimes\mathbf{C}[L]/\ (\alpha -(\alpha,\alpha)/2)E_\alpha.$$
The algebra structure on $ U(\mathfrak{h})$ is the usual one, the structure on $\mathbf{C}[L]$ is almost the usual one
$$E_\alpha\cdot E_\beta\ =\ \text{const.}\cdot E_{\alpha+\beta}$$
(for the constants see equations 2.9 and 2.10 of arxiv.org/abs/q-alg/9605032), which together with
$$[\alpha, E_\beta]\ =\ (\alpha,\beta) E_\beta$$
give the algebra structure on $U(\mathfrak{h})\otimes\mathbf{C}[L]$.
It is a finitely generated algebra, and contains a copy of $U(\mathfrak{h})$ within it.
|
2025-03-21T14:48:31.473516
| 2020-07-11T20:07:04 |
365428
|
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|
Stack Exchange
|
Gibbs distribution as a pushfoward
Let $\Omega_{0}:=\{-1,1\}$ be a single spin space. If $\Lambda \subset \mathbb{Z}^{d}$ is a fixed finite set, take $\mathcal{F}_{0}$ to be the $\sigma$-algebra $2^{\Omega_{0}}$ on $\Omega_{0}$. We define $\Omega_{\Lambda} :=\{-1,1\}^{\Lambda}$ and $\mathcal{F}_{\Lambda}$ to be the associate product $\sigma$-algebra, i.e. $\mathcal{F}_{\Lambda} = 2^{\Omega_{\Lambda}}$. Moreover, take $\Omega:= \Omega_{0}^{\mathbb{Z}^{d}}$ and $\mathcal{F} := 2^{\Omega}$. If $\eta \in \Omega$, we can define the Hamiltonian for ferromagnetic models $\mathcal{H}_{\Lambda,h, \eta}:\Omega_{\Lambda}\to \mathbb{R}$ to be:
$$H_{\Lambda,h,\eta}(\omega_{\Lambda}) := -\sum_{x,y \in \Lambda}J(x,y)\omega_{x}\omega_{y}-\sum_{x\in \Lambda}\bigg{(}h+\sum_{y\in \Lambda^{c}}J(x,y)\eta_{y}\bigg{)}\omega_{x} $$
Here, $J=J(x,y)$ are coupling strenghts which are assumed to be non-negative and translation invariant. The above defined Hamiltonian is often referred as the Hamiltonian with boundary condition $\eta$.
Suppose $\nu_{0}$ is a counting measure on $\Omega_{0}$ given by $\nu_{0}(\emptyset)=0$, $\nu_{0}(\{\pm 1\}) = 1/2$. Take $\nu_{\Lambda}$ to be its associate product measure. The finite volume Gibbs state $\mu_{\Lambda,\beta, h, \eta}$ is a probability measure on $\Omega_{\Lambda}$ defined to be:
$$\mu_{\Lambda,\beta,h,\eta}(A) := \frac{1}{Z_{\Lambda, \beta, h, \eta}}\int_{A}e^{-\beta H_{\Lambda,h,\eta}(\omega_{\Lambda})}d\nu_{\Lambda}(\omega_{\Lambda})$$
where $\beta > 0$ is the inverse absolute temperature and $Z_{\Lambda,\beta,h,\eta}$ is a normalization constant called the partition function.
Now, as discussed in Friedli & Velenik's book (see chapter 3, page 81) it is convenient to extend some of the above definitions to subspaces of $\Omega$. Thus, let $\Omega_{\Lambda}^{\eta}:=\{\omega \in \Omega: \hspace{0.1cm} \mbox{$\omega_{x}=\eta_{x}$ if $x \in \Lambda^{c}$}\}$ and $\mathcal{F}_{\Lambda}^{\eta}$ to be the $\sigma$-algebra $\mathcal{F}_{\Lambda}^{\eta} = 2^{\Omega_{\Lambda}^{\eta}}$ on it. We define the Gibbs state $\mu_{\Lambda,\beta, h}^{\eta}$ on $\Omega_{\Lambda}^{\eta}$ by its evaluation on singletons:
$$\mu_{\Lambda,\beta,h}^{\eta}(\{\omega\}) = \frac{1}{Z_{\Lambda,\beta,h}^{\eta}}e^{-\beta H_{\Lambda,h}^{\eta}(\omega)}$$
where, here $H_{\Lambda, h}^{\eta}(\omega) := H_{\Lambda,h,\eta}(\Omega_{\Lambda})$ with $\omega_{\Lambda}$ being the projection of $\omega$ to $\Omega_{\Lambda}$ and $Z_{\Lambda,\beta,h}^{\eta}$ the new normalizing constant.
Question: Is $\mu_{\Lambda, \beta,h}^{\eta}$ just the pushfoward of $\mu_{\Lambda,\beta,h,\eta}$? In other words, consider the injection $i_{\eta}: \Omega_{\Lambda}\to \Omega_{\Lambda}^{\eta}$ given by $i_{\eta}(\omega_{\Lambda}) = \omega_{\Lambda}|\eta_{\Lambda^{c}}$ where $\omega_{\Lambda}|\eta_{\Lambda^{c}}$ is the sequence $\sigma$ on $\Omega$ with $\sigma_{x}=\omega_{x}$ if $x\in \Lambda$ and $\sigma_{x} = \eta_{x}$ otherwise. Is it true that $\mu_{\Lambda,\beta,h}^{\eta}(A) = \mu_{\Lambda,\beta,h,\eta}(i_{\eta}^{-1}(A))$? See, I think this is true because it seems these two measures coincide on singletons, but I might be misunderstanding something.
ADD: I think I should elaborate my question a little more. If these two measures agree on singletons, they surely will agree on all $\Omega_{\Lambda}$ since this set is finite. The point is that I don't know for sure if these measures indeed agree on singletons. But let me elaborate the reasoning that led me to this belief.
Because $\mathcal{F}_{\Lambda}=2^{\Omega}$, every singleton is measurable. Thus, take a singleton $\{\omega_{\Lambda}\} \in \mathcal{F}_{\Lambda}$. Then, because $\nu_{\Lambda}$ is a counting measure, we have:
\begin{eqnarray}
\mu_{\Lambda, \beta, h,\eta}(i_{\eta}^{-1}(\{\omega\})) = \mu_{\Lambda,\beta,h\eta}(\{\omega_{\Lambda}\}) = \frac{1}{Z_{\Lambda,\beta,h,\eta}}e^{-\beta H_{\Lambda, h,\eta}(\omega_{\Lambda})} = \frac{e^{-\beta H_{\Lambda, h,\eta}(\omega_{\Lambda})}}{\sum_{\omega_{\Lambda}\in \Omega_{\Lambda}}e^{-\beta H_{\Lambda, h,\eta}(\omega_{\Lambda})}} = \frac{e^{-\beta H_{\Lambda, h}^{\eta}(\omega_{\Lambda})}}{\sum_{\omega \in \Omega_{\Lambda}^{\eta}}e^{-\beta H_{\Lambda,h}^{\eta}(\omega)}} = \mu_{\Lambda,\beta,h}^{\eta}(\{\omega\})
\end{eqnarray}
If two measures on a finite set agree on singletons, doesn’t that mean they’re equal?
@AnthonyQuas it surely does. Maybe my post was not entirely clear on this. The point is that I think but I don't know for sure that these measures agree on singletons. Maybe I should edit it explaining the reasoning behind this belief.
Yes, these two measures indeed agree on singletons. Which point of view is more convenient depends on the application (for instance, in Chapter 6 of our book, the second point of view is better suited for the analysis; in Chapter 8, the first point of view turns out to be more convenient; in Chapter 3, both could be used equally easily). As you notice, it is very simple to go from one to the other.
@YvanVelenik great! Thank you so much!
@MathMath: On $\Omega={-1,1}^{\mathbb{Z}^d}$, taking $\mathcal{F}=2^{\Omega}$ as the sigma-algebra is no good. In general, your sigma-algebras should be Borel sigma-algebras for suitable topologies. For finite sets like $\Omega_0$ or $\Omega_{\Lambda}$ the correct topology is the discrete one. But when considering $\Omega$, you should take the product topology coming from the discrete one on the factors ${-1,1}$. Then $\mathcal{F}$ should be the Borel sigma-algebra for this product topology.
The whole point of the construction, is to consider the limit of probability measures when $\Lambda$ fills the whole lattice. This is a problem of weak convergence of probability measures, which requires topologies for this to make sense.
|
2025-03-21T14:48:31.473984
| 2020-07-11T20:21:33 |
365429
|
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|
Stack Exchange
|
Independence results on pure algebra
I think that the most celebrated result in this direction is Shelah's famous work on Whitehead's Problem:
Is every abelian group $A$ such that $Ext^1(A, \mathbb{Z})=0$ free?
This is known to be indepedent of ZFC (cf. Shelah, Whitehead groups may not be free, even assuming CH I and II), and it seems that there is an interesting connection between certain questions of ZFC and abelian group theory (cf. Eklof The affinity of set theory and abelian group theory).
Another famous work relating algebra and 'higher' set theory is B. L. Osofsky Homological dimension and the continuum hypothesis and subsequent works.
Independence results, Martin's axiom, Continuum Hypothesis and large cardinal axioms have proven to be interesting to 'pratical' questions in many areas of mathematics (point-set topology, functional analysis, order theory, etc).
Is there any current research trend in using sistematically Independence results and other aspects of 'higher' set theory to understand problems of algebra, such as in the above mentioned mathematical disciplines?
The following survey by Scholze talks about cardinals, so potentially it could lead to some independence!!
https://www.math.uni-bonn.de/people/scholze/Condensed.pdf
The other one which comes to the stage again, is strong homology, for example you can consult the following paper
https://arxiv.org/abs/1509.09267
There are also some new papers in this direction, in particular of Jeff himself. This may not be directly related to algebra, but...
@Rahman.M thank you very much!
Have you looked at Eklof and Mekler "almost free modules: set theoretic methods"? You might find something interesting in there. A simple example "there is a nonreflexive free abelian group iff there is a measurable cardinal"
@AlessandroCodenotti this example is impressive. I will check it!
|
2025-03-21T14:48:31.474142
| 2020-07-11T20:47:56 |
365433
|
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|
Stack Exchange
|
Cohomological Brauer group vs classical
Let $X$ be a smooth scheme over $\mathbb{C}$.
A $O_X$-algebra $A$ is called Azumaya algebra on $X$
if locally it's ismorphic to matrix algebra: ie for
every $p \in X$ there exist open $U \subset X$ with
$p \in U$ and $A \vert _U \cong Mat_{r}(O_U) $
for some rank $ r >0 $.
Two Azumaya algebra
$A$ & $B$ are equivalent iff there exist two locally free $O_X$-modules $E,F$ locally of finite rank
with $A \otimes End(E) \cong B \otimes End(F)$ (as Asumaya algebras).
We call $Br(X)$ as the Brauer group defined as
the set of isomorphism classes of Azumaya
algebras modulo the described equivalence relation.
We want to draw analogy of the isomorphism
$Pic(X) \cong H^1(X, O_X^*)$ and endow $H^(X,O_X^*)$ with a
intepretation classifying isomorphy classes
of intersting geometric objects as well.
In
https://en.wikipedia.org/wiki/Brauer_group#The_Brauer_group_of_a_scheme
is remarked that for quasi-compact scheme $X$ the torsion subgroup of
the étale cohomology group $H^2_{et}(X, O_X^*)$ is called
the cohomological Brauer group.
Question: Is there an explicit way to relate $Br(X)$ with
torsion group of $H^2_{et}(X, O_X^*)$. A morphism from $Br(X)$ to this group ? If yes,
how this construction work? Could anybody sketch the idea or give a reference
where this construction (if it exist) is explaned?
From the same wikipedia page you link: "The Brauer group is always a subgroup of the cohomological Brauer group. Gabber showed that the Brauer group is equal to the cohomological Brauer group for any scheme with an ample line bundle (for example, any quasi-projective scheme over a commutative ring)." The map from the Brauer group to the cohomological Brauer group is described, for fields, a little higher up on that Wikipedia page. The idea in etale cohomology is similar. Full details are given in Milne's etale cohomology group, for example.
See section 3.3 of http://wwwf.imperial.ac.uk/~anskor/brauer.pdf
This is explained in great details in the Bourbaki lectures of Grothendieck, reprinted in "10 exposés sur la cohomologie des sch'emas" (North-Holland).
Briefly, one uses the exact sequence
$$
H^{1}(X,GL_{n})\rightarrow H^{1}(X,PGL_{n})\rightarrow H^{2}(X,\mathbb{G}_{m})
$$
(etale cohomology). The set $H^{1}(X,PGL_{n})$ classifies the isomorphism classes of Azumaya
algebras of degree $n^{2}$ over $X$, the set $H^{1}(X,GL_{n})$ classifies the
isomorphism classes of
locally free $\mathcal{O}_{X}$-modules of rank $n$, and the first map sends $V$ to
$End(V)$. See the references in the comments.
|
2025-03-21T14:48:31.474333
| 2020-07-11T21:02:46 |
365434
|
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Stack Exchange
|
Invariant theory in universal algebra
Let $\mathcal{L}$ be a finite first-order language with no relation symbols and let $\mathcal{K}:=\mathcal{V}(\Theta)$ be a variety in this language defined by a set of identities $\Theta$.
My questions are motivated, of course, by Noether's Theorem in invariant theory, and the Chevalley-Shephard-Todd theorem (which can be seen as statements about the variety of commutative associative algebras) for algebraic systems (using Malcev's terminology).
Question 1
Let $A$ be a finitely generated algebraic system in $\mathcal{K}$ and let $G$ be a finite group acting by $\mathcal{K}$-automorphisms on $A$. When is $A^G:=\{a \in A| g\cdot a=a, \forall g \in G\}$ again finitely generated? A subcase of important interest is when $A$ is a relatively free algebraic system of $\mathcal{K}$.
Question 2
Let $A$ be a finitely generated and relatively free algebraic system in $\mathcal{K}$ and let $G$ be a finite group of automorphisms, as above. When is $A^G$ a (possibly infinitely generated) relatively free algebraic system of $\mathcal{K}$ itself?
For varieties of associative algebras, the work on Questions 1 and 2 has reached a very mature form (cf. Formanek, Noncommutative invariant theory, MR0810646). There has been an intensive study of these questions for varieties of Lie algebras by a number of people (V. Drenksy, V. Petrogradskii, etc) but I know of no work for other important varieties of non-associative algebras (such as Jordan algebras, alternative algebras, Malcev algebras, etc), and also nothing about general algebraic systems.
Refinement of Questions 1 and 2 Is there any general result in Universal algebra (or Model theory) that is relevant for the purposes of these questions? Is something known about varieties of non-associative algebras (other than Lie)?
Can you please explain what do you mean by the symbol $A^G$?
@JakubOpršal $A^G = { a \in A| g a = a, \forall g \in G }$. Thanks, I will edit the question for clarity
Refinement of Questions 1 and 2: Is there any general result in Universal algebra (or Model theory) that is relevant for the purposes of these questions?
I think that this is too general of a question to expect a complete and satisfying answer. Let me make a few remarks anyway.
Remark 1.
The Math Subject Classification numbers for General Algebraic Systems all have the form 08XXX. The
Math Subject Classification numbers for invariant theory (in the sense of this question) are 13A50 (actions of groups on commutative rings), 14L24 (geometric invariant theory), 16R30 (trace rings and invariant theory), and 16W22
(actions of groups and semigroups; invariant theory (associative rings and algebras)). If you search the MathSciNet database for papers with Primary or Secondary MSC number 08XXX and at least one
Primary or Secondary MSC number from 13A50, 14L24, 16R30, or 16W22, you get exactly three papers. None of them seems to be related to this problem.
You can examine the papers yourself, if you want to check this, they are:
Khrypchenko, Mykola; Novikov, Boris
Reflectors and globalizations of partial actions of groups.
J. Aust. Math. Soc. 104 (2018), no. 3, 358-379.
MSC: 18A40 (08A02 08A55 08B25 08C05 16W22)
Dokuchaev, M.
Recent developments around partial actions.
São Paulo J. Math. Sci. 13 (2019), no. 1, 195-247.
MSC: 6W22 (08A02 16S10 16S35 20C25 46L55 54H15)
Brookes, Matthew D. G. K.
Congruences on the partial automorphism monoid of a free group action.
Internat. J. Algebra Comput. 31 (2021), no. 6, 1147-1176.
MSC: 08A30 (16W22 20M18 20M20)
So, it seems that no one has combined the types of ideas
mentioned in the problem statement. (Perhaps in very old papers
this has been done, and the results are not
reflected well in the MSC system.)
Remark 2.
The original question asks about more than just papers/results on this
topic, but also which universal algebraic
results might be relevant. As I wrote above,
I think that this is far too general
a question. The field of UA is approximately 90 years old.
MathSciNet lists 13790 papers in MSC 08XXX stretching back to 1931.
It is possible that something in one of those papers is relevant.
Today there are 536 papers in the database which have
MSC Primary or Secondary number 08B20, which is the number
for Free Algebras. Any of those might be relevant.
Remark 3.
I think that questions of this type could have interesting
answers. I could imagine posing the following project
to someone.
Let Property $P(G)$ be the property of a variety $\mathcal V$
that, for a finite group $G$, the subalgebra
$\mathbf{F}_{\mathcal V}(n)^G$ of the $n$-generated
$\mathcal V$-free
algebra $\mathbf{F}_{\mathcal V}(n)$
is also $\mathcal V$-free algebra.
The project is:
examine all varieties generated by $2$-element algebras and determine
which of them have property $P(G)$ for some/all finite groups $G$.
It is easy to see that the variety generated by the $2$-element
semilattice has the property $P(G)$ for any finite $G$.
It is not too hard to see that the variety
generated by the $2$-element
lattice does NOT have the property $P(G)$ for the $2$-element group $G$.
Perhaps by examining all the varieties generated
by $2$-element algebras, some patterns might emerge.
I think the project is feasible, because varieties
generated by $2$-element algebras have been classified.
In order to add some math to this answer, let me
close by explaining why the variety
generated by the $2$-element
lattice does NOT have the property $P(G)$ for the $2$-element group $G$.
The variety in question is the variety of distributive lattices.
The sizes of the small free algebras are
$|\mathbf{F}_{\mathcal V}(0)|=0$,
$|\mathbf{F}_{\mathcal V}(1)|=1$,
$|\mathbf{F}_{\mathcal V}(2)|=4$,
$|\mathbf{F}_{\mathcal V}(3)|=18$.
Let $G$ be the group of automorphisms generated
by the automorphism $\alpha$ of
$\mathbf{F}_{\mathcal V}(x,y,z)$
which fixes $x$ and swaps $y$ and $z$.
It is not hard to see that
$|\mathbf{F}_{\mathcal V}(3)^G| = 8$,
so this subalgebra of
$\mathbf{F}_{\mathcal V}(3)$ cannot be a relatively free algebra.
In fact, the universe of
$\mathbf{F}_{\mathcal V}(3)^G$ is generated
by $\{x, y\wedge z, y\vee z\}$ and consists of
$\{x\wedge y\wedge z, y\wedge z, x\wedge (y\vee z),
(x\wedge (y\vee z))\vee(x\wedge y), x, y\vee z,
x\vee (y\wedge z), x\vee y\vee z\}$.
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2025-03-21T14:48:31.474693
| 2020-07-11T21:22:19 |
365435
|
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|
Stack Exchange
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Extending etale morphisms
Let $Y$ be an affine, integral, Gorenstein surface. Let $y \in Y$ be a closed point such that there exists a finite, etale morphism $f: X \to Y\backslash \{y\}$ from an integral variety $X$ to the open subvariety $Y \backslash y$. Does there exist an integeral variety $\overline{X}$ containing $X$ as an open dense subscheme and a finite morphism $\overline{f}: \overline{X} \to Y$ restricting to $f$ over $X$ such that the fiber over $y$ consists of exactly one closed point (not necessarily reduced)?
Yes: first choose any finite morphism $g:Z \to Y$ extending $f$ with $Z$ integral (e.g., by normalizing $Y$ in $X$), and then construct $\overline{X}$ by crushing the finite closed subscheme $g^{-1}(y) \subset Z$ to a single point (so take a suitable fibre product at the level of rings).
@Anonymous Thank you. Could you put this in the answer. Especially, could you elaborate a little (or give reference) for what you mean by "crushing". Is it something like blowing down? If so, why does a "crushing" always exist? I think the answer lies in what you say "suitable fibre product at the level of rings", but this statement is not clear to me.
For the existence of the crushing (also called "pinching", or Ferrand pushout) see this paper, or this section of the Stacks Project.
In this case the construction is very simple: take $g$ as in Anonymous's comment. Then $\ol{X}=\mathrm{Spec}(\mathscr{A}))$ where $\mathscr{A}\subset g_*\mathscr{O}_Z$ is the subalgebra of sections whose restriction on $g^{-1}(y)$ is constant (i.e. comes from $\kappa(y)$).
@LaurentMoret-Bailly and Anonymous; Thank you for the answer
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2025-03-21T14:48:31.474838
| 2020-07-11T21:26:59 |
365436
|
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|
Stack Exchange
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Meaning of "combinatorial data"
I saw several times that often some data describing certain
algebraic objects,
eg the set of cells of a simplical complex
or a Cech cycle of a chosen coving of a variety
are called "combinatorial data" or "encoding combinatorial data".
My question is simply what is the justification of the
name "combinatorial" here? what is combinatorial on these data?
More generally can it be precised when a certain data
describing an algebraical object is called "combinatorial"?
A short UPDATE to precise what I mean (literally that's the comments below):
The meaning of "combinatorial" for abstract simplicial complexes is pretty easy to see: indeed, a data consisting of set of vertices $V=\{ v_1,v_2,...,v_n \}$ and a $m$-simplex of $S$ is defined as a subset $\{ v_{i_1},...,v_{i_m} \}$ of $V$. then a subset $S$ of power set $P(V)$ of $V$ is called abstract simplicial complex if for every $m$-simplex $\{ v_{i_1},...,v_{i_m}\}$ contained in $S$ every subset $\{ v_{i_{j_1}},...,v_{i_{j_d}} \}$ is (as a $d$-simplex) is contained in $S$ as well.
Therefore obviously not every subset of the power set $P(V)$
of $V$ is a abstract simplicial complex. So to determine which subsets of $P(V)$ can occure as abstract simplicial complexes is a combinatorial problem. That's where I see the "combinatorial flavour" here. So I think that this is precisely the justification for the word "combinatorial" if one can associate to an algebraic or topological object an abstract simplical complex
encoding sometimes a lot of information about the original object. (the most prominent example is surely the Nerve theorem which desides when this "combinatorial data" essentially suffice to reconstruct the original object up to homotopy).
But how to draw the same analogy to Cech cycles isn't clear to me. Is it possible to associate abstractly a abstract simplicial complex to a Cech cycle in order to "make" it combinatorial?
I don't think there's a precise formal meaning. I understand it to mean something Kronecker would approve of, i.e., an essentially finistic mathematical object, something you could encode on a computer (e.g., a finite graph or hypergraph, as opposed to something like a differentiable manifold).
the origin of the name as well as a general definition is discussed in Wikipedia --- the combinatorial map tells you how to combine simplices to form a simplicial complex.
yes I thought that orginally this arose from the concept of abstract
simplicial complex: a data $S$ consisting of vetrices $V={v_1,v_2,...,v_n}$
an a $m$-simplex of $S$ is a subset ${v_{i_1},..., v_{i_m}}$ of $V$.
$S$ is called abstract simplicial complex if for every $m$-simplex
${v_{i_1},..., v_{i_m}}$ contained in $K$ every subset
${v_{i_{j_1}},..., v_{i_{j_d}}}$ is (as a $d$-simplex) is contained in $S$.
Therefore not every subset of the power set $P(V)$ of $V$ is a abstract simplicial complex. So to determine which subsets of $P(V)$ can occure as abstract simplicial complexes is a combinatorial problem. But how to draw the same analogy to cech cycles isn't clear to me. Is it possible to associate abstractly a abstract simplicial complex to a Cech cycle in order to "make" it combinatorial?
I think this is substantially an issue of tradition. Some decades ago (e.g. 1970), when an otherwise finitistic thing seemed to admit no simpler description than ... itself (e.g., in terms of Kolmogorov-Solomonov-Chaitin complexity, for example), it was referred to as "combinatorial".
Yes, that sense of the word is not closely related to usage in 2020.
In particular, in older work, there is probably no reward in looking for significant subtler meaning, other than "it is what it is"... as opposed to being "compressible" in any way in the larger context of mathematics.
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2025-03-21T14:48:31.475075
| 2020-07-11T22:40:18 |
365438
|
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|
Stack Exchange
|
Non-zero divisors on an $I$-completely flat module
Let $A$ be a commutative ring (not necessarily Noetherian), $I=(f_1, f_2, \dots, f_n) \subseteq A$ a finitely generated ideal that is generated by a regular sequence. Let $M$ be an $A$-module, and assume that both $M$ and $A$ are $I$-adically complete. Finally, suppose further that $M$ is $I$-completely flat (sometimes also called $I$-adically flat), meaning that $\mathrm{Tor}_{>0}^{A}(A/I, M)=0$ and $M/IM$ is a flat $A/I$-module.
My question is
If $x \in A$ is a non-zero divisor of $A$, is $x$ necessarily a non-zero divisor on $M$?
This seems true at least under some further assumptions on $x$, such as $xA \cap I^k=xI^k$ for all $k$, but I am actually interested in the somewhat perpendicular case when $x \in I$, or better yet, the general question whether such $I$-completely flat modules are torsion-free. (The claim is also true for $x=f_1, f_2, \dots, f_n$ I believe.)
Some details:
One can look at the exact sequences
$$0 \rightarrow K_n/I^n \rightarrow A/I^n \stackrel{x}{\rightarrow} A/I^n $$
where $K_n=(I^n: x)$.
Taking $\varprojlim_n,$ from the fact that $x$ is a non-zero divisor on $A$ it follows that $\varprojlim_n K_n/I^n=0$.
It can be shown that the $I$-complete flatness is equivalent to $I^n$-complete flatness. Thus, applying $-\otimes_AM=-\otimes_{A/I^n}M/I^nM$ to the above exact sequence, the sequence
$$0 \rightarrow K_nM/I^nM \rightarrow M/I^nM \stackrel{x}{\rightarrow} M/I^nM$$ remains exact, and again, taking $\varprojlim_n$ one obtains that $M[x]=\varprojlim_n K_nM/I^nM$.
So the question is, is $\varprojlim_nK_n/I^n=0$ enough to guarantee $\varprojlim_nK_nM/I^nM =0$ in this situation?
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2025-03-21T14:48:31.475206
| 2020-07-11T23:51:51 |
365441
|
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|
Stack Exchange
|
Simultaneous small fractional parts of polynomials
Fix $\epsilon>0$. For a finite set of arbitrary-degree polynomials with integer constant term, $p_1(x), ..., p_m(x)\in \mathbb{R}[x]$ is it possible to find an $n\in \mathbb{N}$ such that $$\max_{i=1,...,m}||p_i(n)||<\epsilon$$ where $||\cdot||$ denotes the distance to the nearest integer?
In the description of the book Small Fractional Parts of Polynomials a problem of this type is listed as an open question for degree greater than 4, but the book is from 1977. Research has continued since then on small fractional parts of polynomials. Has progress improved on this problem, or is it still open?
Updates: Discussion in the comments indicates that this can be proved by Weyl's equidistribution criterion. While this proves the result for $\mathbb{Q}$-linearly independent collections $\{p_1(n),...,p_k(n), 1, n, n^2,...\}$ it does not prove it when the collection is linearly dependent, since equidistribution no longer holds in that case. How can we generalize to when the polynomials may be $\mathbb{Q}$-linearly dependent?
Not if each polynomial is of the form $P(x) + 1/2$ for some $P \in {\bf Z}[X]$. So some hypothesis is necessary. What is condition imposed in the $p_i$
in your source?
Ah of course, sorry! It is imposed that the polynomials have integer constant term. I will edit the post.
OK, but something still seems to be missing.
For example, if $p_1$ never takes integer values on $\mathbb N$,
and $p_2 = -p_1$, then ${ p_1(n) } + { p_2(n) } = 1$
for all $n \in \mathbb N$, so the fractional parts cannot both be small.
Great catch. This shouldn't be fractional part, but the nearest integer function instead.
*distance to the nearest integer
OK, but that must be implied by Weyl's equidistribution criterion on a torus.
Schmidt surely knows this, so I expect that there's some other issue,
such as speed of convergence.
@NoamD.Elkies do you mean Weyl's criterion for a single polynomial?
So it is equivalent to the analogous question about several polynomials of the form $cx^k$, $k>0$, right?
@Fedor Petrov: There's a natural generalization from
the circle ${\bf R} / {\bf Z}$ to the torus ${\bf R}^m / {\bf Z}^m$.
A sequence of vectors $v_n \in {\bf R}^m / {\bf Z}^m$
is asymptotically equidistributed iff for even nonzero
$a \in {\bf Z}^m$ we have
$$
\sum_{n=1}^N \exp 2\pi i (a \cdot v_n) = o(N)
$$
as $N \to \infty$. This does in the end come to the estimate for
a single polynomial. (If $p_1,\ldots,p_m$ are $\bf Q$-linearly dependent
then they're equidistributed in some closed subgroup of the torus.)
@NoamD.Elkies Would you be able to point to a resource proving that $\mathbb{Q}$-linearly dependent collections are equidistributed on a closed subgroup?
The claim is true as stated, and can be arrived at using Weyl's criterion which was pointed out in the comments. As I post this answer, there is no consensus on the rate of convergence of the $N$.
We proceed by induction on $k$, the number of polynomials. For $k=1$, either $p_1$ has only rational coefficients or it has at least one irrational. If it has an irrational one, Weyl's criterion shows that $p_1(n)$ is an equidistributed sequence, and thus has infinitely many $N$ for which $||p_1(N)||<\epsilon$. If $p_1$ has only rational coefficients, we can simply choose $N$ to be the product of every coefficient's denominator to get $||p_1(N)||=0$.
Now suppose the theorem holds for some fixed $k$. Consider a polynomial list $p_0, p_1, ..., p_k$ of length $k+1$. If $\{p_0(x),p_1(x),...,p_k(x), 1, x, x^2,...\}$ is $\mathbb{Q}$-linearly independent, then Weyl's criterion on the $k+1$ torus applies and equidistribution gives the existence of infinitely many $N$ for which $||p_i(N)||<\epsilon$ for all $i=0,...,k$.
If, on the other hand, the aforementioned collection is instead $\mathbb{Q}$-linearly dependent, Weyl's criterion does not apply. However, $$p_0(x)=q_1p_1(x)+\cdots +q_kp_k(x)+r_0+r_1x+\cdots +r_sx^s$$ for $q_i,r_j\in \mathbb{Q}$. If we let $q$ be the product of all denominators of the $r_j$, we have that at any natural number $n$, $$||p_0(qn)||=||q_1p_1(qn)+\cdots +q_kp_k(qn)||.$$
Apply the inductive hypothesis to $q_1p_1(qx),...,q_kp_k(qx)$. Choose $M$ so that $||q_ip_i(qM)||<\epsilon/2k$ for all $i=1,...,k$. At this small scale, the $||\cdot||$ function is subadditive in $k$ arguments, so $$||p_0(qM)||\leq ||q_1p_1(qM)||+\cdots +||q_kp_k(qM)||<k(\epsilon/2k)<\epsilon.$$
Therefore, at $N=qM$, $||p_i(N)||<\epsilon$ for all $i=0,...,k$. We are finished by induction.
|
2025-03-21T14:48:31.475855
| 2020-07-12T04:37:38 |
365447
|
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|
Stack Exchange
|
Loop spaces motivation
I read that one of the main goals of utilization simplicial methods is to prove that a space is a loop space. On the other hand where lies the main importance to recognize topological spaces as loop spaces? Surely, if a space is a loop space then its connected components obtain a magma structure via concatenation because if $X$ is homotopy equivalent to a loop space $\Omega(Y)$ then $\pi_0(X)= \pi_1(Y)$ and thus there is a map $*: \pi_0(X) \times \pi_0(X) \to \pi_0(X)$.
Is this the only motivation behind the study of loop spaces? Or what other interesting aspects can be gained from studying loop spaces from the viewpoint of homotopy theory?
The rational cohomology of an infinite loop space is very nice: it is generated as a free graded commutative algebra by a basis for the rational cohomology of the associated spectrum. Why is this useful? Here’s one beautiful use of it. The Mumford conjecture is about the rational cohomology of the mapping class group, and it turns out the classifying space of the mapping class group is an infinite loop space. So figure out what the associated spectrum is and bam you’ve solved a huge problem just by knowing cohomology of loop spaces.
You might be interested in https://math.stackexchange.com/q/164118/10014
I should say the classifying space of the mapping class group is homology equivalent to an infinite loop space, it can’t be an infinite loop space because the mapping class group is not abelian.
Being equivalent to a loop space is much stronger than \pi_0 being a group. By considering loops parametrized by [0, t] for all t\geq 0, you see that a loop space is a deformation retract of a (strictly unital and associative) topological monoid. But a strict algebra structure is not preserved by homotopy equivalence, so topological monoids are not a correct notion of 'homotopical associative monoid.' The definition of A-infinity algebra, which is the homotopically correct intrinsic notion of associative monoids, takes a cue from the recognition of algebraic structures present in loop spaces.
@crispr, well, @NarukiMasuda did say it was a weak(er) property …. But, for what it's worth, note that only countable sets can be cyclic groups.
@crispr What I meant is that, while the magma structure on $\pi_0(X)$ in the question (so not an arbitrary one) is a group when $X$ is a loop space (which is already stronger than what OP said), we can even lift the multiplication on $X$ which is associative and unital "up to coherent homotopy" (whatever it means), and the existence of such a multiplication on $X$, in turn, characterizes loop spaces. This is the so-called 'recognition principle of the loop spaces' which probably OP had in mind.
There are several useful points in the comments, but I want to go beyond them and try to give a more comprehensive answer, so this question doesn't linger unanswered. Some great sources are May's Geometry of Iterated Loop Spaces (GILS) and A Concise Course in Algebraic Topology (CCAT). As the OP points out, if $X$ is a loop space, then concatenation makes $\pi_0(X)$ a group.
Is this the only motivation behind the study of loop spaces? Or what other interesting aspects can be gained from studying loop spaces from viewpoint of homotopy theory?
No, there is much more to loop spaces than the observation that concatenation yields a group structure on $\pi_0(X)$. First of all, motivation is easy. Since homotopy theory is intimately tied to paths, loop spaces themselves are fundamental objects, e.g., because they allow you to shift dimension as $\pi_i(\Omega X)\cong \pi_{i+1}(X)$. The operation of taking loops is connected to the suspension operation as follows. Let $X$ and $Y$ denote based spaces, $F(X,Y)$ denote the space of based maps between them (so $\Omega X = F(S^1,X)$), and $\Sigma X = X\wedge S^1$ denote the suspension of $X$. The usual hom-tensor adjunction tells you that $F(\Sigma X,Y) \cong F(X,\Omega Y)$. Taking $\pi_0$, we have $[\Sigma X,Y] \cong [X,\Omega Y]$, and composition of loops turns this set into a group. Hence, $\Omega Y$ is a cogroup object in the homotopy category of pointed spaces, and this is used in Hovey's book to set up the much more general homotopy theory encoded by model categories.
Since the theory of $\Omega$-spectra starts with a sequence of based spaces $T_n$ and weak equivalences $T_n\to \Omega T_{n+1}$, loop spaces are also foundational to stable homotopy theory. They pop up in the long exact sequences induced by fiber and cofiber sequences, that allow us to compute things in stable homotopy theory. They pop up in Postnikov towers and Brown representability (since $K(A,n) = \Omega K(A,n+1)$). One way to prove Bott periodicity is to study the homotopy equivalence of $H$-spaces $\beta: BU \times \mathbb{Z} \to \Omega^2(BU\times \mathbb{Z})$. So, there's plenty to motivate the study of loop spaces. Let's say more about "interesting aspects."
The introduction to Adams' book Infinite Loop Spaces mentions work of Morse and Serre computing the number of geodesics on a Riemannian manifold using loop spaces, work of Serre on $H$-spaces and the Pontryagin product on $H_*(X)$, and the development of the Leray-Serre spectral sequence and its resulting homology calculations. Loop spaces give us more to compute with, and double loop spaces, $n$-fold loop spaces, and infinite loop spaces, give us even more.
As has been pointed out in the comments, the recognition principle says that $n$-fold loop spaces $\Omega^n Y$ are (up to homotopy) the same thing as $E_n$-algebras. For $n=1$, these are the same as $A_\infty$-spaces, as discussed in the preface to GILS. The Moore path space trick Naruki mentioned (parametrizing loops by $[0,t)$) gives a model for a strictly associative and unital topological monoid of loops, and the usual loop space is a deformation-retract, which is one way to understand the $A_\infty$-space structure. It's easy to show that $\Omega Y$ is a grouplike homotopy associative $H$-space, but the $A_\infty$-structure is better.
As pointed out in the link Najib provided, the $i^{th}$ stable homotopy group of $X$ is equal to $\pi_{i+k} \Sigma^{k} X = \pi_i \Omega^k \Sigma^k X$ for sufficiently large $k$, so spaces of the form $\Omega^k \Sigma^k X$ for $1\leq k \leq \infty$ contain a tremendous amount of information about $X$. As pointed out in the preface to GILS, this leads you naturally to the James construction and to Dyer-Lashof operations, which are essential for understanding the algebraic structure of the (co)homology of $X$, for an understanding of power operations, and for computations in the Adams spectral sequence. In GILS, May finds geometric approximations to these spaces, and descriptions of $H_*(\Omega^n \Sigma^n X)$ as functors of $H_*(X)$. The resulting understanding of Dyer-Lashof operations is the foundation upon which much computational work has been done, as wonderfully summarized in an article of Tyler Lawson.
Thank you a lot for your great answer. One aspect I would like to understand deeper: You wrote It's easy to show that $\Omega Y$ is a grouplike homotopy associative $H$-space, but the $A_{\infty}$-structure is better. Could you explain a bit more precisely what you mean by "better"?
Sure. To be "homotopy associative" means associative in the homotopy category. That is, the associativity diagram (the two ways to get from $X\wedge X\wedge X$ to $X$) commutes up to homotopy. To be "associative" means it commutes on the nose (that the maps are literally equal). To be $A_\infty$ means it commutes up to infinitely coherent homotopy, e.g., a homotopy for the associativity diagram, plus homotopies between such homotopies, plus homotopies between those, etc. Way more structure than just the lowest level homotopy. More data/structure = better calculations.
|
2025-03-21T14:48:31.476464
| 2020-07-12T05:14:57 |
365448
|
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|
Stack Exchange
|
Are cyclic orbitopes of permutahedra necessarily simplicies?
Suppose that $v=(v_1,\ldots, v_d)\in \mathbb{R}^d$ lies in the linear subspace $v_1+\cdots +v_d=0$, and moreover that the coordinates are pairwise distinct. The permutahedron \begin{equation} P(\mathcal{S}_d;v)=Conv(\mathcal{S}_d\cdot v) \end{equation} is the convex hull of $v$ under the symmetric group action on coordinates. It is a $(d-1)$-dimensional polytope.
Now consider the cyclic subgroup $C_d$ of $\mathcal{S}_d$ generated by the permutation $(123\cdots d)$ and consider the corresponding orbitope $P(C_d; v)=Conv(C_d\cdot v)$.
Question: Is $P(C_d;v)$ necessarily a $(d-1)$-dimensional simplex?
Let $M$ be the circulant matrix whose rows are given by cyclic shifts of $(v_1,\dots v_d)$ and let $P(x)=v_1+v_2x+\cdots+v_dx^{d-1}$ be the associated polynomial. Moreover, let $s$ be the degree of $\gcd(P(x),x^{d}-1)$.Then the rank of $M$ is equal to $d-s$, so it is possible to come up with examples of vectors $v$ such that the $C_d$ orbit is not $(d-1)$-dimensional by making $s$ large.
For example the matrix
$$\begin{bmatrix}
2 & 1 & -2 & -1 \\
1 & -2 & -1 & 2 \\
-2 & -1 & 2 & 1 \\
-1 & 2 & 1 & -2
\end{bmatrix}$$
has rank 2, so the associated orbit polytope has dimension $2$ rather than $3$.
Thank you! Given the gcd interpretation (thank you for altering me to that), It doesn't seem possible to find linear conditions so that the circulant matrix has rank $d-1$ provided $(v_1,\ldots, v_d)$ lies outside the resulting linear subspaces (that's really what I needed for my purposes)
|
2025-03-21T14:48:31.476602
| 2020-07-12T05:28:43 |
365449
|
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|
Stack Exchange
|
Almost sure stochastic equicontinuity
Suppose $\mathcal{G}$ is a normed closed class of functions with finite entropy and envelope with a finite second moment (details below), and $g_0$ is a function in the interior of that class. Let $Z_i$, $i=1,\dots,n$ be $i.i.d$ random variables sampled from the distribution $P$ on $\mathbb{R}^d$ and suppose that $E[g(Z_i)]=0$ for all $g\in \mathcal{G}$. Finally, let $(\delta_n)$ be a sequence of positive numbers such that $\delta_n\rightarrow 0$ as $n\rightarrow\infty$. Then, does
\begin{equation*}
\sup_{||g-g_0||\leq \delta_n}\left|\frac{1}{\sqrt{n}}\sum_{i=1}^n g(Z_i)-g_0(Z_i) \right|\rightarrow_{a.s.} 0?
\end{equation*}
Some details:
By finite entropy, I mean that $\int_0^\infty \sqrt{\log N(\varepsilon,\mathcal{G},||\cdot||)}d\varepsilon<\infty$, where $N$ is the covering number for balls of size $\varepsilon$.
Let the envelope function be $M$, the the second moment is $\int M(z)^2P(dz)<\infty$.
The norm $||\cdot||$ may be either $\sup_z|g(z)|$ or $\left(\int |g(z)|^pP(dz)\right)^{1/p}$ for any $p\in \mathbb{N}$.
If convenient, you may assume $\delta_n$ to be a decreasing sequence, and/or $\delta_n=o(n^{1/2-\alpha})$ for some $\alpha>0$.
Note that the result above holds for convergence in probability. (Lemma 2.17 in Pakes and Pollard (1989) http://www.stat.yale.edu/~pollard/Papers/PakesPollard89Econometrica.pdf or Theorem 1 in Andrews (1993) https://cowles.yale.edu/sites/default/files/files/pub/d10/d1059.pdf ).
Can anybody help? I have attempted a lot of things, but nothing is working. I have also tried to ask a similar but apparently confusing question here and in Math Stack Exchange. Even if you know some result which may be related, it will help. Even if the result has stronger conditions than the ones stated here, it will help. Thanks.
I think the answer depends on how fast $\delta_n$ goes to $0$. The answer should be yes if $\delta_n$ goes to $0$ fast enough, and no otherwise. To determine an exact threshold, I think one needs a full-blown paper. However, I think one would have problems publishing such a paper, as the a.s. convergence does not seem natural here.
@losif Pinelis. Indeed, it may come to a speed of convergence issue. For convergence in probability the speed is irrelevant, as the result exists for any series $\delta_n=o(1)$. It could also come down to restrictions in the function space. You are right that the almost sure convergence does not seem natural and primitives for almost sure stochastic equicontinuity are not showing up anywhere in my lit searches. Unfortunately this condition is sometimes required in papers about the convergence of bootstrap procedures, which is what I need this for. I'm losing hope.
@losif Pinelis: I should mention that establishing the result for a rate any faster than the one in my point 4 would not help me though.
|
2025-03-21T14:48:31.476839
| 2020-07-12T05:54:16 |
365451
|
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|
Stack Exchange
|
Tiling with similar tiles
Question 1: Is there a polygon $P$ that
cannot tile the plane
and
tiles the plane when copies of $P$ and some other polygon(s) all similar in shape to $P$ but of different size(s) can be used?
Basically, with copies of $P$ alone we should be able to form a layout with gaps which are all similar in shape to P and of different size(s).
Motivation: In basic tiling, we are constrained to use congruent copies of a candidate polygonal region (congruent up to some isometry) to fill the plane without gaps. Here, we consider relaxing the constraint to allow scaled copies of the candidate region and try to see whether this relaxation can non-trivially improve the chance a candidate region has of being a tile.
Note: Two cases to this question – $P$ is convex and not necessarily so.
Question 2: Is there a $P$ such that it is not a rep-tile but a large copy of $P$ can be tiled by several units all similar to $P$?
Note: By rep-tile, we mean a polygon that can be cut into some finite number of equally scaled down copies of itself. So, the $P$ one is looking for cannot be tiled by any finite number of equally scaled down copies of itself but can be tiled with copies of itself which have been scaled down by factors that are different among themselves.
If you say "it tiles the plane" do you also allow mirrored copies? Or would you consider a polygon, that does not tile the plane, but does so if we allow its mirror image, an example for Question 1?
Mirrored copies are of same area - so one would want P and scaled copies of P to be needed in tiling the plane. Considering a mirrored copy of P as different from P could lead to another variant of the question though.
So the answer to @MWinter is yes?
No. As far as the scope of question 1 as stated goes, we need a P which needs copies of itself and similar but scaled down copies of itself to tile the plane.
Non-convex solutions to Question 1
Consider the following polygon (the outward angle on the right is the same as the inward angle at the top)
Since I didn't know any better way to show it does not tile the plane, I brute-forced my way through some case distinctions.
The only non-convex corner of any tile must meet a corner of another tile. It can't be either of the two bottom corners or the top right corner (the "unoccupied angle" would be too small to fit another corner in). If it is the top left corner, then we end up in the situation sketched below in the left picture. If it is the rightmost corner, then the result is the right picture below.
In both cases we clearly cannot complete the partial tiling to a tiling of the whole plane.
On the other hand, we can tile a strip in $\mathbb R^2$ with scaled copies of our polygon as follows.
Edit: Here's another shape (essentially based on the same principle).
The proof that it doesn't tile the plane is similar to above, but we can get rid of most of the casework due to symmetry. Tiling with two different sizes is again possible.
Convex solution
As noted in the comments, cutting the "bowtie" tile along the central symmetry axis solves the convex version of Question 1. Also note that Rao's preprint shows that only pentagons from belonging to one of 15 families tile the plane, and we can choose the bowtie such that the resulting pentagon belongs to none of them.
Edit 2: I just found out that this convex solution is also presented in Figure 3 in this paper from 1982.
Thanks for this very intuitive answer! And I would think the convex case answer you give works fine. The nice thing is you seem to have arrived at a convex case answer via the nonconvex answer.
Your convex pentagon is indeed a solution, i.e., it does no tile space. At least we can choose an instance of that shape that does not. Your pentagon has a degree of freedom: its interior angles are $\pi/2$, $\alpha$ and $\beta:=\pi-\alpha/2$. We can choose $\alpha$ as an irrational multiple of $\pi$, and then the only combinations of these angles that give $2\pi$ are either $4\cdot \pi/2$, $2\cdot \pi/2 + \pi$ (if a vertex meets an edge) or $\alpha+2\beta$. It remains some straight-forward case analysis to show that these cannot be the only types of vertices of a tiling.
@M.Winter Thanks for this argument. I actually found another proof based on a different degree of freedom: if we make the "long sides" long enough, then the side where we cut the bowtie becomes the shortest side, and from there it's not hard to show that this side must always map to another shortest side (hint: all angles are at least $\pi/2$).
Here is an answer to Question 2. The following shape (attributed to Karl Scherer on this website) tiles into similar shapes of different sizes.
Convincing myself that it is not a rep-tile took me several case distinctions - I found it easiest to start with one of the right angles and construct the tiling from there until deriving a contradiction (angles of $\pi/3$ can only be "filled" in one way, right angles and angles of $2\pi/3$ can be "filled" in two different ways).
That is a nice and simple example which has mostly settled the question except for the non-convex non- polyomino non-fractal case.. With reference to both questions, one could go farther and ask if there are other types of quadrilaterals (quads with different angle sets) which have this 'similar tiling' property, whether they have any shared property (properties) like say, being cyclic and so forth... And pentagons with the property ….
The website behind the link in the answer has many more examples of polygons that can be decomposed into similar tiles of different sizes (including several non-convex ones); I wouldn't be surprised if some of them were non rep-tiles as well.
This tile $P$ tiles an open half-plane in a hyperbolic fashion. It's "tiled" in the sense that it's a union of copies of tiles similar to $P$ with disjoint interiors. So you can sort of tile the plane with tiles similar to it, you miss just one line. More interesting (and probably what you meant) is what happens if you restrict to finitely many similar copies of $P$, since I guess any polygon's similar copies tile a full measure subset of the plane.
I should probably also add Paint as an answer to
Time-saving (technology) tricks for writing papers .
In case of tiling problems it is implicitly assumed because of such examples that you cannot have infinitely many copies in a bounded region. But I agree that it would be useful to have more definitions in the question.
In the hyperbolic metric you do have finitely many tiles in a bounded region. Also, the open half plane is not tiled in the sense I define, by for instance a typical pentagon. (I think.)
Actually, I'm not sure about that comment or my answer. Does this tile a half plane in a more interesting sense than any other polygon?
Yes, the cover is locally finite.
@VilleSalo in the hyperbolic metric all tiles are congruent
My hyperbolic point was a bit tongue in cheek. But I agree
It seems you did not restrict to finitely many scaled down copies of the tile. So here is one interesting tile for which you need infinitely many scaled down copies in order to tile the plane. On the left is the tile itself, and on the right is how to use scaled copies to tile an L tromino, after which it is trivial to tile the plane.
I had not thought of infinitesimal tiles. But what you point out seems quite interesting!
Are you willing to allow tiles with fractal boundary? (I can see that you write "polygon" throughout, so maybe not?).
If so, then another example is the so-called "Koch snowflake". See https://en.wikipedia.org/wiki/Koch_snowflake#Tessellation_of_the_plane . If you allow such tiles, then this is also a positive answer to your Question 2.
I hadn't thought of fractals. But what you point out is certainly of interest.
Another family of fractal examples is provided by Thurston's famous unpublished notes [Groups, tilings, and finite state automata: Summer 1989 AMS Colloquium lectures]:
http://timo.jolivet.free.fr/docs/ThurstonLectNotes.pdf .
Look at Figure 9.6.
|
2025-03-21T14:48:31.477418
| 2020-07-12T06:51:25 |
365453
|
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|
Stack Exchange
|
Categorified probability and statistics?
To put it simply, what if the sample space underlying our probability space is a category instead of a mere set. Has a theory or probability and statistics been developed for such a situations in which the samples can have relations between each other?
This questions is somewhat similar to this one however I do not find the answer to my question in there.
Would you be satisfied with a groupoid instead of a category? I don't know of any such development offhand, but this case seems more likely if it has been done.
There is a highly upvoted earlier question that asks for an introduction to probability theory from a categorical perspective: https://mathoverflow.net/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical
Posets are the special case of categories where there is most one morphism between any two objects. See e.g. https://www.sciencedirect.com/science/article/pii/S0020025519311338 for one recent paper on random posets.
REVISED ANSWER
First Approach, from Random Graphs
Never seen it anywhere. Anyway, here is my two-cents:
First of all, there is a well established theory of RANDOM GRAPHS (see here for a ref. Of course, there are many choices available, you may make nodes, edges or both random, and you may confine yourself to certain classes of probability distributions).
Now, a a category is a (directed) graph, in the precise sense that there is an obvious forgetful functor $ F: Cat \rightarrow Graph$.
The first step toward defining a random category would be to say that it "forgets" to a random graph. In other words, just like the category of (deterministic) Graphs embeds as a subcategory of Random Graphs, the category of standard deterministic categories CAT would embed in the cat of Random Categories, in such as way as to preserve the respective forgetful functors.
A category has more structure than a directed graph, it has composition, identity, and also commuting diagrams, so you need to add those to the picture:
composition is easily taken care of, for instance by stipulating that the probability of a composite arrow is the product of the probability of its components. Similarly, identity morphisms could be assigned the same probability of their corresponding nodes.
Now you are left with the diagrams (I am talking now of only those diagrams which are not forced by the laws of composition). There you have a certain degree of freedom. My intuition would be: first play with random free categories constructed from random graphs by free completion, then stipulate an assignment of probabilities to the space of all generating diagrams in the category.
Second Approach, from Random Simplicial Sets.
Another (broader) take on the above: categories are, in a sense, particularly simple types of SIMPLICIAL SETS, namely the ones which satisfy the horn condition.
So, rather than starting from random graphs and building your way up to categories, start with a notion of random simplicial set and by restriction you will have your random categories.
How? Don't know if the notion of random simplicial sets has been explored, but certainly there is something on random simplicial complexes, see for instance here). This could be used as motivating example.
Hope it helps
PS Final comment: in fact approaches 1 and 2 are in the end the same, 1 is part of 2, because graphs are simplicial sets of dimension 1.
Thank you. Actually random graphs would wotk for me too, I just need a model with variable numbers of nodes and edges at the same time.
|
2025-03-21T14:48:31.477700
| 2020-07-12T07:21:05 |
365454
|
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"darij grinberg",
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|
Stack Exchange
|
Is the group Hopf algebra left and right adjoint?
Suppose that $G$ is a group and $k$ is a field. Then it is well known that the group ring (group algebra) functor $k[\bullet]$ is left adjoint to the group of units functor, the latter of which associates the group of units to each ring. This implies that every group morphism from $G$ into the group of units of an algebra $A$ can be uniquely extended into an algebra morphism from $k[G]$ to $A$.
Now, since $k$ is a field, $k[\bullet]$ can also be seen as the Hopf group algebra. My question is now twofold:
1.) Is the Hopf algebra functor still left adjoint to something? Hence inherits similar extension properties?
2.) Is the Hopf algebra functor (also) right adjoint to something?
1.) Yes. The functor from {groups} to {Hopf algebras} that sends $G$ to $k\left[G\right]$ is left adjoint to the functor from {Hopf algebras} to {groups} that sends $H$ to the group of all grouplike elements of $H$.
|
2025-03-21T14:48:31.477793
| 2020-07-12T07:29:42 |
365455
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365455"
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|
Stack Exchange
|
Bounding the product of lipschitz function
Assume we know that $f(x,y): \mathbb{R}^{a+b} \to \mathbb{R}^d$ is lispchitz w.r.t. $y$, i.e.
$$\|f(x,y) - f(x,y')\| \leq L \|y-y'\|.$$
Is there a way to bound the product $f(x,y)\cdot f(x,y')^T \in\mathbb{R}^{d\times d}$?
Say $f(x,y)=y$. Then $f(x,y)^2=y^2$, which is not Lipschitz in $y$. So the answer is negative.
|
2025-03-21T14:48:31.477855
| 2020-07-12T07:54:11 |
365457
|
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"authors": [
"Igor Belegradek",
"Peter Michor",
"Sak",
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|
Stack Exchange
|
Geodesic line with endpoints in interior of Riemannian manifold or Alexandrov space
Let $X$ be a finite dimensional Alexandrov space with curvature bounded below and non-empty boundary. Let $\gamma$ be a shortest geodesic path in $X$ whose endpoints belong to the interior of $X$.
Is it true that $\gamma$ is contained in the interior of $X$? Is that true at least under the assumption that $X$ is smooth Riemannian manifold with smooth boundary (thus $X$ must be locally geodesically convex)?
The case when $X$ is a smooth compact domain in a Riemannian manifold follows from elementary arguments once you know that $X$ is geodesically convex (which follows from the assumption that $X$ is an Alexandrov space as I mention in the comments to Michor's answer). Namely, if $[x,y]$ is a segment in $X$ with $x,y$ in the interior, then the segments from $x$ to points of a small neighborhood of $y$ fill a neighborhood of $(x,y)$, and since by convexity of $X$ all such segments must lie in $X$, no point of $(x,y)$ lies in $\partial X$.
No: Take a banana shaped domain with smooth boundary in the plane.
Edit: This answer is wrong. I leave it here for the very instructive comments by Igor Belegradek.
I think I see geodesic triangles near the concave boundary in the banana where Gauss-Bonet implies large negative curvature.
The banana is not an Alexandrov space.
@IgorBelegradek Why is the banana not an Alexandrov space? It seems to me that by compactness, Toponogov's theorem should hold possibly with a very very negative lower curvature bound.
@Sak: see https://mathoverflow.net/questions/223934/when-a-riemannian-manifold-with-boundary-is-an-alexandrov-space.
@IgorBelegradek wow, the more you know! Thanks!
Thanks. I did not know this. I will delete the answer soon.
@PeterMichor: please keep it. It is instructive. I even upvoted it. :)
I wish to add that a smooth compact domain in a Riemannian manifold must be convex if the intrinsic metric on the domain is $CD(K,\infty)$, see Theorem 1.5 in https://arxiv.org/abs/1902.00942. This applies to every finite-dimensional Alexandrov space by https://arxiv.org/abs/1003.5948.
|
2025-03-21T14:48:31.478016
| 2020-07-12T10:04:29 |
365462
|
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|
Stack Exchange
|
How many squares can be determined using $n$ points?: revisited
The question was recently asked How many squares can be determined by $n$ points in $\mathbb{R}^3$?
The main observations were:
In $\mathbb{R}^2$ at most $O(n^2)$ as every pair of points determines at most 3 squares.
In $\mathbb{R^3}$ at most $O(n^3)$ as every triple of points determines at most one square
This can be improved to $O(n^{7/3}$) as it is known that this is sharp for the number of right triangles determined by $n$ points in $\mathbb{R}^3.$
However it would be more relevant to have a bound on the number of isosceles right triangles.
I would guess that in the long run the optimum arrangement in $\mathbb{R}^3$ is a square grid of side $s=\sqrt{n}.$
If I calculate correctly, a square grid of $n=s^2$ points determines $\frac{s^4-s^2}{12}$ squares, so indeed $O(n^2)$.
I would also guess that this might be best for $\mathbb{R}^3$ and, if not , then a cubic grid of side $s=\sqrt[3]{n}$ is best.
These aren't very educated guesses. So my main question (in two parts) is
Suppose $n=t^6$,
Does any arrangement of $n$ points in $\mathbb{R}^2$ determine more squares that a square grid of side $t^3?$
Does any arrangement of $n$ points in $\mathbb{R}^3$ determine more squares that a square grid in a plane of side $t^3?$
I picked $n=t^6$ to allow for a cube of side $t^2$ as well.
I think it best to ask one main question, but those seem related enough. Secondary questions are:
Is anything known about the arrangements in $\mathbb{R}^3$ which achieve $O(n^{7/3})$ right triangles?
Is anything known about bounds and arrangements for the number of isosceles right triangles?
Are formulas or bounds known for the number of squares determined by the points of a cubic grid of $s^3$ points?
An answer to that last one might show that the answer to the second part of the main question is NO.
In a cubic grid of $s^3$ points one has $3s$ squares of side $s$ parallel to coordinate planes. So there are $3s\frac{s^4-s^2}{12}$ squares in those planes. That is only $O(n^{5/3})$. However any two orthogonal vectors of equal length determine a family of planar lattices with a certain number of squares. Determining the number of such squares within a cube may be known but is certainly more delicate that the planar case.
A particularly productive length is $15:$
The integer vectors of length $15$ are $(15,0,0),(14,5,2),(12,9,0),(11,10,2),(10,10,5)$ along with their permutations and changing the sign on some coordinates.
A number of orthogonal pairs are possible such as
$(15,0,0),(0,15,0)$
$(10,10,5),(10,-5,-10)$
$(12,9,0),(-9,12,0)$
$(15,0,0),(0,9,12)$
$(14,5,2),(2,-10,11)$
$(14,5,2),(-5,10,10)$
$(11,10,2),(-10,10,5)$
And again the coordinates can be permuted and signs flipped.
|
2025-03-21T14:48:31.478202
| 2020-07-12T10:29:41 |
365463
|
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|
Stack Exchange
|
Surjective homomorphisms between braid groups
There cannot be a surjective homomorphism $B_2 \to B_n$ for any $n > 2$ because $B_2$ is commutative and $B_n$ is not. It seems plausible that if $m < n$, there cannot be a surjective homomorphism $B_m \to B_n$.
If $m>n$, there are surjective maps $B_m \to B_n$ obtained by forgetting a certain number of strands, but these are not homomorphisms: e.g., if $f: B_3 \to B_2$ is the map forgetting the third strand, then $f(\sigma_1 \sigma_2 \sigma_1) = \sigma_1$, but $f(\sigma_1) \cdot f(\sigma_2) \cdot f(\sigma_1) = \sigma_1^2$. Here, $\sigma_i$ is the braid swapping the $i$-th and the $(i+1)$-st strand using a single positive crossing.
How do these observations generalize? I.e., under what conditions on $m,n \geq 2$ does there exist a surjective homomorphism $B_m \to B_n$? My suspicion is that this requires $m = n$ or, similar to the case of symmetric groups, $(m,n) = (4,3)$, is this true?
I originally posted this on MathSE but since this seems to be sort of hard I thought it would be appropriate to post it on MO. I hope this sort of cross-post is acceptable.
Edit: As pointed out to me in the comments by @YCor, in the case $4 \neq m > n > 2$ there are indeed no surjective homomorphisms $B_m \to B_n$ by Theorem 3.1 in Lin (integral homomorphisms have cyclic image). The seemingly easier case $m < n$ is what remains unanswered.
Isn't the rank of the braid group $B_m$ given by $\dim H_1(\mathrm{Conf}_m(\mathbb{R}^2); \mathbb{Q}) = \binom{m}{2}$? That settles the question for the existence of surjective morphisms $B_m \to B_n$ for $m < n$ (not the other way around obviously). Edit: well, no, of course, that's the pure braid group. (For which the "forgetting strands" maps are surjective homomorphisms, BTW.) I just leave this here in case someone thinks the same mistake.
You probably get information and restrictions on homomorphisms $B_m\to B_n$, $m>n$ from https://arxiv.org/abs/math/0404528 (because for $n\ge 3$ and large $m$ $B_m$ should not surject onto $S_n$). See also https://arxiv.org/abs/1910.07177
@YCor Thanks for pointing out these sources. Lin's Theorem 3.1 b) states that for $4 \neq m > n$, homomorphisms $B_m \rightarrow B_n$ are integral (i.e., their image is torsion-free cyclic), so this actually settles my question for $4 \neq m > n > 2$, without any discussion of the cardinalities of finite quotients as in Chudnovsky et al. Do you agree?
Probably, I hadn't looked in detail, hoping you would do so, and you did!
If there is a surjection $B_m \rightarrow B_n$ then there is a surjection $B_m\rightarrow S_n$. When $m < n$ this already seems unlikely to me.
You might also want to look at this paper of Chen, Kordek and Margalit. I think it’s the state of the art: https://arxiv.org/abs/1910.00712v1 .
|
2025-03-21T14:48:31.478536
| 2020-07-12T11:02:50 |
365464
|
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|
Stack Exchange
|
Graph chromatic numbers defined by interactive proof
Edit (2020-07-15): Since the discussion below is perhaps a bit long, let me condense my question to the following
Short form of the question: Let $G$ be a finite graph (undirected and without self-loops), and $0\leq p\leq 1$ be a real number. Is there a standard name for the smallest $n$ for which there exists a probability distribution on the set of all maps $c\colon V(G) \to \{1,\ldots,n\}$ (where $V(G)$ is the vertex set of $G$) such that for every edge $(x,y)$ of $G$ the probability of $c(x)\neq c(y)$ is $\geq p$ (so when $p=1$ this is just the chromatic number of $G$)? Is it discussed somewhere in the literature?
Long version:
Let $G$ be a graph (undirected and without self-loops; and I'll mostly be thinking finite even though the definition doesn't require it), $n$ a natural number (number of colors) and $0\leq p\leq 1$ real. Let us say that $G$ is interactively $n$-colorable with threshold $p$ (for lack of a better term) when Alice and Bob have a strategy in the following game of interactive proof:
The game: Alice and Bob (also collectively known as “the provers”) agree on a strategy in advance, but thenceforth cannot communicate. Vera (also known as “the verifier”) presents each of the provers with a vertex in the graph $G$ and expects an element of $\{1,\ldots,n\}$ (a “color”) in response. Alice and Bob win when they have a strategy which ensures that:
when presented with the same vertex they will always answer the same color,
when presented with vertices connected by an edge in $G$ they will give answers that are distinct with probability $\geq p$ (against any choices made by Vera).
Under the assumptions of local realism (which I make by default), condition (1) requires that Alice and Bob have, in fact, agreed on a choice of color for each vertex, so the game can be simplified to have just one prover who should be able to pick such a choice at random in such a way that each given edge has probability $\geq p$ of having distinct colors. If furthermore $p=1$, this is merely asking that $G$ is, indeed, colorable with $n$ colors (while if $p=0$, the prover(s) win trivially for any graph with $1$ color).
For example, the cycle graph with an odd number $k$ of vertices is interactively $2$-colorable with threshold $1 - \frac{1}{k}$ by choosing a random edge at which to break the coloring.
Let's say that the interactive chromatic number with threshold $p$ of $G$ is the smallest $n$ such that $G$ is interactively $n$-colorable with threshold $p$.
Question: Does this chromatic number have a standard name? (Or equivalently, the sup of the $p$ for which $G$ is colorable for a given $n$.) If so, where can I learn more about it?
Remarks / variants: One reason why I presented it in the form of the interactive game described above is that it admits a quantum variant in which, instead of demanding local realism, we allow Alice and Bob to prepare and share an entangled quantum state (there are possibly nonequivalent ways to define this, though). For $p=1$ this is known as the “quantum chromatic number”, but it makes sense with a more general $p$ as in the case described above. E.g., the triangle graph is quantum interactively colorable with $2$ colors and threshold $\frac{3}{4}$, whereas it is interactively colorable with $2$ colors only up to threshold $\frac{1}{2}$. So I am also interested in knowing of the quantum case has a standard name or has been studied in the literature. (The paper “Non-closure of the set of quantum correlations via graphs” by Dykema, Paulsel and Prakash seems related but doesn't appear to define exactly the concept I just mentioned.) (Also, another point worth mentioning is that if Alice and Bob, instead of a quantum system, have access to an unlimited supply of Popescu-Rohrlich boxes, they can $2$-color any graph with threshold $1$.)
Finally, we could also, instead of fixing $p$ for all vertices, consider not a graph but an edge-weighted graph where each edge is labeled by the threshold for that particular graph. I am also interested in knowing whether this generalization has a name.
This may be (vaguely) related to fractional coloring (1,2)?
@PaucalNumber I have fixed your comment. May I respectfully suggest a meta question next time pointing out the bug? Then any mod can help.
I do not think there is a standard name for this, but I may prefer to call it $p$-random chromatic number....
Have you used this name before this question (or know of anyone who has)? Or are you suggesting a name now?
Nobody used this. I know some similar names such as $p$-flexible chromatic number ($p$ lies between 0 and 1) used for other coloring problems, so I am just suggesting the name of $p$-random chromatic number for your problem. By the way, your game looks somehow like a random version of the graph coloring game https://en.wikipedia.org/wiki/Graph_coloring_game
|
2025-03-21T14:48:31.478895
| 2020-07-12T13:14:28 |
365467
|
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|
Stack Exchange
|
Best known primality test for the whole intervals of integers up to $10^{20}$ — like the sieve of Eratosthenes
What are the best known primality test(s) for the whole intervals of integers up to $N=10^{20}$ ? "Best" means "have minimal amortized time per tested integer".
That is, the algorithm(s) should be fast only when testing a whole interval of integers. Just like sieve of Eratosthenes works well only if we test all integers in the whole interval $[N, N+N^{1/2+\varepsilon})$ and degrades otherwise (say, sieve of Eratosthenes for interval $[N, N+1)$ is the same as method of trial division).
UPDATE:
The question probably should be asked as "What is the best prime generation algorithm for modern computer hardware for integers up to $10^{20}$
Then for the modern CPU the answer probably is "cache-friendly prime generation algorithms" — this polite name is for algorithms that workaround severe non-randomness access time to modern RAM. There should be answer for modern GPGPU too.
I still expect that somebody will answer, not me.
Just a comment since this is a point test rather than an amortized test, but in case it’s useful for others: Michal Foriˇsek and Jakub Janˇcina, Fast Primality Testing for Integers That Fit into a Machine Word.
@GeoffreyIrving Sometimes it may be possible to modify point test into amortized test, just like trial division test may be modified into the sieve of Eratosthenes and have some perfomance gain. But, of course, I am more interested in the "ready to use" tests.
Wheel sieving is ready to use. You might find that in combination with some other tests useful. Gerhard "Sometimes Settles For Pretty Good" Paseman, 2020.07.12.
Since you have roughly 10^18 primes with 20 decimal digits, what are you going to do with them? Gerhard "Please Don't Print Them Out" Paseman, 2020.07.13.
I'm thinking of a GPU version of an algorithm based on https://mathoverflow.net/q/243490 . You can add a comment there if you are interested and would like to contribute. Gerhard "Is Going For Gentle Solicitation" Paseman, 2020.07.13.
Primesieve is good in practice: https://github.com/kimwalisch/primesieve
@GerhardPaseman 1. Wheel sieving is useful, but not a simple thing: it is cache unfriendly, at least by naive implementation. 2. It may be possible to use 10^18 primes to estimate, say, Brun's constant more precisely. And 10^20 is upper limit. 3. GPU version is interesting, but I still do not understand GPU programming (but plan to learn it sometimes). BTW, the sieve of Eratosthenes seems to be GPU unfriendly due to demand for random writes and big cache size. I shall carefully read your question mathoverflow.net/q/243490
@MaxAlekseyev For now this primesieve looks like the best know practical sieve. But I suspect that good algorithm will be 10 times faster for primes starting from $10^{15}$, at least on processors like x5-Z8350.
@MaxAlekseyev I still have not tested it on processors with big cache size.
|
2025-03-21T14:48:31.479143
| 2020-07-12T13:35:58 |
365468
|
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|
Stack Exchange
|
non-associative but commutative algebra
Is it possible(or may be easier) to give an example of non associative algebra but commutative?
At first sight, it seems possible to prove associativity from commutativity but later realised it may no be the case.
The only example of non associative algebra which I know is Octonion but which is non-commutative. Then I started to create an artificial example. Here is my approach(same as to create an commutative):
Let's $\textbf{C}$ denote the category of associative algebra. Suppose free object exists in this category say $F$ be a free associative algebra, then consider the quotient of $F$ by the ideal generated by its commutator elements.
I don't know how much this work but certainly this way we will have lot of such examples.
One could take the semigroup algebra (over a field, say) of the Wedderburn-Etherington monoid. See H. W. Becker, Solution to Advanced Problem 4277, Amer. Math. Monthly 56 (1949), 697-699.
Isn't abelian lie algebra is associative?
In your question, it should be quotient by the ideal generated by the commutators. The result would just be an polynomial algebra, so it does not work.
But how do you know that it's polynomial algebra.
The polynomial algebra is the free object in the category of commutative associative algebras. If you begin with the free associative algebra and mod out the commutators, it is a matter of checking that the resulting algebra will also be free in the category of commutative associative algebras. Since such categories are, in fact, what is known as a variety of algebras (universal algebra parlance), the result follows.
@RichardStanley I can't find any occurrence of this monoid, but a monoid is associative and so is its semigroup algebra. You probably have some non-associative magma in mind rather than a monoid?
@YCor I shouldn't have called it a monoid. It is the free commutative magma with one generator. There is some information at https://en.wikipedia.org/wiki/Wedderburn-Etherington_number.
Answers are useful, I eventually voted to reopen.
The class of Jordan algebras are the most important class of algebras in this direction.
They are defined by the two identities,
(commutativity): $xy=yx$,
(Jordan identity): $(xy)(xx)=x(y(xx))$.
They are an extremely important class of non-associative algebras (check Jacobson's Strucuture and Representation of Jordan algebras or McCrimmon's A taste of Jordan algebras)
The most easy way to produce such an algebra is by considering an associative (alternative suffices) algebra $A$ and changing the product from $.$ to $*$ as follows: $a*b = \frac{ab+ba}{2}$. Some care of course is necessary in case of characteristic two.
Oh nice. Thanks.
The "midpoint" operation $(x,y)\mapsto\frac12(x+y)$ on the reals is commutative and not associative.
The same operation works in any vector space over any field of characteristic $\neq2$.
simplex sigillum veri. Very nice Andreas !
How it's ring operations? What is the unit element?
Nice example. @Sunny Note this isn't bilinear so to make it into an algebra you need to linearize it, or modify it in some other way.
Similarly, you can linearize the NAND operation on ${T,F}$ to get a two-dimensional (non-unital) example. Or adjoin a unit to get a three-dimensional unital example.
It's not a ring. I understood "algebra" in the sense of universal algebra. If you want it to be understood in the sense of rings and bilinearity, then you should linearize it: Take the vector space with my example as a basis, and extend the midpoint operation from basis vectors to arbitrary vectors by bilinearity.
To see what's going on, let's describe the free commutative, non-associative, algebra on $n$ elements $x_1, \dots, x_n$, which we call $A_n$ .
Let $X_n$ be the set of isomorphism classes of binary rooted trees $T$ with a function $f: {\rm Leaves}(T) \to \{x_1, \dots, x_n\}$, which labels the leaves. Then $A_n = k X_n$ is the free vector space with basis $X_n$. The multiplication map $A_n \otimes A_n \to A_n$ is induced by a multiplication $X_n \times X_n \to X_n$, which joins pairs of trees together at a new root.
That is, given binary labelled trees $(T_1,f_1)$ and $(T_2,f_2)$, we may form a new tree $T_1\vee T_2$, with a new root and $T_1$ above one branch of the root and $T_2$ above the other branch. (Sorry if this is unclear-- this is most easily conveyed by a picture). The new labelling is just $f_1 \sqcup f_2$.
Its not hard to check that this algebra is commutative but not associative when $n \geq 2$. E.g. $(x_1 \cdot x_2) \cdot x_2 \neq x_1 \cdot (x_2 \cdot x_2)$ since these correspond to non-isomorphic labelled trees with $3$ leaves.
Alternatively, to check $(x_1x_2)x_2\neq x_1(x_2x_2)$ one can produce a small quotient where it's nonzero: the 3-dimensional (not-assumed-associative) commutative algebra with basis $(a,b,c)$ and product: $a^2=c$, $bc=cb=a$, and all other products zero. Then $(ba)a=0$ while $b(aa)=bc=d$, so $(ba)a\neq b(aa)$.
|
2025-03-21T14:48:31.479497
| 2020-07-12T14:09:41 |
365470
|
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|
Stack Exchange
|
Does the Galois groups $G_1$ and $G_2$ are isomorphic under `some suitable assumptions`?
Let $E_1,~E_2$ be two elliptic curve on the $p$-adic field $K \supseteq \mathbb{Q}_p$.
Consider the $p$-power torsion points and adjoin them with $K$.
Denote $E_i[p^n], ~i=1,2$ to be the set of $p^n$-torsion points of $E_i,~ i=1,2$, where $n$ some positive integers.
Denote the extensions $K_n=K(E_1[p^n])$ and $K_n'=K(E_2[p^n])$.
Case I: Consider the corresponding Galois extensions $G_1=\text{Gal}(K_n/K)$ and $G_2=\text{Gal}(K_n'/K)$.
Case II More generally, If further assume $\Lambda=\bigcup_{p,n} E_1[p^n]$ and $\Lambda'=\bigcup_{p,n} E_2[p^n]$, then consider $G_1=\text{Gal}(K(\Lambda)/K)$ and $G_2=\text{Gal}(K(\Lambda')/K)$.
My question-
For both above two cases,
If $n \to \infty$, does the Galois groups $G_1$ and $G_2$ are isomorphic under some suitable assumptions ?
Does $G_1$ and $G_2$ are map-able by a good map under some suitable assumptions when $n \to \infty$?
Any intuitive idea, discussions are appreciated. Thanks
|
2025-03-21T14:48:31.479582
| 2020-07-12T15:29:39 |
365475
|
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|
Stack Exchange
|
Applications of Tits' alternative in algebraic number theory
I have recently studying Tits' alternative. The theorem statement goes like the following:
Tits' alternative: Let $G$ be any finitely generated linear group over a field. Then one of the following is true,
$(1)$ $G$ contains a solvable normal subgroup of finite index,
$(2)$ $G$ contains a non-abelian free subgroup (of rank at least $2$).
I am in search of applications of this wonderful theorem in algebraic number theory. Any help, resources or reference will be appreciated. Thanks in advance.
The proof of the Tits alternative really consists, assuming that $G$ is not virtually solvable, in constructing a free subgroup (rather, say, that assuming that $G$ contains no free subgroup, that $G$ is solvable). Hence the question is maybe what such free subgroups are useful to?
I don't know how it is applicable in algebraic number theory but it is a very important result in Geometric group theory, a branch deals with the study of finitely generated infinite groups using geometry. Tit's alternative is one of the key step used in proving one of the famous theorem of Gromov's which says that a finitely generated group has polynomial growth iff it has nilpotent subgroup of finite index.
|
2025-03-21T14:48:31.479690
| 2020-07-12T16:11:08 |
365478
|
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|
Stack Exchange
|
Differential birational equivalence
Suppose the base field algebraically closed and of zero characteristic.
There are two fascinating questions in the intersection of ring theory and algebraic geometry (for which an excellent discussion is Y. Berest, G. Wilson, "Differential isomorphism and equivalence of algebraic varieties" MR2079372, whose terminology I use here).
(Differential isomorphism) Let $X$ and $Y$ be two irreducible affine varieties. If they have isomorphic rings of differential operators $\mathcal{D}(X) \simeq \mathcal{D}(Y)$, are they isomorphic? In general, what this says about the relation between $X$ and $Y$?
(Differential equivalence) Let $X$ and $Y$ be two irreducible affine varieties. If $\mathcal{D}(X)$ and $\mathcal{D}(Y)$ are Morita equivalent, what this says about the relation between $X$ and $Y$?
It can be shown without much difficulty that for an affine irreducible $X$, $\mathcal{D}(X)$ is an Ore domain, and hence admits a skew-field of fractions, $\mathbb{D}(X)$.
Question 1: If $X$ and $Y$ are two irreducible affine varieties and $\mathbb{D}(X) \simeq \mathbb{D}(Y)$, are they birationally equivalent? In general, what this implies about the relation between $X$ and $Y$?
There are only two facts I know regarding this question.
The first one is that if $\mathbb{D}(X) \simeq \mathbb{D}(Y)$, then $X$ and $Y$ must have the same dimension, as expected. For smooth varieties this is a well known result with different proofs in the literature, but if we do not assume the varieties smooth, the only proof I know uses some rather non-trivial facts about the lower transcendece degree by J. J. Zhang, which is a noncommutative analogue of the commutative transcendence degree.
The only other work I know of in this direction is J. P. Bell and A. Smoktunowicz, "Rings of differential operators on curves", MR3004084, for $X$ and $Y$ smooth projective curves. In the paper it is shown that if $\mathbb{D}(X) \simeq \mathbb{D}(Y)$, then $X$ and $Y$ should have the same genus. However, if $X$ and $Y$ are non-isomorphic curves of the same genus, it is unkown if $\mathbb{D}(X)$ should be different of $\mathbb{D}(Y)$.
Edit: adding more context The above question about curves is particularly relevant in view of Artin's conjecture about the birational classification of noncommutative projective surfaces. Namely, Artin has given an finite and small list of, conjecturally, all division algebras that can appear as "noncommutative funtion fields" of such surfaces. In the list we have the division algebras $\mathbb{D}(X)$, where $X$ is a smooth projective curve. Hence, the problem of determining when, give two curves $X$ and $Y$, $\mathbb{D}(X)$ and $\mathbb{D}(Y)$ are isomorphic is a very important one.
Question 2: How we can distinguish $\mathbb{D}(X)$ from $\mathbb{D}(\mathbb{A}^n)$; i.e., show that it is not a Weyl field (the skew-field of fractions of the Weyl algebra)?
I know of something about Question 2 related to the famous Gelfand-Kirillov Conjecture (cf. J. Alev, A. Ooms, M. Van den Bergh, "A class of counterexamples to the Gel'fand-Kirillov conjecture", MR1321564).
I assume that the usual function field $K(X)$ would be embedded into $\mathbb{D}(X)$. Would it be a maximal commutative subfield? If so, then you should be answer Q1.
@DonuArapura $\mathbb{D}(X)$ has many maximal comutative subfields, and given knowledge of $\mathbb{D}(X)$ only, there is no way to recover the particular subfield $k(X)$. For instance $\mathbb{D}(\mathbb{A}^1)$ have subfields of the the form $k(C)$, where $C$ is a curve of arbitrary genus.
OK, I guess I had the wrong picture.
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2025-03-21T14:48:31.479922
| 2020-07-12T16:46:14 |
365481
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"Manuel Norman",
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Stack Exchange
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Logical axioms used in the construction of counterexamples to ISP
In many cases, some problems are either solved in an affirmative way, or in a negative way. However, in some cases it turns out that some logical axioms lead to a proof of a certain statement, while other axioms can lead to the construction of counterexamples. A well known example is the Whitehead problem: using the axiom of constructibility $V=L$, every Whitehead group can be shown to be free; using Martin's axiom and the negation of CH, examples of non-free Whitehead groups can be constructed. Now suppose that we can prove (maybe under ZFC?) that the invariant subspace problem (ISP) holds for all infinite dimensional complex Banach spaces. Then, this would mean that the construction of the counterexamples by, for instance, Enflo and Read, involves different axioms from the ones in the supposed proof. So my question is: which logical axioms have been used in the constructions by Enflo and Read that show the existence of operators with no nontrivial invariant subspaces for some non reflexive Banach spaces?
I am almost certain that Read's construction on $\ell_1$ works in ZFC and I am reasonably confident that Enflo's does. Note that the ISP is only interesting for separable Banach spaces, and is in some sense really a statement about separable subalgebras of B(E), so I do not see any reason to expect independence from ZFC
Put it this way: if I want to "suppose that we can prove (maybe under ZFC?) that the invariant subspace problem (ISP) holds for all infinite dimensional complex Banach spaces" I think it would just be quicker for me to suppose 1=0 (ex falso quodlibet)
Thank you very much!
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2025-03-21T14:48:31.480055
| 2020-07-12T16:51:23 |
365482
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"Gustave",
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Stack Exchange
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Observability inequality for the 1D transport equation
Let $(a,b) \subset (0,1)$. Consider the following transport equation
$$z_t+z_x=0, \ (t,x)\in(0,T)\times(0,1), \\z(t,0)=0, \ z(0,x)=z_0(x).$$
It is clear that the solution to the above equation is given by $z(t,x)=z_0(x-t),\ \text{if} \ x-t\in (0,1)$ and $0$ otherwise.
I want to prove the following observability inequality: There exists a positive constant $C$ such that
$$\int_0^T \int_a^b z_0^2(x-t)dxdt\geq C\int_0^1z_0^2(x)dx.$$
I know that this inequality is satisfied if and only if $T \geq 1-b$ and $a=0$ but I don't see how to prove it. Any ideas or references?.
Thank you.
We have $0\le a\le b\le1$ and $T\in(0,\infty)$. We want to know when there is a positive constant $C$ such that
$$\int_0^T dt\, \int_a^b dx\, u^2(x-t)\geq C\int_0^1 dx\,u^2(x) \tag{1}$$
for all measurable functions $u\colon\mathbb R\to\mathbb R$ such that $u(x)=0$ for $x\notin(a,b)$.
The answer is: never. Indeed, without loss of generality $a<b$. The left-hand side of (1) is
$$\int_0^T dt\, \int_a^b dx\, u^2(x-t) \\
=\int_{\mathbb R} ds\, u^2(s)\int_{\mathbb R} dt\,1\{s\in(0,1),t+s\in(a,b),t\in(0,T)\} \\
=
\int_0^1 ds\,u^2(s)w(s),$$
where
$$w(s):=\max[0,\min(T,b-s)-\max(0,a-s)].$$
Clearly, the weight function $w$ is continuous, so that $w(b-)=w(b)=0$.
Letting now $u:=1_{(b-h,b)}$ with $h\downarrow0$, we see that the left-hand side of (1) is $\int_{b-h}^b ds\,w(s)=o(h)$, whereas the right-hand side of (1) is $C\int_{b-h}^b ds=Ch$, so that (1) fails to hols for any real $C>0$.
Added in response to a comment by the OP: Note that
$$w(s)=\max[0,\min(T,b-a,b-s,T-a+s)]\ge\min(b-1,T-a)=:m$$
for all $a\in[0,1]$ and $s\in[0,1]$.
So, (1) will hold with $C=m$. If you now want $C$ to be $>0$, just require that $m$ be $>0$; that is, (in addition to the condition $a\in[0,1]$) require that $b>1$ and $T>a$.
Thank you Mr. Pinelis for this clarification. Does the inequality any chance to hold for some $T, a, b$ ?. Because in control theory, this is a classical result but I can not find it in any book. Thank you again sir.
@Gustave : As my answer shows, the inequality cannot hold for any $T,a,b$ satisfying your conditions $0\le a\le b\le1$ and $T\in(0,\infty)$. The most probable reason why you cannot find this "classical result" in any book is that you remember the result incorrectly. However, as now shown in the addition to my answer, your inequality will hold if $a\in[0,1]$,$b>1$, and $T>a$.
Thank you sir for the great answer. I'm very thankful for you.
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2025-03-21T14:48:31.480226
| 2020-07-12T17:06:29 |
365484
|
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|
Stack Exchange
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Computational complexity of proof verification
Let $\mathcal{L}$ be a recursive first-order theory, with a deductive system $\Xi$ (for instance, Hilbert-Ackerman proof system). Let $\phi$ be a formula and let $l=(\psi_1, \ldots, \psi_n=\phi)$ be a sequence of formulas.
Question 1: Suppose we what want to discuss the (asymptotical) computational complexity cost of checking wether $l$ constitutes a proof for the pair $(\mathcal{L}, \Xi)$. What are the relevant numerical parameters, depending on $L$, involved in such a complexity function, and to which complexity class it belongs (P, NP, etc)?
Question 2: How much the complexity of verifying $l$ is a proof changes if we change the deductive system (Gentzen's style, for instance), or consider a suitable higer-order theory, or etc?
I apologize in advance about question 2, I hope it makes sense (albeit it is a somewhat non-rigorous question).
The motivation of these questions are the very famous work of Gödel, On the lenght of proofs, and naturally, P=NP? problem.
Your setup doesn't provide any complexity restrictions on determining whether a formula is an axiom or not, beyond demanding that this is computable. Thus, you won't be able to limit the complexity of proof verification either.
Determining the axioms is going to be the only issue, though. Any reasonable proof system will make verifiying proofs a polynomial-time problem relative to an oracle for the axioms. Typically, quadratic time would do: You read through the proof step by step, check whether the pre-requisites are really there; and if so, if the deduction step matches the rules.
Thank you very much. This is great. Since you pointed out, and I neglected, if we consider some canonical axiom system (ZFC, PA, Groups, modules, etc), the axioms can be decided polynomially, right?
Yes, for those examples deciding the axioms is easy. Take for instance ZFC. There are finitely many special axioms, and then the axioms schema replacement and separation. Checking whether a given formula is an instace of the schema is just a pattern-matching issue.
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2025-03-21T14:48:31.480525
| 2020-07-12T17:25:59 |
365487
|
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|
Stack Exchange
|
A syntax independent theory of categories
The classic way to encounter the theory of categories is via Set Theory via the typical definition we see for categories. We see all kinds of categories that are equivalent to the category of small categories. I wonder about presentations of the theory of categories. Is there any work done on cataloguing the various presentations. To facilitate an answer we may need to define a presentation of the theory. It may consist of a logical language or even the standard presentations of algebraic structures. For instance, a presentation of the theory of partial monoids would count as a presentation of Categories. The presentation should come with enough structure to analyze all small Categories.
What I am really looking for more is something like a syntax independent theory of categories. For instance, I had a thought a while ago about monads and string diagrams. Awodey puts forth monads as syntax independent theories (Awodey, Category Theory 2006). I am likely misreading this, or reading too much into it, but in his chapter on monads he states "...it turns out that every adjunction describes, in a "syntax invariant" way, a notion of an "algebra" for an abstract "equational theory". This is just about all he says. The way I see it, we are not using a logical language when presenting an algebra with either an adjunction or the monad it generates. You might say that string diagrams have a syntax, since string actions are equivalent to various algebraic axioms, but that is not very precise.
Then, I saw Marsden put together a presentation of categories in terms of string diagrams.
I like to think that string diagrams can be seen as containers. This is a paper about containers. So the idea is that you have a (co)monad that encodes the container for the theory of categories. That is more what I would say is a syntax independent theory of Categories. I am not sure how this would relate to all presentations of the theory of categories.
Can you give a link for that Awodey claim? It seems to me that's the only thing that would help me know what "syntax-independent" means here.
It's in his this text book. I will link it.
Since I don't have that book, and I'm not sure it's electronically available, maybe you should summarise the presentation and explain why (or quote Awodey explaining why) it should be considered syntax-independent. This is just part of writing a self-contained question.
See Hyland, Power, The Category Theoretic Understanding of
Universal Algebra: Lawvere Theories and
Monads - https://doi.org/10.1016/j.entcs.2007.02.019
Yeah, I was thinking you probably want to look at sketches, since these are ways of presenting the analogue of Lawvere theories that work for defining small categories (equational theories/Lawvere theories aren't sufficient, due to the partially-defined composition on the set of all arrows).
For future reference, I would have recommended asking what Awodey means here and for references to more about that, before asking if it would fit the thing you are thinking about. If it were suitable, you have to understand about what he is talking about in a more in-depth way than the passing reference; as it is not suitable, you still learn something useful, but you could have seen why it's not appropriate. It's taken two questions, a now-deleted answer and a more than one comment discussion to get to the point where we are getting the references you need (in François' and my comments).
See also the references at https://ncatlab.org/nlab/show/syntax-semantics+duality.
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