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2025-03-21T14:48:30.961151
| 2020-05-14T16:06:18 |
360350
|
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"Focus",
"Joshua Mundinger",
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|
Stack Exchange
|
Algebraic analog of a geometric result
There is a famous topological result:
Let $X$ be a smooth manifold of dimension $n$, $E$ be a vector bundle of rank $k > n$, then $E$ contains a trivial line bundle.
So, I guess that (enlightened by Hartshorne's Exercise 2.8.2):
Let $A$ be a ring (some more conditions are needed, say Noetherian, universally catenary,...) of Krull dimension $n$ and let $P$ be a projective module of rank $k > n$. Then there exists a split injection $A \hookrightarrow P$
I wonder if there's a result similar to this one.
You want $A$ to be a direct summand of $M$. Since a projective module is torsion free, any nonzero element gives rise to an injection $A\hookrightarrow M$.
Serre proved that any vector bundle on an affine scheme X of rank larger than the dimension of X has a trivial bundle as a summand; this is Theoreme 1 in http://www.numdam.org/article/SD_1957-1958__11_2_A9_0.pdf.
@abx,sure you are right.I‘ve edited, thanks.
@skd, Thanks a lot
@skd are there some global results on schemes？
Serre proved his result for arbitrary varieties over an infinite filed. If $E$ is a globally generated vector bundle of rank greater than the dimension, a general section is nowhere vanishing. Of course, this does not give a splitting in general, except for affine varieties.
Here's a concrete example: let $E/\mathbb C$ be an elliptic curve. A vector bundle of rank $2$ is equivalent to a representation $\pi_1(E) \to GL_2(\mathbb C)$ up to conjugacy, that is, two commuting matrices up to conjugacy. These have a common eigenvector, but the vector bundle does not split unless the commuting matrices are simultaneously diagonalizable.
@JoshuaMundinger: "A vector bundle of rank 2 is equivalent to a representation" ― this doesn't seem quite right. On $\mathbf P^1$ there are interesting vector bundles such as $\mathcal O \oplus \mathcal O(1)$, but $\pi_1(\mathbf P^1)$ is trivial...
Focus, did you see that the comment by @skd completely answers your question? Why did you edit it?
@R.vanDobbendeBruyn Yes, you're right; representations of $\pi_1$ are local systems, not vector bundles.
|
2025-03-21T14:48:30.961331
| 2020-05-14T17:00:52 |
360355
|
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|
Stack Exchange
|
Critical points of polarized endomorphisms of algebraic varieties
My main question is the following:
Let $f: \mathbb{CP}^n \to \mathbb{CP}^n$ be a holomorphic endomorphism of degree $d \ge 2$ of $\mathbb{CP}^n$ .
1. Let $X \subset \mathbb{CP}^n$ be an irreducible algebraic set such that $f(X) = X$. Set $g:= f|_X$. Denote by $V \subset X$ the minimum (with respect the inclusion) algebraic set such that
$$g: X \setminus g^{-1}(V) \to X \setminus V$$
is a covering, i.e. $V$ is the ramification locus of $g$. Is $V$ always an algebraic set of codimension one in $X$ ?
Two partial questions, whose answer are equally interested to me as the main question.
2. Do there exist an irreducible algebraic set $X$ such that $f: X \to X$ and an algebraic set $V \subset X$ of codimension at least $2$ in $X$ such that $f: X \setminus f|_X^{-1}(V) \to X \setminus V$ is a covering ?
3. Does there exists an irreducible algebraic set $X \subset \mathbb{CP}^n$ such that $f(X) = X$ and $f: X \to X$ is a local biholomorphism ?
Motivation:
My motivation for this question is the class of post-critically finite endomorphisms of $\mathbb{CP}^n$. More precisely, let $f : \mathbb{CP}^n \to \mathbb{CP}^n$ be a holomorphic endomorphism. The critical value set $V_f$ of $f$ is an algebraic set such that
$$f: \mathbb{CP}^n \setminus f^{-1}(V_f) \to \mathbb{CP}^n \setminus V_f$$
is a covering. The map $f$ is called post-critically finite if $$PC(f):=\bigcup\limits_{j \ge 0} f^{\circ j}(V_f)$$ is an algebraic set of codimension one of $\mathbb{CP}^n$, where $f^{\circ m}:= f \circ \ldots \circ f$ is the $n$-th iterate of $f$. In other words, for every irreducible component $\Gamma$ of $PC(f)$, $f^{\circ k}(\Gamma)= f^{\circ (k+m)}(\Gamma)$ for some $k \ge 0, m\ge 1$, i.e. $f^{\circ k}(\Gamma)$ is invariant by $f^{\circ m}$. I want to give the same notion of being post-critically finite for the restriction $f^{\circ m}$ to $X:=f^{\circ k}(\Gamma)$. Set $g:= f^{\circ m}|_{X}$.
So the first step is to define what is the critical value set $V_g$ of $g$. Naturally, we can choose a set $V_g$ such that
$$g: X \setminus g^{-1}(V_g) \to X \setminus V_g$$
is a covering. Since $g$ is the restriction of $f^{\circ m}$, one candidate can simply be $X \cap V_{f^{\circ m}}$. However, $V_{f^{\circ m}}$ and $X$ have essentially no relation, the intersection can be very wild. For example, when $X \subset V_{f^{\circ m}}$ or when it can have irreducible component of several dimension. I want to get a codimension set in $X$. However, at this point, I wonder whether can indeed find the case of codimension higher than one
2. Do there exist an irreducible algebraic set $X$ such that $f: X \to X$ and an algebraic set $V \subset X$ of codimension at least $2$ in $X$ such that $f: X \setminus f|_X^{-1}(V) \to X \setminus V$ is a covering ?
Another choice of definition is that, the critical value set is the image of the critical set. Precisely, the critical set $C_f$ is the set of points where the derivative of $f$ is not invertible. Then, $V_f = f(C_f)$. The fact that $C_f$ is an algebraic set of codimension one of $\mathbb{CP}^n$ is non trivial. The reason is that, $C_f$ is the locus of vanishing the Jacobian determinant of $f$. Additionally, the Jacobian determinant of an endomorphism of degree $d$ is a polynomial of degree $(n+1)(d-1)$, thanks to the computation of the classical endomorphism $[x_1:\ldots:x_n] \mapsto [x_1^d:\ldots:x_n^d]$ and the fact that holomorphic endomorphisms form an open connected set in the space of meromorphic endomorphisms of $\mathbb{CP}^n$. When $X$ is a smooth, the critical set $C_g$ of $g$ can be defined in the same fashion, i.e. the vanishing of the Jacobian determinant. In the paper DYNAMICS OF POST-CRITICALLY FINITE MAPS IN HIGHER DIMENSION, Mathieu Astorg, 2018, the author showed that $C_g$ is included in the intersection of $X$ and irreducible components of $C_f$ other than $X$. When $X$ is singular, we can still define the derivative map on the Zariski tangent space, and we can still define the critical locus is the set of points where the derivative is not invertible. Can it be empty ? This leads to the third question.
3. Does there exist an irreducible algebraic set $X \subset \mathbb{CP}^n$ such that $f(X) = X$ and $f: X \to X$ is a local biholomorphism ?
It may be an easy question. Any comment, and especially, reference suggestions about holomorphic maps on algebraic/analytic varieties are welcomed.
Edit: Thanks to the example pointed in a comment by Lucas Kaufmann, the class of endomorphisms with an elliptic curve should provides the answer when $n =2$ and the algebraic set is smooth. Therefore, I would particularly interested in a kind of results in higher generality, say, algebraic set of dimension 2, possibly singular, in $\mathbb{CP}^n$ with $n \ge 3$.
Hi! For your question 3, there are maps in P^2 leaving an elliptic curve invariant (check the paper by Bonifant-Dabija-Milnor). By Riemann-Hurwitz the restriction of the map to this curve is étale.
Thanks Lucas! if i am not mistaking, the tangent-process on the example of BDM lifts to the double map on the normalization, right ?
An incomplete answer, too long for a comment:
Such questions deal with ``purity of the branch locus''. Let
$f: X \to Y$ be a finite surjective morphism between irreducible (complex) projective varieties, with $Y$ smooth and $X$ normal. A result by (separately) Zariski znd Nagata from 1950s implies that the ramification locus of $f$ is of pure codimension 1 in $Y$ when nonempty (the Wikipedia page on purity links to both papers).
The problem with straightforward application in the situation when $Y=X$ is that interesting examples (=other than projective spaces of lower dimension) of varieties $X$ admitting a morphism $f: X \to X$ of degree greater than 1 are often not smooth (see e.g. A. Beauville, Endomorphisms of hypersurfaces and other manifolds, International Mathematics Research Notices, 1 (2001), pp. 53-58.) I am not an expert on purity theorems, so I cannot point you to a suitable generalization which would give a negative answer to your Question 2, but this looks promising: Hansen, Johan P.
Higher order singularities of morphisms to projective space.
Proc. Amer. Math. Soc. 97 (1986), no. 2, 226-232
Thank you for the references. It seems that this purity question remains vastly open when $Y$ is singular.
|
2025-03-21T14:48:30.961710
| 2020-05-14T18:12:58 |
360361
|
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|
Stack Exchange
|
Characteristic polynomial of checker matrix
For every integer $n > 0$, let $C_n$ be the $4n \times 4n$ matrix having $1$'s in all positions $(i, j)$ such that $i - j$ is even, $3$'s in the two diagonals determined by $|i - j| = 2n + 1$, and $0$'s everywhere else. For example, we have
$$C_2 = \begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 3 & 1 & 0 \\
0 & 1 & 0 & 1 & 0 & 1 & 3 & 1 \\
1 & 0 & 1 & 0 & 1 & 0 & 1 & 3 \\
0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\
1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\
3 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\
1 & 3 & 1 & 0 & 1 & 0 & 1 & 0 \\
0 & 1 & 3 & 1 & 0 & 1 & 0 & 1\end{bmatrix} .$$
I'd like to prove a formula for the characteristic polynomial of $C_n$. From some numerical experiments, I believe that it is
$$(\lambda - 3)^{2n - 2} (\lambda + 3)^{2n - 2} (\lambda^2 - (2n-3)\lambda - 3) (\lambda^2 - (2n+3)\lambda + 3),$$
but I failed to prove that.
Any suggestion is welcome. Thanks.
Note 1. What makes things difficult are the $3$'s. If instead of them there were $0$'s, then we would have a circulant matrix, and using the theory of circulant matrices the characterist polynomial would be easily proved to be $\lambda^{4n - 2}(\lambda - 2n)^2$.
Note 2. Following Pat Devlin's suggestion, I checked the eigenspace of $\lambda = 3$ and it seems to be spanned by the row vectors of the following matrix $(2n-2)\times 4n$ matrix
$$\begin{bmatrix}\begin{matrix}-1 \\ -1 \\ \vdots \\ -1\end{matrix} & I_{2n-2} & \begin{matrix}0 & 0 & -1\\ 0 & 0 & -1 \\ \vdots \\ 0 & 0 & -1\end{matrix} & I_{2n-2}\end{bmatrix} .$$
This shouldn't be difficult to prove, and similarly for the eigenspace of $\lambda = -3$. But I have no idea how to deal with the eigenvalues related to the factor $(\lambda^2 - (2n-3)\lambda - 3) (\lambda^2 - (2n+3)\lambda + 3)$.
(If your conjecture is true, then the eigenspaces are probably nice. There’s also probably some nice involution swapping the eigenspaces for 3 and -3 and swapping the two-dimensional eigenspaces as well.)
In fact, let $H_k$ denote the $4k \times 4k$ matrix consisting of the 0’s and 1’s of your matrix (with the 3’s replaced by 0’s). And let $W$ be the vectors in the kernel of $H_k$ which also satisfy that the middle coordinates of the vector are both 0. Then your matrix sends $W$ to itself in a nice way, and $W$ has codimension 4. I bet $W$ is the union of your two big eigenspaces.
@PatDevlin The eigenspaces for 3 and -3 seem nice indeed. Thanks. I added a note about that. However, I don't see how to deal with the other four eigenvalues.
Following your edit, we’ve now found all but four eigenvectors. By the spectral theorem, these last four will be orthogonal to $W$, which is to say, they’re contained in the set of vectors such that except for the middle two coordinates, all the odd coordinates are equal, and all the even coordinates are equal. (This is a four-dimensional space, and its the orthogonal complement of $W$).
One silly hack would be to try to deduce just enough about the eigenvalues to finish it off. For instance, we have all but four eigenvalues. We also know the trace of the matrix, which gives us one equation relating the other four (namely, they add to $4k$). You could look at a few powers of your matrix to get other stuff, but that feels kind of silly. Better would be to say “your matrix is similar to the following,” which is more or less the spirit of these comments so far.
Ok! Your conjecture is true.
Let $W$ be the space spanned by the eigenvectors for $\lambda \in \{-3, 3\}$ as described in my comments. Let $V$ be the subspace of $\mathbb{R}^{4n}$ consisting of vectors of the form
$$V = \{(a,b,a,b,a,b, \ldots, a, x, y, b, a, b, \ldots, a,b)\},$$
where the entries corresponding to $x,y$ are in positions $2n$ and $2n+1$ of the vector. (So $V$ is the orthogonal complement of $W$.)
Let $T : V \to \mathbb{R}^4$ by $T(\vec{v}) = (a,b,x,y)$ in the obvious way (so $T$ is an isomorphism).
We can check that $V$ is invariant under the action of $C_{n}$. And moreover, that $$T \circ C_{n} \circ T^{-1} \begin{pmatrix}a\\b\\x\\y \end{pmatrix} = \begin{pmatrix}(2n-1)a +y + 3b\\(2n-1)b+x+3a\\(2n-1)b+x\\(2n-1)a+y \end{pmatrix}.$$
Thus, $C_{n}$ restricted to $V$ is isomorphic to the above linear map on $\mathbb{R}^4$, namely
$$\begin{pmatrix}a\\b\\x\\y\end{pmatrix} \mapsto \begin{pmatrix}2n-1 & 3 & 0 &1\\
3 & 2n-1 & 1 &0\\
0 & 2n-1 & 1 &0\\
2n-1 & 0 & 0 &1\end{pmatrix} \begin{pmatrix}a\\b\\x\\y\end{pmatrix},$$
and this map has the desired remaining four eigenvalues as in your conjecture.
The methods in the paper by Junod:
Junod, Alexandre, Hankel determinants and orthogonal polynomials., Expo. Math. 21, No. 1, 63-74 (2003). ZBL1153.15304.
Answer your question (and much more general ones, too).
Could you be more specific please? Because Junod's paper is about Hankel matrices, but my matrix isn't Hankel, it is Toeplitz.
@MargeL I am not sure what I was thinking (not enough sleep...)
|
2025-03-21T14:48:30.962067
| 2020-05-14T18:58:27 |
360363
|
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"Iosif Pinelis",
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|
Stack Exchange
|
Is there a generalised version of the Donsker invariance principle for a "sort-of continuous-time-random-walk"?
(The following question arises from my Math.SE question https://math.stackexchange.com/questions/3643865.)
Let $\rho$ be a probability measure on $\mathbb{R} \times (0,\infty)$, and writing $\ \pi_1 \colon \mathbb{R} \times (0,\infty) \to \mathbb{R}\ $ and $\ \pi_2 \colon \mathbb{R} \times (0,\infty) \to (0,\infty)\ $ for the coordinate projections, suppose that $\int \pi_2^2 \, d\rho < \infty$, $\int \pi_1^2 \, d\rho=\int \pi_2 \, d\rho$ and $\int \pi_1 \, d\rho = 0$.
For each $\lambda>0$, consider the one-dimensional stochastic process $(W^{(\lambda)}_t)_{t \geq 0}$ which, for each sample point $\omega$, linearly interpolates the discrete-point-mapping
\begin{align*}
t_n(\omega) \ \mapsto \ &\frac{1}{\sqrt{\lambda}} \sum_{i=1}^n X_i(\omega) \\
t_n(\omega) \ := \ &\frac{1}{\lambda} \sum_{i=1}^n \Delta_i(\omega) \quad \text{for each $n \geq 0$}
\end{align*}
where the random vectors $\begin{pmatrix} X_i \\ \Delta_i \end{pmatrix}$, $i \geq 1$, are i.i.d. with law $\rho$.
Equipping $C([0,\infty),\mathbb{R})$ with the topology of uniform convergence on bounded sets, is it the case that the $C([0,\infty),\mathbb{R})$-valued random variable $(W^{(\lambda)}_t)_{t \geq 0}$ converges in distribution to a Wiener process as $\lambda \to \infty$?
The case where $\pi_2$ projects $\rho$ onto a Dirac mass is essentially Donsker's invariance principle; so I am wondering about the more general case. I emphasise that I do not wish to assume that $\pi_1$ and $\pi_2$ are independent under $\rho$.
I realise it would probably be good to say a bit more about the motivation behind this question.
The Wiener process has some nice properties, such as increments that are stationary and memoryless. Donsker's theorem describes one way in which a Wiener process can physically arise, namely as a random walk with small step distance $\sqrt{\Delta}$ and high step frequency $\frac{1}{\Delta}$. But as a continuous-time process, this random walk does not have increments that are both stationary and exhibit decay of correlations.
There may be situations in which one wishes to work with a Donsker-like approximation to Brownian motion (e.g. SDEs driven by "bounded noise", to avoid extreme events in the long-term behaviour of the system), but keeping some of the nice properties of Brownian motion, such as increments that are stationary and exhibit decay of correlations. The raw form of Donsker's theorem will not achieve this, and so some modified version such as described above can be used.
@IosifPinelis For each $\lambda$, if we define the set $S_\lambda(\omega):=\left{\frac{1}{\lambda}\sum_{i=1}^n\Delta_i(\omega):n\in\mathbb{N}\cup{0}\right}\subset [0,\infty)$, then the map $t\mapsto W^{(\lambda)}t(\omega)$ on $[0,\infty)$ is defined to be linear between each consecutive pair of points in $S\lambda(\omega)$, with $W^{(\lambda)}{\frac{1}{\lambda}\sum{i=1}^n\Delta_i(\omega)}(\omega)=\frac{1}{\sqrt{\lambda}}\sum_{i=1}^n X_i(\omega)$ for each $n \in \mathbb{N}\cup{0}$. Do you think that how I phrased it in the post wasn't clear enough?
So in words, you want to start with a process $Z_t$ that waits for time $\Delta_i$ and then jumps by an amount $X_i$, where $\Delta_i$ is not necessarily independent of $X_i$? Then $W_t^{(\lambda)}$ is the linearly interpolated and rescaled version of this process, and you want to see if this converges weakly to Brownian motion in the scaling limit?
@JulianNewman : I had misread some of your post, sorry. However, I think the limit process should be $t\mapsto(B_t,t)$, rather than a Brownian motion $t\mapsto B_t$.
@NateEldredge Exactly!
@IosifPinelis No $(W_t^{(\lambda)})$ is not an $(\mathbb{R} \times (0,\infty))$-valued process, it is simply an $\mathbb{R}$-valued process.
I think Iosif Pinelis is correct, but his comment should be expanded as follows.
Notation: let
$$ p(t) = \lfloor t + 1\rfloor - t , \qquad q(t) = t - \lfloor t\rfloor . $$
Whenever we have a discrete-time process $Z_n$, we extend it into a continuous one, piecewise linear, defined by:
$$ Z_t = p(t) Z_{\lfloor t\rfloor} + q(t) Z_{\lfloor t + 1\rfloor} . $$
With no loss of generality assume that $\mathbb{E} \Delta_n = 1$ to simplify the notation.
Step 1. The bi-variate process $$(Y_n,Z_n) = (\sum_{i=1}^n (\Delta_i - 1), \sum_{i=1}^n X_i)$$ satisfies Donsker's invariance principle, that is,
$$ (Y^{(\lambda)}_t, Z^{(\lambda)}_t) = (\lambda^{-1/2} Y_{\lambda t}, \lambda^{-1/2} Z_{\lambda t}) $$
converges weakly (in the Banach space of continuous functions on $[0,T]$, for any given $T > 0$) to a two-dimensional Brownian motion $(\tilde Y_t, \tilde Z_t)$.
Step 2. Let $$(T_n,Z_n) = (\sum_{i=1}^n \Delta_i, \sum_{i=1}^n X_i) = (Y_n + n, Z_n) $$ and consider
$$ (T^{(\lambda)}_t, Z^{(\lambda)}_t) = (\lambda^{-1} T_{\lambda t}, \lambda^{-1/2} Z_{\lambda t}) = (\lambda^{-1/2} Y^{(\lambda)}_t + t , Z^{(\lambda)}_t) .$$
It is then easy to see that $(T^{(\lambda)}_t, Z^{(\lambda)}_t)$ converges weakly (in the same sense) to $$(\tilde T_t, \tilde Z_t) = (t, \tilde Z_t),$$ where $\tilde Z_t$ is the Brownian motion.
This is somewhat technical, but relatively standard, I think: one can use the characterisation of weak convergence of processes in terms of convergence of finite-dimensional distributions and equicontinuity (or $J$-compactness if one prefers the more general approach via Shorokhod topology). See Theorem 1.6.2 in [Silvestrov D.S. (2004) Weak convergence of stochastic processes. In: Limit Theorems for Randomly Stopped Stochastic Processes. Probability and its Applications. Springer, London] for details.
Step 3. The process $W_t^{(\lambda)}$ is a continuous, piecewise linear process defined by
$$ W^{(\lambda)}_t = Z^{(\lambda)}_{(T^{(\lambda)})^{-1}_t} .$$
Our claim is: this process converges weakly
$$ \tilde W_t = \tilde Z_{\tilde T^{-1}_t} = \tilde Z_t. $$
Here we use the very definition of weak convergence. Let $\Phi$ be a continuous functional on the space of paths $\omega_t$ of $W^{(\lambda)}_t$ or $\tilde W_t$. Then $\Phi$ induces a functional $\Psi$ on the space of paths $(\tau_t, \zeta_t)$ of $(T^{(\lambda)}_t, Z^{(\lambda)}_t)$ and $(\tilde T_t, \tilde Z_t)$, in the following sense:
$$ \Psi((\tau_t), (\zeta_t)) = \Phi((\tau_{\zeta^{-1}_t})) , $$
where $\zeta^{-1}_t = \inf \{ s : \tau_s \ge t \}$ is the generalised inverse of $\zeta_t$.
Of course $\Psi$ is no longer continuous. However, it is easy to see that the set of discontinuities of $\Psi$ has zero measure with respect to the law of $(\tilde T_t, \tilde Z_t)$. Roughly speaking: the paths of $(\tilde T_t, \tilde Z_t)$ are of the form $((\tilde \tau_t), (\tilde \zeta_t))$, where $\tau_t = t$ for all $t$. Given such a path and $\varepsilon > 0$, choose $\delta > 0$ so that the $\delta$-modulus of continuity of $\tilde \zeta_t$ is less than $\varepsilon$. If $(\tau_t)$ is $\delta$-close to $(\tilde \tau_t) = (t)$ and $(\zeta_t)$ is $\varepsilon$-close to $(\tilde \zeta_t)$, then $|\tau_t^{-1} - t| < \delta$ (by a simple calculus exercise), and hence
$$ |\omega_t - \tilde \omega_t| = |\zeta_{\tau_t^{-1}} - \tilde \zeta_t| \leqslant |\zeta_{\tau_t^{-1}} - \tilde \zeta_{\tau_t^{-1}}| + |\tilde Z_{\tau_t^{-1}} - \tilde Z_t| \le 2 \varepsilon . $$
This proves continuity of $\Psi$ at $((\tilde \tau_t), (\tilde \zeta_t))$, as desired.
This is perfectly sufficient for our needs: weak convergence of $(T^{(\lambda)}_t, Z^{(\lambda)}_t)$ to $(\tilde T_t, \tilde Z_t)$ implies that
$$ \mathbb{E} \Psi((T^{(\lambda)}_t), (Z^{(\lambda)}_t)) = \mathbb{E} \Phi((W^{(\lambda)}_t)) $$
converges to
$$ \mathbb{E} \Psi((\tilde T_t), (\tilde Z_t)) = \mathbb{E} \Phi((\tilde W_t) , $$
and consequently $W^{(\lambda)}_t$ converges weakly to $\tilde W_t$.
I am sure the above argument is well-known, for instance, in the literature on "Lévy flights", "continuous-time random walks", or similar. There are still some details missing, but I hope the idea is now clear.
Thank you. I think my intuition had already pretty much arrived at where you got to, and my real trouble is justifying the last step - which you say you believe should be easy? Essentially your logical format is "we know $f_n(x_\ast)\to a$ and $x_n\to x_\ast$, and we need to deduce $f_n(x_n)\to a$". My crude visual intuition is that this holds in our case, but it's certainly not obvious to me; when we multiply the deviation of $T_{\lfloor\lambda t\rfloor/\lambda}$ away from $t\mathbb{E}\Delta_1$ by the unsigned gradient $\sqrt{\lambda}$ of $W^{(\lambda)}$, the result doesn't tend to $0$.
Time permitting, I will try to expand my comment later this week.
I expanded the answer, but there are still details missing. Sorry for that, this is the best I can do now.
Thank you very much, I'm now certainly satisfied. I have added an answer that essentially follows your argument with a slightly different structure.
Having read Mateusz Kwaśnicki's answer, I will now write it in my own way:
Lemma. Let $S_\infty$ and $T$ be separable metric spaces, and let $(S_j)_{j \in \mathbb{N}}$ be a sequence of Borel subsets of $S_\infty$. Let $\{F_j\}_{j \in \mathbb{N} \cup \{\infty\}}$ be a family of Borel-measurable functions $F_j \colon S_\infty \to T$, with $F_\infty$ being continuous. Suppose that for every $(x_j)_{j \in \mathbb{N} \cup \{\infty\}} \in \prod_{j \in \mathbb{N} \cup \{\infty\}} S_j$, if $x_j \to x_\infty$ as $j \to \infty$ then $F_j(x_j) \to F_\infty(x_\infty)$ as $j \to \infty$. Then for any family $(\Xi_j)_{j \in \mathbb{N} \cup \{\infty\}}$ of $S_\infty$-valued random variables $\Xi_j$ with $\Xi_j \in S_j$ almost surely, if $\Xi_j$ converges in law to $\Xi_\infty$ then $F_j(\Xi_j)$ converges in law to $F_\infty(\Xi_\infty)$.
Proof of the result given the Lemma.
Let $M=\mathbb{E}[\Delta_i]=\mathbb{E}[X_i^2]$. Let $S_\infty=C([0,\infty),\mathbb{R}^2)$, and for each $\lambda \in (0,\infty)$ let
$$ S_\lambda \ = \ \left\{ t \mapsto \begin{pmatrix} \sqrt{\lambda}(w(t)-Mt) \\ v(t) \end{pmatrix} \, : \, w \in \mathrm{Homeo}([0,\infty)), v \in C([0,\infty),\mathbb{R}) \right\}. $$
For each $\lambda \in (0,\infty)$ and $n \in \mathbb{N}$, let
$$ \Xi_\lambda(\tfrac{n}{\lambda}) \ = \ \begin{pmatrix} \frac{1}{\sqrt{\lambda}} \sum_{i=1}^n (\Delta_i - M) \\ \frac{1}{\sqrt{M\lambda}} \sum_{i=1}^n X_i \end{pmatrix} $$
and extend by linear interpolation: for $r \in (0,1)$, writing $t=\frac{n+r}{\lambda}$, we define
\begin{align*}
\Xi_\lambda(t) \ &= \ r\Xi_\lambda(\tfrac{n+1}{\lambda})+(1-r)\Xi_\lambda(\tfrac{n}{\lambda}) \\
&= \begin{pmatrix} \left( \frac{1}{\sqrt{\lambda}} \left( r\Delta_{n+1} + \sum_{i=1}^n \Delta_i \right) \right) - \sqrt{\lambda}Mt \\ \frac{1}{\sqrt{M\lambda}} \left( rX_{n+1} + \sum_{i=1}^n X_i \right). \end{pmatrix}
\end{align*}
Hence, for each $\lambda \in (0,\infty)$, we have that $\Xi_\lambda\overset{\mathrm{def}}{=}(\Xi_\lambda(t))_{t \geq 0}$ belongs almost surely to $S_\lambda$, with the corresponding $w \in \mathrm{Homeo}([0,\infty))$ given by
$$ w(\tfrac{n}{\lambda}) \ = \ \frac{1}{\lambda} \sum_{i=1}^n \Delta_i . $$
The Donsker invariance principle gives that as an $S_\infty$-valued random variable, $\Xi_\lambda$ is convergent in distribution as $\lambda \to \infty$, to a two-dimensional Brownian motion $\Xi_\infty\!=\!(\Xi_\infty(t))_{t \geq 0}$ whose second-coordinate projection is a one-dimensional Wiener process (i.e. standard Brownian motion).
Now for each $\lambda \in (0,\infty)$, define $F_\lambda \colon S_\lambda \to C([0,\infty),\mathbb{R})$ by
$$ F_\lambda\!\begin{pmatrix} u \\ v \end{pmatrix} \ = \ \sqrt{M}v \circ \left( M.\mathrm{id}_{[0,\infty)} + \tfrac{1}{\sqrt{\lambda}} u \right)^{\!-1}, $$
and define $F_\infty \colon S_\infty \to C([0,\infty),\mathbb{R})$ by
$$ F_\infty\!\begin{pmatrix} u \\ v \end{pmatrix}(t) \ = \ \sqrt{M}v\!\left( \tfrac{1}{M} t \right). $$
So $F_\lambda(\Xi_\lambda) = (W_t^{(\lambda)})_{t \geq 0}$ for each $\lambda \in (0,\infty)$, and $F_\infty(\Xi_\infty)$ is a one-dimensional Wiener process.
Now for any $\lambda_n \nearrow \infty$, any $(x_n)_{n \in \mathbb{N}}=\!\left( \begin{pmatrix} u_n \\ v_n \end{pmatrix} \right)_{\!\!n \in \mathbb{N}} \in \prod_{n \in \mathbb{N}} S_{\lambda_n}$ and any $x_\infty=\!\begin{pmatrix} u_\infty \\ v_\infty \end{pmatrix} \in S_\infty$, if $x_n$ converges uniformly on bounded sets to $x_\infty$, then $M.\mathrm{id}_{[0,\infty)} + \frac{1}{\sqrt{\lambda_n}} u_n$ converges uniformly on bounded sets to $M.\mathrm{id}_{[0,\infty)}$, so $\left( M.\mathrm{id}_{[0,\infty)} + \frac{1}{\sqrt{\lambda_n}} u_n \right)^{\!-1}$ converges uniformly on bounded sets to $\frac{1}{M}\mathrm{id}_{[0,\infty)}$; and also $v_n$ converges uniformly on bounded sets to $v_\infty$; and therefore $F_{\lambda_n}(x_n)$ converges uniformly on bounded sets to $F_\infty(x_\infty)$.
Hence the Lemma gives the desired result.
Proof of the Lemma.
For convenience of notation, assume that all the random variables are over a probability space $(\Omega,\mathcal{F},\mathbb{P})$.
Let $d$ be the metric on $T$. For each $\varepsilon>0$, define the decreasing sequence $(G_j(\varepsilon))_{j \in \mathbb{N}}$ of subsets of $S_\infty$ by
$$ G_j(\varepsilon) \ = \ \overline{\bigcup_{i=j}^\infty \{x \in S_i : d(F_i(x),F_\infty(x)) \geq \varepsilon \}}. $$
We first show that for all $\varepsilon>0$, $\bigcap_{j \in \mathbb{N}} G_j(\varepsilon)=\emptyset$. If we have $x_\infty \in \bigcap_{j \in \mathbb{N}} G_j(\varepsilon)$, we can construct a sequence $n_j \nearrow \infty$ and a sequence $x_{n_j} \in S_{n_j}$ converging to $x_\infty$ such that $d(F_{n_j}(x_{n_j}),F_\infty(x_{n_j})) \geq \varepsilon$ for all $j \in \mathbb{N}$; and since $F_\infty(x_{n_j}) \to F_\infty(x_\infty)$ (by continuity of $F_\infty$), it follows that $F_{n_j}(x_{n_j})$ does not converge to $F_\infty(x_\infty)$, contradicting our assumption.
Now fix any family $(\Xi_j)_{j \in \mathbb{N} \cup \{\infty\}}$ of $S_\infty$-valued random variables $\Xi_j$ with $\Xi_j \in S_j$ almost surely, such that $\Xi_j$ converges in law to $\Xi_\infty$. For each $\varepsilon>0$ we have that $\mathbb{P}(\Xi_j \in G_j(\varepsilon)) \to 0$, since
\begin{align*}
\limsup_{j \to \infty} \mathbb{P}(\Xi_j \in G_j) \ &\leq \ \lim_{m \to \infty} \limsup_{j \to \infty} \mathbb{P}(\Xi_j \in G_m) \\
&\leq \ \lim_{m \to \infty} \mathbb{P}(\Xi_\infty \in G_m) \quad \text{since $\Xi_j$ converges in law to $\Xi_\infty$} \\
&= \ 0 \quad \text{since } \bigcap_{m \in \mathbb{N}} G_m=\emptyset.
\end{align*}
Now to show the desired convergence, fix a bounded Lipschitz function $g$ on $T$ which, without loss of generality, we assume to map into $[0,1]$ and to have Lipschitz constant $1$. Fix $\varepsilon>0$. Let $N$ be sufficiently large that for all integers $j \geq N$,
$$ \mathbb{P}(\Xi_j \in G_j(\tfrac{\varepsilon}{3})) \ < \ \tfrac{\varepsilon}{3} $$
and
$$ \big| \mathbb{E}[g(F_\infty(X_j))] - \mathbb{E}[g(F_\infty(X_\infty))] \big| < \tfrac{\varepsilon}{3}, $$
where the latter is possible by the convergence in law of $X_j$ to $X_\infty$, since $g \circ F_\infty$ is a bounded continuous function on $S_\infty$. Using the former of these two statements, an easy calculation yields
$$ \big| \mathbb{E}[g(F_j(X_j)) - g(F_\infty(X_j))] \big| < \tfrac{2\varepsilon}{3}, $$
and combining this with the latter gives
$$ \big| \mathbb{E}[g(F_j(X_j))] - \mathbb{E}[g(F_\infty(X_\infty))] \big| < \varepsilon $$
as required.
|
2025-03-21T14:48:30.963017
| 2020-05-14T19:16:39 |
360365
|
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|
Stack Exchange
|
Relation between expected values of eigenvalues of Laplacian matrix of a graph and eigenvalues of expected Laplacian matrix of that graph?
Particularly, I am dealing with Erdős–Rényi random (,), so the expected Laplacian matrix of (,) is (−), where and are one and identity matrices, respectively.
In addition,if the distribution (unsure, but might be power law) of the eigenvalues of Laplacian matrix of the graph (,) is known, then it seems to me that expected value of the eigenvalues has some closed form formula depending on and in the asymptotic case.
See the papers by Erdos (no relation) and collaborators, e.g.:
Erdős, László; Knowles, Antti; Yau, Horng-Tzer; Yin, Jun, Spectral statistics of Erdős-Rényi graphs. I: Local semicircle law, Ann. Probab. 41, No. 3B, 2279-2375 (2013). ZBL1272.05111.
Erdős, László; Knowles, Antti; Yau, Horng-Tzer; Yin, Jun, Spectral statistics of Erdős-Rényi graphs II: eigenvalue spacing and the extreme eigenvalues, Commun. Math. Phys. 314, No. 3, 587-640 (2012). ZBL1251.05162.
|
2025-03-21T14:48:30.963448
| 2020-05-14T20:07:20 |
360367
|
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|
Stack Exchange
|
On submodules of vector fields
I don't know much about modules aside from their basic definition and that they are more complicated than vector spaces.
I am asking this question because I wish to have a more "algebraic" understanding of Frobenius' theorem in differential geometry.
It seems to me that Frobenius' theorem can be stated using submodules of vector fields and ideals of the exterior algebra without necessarily invoking distributions.
If we do invoke distributions, then suppose that $M$ is a smooth $n$ dimensional manifold, $\Delta:x\mapsto\Delta_x$ is a $k$-dimensional distribution in $M$, and $\Gamma_M(\Delta)$ is the set of smooth sections of the distribution. Then $\Gamma_M(\Delta)\subseteq\mathfrak X(M)$ is a submodule of $\mathfrak X(M)$ the module of all (smooth vector fields) on $M$. Some statements can be made about this submodule:
If $\Delta$ has a fixed dimension $k$, then $\Gamma_M(\Delta)$ is "locally finitely generated and of rank $k$", where I put quotes on this statement as I have no idea how this is called in the literature if at all, but what I mean is that around any $x\in M$ there is an open set $U$ such that the restriction $\Gamma_U(\Delta)$ is generated by $k$ pointwise linearly independent vector fields $E_1,...,E_k$, and this number $k$ is is minimal in the sense that there are no open sets (aside from $\varnothing$ I guess) where there are less linearly independent generators, like, say $F_1,...,F_l$ where $l<k$.
I assume that if $\Delta$ has no fixed dimension, and its elements have $\dim\Delta_x\le k$, then every $x\in M$ has an open neighborhood $U$ where one can define $k$ generators, and not less in general, but these will not be pointwise linearly independent at all points.
Let us call this submodule $\Gamma_M(\Delta)$ differentially closed if $X,Y\in\Gamma_M(\Delta)$ implies $[X,Y]\in\Gamma_M(\Delta)$. Frobenius' theorem is closely related to the fact - at least when $\Delta$ is of fixed dimension - that a differentially closed submodule (of locally rank $k$) always has $k$ commuting local generators.
The only proofs I know of Frobenius' theorem is valid only for the case of distributions with fixed dimension, where the existence of local, pointwise independent generators of a fixed number is used heavily in the proof.
Moreover, if we assume that $\Gamma_M(\Delta)$ is locally finitely generated and of rank $k$, then one can easily show that the set $\mathcal I\subseteq\Omega(M)$ consisting of differential forms that vanish when all of their arguments are elements of $\Gamma_M(\Delta)$ is an ideal of the exterior algebra that is locally generated by $n-k$ pointwise independent $1$-forms, the ideal is differential ($\mathrm d\mathcal I\subseteq I$) if and only if $\Gamma_M(\Delta)$ is differentially closed, and in this case $\mathcal I$ is always locally generated by $n-k$ exact $1$-forms.
Essentially, I am interested in statements (in-answer explanations, papers, textbooks) about submodules of $\mathfrak X(M)$ where these submodules are not defined by distributions from the get-go, such as
Given an arbitrary submodule $\mathfrak M\le\mathfrak X(M)$, when does it have a well-defined "local rank"?
Does a general submodule always determine a (possibly degenerate) distribution?
How does duality with an ideal of the exterior algebra work in the general case? I relied heavily on local generators for proving dual statements. In particular, if there is always a corresponding distribution then I guess the differential closedness of the submodule/ideal implies complete integrability of the distribution, but what about the statements about commuting/exact generators?
Etc.
I don't expect a complete answer here, but at least some paper or advanced textbook that treats Frobenius' theorem from such a heavily algebraic point of view would be nice and illuminating for me.
H. Sussmann, Orbits of families of vector fields and integrability of distributions, Trans. A.M.S., vol. 180, June 1973, discusses these issues at length. Any module of vector fields has a notion of orbit, a minimal nonempty subset closed under the flows of the vector fields, and the orbits are submanifolds, but perhaps of different dimensions. If the module is closed under its own flows, and of constant rank, the Frobenius theorem applies. Consider the module of all smooth vector fields on the plane $f(x,y)\partial_x+g(x,y)\partial_y$ so that $g=0$ for $x<0$. These vector fields have as orbits the entire plane, so a local picture of the module involved is never enough. Clearly there is no local rank defined near the vertical axis. For relations between differential form descriptions of the Frobenius theorem and the vector fields descriptions, you can see p. 6 of my lecture notes on exterior differential systems.
For the second question, since a distribution is a subbundle of the tangent bundle, and therefore a vector bundle itself, if the manifold is compact and Hausdorff, you need the submodule $\mathfrak{M}$ to be a projective module and finitely generated. This is because of Serre-Swan theorem. I think that this paper could be very useful for you (see section 3).
Sorry for this highly incomplete answer.
|
2025-03-21T14:48:30.963778
| 2020-05-14T20:59:32 |
360370
|
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|
Stack Exchange
|
How to solve optimization problem with matrix constraint?
I'm working with the following optimization problem below.
$$\min_{\Pi} \left[
\frac{1}{4 \lambda
}\left((\Pi\vec{1}-s)^T K(\Pi\vec{1}-s) + \left(\Pi^T \vec{1}-t\right)^T K \left(\Pi^T \vec{1}-t\right)\right)-\text{Tr}[\Pi K]
\right]$$
Here, $\lambda$ is a scalar, $\Pi$ and $K$ are $n\times n$ matrices, and $s,t$ are $n\times 1$ probability vectors (i.e. the components sum up to $1$). Also $K$ is PSD and $\Pi$ has elements summing up to $1$.
I was able to empirically show that the diagonal elements of $\Pi^{*}$, the optimal $\Pi$, correspond to $\frac{s+t}{2}$ regardless of the choice of $\lambda$, but I am having difficulty showing this to be the case theoretically. Does anyone have any advice? I tried for the case when $n=2$ using Lagrange multipliers and even then it gets very messy.
I set up a Lagrange multipier where the matrix constraint $1^T \Pi 1=1$, and I end up getting as my first order conditions
$$2K [(\Pi+\Pi^T) \vec{1}-(s+t)] \vec{1}^T - K - \lambda \vec{1}\vec{1}^T = 0$$
and
$$1-\vec{1}^T \Pi \vec{1}=0.$$
I just don't see where to go from here. I'd like to use these to show $\vec{diag(\Pi^*)} = \frac{1}{2}(\vec{s+t})$. Also, how do I come up with KKT duals when every element of $\Pi$ must be nonnegative?
Anyone have any ideas?
For $n \ge 2$, the problem is generally unbounded, unless there is an additional constraint, such as all elements of $P \ge 0$.. Presuming this additional constraint is added, I did not find agreement with your assertion on any of several instances I generated with random s and t probability vectors and random PSD K, and numerically solved as QP. However, your assertion appears to be true, if also K is a (positive) multiple of the Identity matrix, but not true if K is a general PSD diagonal matrix.
Oh so yeah also all the elements of $\Pi$ are nonnegative, yes.
Can you tell me which instances gave a counterexample?
Every instance I tried, except for the exceptions I mentioned. Maybe all your inputs are more structured specials cases. What instances did you solve and how do they map against the special cases I mentioned?
I used K= [1,4; 4,1], s=[.5,.5], t=[.2,8] for a 2D case and K=[1,0,0;0,1,0;0,0,1], s=[.1,.4,.5], t=[.4,.2,.4] for the 3D case. Both of these got me a diagonal with s+t / 2.
And how would I incorporate the constraint that all elements are greater than or equal to 0 in a Lagrangian?
KKT conditions https://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_conditions. QP, so satisfies Linearity constraint qualification. I am doubting you will find a closed form solution.
Does that mean there isn't a way to theoretically show the diagonal would be the average of s and t?
Well, certainly not possible if there are counterexamples. Unless I made a mistake, or the inputs are further restricted than what I did. But your hypothesis seems to hold for K being a positive multiple of the Identity matrix,
Can you please give me a specific counterexample? I still have not found any counter-examples on my end so I'd like to check myself.
In MATLAB notation: s=[.5;.5];t=[1;0];K=diag([1;2]); results in optimal P=diag([0;1]), optimal objective =-1.90625. However, using P=diag((s+t)/2)=diag([.75;.25]) results in objective=-1.240625, which is inferior, so not optimal.
Forgot to mention that counterexample is with $\lambda=10$. Same optimal solution with $\lambda=1$, but different optimal solution for $\lambda=0.1$.
Would you mind sharing your Matlab code with me? Greatly appreciate it! Email is<EMAIL_ADDRESS>
|
2025-03-21T14:48:30.964021
| 2020-05-14T21:55:10 |
360376
|
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|
Stack Exchange
|
Does this strong form of being almost 1-to-1 imply injectivity?
Let $\pi\colon(X,T)\to (Y,T)$ be a factor map between minimal subshifts. Suppose there exists $\tilde{Y} \subseteq Y$ such that
$\# \pi^{-1}(y) = 1$ for all $y \in \tilde{Y}$.
$\tilde{Y}$ is a residual subset of $Y$ i.e. $\tilde{Y} = \bigcap_{n=1}^\infty\tilde{Y}_n$ for some collection $\{\tilde{Y}_1,\tilde{Y}_2,\dots\}$ of dense open subsets of $Y$.
$\mu(\tilde{Y}) = 1$ for every $T$-invariant probability measure in $Y$.
Is it true that $\pi$ is injective?
No. Consider an irrational rotation $R$ of the circle (which I identify with [0,1)) by an angle $\alpha$. Let $\alpha<\beta<1$ be a point not lying in the orbit of 0 under $R$. Set $A_1=[0,\alpha)$, $A_2=[\alpha,\beta)$, $A_3=[\beta,1)$ also $B_1=A_1$ and $B_2=A_2\cup A_3$.
Consider the partitions $P=\{A_1,A_2,A_3\}$ and $Q=\{B_1,B_2\}$.
For $x\in S^1$, let $\epsilon_1(x)\in \{1,2,3\}^{\mathbb Z}$ be such that $R^n(x)\in A_{\epsilon_1(x)_n}$ and similarly let $\epsilon_2(x)\in \{1,2\}^{\mathbb Z}$ be such that
$R^n(x)\in B_{\epsilon_2(x)_n}$.
Let $X$ be the (minimal) subshift of $\{1,2,3\}^{\mathbb Z}$ consisting of the orbit closure of $\{\epsilon_1(x)\colon x\in [0,1)\}$ and $Y$ be the orbit closure of $\{\epsilon_2(x)\colon x\in [0,1)\}$. There is an obvious factor map from $X$ to $Y$ (replacing 3 symbols by 2's). This factor map is 1-1 off a countable set (corresponding to orbits passing through $\beta$).
Nothing wrong with Anthony Quas's answer, and indeed the answer is "no", but I'll just remark that you already asked essentially the same question before, and the same answer works directly, with $X = Y$. Let me recall my answer to this question of yours.
"The answer is yes [the previous question was phrased differently]. In this paper Downarowicz proves the following theorem
There exists a regular Toeplitz sequence $\omega$ such that the induced
Toeplitz flow $(\bar O(\omega), S)$ is noncoalescent, more precisely, it admits an endomorphism $\pi : \bar{O}(\omega) \to \bar{O}(\omega)$ of the first type.
Here, $\bar O(\omega)$ is the orbit closure (so a Toeplitz subshift since $\omega$ is Toeplitz), $S$ the shift map, noncoalescent means not injective, and first type means every Toeplitz point has a unique preimage, in particular some point does."
To see that this answers your question, set $X = Y$, and $\tilde Y$ the Toeplitz points of $Y$, so that every point in $\tilde Y$ has unique preimage. What we need is that the Toeplitz points are a residual set, and that their measure is $1$ in every invariant probability measure. Both facts are well-known. The first fact is mentioned e.g. on page 23 of this paper (it works for any Toeplitz subshift), and the second is Theorem 13.1. in the same reference (it works for any regular Toeplitz subshift).
|
2025-03-21T14:48:30.964218
| 2020-05-14T22:07:34 |
360378
|
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|
Stack Exchange
|
Galois group of polynomials related to Fibonacci and Catalan numbers
Let $F_n$ be the Fibonacci and $C_n$ the Catalan numbers.
Define a polynomial by $F_n(x):=\sum\limits_{k=1}^{n}{F_k x^{n-k}}$.
For example $F_8(x)=x^7+x^6+2x^5+3x^4+5x^3+8x^2+13x+21$.
And another polynomial $C_n(x):=\sum\limits_{k=1}^{n}{C_k x^{n-k}}$.
For example $C_8(x)=x^7+2x^6+5x^5+14x^4+42x^3+132x^2+429x+1430.$
Question 1: Is $F_n(x)$ irreducible for all $n \geq 3$ and with Galois group $S_{n-1}$?
Question 2: Is $C_n(x)$ irreducible for all $n \geq 3$ and with Galois group $S_{n-1}$?
Both questions have a positive answer for $n \leq 25$.
Question 3: Is there an official english name for irreducible polynomials of degree $n$ with Galois group $S_n$? In German they are called "affektlos", see for example this article by Schur : https://www.degruyter.com/view/journals/crll/1931/165/article-p52.xml .
Just a quick note: I wrote a quick GAP script that verifies that the questions have a positive answer for $n \leq 32$, which is as far as GAP can go with this at the moment.
This might be totally irrelevant but Schur showed that all truncated exponential polynomials are irreducible. See https://kconrad.math.uconn.edu/blurbs/gradnumthy/schurtheorem.pdf
@Carl-FredrikNybergBrodda That is true for the generating function of any rational function...
It turns out that Magma can compute Galois groups far quicker than GAP can. For example, checking the Fibonacci polynomials up to $n \leq 100$ only takes a few minutes. This might be of interest to you.
|
2025-03-21T14:48:30.964339
| 2020-05-14T23:32:34 |
360381
|
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Stack Exchange
|
To show holomorphicity of a certain infinite series of functions
I have the following sequence of holomorphic functions on $(f_n(s))_{n \geq 1}$ on the closed region $R:= \{ s \in \mathbb C : \Re(s) \geq 1\}$ by
$$f_n(s) :=
\begin{cases}
\log(1+p^{-s})-p^{-s}\text{, if }n=p\text{ is prime}\\
0\text{, otherwise }
\end{cases}$$
(where we have taken the principal branch of the logarithm) and I want to show that the sum $\sum_{n \geq 1} f_n(s)$ is holomorphic on $R$. Now, I read here (https://math.stackexchange.com/questions/3675324/weierstrass-theorem-for-closed-domains/3675344#3675344) that Weirstrass' Theorem doesn't apply for this closed region $R$, so I have no idea on how I could get started to show this result. I would be really grateful for a proof or some hints in that regard. Thanks a lot.
If $\Re(s) > 1/2+\epsilon$ for $\epsilon > 0$ we have $\log(1+p^{-s}) = \sum_{j=1}^\infty p^{-sj}/j$ so
$|f_p(s)| \le \sum_{j=2}^\infty p^{-\Re(s)j}/j \le p^{-1-2\epsilon}/(1-p^{-1/2-\epsilon})$. Since $\sum_p p^{-1-2\epsilon}$ converges, your series converges uniformly on $\{s: \Re(s) > 1/2 + \epsilon\}$, and therefore the sum is holomorphic there, and in particular this is true on $R$.
|
2025-03-21T14:48:30.964446
| 2020-05-15T00:07:05 |
360383
|
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|
Stack Exchange
|
Computational complexity of computing the trace of a matrix product under some structure
I have two problems related to computing some trace, and some (possibly suboptimal) answers. My question is about a potential more efficient algorithm for each one. (More interested in an answer to question 1.)
Let $U, V$ and $F$ be three real matrices. All three matrices have size $d \times r$, with $r \ll d$ (that is, $U, V$ and $F$ are ``tall''). I want to compute $\mathrm{trace}(U V^\top F F^\top)$. Computing $A = F^\top U$, $B = V^\top F$ and the trace of $AB$ has complexity $\mathcal{O}(r^2 d)$. Is there any faster algorithm (taking into account $r \ll d$)? Can we get $\mathcal{O}(r d)$?
Let $U, V$ and $M$ be three real matrices. $U$ and $V$ have size $d \times r$ (with $r \ll d$), and $M$ is lower triangular (with positive elements in its diagonal) of size $d \times d$. I want to compute $\mathrm{trace}(U V^\top M M^\top)$. The simple algorithm of computing $A = M^\top U$, $B = V^\top M$, and then the trace of $AB$ has a complexity $\mathcal{O}(r d^2)$. Is there any faster algorithm (taking into account $r \ll d$)?
If this question does not belong here please let me know! (If so, also where can I post it.)
Thanks!
Edit: just for reference, typical values are $d \in [200, 1700]$ and $r \in [3, 15]$.
The answer may depend on how "tall" your matrix is; if $r > O(d^{0.7})$ (not a tight bound), then a naive application of the Coppersmith-Winograd algorithm gets the answer faster than the naive algorithm for part 1; for part 2, the restriction is $r > O(d^{0.4})$ (both if I've done my arithmetic right). https://en.wikipedia.org/wiki/Coppersmith%E2%80%93Winograd_algorithm
Thanks. For what I've read about Coppersmith-Winograd algorithm (not much to be honest), it seems to be quite impractical. This is something I'll implement as a part of a bigger algorithm. As a reference, I typically have $d \in [200, 1700]$ and $r \in [3, 15]$. Using $r \ll d$ in the question might have been a slight abuse (maybe?). My biggest concern is for question 1, can I go from $r^2d$ to $rd$? Maybe there's a simple way of verifying that's not possible.
If $r$ is at most $15$ in the instances you're dealing with, the difference between $O(r^2 d)$ and $O(r d)$ may not be of much practical importance: a $O(r^2 d)$ algorithm might outperform a $O(r d)$ algorithm, depending on details of implementation.
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2025-03-21T14:48:30.964619
| 2020-05-15T01:22:22 |
360384
|
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|
Stack Exchange
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Unknown work of Nöbeling on topological/Hausdorff dimension
Let $\mathcal{H}^n$ denote the Hausdorff measure, $\dim_H X$ the Hausdorff dimension, and $\dim X$ the topological dimension of $X$.
A well known result of
Szpilrajn (He changed his name to Marczewski while hiding from Nazi persecution) proved in [S] asserts that
if $\mathcal{H}^{n+1}(X)=0$, then the topological dimension of $X$ is at most $n$.
Szpilrajn's proof is reproduced in [Theorem 7.3, HW] and [Theorem 8.15, H].
Szpilrajn however, mentions in [S] that his argument is based on Nöbeling's proof of a weaker result that topological dimension is bounded from above by the Hausdorff dimension of a metric space. However, he did not provide any reference to Nöbeling's work.
There is also no reference to Nöbeling's work in the book by Hurewicz and Wallman.
Question. Does anybody know the reference to the original work of Nöbeling?
[H] J. Heinonen,
Lectures on analysis on metric spaces. Universitext. Springer-Verlag, New York, 2001.
[HW] W. Hurewicz, H. Wallman, Dimension Theory. Princeton Mathematical Series, v. 4. Princeton University Press, Princeton, N. J., 1941.
[S] E. Szpilrajn, La dimension et la mesure, Fund. Math. 28 (1937), 81--89.
@FrancoisZiegler yes, Szpilrajn references this paper in the introduction to [S], and the result quoted by OP is the first theorem in this paper.
The reference given by Francois Ziegler is: Nöbeling, G.
Hausdorffsche und mengentheoretische Dimension. (German) JFM 57.0749.02
Ergebnisse math. Kolloquium Wien 3, 24-25 (1931).
@FrancoisZiegler your second "reprint" link is a Google books link, which in my case only yields the 1st page (the second is referred as "not available" — I guess availability means Google decides when and to whom it makes it available)
So, the sought for paper is:
Nöbeling, G., Hausdorffsche und mengentheoretische Dimension, Ergebnisse math. Kolloquium Wien 3, 24-25 (1931).
And here is a ``translation" (to English and to modern math exposition standards, if such a thing exists.) What is called the set-theoretical dimension is defined inductively: dim of the empty set is $-1$, and we set $\dim M = k$ if $k$ is the least integer with the property that every point of $M$ has arbitrarily small (open) neighborhoods whose dim is $k-1$.
Hausdorff and set-theoretical dimension. By George Nöbeling.
Let $M$ be a subset of the Euclidean $\mathbb{R}^n$. One covers $M$ by finitely or countably (infinitely) many balls $K_j$, with diameters $d_j < \rho$, for $p \leq n$ (any non-negative number) form the sum $ \sum d_j^p $. Let $ L_p(\rho, M)$ be the infimum of such sums for all such coverings. Put
$$
L_p (M) = \lim_{\rho \to 0} L_p (\rho, M) .
$$
Obviously there is exactly one number $p= p (M)$ such that for every $q > p$, $L_q (M) = 0$ and for every number $q <p$, $L_q (M) = \infty$. We call this clearly defined number $p$ the Hausdorff dimension of the set $M$ and claim that
THEOREM: For any set $M$, the Hausdorff dimension is at least equal to the set-theoretical dimension.
PROOF. The proof is done by induction on the set-theoretical dimension $\dim M$. For every $k \in \{0,1,2,\cdots\} $ we prove that if $M$ is a set with $\dim M \geq k$ then $p(M) \geq k$.
For $k=0$ this is obvious thanks to $ p (M) \geq 0.$ Suppose the claim is true for $k$. We must thus prove that if $M$ is any set with $\dim M \geq (k + 1)$, then $p(M) \geq k+1$.
Since $M$ is at least $(k+1)$-dimensional, there exit a point $P$ of $M$ and a number $r_0$, such that for every $(n-1)$-dimensional sphere $S_r$, with the radius $r \leq r_0$ the intersection $M \cap S_r$ is at least $k$-dimensional. [Otherwise, every point of $M$ would have arbitrarily small open neighborhoods whose boundaries had dim $k-1$ or less, and so by definition $M$ would have dimension less than or equal to $k$.], and therefore according to the induction hypothesis that
$$
\forall r \leq r_0, \; p (M \cap S_r) \geq k \, .
$$
For each $i \in \mathbb{N}$, let $\{K_{ij}\}_j$ be a covering of the set $M$ by spheres of diameter $d_{ij} < \frac{1}{i}$. For a number $q < k +1$ and an $0 < x \leq r_0$ we set
$$
f_{ij} (x) =
\begin{cases}
d_{ij}^{q-1} & \text{if $S_x \cap K_{ij} \neq \emptyset $ ,}\\
0 & \text{Otherwise.}\\
\end{cases}
$$
We also set
$$
s_i (x) = \sum f_{ij} (x).
$$
Obviously,
$$
s_i(x) \geq L_{q-1}(1/i,S_x \cap M) \, .
$$
Since $ p (M \cap S_r) \geq k > q-1 $, it follows from the induction hypothesis that
$$
\forall x \in (0,r_0], \; \lim_{i \to \infty} s_i(x) = \infty \, .
$$
Thus,
$$
\sum_j \int_0^{r_0} f_{ij}(x) \, dx = \int_0^{r_0} s_i(x) \, dx \xrightarrow{i \to \infty} \infty \, .
$$
Now observe that $f_{ij} (x) = d_{ij}^{q-1}$ for $x$ from an interval whose length is at most $d_{ij}$ -- the diameter of $K_{ij}$ -- and otherwise $f_{ij} (x) = 0.$ Therefore,
$$
\sum_j d_{ij}^{q} = \sum_j \int_0^{d_{ij}} d_{ij}^{q-1} \, dx \geq \sum_j \int_0^{r_0} f_{ij}(x) \, dx \, .
$$
and hence,
$$
\sum_j d_{ij}^{q} \xrightarrow{i \to \infty} \infty.
$$
Since this is true for any coverings $K_{ij}$, we conclude
$$
L_q(M) = \infty \implies p(M) \geq q \, .
$$
Since $q <k + 1$ was arbitrary we have shown that $p (M) \geq k + 1$. This concludes the induction and proves our theorem. $\Box$
It seems to be:
Nöbeling, G., Hausdorffsche und mengentheoretische Dimension, Ergebnisse math. Kolloquium Wien 3, 24-25 (1931). ZBL57.0749.02.
Google shows the first and sporadically the second page (19 lines) from this reprint:
Menger, Karl, Results of a mathematical colloquium, Wien: Springer. ix, 470 S. (1998). ZBL0917.01024.
|
2025-03-21T14:48:30.965014
| 2020-05-15T01:57:21 |
360386
|
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|
Stack Exchange
|
Reverse mathematics of Cousin's lemma
This paper by Normann and Sanders apparently caused a stir in the reverse mathematics community when it came out a couple years ago. It says that Cousin's lemma, which is an extension of the Heine-Borel theorem, requires the full strength of second-order arithmetic (SOA) to prove. It also says this lemma is helpful in mathematically justifying the Feynman path integral. So this is an apparent counterexample to both
a) The reverse mathematics precept that theorems of classical analysis can (usually?) be proved using one of the "Big Five" subsystems of SOA, with the strength of subsystem required forming a useful classification of such theorems; and
b) Solomon Feferman's argument that scientifically useful mathematics can generally be handled by relatively weak axioms, generally not stronger than PA / ACA0.
I don't exactly have a mathematical question about the Normann-Sanders paper, but would like to know if it has impacted the reverse mathematics program, and what its significance is seen as. Could path integrals really require such powerful axioms?
Also, Cousins' lemma is traditionally fairly easily proved using the completeness property of the real numbers. The issue is that the completeness property is a second-order property of the reals (i.e. it uses a set quantifier), and SOA is a first-order theory of the reals, that doesn't have sets of reals. In classical analysis though, the completeness axiom really does refer to sets of reals, and this result shows that converting a completeness-based proof to a first-order proof isn't so easy (I haven't read the paper closely and have no idea right now how to prove Cousin's lemma in SOA). Is that significant?
I can understand that the (second order) induction axiom from the Peano axioms translates naturally to the induction schema in first order PA, making induction proofs work about the same way as before. I'd be interested to know why analysis is identified with SOA instead of something that allows sets of reals (needed for functions anyway), since there's not such a straightforward translation of the completeness axiom. Analysis=SOA goes back a long way, since the Hilbert program aimed to prove CON(SOA) once it was done with the consistency of arithmetic. Reverse mathematics came much later.
Thanks!
That some results don't fit into the Big Five framework is not exactly news; see this and this for example. But examples like this are always interesting. As for what Feferman would say, we can't know for sure, but he might challenge the claim that Cousin's Lemma is indispensable for physics. As I understand it, the "correct" way to formalize the Feynman path integral mathematically is still an open question, so it's still unclear that full SOA is required for path integrals.
Tim Chow's comment seems accurate: BOOT (or RANGE) in my below answer is not stronger than $ACA_0$ (in isolation) and implies HBU. One can formalise mathematics using BOOT and stay at the level of $ACA_0$ as long as one avoids $(\exists^2)$. Due to finite measurement precision, we cannot know whether nature involves discontinuous functions (like $(\exists^2)$), i.e. we can avoid those in applications. However, BOOT is "anti-predicativst" as it (essentially) states the existence of the range of an arbitrary third-order functional, i.e. one explicitly quantifies over N^N in its full glory.
@TimothyChow Given that Feynman's path integral can be made to work with an innocent-looking refinement of the Riemann integral (which nonetheless requires some more logical strength as Normann and Sanders show), I am surprised that Feynman's integrals remain so mysterious and controversial.
@MonroeEskew: Physicists invoke path integrals in many different contexts, and there is no uniform recipe for making all their calculations mathematically rigorous. In specific cases of interest, one usually takes advantage of some special feature to make everything rigorous. But if ad hoc reasoning is permissible, then one can argue that the uses that are indispensable for physics may not really need the full strength of the gauge integral that Normann and Sanders analyze. See here and here for a bit more info.
@TimothyChow It’s weird that these stackexchange discussions you linked do not mention the gauge integral, which seems conceptually much simpler than the other constructions. According to this wikipedia entry, people have proposed that it replace the standard definition in analysis textbooks because it is so simple. https://en.wikipedia.org/wiki/Henstock–Kurzweil_integral
@MonroeEskew : I don't know the area well enough to be sure, but my understanding is that formalizing the physicists' calculations involves a lot more than just giving a rigorous definition of the path integral itself. It may be that the gauge integral gets you off to a good start, but obstacles arise down the road. I found this paper which I admittedly don't understand, but which seems to claim that the gauge integral has some limitations when applied in the way physicists want to apply the Feynman path integral.
As shown by Pat Muldowney (see google), the gauge integral can handle a special case of the Feynman path integral. The infinite dimensional case turns out to be quite subtle, and even then one does not cover all of the applications.
@TimothyChow Measure theory is firmly engrained and will remain so due to inertia. As an argument against the gauge integral, it is often said that the step to the multi-dimensional gauge integral is harder than in the case of the Lebesgue integral, though this seems to be (mostly) a matter of presentation. By contrast, what is great about the gauge integral is Hake's theorem: for any function $f$, if $\lim_{a\rightarrow b} \int_a^c f $ or $\int_b^c f$ exists, then both exist and are equal. This is how physicists (want to) use limits: take limits first and ask questions later/never.
@MonroeEskew The "use the gauge integral" approach is for the path integral in quantum mechanics. I don't think the path integral in quantum mechanics is very controversial, there are many rigorous versions of it, some of which have been around a very long time (e.g. using analytic continuation). The reason it's not used in textbooks so much is that it takes a lot more effort to set up and is not so useful on certain problems (e.g. the hydrogen atom). When people talk about difficulties with rigour and path integrals, they are almost always referring to quantum field theory ...
@MonroeEskew As has been explained better elsewhere on this site, the difficulty there is that the Lagrangian is defined on smooth functions and involves nonlinear terms, while the space the measure is naturally defined on is a space of tempered distributions, which don't allow multiplication (in a rather strong way). Existence results are usually proved by taking a lattice in a box and letting the spacing go to zero and the box containing the lattice fill up $\mathbb{R}^n$, and having a measure converge weakly to a measure on the space of distributions.
@SamSanders I have never found one of these "QWERTY arguments" for why people keep using something to actually be correct, but am prepared to be shown wrong. I think the reason people still use measure theory because it gives a single formulation that can be used for i) Calculus on $\mathbb{R}^n$, ii) Probability theory on spaces of sequences of trials (where mutually singular measures come up very naturally) iii) Haar measure for representation theory and the structure of locally compact groups. iv) The Riesz representation theorem and integral operators in functional analysis ...
@SamSanders As far as I know, the gauge integral doesn't do (ii)-(iv).
@RobertFurber An elementary result in gauge integration is "a function $f$ is Lebesgue integrable IFF $f$ and $|f|$ are gauge integrable. Thus, one can readily recover the Lebesgue integral in this case, and I believe one can recover measure theory in general. Hence, the gauge integral can do all the things you mention (being a generalisation).
Sam Sanders here, one of the authors of the paper you mention. Thanks for the nice words.
I will answer your questions based on my personal opinion.
You write:
[...] would like to know if it has impacted the reverse mathematics (RM) program, and what its significance is seen as. Could path integrals really require such powerful axioms?
First of all, I cannot speak for the RM community. What I can tell you is that we have gotten all sorts of comments, negative and positive, from lots of (senior) people. I believe it is also fair to say that our paper (and related results) shows that coding even e.g. basic analysis in second-order RM does not accurately represent mathematics. There are those that disagree, as one would expect, but I would speculate people in RM have (on average) started working on "less coding intense" topics. Part of the problem is that second-order arithmetic cannot directly deal with functions from $\mathbb{R}$ to $\mathbb{R}$ that are discontinuous, which is discussed in detail in my "splittings and disjunctions" paper (see arxiv/NDJFL).
Secondly, regarding your question on "powerful" axioms. This ties in nicely with recent results Dag Normann and I have. In a nutshell, the usual scale for measuring logical strength (based on comprehension and discontinuous functionals) is not satisfactory for e.g. Cousin's lemma and we need a second scale, based on (classically valid) continuity axioms from Brouwer's intuitionistic mathematics. This new scale is a Platonist's dream: the canonical ECF embedding maps part of this new scale and equivalences to the 'Big Five' and their equivalences. In other words, the Big Five are merely a reflection of a higher-order truth!
Let me first sketch the main results in our paper pertaining to Cousin's lemma.
We work in the language of higher-order arithmetic. This means that all the below should be interpreted in Kohlenbach's higher-order RM and Kleene's higher-order computability (S1-S9). Not much specific knowledge of these frameworks is needed, however.
Let HBU be Cousin's lemma for [0,1], i.e. for any $\Psi:[0,1] \rightarrow \mathbb{R}^+$, there are $y_0, ..., y_k \in [0,1]$ such that $\cup_{i\leq k} B(y_i, \Psi(y_i))$ covers $[0,1]$.
In other words, the reals $y_0, ..., y_k \in [0,1]$ provide a finite sub-covering of the uncountable covering $\cup_{x\in [0,1]}B(x,\Psi(x))$.
Let $Z_2$ be second-order arithmetic with language $L_2$. The systems $Z_2^\omega$ and $Z_2^\Omega$ are known conservative extensions of $Z_2$. Then the former cannot prove HBU while the latter can. This is what we mean by "a proof of HBU requires full second-order arithmetic", as $Z_2^\omega$ cannot prove HBU, but $Z_2^\Omega$ can.
Clearly, HBU is a statement in the language of third-order arithmetic. The system $Z_2^\omega$ is also third-order in nature: it includes, for any $k \geq 1$, a third-order functional $S_k$ that decides $\Pi_k^1$-formulas from $L_2$ in Kleene normal form (see the work of Sieg & Feferman). The system $Z_2^\Omega$ is fourth-order, as it is based on Kleene's (comprehension) quantifier $\exists^3$ (see the work of Kleene on higher-order recursion theory). Note in particular that HBU is provable in ZF: countable choice is not needed.
There are many statements that exhibit the same (or similar) behaviour as HBU. I refer to e.g. our paper on Pincherle's theorem (APAL) and open sets (JLC), where whole lists can be found, as well as the original paper. Convergence theorems for nets in $[0,1]$ indexed by Baire space also behave like HBU (see my 2019 CiE and WolliC papers).
Now that we have established the results, let me explain what these results mean.
Indeed, there is an apparent contradiction here: on one hand, HBU should be intuitively weak, but we need absurdly strong comprehension axioms to prove HBU. This feeling you express in your posting, I believe.
The fundamental problem is that we are comparing apples and oranges as follows:
The aforementioned comprehension functionals $\exists^3$ and $S_k$ are discontinuous (in the usual sense of mathematics). By contrast, HBU does not imply the existence of a discontinuous function (say on $\mathbb{R}$ or Baire space). Let us call a (third-order) theorem 'normal' if it implies the existence of a discontinuous function on $\mathbb{R}$, and 'non-normal' otherwise.
It is an empirical observation that there are many non-normal theorems (like HBU) that cannot be proved in $Z_2^\omega$, but can be proved in $Z_2^\Omega$. In other words, the usual 'normal' scale based on comprehension functionals is not a good scale for analysing the strength of non-normal theorems.
In a nutshell: normal theorems = apples and non-normal theorems = oranges.
An obvious follow-up question is:
What is a good scale for analysing non-normal theorems?
As explored in the following paper (see Section 5), the neighbourhood function principle NFP provides the right scale.
https://arxiv.org/abs/1908.05676
NFP is a classically valid continuity axiom from Brouwer's intuitionistic mathematics.
Fragments of NFP are equivalent to e.g. HBU and other milestone non-normal theorems, like the monotone convergence theorem for nets in [0,1] (called $\textsf{MCT}_{\textsf{net}}$ in the above paper). Note that NFP was introduced under a different name by Troelstra-Kreisel, and is mentioned in Troelstra & van Dalen.
Finally, the Kleene-Kreisel 'ECF' embedding is the canonical embedding of higher-order into second-order arithmetic. It maps third-order and higher objects to second-order associates/RM-codes, which reflects the 'coding practise' of RM.
What is more, the ECF embedding maps equivalences involving HBU to equivalences involving WKL, as follows:
HBU $\leftrightarrow$ Dini's theorem for nets (indexed by Baire space).
is mapped by ECF to
HBC $\leftrightarrow$ Dini's (usual) theorem (for sequences),
where HBC is the Heine-Borel theorem for countable coverings of intervals.
Another example is the following:
RANGE $\leftrightarrow$ Monotone convergence theorem for nets (indexed by Baire space)
is mapped by ECF to
range $\leftrightarrow$ Monotone convergence theorem (for sequences),
where RANGE states that the range of a third-order function exists, while range
states that the range of a (second-order) function exists; it is well-known that range $\leftrightarrow \textsf{ACA}_0$.
In general, the Big Five equivalences are a reflection of higher equivalences under ECF as in the following picture:
Since ECF is a lossy translation, this pictures resembles -in my not so humble opinion- the allegory of the cave by Plato. This observation is inspired by Steve Simpson's writings on Aristotle that can be found in the (RM) literature.
I would like to finish this answer with a history lesson: I have heard wildly inaccurate claims from very smart (RM) people about the history of mathematics. These claims are often used to justify the coding practise of RM. So let us set the historical record straight, while it still matters.
1) Hilbert and Bernays did NOT introduce second-order arithmetic. In the "Grundlagen der Mathematik", they formalise a bunch of math in a logical system $H$ (see esp. Supplement IV). This system involves third-order parameters in an essential way, as has been observed before by e.g. Sieg (see e.g. his book on Hilbert's program). Hilbert-Bernays vaguely sketch how one could perhaps do the same formalisation with less.
I was told that Kreisel then introduced second-order arithmetic based on the above.
2) Riemann's Habilschrift established discontinuous functions as part of the mathematical mainstream around 1850. Thus, discontinuous functions definitely predate set theory.
3) The modern concept of function was already formulated by Dedekind and Lobasjevski in the 1830ies. (This view is not without its critics)
4) The gauge integral is more general than the Lebesgue integral. The main theorems (Hake and FTC) of the former in particular apply to any (possibly non-measurable) function. In this way, the development of the gauge integral does not need measure theory, but can instead be done similarly to the Riemann integral. To study the gauge integral restricted to measurable functions goes against its generalist/historical spirit, to say the least.
5) There are a number of formalism to give meaning to Feynman's path integral. The gauge integral is heralded as one of the few that can avoid 'imaginary time', a desirable feature from the pov of physics. This is briefly discussed on page 20 here:
https://arxiv.org/abs/1711.08939
References are provided, of course.
Thanks! For those following along, "Splitting and Disjunctions" is here and this paper also looks relevant. I'm still reading them.
The main theme of the "Splittings and ..." paper is the following: on one hand there are only few results in second-order Reverse Mathematics where one can prove $A\leftrightarrow [B \wedge C]$ (a splitting), where $B$ and $C$ are independent and natural. There are only very few results in second-order Reverse Mathematics where one can prove $D\leftrightarrow [E \vee F]$ (a disjunction), where $E$ and $F$ are independent and natural. On the other hand, I identify a plethora of such splittings and disjunctions in higher-order Reverse Mathematics. I also discuss the cause of this observation.
Sam's answer is, of course, the definitive one. For curiosity's sake, I'll mention that Rod Downey, Noam Greenberg and I have recently been looking at Cousin's lemma for restricted (i.e. cardinality $\mathfrak{c}$) classes of functions.
We have a proof of Cousin's lemma in $\mathsf{ATR}_0 + \Delta^1_2$-induction that works for "any function", to the extent that "any function" can be formalised in this system. This effectively gives an upper bound for Cousin's lemma for functions appropriately definable in second-order arithmetic. This doesn't contradict what Sam said above, since you need third-order arithmetic to talk about all functions, and then $\mathsf{ATR}_0 + \Delta^1_2$-induction is not enough.
We've specifically focused on continuous functions, Baire functions and Borel functions. So far, we've proved:
Cousin's lemma for continuous functions is equivalent to $\mathsf{WKL}_0$;
Cousin's lemma for Baire 1 functions is equivalent to $\mathsf{ACA}_0$;
Cousin's lemma for Baire $2$ functions (and hence for Borel functions) implies $\mathsf{ATR}_0$.
These results are in my thesis as well as a joint paper we've written.
To answer the OP's questions, Dag Normann and Sam Sanders' results do contradict (a). But to me, this is not such a surprise. As Sam notes, arbitrary discontinuous functions can't be formalised in second-order arithmetic. So, to the extent that classical analysis deals with discontinuous functions, we can't even formalise it in second-order arithmetic, let alone prove its theorems in, say, $\Pi^1_1$-$\mathsf{CA}_0$. This just shows that SOA is sufficient for most, but not all, of mathematics (it can't really handle topology, set theory, etc).
As for (b), it also contradicts this, if you assume that it is really necessary to consider Cousin's lemma for arbitrary gauges. As Russell A. Gordon shows in his book, "The Integrals of Lebesgue, Denjoy, Perron, and Henstock", it is enough to consider measurable gauges, and maybe this can be further restricted to Borel gauges (I don't know). In that case, maybe $\mathsf{ATR}_0 + \Delta^1_2$-induction is enough for the physicists.
Updated to add new results 2021/05/10
I should add that Cousin's lemma for measurable gauges already has the same power/strength as the general case. There are theorems about "how complex" gauges are relative to the function being integrated.
Moreover, you claim "This just shows that SOA is sufficient for most, but not all, of mathematics (topology, set theory, etc)." is also incorrect. As shown by Dag Normann and myself (https://arxiv.org/abs/2007.07560), basic theorems about Riemann integration (Arzela's convergence theorem, 1885) are extremely hard to prove when not formulated with codes for Riemann integrable functions. Coding in SOA only works reliably for continuous functions, but no longer so for continuous almost everywhere functions.
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2025-03-21T14:48:30.966431
| 2020-05-15T02:55:56 |
360387
|
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|
Stack Exchange
|
A question on multiplicity of complex polynomial
This is not a research level question. But due to some reason I can't ask this question on Math Stack Exchange. So, I am asking this question here.
By definition we know that we can measure the multiplicity of a root of a function $f(z)$ as follow.
If the a root of some function $f(z)$ is $\alpha_i$, then the multiplicity of that root is $a_i$ for which $\lim \limits_{x \to \alpha_i} |\frac{f(z)}{(z-\alpha_i)^{a_i}}|$ has a finite value.
So, following this we can see that all the roots ($(2n+1)i\pi$,
) of the polynomial $e^z+1$ has multiplicity $1$.
So, $$e^z+1= 2\prod_{n=-\infty}^{\infty} \left(1-\frac{z}{(2n+1)i\pi}\right)$$.....(1)
Which further can be written as
$$e^z+1= 2\prod_{n=0}^{\infty} \left(1+\frac{z^2}{((2n+1)i\pi)^2}\right)$$
But this only consists of $z^{2k}$ not the odd powers.
If we directly check this from (1) we see the coefficients of $z^3$, we see they are $0$ which can be directly concluded from (1).
But actually $e^z+1=2+z+\frac{z^2}{2!}+....$. So, what am I doing wrong?
...$e^z + 1$ is not a polynomial. Some food for further thought: what are the zeroes of $e^z$, and can it be represented in a similar way?
I got my answer. And voted to delete the post.
Your product on the right does not converge for any $z \neq 0.$ Never fear, the Weierstrass factorization theorem is your friend.
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2025-03-21T14:48:30.966569
| 2020-05-15T03:04:56 |
360388
|
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|
Stack Exchange
|
Half integral weight modular forms that reduce to a nonzero constant modulo a given prime
Let $B_k = \frac{N_k}{D_k}$ be the reduced numerator and denominator of the $k$-th Bernoulli number. For a given prime $p>2$, the (unconventionally normalized) Eisenstein series $E_{p-1}(z) = N_{p-1} − 2(p-1)D_{p-1}\sum_{n=1}^{\infty} \sigma_{p-2}(n)q^n$ has the property that its Fourier coefficients are all integers, and the constant coefficient is the only coefficient not divisible by $p$.
Given a prime $p$, I am interested in the existence of a modular form $f_p$ with analogous properties but whose weight is half integral. In other words, $f_p$ should have integral Fourier coefficients, the constant coefficient should be prime to $p$, and all other coefficients should be divisible by $p$.
The exact weight and level are unconstrained, but preferably should be kept small enough to compute with for moderately small values of $p$.
The only example I know of is the theta function $\theta(z) = 1 + 2q + 2q^4 +...$ for $p=2$.
Note once you have one such form you get infinitely many by adding appropriate cusp forms. You can also construct examples (at least in weight 2) from certain Eisenstein congruences. For half-integral weight, have you looked at Cohen-Eisenstein series? My guess is that you can either use these directly, or possibly combined with a congruence.
@Kimball I had not heard of Cohen-Eisenstein series before, thank you for bringing them to my attention. From the definition it's not clear to me how they might be useful. Would you mind elaborating a bit on your thoughts?
By using Rankin-Cohen brackets you can build many other examples of half-integral modular forms of higher weight which are 1 mod 2 by bracketing with the theta function you have. Similarly you can bracket $E_{p-1}$ with a half-integral modular form to get something that is constant modulo p.
Cohen-Eisenstein series are a half-integral weight version of the classical even weight Eisenstein series. See Cohen's 1975 Math Ann paper, or search "Cohen-Eisenstein" to find some more recent references.
@GTA Naively it seems to me as though the only way to get a form this way which is constant mod $p$ would be to bracket $E_{p-1}$ with a half integral weight form that is already constant mod $p$, which isn't helpful. I only need a single form for each prime.
If such a form does exist, then its level must be a multiple of $p$.
If $f = \sum a_{n} q^{n}$ is a half-integer weight modular form with integer coefficients with $a_{i} \equiv 0 \pmod{p}$ for all $i > 0$ and $\gcd(a_{0},p) = 1$, then by multiplying $f$ by an integer relatively prime to $p$, one can assume $a_{0} \equiv 1 \pmod{p}$. If $k+1/2$ is the weight of the modular form, then $f^{2}$ would be an integer weight modular form for $\Gamma_{1}(N)$ (for some $N$) with the property that $f^{2} \equiv 1 \pmod{p}$. The form $f^{2}$ would have weight $2k+1$.
There's a well-developed theory of integer-weight modular forms modulo $p$ for $\Gamma_{1}(N)$ with $p \nmid N$ and the best reference is the paper ``A tameness criterion for Galois representations associated to modular forms (mod p)'' by Benedict Gross (Duke Math Journal, 1990, pages 445-517). In this paper it is proven that if $f$ and $g$ are modular forms for $\Gamma_{1}(N)$ with $f \equiv g \pmod{p}$, then the weights of $f$ and $g$ must be congruent modulo $p-1$. This creates a contradiction in the situation above because we cannot have $2k+1 \equiv 0 \pmod{p-1}$, since $p-1$ is even and $2k+1$ is odd.
Unfortunately, I do not currently have access to the paper of Gross (the link above is behind a paywall), and there may be some caveats in the theorem I've stated above (EDIT: The OP has confirmed that one needs the hypothesis that $p \nmid N$.). In case it's useful, here's another MO post asking a question about modular forms modulo $p$ in level $> 1$. The $N = 4$ case of the theorem I quoted above was independently proven by Tupan. (See the paper "Congruences for $\Gamma_{1}(4)$ modular forms of half-integral weight" in the Ramanujan Journal in 2006.) In particular, level $4$ Cohen-Eisenstein series and Rankin-Cohen brackets of level $4$ forms with $E_{p-1}$ cannot produce examples of half-integer weight forms $\equiv 1 \pmod{p}$. (Thanks to the OP for looking up the paper and pointing out the necessity of the hypothesis that $p \nmid N$.)
If $p \geq 5$ is prime, there is a form of weight $\frac{p-1}{2}$ for $\Gamma_{1}(p)$ that is $\equiv 1 \pmod{p}$, namely
$\frac{\eta^{p}(z)}{\eta(pz)} = \prod_{n=1}^{\infty} \frac{(1-q^{n})^{p}}{1-q^{pn}}$. One can search for examples and come tantalizingly close. For example, there's a weight $9/2$ form for $\Gamma_{0}(20)$ whose Fourier expansion beings
$$
F(z) = 1 - 20q^{13} - 40q^{14} + 90q^{16} - 40q^{17} - 40q^{18} + \cdots.
$$
The coefficient of $q^{n}$ is a multiple of $5$ for $1 \leq n < 100$, but the coefficient of $q^{100}$ is $15292$. (If $a_{n} \equiv 0 \pmod{5}$ for $1 \leq n \leq 108$, Sturm's theorem would force it to always be true.)
I'm not sure if you were misled by the title (which isn't clear), but the OP seems to be looking for forms with a_n = 0 mod p for all n > 0 but a_0 not 0 mod p.
Thank you very much for the answer! In the Gross paper the result you state is proven in proposition $4.9$, by showing that the kernel of reduction mod $p$ is the principal ideal $(1-A)$, where $A$ is the Hasse invariant. The author does indeed assume $p$ does not divide $N$ generically throughout section $4$, but determining whether or not this assumption is actually needed for proposition $4.9$ is well beyond my capabilities.
Could you explain how you found that example form? I tried to write some Sage code to search for these forms by generating a basis and then reducing mod p and solving a matrix equation, but when I ran it on weight $9/2$, $\Gamma_1(80)$ (Sage requires $16|N$ for some reason), it didn't find the form you mentioned, so I must be doing something wrong.
I used Magma with can compute the spaces $M_{k/2}(\Gamma_{0}(N),\chi)$ and $S_{k/2}(\Gamma_{0}(N),\chi)$. It appears that Sage actually computes bases for spaces of cusp forms. There's a trick you can use to verify the presence of this form - the function $g(z) = \eta^{2}(4z) \eta^{2}(20z) \in S_{2}(\Gamma_{0}(80))$. You can then look in the space of weight $13/2$ forms for $\Gamma_{1}(80)$ (with trivial character) for a form that is congruent to $g$ (up to $q^{102}$) modulo $5$.
|
2025-03-21T14:48:30.967051
| 2020-05-15T03:18:50 |
360389
|
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|
Stack Exchange
|
Exponential or sub-exponential ergodicity?
Consider the one-dimensional stochastic differential equation $$d X(t) = -sgn(X(t))dt + dW(t),$$ where $W$ is a standard Brownian motion, and $sgn(x) = 1$ if $x > 0$ and $-1$ if $x\le 0$. It can be shown that this process possesses an invariant distribution $\pi$ with density $m(x) = C e^{-|x|}$ for some normalizing constant $C> 0$. Denote by $P_t(x,\cdot)$ the transition probabilities of the process.
My question is: does $P_t(x,\cdot)$ converge to $\pi$ in total variation norm at exponential rate or strictly sub-exponential rate? Thanks.
Answer given is already nice; alternatively, exponential convergence in the total variation distance can be verified by an application of Harris's Theorem with Foster-Lyapunov function $\mathcal{V}(x)=\exp(\varphi_{\epsilon}(|x|^2))$ where $\varphi_{\epsilon}(\cdot)$ is a $C^2$ approximation to the square root, $$
\varphi_{\epsilon}(x) = \begin{cases} \sqrt{x} & x \ge \epsilon^2 \;, \\
-\frac{1}{8 \epsilon^3} x^2 + \frac{3}{4 \epsilon} x + \frac{3 \epsilon}{8} & x \le \epsilon^2 \;. \end{cases}
$$
For a $C^2$ function $f$, let $$\mathscr{L} f(x) = -sgn(x) \frac{\partial f}{\partial x}(x) + \frac{1}{2} \frac{\partial^2 f}{\partial x^2}(x)$$ denote the action of the infinitesimal generator of the process $X$ on the function $f$. Then for $x \ge \epsilon^2$ $$
\mathscr{L} \mathcal{V}(x) = - \frac{1}{2} \mathcal{V}(x)
$$ and similarly for $x \le -\epsilon^2$. Thus, globally, $$
\mathscr{L} \mathcal{V}(x) \le - \frac{1}{2} \mathcal{V}(x) + K \;, \quad \text{where $K = \sup\left\{ \mathscr{L} \mathcal{V}(x) + \frac{1}{2} \mathcal{V}(x) : |x| \le \epsilon^2 \right\}$}\;.
$$ Combining this with a minorization condition (which follows from ellipticity of the SDE), Harris's theorem can now be invoked to obtain the required exponential convergence in the total variation distance. For a reference see, e.g., Theorem 2.5 of the following paper.
Mattingly, J. C.; Stuart, A. M.; Higham, D. J., Ergodicity for SDEs and approximations: locally Lipschitz vector fields and degenerate noise., Stochastic Processes Appl. 101, No. 2, 185-232 (2002). ZBL1075.60072.
Here are some additional references.
Meyn, Sean; Tweedie, Richard L., Markov chains and stochastic stability. Prologue by Peter W. Glynn., Cambridge Mathematical Library. Cambridge: Cambridge University Press (ISBN 978-0-521-73182-9/pbk; 978-0-511-62663-0/ebook). xviii, 594 p. (2009). ZBL1165.60001.
Hairer, Martin; Mattingly, Jonathan C., Yet another look at Harris’ ergodic theorem for Markov chains, Dalang, Robert C. (ed.) et al., Seminar on stochastic analysis, random fields and applications VI. Centro Stefano Franscini, Ascona, Italy, May 19–23, 2008. Basel: Birkhäuser (ISBN 978-3-0348-0020-4/pbk; 978-3-0348-0021-1/ebook). Progress in Probability 63, 109-117 (2011). ZBL1248.60082.
Thanks a lot. I should have thought about the Lyapunov function $V(x) = exp(|x|)$ (for $x$ away from the origin) before!
The rate is exponential. First, let me show that $|X(t)|$ (which is a reflected BM with constant drift towards the origin) has exponential rate of convergence. I do it by a coupling argument: you run two independent copies of the process, one started from $x$ and another from the stationary distribution, until they first collide, and therefrom run them together. The TV distance between their distributions at time $t$ is bounded by twice the probability that they do not collide by that time. But, if by time $t$ each process has hit the origin, they must have collided. Hence, it suffices to show that for each process, the probability that it does not hit the origin before time $t$ decays exponentially. This is simply a large deviation estimate for BM.
It's not hard to modify this argument so that it works for $X(t)$. For example, I can run the two processes independently until their absolute values coincide, then run them as mirror reflections of each other until they hit the origin, and therefrom run them together.
Alternatively, you could probably compute the transition density for $|X(t)|$ explicitly, from Kolmogorov's equation. The spectrum of the operator in the RHS (once you divide $P_t(x,y)$ by $e^{-y}$ to bring it into symmetric form) has eigenvalue $0$ and a continuous part $(-\infty,-1]$, hence there's a spectral gap. But the reflecting boundary conditions are somewhat subtle, so I am not sure how to cleanly write the relevant self-adjoint extension...
Many thanks! This is a very elegant proof.
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2025-03-21T14:48:30.967479
| 2020-05-15T07:13:38 |
360395
|
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|
Stack Exchange
|
Extension Sobolev functions across of lower dimensional subset
This question may be well-known to experts, but I am trying to get myself a rigorous proof. Consider open set $\Omega=B^n_1(0)\setminus B_1^k(0)$ in $\mathbb{R}^n$. If function $u$ is in $H^1(\Omega)$, can we make it in $H^1(B_1^n(0))$ by arbitrarily adding value on $B_1^k(0)$. I think this problem can be similarly stated in a more general setting instead of balls.
What I have tried: take any $f\in C_c^\infty(B_1^n)$ and define a cutoff function $\varphi_\epsilon$ that is 1 when distance r from $B_1^k$ is bigger than $\epsilon$ and decreasing in the $\epsilon$-neighborhood with derivative smaller than $1/\epsilon$, then
$$-\int_{B^n_1} (\varphi_\epsilon f)_iu=\int_{B^n_1}(\varphi_\epsilon f)u_i.$$
Then I want to use the DCT to take the limit as $\epsilon\rightarrow 0$
Two questions: (1) On the RHS: it seems that we need $u_i$ is bounded, but we don't have this. Maybe we can use that $\int_\Omega u_i^2$ is bounded, $u_i$ is integrable by holder inequality? However, it is not even defined on $B_1^k$, can we just assume it is defined because we are using integration here?
(2) The LHS becomes $-\int_{B_1^n}(\varphi_\epsilon)_ifu-\int_{B^n_1}\varphi_\epsilon f_iu$. In terms of the second term, I have the same question as above. Regarding the first term, this is the main question I have:
$$|\int_{B_1^n}(\varphi_\epsilon)_ifu|\leq \|u\|_{L^2}\frac{C}{\epsilon^2}\times Volume(\epsilon-Neighborhood)\leq C \frac{1}{\epsilon^2}\epsilon^{n-k}=C\epsilon^{n-k-2}$$
where I used Holder inequality in the first inequality. That means codimension $n-k$ must be strictly bigger than 2 so that we can make it vanish. Does that sound correct? Because in my intuition codimension 2 should be fine. I think if $u$ is bounded in the neighborhood of that subset, it will give what I want, however, we don't have that, right?
Maybe more general question is: what is the difference between $H^1(K\setminus C)$ and $H^1(K)$ if $K$ is a bounded domain and $C$ is a subset.
Maybe what I have done doesn't make any sense. If someone can answer my questions or direct me to some related references, it will be very much appreciated.
|
2025-03-21T14:48:30.967638
| 2020-05-15T07:42:34 |
360397
|
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Stack Exchange
|
Proof of universality of Toeplitz algebra
It is well-known that the Toeplitz algebra $\mathcal{T}$ (that I view as concrete subalgebra of $\mathbb{B}(\ell^2(\mathbb{N})$) is the universal algebra generated by an isometry, that is, for any $C^*$-algebra $A$ and an isometry $v \in A$, there exists a unique algebra $\mathcal{T} \rightarrow A$ such that $S \mapsto v$, where $S$ denotes the shift operator in $\mathcal{T}$.
The proof of this fact that I know is done using the Wold decomposition (explained e.g. in Wegge-Olsen, 3.F). He then writes "there are probably easier ways (of which I'm ignorant) to do this". Is anyone here not ignorant of easier ways? In particular, is there a "more $C^*$-algebraic" proof?
Turning things around, we can just define $\mathcal{T}$ as the universal $C^*$-envelope of the $*$-algebra generated freely by $S$ and $S^*$. This then has the universal property by abstract nonsense, but the proof I know that this algebra is isomorphic to the "concrete one" again uses Wold decomposition.
The proof for Cuntz-Toeplitz algebras, i.e. algebras generated by some number of isometries with orthogonal ranges, is pretty $C^{\ast}$-algebraic. Let $s$ be the universal isometry and let $\mathcal{T}_{u}$ be the universal $C^{\ast}$-algebra generated by an isometry. We want to show that the $\ast$-homomorphism $\Phi: \mathcal{T}_{u} \to \mathcal{T}$ defined by $s \mapsto S$ is injective. There are circle actions on both of these actions, given just by multiplication by a scalar. They commute with $\Phi$, so there is an induced map between the fixed-point subalgebras. Moreover, there are faithful conditional expectations onto these fixed-point subalgebras, so if the restriction of $\Phi$ is injective, then $\Phi$ itself is injective as well. The fixed point subalgebras are spanned by the elements $t_n := s^{n} (s^{\ast})^{n}$ ($T_n:= S^{n} (S^{\ast})^{n}$). The span of first $m$ $t_n$'s is actually a finite dimensional subalgebra and just by checking the dimension you may prove that the restriction of $\Phi$ to such a subalgebra is injective, hence isometric. It follows easily that $\Phi$ restricted to the fixed-point subalgebra is isometric.
I don't think this is really an easier way, but it definitely works in more general situations.
|
2025-03-21T14:48:30.967823
| 2020-05-15T07:50:18 |
360399
|
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|
Stack Exchange
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Initial value problems on manifolds around submanifolds (reference)
I am looking for a reference on initial value problems formulated on smooth manifolds with initial conditions on submanifolds. More precisely, let $X$ be a smooth manifold and $Y\subset X$ a embedded submanifold. I am interesting in solutions of PDE's (any kind) formulated on $X$ (or a arbitrary small neighborhood of $Y$ in $X$) with initial conditions on $Y$. The solutions should also exist on an arbitrary small neighborhood of $Y$ (maybe smaller that the neighborhood on which the PDE is formulated). I am not interested in a particular PDE, but rather I would like to know if there is a general theory on this kind of problems.
Cheers,
Ben
The theory is usually not general, but depends on the properties of specific PDEs. The Einstein equations is an example that is often treated on manifolds (rather than just on open domains in $\mathbb{R}^n$): The Cauchy Problem in General Relativity (EMS, 2009).
Do you assume $Y$ to have codimension 1? If so, the Cauchy-Kovalevski theorem is the most general theorem if you don’t assume the PDE is hyperbolic or parabolic.
Not necessary. $Y$ could be of any codimension.
Can I advertise my lecture notes? https://arxiv.org/abs/1706.09697
I think you might want to look into the theory of exterior differential systems. Unfortunately, the theory is only in good shape for real-analytic PDE systems, though there are some situations in which some version of a smooth theory is known to have positive and useful results.
Typically, the theory of EDS is used for systems whose initial value problems are 'naturally' posed on higher codimension submanifolds, and this usually means that the systems involved are overdetermined, at least in their given formulation.
For a simple example, the Cauchy-Riemann equations for a holomorphic function $h$ of two complex variables are overdetermined: They are four first-order PDE for 2 functions of four variables, and the 'natural' initial value problem is to specify the real and imaginary parts of $h$ on a 'totally real' 2-dimensional surface $R\subset\mathbb{C}^2$. In order for this IVP to have a good local solution, you typically need $R$ to be a real-analytic surface and the specified real and imaginary parts of $h$ to be real-analytic functions on $R$.
For another example, consider the so-called "Björling problem": Given a space curve $c\subset\mathbb{R}^3$ and a unit normal vector field $n$ along $c$, find a minimal surface $S\subset\mathbb{R}^3$ that contains $c$ and has $n$ as its unit normal vector field along $c$. This can be formulated as a first order PDE system for a surface in $X^5 = \mathbb{R}^3\times S^2$ that contains the curve $\bigl(c(t),n(t)\bigr)$ in $X$ (and where this 'initial condition' already is supposed to satisfy the ODE $n(t)\cdot c'(t) = 0$). Again, you only get a good existence theory when $c$ and $n$ are real-analytic, but then you do get existence and local uniqueness.
There are many applications of exterior differential systems in differential geometry. In fact, the theory was initiated by Élie Cartan (and developed further by Erich Kähler) specifically for applications to geometric problems.
There is a certain amount of 'overhead' needed to develop in the theory, so if you are looking for a tool to use on a particular problem, you might want to consult with someone who already knows the theory to determine whether EDS is likely to be of use to you for the class of problems you have in mind before you invest in learning the subject (unless, of course, you are just curious about the theory in general).
|
2025-03-21T14:48:30.968114
| 2020-05-15T07:57:52 |
360400
|
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Stack Exchange
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Injective maps and direct limits
I have the following question. Assume you have a category $\mathcal{C}$ such that direct limits exists. Let $(C_n)_{n\in\mathbb{N}}$ be a sequence of objects in $\mathcal{C}$ and consider the following diagram
$$
C_1\rightarrow C_2\rightarrow C_3\rightarrow\cdots
$$
and suppose that all maps $f_i:C_i\to C_{i+1}$ are injective. Let us denote by $C$ the direct limit of this diagram. May we conclude that the maps $g_i:C_i\to C$ are injective as well? Thank you very much in advance.
What does it mean that an arrow is "injective"? why do you call arrow "maps"?
Okay let me be more specific. I am actually considering categories like Sets, Semigroups, etc. where injective makes sense.
@YCor, I understand the question about 'injective', but surely it is just a matter of taste whether to refer to arrows, maps, or morphisms in a category. (After all, I've only ever seen $\operatorname{Mor}(C, C')$, never $\operatorname{Arr}(C, C')$.)
@LSpice yes, but in this precise case it doesn't apply, as the question makes sense only for maps in the sense "functions between sets" and hasn't seriously thought about what the right setting for the question is. (I should add that the fact that people use "map" for arrows in arbitrary categories might be misleading to students/beginners, as reflected by this very question.)
Suppose your category is generated under colimits by compact objects (this is the case in algebraic categories such as sets, semigroups etc.).
Then if you replace "injective" by "monomorphism" (which again, is the same in algebraic categories such as sets, semigroups etc. ) any sequence of monomorphisms satisfies the desired property.
Indeed, we wish to prove that for any $X$, $\hom(X,C_i)\to \hom(X,C)$ is injective. Writing $X$ as a colimit of compact objects, and using the fact that limits preserve injections in $\mathbf{Set}$, we may assume that $X$ is compact.
But if $X$ is compact, $\hom(X,C) = \varinjlim_j\hom(X,C_j)$, and so if $\alpha,\beta : X\to C_i$ become equal in $C$, they must become equal at some stage : there is $j\geq i$ such that $\alpha, \beta : X\to C_i\to C_j$ are equal, and therefore, since $C_i\to C_j$ is a monomorphism, $\alpha = \beta$.
Another situation in which the result is true is that of an Abelian category with enough injectives. Here I take "injective map" to mean monomorphism. In such setting, choose a monomorphism $m_i \colon C_i \to I$, where $I$ is injective. Since $I$ is injective and $C_i \to C_{i + 1}$ is a monomorphism, you can lift $m_i$ to a morphism $m_{i+1} \colon C_{i+1} \to I$, and so on. By putting together all $m_i$, you get a morphism $m \colon C \to I$ such that $m_i$ factors as $C_i \to C \to I$. Since $m_i$ is a monomorphism, the morphism $C_i\to C$ is a monomorphism as well.
I learned this nice argument in Neeman - A counterexample to a 1961 "theorem" in homological algebra, where he also construct an example of an Abelian category and a chain of monomorphism such that the direct limit is $0$ - showing that the analogue of this result is not always true!
Where does this argument use the assumption that the category is abelian?
Actually it doesn't, it is just that injective objects are more commonly used in that context
$\def\I{\mathcal{I}}\def\C{\mathcal{C}}\def\c{\operatorname{colim}}$Another example is when $\C$ is an abelian category in which direct limits are exact. Let ${C_i\mid i\in \I}$ be a direct system in $\C$ such that $C_i\to C_j$ is injective for all $i\leq j$. Let $i\in\I$. We want to see that $C_i\to \c_{j\in\I}C_j$ is injective. We may assume that $i$ is initial in $\I$ (replace $\I$ by the full subcategory ${j\in\C\mid i\leq j}$, which is cofinal). The components of the functor morphism $C_i\Rightarrow C$ are all injective. Hence $\c(C_i\Rightarrow C)=C_i\to\c_{j\in\I}C_j$ is injective.
(Here $C_i\Rightarrow C$ is the obvious natural transformation of functors $\mathcal{I}\to\mathcal{C}$ from the constant functor $j\in\mathcal{I}\mapsto C_i$ to our direct system $C:j\in\mathcal{I}\mapsto C_j$.)
Your question is a bit imprecise, but based on what you said in your comment, I think for your situations the answer will be "yes". For Sets, it's an easy exercise to prove that all the $g_i$ are injective. Simply write down the formula for the colimit and work at the level of elements in $C_i$ and $C$. A reference for this fact is Proposition 9.3 in Gaunce Lewis's thesis, which Peter May gave me a long time ago in this question. Lewis also proves the result for compactly generated spaces. You can follow his model to prove it for a wide class of categories built from sets, including semi-groups. Many of these categories are "sets with extra structure" encoded via a monad $T$, and I suspect you could formulate a general result about directed colimits of injections in categories of $T$-algebras over Set.
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2025-03-21T14:48:30.968452
| 2020-05-15T08:51:25 |
360406
|
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Stack Exchange
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distribution on the inverse Wishart matrix eigenvalues summation
Let $\lambda_1>\lambda_2>....>\lambda_N$ be the ordered eigenvalues of Wishart matrix my objective is to find if it is possible the distribution of:
\begin{align}
s = \sum\limits_{i = 1}^N {\frac{1}{{1 + a{\lambda _i}}}}
\end{align}
where a is positive and $N\ge 2$.
For small $N$ explicit expressions are cumbersome. For $N\gg 1$ the distribution $P(s)$ is a Gaussian. The mean is given by integration with the Marchenko-Pastur distribution, the variance is given by integration with a formula given in arXiv:9310010, Equation 17. Let me work this out:
The Wishart matrix is $X=WW^T$, with $W$ an $N\times M$ real matrix, $N\leq M$ and $y=N/M$. I rescale the eigenvalues $\lambda_i$ of $X$ by $x_i=\lambda_i/M$, and define $\alpha=aM$. We seek the distribution of $s=\sum_{i}(1+\alpha x_i)^{-1}$. The support of the eigenvalue density $\rho(x)$ is the interval $(a_-,a_+)$, with $a_\pm=(1\pm\sqrt y)^2$. For $N\gg 1$ one has the Marchenko-Pastur distribution
$$\rho(x)=\frac{1}{2\pi}\frac{N}{yx}(x-a_-)^{1/2}(a_+-x)^{1/2},$$
normalized to $\int\rho(x)dx=N$. The mean of the Gaussian is then equal to
$$\mathbb{E}[s]=\int_{a_-}^{a_+}\frac{\rho(x)}{1+\alpha x}\,dx=N\frac{\sqrt{\alpha^2 (y-1)^2+2 \alpha (y+1)+1}+\alpha (y-1)-1}{2 \alpha y}.$$
For the variance we apply Eq. 17 of the cited paper,
$${\rm var}\,s=\frac{1}{\pi^2}\int_{\alpha_-}^{a_+}d\lambda\int_{a_-}^{a_+}d\mu\frac{\sqrt{(\mu-a_-)(a_+-\mu)}}{\sqrt{(\lambda-a_-)(a_+-\lambda)}}\frac{1}{\lambda-\mu}\frac{1}{1+\alpha\lambda}\frac{d}{d\mu}\frac{1}{1+\alpha\mu},$$
the integrals being Cauchy principal values.
I have not succeeded in evaluating the integrals in closed form. Here is a numerical calculation of the variance for $\alpha=1$ as a function of $y$,
and a plot of the variance for $y=0.5$ as a function of $\alpha$,
For the numerical evaluation it is convenient to rewrite the integral in the form
$${\rm var}\,s=\frac{1}{\pi^2}\int_{a_-}^{a_+}d\lambda\int_{a_-}^{a_+}d\mu\ln|\lambda-\mu|\frac{d}{d\mu}\left(\frac{\sqrt{(\mu-a_-)(a_+-\mu)}}{\sqrt{(\lambda-a_-)(a_+-\lambda)}}\frac{1}{1+\alpha\lambda}\frac{d}{d\mu}\frac{1}{1+\alpha\mu}\right).$$
You wrote that the support of the eigenvalue density $\rho(x)$ is a specific (finite) interval, but the Marchenko-Pastur theorem gives us only limiting results, that is, the distribution converges to one with finite support. In practice, it could be that for any $M,N$, the support is infinite, right? Specifically, I am interested in proving that $\mathbb{E}\lambda_max$ is finite (for large enough $N,M$), and I can't seem to find such a result.
the probability density function of the largest eigenvalue is given by the Tracy-Widom law; it is sharply peaked around the large-$N$ limits $a_\pm$, with a standard deviation that scales as $N^{-1/6}$.
Sorry, I did not clarify enough, I meant $\lambda_{max}$ of the inverse-Wishart, not the Wishart. That is, I am interested at $\mathbb{E} \lambda_{max}(W_n^{-1}(m,I))$
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2025-03-21T14:48:30.968655
| 2020-05-15T10:33:09 |
360411
|
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Stack Exchange
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Difference of two functions with constant mean curvature
Define the set $\Omega := (-\epsilon,\epsilon) \times (-1,1)^{n-1}$, and define
$\Gamma := \{-\epsilon,\epsilon\} \times (-1,1)^{n-1} \subset \partial \Omega$.
Suppose I have two functions $u,v \in C^\infty(\Omega) \cap C^1(\overline{\Omega})$,
with the following properties:
$u = v$ and $\nabla u=\nabla v$ on $\Gamma$
$u > v$ on $\Omega$
$u$ has constant positive mean curvature $K$
$v$ has constant negative mean curvature $-K$.
That is, $u$ and $v$ curve 'away from' each other, but they still agree
along with their first derivatives on $\Gamma$. (I would be happy if I could prove
no such pair of functions exists, but I suspect they do exist, possibly resembling
the surfaces in Figure 4 of Triply periodic constant mean curvature surfaces.)
My question is, can I find a lower bound on $\sup_\Omega |u-v|$, in terms of
$n$, $\epsilon$ and $K$, which doesn't converge to 0 as $\epsilon \to 0$?
My intuition is that the mean curvature difference implies $u-v$ has at least
some positive second derivatives. When those are integrated over the intervals $(-1,1)$,
there should be some lower bound on the maximum distance between them.
More generally, does there exist a similar lower bound if we allow $\Omega$ to range
over all subsets of $(-1,1)^n$ such that $\Omega \cap (-1/2,1/2)^n \not=\emptyset$,
but we restrict the $n$-dimensional volume of $\Omega$ to be less than $\epsilon$?
(If it helps, this paper's Notation and Methods section gives a formula for
the mean curvature of the graph of a function.)
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2025-03-21T14:48:30.969030
| 2020-05-15T12:27:59 |
360417
|
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|
Stack Exchange
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Galois groups of special polynomials
This question is motivated by long experiments with GAP.
Call a monic polynomial with integer coefficients special in case it is irreducible and has only coefficients $-1$, $0$ or $1$. Let $n \geq 5$.
Question 1: In case $n=p$ is a prime number, is it true that any special polynomial of degree $n$ is not solvable by radicals, or equivalently has non-solvable Galois group?
This was true for all primes $\leq 17$, in fact all such polynomials has Galois group $S_n$ (and there are over 100000 such polynomials for primes less than or equal 13), which motives the stronger question:
Question 1': In case $n=p$ is a prime number, is it true that any special polynomial of degree $n$ has Galois group $S_{n}$?
This is also true for $p=2,3$ by the way.
|
2025-03-21T14:48:30.969103
| 2020-05-15T12:39:43 |
360419
|
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|
Stack Exchange
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Is there a flaw in this proof of the validity of the Palais-Smale condition?
In Chapter 3 of his monograph (available on Researchgate), Kavian applies the Mountain Pass Theorem to a semilinear elliptic equation. To this aim, he needs to check that a functional satisfies the Palais-Smale condition.
On page 159, he introduces a technical Lemma 6.6, in which a weighted inequality is proved in the Sobolev space $H_0^1(\Omega)$. The weight function $m$ must satisfy a condition of the form $$m(x)|s|^\theta \leq b_0(x)|s|+b_1 \left(1+|s|^{p+1} \right), \tag{1}$$ where $b_0 \in L^{p_0}$ with $p_0 \geq \frac{2N}{N+2}$ and $b_1$ is a real number.
Then he proves Proposition 6.7, namely the validity of the Palais-Smale condition under suitable assumptions on the nonlinearity $g$. For $G(x,s)=\int_0^s g(x,t)\, dt$, Kavian shows that (the number $R$ is fixed, no problem here)
$$
G(x,s) \geq \min \{G(x,-R),G(x,R)\} R^{-\theta} |s|^\theta - c_1(x),
$$
for some $c_1 \in L^{p_0}$. Then he defines $m(x) = \min \{G(x,-R),G(x,R),1\} R^{-\theta}$ and applies Lemma 6.6.
Here comes my question: is it legitimate to apply the technical Lemma? My feeling is that $m$ does not satisty (1), unless $b_1$ is allowed to be a function of $x$. I can see no way to get rid of $c_1$ with a uniform constant.
|
2025-03-21T14:48:30.969229
| 2020-05-15T12:46:54 |
360421
|
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|
Stack Exchange
|
Does the category of local rings with residue field $F$ have an initial object?
Let $F$ be a field. Does the category $C_F$ of local rings $R$ equipped with a surjective morphism $R\longrightarrow F$ have an initial object?
This is, for instance, true if $F=\mathbb{F}_{p}$ for some prime $p$: If $R$ is a local ring with residue field $\mathbb{F}_{p}$, then any $x\in\mathbb{Z}\setminus(p)$ must map to something invertible under the morphism $\mathbb{Z}\longrightarrow R$. Hence that morphism factors as $\mathbb{Z}\longrightarrow\mathbb{Z}_{(p)}\longrightarrow R$; thus $\mathbb{Z}_{(p)}$ is the initial object.
But what happens in the more general case? I guess it should be true at least if $F$ is of finite type over either $\mathbb{Q}$ or $\mathbb{F}_{p}$ (where $p$ is a prime), but I have no idea how to prove it.
I guess that "$F$ of finite type over $\mathbf{Z}$" should be "$F$ is a finitely generated field"? as formulated now it looks like it's a field that is a finitely generated $\mathbf{Z}$-algebra, but this would then be a finite field. Still a first interesting case is that of non-prime finite fields.
@YCor thanks, fixed that.
But I'm unsure what you mean by "finite type". Probably "finitely generated field" is more clear, or "finite field or number field" if you really mean finite type (= finitely generated ring)
I will show that this is not the case for $F=\mathbb{F}_9$. The proof generalize to any $\mathbb{F}_{p^k}$, with $k >1$.
I'm starting from the observation that:
$\mathbb{F}_9 \simeq \mathbb{Z}[i]/(3) \simeq \mathbb{Z}[\sqrt{2}]/(3)$. where I'm using the isomorphism that identifies $i$ and $\sqrt{2}$ to identify $\mathbb{Z}[i]/(3)$ and $\mathbb{Z}[\sqrt{2}]/(3)$.
I'm now considering $A= \mathbb{Z}[i,\sqrt{2}]$. It has a surjective map $\phi:A \to \mathbb{F}_9$ induced by the two maps $\mathbb{Z}[i] \to \mathbb{F}_9$ and $\mathbb{Z}[\sqrt{2}] \to \mathbb{F}_9$ above.
I can localize $A$ at $\ker \phi$ to make it an object of $C_F$.
So, in the category $C_F$ I have a diagram:
$$ \mathbb{Z}[i]_{(3)} \to A_{\ker \phi} \leftarrow \mathbb{Z}[\sqrt 2]_{(3)} $$
If there was an initial object $B$ in $C_F$, its unique map to $A_{\ker \phi}$ should factors through both $\mathbb{Z}[i]_{(3)} $ and $\mathbb{Z}[\sqrt 2]_{(3)}$ hence, it should factor through their in $A_{\ker \phi}$. But this intersection is reduced to $\mathbb{Z}_{(3)}$, so we should have a map $B \to \mathbb{Z}_{(3)}$ compatible with the map back to $\mathbb{F}_9$, but as the map $\mathbb{Z}_{(3)} \to \mathbb{F}_9$ is not surjective, and the map $B \to \mathbb{F}_9$ needs to be, we have a contradiction.
The pullback of $\mathbb{Z}[\sqrt{2}]{(3)}\longrightarrow\mathbb{F}{9}$ and $\mathbb{Z}[i]{(3)}\longrightarrow\mathbb{F}{9}$ is $\mathbb{Z}{(3)}$ when viewed as morphisms in the category of rings. As morphisms in the category $C{F}$, their pullback is $\mathbb{Z}[x]{(3,x^{2}+1)}$ - and that lies in $C{F}$.
The argument is not about pullbacks : it looks at the map $R\to A_{\ker\phi}$ (where $R$ is your initial object). Since this map factors through both of the subrings indicated (by uniqueness), its image (as a map of rings) is contained in their intersection; and that is not possible, as the restriction of the quotient map to said intersection is not surjective.
Well, it is about pullback, but pullback over $A_{\ker \phi}$ not pullback over $\mathbb{F}9$. For a general field $F$ this category has all products (given by fiber product over $F$), but it can be shown by abstract category theory that this category will have an initial object if and only if has all limits, i.e. if and only if it has fiber products. Then I started from an example of a fiber product that do not exists (the fiber product over $A{\ker \phi}$) and back tracked it to a counter example to the original question.
@MyGrandmother'sCobblestone: By the way, your computation of the product in $C_F$ (or equivalently, fiber product over $\mathbb{F}_9$) is not quite right: you also need to quotient out by the kernel of $\mathbb{Z}[X] \to \mathbb{Z}[\sqrt 2] \times \mathbb{Z}[i]$, that is by $(X^2+1)(X^2-2)$.
|
2025-03-21T14:48:30.969503
| 2020-05-15T12:51:51 |
360422
|
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Stack Exchange
|
Reflexive vs. pseudo-coherent abelian groups
Recall that a module M over some ring R is pseudo-coherent if it admits a resolution whose terms are finitely generated projective R-modules. Another notion is to ask whether M is reflexive when regarded as an object in the derived category $D(Mod_R)$, i.e.,
$$M \to RHom(RHom(M, R), R))$$
is a quasi-isomorphism, where RHom denotes the derived functor of the usual homomorphisms of R-modules.
For what rings does pseudo-coherence imply reflexivity in this derived sense?
When does the converse hold? (Maybe for nice rings, even for the integers?), i.e., are reflexive modules (where reflexivity is always understood in the derived sense) pseudo-coherent?
Small comment: if $R$ is Noetherian, then pseudo-coherent just means finitely generated. In the non-derived sense, reflexive does not imply finitely generated; for example $\mathbf Z^{(\mathbf N)}$ is reflexive. This depends highly on the ring; for example for a DVR $R$ the analogous result is true if and only if $R$ is not complete; see e.g. this post.
@Yai0Phah: Thanks - I think you are right, the assertion I made in the question is unclear. It might still be correct if R has finite global dimension. I will edit the question accordingly.
|
2025-03-21T14:48:30.969618
| 2020-05-15T14:22:05 |
360429
|
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|
Stack Exchange
|
Do Poincaré duality algebras need to be defined over a field?
I asked the below question here on MSE, but after some time and a bounty offering I have not received an answer.
A graded commutative, connected $\mathbb{k}$-algebra $A$ is called a Poincaré duality algebra if it is finite dimensional over $\mathbb{k}$, so that $$A = \bigoplus_{i=1}^d A^i,$$ and the $\mathbb{k}$-linear maps
$$A^i \rightarrow Hom_{\mathbb{k}}(A^{d-i},A^d), \quad a \mapsto \phi_a, \text{ where } \phi_a(b)=ab$$
are isomorphisms.
For example, it can be stated that the cohomology algebra of a manifold is a Poincaré duality algebra.
Every definition that I've seen of a Poincaré duality algebra (for example in this paper by Stanley and Lambrechts) assumes that $\mathbb{k}$ is a field. Is this necessary, or just convenient? Is there a problem with simply having $\mathbb{k}$ a ring, e.g. $\mathbb{Z}$? Would we then require some additional assumption such as zero torsion?
I mean, you can always make any definition you want, the question is how useful it is. Over a general ring it's a bit annoying because the condition "the pairing $A^i \otimes A^{d-i} \to \Bbbk$ is nondegenerate" is not equivalent to "$A^i \to \hom_\Bbbk(A^{d-i}, \Bbbk)$ is an isomorphism". What is the motivation behind this question?
Also your definition is not really the usual one. Usually you have to fix some linear form $\epsilon : A^d \to \Bbbk$ and the isomorphism is $a \mapsto (b \mapsto \epsilon(ab))$. With your definition it's not clear that $A^d = A^0$ for example, just that $A^0 = \hom(A^d,A^d)$ for example. (But again, maybe your applications call for this?)
As you said, torsion phenomena make it so that $H^(X;\mathbb Z)$ is not a Poincaré algebra in the above sense if $X$ is a compact oriented manifold. One possible way out is to work in a derived way, i.e. with the chain complex $C^(X;\mathbb Z)$ and not just its homology; the resulting notion of a self-dual chain complex (really, a stable infinity category equipped with a "Poincaré structure) is developed in an ongoing lecture by Thomas Nikolaus, although the relevant course notes have not been published yet: https://www.uni-muenster.de/IVV5WS/WebHop/user/nikolaus/Surgery.html
Thank you for your comments. My motivation lies in understanding some statements regarding the homology/cohomology algebras of "Poincaré duality spaces" - where by this terminology I mean spaces which have Poincaré duality in the sense of manifolds, but which are not necessarily manifolds. One definition in particular defines such spaces to be those for which their cohomology algebras are Poincaré duality algebras in the above sense (without specifying $\mathbb{k}$ a field), but as Bertram points out, if working with the above definition this is only the same if $\mathbb{k}$ is a field.
You can find an example useful in the context of Koszul homology in definition 1 in https://www.cambridge.org/core/journals/glasgow-mathematical-journal/article/ascent-and-descent-of-gorenstein-property/D7E9D1C81352123F2676BFDA7B3DC883
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2025-03-21T14:48:30.970013
| 2020-05-15T14:59:13 |
360432
|
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|
Stack Exchange
|
The existence of rational points
Solving some problem parametrically, I got the following answer:
$$ \dfrac{5x}{4} + \sqrt{\dfrac{y^2}{4} - \dfrac{x^2}{16}} + \dfrac{1}{10} \sqrt{10x^2 + 9y^2} + \dfrac{1}{5} \sqrt{5x^2 + 16y^2} $$
And I had a question: can there be a rational answer for some $ x, y \in \mathbb {R_+} $? Well, or what “maximally” simple answer can you get where each term will be positive? I suspect that there will still be one root always for any values of $ x, y $. By a simple answer, I understand the number of radicals used after all the simplifications, for some reason you can somehow call it scientifically, but I don’t know how.
Aside from the confusion about the meaning of maximal simplicity, I don't understand what "can there be a rational answer" means. Do you mean "does the formula ever take a rational value"? I guess that $\mathbb R_+$ is strictly positive reals, or else the answer is clear.
Yes, does the formula take rational values for strictly positive arguments, and so that each term is positive.
$x=1$, $y=\sqrt{41}/10$
It rather obviously takes every positive real value, and therefore every positive rational value. Somewhat more interesting would be the case where $x$ and $y$ are rational.
Amazing result! Show how you found these values? No, I had enough material positives. @duje
$x=2y$ for any rational $y$.
if $x = 2y$ 2 term zero.
Take $x=1$ and $y^2=z$. Then conditions are that $z/4-1/16$, $10+9z$ and $5+16z$ are squares. By multiplying these conditions we get the elliptic curve $u^2 = (z/4-1/16)(10+9z)(5+16z)$. It has torsion group $Z/2Z \times Z/4Z$ and rank $1$. A point $P$ of infinite order corresponds to $z=-1/3$. One of the torsion points $T$ corresponds to $z=1/4$ and this value in fact satisfies the starting three conditions. However, the first factor is zero. Other solutions of the system correspond to the points of the form $T+2kP$. For $T+2P$ we get $z=41/100$, hence the solution mentioned in the comment $y=\sqrt{41}/10$. For $T+4P$ we get $z=38904227209/31543891236$ and hence $y=\sqrt{38904227209}/177606$.
|
2025-03-21T14:48:30.970192
| 2020-05-15T15:29:05 |
360434
|
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|
Stack Exchange
|
The expected size of a subtree of any labelled rooted tree
Consider the set of labelled rooted trees of size $n$, $\mathcal{T}_n$. Let $r$ be the root of each tree $T=(V,E)\in\mathcal{T}$, $r\in V(T)$, and let $n(u)$ be the number of vertices of the subtree $T(u)$ of $T$ (then, $n(r)=n=|V(T)|$, and $n(l)=1$ if $l\in V$ is a leaf of $T$). For any such tree $n(u)$ is easy to compute.
I would like to know whether there are results on the expected size of a subtree (chosen uniformly at random) of a labelled rooted tree, over the space of uniformly random labelled rooted trees. I'm interested in both closed-form formulae and the methods used for their derivation since I need to evaluate other expected values over the same space of trees. I'm mostly interested in exact formulae, but asymptotic results are also welcome. My search on the internet has been fruitless so far, so any help is appreciated.
How are you choosing $u$? Is it that you are choosing a uniformly random $T$ and then choosing a uniformly random $u$ from it?
Yes, I forgot to mention this. Both the tree $T$ and the vertex $u$ are chosen uniformly at random.
Ok. Just a quick note, in case it helps your searching, is that this is equal to the expected distance from a uniformly chosen vertex to the root: \begin{align*}\sum_T \sum_{v \in T} \sum_{w \geq v} 1 &= \sum_T \sum_{w \in T} \sum_{v \leq w} 1 \ &= \sum_{T} \sum_{w \in T} |w| \end{align*}
Yes. But be careful. For a path graph of 3 vertices rooted at the central vertex, we have that $\sum_{u\in V} n(u)=5$. If we use distances $d(u)$ to the root expressed in edges the summation becomes $\sum_{u\in V} n(u)= \sum_{u\in V} (1 + d(u))$. We could also express the distance in number of vertices, considering the vertex itself, i.e., define the distance from $u$ to $r$ as the number of edges from $u$ to $r$ plus 1 (vertex $u$ -- or vertex $r$).
You're right, I'm off by $1$, so it's the expected distance plus $1$.
The limiting distribution of the heights of vertices of a random labelled tree was found by Rényi and Szekeres. See https://www.cambridge.org/core/services/aop-cambridge-core/content/view/C30DF2A9C26541526D847E0A0D46F2A8/S1446788700004432a.pdf/on_the_height_of_trees.pdf.
Thank you for the reference. I'll get into the details as soon as I can.
|
2025-03-21T14:48:30.970372
| 2020-05-15T15:46:09 |
360436
|
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|
Stack Exchange
|
Why $E_1(\mathbb{Q}_p)\cong\mathbb{Z}_p$
I read an article where it is said: $E_1(\mathbb{Q}_p)\cong \mathbb{Z}_p$ where $E$ is an elliptic curve over $\mathbb{Q}_p$ and $E_1(\mathbb{Q}_p)=\{P\in E(\mathbb{Q}_p):\tilde{P}=\tilde{O}\}$.
The author says that the proof is in "Arithmetic of elliptic curves" by J. Silverman, at page 191, but there it is said:
If $E$ is an elliptic curve over $\mathbb{Q}_p$ and $\hat{E}$ is the formal group, then:
$$E_1(\mathbb{Q}_p)\cong \hat{E}(p\mathbb{Z}_p)$$
So I don't know a good reference for the proof of $E_1(\mathbb{Q}_p)\cong \mathbb{Z}_p$.
The isomorphism between the formal group (the right-hand side in your displayed equation) and the p-adic integers must be in Chapter 4 of the same book, towards the end. It may be false for p = 2, by the way.
Crossposted at MSE. When cross-posting, it is important to link all versions of the question to prevent needlessly duplicating work.
As RP says, there's a chapter in The Arithmetic of Elliptic Curves that discusses formal groups, and in particular the points of a formal group defined over a complete local ring. The specific result that you want is Chapter IV, Theorem 6.4(b), in the special case that $K=\mathbb Q_p$ and $R=\mathbb Z_p$ and $\mathcal M=p\mathbb Z_p$. That theorem says that there the formal logarithm gives an isomorphism
$$ \log_{\mathcal F} : \mathcal F(\mathcal M^r) \longrightarrow \hat{\mathbb G}_a(\mathcal M^r), $$
provided that $r$ is an integer satisfying $r>v(p)/(p-1)$. For your case, $v(p)=1$, so the isomorphism is valid for all $r\ge1$ except, as noted by RP, when $p=2$, in which case you'll need $r\ge2$. And indeed, for $p=2$ you may need $r\ge2$, since there are formal groups over $\mathbb Z_2$ in which $\mathcal F(2\mathbb Z_2)$ has an element of order 2, hence it cannot possibly be isomorphic to the additive group.
thank you very much!!
|
2025-03-21T14:48:30.970536
| 2020-05-15T18:05:00 |
360441
|
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|
Stack Exchange
|
Hilbert series of special linear sections of Grassmannian $Gr(2,n)$
Consider the Grassmannian $\operatorname{Gr}(2,n)$. I want to know Hilbert series of $H_1 \cap H_2 \dots \cap H_m \cap \operatorname{Gr}(2,n)$ in the Plücker embedding of $\operatorname{Gr}(2,n)$, here $H_1, H_2, \dots ,H_m$ are very special hyperplanes, that is, each defined by vanishing of one Plücker coordinate.
The answer depends on he choice of $H_i$. For instance, for $n = 4$, $m = 3$, if you consider the Plucker equations $x_{12} = x_{13} = x_{14} = 0$, the corresponding linear section is $\mathbb{P}^2$, and if instead you consider $x_{12} = x_{23} = x_{34} = 0$, the corresponding linear section is a reducible conic. They have different Hilbert polynomials.
Thanks Sasha. I want to know if there are some results (with some conditions on coordinates, if needed) for Gr(2,n). Or any progress in this direction.
if the linear section is dimensionally transverse (i.e., its dimension is equal to $2(n-2) - m$), then the Hilbert polynomial is equal to $\sum (-1)^i\binom{m}{i} h(t - i)$, where $h(t)$ is the Hilbert polynomial of $Gr(2,n)$.
Thanks Sasha. But I am interested in knowing the case when the hyperplanes are very special.
Then the answer depends on how special they are. Please, try to be more precise.
|
2025-03-21T14:48:30.970638
| 2020-05-15T19:13:23 |
360451
|
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|
Stack Exchange
|
Explicit formula for the moment map of toric manifold
Let $P$ be a Delzant polytope in $M\otimes{\mathbb R}\cong \mathbb R^n$, and it is well-known that we can associate to it a toric manifold $X=X_P$ with the moment map $\pi: X\to P$.
I would like to explicitly describe the restriction of the moment map $\pi_0:=\pi|_{(\mathbb C^*)^n}$ on the torus $(\mathbb C^*)^n\subset X$.
Denote all the vertices of $P$ by $m_1,\dots, m_k\in M_{\mathbb R}$. Suppose all $m_i\in M$, then each $m_i$ corresponds to a function $\chi^{m_i}: (\mathbb C^*)^n\to \mathbb C$, and the moment map is given by the formula
$$
x \mapsto \frac{\sum_i|\chi^{m_i}(x)|\cdot m_i}{\sum_i |\chi^{m_i}(x)|^2} \in M_{\mathbb R}
$$
Question:
(1) When we drop the condition $m_i\in M$, namely $m_i\in M_{\mathbb R}\setminus M$, what will happen? I think the issue is $\chi^{m_i}$ is not well-defined.
(2) When we allow $P$ to be unbounded, can we find the moment map explicitly?
Can we obtain a similar formula for either Case (1) or (2) above? This might be standard, but I fail to find a reference.
|
2025-03-21T14:48:30.970737
| 2020-05-15T20:02:38 |
360456
|
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|
Stack Exchange
|
Why does the inverse Alexander polynomial appear in the MMR conjecture?
In an attempt to better understand why the inverse Alexander polynomial appears in the MMR conjecture, I was reading the paper [1] of Bar-Natan and Garoufalidis giving their proof of the conjecture using weight systems. In particular, they discuss Rozansky's path-integral "proof" of the conjecture in section 1.4. I mostly follow this argument, except for the following claim:
Cheeger and Mueller proved that the Ray-Singer torsion is equal to the Reidemeister torsion, which by Milnor and Turaev was shown to be proportional to the inverse of the Alexander polynomial $A(K)$ of $K$.
My understanding was that Milnor and Turaev show that the torsion is proportional to the Alexander polynomial, not its inverse.
Am I wrong? Is it a different torsion than the one I'm thinking of?
The torsion involves an alternating product, so it could just be different conventions.
In that case understanding why this convention is right probably means digging deeper into the proof.
[1] On the Melvin–Morton–Rozansky conjecture
I think that Turaev and Milnor have different conventions for Reidemeister torsion. One relates Reidemeister torsion to the Alexander polynomial, the other to its inverse.
You're right: usually in the topology literature the torsion is normalized so it agrees with the Alexander polynomial, and in the analysis literature the inverse is used instead.
However, that brings up the question of why the asymptotics of the colored Jones polynomial (a.k.a. the semiclassical behavior of quantum Chern-Simons theory) corresponds to the analytic/Ray-Singer torsion and not its inverse.
On reason this is confusing to me is that the regular $n=2$ Jones polynomial is related to the Alexander polynomial, not its inverse.
|
2025-03-21T14:48:30.970884
| 2020-05-15T20:56:26 |
360460
|
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"Dmitri Pavlov",
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|
Stack Exchange
|
Why does this construction give a weak factorization system in the category of span diagrams?
In Dwyer and Spalinski's classic paper Homotopy Theories and Model Categories, they describe homotopy pushouts by defining a model structure on the category of span diagrams in a given model category $C$. In their Prop 10.6, in proving this is a model category, specifically in proving the weak factorization axiom, they seem to implicitly use the fact that (in their notation) if a map $f$ is such that $f_a$, $f_b$, and $f_c$ are acyclic cofibrations, then $i_a(f)$ and $i_c(f)$ are again acyclic cofibrations. However, it is far from clear to me why that ought to be the case. It is easy to see that they are weak equivalences using 2-out-of-3, but I can't seem to find a reason they should be cofibrations.
Please do not cross-post simultaneously to two different sites without giving a link to the cross-post. Cross-post: https://math.stackexchange.com/questions/3676771/why-does-this-construction-give-a-weak-factorization-system-in-the-category-of-s
they seem to implicitly use the fact that (in their notation) if a map f is such that fa, fb, and fc are acyclic cofibrations, then ia(f) and ic(f) are again acyclic cofibrations.
No, that's not what they're using. They're establishing the lifting property
of acyclic cofibrations with respect to fibrations.
Acyclic cofibrations are by definition cofibrations that are weak equivalences.
They define cofibrations as maps $f$ such that $i_a(f)$, $i_b(f)$, and $i_c(f)$
are cofibrations in $C$.
Weak equivalences are defined indexwise.
Thus, acyclic cofibrations are precisely thos maps $f$ for which
$i_a(f)$, $i_b(f)$, and $i_c(f)$ are acyclic cofibrations in $C$,
as follows from the 2-out-of-3 property for weak equivalences.
This model structure is a special case of the model structure
on directed diagrams (such as $C^D$),
which itself is a special case of a Reedy model structure.
In this case, both nonindentity arrows in $D$ are positive.
The (acyclic) cofibrations are precisely
Reedy (acyclic) cofibrations in the Reedy model structure,
and the functors $i_a$, $i_b$, $i_c$ are precisely the Reedy latching maps.
|
2025-03-21T14:48:30.971030
| 2020-05-12T20:07:06 |
360164
|
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|
Stack Exchange
|
prove a bondle is an indefinite Hermitian manifold which is Kahler if and only if the manifold is locally flat
Let $M(J,g)$ be an indefinite Kahler manifold, then $%
TM(J^{H},g^{D})$ is an indefinite Hermitian manifold which is Kahler if and
only if $M$ is locally flat. Here $J^{H}$ denotes the horizontallift of $J$ and $g^{D}$ is the sasaki metric
You might want to try asking this question, and your other questions, at math.stackexchange first, before asking them here, unless you are certain that they are open research problems, or otherwise unsuitable for math.stackexchange. To me, this looks like it could perhaps be an exercise from a graduate level textbook. Why do you think it is an open research problem?
you're right, I should probably post it on math.stackexchange first
Add a comment with a link, when you do ask it there.
https://math.stackexchange.com/questions/3671933/prove-a-bondle-is-an-indefinite-hermitian-manifold-which-is-kahler-if-and-only-i
|
2025-03-21T14:48:30.971126
| 2020-05-12T21:09:06 |
360167
|
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|
Stack Exchange
|
Similar to inverse plus rank 1
Given a real, invertible matrix $A$. For which vectors $b$ and $c$ is
$$
A^{-1} + bc^T
$$
similar to $A$? And is the rank-1 matrix $bc^T$ unique?
Let's assume for simplicity that $A$ has distinct eigenvalues, so similarity to $A$
just means having the same characteristic polynomial.
The Matrix determinant lemma
says that the characteristic polynomial
$$ \eqalign{\det(A^{-1} + b c^T - \lambda I) &= (1 + c^T (A^{-1}-\lambda I)^{-1} b) \det(A^{-1}-\lambda I)\cr
&= (1 + c^T A (I - \lambda A)^{-1} b) \det(A)^{-1} \det(I - \lambda A)}$$
which we want to be the same as the characteristic polynomial $\det(A - \lambda I)$ of $A$.
Thus we want $$
1 + c^T A (I - \lambda A)^{-1} b = \frac{\det(A) \det(A - \lambda I)}{\det(I-\lambda A)}$$
Writing $b$ as a linear combination $\sum_j b_j v_j$ of the eigenvectors of $A$
and $c$ as a linear combination $\sum_j c_j u_j$ of the left eigenvectors of $A$
(where $Av_j = \lambda_j v_j$, $u_j^T A = \lambda_j u_j^T$, and $u_i^T v_j = \delta_{ij}$), the left side is
$$ 1 + \sum_j c_j b_j \frac{\lambda_j}{1 - \lambda \lambda_j}$$
and the equation will be true as long as $-c_j b_j $ is the residue of the
right side at $\lambda = \lambda_j^{-1}$.
We will not have uniqueness because only the products $c_j b_j$ matter.
|
2025-03-21T14:48:30.971233
| 2020-05-12T22:13:50 |
360171
|
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|
Stack Exchange
|
When does QCoh have 'enough perfect complexes'?
Let $X$ be a derived fpqc stack on the $\infty$-category of connective spectral affine schemes $\mathbf{Aff}^{\mathrm{cn}}=(\mathbf{Ring}^{\mathrm{cn}}_{E_\infty})^{\mathrm{op}}$, that is to say, a functor $X:(\mathbf{Aff}^{\mathrm{cn}})^{\mathrm{op}}\to \mathcal{S}$ satisfying fpqc descent. Then we can define its $\infty$-category of quasicoherent sheaves formally by a Kan extension.
Say that a symmetric monoidal stable $\infty$-category $\mathcal{C}$ 'has enough perfect objects' if its full subcategory of dualizable objects is dense (that is to say the induced functor $\mathcal{C}\to \operatorname{Ind}(\operatorname{Perf}(\mathcal{C}))$ is fully faithful).
Are there examples of fpqc stacks $X$ as above for which $\operatorname{QCoh}(X)$ does not have enough perfect objects?
What about if we restrict our question to ((Quasi)-Geometric Stacks, Artin Stacks, Deligne-Mumford Stacks, Algebraic Spaces, Schemes)?
For certain, this is true for quasicompact quasiseparated schemes and algebraic spaces, as well as quasi-Geometric stacks $X$ such that $\operatorname{QCoh}(X)$ is compactly generated and the structure sheaf is a compact object (proven in Lurie, Spectral Algebraic Geometry).
I don't know that much about this, but there's some relation to the resolution property, right?
Consider $X=\textrm{Spec}(k[x_1,x_2,...])$ the infinite dimensional affine space, $U=X-0$ punctured, and $Y=X\cup_U X$ with the doubled origin. The first is affine, the second not quasi-compact, the third not quasi-separated. They all have the same perfect complexes, but different QCoh. The third has extra QCoh, so fails your hypothesis. The second might be OK. @TimCampion only mildly. That is a finite question, which is automatically true in the derived setting, whereas this question is about the difference between finite and infinite.
@BenWieland You should give this as an answer! Thanks! If you can also come up with cases where QCoh "doesn't have enough almost-perfects", that would also be neat, but this answer is enough if you don't feel like it or can't think of anything =).
If you want to go stacks, there are serious obstructions against being generated by perfect complexes. See this nice paper of Hall, Neeman and Rydh: arxiv1405.1888
@Denis-CharlesCisinski I was looking at that paper, a bit earlier today. I thought that their theorem is about obstructions to being generated by compact objects (which would be equivalent in the case of a quasi-geometric stack to having QCoh(X)=Ind(Perf(X))=Ind(QCoh(X)^ω). I'm asking for something a lot weaker, namely that the functor QCoh(X)→Ind(Perf(X)) is fully faithful (and with no assumption that the perfects are compact). Is there a part of the paper where they address this question too? At least from the part I read so far, they're showing that QCoh(X)^ω is often trivial.
I'm in a situation where even having a localization Ind(Perf(X))→QCoh(X) is useful to me.
Robert Thomason was the first person to draw attention to this question, before derived schemes and infinity categories. I believe that he proved that for a quasi-compact and quasi-separated scheme that $D_{qc}=\textrm{Ind}(\textrm{Perf})$. For example, see Thomason-Trobaugh section 2.3, though at first glance it appears that only proves the weaker statement that it has enough perfect complexes.
Somewhere he gives two examples to show the necessity of the two hypotheses. Consider an affine scheme with a point of infinite codimension, say, $X=\textrm{Spec}\;k[x_1,x_2,…]$. Let $U$ be the complement of the point. It is not quasi-compact. Let $Y=X\cup_U X$ be $X$ with the point doubled. It is not quasi-separated. A perfect complex is built from finitely many operations, so its support has finite codimension, so they do not notice the points of infinite codimension, so the three schemes all have the same perfect complexes. But they have different quasi-coherent complexes, such as the skyscrapers on the origins. In particular, $Y$ has two such sheaves, but they cannot be distinguished by perfect complexes. Whereas $U$ has too few sheaves, so it fails the strong hypothesis of equivalence of categories, but it still has enough perfects: $\textrm{QCoh}(U)\subset \textrm{QCoh}(X)=\textrm{Ind}(\textrm{Perf(X)})=\textrm{Ind}(\textrm{Perf(U)})$
Is there a finite dimensional example? For example, consider this non-quasi-compact scheme built from varieties. Let $Z_0=\mathbb A^2$ and $x_0=0\in Z_0$. Let $Z_{n+1}$ be the blow up of $Z_n$ at $x_n$ and let $x_{n+1}$ be a point in the exceptional fiber. Let $U_n=Z_n-\{x_n\}$. Then $U_n$ is open in $U_{n+1}$ and let $U'=\bigcup U_n$. Does it satisfy $D_{qc}=\textrm{Ind}(\textrm{Perf})$? I believe that it can be compactified by adding a 2-dimensional valuation ring. If so, we could double that point to get a non-quasi-separated scheme. Would it fail to have enough perfects?
For $n < \infty$, the spectrum of a valuation ring of dimension $n$ has exactly $n+1$ points (as the primes are totally ordered), so any open subset is trivially quasi-compact.
@Anonymous thanks, edited to be more hypothetical
|
2025-03-21T14:48:30.971603
| 2020-05-12T22:16:52 |
360172
|
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|
Stack Exchange
|
A transversal for the $\operatorname{Ad}(K)$ action on a sphere in $\mathfrak{p}$
This exercise level question has been unanswered on MSE for a few years. I hope you can answer it either there or here.
$G$ is a semisimple Lie group with a choice of Cartan decomposition on its Lie algebra $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$.
The Cartan involution also induces an inner product on $\mathfrak{g}$ and we denote the sphere in $\mathfrak{p}$ by $\mathfrak{p}_1$.
Let $K$ be the connected subgroup associated to $\mathfrak{k}$ and let $\mathfrak{a}\subset \mathfrak{p}$ be a maximal abelian subalgebra.
The action of $\mathfrak{a}$ on $\mathfrak{g}$ gives rise to a root system $\Sigma$ on $\mathfrak{a}^*$.
Fix a notion of positivity here and let
$$\mathfrak{a}^+ := \lbrace H \in \mathfrak{a}: \lambda(H)>0 \text{ for all } \lambda \in \Sigma^+\rbrace$$
Question: Why is it that the unit vectors in ${\mathfrak{a}^+}$ is a transversal for the adjoint action of $K$ on $\mathfrak{p}_1$?
More precisely, for every $X \in \mathfrak{p}_1$, there is a unique unit vector $H$ in $\overline{\mathfrak{a}^+}$ which is in the $K$-orbit of $X$.
$\mathfrak{p} = \bigcup \operatorname{Ad}(k)\mathfrak{a}$, coincidence of the analytic Weyl group and algebraic Weyl group, and the fact that $\overline{\mathfrak{a}^+}$ is a fundamental domain for the algebraic Weyl group acting on $\mathfrak{a}$, together imply that each $K$ orbit in $\mathfrak{p}$ passes through at least once. But it's not yet clear to me why two unit vectors in $\overline{\mathfrak{a}^+}$ cannot be related by some $k \in K\smallsetminus N_K(\mathfrak{a})$.
Thank you for reading.
As an intersection of 2 open subsets
Open in $\mathfrak{a}$? $g\mathfrak{a}^+$ might not be contained in $\mathfrak{a}$, right?
Oops sorry, it's not open. I'm not used to "fundamental domain" in such context, would you way what it means exactly for $D=\overline{\mathfrak{a}^+}$ to be a fundamental domain? It ought to mean that for every $g\in K-{1}$ we have $gD\cap D$ is small, but in which sense exactly? (it's not empty, since it contains zero and I'd guess it may contain more than zero) does it mean it's contained in the relative boundary of $D$ (boundary of $D$ in $\mathfrak{a}$)?
I made an edit. Does the question make sense?
It seems that you are acting on a cone, not a sphere, right?
@LSpice both. OP passed to the 1-sphere to avoid intersections, but in the original setting, the question can currently be stated as: is it true that every $K$-orbit in $\mathfrak{p}$ meets the closure of $\mathfrak{a}^+$ in a singleton.
Things are clear to me now. Just to clarify my previous confusion: if a group $G$ acts on a set $X$, a transversal is a subset $Y$ meeting each orbit in a singleton (that is, $Y$ meets every orbit and no two elements of $Y$ are in the same orbit). Strictly speaking, a fundamental domain is $Y\subset X$ such that the $(gY)_{g\in G}$ partition $X$ (this is much stronger and does not always exist). Practically, one uses to relax the condition that the $gY$ are disjoint to: they have "small" intersection. Here $Y$ is closed convex; small can mean: of dimension $<\dim(Y)$.
I understand now. This is why you previously made the comment ‘an uncountable group can’t have a fundamental domain of positive measure’? I’ll now use the word ‘transversal’. Also, I wasn’t sure whether to state the question in terms of the full cone or the unit vectors there. Both are equivalent since Ad(k) preserves length.
It's a good question. I'd suggest not to delete and rather post and accept a community-wiki answer linking to the MSE answer.
This has been answered here on MSE.
FYI, this is Theorem 4.21 (without proof) on page 74 in Bekka and Mayer - Ergodic theory and topological dynamics of group actions on homogeneous spaces.
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2025-03-21T14:48:30.972002
| 2020-05-12T23:42:48 |
360175
|
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|
Stack Exchange
|
Size of the smallest union chain containing a family
A family of sets $\mathcal{F}$ is a union chain if each set in $\mathcal{F}$ of size at least $2$ is the union of two other sets in $\mathcal{F}$. That is, $X\in \mathcal{F}$ and $|X|\geq 2$, then $X=Y\cup Z$ for some $Y,Z\in \mathcal{F}$ and $Y,Z\neq X$.
Let $\mathcal{A}$ be a family of sets.
Let $|\mathcal{A}|=n$ and $\left|\bigcup_{A\in \mathcal{A}} A\right|=m$.
What are some upper bounds for $|\mathcal{F}|$ in terms of $n$ and $m$, where $\mathcal{F}$ is the smallest union chain that contains $\mathcal{A}$? Is $|\mathcal{F}| = O(m+n)$?
The terminology comes from addition chain, but instead of addition we have set unions.
By binary division, each set in A having k elements can be represented as a union chain of about 2k many sets, so there is an upper bound on the sum of the sizes of the sets. Some savings can be realized by using shared sets of size 2,4,8 and so on. I would be surprised if O(n+m) is always achievable. Gerhard "Chain Isn't The Best Word" Paseman, 2020.05.12.
|
2025-03-21T14:48:30.972114
| 2020-05-12T23:48:30 |
360177
|
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|
Stack Exchange
|
Measure Theories with a different convention to $\infty\cdot 0 =0$
As we all know in a first course in measure theory we define a symbol $\infty$ to satisfy $\infty \cdot 0=0$, but there are more two possible choices for a convention as someone has shown to me; one is $\infty\cdot 0 =\infty$ and the other $\infty \cdot 0 =-\infty$.
Can someone please provide references for those measure theories with a different convention than $\infty \cdot 0 = 0$?
Thanks in advance!
Arguably $\infty \cdot 0 = \infty$ is consistent with usual measure theory: the (usual Lebesgue) measure of a singleton subset of $\mathbb{R}$ is 0 but the measure of all of $\mathbb{R}$ is infinity.
@SamHopkins $\mathbb{R}$ is not a countable union of singletons, though.
I feel like any measure theory where you get rid of countable additively and insist all infinite sets have infinite measure is gonna be a very trivial theory...
@SamHopkins either way I'd like to see those theories. Some say all of maths when done correctly is trivial. cheers!
Finitely additive functions are certainly studied: see e.g. https://en.wikipedia.org/wiki/Sigma_additivity#Additive_(or_finitely_additive)_set_functions
Precisely where in measure theory you find this convention?
$\infty\cdot0 = \infty$ I can imagine, but $\infty\cdot0 = -\infty$? Then $-\infty = \infty\cdot(0\cdot0) = (\infty\cdot0)\cdot0 = (-\infty)\cdot0 = -(\infty\cdot0) = \infty$, if multiplication is still associative (and, if not, then why not, say, $\infty\cdot0 = 1$?).
@LSpice so it seems that my friend who proved his argument was partially correct. Thanks for pointing it to me LSpice!
I think the precise convention needs to be explained a bit more; is it saying that the integral of the zero-function over a set of measure infinity is 0? Or that the integral of the infinite function over a set of measure 0 is 0? Or something else entirely?
@Alan I didn't vote to close but yes, it is not research level and should have been asked at math.stackexchange.
@NikWeaver ok, I won't ask anymore questions on overflow; I know when I am not welcomed.
@Alan I checked your activity history and it seems to me that you contribute a lot of value to mathoverflow, and the community here has shown its appreciation by giving you many upvotes. The current question isn't appropriate but I wish you wouldn't take that judgement too personally.
One of my answers was downvotes to oblivion, far worse than -2. Don't place more importance on this than it deserves.
So @LSpice am I right in saying we can take $\infty \cdot 0$ to be any number and thus we might get uncountable number of measure theories?
@NikWeaver sorry, this is my last post or thread in this cursed QA website. I will still ask and perhaps also answer questions in MSE, but for me mathoverflow is a forgotten memory.
@NikWeaver there are many questions here that are not research level though they don't get closed!
It's not a convention, it's a theorem. Let's say I have a measure space $X$ and a function $f: X \to \overline{\mathbb{R}}$ which is identically zero off of a null set $N$, and constantly $+\infty$ on that null set. You can say that $\int f = 0$ ``by convention'', but you can also say that $\int f$ equals the area under the graph of $f$, i.e., the measure of the set $N\times [0,\infty)$. And that area has to be zero by countable additivity because the measure of $N\times [n, n+1)$ is $0\cdot 1 = 0$ for all $n$. The sum of infinitely many zeros has to be zero because that is what the partial sums converge to.
An instructive special case is the line $\{0\}\times [0,\infty) \subset \mathbb{R}^2$. Its measure is $0\cdot\infty$, right? Now for any $\epsilon$ find a sequence of rectangles which covers the line and whose areas sum to $\epsilon$. So this $0\cdot\infty$ has to be zero.
Another way to say it is that the integral of the given $f$ is zero no matter how you define $0 \cdot \infty$. Adopting that convention just means that you can write a more compact expression for the integral of $f = c \cdot 1_A$, one which doesn't require $c=\infty$ or $m(A)=\infty$ to be special cases. But measure theory doesn't change in any way if you use a different convention.
@NateEldredge right, exactly.
@NikWeaver so we cannot have a measure theory with $\infty \cdot 0 = \infty$? I don't see how do you preclude it?
|
2025-03-21T14:48:30.972456
| 2020-05-12T23:51:47 |
360178
|
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|
Stack Exchange
|
Cartesian dissimilarity of a function $\ f:A^3\to A^3\ $ and its inverse
Let $\ A\ $ be an arbitrary set. Let $\ |A|>1\ $ (to avoid triviality). Let each of the functions $\ f_k:A^{\{1\ 2\ 3\}}\to A\ $ depend on all three arguments for $\ k=1\ 2\ 3,\ $ while each of the functions $\ g_k:A^{\{1\ 2\ 3\}}\to A\ $ does not depend on $k$-th variable, for each $\ k=1\ 2\ 3.$
I'll explain "dependent/independent" at the end of this note. It'll be preceded by "diagonal product $\ \triangle\ $ of functions (here, of three of them).
Assume also that
$$ f:=f_1\triangle f_2\triangle f_3\ \ \text{and}
\ \ g:=f_1\triangle f_2\triangle f_3\, :\,A^3\to A^3 $$
are inverse one to another. There are (historic) examples when $\ A\ $ is countable or of cardinality continuum, and for all infinite cardinals except, possibly, for the weird ones. There are examples (by analogy) whenever $\ |A|\ $ is finite and odd; there should be some examples when $|A|$ is not a power of $2$, but -- I conjecture -- never when it is:
..............................................................
CONJECTURE: cardinality $\ |A|, $ when it's finite, is not a (non-trivial) power of $\ 2\ $ (is different from $\ 2^k\ $ for any natural $k=1\ 2\ \ldots$).
-----------------------------------------------------------
EXAMPLES:
Let $\ A = \Bbb Q\ $ or $\ \Bbb R\ $ or $\ \Bbb Z_n\ $ for arbitrary odd $\ n>1.\ $ Define:
$\ f_1(a\ b\ c)\ :=\ b+c-a $
$\ f_2(a\ b\ c)\ :=\ a+c-b $
$\ f_2(a\ b\ c)\ :=\ a+b-c $
and
$\ g_1(a\ b\ c)\ :=\ \frac{b+c}2 $
$\ g_2(a\ b\ c)\ :=\ \frac{a+c}2 $
$\ g_2(a\ b\ c)\ :=\ \frac{a+b}2 $
Then, let $\ f\ $ and $\ g\ $ be defined as above. This establishes the odd cardinality case.
Remark: Every finite abelian group $\ X\ $ of odd order
($\ |X|\, $ -- odd) will do in place of $\ Z_n\,$ (with odd $n$).
The finite even cardinalities different from $\ 2^n\ (n\in\Bbb N),\ $ present a mixed story in an analogy to other combinatorial themes which admit exotic examples. This would be the complementary conjecture -- the mixed picture.
-----------------------------------------------------------
Diagonal product of functions (or morphisms)
Consider set $\ X\ $ and sets $\ Y_q\ $ and functions
$\ f_q:X\to Y_q\ (q\in Q).\ $ Then, the diagonal product
$\ f:=\triangle_{q\in Q} f_q :X\to\prod_{q\in Q} Y_q\ $
is given by:
$$ \forall_{q\in Q}\quad \pi_q\circ f\ := f_q $$
i.e.
$$ \forall_{q\in Q}\,\forall_{x\in X}\quad (f(x))(q)\ := f_q(x) $$
-----------------------------------------------------------
Dependent / independent
Let $X\ Y\ T\ $ be arbitrary sets, and $\ s\in T.\ $ Elements $\ x\in X^T\ $
are functions $\ x: T\to X$.
A function $\ f:X^T\to Y\ $ does not depend on (is independent of) variable
$\ s\ \Leftarrow:\Rightarrow$
$$ \exists_{f_s\in X^{T\setminus\{s\}}}\,\forall_{x\in X^T}\quad
f(x)=f_s(x|T\setminus\{s\}) $$
Otherwise, $\ f\ $ depends on variable $\ s,\ $ i.e.
$$ \exists_{w\ x\,\in\,X^T}\ \ (\, w|T\!\setminus\!\{s\}\,=\,
x|T\!\setminus\!\{s\}\quad\text{and}\quad f(w)\ne f(x) \,) $$
====================================
PS. In the style of Q&A, I've provided only the special case of the general question about the number of independent variables of a function and its inverse, $\ f\ $ and $\ g.\ $ Actually, we want to know the whole structure of sets of independent variables for $\ f\ $ and $\ g.\ $ This question is as fundamental as it goes, hence it belongs to the theory of the Foundations of Mathematics. (Please, someone reattach the related tag to my question).
can you define the operation $\Delta$?
I assume $(f_1\triangle f_2\triangle f_3)(x_1,x_2,x_3)$ just means $(f_1(x_1,x_2,x_3),f_2(x_1,x_2,x_3),f_3(x_1,x_2,x_3))$.
Could you give the examples you have in mind when $|A|$ has an odd prime divisor?
@kodlu, yes, as YCor said; I've added the definition anyway.
@YCor, I gave the odd case examples. The rest is unclear. Earlier I was a bit sloppy about the even case when there is an odd divisor > 1.
@WlodAA: So your question is the conjecture? If so, then please notice that conjectures are considered off-topic on MO (even if they are of research level), so your question will likely get closed (even though it might be mathematically interesting).
@AlexM., I don't care. I am into mathematics. That's why somehow I am still here but just barely. Perhaps I'll leave MO after this incident for good.
(I still can't believe it that an original conjecture can be against MO rules, how ridiculous!)
@WlodAA: It is very reasonable, in fact, if you think about it for a minute. MO is a place that attempts to produce, and then store, definite answers. A conjecture, on the other hand, might not have answers, by its very nature. A conjecture might generate an interesting discussion, but no definite answer. Therefore, it would be more appropriate for a forum, which MO is not. Furthermore, there are plenty of wannabe researchers who would flood MO with conjectures if there were allowed (think of how easy it is to produce them in additive number theory). This is why they are not accepted.
@AlexM. this "conjecture ban" is ridiculous in this case. It concerns "famous conjectures" and variants thereof. In a question like this, calling it "conjecture" is like saying "I strongly believe that the answer to my question is yes", and I see no point in your personal initiative to discourage this.
With computer (Sagemath) if not mistaken, I checked that there is no solution for $n=2$. Just by brute force checking for all $4096$ triples of functions $u,v,w,2^2\to 2$, whether $(x,y,z)\mapsto (u(y,z),v(z,x),w(x,y))$ is injective and the inverse function satisfies that its coordinates depends on all arguments. I'll have a look for $n=4$.
@YCor, thank you for the pro-conjecture support and for your computations. #### "I'll have a look for =4." -- where is John von Neumann when we need him?
The problem of existence is solved (by Alexei Kulikov) as there are solutions for all $n\ge 3$. Still counting solutions is natural. Actually, it sounds even more natural to count $u_n$, the cardinal of the set $U_n$ of permutations of $[n]^3$ of the form $(x,y)\mapsto (a(y,z),b(z,x),c(x,y))$, and $v_n$, the cardinal of the subset $V_n$ of those for which no coordinate of the inverse is constant w.r.t some coordinate. For instance $U_n$ is stable under pre and post-composition by permutations of coordinates (which is an action of $Sym(n)^6$ on $Sym([n]^3)$).
OP: "(This question [...] belongs to [...] Foundations of Mathematics. (Please, someone reattach the related tag to my question)." I added "co.combinatorics" (to which it primarily belongs), removed "lo.logic" and later Andres Caicedo removed "set-theory". If there is a motivation from foundations, feel free to put back one of these tags or both (and erase the parenthesis, maybe even put the motivation at the beginning). As far as I'm concerned I won't try to revert (but please keep co.combinatorics since it's interesting for its own sake as combinatorial question).
@YCor, of course, the question belongs to combinatorics, it's most natural; thank you, YCor. (Somehow, I missed the combinatorics tag).
I've accepted @AlekseiKulikov answer. Let's also remember the YCor's contributions. I hope that this thread is only the beginning of the topic of the Cartesian asymmetric of pairs $\ f\ g\ $ of functions inverse one to another. One possible development direction is about the algebra induced by the situation (sub-directions: additional assumptions). Another, let's go beyond dimension $3$.
$\newcommand{\F}{\mathbb{F}}$
Your conjecture is false, I will construct a counterexample for all powers of two $2^n$ with $n \ge 2$.
Let us identify $A$ with $\F_{2^n}$. An example from OP corresponding to the odd numbers, is a linear mapping. Our functions also would be linear. Let $G$ be a matrix constructed from the functions $g_1, g_2, g_3$, as rows and $F$ be its inverse, constructed of the functions $f_1, f_2, f_3$. Then we want the following things: matrix $G$ has zeroes on the diagonal and matrix $F$ does not have zero entries. If we start with the matrix $G$ with the zeroes on the diagonal then the elements of $F$ are product of some two elements of $G$, divided by the determinant. So as long as all non-diagonal elements of $G$ are non-zero and the determinant is non-zero we won. This can be achieved for all $n \ge 2$ by, for example, the following construction:
Let $a \in \F_{2^n}$, $a\ne 0, 1$ and consider functions
$$g_3(x, y, z) = x + y,$$
$$g_2(x, y, z) = x + z,$$
$$g_1(x, y, z) = y + az.$$
then the functions $f$ are
$$f_1(x, y, z) = \frac{1}{a+1}(ay-x+z),$$
$$f_2(x, y, z) = \frac{1}{a+1}(-ay+az+x),$$
$$f_3(x, y, z) = \frac{1}{a+1}(y+x-z).$$
Since $a\ne 1$ we have $a+1\ne 0$ (we are in the field of characteristic two) and so it is a working example.
For the case $N = 2^kM$, $M$ odd, $M > 1$ we can just copy $2^k$ times the construction for $M$ from the OP. It leaves only the cases $N = 1, 2$. For $N = 1$ there are obviously no solutions and for $N = 2$ one can simply bruteforce all the potential cases (as was done by Ycor in the comments while I was writing my answer).
EDIT here is a simpler construction which doesn't need finite fields but only modular arithmetics and work for all $m \ge 3$. We will work in $\mathbb{Z}/m\mathbb{Z}$. Consider $$g_3(x, y, z) = x +y,$$ $$g_2(x, y, z) = x + z,$$ $$g_1(x, y, z) = y-2z.$$ Then the functions $f$ are
$$f_1(x, y, z) = -z+2y+x,$$
$$f_2(x, y, z) = 2z-2y-x,$$
$$f_3(x, y, z) = z-y-x.$$
Actually, my answer works for all rings $R$ with at least one invertible element $b\ne 1$ (just put $a = b -1$), in particular, for all $\mathbb{Z}/\mathbb{Z}_N$, $N \ge 3$.
In the question it's required that $g_1$ doesn't depend on $x$, $g_2$ doesn't depend on $y$, $g_3$ doesn't depend on $z$.
@YCor oh well, yes, you're right, I will fix this issue
If the commutative finite ring has invertible elements $a,b$ with $a-b=1$, then you can get a $3\times 3$ invertible matrix with zero diagonal, whose inverse has only nonzero entries. For given $n$, there exists a ring of cardinal $n$ with this property iff $n\notin 4\mathbf{Z}+2$ (by Chinese theorem one boils down to $n=q\neq 4$ prime-power in which case a field structure works). So powers of 2 are settled. For $6\le n\in 4\mathbf{Z}+2$ however we can't expect to solve the problem with an abelian group structure and an endomorphism, since projecting to the 2-part yields a contradiction.
@YCor I'm not sure what you mean by this comment. We don't need $b$ (in your comment's notation) to be invertible! Only nonzero.
Invertible ensures the non-vanishing of the coefficient. I'll post a partial answer to clarify my points without skipping details.
@YCor sorry, I do not understand at all what you are talking about. My example in the edit of the post works for all $n \ge 3$, including $2 \mod 4$. For the $3\times 3$ matrix that you mentioned to exist we need $a, b$ with $a-b = 1$, $a$ invertible and $b\ne 0$, there is no need for $b$ to be invertible as well.
Yes you're right, sorry. The condition of depending on all coordinates is weak enough to not be affected by projecting on the 2-part. So there are examples for $n$ iff $n\ge 3$ (modulo double-checking my computation) [corrected: there are none for $n=0,1$].
@YCor there are no examples for $n = 0,1$ for vacuous reasons :)
...Namely $\begin{pmatrix}0 & a & b\ 1 & 0 & 1\ 1 & 1 & 0\end{pmatrix}$ for $a+b=1$, $a$ invertible, $b\neq 0$, the determinant being 1 and inverse being $\begin{pmatrix}-1 & a & a\ 1 & -b & b\ 1 & b & -a\end{pmatrix}$.
@YCor actually, your example seems even better because it actually doesn't need $a$ to be invertible, only nonzero! So apparently there's an example in any (commutative) ring with at least three elements...
Aleksei and YCor, this is so nice!
@YCor, visually, you can make your matrices more pleasing(?), i.e. more symmetric (just visually) by applying $\ a\ c\ $ in place of $\ a\ b.\ $ But then, poet Julian Tuwim said that "symmetry is an idiot's aesthetics."
$$\left[\begin{array}{ccc}
-c & c & a\
1 & -1 & 1\
c & a & -a\end{array}\right]\cdot
\left[\begin{array}{ccc}
0 & a & 1\
1 & 0 & 1\
1 & c & 0\end{array}\right]\ =\ \Bbb I_3,\ $$
where a+c=1.
Actually, without the assumption about $\ a+c=1,\ $ the above matrix product equals to $\ (a+c)\cdot\Bbb I_3.\ $ This certainly has potential for number theory and algebraic geometry; in particular, the determinant of the above left matrix is equal to $\ (a+b)^2.$
|
2025-03-21T14:48:30.973216
| 2020-05-13T00:30:47 |
360182
|
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|
Stack Exchange
|
Are there infinitely many primes of the form $x^2+(x+y)y^2$?
Heath-Brown [Acta Math. 186(2001), 1-84] proved in 2001 that there are infinitely many primes of the form $x^3+2y^3$ with $x,y\in\mathbb Z^+=\{1,2,3,\ldots\}$.
In contrast, here I ask the following question.
Question 1. Are there infinitely many primes of the form $x^2+xy^2+y^3=x^2+(x+y)y^2$ with $x,y\in\mathbb Z^+$? Is it possible to provide a positive answer with a proof?
This question is motivated by my following conjecture (cf. http://oeis.org/A232174).
Conjecture 1. Any integer $n>1$ can be written as $x+y$ with $x,y\in\mathbb Z^+$ such that $x+ny$ and $x^2+ny^2$ are both prime.
This conjecture implies that Question 1 has a positive answer.
By a theorem of Heath-Brown and Moroz [Proc. London Math. Soc. 84(2002), 257-288], there are infinitely many primes of the form $x^3+(x+y)y^2$ with $x,y\in\mathbb Z^+$. Here I ask the following further question.
Question 2. Are there infinitely many primes of the form $p^3+(p+q)q^2$ with $p$ prime and $q\in\mathbb Z^+$?
This was motivated by my following conjecture (cf. http://oeis.org/A232186).
Conjecture 2. Any integer $n>2$ can be written as $p+q$ with $q\in\mathbb Z^+$ such that $p$ and $p^3+nq^2$ are both prime.
Conjecture 2 implies that Question 2 has a positive answer.
Heath-Brown and Moroz.
Thanks. By Heath-Brown and Moroz's result, there are infinitely many primes of the form $x^3+(x+y)y^2$. Now we focus on primes of the form $x^2+(x+y)y^2$.
It is possible to show that $x^2 + xy^2 + y^4$ takes on infinitely many prime values (and possibly with $y$ being restricted to be prime as well) by generalizing Friedlander-Iwaniec (and Heath-Brown/Li respectively), but these methods won’t work with $x^2 + xy^2 + y^3$.
I have extended the posting by adding Question 2 and Conjecture 2.
|
2025-03-21T14:48:30.973346
| 2020-05-13T02:41:42 |
360189
|
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|
Stack Exchange
|
Representation of central extension
Let $G$ be a finite abelian group of rank $n$ and $H\rightarrow G$ a central extension with cyclic finite kernel.
Is it true that we can find a faithful representation $H\rightarrow {\rm GL}_{k(n)}(\mathbb{C})$ where $k(n)$ only depends on $n$?
I feel something like this must be true from the fact that $G$ should admit an irreducible projective representation into ${\rm PGL}_{k(n)}(\mathbb{C})$ where $k(n)$ only depends on $n$.
The answer is no.
According to Theorem 1.3 of this paper, if $H$ is a nilpotent group of class $2$ with cyclic commutator subgroup, then the minimal degree $m_\mathsf{f}(H)$ of a faithful complex representation of $H$ is given by $$m_\mathsf{f}(H) = \sqrt{|H:Z(H)|} + m_\mathsf{f}(Z(H)) - 1.$$
So, for example, the minimal degree of the group $$G = \langle x,y,z \mid x^n=y^n=[x,z]=[y,z]=1, [x,y]=z \rangle$$ of order $n^3$
is $n$, and $G/Z(G)$ is abelian of rank 2.
Actually, it is enough to evoke Jordan. If one takes the group of upper unipotent triangular $3\times 3$ matrices (Heisenberg) over $\mathbf{Z}/p\mathbf{Z}$, it has rank 2 but no abelian subgroup of index $<p$. Hence the dimension of its smallest faithful representation tends to infinity when $p\to\infty$. (There are explicit sharp bounds for the Jordan theorem, namely $(n+1)!$ for large $n$, which here are very far from the right bound. Actually Wedderburn directly yields that the smallest faithful representation has dimension $p$ and it's maybe easier than Jordan's older theorem.)
Here is an elementary proof of a related general fact. Let $H$ be any finite nilpotent group with $H^{\prime} = Z(H)$ cyclic of order $m$. Let $z$ be a generator of $Z(H)$. Then $\langle z \rangle $ has a (faithful) linear character $\lambda$ such that $\lambda(z)$ is a complex primitive $m$-th root of unity.
Note then that all irreducible constituents of ${\rm Ind}_{Z(H)}^{H}(\lambda)$ are faithful. For let $\chi$ be one such. Then by Frobenius recipirocity (and Clifford' Theorem), ${\rm Res}^{H}_{Z(H)}(\chi) = \chi(1)\lambda$, so that ${\rm Res}^{H}_{Z(H)}(\chi)$ is certainly faithful. On the other hand, if $\chi$ were not faithful, then ${\rm ker} \chi$ contains a minimal normal subgroup $M$ of $H$. Since $H$ is nilpotent, $M \leq Z(H)$ ( for $M \cap Z(H) \neq 1$ and $M$ is minimal), contrary to the fact that $M \leq {\rm ker} \chi$ and ${\rm Res}^{H}_{Z(H)}(\chi)$ is faithful.
Furthermore, ( as is well-known, and may be found in the character theory text of I.M. Isaacs for example), since $H$ is nilpotent, it follows that if $\theta$ is any faithful irreducible character of $H$, then $\theta$ vanishes identically outside $Z(H)$. For choose $a \in H \backslash Z(H)$ and choose $b \in H \backslash C_{H}(a)$ we have
$[a,b] = w$ for some $1 \neq w \in Z(H)$. Then $b^{-1}ab = wa$.
Hence $\theta(a) = \theta(b^{-1}ab) = \theta(wa)$. But by Schur's lemma, $w$ is represented by a scalar matrix in any representation affording character $\theta$, and the scalar, $\alpha$ say, is not $1$ as $\theta$ is faithful . Hence $\theta(a) = \theta(wa) = \alpha \theta(a) $, so that $\theta(a) = 0$.
Hence for our previous faithful irreducible character $\chi$ of $H$, the orthogonality relations yield $|H| = \sum_{ h \in H}|\chi(h)|^{2} = |Z(H)|\chi(1)^{2}$, so that $\chi(1) = \sqrt{[H:Z(H)]}.$
|
2025-03-21T14:48:30.973572
| 2020-05-13T03:46:47 |
360194
|
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|
Stack Exchange
|
Does Merkurjev's argument help Voevodsky's program?
In the talk
Unimath - its present and its future, July 10, 2017. Video and slides of a talk, Isaac Newton Institute for Mathematical Sciences, Cambridge. (abstract)
Voevodsky mentioned that he was still interested in formalising his pre-Univalent foundations work, but on page 29 of the slides says:
Another direction is the one that I have stated in my Bernays lectures at the
ETH in 2014 - to formalize a proof of Milnor’s conjecture on Galois
cohomology.
It has not been developing much. On the one hand, I discovered that
formalizing it classically it is not very interesting to me because I am quite
confident in that proof and in its extension to the Bloch–Kato Conjecture.
On the other hand, when planning a development of a constructive version of
this proof one soon encounters a problem. The proof uses the so called
Markurjev–Suslin transfinite argument that relies on the Zermelo’s wellordering theorem that in turn relies on the axiom of choice for sets.
There is a note to the abstract, which says (I have reformatted slightly):
Toward the end of the talk, Voevodsky says it would be good to find a constructive proof to replace the "Merkurjev–Suslin transfinite argument" in the proof of Bloch–Kato. Merkurjev says they did not use any transfinite induction in their argument, and has provided the following additional details:
"Let $F$ be a field, $S$ the set of all $n$-symbols over $F$ (modulo $\ell$) and $T$ the set of all finite subsets of $S$. We order $T$ by inclusion. For any $A$ in $T$ let $X_A$ be the product of norm varieties $X_a$ for all symbols $a$ in $A$. If $A$ is a subset of a finite set $B$ in $T$, we have projection $X_B \to X_A$ and hence a field homomorphism $F(X_A) \to F(X_B)$. Let $F_1$ be the colimit of this system of fields. Then construct a field extension $F_2/F_1$ the same way (replacing $F$ by $F_1$), etc. We get a tower of field extensions $F$ in $F_1$ in $F_2$ in ... Finally, let $K$ be the union of all $F_i$. Clearly, every $n$-symbol over $K$ is trivial modulo $\ell$."
See also Theorem 1.12 of the book The Norm Residue Theorem in Motivic Cohomology by Haesemeyer and Weibel, which presents a transfinite version of the argument.
So this raises the question:
what problem is left regarding formalising the argument and/or finding the desired constructive proof?
I presume there are a lot of details, but is it known what might be the major obstacle? Or is the argument provided by Merkurjev still insufficient to address Voevodsky's concerns: the "transfinite argument" is not needed, but there are other issues with the simpler construction?
I'm not interested in the merit or otherwise of Voevodsky's choice of approach to a univalent foundations proof of B–K, just whether this specific wrinkle has been smoothed.
|
2025-03-21T14:48:30.973812
| 2020-05-13T05:26:33 |
360197
|
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|
Stack Exchange
|
Example of a non-reflexive Banach space and two sequences
Let $(E,\mathcal {A}, \mu ) $ be a finite measure space and $X$ be a Banach space. The set of all Bochner-integrable functions from $E$ into $X$ is denoted by $\mathcal{L}_X^1$.
If $X$ is reflexive, we have the following theorem
Theorem 1
Let $ (f_n)_{n\geq 1} \subset \mathcal {L}_{X}^1$ is a sequence with : $$\sup_n \int_{E}{\|f_n\| d\mu} < \infty .$$
Then there exist $ h _{\infty} \in \mathcal {L}_{\mathbb {R}}^1 $ and a sub-sequence $ (g_k)_k $ of $(f_n)_n $ such that for every sub-sequence $ (h_m)_m $ of $(g_k)_k$ : $$ \frac{1}{i}\sum_{j=1}^{i}{h_j(t)}\to h _{\infty}(t) \text{ weakly in }X\text{ a.e. }$$
Proof of this result exists in the article "Infinite-dimensional extension of a theorem of Komlos" by Erik J. Balder (Theorem A).
If $X$ is not reflexive, we have the following theorem
Theorem 2
Let $ (f_n)_{n\geq 1} \subset \mathcal {L}_{X}^1$ is a sequence with :
$$
\begin{cases}
\bullet~~ \{f_n(t)\}\text{ is relatively weakly compact a.e.,}\\
\bullet~~ \sup_n \int_{E}{\|f_n\| d\mu} < \infty.\\
\end{cases}
$$
Then there exist $ h _{\infty} \in \mathcal {L}_{\mathbb {R}}^1 $ and a sub-sequence $ (g_k)_k $ of $(f_n)_n $ such that for every sub-sequence $ (h_m)_m $ of $(g_k)_k$ : $$ \frac{1}{i}\sum_{j=1}^{i}{h_j(t)}\to h _{\infty}(t) \text{ weakly in }X\text{ a.e. }$$
Proof of this result exists in the article "Infinite-dimensional extension of a theorem of Komlos" by Erik J. Balder (Theorem B).
My problem:
I want an example of a non-reflexive Banach space and two sequences, such that:
The first sequence is bounded in $\mathcal {L}_{X}^1$ but it does not verify the consequence of Theorem 1.
The second sequence verifying the hypotheses of Theorem 2 and its consequences.
The second point is easy: take $f_n \equiv 0$.
|
2025-03-21T14:48:30.974066
| 2020-05-13T08:14:44 |
360200
|
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"José Figueroa-O'Farrill",
"LSpice",
"Robert Bryant",
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"url": "https://mathoverflow.net/questions/360200"
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|
Stack Exchange
|
Vector field with constant divergence around embedded submanifold
Let $M$ be a smooth $n$-dimensional manifold and $N\subset M$ be a closed embedded submanifold of codimension at least $2$. Furthermore, let $\mu$ be a volume form on $M$.
Question: Does there exists a vector field $X$ on an open neighborhood $U\subset M$ of $N$ (arbitrary small neighborhood of $N$, not fixed!) such that its divergence with respect to $\mu$ is $\equiv1$ (divergence is defined by $\mathcal{L}_{X}\mu = \text{div}(X)\mu$) and $X(p)=0$ for all $p\in N$?
I am not sure about the assumption on the codimension to be at least $2$. What is important for me is that the codimension is "big enough" (definitely it should not be $1$).
NEW QUESTION: Is the volume form $\mu$ on a sufficiently small neighborhood of $N$ exact, i.e. does there exists a $n-1$ form $\eta$ such that $d\eta = \mu$? And if so, is there any chance that this solves the original question? Under what conditions or further assumptions would this "NEW QUESTION" solve the original one?
Greetings,
Stan
Is the open neighbourhood $U$ part of the data or are you asking whether there exists some $U$ where the vector field satisfies the conditions?
actually the open set $U$ should be arbitrary small neighborhood on $N$. Hence not fixed. I will edit the question. Thanks for the advice.
Yes the volume form is exact. The de rham cohomology of a non compact manifold vanishes in top degree.
yes! what further assumptions are necessary so that exactness of $\mu$ solves the orginal question?
then since $\mu=d\eta$ and $\eta$ is a $n-1$ form there exists a vector field $X$ such that $\iota_{X}\mu = \eta$. Hence divergence of $X$ is $1$. But there is no guarantee that $X$ vanishes on $N$. Under what conditions does $X$ vanish on $N$? Or can it be choosen such that it vanishes?
I don't understand your quantification of $U$, but @JoséFigueroa-O'Farrill's makes sense to me. Is it correct to say that you want the divergence to be identically $1$ on some neighbourhood $U$ of $N$?
yes! $U$ is arbitrary small around $N$.
The answer is 'yes, there always is such a vector field $X$' and, in particular, the answer to your 'new question' is also 'yes'. (In fact, the first 'yes' implies the second 'yes', but the second 'yes' is used in the proof of the first 'yes'.)
Here is an outline of the proof (and it works also in the codimension $1$ case; I'm not sure why you want to exclude that):
First, since things are local around $N$, you can reduce to the case that $M$ is a vector bundle and $N$ is the zero section of that vector bundle (in which case, we can obviously identify $N$ with the base of the vector bundle). This is a standard result in differential topology, and one can find proofs in many places. Basically, you fix a Riemannian metric on $M$ and use the normal exponential map to show that a neighborhood of $N$ in $M$ is diffeomorphic to a neighborhood of the zero section of the normal bundle to $N$.
So suppose that $M^{n+k}$ is a vector bundle over $N^n$ and that $\mu$ is a volume form on $M$. Let $R$ be the vector field on $M$ whose time $t$ flow is scalar multiplication by $e^t$ in the fibers of $M\to N$ (which are vector spaces, so scalar multiplication is well-defined). Note that $R$ vanishes along $N$.
Then I claim that, for $k\ge 1$, there is a unique function $f$ on $M^n$ such that the vector field $X = fR$ has divergence $1$ with respect to $\mu$, i.e., that $\mathcal{L}_X\mu = \mathrm{d}\bigl(f\,\iota_{R}\mu\bigr) = \mu$, where $\iota_R\mu$ is the $(n{+}k{-}1)$-form that is the 'interior product' (aka 'left hook') of $R$ with $\mu$.
This is a claim about the unique solvability of a singular, linear first-order PDE: Let $\mathrm{d}\bigl(\iota_{R}\mu\bigr) = \kappa\,\mu$. Then $\kappa$ is a smooth function on $M$ that satisfies $\kappa(p) = k>0$ for all $p\in N$. I am claiming that there exists a unique smooth (and positive) function $f$ on $M$ that satisfies the linear inhomogeneous equation
$$
\mathrm{d}f(R) + \kappa\,f = 1.
$$
Since $R$ vanishes along $N$, this implies $f(p) = 1/k$ for all $p\in N$.
You can see what makes this a little delicate by examining what happens along a single line in a single fiber: If $v\in M_p$ is a nonzero element in the fiber over $p$, we can parametrize the line $\mathbb{R}v\subset M_p$ by $t\mapsto t{\cdot}v$ for $t\in \mathbb{R}$. In this case, the PDE to be solved becomes an ODE
$$
tf'(t) + h(t)\,f(t) = 1
$$
where $h$ is a smooth function on $\mathbb{R}$ that satisfies $h(0) = k>0$. This is a regular, singular ODE, and while its (unique) smooth solvability is a classical fact, this seems not to be that well-known these days. The uniqueness on each line tells you that there is at most one smooth solution $f$ to the full equation on $M$, and it's not hard to use the uniqueness to show that, in fact, you get global smoothness for $f$ as well.
If you are doubtful, I can supply the proof. While this is a classical ODE/PDE fact, I don't, off-hand, remember a good source, just a proof (which must be the standard one).
Added remark about ODE/PDE fact: First, consider the ODE on the real line $\mathbb{R}$
$$
t\,f'(t) + h(t) f(t) = g(t)
$$
where $h$ and $g$ are given smooth functions on $\mathbb{R}$ and $h(0) = k>0$. We want to show that there is a smooth solution $f$ and that it is unique. Write $h(t) = k - t m'(t)$ for some smooth function $m$ on the real line. When $g = 0$, the only solutions are $f(t) = c\,t^{-k}\mathrm{e}^{m(t)}$, where $c$ is a constant, and hence only when $c=0$ is the solution smooth. For general $g$, we use variation of parameters and look for a solution of the form $f(t) = c(t) t^{-k} \mathrm{e}^{m(t)}$
for some function $c(t)$ that vanishes to order at least $k$ at $t=0$. Substituting this into the above equation, we find that $c'(t) = t^{k-1}g(t)\mathrm{e}^{-m(t)}$, so, since we want $c(0)=0$, we have
$$
c(t) = \int_0^t \tau^{k-1}g(\tau)\mathrm{e}^{-m(\tau)}\,d\tau.
$$
Since $k>0$, this integral vanishes to order $k$ at $t=0$. Thus,
$$
f(t) = \mathrm{e}^{m(t)}t^{-k}\int_0^t \tau^{k-1}g(\tau)\mathrm{e}^{-m(\tau)}\,d\tau
$$
is a smooth solution. It is unique because we have already shown uniqueness when $g=0$. For use below, note that, if $g$ vanishes to order $l>0$ at $t=0$, then so does $f$, and we always have $f(0) = g(0)/k$.
Now, returning to the general case of $\mathrm{d}f(R) + \kappa f = 1$,
we see, by applying the above argument to each line in $M_p\subset M$
for $p\in N$, that there is a unique function $f$ on $M$ that satisfies this equation and that $f$ is smooth on $M$ except possibly along the zero section $N\subset M$ itself. It is clearly smooth on every line through $0_p\subset M_p$, but one might worry that it is not smooth on a neighborhood of $N$ in $M$.
However, the following argument shows that this is not the case: Let $F_k(M,N)$ denote the space of smooth functions on $M$ that vanish to order at least $k$ along $N$. This is a decreasing filtration of $C^\infty(M) = F_0(M,N)\supset F_1(M,N)\supset\cdots$. One easily sees that the linear operator $D(f) = \mathrm{d}f(R) + \kappa\,f$ maps $F_i(M,N)$ into $F_i(M,N)$ and hence induces a linear operator $D_i:G_i(M,N)\to G_i(M,N)$ on the associated graded
$$
G_i(M,N) = F_i(M,N)/F_{i+1}(M,N)\simeq S^i(M^*)
$$
(bearing in mind that $M$ is a vector bundle over $N$). Since $R$ is the Euler (radial) vector field on each fiber $M_p$ and since $\kappa(p)=k$ for all $p\in N$, it follows that $D_i$ is simply multiplication by $i{+}k$ for $i\ge0$ and hence is an isomorphism of $G_i(M,N)$ with itself for $i\ge0$.
Consequently, for every $i\ge0$, there exists a smooth $f_i$ on $M$
such that $\mathrm{d}f_i(R) + \kappa\,f_i = 1 - h_i$ where $h_i\in F_{i+1}(M,N)$. Using the above 'integration construction on each line', we can find a function $u_i$ on $M$ that is smooth away from $N$, vanishes to order $i{+}1$ along $N$ and satisfies $\mathrm{d}u_i(R) + \kappa\,u_i = h_i$. It follows that $f = f_i + u_i$. Thus, for every $i\ge0$, $f$ can be written as the sum of a function $f_i$ that is smooth on $M$ and a function $u_i$ that vanishes to order $i{+}1$ along $N$ and is smooth away from $N$. Consequently, $f$ is differentiable to all orders along $N$ and hence is smooth on all of $M$ as desired.
@Robert Bryant:
thank you very much for the answer. Unfortunately I have not been able to register completely and my account details have been deleted. therefore i cannot mark the question as "answered". the one I'm writing about is a new accout. I am very sorry. I also had a similar idea about the existence of the solution, but I didn't manage to finish. I would be very happy about details uniqueness and smoothness of the solutions.
PS. with this new accout I am not able to write this as a comment.
Greetings,
Stan
I believe that you can ask the moderators to merge the two accounts. (These kinds of mistakes happen from time to time, and the only way to fix them is to ask the moderators to do it.)
StanleyT, please go here: https://mathoverflow.net/help/merging-accounts
|
2025-03-21T14:48:30.974921
| 2020-05-13T08:42:29 |
360204
|
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|
Stack Exchange
|
Can the joint law $P \circ (X,Y)^{-1}$ of two random variables $X$ and $Y$ be written as $P \circ (X,\phi(X,U))^{-1}$ for $U$ uniform in $[0,1]$?
I want to know whether there is some general assumpitons we can make on two measurable spaces $E$ and $F$ (e.g. polish, complete, separable,...) such that we can ensure that the following "Theorem" holds:
Theorem: Suppose that $X: \Omega \rightarrow E$ and $Y: \Omega \rightarrow F$ are two random variables on a probability space $(\Omega, \mathcal{F},P)$. Then there exists a probability space $(\hat{\Omega}, \hat{\mathcal{F}}, \hat{P})$, random variables $\hat{X}:\hat{\Omega} \rightarrow E$ and $\hat{U}: \hat{\Omega} \rightarrow [0,,1]$ with $\hat{P} \circ X^{-1} = P \circ X^{-1}$ and $\hat{U}$ uniformly distributed in $[0,1]$, and a product-measurable mapping $\phi : E \times [0,1] \rightarrow F$ such that
$$ P \circ (X,Y)^{-1} = \hat{P} \circ (\hat{X}, \phi(\hat{X}, \hat{U}))^{-1}.$$
I saw something similar when proving existence of discrete time markov chains, where $E$ was assumed to be at most countable and endowed with the discrete $\sigma$-algebra. And I also came across this paper by Skorohod which assumes (in the above setting) that $E=F$ is a Polish space and proves an even stronger result then the above. But what general assumptions can we make about $E$ and $F$ such that the above can hold?
Thanks a lot in advance!
Lemma. Let $(\Omega,\mathcal{A},\mathbb{P})$ be a probability space, $X$ a real-valued random variable on this space, and $\mathcal{C} \subset \mathcal{A}$ a $\sigma$-field. Let $F$ be the cumulative distribution function of the conditional law $\mathcal{L}(X \mid \mathcal{C})$. Let $\xi$ be a random variable with uniform law on $[0,1]$ and independent of $\mathcal{C}\vee\sigma(X)$. Define
$$
U = F^-(X) + \xi\bigl(F(X) - F^-(X)\bigr).
$$
Then $U$ is a random variable independent of $\mathcal{C}$, uniformly distributed on $[0, 1]$, and one has $X = G(U)$ where $G$ is the right-continuous inverse function of $F$.
Proposition. Let $(\Omega,\mathcal{A},\mathbb{P})$ be a probability space. Let $\mathcal{C}\subset\mathcal{A}$ be a $\sigma$-field, $V$ be a real-valued random variable, and define $\mathcal{B} = \mathcal{C}\vee\sigma(V)$. Then there exist an embedding $\iota$ from $(\Omega,\mathcal{A},\mathbb{P})$ to a probability space $(\hat{\Omega},\hat{\mathcal{A}},\hat{\mathbb{P}})$ and a uniform random variable $\hat U$ independent of $\hat{\mathcal{C}}$ such that $\hat{\mathcal{B}} \subset \hat{\mathcal{C}} \vee \sigma(\hat U)$.
The proposition is proved with the help of the lemma. The proofs are given in my paper On standardness and cosiness (Lemma 3.24 and Proposition 3.25).
Apply this proposition to $\mathcal{C} = \sigma(X)$ and $V$ real-valued such that $\sigma(V) = \sigma(Y)$, which exists assuming $E_2$ is a standard Borel space.
Tell me if you need more details.
|
2025-03-21T14:48:30.975132
| 2020-05-13T09:39:19 |
360207
|
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|
Stack Exchange
|
Flatness of the Hitchin system?
The Hitchin fibration is a central topic of study in modern geometry. It seems to be folklore knowledge that the morphism from the coarse moduli space of semi-stable Higgs bundles to the Hitchin base (the direct sum of spaces of global sections of powers of the canonical bundle) is flat. Where is this proven?
If you are okay working with the Hitchin map defined on the moduli stack of Higgs bundles rather than the coarse moduli space, then you can find a proof of this statement in the paper The global nilpotent variety is Lagrangian by V. Ginzburg, at least in the case that the genus of the base curve is at least $2$. The precise reference is Corollary 9.
The idea is to show that the global nilpotent cone (the most singular fiber of the Hitchin map) has the same dimension as $Bun_G$. This implies that the Hitchin map is equidimensional and that the stack of Higgs bundles $T^*Bun_G$ is lci. Since the Hitchin base is non-singular, this implies flatness by miracle flatness (Stacks Project Lemma 00R4). Now the stack of semi-stable Higgs bundles is an open substack so the restriction of the Hitchin map to this locus is also flat.
Edit: I think the result also holds for the coarse moduli space of semi-stable Higgs bundles but it seems to be a bit subtle. I think equidimensionality of the coarse Hitchin map follows from that on the stack. Then by miracle flatness the map is flat if and only if the coarse space is Cohen-Macaulay. Locally the coarse space of semi-stable Higgs bundles looks like a quotient $V//H$ where $H$ is reductive and $V$ is affine chart for the stack (in the smooth topology). Thus the question becomes when is $V//H$ Cohen-Macaulay?
For $V$ non-singular this is the Hochster-Roberts theorem. However, it can fail in general even when $V$ itself is Cohen-Macaulay (in fact even when $V$ is a complete intersection). See for example the last paragraph of example $I$ here. In this case we are saved by the fact that moduli space of semi-stable Higgs bundles has symplectic singularities which are in particular Cohen-Macaulay. See for example this paper.
It seems to me then that being in the symplectic setting is used not just for the dimension bounds but also to ensure that the moduli space is Cohen-Macaulay so I'm not sure what to expect for Higgs bundles valued in an arbitrary line bundle $L$.
Thank you! It actually seems like the same argument also works for the coarse moduli space of semi-stable Higgs bundles, no? Also, flatness should be true as well for the Hitchin fibration on the moduli space of Higgs bundles that take value in a line bundle other than the canonical bundle (one more positive). The symplectic context is missing in that set-up. Would it be possible to prove the key result (Prop. 8 in Ginzburg) without Lemma 7 (the fact that the stack of globally nilpotent Higgs bundles is isotropic)?
@Johan I added an edit addressing these questions above since it was too long for a comment.
|
2025-03-21T14:48:30.975373
| 2020-05-13T09:41:52 |
360208
|
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|
Stack Exchange
|
Reference needed for proof of a Tauberian theorem
I was reading the following paper by Hubert Delange: http://www.numdam.org/article/ASENS_1956_3_73_1_15_0.pdf 1. In page 26, he provides a proof of Theorem b, the bulk of which relies on a result in some reference provided by him in footnote 5. However, despite my best attempts, I have been unable to find the aforementioned reference. Any proof or reference with a complete proof of the theorem would be really appreciated. Thanks a lot.
1Delange, Hubert, Sur la distribution des entiers ayant certaines propriétés, Ann. Sci. Éc. Norm. Supér., III. Sér. 73, 15-74 (1956). ZBL0072.27501, MR0083519.
As I don't speak French, this is partially a guess. But I think this is paper you're looking for. (Maybe somebody else will be able to suggest other references for the same result.)
Looking at the footnote 5 saying "Loc. cit., p. 235-236 et 239." and the footnote 4 saying "Ann. Se. Ec. Norm. Sup., (3), l. 74, 1934' p. '213-242." - this paper seems as a reasonable guess.
Delange, Hubert, Généralisation du théorème de Ikehara, Ann. Sci. Éc. Norm. Supér., III. Sér. 71, 213-242 (1954). ZBL0056.33101, MR0068667.
The paper you linked says about Theorem b: "Ces deux théorèmes se déduisent immédiatement des théorèmes III et IV de notre travail cité plus haut5". The paper "Généralisation du théorème de Ikehara" indeed contains Théorème III on pages 235--236 and Théorème IV on page 239.
Archive with old issues Annales Scientifiques de l'École Normale Supérieure is available at NUMDAM.
|
2025-03-21T14:48:30.975502
| 2020-05-13T10:02:55 |
360210
|
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|
Stack Exchange
|
Studying the vast world of Number Theory
I'm a high school student, interested in mathematics, especially in number theory.
While preparing for the IMO test, and thinking about generalizations or the root of many olympiad problems led me to high level NT.
(for instance, ISL 2011 A8 asks if $x^5 +y^2 $ forms a complete residue system in $mod p$, which is actually killed by Hasse-Weil bound, and a classic problem asking to prove $x^3 -2y^3 =1$ has finite solutions is a special case of Thue's theorem.)
Like this, my NT study has been definitely not systematic. To start with, I've read Analytic number theory(Apostol), and my knowledge in this field is round about understanding the proof of PNT and being familiar with analytically calculating some series related to primes and sieve methods.
I've also seen a bit of algebratic number theory. Not much, but I've read the part about quadratic fields in 'Introduction to number theory(Hardy)'. These days, I'm studying about transcendental number theory, and now I'm reading about baker's theorem.
Is my way of studying NT Ok? I feel like I'm studying too randomly, and what I've read doesn't seem to be properly organized inside my brain. Another disadvantage is that studying in this way, I don't see the end;there are so much NT papers, and I feel that I will not be able to cover modern NT even if I study for my whole lifetime.....
Could you give me some advice about how to study NT, and could somebody tell me about roughly how much modern Number theory has developed in this point(i.e. how much should I have to study to study to get on a level to be able to understand most of the papers about pure NT?)
I fear your question will get closed on this site as it is not specifically about mathematical research, but about learning basic text in number theory. Probably it is better suited for an other forum. Also you may want to narrow the question and make it more specific. It is hard to answer such a vast question about all number theory. I don't know many mathematicians that could "understand most of the papers about pure number theory", for instance.
Your intuition is correct: no person will likely learn all about number theory in their lifetime, even if they are as fortunate enough to have so much background and skill at a very early age like yourself to be able to understand Baker’s methods in high school. You have to narrow your focus.
|
2025-03-21T14:48:30.975702
| 2020-05-13T10:18:47 |
360214
|
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|
Stack Exchange
|
Bound between distance between Rotation Matrices
Let $\|\cdot\|_F$ denote the Fröbenius norm on the set of $d\times d$ matrices. By restriction this induces a metric on $SO(n)$.
Let's make an observation.
Since $X\in SO(n)$ is a rotation matrix then it is an isometry hence if $\lambda$ is an eigenvalue of $A$ with corresponding eigenevector $x$ we have that
$$
\|x\|=\|Ax\|=\|\lambda x\|= |\lambda| \|x\|
\,\Rightarrow\,
|\lambda|=1.
$$
Therefore, we get the crude bound
$$
\begin{aligned}
\sup_{X, Y \in SO(n)} \|X-Y\|_F
\leq &
\sup_{X,Y \in SO(n)} \|X\|_F + \|Y\|_F
\\= &
\sup_{X, Y \in SO(n)} \sqrt{
\sum_{i=1}^n \lambda_i(X)
}
+
\sqrt{
\sum_{i=1}^n \lambda_i(Y)
}
\\= & 2\sqrt{n}
,
\end{aligned}
$$
where I use $\lambda_i(X)$ to emphasize the $i^{th}$ eigenvalue of $X$.
However, here are my two issues with this bound:
It is not specific to $SO(n)$ and applies to any set of linear isometries of $\mathbb{R}^n$,
It is clearly crude since it entirely disregards the distance between $X$ and $Y$ and only looks at their "norm" individually..
Is a sharp(er?) estimate for
$$
\sup_{X,Y \in SO(n)} \|X-Y\|_F,
$$
known? Specifically, can we bound this quantity by $1$?
What is $|\cdot|_F$?
@YCor I added a host of details...
What is the Frobenius norm?
sq. root of sum of square of all entries of the matrix
Well, but the bound is achieved on $Y=-X$ if $n$ is even...
And for odd $n$ we can use that the inequality is invariant w.r.t simultaneous left or right multiplication of X and Y and then reduce the problem to the even case. If follows that the bound for $n=2m+1$ is the same as for $n=2m$.
First Point: The bound isn't sharp
Consider the case where $n=2$. The every matrix in $SO(n)$ is of the form
$$
A_{\theta} \triangleq \begin{pmatrix}
cos(\theta) & -sin(\theta)\\
sin(\theta) & cos(\theta),
\end{pmatrix}
$$
for some $\theta \in [0,2\pi]$ (note: fun easy proof of compactness of $SO(2)$). In particular,
$$
\|A_0 - A_{\frac{\pi}{2}}\|_F =
\left\|\begin{pmatrix}
1 & -1\\
1 & 1,
\end{pmatrix}\right\|_F= \sqrt{4} = 2.
$$
So $2\sqrt{2}$ is not sharpe.
Second point: $1$ cannot be achieved
This also shows that $1$ cannot be achieved if $SO(n)$ is metrized by the Fröbenius norm, since we just got $2$...
Third point/question: $1$ can maybe be Achieved if we instead consider the Spectral Norm
Recall that the spectral (or Operator norm) of a $n\times n$ matrix is given by
$$
\|X\|_{\infty} = \max_{i=1,\dots,n} |\sigma_i(A)|.
$$
Therefore, by your remark on the eigenvalues of any $A \in SO(n)$ we have that
$$
\sup_{X,Y \in So(n)}\, \|X-Y\|_{\infty} \leq 2.
$$
However, a quick computation shows that
$$
\|A_0 - A_{\frac{\pi}{2}}\|_{\infty} = \sqrt{2}>1.
$$
So $1$ cannot be achied..
Suggestion: If you're willing to take any metric induced by a norm on the set of $n\times n$ matrices then I would just use
$$
\|X-Y\|_n' := \frac1{2\sqrt{n}} \|X-Y\|_F.
$$
Note that it generates the same topology on $SO(n)$ since all norms are equivalent on finite-dimensional normed spaces...
So, if you can use this, then your bound will give you a metric induced by a norm which is uniformly bounded by $1$ on $SO(n)$!
Hopefully this works for you.
I encourage you to follow-up and find out if the spectral norm can achieve value $1$, this would be interesting?
|
2025-03-21T14:48:30.975956
| 2020-05-13T10:36:27 |
360216
|
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|
Stack Exchange
|
Decomposing a polynomial ring into Specht Modules
Let $S_{\pi}$ where $\pi$ is an integer partition of $n$, denote the Specht module corresponding to $\pi$.
I am trying to decompose the set of all homogeneous polynomials in $x_1,x_2,...,x_n$ generated linearly (over any field of characteristic zero) by the monomials of the form $x_i^2x_jx_k$ ($i,j,k$ are distinct), into Specht modules. I managed to do it for the polynomials generated by each of the following classes of monomials with $i,j,k,l$ distinct: $x_i^3x_j,x_i^2x_j^2,x_i^4,x_ix_jx_kx_l$.
Once it is achieved for ${x_i}^2x_jx_k$ a decomposition is successfully found for the space of degree 4 homogeneous polynomials in $n$ variables where $n$ is large enough, say $n\ge20$. This is the aim.
First $x_i^3x_j$ $(i\ne j)$: We know that $x_i^3x_j=\displaystyle \frac{x_i^3x_j+x_ix_j^3}2+\frac{x_i^3x_j-x_ix_j^3}2$
(a) The terms of the form $\displaystyle \frac{x_i^3x_j+x_ix_j^3}2$ generate linearly a space isomorphic as $S_n$-modules (the module action is by permuting indices) to the homogeneous square-free degree 2 polynomials. This is isomorphic to $S_{(n-2,2)}\oplus S_{(n-1,1)}\oplus S_{(n)}$.
(b) The terms of the form $\displaystyle\frac{x_i^3x_j-x_ix_j^3}2$ generate linearly a space isomorphic as $S_n$-modules to the second exterior power of a vector space generated by $\{x_1,x_2,...,x_n\}$ via $x_i\wedge x_j\mapsto\displaystyle\frac{x_i^3x_j-x_ix_j^3}2$. Thus this is isomorphic to $S_{(n-2,1,1)}\oplus S_{(n-1,1)}$.
So the decomposition is
$\displaystyle S_{(n-2,2)}\oplus S_{(n-2,1,1)}\oplus 2S_{(n-1,1)}\oplus S_{(n)}$
where "$2$" indicates that we have two copies of $S_{(n-1,1)}$.
$x_i^4$: This is simply a vector space generated by $x_i^4$, and is a direct sum of the standard and the trivial representations of $S_n$ that is $S_{(n-1)}$ and $S_{(n)}$. Thus the decomposition is $S_{(n-1,1)}\oplus S_{(n)}$.
$x_ix_jx_kx_l$: These generate the module isomorphic to module $M_\lambda$ as in Bruce Sagan's book "The Symmetric Group" where $\lambda=(n-4,4)$ which one figures is just $S_{(n-4,4)}\oplus S_{(n-3,3)}\oplus S_{(n-2,2)}\oplus S_{(n-1,1)}\oplus S_{(n)}$.
$x_i^2x_j^2$: These generate the module isomorphic to module $M_\lambda$ where $\lambda=(n-2,2)$ which one figures is just $S_{(n-2,2)}\oplus S_{(n-1,1)}\oplus S_{(n)}$.
The reason behind showing above is that these are alarmingly simple deductions though I can't seem to find one nearly as slick for the class $x_i^2x_jx_k$. I found that there is a submodule isomorphic to $M_{(n-3,3)}$ inside this class of polynomials. But that is the closeset I could get.
I tried dimension count as well, because the homogeneous degree-4 polynomials are of dimension ${n+3\choose 4}$.
The terms in the decompositions including the $M_{(n-3,3)}$ above sum up to
$\displaystyle S_{(n-4,4)}\oplus 2S_{(n-3,3)}\oplus 4S_{(n-2,2)}\oplus S_{(n-2,1,1)}\oplus 6S_{(n-1,1)}+5S_{(n)}$.
This has a total dimension $n^4-2n^3+35n^2+38n$ which, subtracted from ${n+3\choose4}$ is $\frac13(n^3-3n^2-4n)$ which should be the sum of the dimensions of the remaining irreducible components (that is, copies of Specht modules).
But this counting technique leads to many possible decompositions and I am kind of out of ideas on this. Could anyone help?
The span of the monomials of the form $x_i^2x_jx_k$ is the Young permutation module $M^{(n-3,2,1)}$. (Proof. Observe that $x_1^2x_2x_3$ has stabiliser $\langle (2,3)\rangle \times S_{\{4,\ldots,n\}}$, so the relevant Young subgroup is $S_{n-3} \times S_2 \times S_1$.) Using Kostka numbers (equivalently, multiplicities of Schur functions in complete symmetric funtions) this decomposes as
$$M^{(n-3,2,1)} \cong S^{(n-3,2,1)} \oplus S^{(n-3,3)} \oplus S^{(n-2,1,1)} \oplus 2S^{(n-2,2)} \oplus 2S^{(n-1,1)} \oplus S^{(n)}$$
provided that $n \ge 6$. (I'm using superscripts for Specht modules since this is the notation I'm used to.)
To make this explicit, Specht's original construction of Specht modules shows that $S^{(n-3,2,1)}$ is generated by the product of Vandermonde determinants
$$\left| \begin{matrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ x_1^2 & x_2^2 & x_3^2 \end{matrix} \right| \left| \begin{matrix} 1 & 1 \\ x_4 & x_5 \end{matrix}\right| $$
and it's clear that the unique trivial submodule is spanned by
$$x_1^2x_2x_3 + x_1^2x_2x_4 + \cdots + x_{n-2}x_{n-1}x_n^2,$$
i.e. the sum of all monomials whose exponents are $2$, $1$, $1$, $0, \ldots, 0$ in some order.
For the other factors it is harder to make them explicit, but this can be done by using semistandard homomorphisms (see for instance James' lecture notes). Since we're working in characteristic zero, any non-zero homomorphism from a Specht module must be injective. I'll give some details here. By this theory,
$$\mathrm{Hom}_{\mathbb{C}S_n}(S^{(n-1,1)}, M^{(n-3,2,1)})$$
is spanned by the semistandard homomorphisms for the two semistandard tableaux of shape $(n-1,1)$ and content $(n-3,2,1)$: these have rows $1\ldots122,3$ and $1\ldots123$, $2$, respectively. Each such homomorphism extends to $M^{(n-1,1)}$. Taking as a model for $M^{(n-1,1)}$ the natural permutation module $\langle e_1,\ldots, e_n\rangle$, the corresponding extended homomorphisms are
$$e_n \mapsto (x_1x_2^2 + x_1x_3^2 + \cdots + x_{n-2}x_{n-1}^2)x_n$$
and
$$e_n \mapsto (x_1x_2+x_1x_3+\cdots + x_{n-2}x_{n-1})x_n^2,$$
respectively. One then has to restrict these homomorphisms to $S^{(n-1,1)} \subseteq M^{(n-1,1)}$ (for instance it is generated by $e_n-e_1$) to get two submodules of $M^{(n-3,2,1)}$ isomorphic to $S^{(n-1,1)}$, as in the claimed decomposition.
Also, should that determinant not be $\det\begin{vmatrix}1&1&1\x_1&x_2&x_3\x_1^2&x_2^2&x_3^2\end{vmatrix}x_i$? where $i=1,2,3$?
Thanks so so much Mark, specially for reminnding me of the Kostka numbers, I should have known that! The $M_{(n-3,2,1)}$ just occurred to me this morning and I was on the right track!
I've corrected the determinant: it comes from the standard tableaux of shape (n-3,2,1) with first column $1,2,3$ and second column $4,5$. For example, expanding both on the main diagonal gives $x_2x_3^2x_5$ which is a monomial of the right 'shape'.
I might have asked already, but: has there been any more modern exposition of semistandard homomorphisms than in James's LNM booklet? I've tried a couple times to generalize them to pictures, and each times I've gotten lost in the forest of non-canonical identifications that at least the definition in James seems to get into.
I don't know of anything that I'm confident will help you. Semistandard homomorphisms have been generalised to Hecke algebras. I think this is Mathas' book and another version is in a paper of Lyle: https://arxiv.org/abs/1101.3192. The extra 'rigidity' in Hecke algebras might perhaps help with the non-canonicity you mention. Just for the record, my notes http://www.ma.rhul.ac.uk/~uvah099/Maths/doubleRevised2.pdf have some relevant ideas. (This is of course known to you: thank you darij for your many helpful corrections.)
In general, these decompositions can be computed using the Pieri rule. (This is essentially the same as Mark's answer)
Fix an integer partition of $d$ as $\lambda = 1^{e_1} \dots r^{e_r}$. Then consider the $S_n$-set of ordered partitions into $r+1$ blocks $[n] = b_0 \sqcup \dots \sqcup b_r$ such that the $i$th block has size $e_i$ for $i > 0$ and the $0$th block has size $n - \sum_{i} e_i$. This set has a canonical bijection with a set of monomials, given by $$b_0 \sqcup \dots \sqcup b_r \mapsto (\prod_k (\prod_{i \in b_k} x_i)^k.$$
The case you are asking about is $\lambda = 2^1 1^2.$
Let us write $V_\lambda$ for the vector space with basis the $S_n$ set associated to $\lambda$. Then we can write $V_\lambda$ as an induced representation as follows $$V_\lambda = {\rm Ind}^{S_n}_{S_{n - \sum_{i} e_i} \times \prod_{i = 1}^r S_{e_i}} \left(\bigotimes_{i= 0}^r {\rm triv} \right).$$
The effect of this induction product can be computed via the Pieri rule, starting with the integer partition $(n- \sum_i e_i)$ corresponding to the trivial representation of $S_{n - \sum_i e_i}$ and adding horizontal strips of length $e_1, \dots, e_r$. This should translate into a combinatorial rule for multiplicities involving skew tableaux with content specified by the $e_i$, but I have not thought this through carefully.
In your case, we begin with the partition $(n-3)$. Multiplying by the horizontal strip $(e_1) = (2)$ we get $(n-1) + (n-2,1) + (n-3,2)$. Multiplying this by $(e_2) = (1)$ we get $$((n-1) + (n-2,1) + (n-3,2))*(1)$$ $$ = (n) + (n-1,1) + (n-2,2) + (n-1,1) + (n-2,2) + (n-2,1,1) + (n-2,2) + (n-3,3) + (n-3,2,1)$$
This is indeed a very valuable input! Thanks @Phil Tosteson! Moving on to polynomials in variables with double indices such as $x_{ij},x_{jk}$ etc, could the 'Ind' formula mentioned here be used?
Where is $(n-2,1)$ coming from? If we add a horizontal strip, to $(n-3)$ shouldn't the first row be $(n-3)$ or $(n-1)$. Perhaps I don't understand the Pieri rule correctly.
|
2025-03-21T14:48:30.976611
| 2020-05-13T11:19:14 |
360219
|
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|
Stack Exchange
|
Is the reduced group $C^*$-algebra quasidiagonal
Let $G$ be an amenable group. I wonder whether it is true that the reduced group $C^*$-algebra $C_r^*(G)$ is quasidiagonal.
This is true for countable discrete groups by the celebrated Quasidiagonality Theorem of Tikuisis-White-Winter (Quasidiagonality of nuclear C∗-algebras. Ann. of Math. (2) 185 (2017), no. 1, 229–284.)
If G is a uncoutable discrete group, does the above conclusion also hold?
Quasidiagonality is a local property for $C^\ast$-algebras (this follows for instance by Voiculescu's characterisation of quasidiagonality combined with Arveson's extension theorem). By writing your uncountable group as a direct limit of countable subgroups, you also obtain quasidiagonality for uncountable discrete groups.
Thanks, Pro Gabe. Please forgive my stupidity. If the amenable group is not discrete, does there exist a counterexample to show that the associated reduced group $C^*$-algebra is quasidiagonal?
There exist non-discrete amenable groups with non-quasidiagonal C*-algebras, see [Beltiţă, Ingrid; Beltiţă, Daniel, Quasidiagonality of C*-algebras of solvable Lie groups. Integral Equations Operator Theory 90 (2018)] for an example.
Pro Gabe, may I ask you another question? My teacher told me that separable simple nuclear C∗-algebras have been completely classified . I wonder which problems do most operator algebraists concern about nowadays?
@mathbeginner your teacher has omitted some key-adjectives (words like Z-stable, for instance)
@mathbeginner You go to a conference to find out, or you browse sites like https://arxiv.org/
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2025-03-21T14:48:30.976738
| 2020-05-13T11:24:58 |
360220
|
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Stack Exchange
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Motivation for persistent homology with respect to eigenfunctions of distance kernel operator in a recent preprint
I have a question about a recent preprint https://arxiv.org/pdf/1912.02225.pdf by Maria, Oudot, and Solomon. As far as I understand, in Section 8 they prove that persistent homology (persistence diagrams, Betti numbers, Euler characteric) with respect to level sets of a function which is a linear combination of eigenfunctions of the distance kernel integral operator is well-behaved. The distance kernel operator $D^X$ is defined as $D^X(f)(x)=\int_X f(x) d(x,y) d\mu(y)$ where $d$ is the distance function on a compact metric measure space $(X, d, \mu)$.
My question is, why is one interested in persistent homology with respect to such functions (which are not explicit as they are eigenfunctions of some integral operator), as opposed to more explicit functions? Given this, why does one take the distance kernel operator instead of another integral operator (with a different kernel)? Are there other ways to construct functions suitable for taking level sets in the definition of persistent homology? These motivational questions don't seem to be mentioned in the text of this preprint.
Is there any chance that one can actually calculate persistence diagrams, Betti numbers, Euler characteric with respect to the above eigenfunctions in terms of some underlying geometry (to have some explicit formulas for these topological invariants as opposed to just knowing that they exist/are well-behaved)?
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2025-03-21T14:48:30.976860
| 2020-05-13T11:29:33 |
360222
|
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|
Stack Exchange
|
Kolmogorov tightness criterion for stochastic processes
I am searching for the criterion stated above and also here: The question about Kolmogorov tightness criterion.
It should state the following: If a sequence of stochastic processes $(X^n)$ fulfills:
$$\mathbb{E}[|X^n_t-X^n_{t'}|^p]\leq C|t-t'|^\alpha$$
then it is tight.
I don't know if my googling skills are just too bad, but I can't find any source for that. Thanks for any advice!
The body of a question should not depend on the title ("the criterion stated above").
Also, @Glorfindel, I think that blank lines around displayed formulas should be discouraged. Markdown regards them as paragraph breaks, which they aren't semantically. That is, inline $$display$$ inline and inline$$display$$inline are both fine (and display as probably intended), but inline$$display$$inline, while it seems to display OK, is semantically wrong. (I mention this semantic issue because the original edit in that part of the post seems to have been semantic, too.)
I assume that you want to show tightness in a space such as $C([0,1])$, as was the case in the question you link to. In fact, the approach of this answer will show tightness in $C^\beta([0,1])$ for every $\beta \in (0, \frac{\alpha - 1}{p})$.
Firstly, note that the statement you write cannot be sufficient for tightness since if $X^n$ is a sequence of constant processes then your condition trivially holds. Such a sequence need not be tight. The extra condition in your linked question that $(X_0^n)_{n \geq 1}$ is a tight sequence in $\mathbb{R}$ prevents such counterexamples.
The key point is that from the proof of Kolmogorov's Continuity Criterion one can derive control on Holder norms of your process. For $\gamma \in (0, \frac{\alpha - 1}{p})$, one has the bound
$$\mathbb{E}([X^n]_\gamma^p) \leq C(p, \alpha, \gamma) \cdot C$$
where $C(p,\alpha,\gamma)$ is a constant depending only on $p, \alpha$ and $\gamma$ (but is independent of $n$) and $[\cdot]_\gamma$ is the usual $\gamma$-Holder seminorm on $C^\gamma([0,1])$. See this answer for a proof.
Let $\|X\|_\gamma = |X_0| + [X]_\gamma$ denote (a norm equivalent to) the usual $\gamma$-Holder norm. Fix here $\varepsilon > 0$. By tightness of $(X_0^n)$ there is an $M_1$ such that $$\sup_n \mathbb{P}(|X_0^n| > M_1) \leq \varepsilon$$
Also, by Markov's inequality and our above control on the Holder seminorm, we have that for $M_2$ sufficiently large,
$$\sup_n \mathbb{P}([X^n]_\gamma > M_2) \lesssim M_2^{-p} \leq \varepsilon.$$
Hence $$\sup_n\mathbb{P}(\|X^n\|_\gamma > M_1 + M_2) \leq \sup_n\mathbb{P}(|X_0^n| > M_1) + \sup_n \mathbb{P}([X^n]_\gamma > M_2) \lesssim \varepsilon.$$
Finally, by compactness of the embedding $C^\gamma([0,1]) \to C([0,1])$ the closed ball of radius $M_1 + M_2$ in $C^\gamma([0,1])$ is relatively compact in $C([0,1])$ so the above inequality yields tightness of your sequence in $C([0,1])$.
Hi Rhys, first of all thank you very much for your helpful answer! Let me ask two questions to clarify my understanding of your answer: in the closing line, after the "Hence", you show that the sequence is tight in $C^\gamma([0,1])$, correct? And, why do you need that second norm $||\cdot ||_\gamma$? Why can't you just show tightness by using the usual $\gamma$-norm?
(1/2) No, in the closing line I show that sequence is in a closed ball in $C^\gamma$ with high probability. This doesn't immediately show tightness since closed balls in $C^\gamma$ are not compact. That's why the next part where you use the compact embedding is necessary. Of course, one could replace $\gamma$ with $\gamma _ \varepsilon$ for $\varepsilon$ suitably small for the main body of the proof and then compactly embed $C^{\gamma+\varepsilon} \to C^\gamma$ to get tightness in $C^\gamma$.
(2/2) For the second question, $[X]\gamma := \sup{s \neq t}\frac{|X_s - X_t|}{|s-t|^\gamma}$ does not give control on $\sup_t |X_t|$ so you can't hope to get nice properties of a closed ball for this seminorm under the embedding $C^\gamma \to C[0,1]$. $|\cdot|\gamma$ is equivalent to the usual norm (not seminorm) on $C^\gamma$ and is more convenient to work with since tightness of $X_0^n$ gives us immediate control on the first term in $|\cdot|\gamma$. This would take more a bit more work if we worked with the usual norm $|\cdot|\gamma^\ast = |\cdot|\infty + [\cdot]_\gamma$.
Thank you very much, I think I got it!
Sorry, one more question: would it simplify the argumentation, if I assume $\sup\limits_{n\geq 0}\sup\limits_{0\leq s\leq 1}\mathbb{E}[|X_s^n|^p]\leq c$ for some constant c (independent from n)?
Not as far as I can see. Since the $\sup_s$ is outside the expectation, Markov's inequality won't give control on $\mathbb{P}(|X^n|_\infty \geq M_1)$, so you probably end up using that condition to essentially conclude tightness of $X_0^n$.
|
2025-03-21T14:48:30.977178
| 2020-05-13T11:53:39 |
360224
|
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|
Stack Exchange
|
Cyclic inequality for 2 dimensional simplex elements
Let $p=(p_{1},p_{2},p_{3})\in\Delta$, with $\Delta:=\lbrace p\in(0,1)^{3}\ |\ p_{1}+p_{2}+p_{3}=1 \rbrace$. I aim to prove (not knowing whether it is true though) that
\begin{equation}
p_{1}^{p_{3}-p_{2}}p_{2}^{p_{1}-p_{3}}p_{3}^{p_{2}-p_{1}}\le1.
\end{equation}
Indeed, if at least two of the three numbers are equal, then the inequality holds (with equality) (thus we may assume wlog $p_{1}<p_{2}<p_{3}$). I've tried a plenty of examples and couldn't find any for which it is wrong. Yet, I fail to prove the validity of this inequality so far and I am therefore thankful for any help.
Then why $(p_{1}-p_{3})\ln p_{2}\le (p_{2}-p_{3})\ln p_{1} +(p_{1}-p_{2})\ln p_{3}$ ?
Still not sure if I get your point. If we consider $p_{1}<p_{2}<p_{3}$ then in $(p_{3}-p_{2})\ln p_{1}+(p_{1}-p_{3})\ln p_{2} +(p_{2}-p_{1})\ln p_{3}$ clearly the first and third term are negative, I agree, the second term however is positive, so why is the whole term negative?
I don't think your 'wlog' is correct. I think you can only assume 3 distinct, and not an order. If you agree with this, then rewriting as
\begin{equation}
p_{1}^{p_{3}}p_{2}^{p_{1}}p_{3}^{p_{2}}/
p_{1}^{p_{2}}p_{2}^{p_{3}}p_{3}^{p_{1}}
\end{equation}
if you call the numerator $f(p_1,p_2,p_3)$ the the denominator is $f(p_3,p_2,p_1)$ and it can't be bigger in general.
Gosh, thats true. Unfortunately :D. So then, probably the statement works only on the ordered simplex. But even if, I'm not sure whether this would still be helpful for my purposes. Anyway, thank you!
If $p_1=3/100$, $p_2=77/100$, and $p_3=20/100$, then the left-hand side of your inequality is $2.3447\ldots>1$. So, your inequality is false in general.
|
2025-03-21T14:48:30.977325
| 2020-05-13T12:30:56 |
360228
|
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|
Stack Exchange
|
Coinvariants of tensor products of Hopf algebras
Let $G$ be a Hopf algebra, considered as a right $G$-comodule in the obvious way.
The axioms of Hopf algebras imply that
$$
G^{\operatorname{coinv}(G)} == \{g \in G : \Delta(g) = g \otimes 1\} = \mathbb{C}1.
$$
Consider now that tensor product of comodules with $k$-factors
$$
G\otimes \dotsb \otimes G.
$$
Is there an explicit description of
$$
(G\otimes \dotsb \otimes G)^{\operatorname{coinv}(G)}?
$$
If $H$ is Hopf, then $H\otimes H$ with diagonal action is free, of the same dimension of $H$. The map is very explicit, you can surely can find it in any book of Hopf algebras (e.g. S. Montgomery). There is also the version for coaction. The statement is that it is isomorphic to $H\otimes V$ where V is a vector space, the $H$ coaction comes only from the left factor, and V=H as vector space. You can iterate fue procedure (or the explicit map) and get a formula for the coinvariants
Let $\Bbbk$ be a field and $H=\Bbbk G$ its group algebra. Then the regular coaction makes $H$ a right comodule over itself and we use the monoidal structure of right comodules to view $H^{\otimes n}$ as a right $H$-comodule.
Using the fact that $\{g\otimes h\otimes k\;|\;g,h,l\in G\}$ is a basis for $H^{\otimes 3}$ one can proof that
$$(H\otimes H)^{coinv}=\Bbbk^{\oplus |G|}.$$
A basis for the vector space $(H\otimes H)^{coinv}$ is given by the elements $g\otimes g^{-1}$, for $g\in G$.
More generally,
$$(H^{\otimes n})^{coinv}=Span_\Bbbk(\{g_1\otimes \ldots \otimes g_n \;|\; g_1\cdot\ldots\cdot g_n=1\}).$$
The dimension of this vector space is
$$\dim (H^{\otimes n})^{coinv}=|G|^{n-1}.$$
According to my understanding， you regard $G$ as a comodule on itself. So what you write is also Ok. Now I offer a way to construct a coinvariants on tensor product of comodules with k-factors. Without loss of generality, we can assume that $k=2$. Firstly, we should define a comodule structure on $G\otimes G$. This was done by in Susan Montgomery's book ``Hopf algebras and their actions on rings'' on page 14 (Definition 1.8.2). I posted it here.
$$\Delta_{G\otimes G}:=(id \otimes m)(id \otimes \tau \otimes id)(\Delta \otimes \Delta),$$
where $m$ is the multiplication, $\tau$ is the switch map. So we can define
$$(G\otimes G)^{coinv G}
:=\{g_1 \otimes g_2\in G\otimes G\mid \Delta_{G\otimes G} (g_1\otimes g_2):=g_1\otimes g_2 \otimes 1 \}.$$
Yes, this is comodule structure on $G \otimes G$ that I am asking about. But I am looking for an explicit description.
|
2025-03-21T14:48:30.977528
| 2020-05-13T12:56:23 |
360232
|
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|
Stack Exchange
|
Homology of topological manifolds
Let $X$ be a topological manifold of dimension $n$ (assuming perhaps that there is a countable basis of open sets). Do NOT assume that $X$ is compact, or oriented, or triangulable (so do not assume it to be smooth).
Can we still conclude that the homology groups $H_i(X, \mathbb{Z})$ are finitely-generated?
Can we still conclude that $H_i(X, \mathbb{Z})=0$ for $i > n$ ?
The second point is related to the first, as one can show that $H_i(X, \mathbb{Q})=0$ and $H_i(X, \mathbb{F}_p) = 0$ for all primes $p$, when $i>n$. If $H_i(X, \mathbb{Z})$ were known to be finitely-generated, we would conclude by the universal coefficients theorem.
[To see the claim: if $X$ is $k$-oriented, it is part of Poincaré duality that $H_i(X, k)=0$ for $i>n$, and $X$ is always $\mathbb{F_2}$-oriented. Next, consider $Y \to X$ the canonical 2-sheeted cover with $Y$ oriented, and apply the homology Serre spectral sequence to the fibration $Y \to X \to B\mathbb{Z}/2$; this gives $H_i(X, k)=0$ when $i>n$ for each ring $k$ in which 2 is invertible.]
thanks!
Pierre
A countably infinite discrete set of points violates condition $1$.
Point 1 is false, see @user2520938's comment (for a connected example, take an infinite connected sum of genus 1 surfaces). Point 2 is Proposition 3.29 in Hatcher if $X$ is noncompact, and Theorem 3.26 if $X$ is compact. Also you have to be a bit careful with your spectral sequence argument since the base of your fibration is not connected.
I'm pretty sure that answer on question 2 depends a lot on the choice of homology theory. And it seems to me that "tubular neigborhood" of 2-dimensional Hawaiian earring would have nontrivial homology in degrees higher that 2 (Barrat&Milnor, An example of anomalous singular theory).
@Najib: thanks for the reference! as for the spectral sequence, $BC_2$ is certainly connected, but you probably worry about simple-connectedness; however, page 2 of the sequence is just $H_(C_2, H_(Y))$, the homology of the group $C_2$ with nontrivial coefficients, and this vanishes in positive degrees if multiplication by 2 is an isomorphism on $H_*(Y)$.
@Denis: the example in Barratt-Milnor is not a manifold; are you saying that one could produce a manifold example by taking a tubular neighbourhood? This would violate both Hatcher's result and my spectral sequence argument (since they use rational coefficients in this paper anyway).
Is this useful?
It's already been noted in the comments that 1) is false: take for example an infinite genus surface.
2) is true. For an oriented manifold you have $H_i(M)\cong H^{n-i}_c(M)$, the cohomology with compact supports (see Hatcher Theorem 3.35). For $i>n$, the right side is $0$ trivially. If $M$ is not oriented, you have essentially the same theorem but with twisted coefficients in the cohomology. See page 207 of the Springer edition of Bredon's Sheaf Theory.
I actually prefer the reference to 3.29 from Hatcher's book, pointed out by Najib in the comments.
|
2025-03-21T14:48:30.977759
| 2020-05-13T14:10:15 |
360240
|
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|
Stack Exchange
|
A characterization of constant functions
In How to recognize constant functions. Connections with Sobolev spaces (Russian Math Surveys 57 (2002); MSN), H. Brezis recalls the following fact:
Let $\Omega\subset{\mathbb R}^N$ be connected and $f:\Omega\rightarrow{\mathbb R}$ be measurable, such that
$$\int\int_{\Omega\times\Omega}\frac{|f(y)-f(x)|}{|y-x|^{N+1}}\,dx\,dy<\infty.$$
Then $f$ is constant.
He adds
The conclusion is easy to state, but I do not know a direct, elementary, proof. Our proof is not very complicated but requires an “excursion” via the Sobolev spaces.
My question is whether there is such an elementary proof in the special case of one space dimension ($N=1$, $\Omega$ an interval).
I don't know about measurable functions, but here's a relatively easy proof for when $f$ is continuously differentiable. Suppose $f$ is not constant. Then $|f'(y)| > c > 0$ on some small interval. For each such $y$, we have $|f(y) - f(x)| > c|y-x|$ on a neighborhood of $y$. Then your integrand (for that particular $y$) is $>! c/|y-x|$ on a neighborhood of $y$, and therefore goes to $\infty$. Since this happens for an interval of $y$'s, the whole double integral is also $\infty$.
Such a function $f$ is in $L^1_{loc}$. Assuming $\Omega=R^N$ we can take convolutions with smooth cut-off function $\phi_\epsilon$ and the integrability condition is preserved. Then $f*\phi_\epsilon$ is a constant (by the argument of @Will Brian) and letting $\epsilon \to 0$, $f$ is constant, as well. For general $\Omega$ this does not work, but in 1d $\Omega$ is an interval $[a,b]$ and we can extend $f$ outside, using the values at the endpoints. I did not check the details but maybe you see immediately if there is a silly mistake in the argument.
@GiorgioMetafune What worries me in your argument is this: we start with a measurable function $f$ for which the double integral from the statement of the problem converges. How do you conclude, that the same holds for $f\ast\phi_\epsilon$?
@Vahe If $g=f*\phi$, then $\int |g(y)-g(x)|/|y-x|^{N+1} dy dx \le \int |f(y-z)-f(x-z)/|y-x|^{N+1}|\phi (z) dx dy dz$. Next write $|y-x|=|(y-z)-(x-z)|$ and change variables by setting $u=y-z, v=x-z$ in the $dx, dy$ integral. Then you get the original double integral to be integrated against $\phi(z), dz$.
@GiorgioMetafune You're right, thanks!
It is here:https://math.stackexchange.com/questions/488780/if-int-mathbb-r2-frac-vert-fx-fy-vert-vert-x-y-vert2dxdy-infty/545051#545051
Not quite an answer, but too long for a comment.
Let me make my life easier a bit and take $\Omega=\mathbb{R}^N$ while increasing the exponent slightly. Namely, I will assume that
$$
I:=\ \int_{\mathbb{R}^{2N}}\ \frac{|f(x)-f(y)|}{|x-y|^{\alpha}}\ d^Nx d^Ny
$$
is finite, where $\alpha>N+1$.
Then we have, just by the triangle inequality involving the midpoint, $I\le J$ where
$$
J:=\ \int_{\mathbb{R}^{2N}}\
\frac{\left|f(x)-f\left(\frac{x+y}{2}\right)\right|
+\left|f\left(\frac{x+y}{2}\right)-f(y)\right|}{|x-y|^{\alpha}}\ d^Nx d^Ny
$$
$$
=\ 2^{N+1-\alpha}\ I\ ,
$$
by a trivial change of variable. Since $I\in [0,\infty)$ and $I\le 2^{N+1-\alpha}\ I$ with $N+1-\alpha<0$, we immediately get $I=0$.
The OP's case is clearly a borderline/endpoint one where the above argument just happens to break down. Perhaps one can get a logarithmic improvement, by using smarter estimates.
The above simple idea is just a "Sobolev-ish"-flavored (as opposed to "Hölder-ish") adaptation of the classic proof of Hölder with exponent greater than one in 1D implies constant, by subdividing the interval $[x,y]$ into $k$ pieces and taking $k\rightarrow\infty$.
|
2025-03-21T14:48:30.978015
| 2020-05-13T15:11:45 |
360243
|
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|
Stack Exchange
|
Disintegration, conditional probabilities, and conditional expectation
On the Wikipedia page there is a note that conditional probability measures can be described by disintegration. However, I can seem to find a clear exposée of how this construction is related to conditional expectation. Rigorously, how are the two related?
http://galton.uchicago.edu/~lalley/Courses/383/ConditionalExpectation.pdf These are nice notes.
There's a nice discussion of these issues in "Conditionng as Disintegration" by Chang & Pollard: https://onlinelibrary.wiley.com/doi/full/10.1111/1467-9574.00056
The example one should have in mind is the Fubini theorem for the unit square (endowed with the Lebesgue measure) projected onto the horizontal base. The conditional measures are then just the Lebesgue measures on the vertical intervals. According to the Rokhlin theory, this is actually the general case, namely, measure preserving maps between reasonable probability spaces are precisely like this (up to an isomorphism). Have a look at the section "Regular conditional probabilities" of the Standard probability space entry (it is written unusually well for Wikipedia) and the references therein.
|
2025-03-21T14:48:30.978127
| 2020-05-13T15:22:58 |
360246
|
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|
Stack Exchange
|
About the metric and embedding of sphere
Let $S^2$ be the $2$-dimensional sphere with a metric $g$.
Q:
Can we or how to find a smooth map $f:S^2\to \mathbb R^3$, such that
(1) $f$ is diffeomorphic to its image $Im(g)=:M$,
(2) $M$ with the induced metric from $ds^2$ of $\mathbb R^3$, is isometric to $(S^2,g)$, where $ds^2$ is the standard metric on $\mathbb R^3$, i.e. $ds^2=dx^2+dy^2+dz^2$, and the isometric is given by $f^{-1}:M\to S^2$.
I do not know any work about the problem(maybe solved).
Any reference about the problem is welcome
If the Gauss curvature is positive, such an embedding is known to exist, a result of Nirenberg. But if there is a point of zero Gauss curvature, I think that nothing is known. See the book Qing Han, Jia-Xing Hong, Isometric Embedding
of Riemannian Manifolds in Euclidean Spaces for the best known results.
A little clarification: When the Gauss curvature is positive, then it's known as the Weyl problem and was proved independently. by Nirenberg and Pogorelov. For nonnegative Gauss curvature, there is a result by Guan and Li, https://projecteuclid.org/euclid.jdg/1214454874. If the Gauss curvature is negative somewhere, then nothing is known.
Pogorelov showed the following more general result. For a general metric on $S^2$, if the Gaussian curvature is strictly greater than $-\kappa$ for a nonnegative number $\kappa$, then the metric can be isometrically embedded into the simply-connected complete space form of curvature $-\kappa$.
|
2025-03-21T14:48:30.978367
| 2020-05-13T16:03:51 |
360252
|
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|
Stack Exchange
|
Existence of measurable "inclusion" into Euclidean space
Let $(\Omega,\mathfrak{F})$ be a measurable space. When does there exist an injective measurable function $f:(\Omega,\mathfrak{F})\to (\mathbb{R}^n,B(\mathbb{R}^n))$ to some Euclidean space, here $B(\mathbb{R}^n)$ is the Borel $\sigma$-algebra.
Thoughts. Clearly, if $\Omega$ is a Riemannian manifold and $\mathfrak{F}$ is its Borel $\sigma$-algebra then this works but I'm thinking of more general, non-topological criteria. (If they exist)
Such a function (in fact, bijective) exists as soon as $\Omega $ is an uncountable polish space — this is the Borel isomorphism theorem.
In general, we can stick to $n=1$.
@abx: Since only injectivity is requested here, any separable metrizable space will do. In particular, a non-measurable subset of $\mathbb{R}^n$ with its Borel $\sigma$-algebra will do. Actually, submetrizable is enough.
@Nate Eldredge: Yes, right.
@abx I did not know this result, thanks.
@NateEldredge I never heard of submetrizable, but I assume it can be worked out directly from the theorem abx provided?
@AIM_BLB: Yes. A submetrizable space, by definition, has a continuous injection into a metric space, which in turn has a continuous injection into its completion, which as abx noted has a bimeasurable bijection to $\mathbb{R}$ assuming separability in the right places.
(Basically the same as Michael's answer)
Theorem 6.5.7 of Measure Theory by V. Bogachev:
Theorem. The following are equivalent:
$\mathfrak{F}$ is countably separated (Bogachev Definition 6.5.1 (ii)): there exists an at most countable collection of sets $F_n \in \mathfrak{F}$ such that for every two distinct points $x,y \in \Omega$, there is some $F_n$ with $x \in F_n$, $y \notin F_n$;
there is an injective measurable function $f : \Omega \to [0,1]$ (or equivalently, to $\mathbb{R}^n$, since they are Borel isomorphic to $[0,1]$);
The diagonal $\Delta = \{(x,x) : x \in \Omega\}$ is measurable with respect to the product $\sigma$-algebra $\mathfrak{F} \otimes\mathfrak{F}$;
There exists a separable (i.e. countably generated) sub-$\sigma$-algebra $\mathfrak{F}_0 \subset \mathfrak{F}$ which contains all the singletons.
This is the case if and only if there exists a countable subfamily of $\mathfrak{F}$ that separates points. Necessity is straightforward; if such a function exists then the preimages of rectangles with rational coordinates at the endpoints will serve as the countable separating family.
For sufficiency, let $\mathcal{C}=\{C_1, C_2,\ldots\}$ be a countable separating family of measurable sets. Then the Marczewski function $f:\Omega\to [0,1]$ given by $f(\omega)=\sum_n 2/3^n1_{C_n}(\omega)$ will be measurable and injective.
|
2025-03-21T14:48:30.978581
| 2020-05-13T16:07:45 |
360254
|
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|
Stack Exchange
|
Methods of sheaf theory for solving Diophantine equations
What are some examples of sheaf theory used to either provide solutions to Diophantine equations, or to state that no such solutions exist?
Well, the Weil conjectures are proved using sheaf theory, but tht's a long story...
Okay, so did you come to understand them by reading Hartshorne, for instance, or how did you come to understand them? [What references did you use, is what I am asking]
While not explicitly about sheaves, Alex Youcis has a write-up of a talk he gave about restating the study of diophantine equations in modern-ish geometric language which might be what you're looking for.
@ReginaldAnderson Do you want to share your 3 years work on this idea...
OK, let me try to be more explicit. Arithmetic geometry is an important part of algebraic geometry, with stunning applications to number theory. It is based in an essential way on the techniques of sheaf cohomology. Listing examples would take forever.
May be my comment was not clear enough.. You have mentioned that you have spent 3 years on this idea... I was asking you how did you spend those 3 years... For "whether I've learned anything in grad school." I do not know what to say... I do not know how that is relevant to my comment..
You could start with the Wikipedia article on the Weil conjectures, and look at the references given there.
How about Brauer-Manin obstructions to rational points? Those come from Azumaya algebras, which are sheaves of modules (satisfying a bunch of criteria, etc.).
This paper uses sheaves (at least in the form of vector bundles with flat connections) to find all the rational points on an explicit curve. https://arxiv.org/pdf/1711.05846.pdf . My paper with Brian Lawrence https://arxiv.org/abs/2004.09046 uses sheaves and perverse sheaves to prove finiteness of integral points for a family of Diophantine equation. The Brauer-Manin obstruction that RP_ mentioned is a good example - specifically, it can often be used to show a Diophantine equation has no solutions.
@WillSawin's comment: "this paper" is Balakrishnan, Dogra, Müller, Tuitman, and Vonk - Explicit Chabauty–Kim for the split Cartan modular curve of level 13; "My paper with Brian Lawrence" (i.e., Will's paper) is Lawrence and Sawin - The Shafarevich conjecture for hypersurfaces in Abelian varieties.
Your question may be get closed, but before that happens, I will leave an answer that might help you. Taken very literally the answer to your question is likely no. It is known that there can be no algorithm to decide whether a Diophantine equation has an integer solution (Hilbert's 10 problem), which makes it unlikely that there would be a sheaf theoretic test for this. However, as already pointed out in the comments, sheaf theoretical methods have been extraordinarily useful in studying Diophantine equations. For example, to repeat what abx said in the comments, the solution to the Weil conjectures by Grothendieck and Deligne is one particularly striking example. Instead of looking at solutions over $\mathbb{Z}$, one studies the number of solutions in $\mathbb{Z}/(p)$ and its extensions. In some cases, such as for hypersurfaces, the conjectures yield explicit bounds on the number of solutions. To attack the conjectures, one needs a generalization of sheaf theory along with the corresponding cohomology theory, called étale cohomology. This is a very big topic, but perhaps this gives you some idea.
Here is a possible explanation from Chapter 1 section 4 of "The Geometry of Schemes" by Eisenbud and Harris.
The apparently abstract idea of the functor of points has its root in the
study of solutions of equations. Let $X = \text{Spec}( R)$ be an affine scheme, where
$R = \mathbb{Z}[x_1, x_2,...]/(f_1, f_2,...)$. If T is any other ring (one should think of
$T = \mathbb{Z}, \mathbb{Z}/(p), \mathbb{Z}_{(p)}, \hat {\mathbb{Z}}_{(p)}, \mathbb{Q}_p, \mathbb{R}, \mathbb{C},$ and so on), then a morphism from
$\text{Spec}( T)$ to $\text{Spec}( R )$ is the same as a ring homomorphism from $R$ to $T$, and
this is determined by the images $a_i$ of the $x_i$. Of course, a set of elements
$a_i \in T$ determines a morphism in this way if and only if they are solutions
to the equations $f_i = 0$. We have shown that
\begin{equation}h_X(T ) = \left\{\!\begin{aligned}\text{sequences of elements $a_1,... \in T $ that } \\ \text{are solutions of the equations $f_i = 0$.} \end{aligned}\right\}
\end{equation}
Similarly, if $X$ is an arbitrary scheme, so that $X$ is the union of affine
schemes $X_a$ meeting along open subsets, then a map from an affine scheme
$Y$ to $X$ may be described by giving a covering of $Y$ by distinguished affine
open subsets $Y_{f_a}$ and maps from $Y_{f_a}$ to $X_a$ for each $a$, agreeing on open sets
(some of the $Y_{f_a}$ may, of course, be empty). Thus an element of $h_X(Y )$ may
be described even in this general context as a set of solutions to systems of
equations, corresponding to some of the $X_a$, with compatibility conditions
satisfied by the solutions on the sets where certain polynomials are nonzero.
Even with this interpretation, the notion of the functor of points may
seem an arid one: while we can phrase problems in this new language, it’s far from clear that we can solve them in it. The key to being able to
work in this setting is the fact that many apparently geometric notions
have natural extensions from the category of schemes to larger categories
of functors. Thus, for example, we can talk about an open subfunctor of
a functor, a closed subfunctor, a smooth functor, the tangent space to a
functor, and so on. These notions will be developed in Chapter VI, where
we will also give a better idea of how they are used.
@ReginaldAnderson Well, even if all you wanted was to study affine schemes, you will often have to study non affine ones first - for instance the proof of Falting's theorem (or Mordell's conjecture) reduces to statements about abelian varieties and to study projective schemes, you definitely need sheaves (in the simplest case - maps to projective space are given by line bundle+sections but it goes much deeper of course). There seems to be no way around it - sheaves and schemes are crucial to algebraic number theory.
Yes, they will let you solve things. Does proving that any high degree equation has only finitely many rational solutions count as "solving things"? Even if you object that it isn't very explicit, there are other ways of bounding solutions like the Chabauty method which provides very explicit bounds on how many solutions such an equation can have. Surely that counts as solving an equation!
I just saw your edit: Yes, you can often construct sheaves (on the etale site!) that tell you about solutions to equations. This is the idea behind the proof of the weak Mordell-Weil theorem which effectively says that the torsion points of an elliptic curve inject into some Galois/etale cohomology group. There are really too many examples to list but here's one more - a recent paper proves that the average rank of elliptic curves over finite fields is some number by computing the cohomology of an explicit sheaf.
A scheme is a sheaf; you need sheaves to define schemes. I think another text that will help you a lot is Hartshorne https://www.springer.com/gp/book/9780387902449; his advisor was Zariski who invented the Zariski topology which eventually lead to sheaves, which eventually lead to schemes. The book starts with sheaves and then defines schemes in terms of sheaves, "The Geometry of Schemes" does this as well. The language of sheaves was first used by Jean Pierre Serre to attack the problems of algebraic geometry and commutative algebra in a famous paper "Faisceaux algébriques cohérents."
The idea is the following: sheaves were originally invented for other more general problems. Serre had the good idea to apply this theory to algebraic geometry (which is roughly "polynomial geometry") and Grothendieck further refined sheaves to locally ringed spaces which are isomorphic to the spectrum of a ring; i.e. affine schemes. Why sheaves? The stalks of the sheaves carry local information about the "manifold" defined by the polynomial. This is exactly the point in most of geometry (for example in Differential Geometry); i.e. to study the interplay between the local and global geometry.
Finitely many solutions is Falting's theorem. The proof is immensely complicated! Here is an outline of the recent paper: https://web.stanford.edu/~aaronlan/assets/longer-selmer-distribution-talk-notes.pdf
Sheaves allow one to perform the topological analog of the algebraic operation of localization. Indeed this was the intended point; that is why it is called the localization. There are many different viewpoints here are some: 1) ideals and coordinate rings, 2) function fields, and 3) germs, stalks, and sheaves. In number theory, the object is discrete, and its not entirely obvious how to apply geometric notions like locality and closeness but sheaves allow us to apply topological/geometric techniques to something non-continous. This is the easy Babylonian/Greeks/Persians way from ancient time.
|
2025-03-21T14:48:30.979206
| 2020-05-13T16:58:56 |
360258
|
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|
Stack Exchange
|
In theory, how would Oneiric numbers be defined?
Background I am not a professional mathematician. I am researching Surreal numbers & games for fun (I think they are truly beautiful). If this question is not appropriate here, I beg forgiveness & ask that it please be migrated it to MSE.
I read More Infinite Games by John H Conway recently & have not been able to stop thinking about the following line (page 4):
Note that $\uparrow$ is not a number: it is the value of a game, which
is a more subtle concept. Also note that $\frac{1}{\uparrow}$ is not defined since it would be
bigger than all surreal numbers and there are no such numbers. (In fact, it does
exist but is one of the Oneiric numbers.)
Unfortnuately, the only thing on the internet about the Oneiric numbers is this paper. If my current understanding is correct, a part of the problem is that the reciprocal of games is not (yet) defined. As previously stated, I'm not an expert & have no clue if working out the reciprocals of games is an impossible (or just very difficult) task, or even if it has already been done.
It has been suggested by a couple people that I try to find someone who knew Conway & might have some idea of what he was thinking. I would be elated to do so. Alternatively, perhaps someone who is knowledgable about the Surreals might venture to take a crack at a definition.
Thanks for taking the time to read this question, I hope it isn't a waste of anyone's time.
As a sidenote, the adjective "oneiric" (of or pertaining to dreams) is one of the most beautiful adjectives I've seen in front of a mathematical object, and perfectly captures what should come after "surreal".
@Carl-FredrikNybergBrodda I couldn't agree more!
Naively, I would expect Conway's remark to mean that there is a way of treating $1/\uparrow$ as a symbol in a formal manner, admitting all the usual operations except those which would cause the entire building to collapse, and that the resulting calculus is what he means by Oneiric numbers (which I agree is an engaging name). Is there any reason to suppose that he had something bigger in mind? Sometimes a remark such as this (especially when a well-chosen neologism is involved) can be so suggestive that it leads one to expect grand things, while the reality is much more prosaic.
@RP_ I don't disagree, but division (arguably even multiplication) on games that aren't numbers or stars typically "causes the entire building to collapse", so I am curious as to what justification there is for writing $1/\uparrow$.
@MarkS. +1. Fair enough.
I have reached out to Richard J. Nowakowski, the editor of More Games of No Chance (the collection where Conway's "More Infinite Games" appeared), to see if he has any insight or a lead.
@MarkS. Amazing!
It wouldn't surprise me if this were one of Conway's jokes; i.e., such a thing exists only in our dreams, or we can dream that one day someone might make sense of it. Note for example that Conway would sometimes say that something was "cohomology" when it wasn't literally cohomology; it was one of his jokes.
Could @JoelDavidHamkins hypnagogic digraphs be a precursor to oneiric numbers?
The only thought I have to add is that Oneiric numbers almost certainly involve proper classes. Note that "Oneiric" is capitalized, suggesting "On", the class of ordinals. Perhaps they even involve classes of classes, or higher iterations.
I emailed John Conway about this very thing over ten years ago. His response (paraphrasing) was along the lines of RP’s comment; if you treat 1/up as a formal entity nothing breaks, but there wasn’t more to the theory than that. I do not have an account, and I don’t have the email still, so I neither want nor expect the bounty. One other detail I remember is that he represented 1/up with the Roman numeral I.
If no other answer comes along (and if one does, I doubt it will top this one), you will be awarded the bounty anyway. Use this account as you please, but it would be nice to know more of who this correspondent of Conway's (meaning you) is. Gerhard "Or We Can Keep Guessing" Paseman, 2020.05.16.
I’m hardly a correspondent; I emailed him exactly once, and it was to ask this question. In hindsight, I’m kind of surprised he responded to a shallow email from a college student he’d never heard of.
|
2025-03-21T14:48:30.979572
| 2020-05-13T17:52:31 |
360260
|
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|
Stack Exchange
|
Fundamental group of the complement of the arrangement of plane nodal curves
I want to calculate the fundamental group of the complement some collection of plane curves (specifically, two nodal cubics in a general position).
I've read about Severi problem (solved by Harris), which states that complement of every reduced irreducible nodal plane curve has an abelian fundamental group. I've understood the idea of the argument (deforming the picture to the arrangement of lines, which only increases the fundamental group).
I think the same (the fact that the complement has an abelian fundamental group) should hold without irreducibility assumption (i.e., for an arrangement of nodal curves which also intersect transversely). Is it true? Is it known?
The result is actually due to Fulton (On the fundamental group of the complement of a nodal curve, Ann. of Math. 111, no. 2(1980), 407-409), and it is valid with no irreducibility assumption.
@abx Thank you! I would accept this is as an answer.
As suggested, I write my comment as an answer: the result is actually due to Fulton (On the fundamental group of the complement of a nodal curve, Ann. of Math. 111, no. 2 (1980), 407-409), and it is valid with no irreducibility assumption. Fulton works with the algebraic fundamental group; the case of the topological fundamental group was done by Deligne (Séminaire Bourbaki 1979/80, Exp.543, pp. 1-10.
|
2025-03-21T14:48:30.979690
| 2020-05-13T19:22:20 |
360268
|
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|
Stack Exchange
|
A metric naturally arise from the Euclidean symplectic structure?
For $n>1$ let $\omega=\sum_{i=1}^n dx_i\wedge dy_i$ be the standard symplectic structure on $\mathbb{R}^{2n}=\mathbb{R}^n \times \mathbb{R}^n$.
We define the following distribution $D$ on $\mathbb{R}^{2n}\setminus\{0\}$:
For $Z\in \mathbb{R}^{2n}\setminus\{0\}$ we define $D_Z=\{V\in \mathbb{R}^{2n}\mid \omega(V,Z)=0\}$
This is a nonintegrable distribution of codimension $1$. We define a meteic on $\mathbb{R}^{2n}\setminus\{0\}$ as follows: The distance $d(x,y)$ is the infimum of the Euclidean length of all $D$- horizontal curves joining(connecting) $x$ to $y$.
Is this metric well defined(i.e. is this distribution totally non integrable)?Does this metric arise from a Riemannian metric on $\mathbb{R}^{2n}\setminus\{0\} \}$?
What about if we consider the same question but we restrict all necessary structures to $S^{2n-1}$?(Intersection of above D$ with tangent space of spher and and comoutation of length of curves on the standard geometry of sphere)
Suppose we have a non-integrable distribution and a riemannian metric on a manifold. Then, one can define the metric using your construction, the resulting object is called sub-riemannian metric. It does not come from any riemannian or finslerian metric. Sub-riemannian metrics are important in the theory of nilpotent groups (and vice versa), as always with such geometric subjects there is a book of M. Gromov on it; see https://www.ihes.fr/~gromov/wp-content/uploads/2018/08/carnot_caratheodory.pdf
It also arises in some applied area, namely, geometry of vision, on this I can not find exact reference now.
Thank you very much for your very interesting answer.
Considering this applied thematic the thing to google is "sub-riemannian model of visual cortex".
thank you again for your answer. i confess that i did not read the book of Gromov yet. I just brows it but i will read it. But just a few more questions: Is the distribution I considered totally non integrable that is the liealgebra of vector field tangent to D is thewhole algebra of vector field? Moreover is there a 1 dimensional folition of S^3 tangent to D(The intersection of D with S^3) such that the foliation has no closed leaf? On the opposite extrem, is there a 1 dimensional foliation of 3 sphere tangent to D whose all leaves are closed?
Note that non of the the three standard vector fields of S^3, which produce foliation by circles, ,are tangent to D.
I think if you restrict codimension 1 distribution on a sphere you will obtain codimension 1 distribution. For example, consider $S^3$ embedded in $\mathbb{R}^4 = \mathbb{C}^2 = \mathbb{H}$. Then, the distribution $\text{Ker}(\omega(v, *))$ is also euclidean orthogonal to $I(v)$ due to the identity $\omega(v, x) = g(Iv, x)$. Now, it is actually generated by vector fields $Jv$ and $Kv$ ($I J K$ being quaternion imaginary units). Commutator of $J$ and $K$ is indeed $I$, so this distribution is non-integrable and generates the whole tangent space.
Considering similar theory for $1$-dimensional distributions they are always integrable, and so I think you can not define the similar theory for them. You can probably obtain the sequence of metrics by allowing to move transversely to the foliation but making it increasingly costly, but I'm not sure how useful this concept is.
|
2025-03-21T14:48:30.979942
| 2020-05-13T20:43:12 |
360274
|
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|
Stack Exchange
|
A real system of bilinear equations with $2n$ unknown and equations
I have the following system of $2n$ bilinear equations, for a square invertible matrix $A \in \mathbb{R}_{n \times n}$, and $2n$ unknowns organized in vectors $x,y \in \mathbb{R}^n$:
$$
diag(y) A x = {\mathbf 1}_n
$$
$$
diag(x) A^T y = {\mathbf 1}_n
$$
where $diag(x)$ denotes a diagonal matrix with diagonal elements $x$, and ${\mathbf 1}_n$ is a vector of ones of length $n$.
What can be said about the solution of this system? for example, when is there a (unique?) solution?
While in general solving bilinear equations is hard,
this system has a specialized simple symmetric structure that can be used.
We can easily implement an iterative algorithm with iterations:
$$
x^{(t+1)} = (diag(y^{(t)})A)^{-1} {\mathbf 1}_n
$$
$$
y^{(t+1)} = (diag(x^{(t)})A^T)^{-1} {\mathbf 1}_n.
$$
What can be said about the convergence of this algorithm? is there a better approach? any hope for a closed-form solution?
|
2025-03-21T14:48:30.980023
| 2020-05-13T20:54:20 |
360276
|
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|
Stack Exchange
|
A uniform mixture of order statistics
Let $0<k<n$ be integers, and let $X$ be a random variable obtained as follows: sample $n$ points independently and uniformly at random in the unit interval, and select (uniformly) one of the $k$ leftmost points. The distribution of $X$ is therefore a uniform mixture of order statistics, with pdf given by $$f(x) =\frac{1}{k}\sum_{i=1}^{k}n\binom{n-1}{i-1}x^{i-1}(1-x)^{n-i} $$
At the bottom, I draw a picture of the pdf for $n=20$ and $k=9$.
My question is, are there any simpler expressions that approximate this well? It has a natural S shape to it so I wonder if it is similar to a logistic curve, for example.
For $x\in(0,1)$,
$$f(x)=\frac nk\,F_{n-k,k}(1-x),$$
where $F_{n-k,k}$ is the cdf of the beta distribution with parameters $n-k,k$. From here, you can get a number of approximations. See e.g. Wikipedia.
Also, you can use the central limit theorem for the beta distribution (based on the delta method or the asymptotics of the beta integral; cf. MathSE) to get the following: If $n\to\infty$, $k\sim an$ for some $a\in(0,1)$, and $(a-x)\sqrt n\to y\in\mathbb R$, then
$$f(x)\to\frac1a\,\Phi\Big(\frac y{\sqrt{(1-a)a}}\Big),$$
where $\Phi$ is the standard normal pdf.
Here are the graphs $\{(y,f(a-y\sqrt n))\colon|y|<3\sqrt{(1-a)a}\}$ (red) and $\{(y,\frac1a\,\Phi\big(\frac y{\sqrt{(1-a)a}}\big))\colon|y|<3\sqrt{(1-a)a}\}$ (blue) for $n=20$, $k=6$, and $a=k/n=.3$:
|
2025-03-21T14:48:30.980145
| 2020-05-13T21:04:07 |
360277
|
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|
Stack Exchange
|
Neighborhood of an orthogonal matrix
Let $A\in O(n)$ be an orthogonal matrix and let $\vec{a}_1,\dots,\vec{a}_n$ be its rows. For a vector $\vec{v}=[v_1,\dots,v_n]$, let $\max(\vec{v})=\max\{|v_1|,\dots,|v_n|\}$. Prove or disprove that if $\max(\vec{a}_i)<1$ for every $i$, then for every small neighborhood of $A$ there exists $B\in O(n)$ such that $\max(\vec{b}_i)>\max(\vec{a}_i)$ for every $i$.
For me the general approach to this kind of problem is to form a map $f:O(n)\to\mathbb{R}^n$ where $f(A)=[\max\{\vec{a}_1\},\dots,\max\{\vec{a}_n\}]$ and analysis the induced map between tangent spaces, but the $\max$ map here is kind of painful. Any thoughts?
Why not consider $(1+\epsilon)A$. This would also be orthogonal with $\max(\vec{b}_i)>\max(\vec{a}_i)$ for every $i$, and can be made to lie within an arbitrary small ball around $A$. Hope I follow your question correctly.
@DSM Why would this be an orthogonal matrix though? The length of every row would be $1+\epsilon$ so each row would not be a unit vector.
I am sorry, I assumed you wanted only orthogonality, and not orthonormality. In that case, my comment does not mean much.
A naive constructive approach that comes to mind: of course, a permutation matrix $P$ is a matrix that satisfies the requirements of $B$ except that it might fail to be in a sufficiently small neighborhood of $A$. If we look at the path $p:[0,1] \to \Bbb R^{n \times n}$ defined by $p(t) = (1-t)A + tP$ and project onto $O(n)$ (via the "orthogonal procrustes" polar decomposition method), then it must hold that $p(t)$ eventually satisfies the necessary requirements. Perhaps it is possible to guarantee that this happens for sufficiently small $t$ for the right choice of $P$.
@Omnomnomnom I have thought about that, but there exists a counterexample since the max of two rows may occur at the same column (I got an explicit 4$\times$4 counterexample but probably too large for a comment).
@student Could you elaborate more on your comment? Thanks!
It's not a constructive argument and I'm too shaky on my differential geometry to flesh it out completely off the top of my head, but I believe the following approach might be fruitful.
Fix a matrix $A$. For each $i$, let $m_i$ denote an index for which $|a_{i,m_i}| = \max(\vec a_i)$. Without loss of generality, suppose that $a_{i,m_i}>0$ for all $i$. Let $U$ denote the relatively open set
$$
U = \{B : \text{for all }i, \max(\vec b_i) = b_{i,m_i}; b_{i,m_i} > |b_{ij}| \text{ for } j \neq m_i; \text{ and } b_{im_i} > a_{im_i}\}.
$$
Consider the function $f:U \to \Bbb R^n$ given by
$$
f(B) = [\max(\vec b_1), \dots, \max(\vec b_n)] = [b_{1,m_1},\dots,b_{1,m_n}].
$$
$f$ is the restriction of a linear and therefore differentiable, and (I think) the differential $df$ has full rank over $U$. It follows by [differential geometry argument of some kind] that for a sequence in $f(U)$ converging to the boundary point $f(A)$ of $f(U)$, there is a corresponding sequence in $U$ converging to $A$.
Let's fill in the details to see if my understanding is correct: $df$ has full rank because $df$ is a $n\times 2n$ matrix (Jacobian of $f$), all rows has only one non-zero entry, and such non-zero entry occurs at different columns for each row. Hence $f$ is a smooth submersion and $f:U\to f(U)$ is surjective. Then we are done as we can select a sequence in $f(U)$ converging to $f(A)$, and there is a corresponding sequence in $U$ converging to $A$.
I am not sure about the last part (sequence) though. Could you provide a reference on that? I think it is true because constant rank maps are projection maps up to a coordinate change, and for projection maps it would be obvious.
@LoCelso As I said, very shaky when it comes to differential geometry; I'm honestly not sure where to look. If I find something to connect the dots there, I'll let you know.
|
2025-03-21T14:48:30.980787
| 2020-05-13T21:49:20 |
360279
|
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|
Stack Exchange
|
Reference request: placing a set with respect to the integer grid
For $x=(x_1,...,x_n)\in \mathbb{R}^n$, let $Q_x=(x_1,x_1+1)\times ...\times (x_n,x_n+1)$ - the open cube having $x$ in its "bottom left" corner. It seems, I can prove (see a draft here) the following
Theorem. Let $K\subset\mathbb{R}^n$. There is $v\in \mathbb{R}^n$ so that
for any $m\in\mathbb{Z}^n$, either $v+K\cap Q_m\ne\varnothing$, or $v+K\cap \overline{Q_m}=\varnothing$. In fact, such $v$'s for a co-meager set in $\mathbb{R}^n$.
This means that we can shift the integer grid so that there is no cube that intersects $K$ only by the boundary (of the cube).
My proof is not very clever, but rather lengthy, and I am not sure I like the way it is written.
Is this fact known?
Edit: Based on Mathieu Baillif's comment it was possible to give a rather simple proof of a similar fact in the context of general Baire topological group. I am however still looking for a reference.
What if you consider the set $K=\cup_{n\in Z}{x_1=0.5 n}$ (assuming no compactness). The cube $Q_m=(m_1,m_1+1)\times ...\times (m_n,m_n+1)$ and its complement will have non-empty intersection with any translate of the set $K$, given by $K+v$. Hope I understood your problem correctly.
@DSM in turn, if i understand your comment, you interpreted $\overline{Q}$ as the complement, whereas i meant the closure.
Thank you, my bad. :)
Would it work to consider measure instead of density? For any face there is some unit vector $w$ such that if $v$ is a vector such that $v + k$ intersects that face but not the interior, then $v+ k + tw$ intersects the interior for every $0<t < 1$. But if $v_1$ and $v_2$ are two vectors as above ${v_1 +tw: 0<t<1} \cap {v_2 +tw: 0<t<1} = \emptyset$. So the measure of all such $v$ must be 0.
Assume $K$ is convex and compact. There exists a $Q_m$ only if one of the supporting hyperplanes has the form $x_k=z$ for some $k$ and for some integer $z$. Suppose in $k$-dimension, $x_k=l_k$ and $x_k=r_k$ are the nearest integer hyperplanes which sandwich $K$. If neither touch $K$, do nothing. If both touch K, translate it by $0.5e_k$ ($e_k$, standard basis vector). If $x_k=l_k$ touches, while $x_k=r_k$ does not, then translate $K$ by $K+\delta/2 e_k$, where $\delta$ is the difference between $r_k$ and $z$ and $x_k=z$ is the supporting hyperplane sandwiching $K$ between itself & $x_k=l_k$.
A similar argument can be made if $x_k=r_k$ touches $K$, while $x_k-l_k$ does not.
The above procedure can be done for each dimension. Finally, by construction, no supporting hyperplane of the form $x_k=z$ will have an integer $z$.
@cha21 assuming i understood your idea: if a vector "doesn't work", then any small alteration does. I don't think this is true, because the problems can arise in some other region of the set
@DSM what if in dimension $2$ a cube only intersects $K$ by the corner? then there is no horizontal or vertical support line. On the other hand, even though i think that this argument can be fixed, compact convex is way too narrow
What I was saying was to decompose it into the union of vectors that don't work for each face/corresponding cube. And it is the countable union of sets of measure zero.
@cha21 but the problem can occur arbitrarily close
@erz, you have a point there. Will think about it.
Let me drop convexity, & keep compactness (and perhaps, connectedness). I'll also ignore the "corner" intersection for now. For a dim $k$, let $T_k$ be the set of tangent planes $x_k=z$. Let $T^Z_k$ be the finite set of closest integer hyperplanes on either side of the tangent planes. For each plane in $T_k^Z$, one can find the distance by which it can be moved on either side till it becomes tangent to $K$. Let this set of distances be called $S$. Choose $\delta_k=\min{s\in S| s>0}$. if $delta_k>0$, translate $K$ by $\delta_k/2 e_k$. Else translate by $0.5e_k$. Do this for every dimension.
@DSM to be honest, i don't quite understand your idea. Do you deal with every dimension independently? Then fixing the next you break the previous.
Am I right in thinking that if you fix $m$ then the set $T_m$ of $v$ such that $K+v$ intersects $Q_m$ only in its boundary is nowhere dense ? If you take a point v in the closure of $T_m$ and small ball $B$ around it, then there is some member $w$ of $T_m$ in this ball and hence a point $x$ in $K$ such that $x+w$ lies in the boundary of $Q_m$. Take $u$ in $B$ very near $w$ such that $x+u$ is in $Q_m$, then there is a small neighborhood $U$ of $u$ such that $x+U\subset Q_m$. It shows that $u$ is not a member of the closure of T_m$. Hence $T_m$ is nowhere dense. Applying Baire gives your result.
(I wrote this comment late at night on my phone, I hope I did not write something silly due to this late your.)
@MathieuBaillif yes, this works! I was trying to find something along these lines, but my proof is 1.5 pages long and exploits geometrical specifics, whereas yours is just a paragraph and is very generalizable. I took liberty at making an answer based on your comment. Hope you don't mind. Thank you!
Great. I don't mind at all. Nice generalization !
Based on Mathieu Baillif's comment we can actually prove a much more abstract
Theorem. Let $G$ be a Cech-complete topological group, let $U_n$ be a sequence of non-empty open sets in $G$, and let $K\subset G$. Then there is $g\in G$ such that for every $n$ either $gK\cap U_n\ne \varnothing$ or $gK\cap \overline{U_n}= \varnothing$. Moreover, such $g$'s form a co-meager set.
Of course, in the context of the original question $G$ is the Euclidean space, and $U_n$ is the sequence of the cubes formed by the integer grid.
Proof. Observe that $A,B\subset G$ we have that $\{g\in G: gA\cap B\ne\varnothing\}=BA^{-1}$. Indeed, $gA\cap B\ne\varnothing$ iff there are $a\in A$ and $b\in B$ such that $ga=b$ or $g=ba^{-1}$.
Let $H_n=\{g\in G,~ gK\cap U_n= \varnothing,~ gK\cap \overline{U_n}\ne \varnothing\}$. Then, $H_n=\overline{U_n}A^{-1}\backslash U_nA^{-1}$. Since $U_n$ is open, so is $U_nA^{-1}$ as a union of open sets. Hence, $\overline{H_n}\subset\overline{\overline{U_n}A^{-1}}\backslash U_nA^{-1}$.
Since the product is continuous from $G\times G$ into $G$ it follows that $$\overline{U_nA^{-1}}\subset \overline{\overline{U_n}A^{-1}}\subset \overline{\overline{U_n}\cdot\overline{A^{-1}}}=\overline{U_n\cdot A^{-1}},$$ from where $\overline{H_n}\subset\overline{U_nA^{-1}}\backslash U_nA^{-1}=\partial (U_nA^{-1})$. A boundary of an open set is nowhere dense, and so $H_{n}$ is nowhere dense.
The set $G\backslash \bigcup H_n$ is co-meager (and so non-empty). If $g\in G\backslash \bigcup H_n$, then for every $n$ $g\notin H_n$, from where either $gK\cap U_n\ne \varnothing$ or $gK\cap \overline{U_n}= \varnothing$. $\square$
|
2025-03-21T14:48:30.981226
| 2020-05-13T23:37:29 |
360288
|
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|
Stack Exchange
|
Write $2n+1=p+q$ with $p$ prime and $q$ practical
Recall that a positive integer $n$ is callled practical if every $m=1,\ldots,n$ can be written as the sum of some distinct divisors of $n$. The only odd practical number is $1$.
In 1996 G. Melfi [J. Number Theory 56(1996), 205-210] proved that any positive even number is the sum of two practical numbers, which can be viewed as an analogy of Goldbach's conjecture. He also proved that there are infinitely many practical numbers $q$ with $q\pm2$ also practical.
In Jan. 2013 I conjectured that (cf. http://oeis.org/A209253) for each positive integer $n$ we can write $2n+1$ as the sum of a practical number and a (Sophie Germain) prime. This looks quite challenging. Let's turn to a weaker version which is an analogue of Chen's theorem for Goldbach's conjecture.
Question 1. Is it possible to prove that any sufficiently large odd number can be written as $q+P_2$
where $q$ is a practical number and $P_2$ is a prime or the product of two primes?
If this question is still difficult, then we may consider the following one.
Question 2. How to prove that any integer $n>1$ is the sum of a practical number and a positive squarefree number?
A.W. Dudek [Ramanujan J. 42(2017), 233-240] showed that any integer $n>2$ is the sum of a prime and a positive squarefree number. I conjecture that any integer $n>2$ can be written as the sum of a positive squarefree number and a practical number $q$ with $q+2$ also practical.
Your comments are welcome!
Corollary 2 of https://arxiv.org/abs/2007.11062 states that every sufficiently large odd integer is the sum of a prime and a practical number.
|
2025-03-21T14:48:30.981349
| 2020-05-13T23:37:56 |
360289
|
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|
Stack Exchange
|
Inequalities for trace/eigenvalues of product of multiple 2x2 matrices
Consider the matrix product $\prod_i^n A_i$,
where each $A_i$ is a $2\times2$ matrix having the form $A_i = \left( \begin{smallmatrix} \lambda + \alpha_i & -\beta_i \\ 1 & 0\end{smallmatrix}\right)$ with
$\lambda \in \mathbb{C}$ and obeys $\Re \lambda \le 0$ and $\Im \lambda\ge 0$.
Each $\alpha_i\in\mathbb{R}$ and obeys $\alpha_i\ge 0$
Each $\beta_i \in \mathbb{R}$ and obeys $\beta_i \ge 0$.
It is easy to check that the determinant of the product obeys $\det \big(\prod_i^n A_i\big)=\prod_i\beta_i$.
Are there any upper or lower bounds (or any other results) on the trace $\mathrm{tr} \prod_i^n A_i$ that can be written in terms of $\lambda$ and the entries $\alpha_i$ and $\beta_i$? Equivalently, I am interested in any results that say something about the two eigenvalues of the product $\prod A_i$, stated in terms of $\lambda$ and the entries $\alpha_i$ and $\beta_i$.
By the way, based on other considerations, I can constrain this trace to be real and strictly positive.
I am particularly looking for bounds that (1) depend on $\lambda$ and (2) become tight in the simple case when all of the $A_i$ are equal to each other (i.e., when $\mathrm{tr}(A^n) = \lambda_1^n + \lambda_2^n$, where $\lambda_1$ and $\lambda_2$ are eigenvalues of $A$). Bounds stated in terms of maximal/minimum elements of $\alpha_i$ or $\beta_i$ are fine.
I know some simple bounds can be derived from the determinant, but such bounds generally do not obey (1) and (2). I have also derived the bound $$
\left\vert \mathrm{tr}\Big[\prod_i^n A_i\Big]\right\vert \le \left\Vert \prod_i^n A_i\right\Vert \le \prod_i^n\left\Vert A_i\right\Vert,
$$
where $\left\Vert \cdot \right\Vert $ is the trace norm. Unfortunately, I don't think this is tight when all the $A_i$ equal each other.
Note that each matrix can be written as a product of symmetric matrices $A_i = \left( \begin{smallmatrix} \lambda+\alpha_i & 1 \\ 1 & 0\end{smallmatrix}\right)\left(\begin{smallmatrix} 1 & 0 \\ 0 & -\beta_i\end{smallmatrix}\right)$. So if it is helpful, the problem can be treated as bounding the trace of a product of symmetric $2\times 2$ matrices.
As a side note, this problems arises in applying the transfer matrix method in physics.
@AlexRavsky - The second equality is not true. The trace is commutative on symmetric matrices when $n \leq 3$ or when the matrices are permuted cyclicly, but not in general. Choosing $\lambda = 0$, $\alpha_i = \beta_i = i$ for $1 \leq i \leq n = 2$ gives the left trace equal to $-1$ and the right trace equal to $5$.
|
2025-03-21T14:48:30.981538
| 2020-05-14T01:35:52 |
360293
|
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|
Stack Exchange
|
Uniform boundedness of resolvents on the imaginary axis
Let $A \colon \mathcal{D}(A) \subset \mathbb{H} \to \mathbb{H}$ be a closed linear operator in a Hilbert space $\mathbb{H}$, which generates a $C_{0}$-semigroup. Suppose that in a $\varepsilon$-neighbourhood of the imaginary axis the operator $A$ does not have spectrum. Is it true that the resolvents $(A-i\omega I)^{-1}$ are bounded in $\mathcal{L}(\mathbb{H})$ uniformly for $\omega \in \mathbb{R}$? Can the same be said for the boundedness in the norm of $\mathcal{L}(\mathbb{H};\mathcal{D}(A))$, where $\mathcal{D}(A)$ is endowed with the graph norm. If not, what are additional conditions to ensure the boundedness.
For example, if $A$ generates an exponentially stable $C_{0}$-semigroup then the boundedness of resolvents follows from the resolvent estimate in the Hille-Yosida theorem.
While the answer is, in general, negative (see Michael Renardy's answer), a sufficient condition for a positive answer is that $\mathbb{H}$ is an $L^2$-space and the semigroup is positive; see for instance Corollary C-III-1.3 on page 294 in "Arendt et. al.: One-parameter Semigroups of Positive Operators (1986)".
By the way, a rather comprehensive overview about the relation between spectral bounds, pseudo-spectral bounds and growth bounds of $C_0$-semigroups can be found in Secctions 5.1 and 5.2 of "Arendt, Batty, Hieber, Neubrander: Vector-valued Laplace Transforms and Cauchy Problems (2011)". Using the notation from this book, your question is whether $s(A) = s_0(A)$, and an explicit counterexample to this is given in Example 5.1.10 (according to the notes at the end of Chapter 5, this example is a modification, due to Wrobel, of Zabczyk's counterexample mentioned in Michael Renardy's answer.
This is essentially equivalent to asking whether the spectrum determines the growth bound. This is well known to be false. A pioneering counterexample is due to Zabczyk.
It might be worthwhile to add that, in the general Banach space setting, the question is equivalent to whether the spectral bound equals the pseudo-spectral bound; since the OP restricted the question to the Hilbert space setting, the pseudo-spectral bound always coincides with the growth bound (by the Gearhart-Prüss theorem), which leads to the statement in the answer.
|
2025-03-21T14:48:30.981727
| 2020-05-14T01:53:42 |
360295
|
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|
Stack Exchange
|
Can chain homotopy induce space homotopy at $E_\infty$ level?
Space-level homotopy induces (co)chain homotopy, but I've never heard of the converse. I am not sure if it is simply not true?
However, for good enough spaces (finite type nilpotent), Mandell proved that weak homotopy types are determined by cochain complexes as an $E_\infty$-algebras up to quasi-isomorphism. Does this suggest that there are (co)chain homotopies in the sense of $E_\infty$, such that all of them backward induce homotopies between maps between spaces?
I read this question as: "What does the functor $C^*(-;\mathbb{Z})$ from finite,type nilpotent spaces to E_infty algebras do on morphism sets?. Is it injective and or surjective?" and I think it is a good question.
In fact Mandell answers this question in his paper. This functor is split injective on morphisms. Moreover, for any map of $E_\infty$-algebras $f:C^(Y)\to C^(X)$, one can construct a map $g:X\to Y$ such that $H^(g)=H^(f)$ (but $C^*(g)$ is different from $f$ in general).
@GeoffroyHorel I don't have the paper by my hand. Does that include chain homotopies $f$, with a degree shift? Also.. though $H^(g) = H^(f)$, I think it's still weaker than the statement that f and g are homotopic..
chain maps that are not of degree 0 can't be compatible with $E_\infty$-structure. Yes I agree that $H^(g)=H^(f)$ does not imply that the two maps are homotopic. In fact there are in general many maps of $E_\infty$-algebras $C^(Y)\to C^(X)$ that are not homotopic in the $E_\infty$-sense to maps of the form $C^*(g)$ for $g:X\to Y$ a map of spaces.
|
2025-03-21T14:48:30.981859
| 2020-05-14T01:59:22 |
360297
|
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|
Stack Exchange
|
On norms of Boolean functions
Let $f: \mathbb{F}_2^n \rightarrow \{-1,1\}$ be a Boolean function, represented by a $N=2^n$ dimensional vector, $f \in \{-1,+1\}^N$.
Define the Fourier transform of $f$ to be $\hat{f}$, where $$\hat{f}(x) = \frac{1}{N} \sum_{y \in \{0,1\}^n} (-1)^{x^T.y} f(y).$$
And lastly, define a function $W: \{-1,+1\}^N \rightarrow \mathbb{R}$, such that $W(f)=\sum_i |\hat{f}(i)|$. (See $W$ is the 1-norm of $\hat{f}$)
Now the question is
Given a vector $f \in \{-1,1\}^N$ such that $W(f) = \delta \sqrt{N}$, does there always exists a vector $h \in \{-1,1\}^N$, that differs from $f$ on atmost $\epsilon N$ coordinates and $W(h) \geq W(f)+ \Omega(\epsilon) \sqrt{N}$, for all $\epsilon \leq 1 - \delta$?
(This is asking can I always increase $W(f)$ (when $W(f)$ is not already the maximum) by changing any constant fraction of coordinates of $f$?)
I can already prove that there exists such $h$ with $W(h) \geq \Omega(\epsilon) \sqrt{N}$. I conjecture the stronger version is also true.
Equivalently,
Let B being the Boolean hypercube, where each vertex $u \in \{-1,1\}^N$. And $u$ be a vertex with $W(u)=\delta \sqrt{N}$. For all vertices $u$, does there always exist a path of length $\epsilon N$ to a vertex $v$ with $W(v) \geq W(u)+ \Omega(\epsilon) \sqrt{N}$. for all $\epsilon<1-\delta$.
interesting. can you indicate how your proof proceeds?
|
2025-03-21T14:48:30.981998
| 2020-05-14T03:41:04 |
360301
|
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"Ville Salo",
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|
Stack Exchange
|
Hausdorff metric selectors
Let $\ M\ $ be the family of all non-empty bounded regular open subsets of
$\ \Bbb R,\ $ where regular means that every $\ G\in M\ $ is equal to the interior
of its closure.
Let distance $\ d(G\ H)\ $ be the Hausdorff distance between the closures of $\ G\ $
and $\ H,\ $ for every $\ G\ H\,\in\,M.$
QUESTION: does there exist a function $\ s:\, M\to\Bbb R\ $ that is a metric selection, meaning that:
$\forall_{G\in M}\quad s(G)\in G;$
$\forall_{G\ H\,\in\,M}\quad |s(G)-s(H)|\ \le\ d(G\ H);$
?
If yes,
can selection $\ s\ $ be injective?
==============================
There are a plethora of similar questions. For instance, one may consider metric spaces different from $\ \Bbb R,\ $ e.g. open interval $\ (-1;1)\ $ or perhaps more interestingly, the two-dimensional Euclidean sphere $\ \Bbb S^2,\ $ etc.
It'd be exciting to know how the existence of Hausdorff metric selector depends on the metric space -- say, would the answer be different for different but topologically equivalent metrics of the same topological metrizable space?
==========================
One could also ask about Lipschitz selectors (with a fixed constant or arbitrary Lipschitz; or even all continuous, etc.) rather than metric. In particular, condition $\ Lip_2\ $ would provide a much larger family of selectors, when the above metric constrain on selector $\ s\ $ is relaxed to:
$$\forall_{G\ H\,\in\,M}\quad |s(G)-s(H)|\ \le\ 2\cdot d(G\ H).$$
Why do you include the "regular" hypothesis?
@EthanDlugie, it's cleaner this way. For instance, the Hausdorff distance between different open sets could be zero. It'd be a nuisance.
I see, e.g. deleting a point from an open interval yields a distinct open set which is not distinct in the Hausdorff metric. Regularity forbids this.
I guess to this point, no selector exists if you remove the regularity hypothesis. For a selector must assign the same value to sets with Hausdorff distance $0$, but the open sets $G=(-1,1)$ and $H=G-{s(g)}$ have Hausdorff distance $0$.
If you consider the weaker question, without demanding injectivity, then the difference between the two versions (all versus regular) doesn't matter.
A general way to produce counterexamples: Take two balls far from each other. Swap them by moving them continuously. The choice must follow the ball where it started, a contradiction.
(I mean, one you go beyond $\mathbb{R}$ and e.g. have a circle in there.)
@VilleSalo, please write an "Answer" about your continuity argument. Please, stress that this is about the "injective version" Your continuity argument is worth to dwell on. It works for all continuous selectors. And there must be still more to it, when we look at more general spaces.
Never mind my mentioning of injectivity under @VilleSalo comment (that would matter only under some special circumstances).
No such selector exists. Retooling my comment above, let $G$ be the open interval $(-1,1)$. Take $\epsilon>0$ such that $\ \epsilon<\min(1-s(G), s(G)-1)\ $ hence the closed $\epsilon$-neighborhood of $s(G)$ is contained (comfortably) in $G$. Then $H=G-[s(G)-\epsilon,s(G)+\epsilon]$ is a union of two non-empty open subset of $G$, is regular as per your definition, and $d(G,H) = \epsilon.\ $ But by construction, no element of $H$ lies within distance $\epsilon$ of $s(G)$.
Clearly the problem here is that you're working with open bounded sets, so they don't contain their boundary points. Maybe you can have better luck with compact sets? I thought that was usually the class of subsets to which one applies the Hausdorff metric anyway.
For compact sets, the $\min$ function works.
Very good! (at the time of my first reading of your solution, I had my private comprehension problems, sorry). #### <And still, your "clearly" is clearly not clear at all. I mean your general final comment which is false>.
edit
Theorem Let $X$ be a metric space containing a homeomorphic copy of the interval $(0, 1)$. Then the regular open sets of $X$ do not admit a uniformly continuous choice function.
I'll show just the case $X = (-2, 2)$ (the interval) and skip the epsilon-delta details and the fact there could be stuff around the embedded path, since the details of this are very similar to the original (see below).
(Note that a choice function admitting an $f$-metric choice function just means uniform continuity from $(S, d_H|_{S \times S})$ to $X$ with $f$ the modulus of continuity.)
For $n \in \mathbb{Z}$ define
$$ U_n = (\arctan(n)/\frac{\pi}{2} - \epsilon_n, \arctan(n)/\frac{\pi}{2} + \epsilon_n) $$
where $\epsilon_n$ are sufficiently small so that these sets are disjoint. So we have "order type $\zeta$ many" open intervals side by side inside $(-1,1) \subset X$. Each $U_n$ is a regular open set in $(-1,1) \subset X$, and $U_n \cap U_m = \emptyset$ if $n \neq m$. The union of all these, $U = \bigcup_n U_n$, is also easily seen to be regular open.
Now suppose $g$ is a choice function for regular opens. Then $g(U) \in U_n$ for some $n \in \mathbb{Z}$. Slide $U_L = \bigcup_{m \leq n} U_m$ continuously to the left side of $X$, join it to a single component and morph it into the interval $V_L = (-5/3,-4/3)$. Slide then $U_R = \bigcup_{m > n} U_m$ to the right side, join it to a single component and morph it to $V_R = (4/3,5/3)$. The choice must follow along, i.e.
$$ g(U) \in U_L \implies g(V_L \cup V_R) \in V_L. $$
But if we define
$U_L' = \bigcup_{m < n} U_m$ and $U_R' = \bigcup_{m \geq n} U_m$, and do the exact same with these sets, we get
$$ g(U) \in U_R' \implies g(V_L \cup V_R) \in V_R. $$
That is the contradiction that squares up the proof.
original
OP has suggested that I write an answer based on my comment. Here's one possible statement you get from that idea, quick write-up, I'll fix later if I screwed up the epsilons.
Let $X$ be a metric space and $S \subset \mathcal{P}(X)$ a set of sets in $X$. Let $f : \mathbb{R}_+ \to \mathbb{R}_+$ be a function. A function $g : S \to X$ is an $f$-metric choice function for $S$ if $g(A) \in A$ for all $A \in S$, and $d(g(A), g(B)) \leq f(d_H(A, B))$ for all $A, B \in S$. We say $S$ then admits an $f$-metric choice function.
Theorem Let $f : \mathbb{R}_+ \to \mathbb{R}_+$ satisfy $\lim_{x \to 0} f(x) = 0$ and let $X$ be a metric space containing a homeomorphic copy of $S^1$. Then the regular open sets of $X$ do not admit an $f$-metric choice function.
Proof. Let $h : S^1 \to X$ be the embedding of $S^1$ into $X$, and let $\epsilon > 0$ be such that opposite points on $S^1$ map at least distance $\epsilon$ apart form each other in the map $h$. Let $0 < \delta < \epsilon/10$ be such that $f(x) < \epsilon/10$ for $x < 3\delta$.
Identify $S^1$ as $\mathbb{R}/\mathbb{Z}$. To each $a \in S^1$ associate the set
$$ k(a) = k_1(a) \cup k_2(a) $$
where
$$ k_1(a) = \overline{B_{\delta}(h(a))}^\circ $$
and
$$ k_2(a) = \overline{B_{\delta}(h(a+1/2))}^\circ \subset X. $$
If $\delta > 0$ is small enough, $k(a)$ is regular open for all $a$. (The interior of the closure of an open set is regular open, so $k_i(a)$ is. The union of two regular opens may not be regular open in general, but since $\delta < \epsilon/10$ this happens.) Again because $\delta < \epsilon / 10$, the sets $k_1(a)$ and $k_2(a)$ are disjoint.
Suppose we had a choice function $g$ for the regular opens that is $f$-continuous. W.l.o.g. we may assume $g(k(a)) \in k_1(a)$ for some $a \in S^1$. Then by picking small enough increments, it is easy to see that in fact $g(k(a)) \in k_1(a)$ for all $a \in S^1$.
(Here's some algebra to show that in case it's not obvious: If the distance between $h(a)$ and $h(a')$ is at most $\delta$, then the distance between $k_1(a)$ and $k_2(a')$ is at least $\epsilon - 3\delta > \epsilon/10$, and
$$ d_H(k(a), k(a')) \leq \max(d_H(k_1(a), k_1(a')), d_H(k_2(a), k_2(a'))) \leq 3\delta, $$
so $g(k(a')) \in k_1(a')$ whenever $g(k(a)) \in k_1(a)$ and $|a'-a|$ is small enough.)
But now we have a contradiction since
$$ g(k(a)) \in k_1(a) $$
and
$$ g(k(a)) = g(k(a+1/2)) \in k_1(a+1/2) = k_2(a) $$
and $k_1(a) \cap k_2(a) = \emptyset$. Square.
Your assumption about $\ X\ $ containing a topological circle is painful since it excludes $\ \Bbb R\ $ (and the homeomorphic images of $\ \Bbb R).\ $ Fortunately, you do not need the full force of your assumption. It's enough for your argument to assume that $\ X\ $ contains a non-degenerated arc, a homeomorphic image of $\ [0;1].\ $ Indeed, this will allow you to create a "similar" circle in the space of sets, which circle you can use for your modified proof. ("Similar" with respect to your goal).
To be fair, I've mentioned $\ S^2\ $ which contains circles.
Of course, the larger a space the harder to find selectors.
Not sure how a circle in the space of sets helps. I have a different trick for an arc (for continuous choice) which I can add in a few hours.
Ville, you use the circle in $\ X\ $ just to use an induced circle (or closed orbit) in the space of the sets -- that's how your proof works. But when you have that circle in the space of sets (by whichever way), you don't need another one in $\ X.$
Well, actually I use the circle to have a copy of one-dimensional projective space in there. Usually that's $S^1$ but the quotient part is what fools choice.
(That's my interpretation of my proof anyway.)
|
2025-03-21T14:48:30.982568
| 2020-05-14T04:08:51 |
360303
|
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|
Stack Exchange
|
Approximating expectation of exponential of Wishart matrix
I am trying to obtain an Approximating expectation of exponential of Wishart matrix $X (N,N)$ with $\operatorname{rank} (X)=K$defined as:
\begin{align}
J = E[{e^{{v^H}Xv}}]
\end{align}
where $v$ is $(1,N)$ given vector.
I have used Wishart distribution defined by:
\begin{align}
f = {e^{\sum\limits_{i = 1}^K {{\lambda _i}} }}\prod\limits_{i = 1}^K {\lambda _i^{N - K}\prod\limits_{j > i}^K {{{({\lambda _i} - {\lambda _j})}^2}} }
\end{align}
and using the eigenvector written of $X$ as $ X = U\Sigma {U^H}$
so the problem can be written as:
\begin{align}
E[{e^{\sum\limits_{i = 1}^K {{b^i}{\lambda _i}} }}]
\end{align}
For smaller value on $N$ and$K$ I found finite expression bat as function of $b^i$ that are obiened by multiplying the vectors of unitary random matrix $U$ by my vector $t$.
can I say that $E[b_i]=||t||^2 $?
if it's possible can same person help me or give me another approach Thanks
I have corrected some typos, but since I wasn't entirely sure I skipped these two: Is "obiened" meant to be "obtained"? Is "bat" meant to be "but"?
The expectation value diverges, at least in the special case discussed below, but likely in general.
Take $v=(1,0,0,\ldots 0)$, so only a single element of $v$ is nonzero. The distribution of $y=v^H Xv=X_{11}$ follows from the known marginal distribution of Wishart matrix elements,
$$P(y)=\frac{1}{\sqrt{2\pi y}}e^{-y/2},\;\;y>0.$$
Then the desired expectation value $J$ of $e^y$ diverges.
sir Can I find the expectation of $b_i$ as a function of $t$?
$\mathbb{E}[b]=\mathbb{E}[U\cdot t]=\mathbb{E}[U]\cdot t=0$
|
2025-03-21T14:48:30.982807
| 2020-05-14T04:13:59 |
360304
|
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|
Stack Exchange
|
Homotopy invariant analogues of localizing invariants
Given a localizing invariant, $E$, valued in spectra, by following the recipe prescribed in 3.13 of https://arxiv.org/abs/1808.05559, we can define a homotopy-invariant version of $E$ on $H\mathbb{Z}$-linear categories:
$$EH(\mathcal{C}) = \underset{\Delta^{op}}{\text{colim}} \ E(\mathcal{C} \otimes_{\mathbb{Z}} \text{Perf}_{\mathbb{Z}[\Delta^{\bullet}]}),$$
where $\mathcal{C}$ is an $H\mathbb{Z}$-linear category and $\mathbb{Z}[\Delta^{\bullet}]$ is the simplicial ring of algebraic simplices, as defined in IV.11.3 of Weibel's K-book (https://sites.math.rutgers.edu/~weibel/Kbook.html), for example.
Variation: One could perform this type of construction with $\mathbb{S}[t] = \Sigma_{+}^{\infty}(\mathbb{N})$, the free $E_{1}$-algebra on a single generator, or $\mathbb{S}\{t\}$, the free $E_{\infty}$-algebra on a single generator.
$KH$ has many nice features not shared by algebraic K-theory. For example, it is a truncating invariant, meaning that for $A$ a connective $E_{1}$-ring spectrum, $KH(A) \simeq KH(\pi_{0}A)$ (note, of course, that this is true for K-theory if we look at the negative K-groups). In fact, according to Remark 5.3.5 of https://arxiv.org/abs/1705.03340, this result, attributed to Antieau-Gepner-Heller, is true if we use $\mathbb{S}[\Delta^{\bullet}]$ as opposed to $\mathbb{Z}[\Delta^{\bullet}]$.
Questions
Are there nice structural features that certain homotopy invariant analogues of localizing invariants posses that the original invariants do not? I would be very interested to know some examples aside from $KH$. A back-of-the-envelope calculation suggests homotopy invariant $THH$ will fail to be a truncating invariant.
Is there anything that can be said from the motivic point of view? For example, homotopy K-theory is representable by the motivic spectrum $KGL$. I am quite ignorant of the motivic side of things, but it seems like one would have to apply a result of Robalo (Corollary 5.11 of https://arxiv.org/abs/1206.3645) to obtain something representable. I think this would cut down the collection of localizing invariants which produce anything motivic.
What types of comparison results might hold between $E$ and $EH$? For instance, $K$ and $KH$ agree on regular Noetherian rings, and for rings where $\ell$ is invertible, provided we use mod $\ell$ coefficients.
How does the Antieau-Gepner-Heller result mentioned above work? Is the proof similar to that of Proposition 3.14 in https://arxiv.org/abs/1808.05559?
|
2025-03-21T14:48:30.982987
| 2020-05-14T05:52:10 |
360308
|
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|
Stack Exchange
|
Action of a toral automorphism on a Markov partition
Let $$A = \begin{bmatrix}
1 & 1 \\
1 & 0\\
\end{bmatrix}.$$
Then the eigenvalues of $A$ are $1/2(1+\sqrt{5})$ and $1/2(1-\sqrt{5})$. The eigenvector corresponding to the unstable eigenvalue is the line $y = 1/2(-1+\sqrt 5)x$, whereas the stable eigenvalue has eigenvector $y = -1/2 (1+\sqrt 5)x$.
Let $L$ be a hyperbolic linear automorphism of the torus induced by $A$.
Could someone explain how we get the following graphs? To simplify the question, let's jsut discuss how $L$ affects $R_1$. The upper left picture is a Markov partition of the torus. The dark regions represent $R_1$, $L(R_1)$ and $L^{-1}(R_1)$.
The closely spaced line segments in $R_1$ get spaced out more widely because $L$ expands distances along the unstable eigenline. These line segments get shorter because they are parallel to the stable eigenline, along which $L$ contracts distances. Finally, $R_1$ turns into three pieces rather than two because it gets wrapped around the torus. Specifically, the small, top-left piece of $R_1$ turns (mostly) into the small, bottom-right piece of $L(R_1)$, while the big strip of $R_1$ turns into the middle-right piece of $L(R_1)$ plus most of the long strip of $L(R_1)$.
By computing the coordinates of the corners of the various rectangles and doing the linear algebra explicitly, you can make this qualitative explanation quantitative. I'm happy to add a sample calculation to this answer if that would be helpful.
Thank you, it is a bit clearer to me now. It would be great if you can add a sample calculation because I find it difficult to see directly how after stretching and contracting we will get those divided regions of $L(R_1)$.
Sophie, could you take the action of $L$ on a line segment as an example to explain how to do the calculation? For example, the line segment of $y=0$ in the unit square.
|
2025-03-21T14:48:30.983140
| 2020-05-14T06:32:12 |
360310
|
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"Denis Serre",
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|
Stack Exchange
|
Is the matrix positive definite given the Gauss-Seidel method converges?
I know that the Gauss-Seidel method converges given that the matrix you want to solve is symmetric positive definite. However, I'm wondering if the "converse" of the statement is true. Namely, if $A$ is a symmetric matrix with positive diagonals and the Gauss-Seidel method converges for all initial guess $x_0$, is $A$ guaranteed to be positive definite?
There is an interesting though partial answer to your question:
Under your assumptions, it is not possible that the signature of $A$ be $(n-1,1)$ (exactly one negative eigenvalue).
Proof: With standard notations, $A=D-E-E^T$ where $E$ is strictly triangular and $D$ diagonal. By assumption, $D>0_n$. Notice that the assumption of convergence implies that $A$ is invertible.
The iteration matrix is $G=(D-E)^{-1}E^T$. The quadratic form $q(x):=x^TAx$ satisfies the identity
$$q(Gx)+y^TDy=q(x),\qquad y:=x-Gx=(D-E)^{-1}Ax.$$
This implies that $G$ preserves the set $q<0$. If the signature of $A$ is $(n-1,1)$, this set is the union of two opposite convex cones $\pm K$. Thus $\pm G$ preserves $K$. One deduces from Brouwer fixed point theorem that $G$ admits an eigenvector $x\in K$, $Gx=\mu x$. Then $\mu^2q(x)<q(x)<0$ gives $|\mu|>1$, which contradicts the convergence of the Gauss-Seidel method.
Nota. The identity about $q$ is the one used to prove the convergence of the method when $A$ is positive definite. More generally, it is used to prove that any method with iteration matrix $L:=M^{-1}N$ where $A=M-N$, such that both $A$ and $M^T+N$ are positive definite, is convergent
Thank you for your answer. But I'm wondering what is $q$ in your proof?
@bernard. Oups, $q$ is the quadratic form associated with $A*. I edit the post.
Note that, as Prof. Serre's answer, $q(x)=x^TAx$ decreases as the iterations go on. In particular, consider $Ax=b$ with $b=0$, we have $x_0^TAx>x_1^TAx_1>\cdots>x^T_nAx_n>\cdots$ and $\lim\limits_{n\to \infty} x_n=0$. If $A$ is not positive definite, there exists $x_*$ such that ${x_*}^{T}Ax_*<0$. Take $x_0=x_*$ then we have $0>x_0^TAx_0>\cdots$, thus $\lim\limits_{n\to \infty} x_n^TAx_n<0$, which contradicts to $\lim\limits_{n\to \infty} x_n=0$.
|
2025-03-21T14:48:30.983306
| 2020-05-14T07:58:04 |
360316
|
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|
Stack Exchange
|
Stein's lemma for Gaussian variables proof
I am reading a paper (https://arxiv.org/abs/1001.3448) and they mentioned Stein's lemma (below) as a useful fact without proof, I also read the reference in the paper but I got nothing. Please help me any material contained proof of this lemma.
(Stein's lemma) For jointly Gaussian variables $Z_1, Z_2$ with zero mean and for any function $\psi: \mathbb{R} \rightarrow \mathbb{R}$, where $\mathbb{E}\{\psi'(Z_1)\}, \mathbb{E}\{\psi'(Z_2)\}$ exist, the following holds
$$\mathbb{E}\{Z_1\psi(Z_2)\} = \text{Cov}(Z_1,Z_2)\mathbb{E}\{\psi'(Z_2)\}$$
You can write $Z_1=\alpha Z_2+\tilde{Z}_1$ with $Z_2,\tilde{Z}_1$ are independent. Then $$\mathbb{E}(Z_1\psi(Z_2))=\alpha\mathbb{E}(Z_2\psi(Z_2))=\frac{\alpha}{\sqrt{2\pi\sigma^2}}\int_{\mathbb{R}}xe^{x^2/2\sigma^2}\psi(x)dx $$ one can conclude by integration by parts.
And why we can write $Z_1 = \alpha Z_2 + \tilde{Z}_1$ with $Z_2, \tilde{Z}_1$ are independent?
I have my own answer now, I choose $\alpha = \frac{\mathbb{E}{Z_1Z_2}}{\mathbb{E}{Z_2^2}}$
|
2025-03-21T14:48:30.983396
| 2020-05-15T20:56:40 |
360461
|
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|
Stack Exchange
|
A question related to newform and irreducible cuspidal representation of $\operatorname{GL}_n$
I was reading adelization of classical automorphic forms and learnt that each cusp form corresponds to an automorphic representation of $\operatorname{GL}_n(\mathbb{A}_\mathbb{Q})$. I understood the proof. But then I found a statement that there is a one-to-one correspondence between newforms of the congruence subgroup $\Gamma_1(N)$ and the irreducible cuspidal representations of $\operatorname{GL}_n(\mathbb{A}_\mathbb{Q})$.
But I couldn't find any proof of it.
Please suggest some references.
Thank you.
Newform theory for $\mathrm{GL}_n$ was originally developed over non-archimedean local fields (at least for $n\geq 3$). The local statements readily yield their global adelic counterparts, since a cuspidal automorphic representation is uniquely a restricted tensor product of local admissible generic representations. It is then straightforward to translate from adelic language to classical language.
So all you need is newform theory for admissible generic representations of $\mathrm{GL}_n$ over a non-archimedean local field. The guiding questions are: what is a newvector in this context, to what extent is it unique, how does it generate the representation, and so on. This theory was developed by Jacquet, Piatetski-Shapiro, and Shalika in their paper "Conducteur des représentations du groupe linéaire", Math. Ann. 256 (1981), 199-214. The paper also has a corrigendum.
A more accessible reference is Section 13.8 in the textbook of Goldfeld and Hundley "Automorphic representations and L-functions for the general linear group" (volume 1; volume 2). In general, reading this book should clarify many of the questions you might have. It filled an important gap in the literature.
Other useful references: this paper of Casselman for $\mathrm{GL}_2$: https://doi.org/10.1007/BF01428197. And another correct proof for $\mathrm{GL}_n$ due to Matringe: https://www.math.uni-bielefeld.de/documenta/vol-18/37.pdf
@PeterHumphries: Thank you, Peter. Perhaps you can also mention somewhere the archimedean analogue you were developing. I am too busy now, e.g. I wrote the above answer in a rush!
Sadly still developing (at least for $\operatorname{GL}_n(\mathbb{R})$; the proofs are complete for $\operatorname{GL}_n(\mathbb{C})$). The theory itself takes too much work to state in a comment, so I will just link to the Oberwolfach Report where I first announced the result: https://www.mfo.de/document/1736/OWR_2017_40.pdf (P.S. It's 4am - go back to sleep!)
All done now! See https://arxiv.org/abs/2008.12406
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2025-03-21T14:48:30.983573
| 2020-05-15T21:31:25 |
360462
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|
Stack Exchange
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Coefficient ring of Satake isomorphism
Let $G$ be a split reductive group over local field $F$, $G^L$ be the (complex) Langlands dual group of $G$. Denote $H$ to be the $\mathbb{Z}$-Hecke algebra of $G$, that is the ring of $G(\mathcal{O}_F)$-biinvariant $\mathbb{Z}$-valued functions on $G(F)$. Let $R(G^L)$ be the Grothendieck ring of finite dimensional representation of $G^L$. It is stated in Gross's survey about Satake isomorphism that $H \otimes \mathbb{Z}[q^{\pm1/2}]$ is isomorphic to $R(G^L)\otimes \mathbb{Z}[q^{\pm1/2}]$, where $q$ is the cardinality of residue field of $F$.
My question is whether $H$ is isomorphic to $R(G^L)$? I suspect that they are non isomorphic when $G$ is some exceptional group.
Is your question whether they can be abstractly isomorphic as rings, or whether the Satake transform is sometimes defined over $\mathbb{Z}$? The latter will never happen when $G$ has Borel subgroups, because you must have the modular character. As you know from Gross's notes, the former does happen for tori.
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2025-03-21T14:48:30.983690
| 2020-05-15T22:03:59 |
360464
|
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"Pat Devlin",
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|
Stack Exchange
|
Linear independence of exponential functions: a reference
Is there a publication containing this obvious fact: For any real $T>0$, any natural $n$, any complex $c_1,\dots,c_n$, and any distinct complex $z_1,\dots,z_n$ such that $\sum_1^n c_k e^{tz_k}=0$ for all $t\in[0,T)$, we have $c_1=\dots=c_n=0$?
Somehow, I cannot find such a publication.
Vandermonde matrix is nonsingular?
Ycart, Bernard (2013), "A case of mathematical eponymy: the Vandermonde determinant", Revue d'Histoire des Mathématiques, 13, arXiv:1204.4716
You will probably find no 20 century publication, except an exercise is some Calculus or Linear algebra book. All these facts were really clarified in 18th century (perhaps by Wronski himself, or Vandermonde), but at that time they did not speak of "linear independence":-)
@ZachTeitler : Thank you for your comment. However, I don't need a proof -- only a reference.
@yarchik : Thank you for your comment. However, I cannot find this fact in that paper by Ycart.
Publication includes textbooks by definition, so it should definitely appear in several of them... I searched Google books [linear independence exponentials] and immediately got this reference: https://books.google.fr/books?id=Y3YSCmWBVwoC&pg=PA618&dq=linear+independence+exponentials&hl=fr&sa=X&ved=0ahUKEwjzo8bZ-rbpAhVID2MBHX_jAXEQ6AEIKjAA#v=onepage&q=linear%20independence%20exponentials&f=false
@AlexandreEremenko : Thank you for your comment. A reference to a textbook would be fine.
@YCor : Thank you for your comment. This may help.
I put this question on my final exam for linear algebra, and I got oodles of cool very different proofs. :-)
@Iosif Pinelis: You may combine probl. 2, and problem 60, of the second volume of Polya Szego, Problems and theorems of Analysis, part 7 "Determinants and Quadratic forms" (The first problem is Vandermonde det, the second is the criterion of linear independence).
See Lemma 3.2 p. 92 in the book Differential Equations, Springer Verlag, 2016 by Viorel Barbu. The very elegant proof there is not the usual proof based on Wronskians. He proves a bit more, namely that the exponentials are linearly independent over the field of rational functions with complex coefficients.
@AlexandreEremenko : Thank you for your latter comment as well. Unfortunately, I don't have the Pólya--Szego book right now. Do they include the case of complex $z_k$ as well?
@LiviuNicolaescu : Thank you for your comment. This is exactly what I needed except that in Barbu's Lemma 3.2 the linear independence is for the exponential functions defined on $\mathbb R$ rather than on a finite nonzero-length interval. However, the proof (which seems the most natural to me and which I had foremost in mind) will of course work for the finite-interval case as well. Would you like to present your comment as a formal answer?
@Iosif Pinelis: It does not matter that $z_k$ are complex: computation of the Vandermonde determinant holds for any field. Frankly speaking I do not understand your difficulty. Polya Szego can be found on line.
@IosifPinelis Glad it helps. I'll keep my answer as a comment. My contribution is minimal
@AlexandreEremenko : I have been unable to find the Pólya--Szego book online; could you please give a link to it? As for complex $z_k$'s and otherwise, my biggest concern is about how ready-to-use the published result is. As I said in the post and a comment, the desired result is obvious and the simplest and most natural proof (for me) is this: "divide by $e^{tz_n}$, differentiate in $t$, and use induction on $n$. $\Box$" -- without using Vandermonde determinants.
Previous comment continued: Yet another proof, of a stronger result -- with $t$ in an $n$-set rather than in an interval but only for real $z_k$'s -- is given on p. 10 of "Tchebycheff systems: with applications in analysis and statistics" by Karlin and Studden. That proof uses Rolle's theorem and thus does not seem to work for complex $z_k$'s.
Apostol's Calculus book has a proof; I believe it is at the beginning of volume 2.
@AndrésE.Caicedo : Thank you for your comment. You are probably referring to Example 7 on page 10 of Section 1.7 of of the book at shorturl.at/cjwxZ . However, the proof there works only for for real $z_k$'s.
somewhat related https://mathoverflow.net/questions/277655/reference-for-exponential-vandermonde-determinant-identity
@AbdelmalekAbdesselam : Thank you for pointing out to this connection.
I will recount the more general statement of linear independence of characters, given in Lang's Algebra book, and credited to Artin. Let $G$ be a group, and $K$ a field. Then distinct homomorphisms $\phi_1, \ldots, \phi_n: G \to K^\times$ are linearly independent.
Proof: Suppose not, and suppose we have a nontrivial linear relation
$$a_1 \phi_1 + \ldots + a_n \phi_n = 0,\qquad (1)$$
where $n$ is taken as small as possible. Clearly $n>1$ and $a_i \neq 0$ for all $i$. Because the $\phi_i$ are distinct, we can find an element $g \in G$ such that $\phi_1(g) \neq \phi_2(g)$. We have
$$a_1 \phi_1(gh) + a_2 \phi_2(gh) + \ldots + a_n \phi_n(gh) = 0$$
for all $h \in G$; by virtue of the $\phi_i$ being homomorphisms, this may be rewritten to say
$$a_1 \phi_1(g)\phi_1 + a_2 \phi_2(g)\phi_2 + \ldots + a_n \phi_n(g)\phi_n = 0, \qquad (2)$$
Dividing $(2)$ by $\phi_1(g)$ and then subtracting (1) from the result, we arrive at a linear relation
$$\left(a_2 \frac{\phi_2(g)}{\phi_1(g)} - a_2\right) \phi_2 + \ldots = 0$$
which has fewer than $n$ summands and is nontrivial by choice of $g$, contradiction. $\Box$
K.Conrad quotes E. Artin's "Galois theory" (1948).
This is very nice, cleaner than the differentiation. Also, (1) does not have to be assumed to hold over all $G$. At least in the case when $G=\mathbb R$, it is enough to assume that (1) holds over a neighborhood of $0$ -- which is what I need.
Right, I'm downvoted because why? This was a reference request.
@ToddTrimble : I don't see a reason for the downvote. I upvoted your answer.
Let $y_k(t)=e^{tz_k}$. Proving by contradiction, suppose that they are linearly dependent, that is
$$\sum_{k=1}^nc_ky_k\equiv 0.$$
Differentiating $n-1$ times we obtain a homogeneous system of linear equations with respect to $c_k$. To have a non-trivial solution,
this system must have non-zero determinant. The determinant is:
$$\left|\begin{array}{cccc}y_1&y_2&\ldots& y_n\\
y_1^\prime& y_2^\prime&\ldots&y_n^\prime\\
\ldots&\ldots&\ldots&\ldots\\
y_1^{(n-1)}& y_2^{(n-1)}&\ldots& y_n^{(n-1)}\end{array}\right|=A(t)
\left|\begin{array}{cccc}1&1&\ldots&1\\ z_1&z_2&\ldots& z_n\\
\ldots&\ldots&\ldots&\ldots\\
z_1^{n-1}&z_2^{n-1}&\ldots&z_n^{n-1}\end{array}\right|,$$
where $A(t)=e^{t(z_1+\ldots+z_n)}\neq 0$.
The determinant in the right hand side is easy to compute.
Consider it as a polynomial with respect to, $z_n$. It is evidently
of degree $n-1$ and has $n-1$ roots at $z_1,\ldots,z_{n-1}$.
Therefore it is of the corm $$C(z_1,\ldots,z_{n-1})(z_n-z_1)\ldots(z_n-z_{n-1}).$$
Looking at the top degree term, we conclude that $C$ is a similar polynomial. So by induction our determinant is
$$\prod_{i<k}(z_i-z_k).$$
this is never zero, since $z_k$ are distinct.
References. Polya Szego, Problems and theorems of analysis, vol II, Part 7, "Determinants and quadratic forms''. Computation of the Vandermonde determinant is problem 2. The Wronskian criterion of linear independence is problem 60.
Remark. Vandermondes's determinant is computed in
ANY undergraduate textbook of
linear algebra, as a first example of determinant. For example, I teach linear algebra with the textbook of Strang, and differential equations with
the textbook of Boyce and di Prima. Both of them have Vandermonde determinant.
Remark 2. Undergraduate textbooks are rarely freely available online. If you insist on a free online reference, you may refer on the proof above.
In my experience, in 99% of the claimed applications of the Vandermonde determinant, one does not actually need the determinant, but only on the fact that the Vandermonde matrix is invertible, and this can be proved directly with a simpler argument: if a row vector is in the left kernel of this matrix, then its entries form the coefficients of a degree-$(n-1)$ polynomial with $n$ distinct zeros.
@Federico Poloni: I agree with this remark.
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2025-03-21T14:48:30.984207
| 2020-05-15T22:18:43 |
360466
|
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|
Stack Exchange
|
An alternative description of normalized cochains in terms of tensor powers of the augmented ideal
I want to know if the following alternative of the normalized non-homogeneous cochains is already know.
Let $G$ be a group and let ${\mathcal I}={\mathcal I}_G$ be its augmentation ideal, ${\mathcal I}=\ker (\varepsilon :{\mathbb Z}[G]\to{\mathbb Z})$.
If $M$ is a $G$-module then let $C^n(G,M)=\{ a:G^n\to M\, :\, a(u_1,\ldots,u_n)=0\text{ if }u_i=1\text{ for some }i\}$ be the normalized non-homogeneous cochains of degree $n$.
Definition We define the ${\mathcal I}$-cochains of degree $n$ as
$$C_{\mathcal I}^n(G,M):={\rm Hom}(T^n({\mathcal I}),M).$$
We have an isomorphism $C_{\mathcal I}^n(G,M)\to C^n(G,M)$ given by $f\mapsto a$, where
$$a(u_1,\ldots,u_n)=f((u_1-1)\otimes\cdots\otimes (u_n-1)).$$
Then the coboundary map $d_n:C^n(G,M)\to C^{n+1}(G,M)$, translated in the language of ${\mathcal I}$-cochains writes in a very convenient form. Namely, we have:
Proposition The coboundary map $d_n:C_{\mathcal I}^n(G,M)\to C_{\mathcal I}^{n+1}(G,M)$ is given by
\begin{multline*}
d_nf(\alpha_1\otimes\cdots\otimes\alpha_{n+1})\\
=\alpha_1f(\alpha_2\otimes\cdots\otimes\alpha_{n+1})+\sum_{i=1}^n(-1)^if(\alpha_1\otimes\cdots\otimes\alpha_i\alpha_{i+1}\otimes\cdots\otimes\alpha_{n+1})
\end{multline*}
for every $\alpha_1,\ldots,\alpha_{n+1}\in{\mathcal I}$.
This formula looks very similar to the one for $d_n:C^n(G,M)\to C^{n+1}(G,M)$, but with the last term of $d_na(u_1,\ldots,u_{n+1})$, $(-1)^{n+1}a(u_1,\ldots,u_n)$, ignored.
The closest thing I found is in Hilton-Stammbach, chapter VI, 13(c). ("Alternative Description of the Bar Resolution".) If we denote by $\bar C^n(G,M)$ the normalized homogeneous cochains and by $\bar C_{\mathcal I}^n(G,M)={\rm Hom}_G({\mathbb Z}\otimes T^n({\mathcal I}),M)$ the cochains resulting from the alternative description of the bar reductions, then the element $\bar a\in\bar C^n(G,M)$ corresponds to $\bar f\in\bar C_{\mathcal I}^n(G,M)$ if $\bar a(u_0,\ldots,u_{n+1})=\bar f(u_0\otimes (u_1-u_0)\otimes\cdots\otimes (u_n-u_{n-1}))$ $\forall u_0,\ldots,u_n\in G$.
The element $f\in C_{\mathcal I}^n(G,M)$ corresponding to $\bar f\in\bar C_{\mathcal I}^n(G,M)$ is given by $f((u_1-1)\otimes\cdots\otimes (u_n-1))=\bar f(1\otimes (u_1-1)\otimes u_1(u_2-1)\otimes\cdots\otimes u_1\cdots u_{n-1}(u_n-1))$ $\forall u_1,\ldots,u_n\in G$. As one can see, there is no nice, simple relation between $f\in C_{\mathcal I}^n(G,M)$ and $\bar f\in C_{\mathcal I}^n(G,M)$, such as, say, $f(\eta )=\bar f(1\otimes\eta )$ $\forall\eta\in T^n({\mathcal I})$.
I mention that I used these ${\mathcal I}$-cochains to solve the problem I described here:
Cohomology of elementary abelian $p$-groups, i.e. $H(G,{\mathbb F}_p)$ with $G\cong{\mathbb F}_p^r$
I have just posted an article on arXiv: https://arxiv.org/abs/2005.11868
The first section deals with these ${\mathcal I}$-cochains. Again, if you saw them somewhere, maybe with some other name and notation, then please let me know.
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2025-03-21T14:48:30.984389
| 2020-05-15T22:30:21 |
360467
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"authors": [
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|
Stack Exchange
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Frobenius splitting for an excellent, non $F$-finite, $F$-pure hypersurface
Let $p$ be an odd prime. Let $k$ be a field of characteristic $p$ such that $[k:k^p]=\infty$ (i.e. $k$ is not $F$-finite ) .
Also assume that $-1$ is not a square in $k$ . Consider the homogeneous polynomial $f(x,y)=x^2+y^2\in k[x,y]$ . Then $f$ is irreducible in $k[x,y]$ , hence $R=k[x,y]/(f)$ is an integral domain (of dimension $1$ , hence Cohen-Macaulay) .
Moreover, $R$ is Excellent. (https://en.m.wikipedia.org/wiki/Excellent_ring)
Also, $R$ is not $F$-finite (as $k$ is not).
We also observe that by Fedder's criteria, $R$ is $F$-pure since $f^{p-1}\notin (x^p,y^p)$ .
Indeed, to see $f^{p-1}\notin (x^p,y^p)$, observe that $f^{p-1}=(x^2+y^2)^{p-1}=x^{p-1}y^{p-1}+\sum_{0\le j\le p-1 , j\ne \dfrac {p-1}2 } (x^2)^{j}(y^2)^{p-1-j}$.
Now $x^{p-1}y^{p-1}\notin (x^p, y^p)$.
Moreover, either $j$ or $p-1-j$ is $\ge \dfrac{p-1}2+1=\dfrac {p+1}2 $ if $j\ne \dfrac {p-1}2$, hence $(x^2)^{j}(y^2)^{p-1-j}\in (x^p, y^p)$ when $j\ne \dfrac {p-1}2$. Thus $f^{p-1} \notin (x^p, y^p)$.
My question is: When can we say that $k[x,y]/(x^2+y^2)$ is $F$-split ?
Your ring is Frobenius split. I also don't think you need the condition that $k$ does not contain a square root of $-1$.
Proof. Denoting by $\overline{k}$ the algebraic closure of $k$, we see that
$$\overline{R} := R \otimes_k \overline{k} \simeq \frac{\overline{k}[x,y]}{x^2+y^2}$$
is $F$-pure by Fedder's criterion, using the same computation you gave. Since $\overline{k}$ is algebraically closed, it is $F$-finite. Thus, $\overline{R}$ is also $F$-finite. Combining these two facts, we see that $\overline{R}$ is Frobenius split.
Now consider the commutative diagram
$$\require{AMScd}\begin{CD}
R @>F_R>> F_{R*}R @>\phi>\exists?> R\\
@VVV @VVV @VVV\\
\overline{R} @>F_{\overline{R}}>> F_{\overline{R}*}\overline{R} @>\bar{\phi}>> \overline{R}
\end{CD}\tag{1}\label{eq:basechange}$$
where the composition in the bottom row is the identity on $\overline{R}$ and the vertical maps are obtained by extending the field extension $k \subseteq \overline{k}$ by scalars along $k \to R$.
We want to show that the map $\phi$ exists such that the composition in the top row is the identity on $R$. Since $k \subseteq \overline{k}$ splits as a map of $k$-vector spaces, the map $R \to \overline{R}$ splits as a map of $R$-modules. Let $f\colon \overline{R} \to R$ be such a splitting. Then, defining $\phi$ to be the composition of the three arrows in the right square of the diagram
$$\begin{CD}
R @>F_R>> F_{R*}R @>\phi>> R\\
& @VVV @AAfA\\
& & F_{\overline{R}*}\overline{R} @>\bar{\phi}>> \overline{R}
\end{CD}$$
we see that the composition in the top row of the diagram \eqref{eq:basechange} is the identity on $R$. Thus, $R$ is Frobenius split. $\blacksquare$
I also want to point out that this is a special case of a result due to Rankeya Datta and myself. The statement for $F$-purity for complete local rings I believe first appeared in [Fedder, Lem. 1.2].
Theorem (Datta–M; see [M1, Thm. B.2.3]). Let $R$ be a ring essentially of finite type over a noetherian complete local ring $(A,\mathfrak{m})$ of prime characteristic $p > 0$. Then,
$R$ is $F$-pure if and only if $R$ is Frobenius split; and
$R$ is strongly $F$-regular if and only if $R$ is split $F$-regular.
Here, $R$ is strongly $F$-regular if every inclusion of modules is tightly closed, following [Hochster, Def. on p. 166], and $R$ is split $F$-regular if for every element $c$ avoiding every minimal prime of $R$, there exists an integer $e > 0$ such that the composition
$$R \overset{F^e_R}{\longrightarrow} F^e_{R*}R \xrightarrow{F^e_*(-\cdot c)} F^e_{R*}R$$
splits as a map of $R$-modules. Split $F$-regularity is the original definition for strong $F$-regularity in the $F$-finite case [Hochster–Huneke, Def. 5.1], although the terminology comes from [Datta–Smith, Def. 6.6.1].
The proof of the theorem uses the gamma construction of Hochster–Huneke [Hochster–Huneke, (6.7) and (6.11)] and [M2, Thm. 3.4], but the idea is to construct a diagram like that in \eqref{eq:basechange}, and use the fact (communicated to Hochster by Auslander) that pure maps from complete local rings split.
In your case, you can set $A$ to be the ground field $k$.
Finally, it is worth mentioning that the theorem does not hold for excellent rings in general: Rankeya and I found excellent regular rings, and even DVR's, that are not Frobenius split [Datta–M]. The rings we consider are Tate algebras and rings of convergent power series over non-Archimedean valued fields. The specific field we use was provided to us by Gabber, but one can also use fields constructed by Blaszczok and Kuhlmann.
Thanks for your answer ... I was indeed looking for an excellent, F-pure domain which is not F-split, so your last paragraph is right on target and I'll carefully take time reading your paper. Also, just wanted to make sure of one thing : $F_{\bar R*} \bar R \cong F_{R*} R\otimes_k \bar k$ in the proof, right ? Oh and btw, the point about $-1$ being not a square in $k$ is indeed just a little nitpicking for my own satisfaction to make sure $k[x,y]/(x^2+y^2)$ is indeed a domain ...
Hi @sde, I'm glad you found those examples interesting! The isomorphism you state does not hold: using identifications of the form $F_{S*}S \cong S^{1/p}$ for reduced rings $S$, the left-hand side is $\overline{k}^{1/p}[x^{1/p},y^{1/p}]/(x^{2/p}+y^{2/p})$, while the right-hand side is $(k^{1/p} \otimes_k \overline{k})[x^{1/p},y^{1/p}]/(x^{2/p}+y^{2/p})$, which is not reduced. To see this last statement, note that the subring $k^{1/p} \otimes_k \overline{k}$ is non-reduced by [Stacks, Tag 030W], for instance.
I see, thanks for explaining ... then do you think $F_{\bar R*}\bar R\cong F_{R*}R\otimes_R \bar R$ holds true ? Actually I'm trying to see why $\bar R$ would still remain $F$-pure via some general possibly characteristic free argument ...
Also, just double checking, that when working with an Algebra, essentially of finite type over a field, there's no need to distinguish between F-purity and F-splitting by the result proved by you and Datta, right ? It's just that this is kind of a cool big leap (but very natural going by your proof) 'cause all the lecture notes by Schwede or Hochster that I've read so far always assumes $F$-finiteness when going from F-purity to F-split while applying Fedder's criteria ..., but as your result shows, there's no need to assume that
Hi @sde, my previous comment says that your isomorphism cannot hold. Also, even if $R$ is $F$-pure, $\overline{R}$ is sometimes not even reduced; see this example of Chevalley. Finally, yes, our result says that when working with essentially of finite type algebras over a field, there is no need to distinguish between $F$-purity and Frobenius splitting, although there are other reasons why one would want $F$-finite hypotheses. I should also mention that Fedder proved they coincide in the complete local case.
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2025-03-21T14:48:30.984913
| 2020-05-15T23:46:49 |
360477
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|
Stack Exchange
|
Is $\int_M\Delta u = 0$ if $u$ is not $C^2$ on a set of measure zero?
Suppose that $M^2$ is a closed Riemannian manifold and that $u$ is a $C^2(M\setminus S),$ where $S$ is a closed measure set consisting possibly on a enumerable amount of points. Can we still conclude that $\int_M\Delta u =0?$
You need to be clear what "$C^2$ at almost every point" means. For example if $M$ is $S^1$ realized as $[0,1]$ with endpoints identified, the function $u(x) = x^2$ is $C^2$ except at one point and clearly doesn't satisfy the desired conclusion.
If you mean instead something like "almost everywhere equal to a $C^2$ function $v$" then the answer is certainly yes, in the sense of weak derivatives, because then $\Delta u = \Delta v$ almost everywhere.
Now I think my first comment is a counterexample again, with $S$ equal to a single point.
@NateEldredge I don't see how your counter example works in dimension greater or equal $2$ since on those it does hold that the integral of the Laplacian of $C^2$ function vanishes.
Oh, I didn't see the added assumption of dimension 2.
If $u$ is well-defined as a distribution on $M$, then the answer is yes, if the integral is properly interpreted, namely. as the evaluation of the distribution on the constant function $1$.
And the Laplacian of $u$ is defined using weak derivatives as a distribution over the full domain.
You have to say what is the meaning of $\Delta u$, and of $\int\Delta u$. For the integral to have a meaning, $u$ has to be a distribution and $\Delta u$ has to be a (signed, Radon) measure. Such distributions are called $\delta$-subharmonic functions.
Then $\int_M\Delta u=0$
without any assumptions about $C^2$ or $S$. (This is called the "Main Theorem" about compact Riemann surfaces
in S. Donaldson, Riemann surfaces, Chap 8, Thm. 5).
But if you understand $\Delta u$ in the naive, classical sense for $C^2$ function, and understand your integral as $\int_{M\backslash S}\Delta u$ then of course this does not have to be $0$.
For example, on the Riemann sphere, let $u(z)=0,\,|z|<1,\; \; u(z)=|z|,\, 1<z<2,\;\; u(z)=2,\, |z|>1$. This function is $C^2$ except on the set of measure zero
consisting of two circles $|z|=1$ and $|z|=2$. But
$$\left(\int_{|z|<1}+\int_{1<|z|<2}+\int_{|z|>1}\right)u(z)>0.$$
Even simpler example is $u(z)=|z|$ on the Riemann sphere, $\Delta u>0$ when
restricted to $0<|z|<\infty$ and $\int_{0<|z|<\infty}u(z)=+\infty$.
|
2025-03-21T14:48:30.985101
| 2020-05-16T01:19:55 |
360481
|
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|
Stack Exchange
|
Arithmetic-geometric mean for rationals?
Let $\operatorname{AGM}(x,y)$ be the arithmetic-geometric mean of $x$ and $y$. Given an error $\varepsilon>0$, a bound $b\in\mathbb R_+$ and a function $f:\mathbb R\rightarrow\mathbb R$ with $f(x)=O(\log x)$ and $f(x)=\Omega(\log\log x)$ for which rationals $\frac pq\in\mathbb Q_+$ with $0<p<b$ and $0<q<b$ is it possible to find $x=\frac{p'}{q'},y=\frac{p''}{q''}\in\mathbb Q$ with $\mathsf{\max}(|p'|,|q'|,|p''|,|q''|)<f(b)$ such that
$$\Bigg|\frac pq-\operatorname{AGM}(x,y)\Bigg|<\varepsilon$$
holds?
Is there explicit methods to write such $x,y$ down?
The density of such representable $\frac pq$ should be tiny. Nevertheless are there special forms where this can be done. So are there special family of rationals?
What's AGM? ${}$
@Wojowu The arithmetic-geometric mean, I think.
This is just some comments.
Since $x,y>0$ for $AGM(x,y)$ to be defined, why the absolute values?
I will assume $f(\cdot)$ takes only integer values and that $p’,p’’,q’,q’’ \leq f(b)$
Let $F(b)$ be the set of fractions with numerator and denominator bounded by $b$. Then $|F(b)|<b^2$ and the largest members are integers until $\frac{b}2$. The smallest are their reciprocals. One would expect things to be densest near $1.$
Fix $b$ And let $c=f(b)$, then the $|F(c)|$ values of $x,y$ are all between $1/c$ and $c$ so $p/q$ better be between $1/c-\epsilon$ and $c+\epsilon$ . Then there are somewhat under $\binom{c^2}2$ values for $AGM(x,y)$ . Each determines an interval of radius $\epsilon$.
Certainly we can get all rationals in the $\epsilon$ neighborhoods of the members of $F(c)$ using $x=y.$ Those alone should allow all rationals in some interval , depending on $\epsilon$, and containing $1$ near its lower bound. Taking $x$ near $y$ would seem to allow a larger interval.
Past that I would suggest starting with $c=10$ or smaller and computing to see what the collection of intervals looks like.
|
2025-03-21T14:48:30.985250
| 2020-05-16T02:42:41 |
360483
|
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|
Stack Exchange
|
expectation of the exponential of the inverse of variable with Marchenko–Pastur distribution
This question is related to another answered before
distribution on the inverse Wishart matrix eigenvalues summation
my question is, is their finite expression for the expectation of
\begin{align}
{\rm E}[{e^{ - \frac{b}{{1 + ax}}}}]
\end{align}
where $x$ is
Marchenko-Pastur distribution variable.
just integrate $\int dx e^{-b/(1+ax)}\rho(x)dx$ with $\rho(x)$ the MP distribution; this integral cannot be evaluated in closed form, but you can evaluate it numerically.
@Carlo Beenakker Sir is there closed for the Laplace transform of MP distribution
no there is not.
|
2025-03-21T14:48:30.985893
| 2020-05-16T07:31:38 |
360492
|
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|
Stack Exchange
|
Singular support of an irreducible perverse sheaf
I was studying Sheaves on Manifolds by Kashiwara and Schapira, and while the singular support seems like a complicated invariant I cannot seem to find a counterexample to the following:
Let $X$ be a smooth complex variety and $\mathcal{F}=IC(U,\mathcal{L})$ be an irreducible perverse sheaf, where $\mathcal{L}$ is a local system on $U\subset X$. Then $SS(\mathcal{F})=T_{\overline{U}}^*X$, where the latter means the conormal bundle at $\overline{U}$.
This seems too easy of an answer to be true, but I still cannot find either a counterexample or a proof, and I cannot think of how to get an explicit answer using the Riemann-Hilbert correspondance either. Any help?
PS. I have asked the question already in Stack Exchange but it wasn't answered and I thought it may be more appropriate here after all?
Take $X=\mathbb C$, $U=\mathbb C^\times$, and $\mathcal L$ a nontrivial rank 1 local system (with monodromy $\mu \neq 1$, say).
Then the singular support of $IC(U,\mathcal L)$ is the union $T^\ast _X X \cup T^\ast_0 X$ of the zero section (which is what your conjecture would give in this case) with the cotangent fiber.
Note that here we have $ j_\ast \mathcal L[1] \cong IC(U,\mathcal L) \cong j_! \mathcal L[1]$. One can compute the singular support either using the sheaf definition from Kashiwwara-Schapira, or by considering the associated $D$-module $D_{\mathbb C}/D_{\mathbb C}(x\partial_x - \log(\mu))$ under the Riemann-Hilbert correspondence.
I didn't see your answer when I wrote mine, but its seems to be the same.
Yes, except your answer has the proper cohomological shifts! I will add those for clarity...
Thank you both very much! I am sorry I could only accept one solution. Another small question, it seems though as if every irreducible component is of that form. Maybe this is true in more generality?
@IoannisZolas It is known that the singular support of a constructible complex is contained in a union of conormal bundles (to the strata). I believe this containment is not necessarily an equality in general, though I don't have a good example off the top of my head.
My impression is that the singular support of a perverse sheaf is the characteristic variety of the regular holonomic $D$-module corresponding to it under Riemann-Hilbert. Assuming that's the case, it is possible to answer this in the negative. Let $X$ be the disk or the affine line if you prefer, and $U=X-\{0\}$. Choose a nontrivial rank one local system $\mathcal{L}$ on $U$. By Riemann-Hilbert, there is regular connection $\nabla$ on $\mathcal{O}_X$ such that $\mathcal{L}=\ker\nabla$ over $U$. Then $IC(U,L)= j_*L[1]$ (or a translate depending on your convention), and the corresponding $D$-module $M$ is the minimal extension of $\nabla$. The characteristic variety of $M|_U$ is the zero section of $T_U^*$. Therefore the characteristc variety of $M$ is either the zero section of $T^*_X$ or the zero section union $T_0^*$. If it's the former, then $M$ would have to be a connection which contradicts our choice, so it must be the latter.
|
2025-03-21T14:48:30.986441
| 2020-05-16T08:44:37 |
360496
|
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|
Stack Exchange
|
A conjecture involving $P_n=\prod_{k=1}^np_k$
For each positive integer $n$ let $P_n=\prod_{k=1}^n p_k$, where $p_k$ is the $k$th prime.
Question. Is my following conjecture true?
Conjecture. For any integer $n>1$, there are $k,m\in\{1,\ldots,n-1\}$ such that
$$P_n\equiv P_k\pmod n\ \ \text{and}\ \ P_n\equiv -P_m\pmod n.$$
For example, $P_{32}\equiv P_{23}\pmod{32}$ and $P_{32}\equiv -P_8\pmod{32}$.
I have verified the conjecture for all $n=2,3,\ldots,70000$. When $n$ is squarefree, the conjecture holds trivially. I'm unable to prove the conjecture fully.
For the motivation of the conjecture, one may look at Conjecture 1.5 and Remark 1.7 in my paper available from http://dx.doi.org/10.1016/j.jnt.2013.02.003.
Your comments are welcome!
As you noted, $P_n$ is always divisible by $\operatorname{rad}(n)$, so dividing everything by $\operatorname{rad}(n)$ gives a question about products of primes modulo $d = n/!\operatorname{rad}(n)$. If $p_i | n$, then $i \log(i) \approx p_i \leq n$, so $i \leq n/W(n)$, so for $n \geq 70000$ this forces $i \leq n/8$. For such $p_i$ you need $i \leq k, m$, so this still leaves $k,m \in {n/8,\ldots,n}$.
That said, according to this paper, not much is known about the distributions of consecutive primes modulo an integer. So you probably really want to exploit that your intervals are rather large compared to the modulus $d = n/!\operatorname{rad}(n)$.
|
2025-03-21T14:48:30.986688
| 2020-05-16T09:53:21 |
360499
|
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|
Stack Exchange
|
Dual varieties and nodal sections
Let $X$ be a(n even dimensional) smooth complex projective variety in $\mathbb{P}^N$, and let $X^{\vee}$ be its dual variety; up to an higher degree Veronese embedding of $X$, I assume that $X^{\vee}$ is a hypersurface (that is, $X$ has no defect).
By construction, for any $H\in X^{\vee}$, the corresponding hyperplane cuts $X$ in a singular section $X_H=X\cap H$.
By Theorem of Reflexivity, for any $(P,H)\in X\times X^{\vee}_{sm}$ such that $H$ is tangent to $X$ at $P$ is equivalent to say the hyperplane corresponding to $P$ in $\overset{\vee}{\mathbb{P}^N}$ is tangent to $X^{\vee}$ at $H$.
Question: assuming that the singular points of $X_H$ are only double ordinary (for simplicity, all singularities are nodes, so $X_H$ is a nodal section), what can I say about $H\in X^{\vee}$?
What kind of property are you asking for? $H$ is a smooth point if and only if $X_H$ has exactly one ordinary double point. What else?
I am not assuming that $X_H$ has exactly only one node. 2) How can I see the equivalence between the smoothness of $H$ and the "nodality" of $X_H$?
One reference is SGA 7, Exp. XVII, Proposition 3.5. If you have $n$ nodes, $X^{\vee}$ has $n$ branches locally around $H$, each branch being smooth.
Perfect! This is very enlightening.
Also see Tevelev's Projectively Dual Varieties.
Tevelev, in that paper\book, remarks principally the case of dual plane curves. Thank you the same.
|
2025-03-21T14:48:30.986814
| 2020-05-16T11:41:56 |
360500
|
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"Bipolar Minds",
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|
Stack Exchange
|
Visualization of higher Bruhat order B(5,2)
I made the following images of the higher Bruhat order B(5,2) (in the sense of Manin/Schechtman) with vZome:
image 1
image 2
image 3
Unfortunately, in vZome its not possible do have regular octagons, so I would like to build it with Stella. As far as I understood, you can't build polyhedra from scratch in Stella, but you have to form them out of basic compounds. Is that correct? If so, then what would be the basic compounds in my case? Sorry for the naive question, I really have no intuition for polyhedra at all.
this seems more like a software question than a math question...
admittedly, part of it, yes.. (the math part is basically: do you understand this polyhedron?) i was hoping to find the experts for this specific type of software in a math forum, but if you would like to close it, it's okay with me
So you just need that graph with regular octagons? Because I have the code to generate the graph in SAGE, but with no control over the octagons. I'm not sure if that helps you. Anyway, let me know if you are interested in the code.
|
2025-03-21T14:48:30.986919
| 2020-05-16T12:05:28 |
360503
|
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"usul",
"verret"
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|
Stack Exchange
|
Graphs which are built from complete graphs : Reference request
Let $V$ be a set of $n$ vertices. Fix $3 \le k \le n$. Let $\binom V k$ be the set of all $k$ element subsets of $V$.
We add the edges in $V$ as follows: Let $\mathcal S \subseteq \binom V k$ be fixed. For each $F \in \mathcal S$, I am making the vertices in $F$ mutually adjacent. Let's call this graph $G_k(\mathcal S)$.
I want to learn how the graph $G_k(\mathcal S)$ looks like?
Is there any name for $G_k(\mathcal S)$ in the literature?
Some references regarding these graphs.
Kindly share your thoughts. Thank you.
This brings to mind the https://en.wikipedia.org/wiki/Clique_complex . In terms of this, every $1$-dimensional face (but not necessarily every $2$-dimensional face) is part of a face of dimension $k-1$. (I admit that does not sound very helpful.)
This seems a little broad, especially Q1. For example, every graph can arise with $k=2$.
@verret I want to start from triangles. Even though $k=2$ is everything I believe even $k=3$ this class is something different and well-studied.
@VilleSalo Sorry. I am not much familiar with the complex. I shall check. Thank you.
Well, for a given $k$, the class of graph one gets is exactly the graphs such that every edge is contained in a $k$-clique. So for $k=3$, it's graphs such that every edge is in a triangle.
Thank you. Is there any classification or a name for graphs satisfying this property?
Especially in computer science / complexity theory, if we were to pick one size-$k$ subset, we would call this "planting a clique" in the graph.
I think that “graphs with every edge in a triangle” is the simplest description of this family, and that no significant or interesting alternative characterisations are widely known. I’ve considered regular graphs (quartic, more precisely) with this property and did not come across anything in general.
@GordonRoyle Thanks for your efforts. I will also check from a similar perspective to regular graphs. Thank you again.
|
2025-03-21T14:48:30.987082
| 2020-05-16T16:10:12 |
360515
|
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|
Stack Exchange
|
A question on preimage of a locally injective meromorphic function
Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be a meromorphic function such that $f'(z) \ne 0$ for all $z \in \mathbb{C}$. Let $\Gamma \subset \mathbb{C}$ be a curve which has no self intersections. If we assume that for any $\omega \in \Gamma$, $f^{-1}(\{\omega\}) \ne \emptyset$, then my question is that, can we find a curve $\gamma \subset \mathbb{C}$ such that $f$ is one-to-one on $\gamma$ and that $f(\gamma)=\Gamma$?
I believe this is true because such function $f$ is locally one-to-one, and even if points on $\Gamma$ can be possibly taken infinitely many times on the complex plane $\mathbb{C}$, we can always choose a certain "branch" such that the restricition of $f$ on such a "branch" is one-to-one. But I'm not sure how to give a rigorous argument to validate the existence of such "branch". Any comments or suggestions will be fully appreciated.
No, this is not true. The simplest example is $f(z)=\int_0^ze^{-\zeta^2}d\zeta$. Preimage of the real line consists of infinitely many curves, each of them is mapped homeomorphically onto one or two intervals of
the three intervals $(-\infty,-\pi/2),\; (-\pi/2,\pi/2),\; (\pi/2,+\infty)$. But none of the curves is mapped on the whole real line.
The picture of these curves is given in Fig. 24, section 253, Chapter XI
of the book Nevanlinna, Analytic functions.
The book does not contain complete proofs; they are given in Nenvanlinna's paper Über Riemannsche Flächen mit endlich vielen Windungspunkten,
Acta Math. 58 (1932), no. 1, 295–373.
Thank you very much! This answer is definitely elegant but certainly beyond my knowledge... Could you elaborate more why $f(z)$ has all of these properties, or just provide a reference for me to check those out?
@student: I added some references.
Thank you very much!
|
2025-03-21T14:48:30.987231
| 2020-05-16T16:13:49 |
360516
|
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|
Stack Exchange
|
Combinatorial proof of invertibility of a symmetric matrix associated to the ring of matrices over a finite field
Let $F$ be a finite field of $q$ elements with characteristic $p$. Let $M_n(F)$ be the ring of $n\times n$ matrices over $F$. We define a $q^{n^2}\times q^{n^2}$ symmetric matrix $L$ over the natural numbers as follows. Let us index rows and columns by matrices in $M_n(F)$. Then $$L_{A,B} = q^{n\cdot \dim \mathrm{Fix}(AB)}.$$ Here $\mathrm{Fix}(X)$ is the set of fixed vectors of the matrix $X$, that is, the eigenspace of $1$. Note that $L_{A,B}$ can alternately be described as the number of matrices $C$ with $ABC=C$, as is easily checked.
Question. Find a combinatorial proof that $L$ is invertible over $\mathbb Q$, or even better over any field of characteristic different than $p$, or even better over $\mathbb Z[1/p]$.
Motivation.
It follows by a result of Putcha and Okninski proving the semisimplicity of $\mathbb QM_n(F)$ that $L$ is invertible over $\mathbb Q$. and I believe it follows by a result of Kovacs that $L$ is invertible over any field of characteristic different than $p$ since Kovacs proved that $M_n(F)$ has a strongly symmetric algebra over such fields. In fact, I believe it is invertible over $\mathbb Z[1/p]$. The matrix $L$ is the matrix of the trace form on the monoid algebra of $M_n(F)$.
The results of Putcha and Okninski and Kovacs go through the representation theory of the multiplicative monoid $M_n(F)$ and are not so trivial. A combinatorial proof of the invertibility of $L$ would give a nice new proof of their results.
Update. my original hope to invert over fields of characteristic different than $p$ was silly because the trace form on general linear groups vanishes if the characteristic divides the order so Kovacs doesn't help. I guess if the characteristic is bigger than $q^n\times q^n$ it should be invertible.
Can the $q$ in the matrix be replaced by a variable?
Is there a $q=1$ version of this result, considering all maps ${1,2,\ldots,n}\to{1,2,\ldots,n}$?
@SamHopkins, the monoid of all self maps does not have a semisimple algebra. But if it's probably not the right q=1 analogue. The Renner monoid for M_n is the monoid of partial permutation matrices or Rook monoid and this has a semisimple algebra in good characteristic
@Darijgrinberg good question.
Let us consider $n=1$. Then the matrix in question is a $q\times q$-matrix with rows and columns indexed by elements of $F$, for which $a_{0,x}=a_{x,0}=1$ for all $x$, and for $x,y\ne0$, $a_{x,y}=q$ for $xy=1$ and $a_{x,y}=1$ otherwise. The determinant of this matrix is easy to compute if you subtract row $0$ from all other rows: the answer is $(q-1)^{q-1}$ times the sign of the permutation of $F^{\times}$ that sends any $x$ to $x^{-1}$. Thus for $n=1$ your expectation about invertibility over a field of characteristic different from $p$, or $\mathbb{Z}[1/p]$ is certainly excessive.
P.S. Of course one may imagine that something magical happens for $n>1$ that $n=1$ does not detect, but I think that it is unlikely in this case.
@VladimirDostenko you are right of course. I would have said the characteristic is larger than the size of the monoid. If the characteristic divides the size of the general linear group its trace form vanishes
|
2025-03-21T14:48:30.987464
| 2020-05-16T17:34:03 |
360520
|
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|
Stack Exchange
|
A time dependent variational problem coming from a second order linear PDE
Fix $u_0\in H^1(\Omega)$ and $f=f(x,y,t)\in L^2(\Omega\times [0,T])$ where $\Omega$ is a sufficiently smooth bounded domain in $\mathbb{R}^2$.
Consider the problem of finding $u:\Omega\times[0,T]\to\mathbb{R}$ satisfying the following variational equation
$$
\begin{cases}
\langle \nabla u, \nabla v\rangle_{L^2(\Omega)}+\langle u,v\rangle_{L^2(\Omega)}=\langle f,v\rangle_{L^2(\Omega)},\;\;\forall v\in H^1(\Omega),\;\forall t\in(0,T], \\\;u(\cdot,0)=u_0, \;
\end{cases}
$$
It is clear that this variational problem comes from the equation
$$\begin{cases}
-\Delta u+u=f, & \text{ on }\Omega\times(0,T],\\
u(0)=u_0.
\end{cases}$$
I did not put boundary conditions, but any boundary condition for me works as long as $\int_{\partial\Omega}(\vec{n}\cdot\nabla u)v=0$ for all $v$.
Note: All of the differential operators seen above are spacial operators, they involve no derivatives in time.
Questions:
What are appropriate function spaces to look for solutions?
Can we have that $u(\cdot,t)\in L^2([0,T])$ for all $t\in[0,T]$?
Can we have that $\text{essup}_{t\in[0,T]}\|u(\cdot,t)\|_{L^2(\Omega)}\leq C$ for some constant $C\in\mathbb{R}$ independent of time and space?
This is just a family of elliptic problems with t as a parameter, so what is an initial condition doing here? Also, you say you did not impose boundary conditions, and then impose a Neumann condition?
As you correctly pointed out, since the operators appearing here are only acting in space your problem amounts to solving an elliptic equation for each time. In particular assigning $u(0)=u_0$ is meaningless: Even if the right-hand side $f$ is continuous in time, and without further compatibility assumptions between $f$ and $u_0$, it is most likely false that the solution of the elliptic problem at time $t=0$ for the datum $f|_{t=0}$ coincides with your "initial datum" $u_0$.
Long story short: since time is not really a variable here, but rather a parameter, you need to figure out a framework to make sense of "regular dependence of the data on the parametere $t$", and then the dependence of the solution will follow. More details below.
In order to give a more specific answer to your question, let $K:=((-\Delta)+ 1)^{-1}$ be the inverse operator mapping the datum $f$ to the solution $u=Kf$ of $-\Delta u+u=f$.
By standard elliptic regularity $K$ is a continuous operator from $L^2$ to $H^2$. Here I'm assuming that your boundary conditions, which you apparently don't prescribe explicitly, are nice enough so that this standard regularity machinery works. If not, the just replace $H^2(\Omega)$ below by $H^1(\Omega)$.
Answers
Bochner space is the key word here. In your first equation you implicitly consider $f(t)=f(.,.,t)$ as an element of $L^2(\Omega)$ for fixed time. (By the way: it seems that you work exclusively in spatial dimension 2, $(x,y)\in\Omega\subset \mathbb R^2$, if so it would be better to edit your question for the sake of clarity.) This is exactly what Bochner spaces do, typically $L^2((0,T);X)$ is the space of squared integrable functions $f:(0,T)\to X$ with values in the Banach space $X$, here you may want to consider of course $X=L^2(\Omega)$.
Just a quick comment, though: whether an $L^2$ function $f(x,y,t)$ in space-time (in the classical sense) can actually be considered as such a Bochner function $t\mapsto f(.,.,t)\in L^2(\Omega)$, and conversely, is actually slightly more delicate than what it may seem at first sight. But this is whole different topics so let me not elaborate too much here, just google "Bochner space" or look up this keyword here on Math.MO you'll find quasillions of related questions and answers.
For your second question the answer is yes. But in order to define $u(t)$ you need to be able to evaluate $f(t)$ at (almost every) time $t$. This is my first answer to your question, so let's assume that indeed your datum $f\in L^2(0,T;L^2(\Omega))$. In this setup it makes sense to evaluate $f(t)\in L^2(\Omega)$ for a.e. $t\in (0,T)$, hence you can define
$$
u(t):= K[f(t)]=(-\Delta +1)^{-1}f(t) \in H^2(\Omega)
\qquad \mbox{for a.e. }t.
$$
So the answer is yes, and moreover you see that $u(t)$ lies in the better space $H^2(\Omega)$.
Actually what this shows is that $u$ can be properly defined as an element of the Bochner space $L^2(0,T;H^2(\Omega))$.
Note here that since $u$ is only defined for a.e. time clearly evaluating $u(0)$ does not make sense (you can always redefine $u$ on a negligible set of times and still get the same $L^2(0,T;H^2(\Omega))$ element)
With these preliminaries out of the way, the answer to your question 3 should now be pretty obvious: No, you cannot hope for such $L^\infty(0,T;L^2)$, unless you improve the time regularity of $f$.
To see this, take any real-valued function of time $\eta=\eta(t)$ such that $\eta\in L^2(0,T)$. Take an arbitrary $F=F(x,y)\in L^2(\Omega)$ and set $f(x,y,t):=\eta(t)F(x,y)$.
Then clearly (the operator $-\Delta +1$ does not depend on time!) the solution of your problem is
$$
u(t)=\eta(t) KF
\qquad\mbox{hence}\qquad
\|u(t)\|_{L^2(\Omega)}=|\eta(t)|\, \|KF\|_{L^2(\Omega)}
$$
and therefore $\sup\limits_{t\in (0,T)} \|u(t)\|_{L^2(\Omega)} = +\infty$ unless $\eta\in L^\infty(0,T)$.
Note here that the same line of thoughts shows actually that the time regularity of $f$ gives the sharp time regularity of the solution, for example $f\in C^k([0,T];L^2)\Rightarrow u\in C^k([0,T];H^2)$ or whatever (you can substitute $C^k$ by whatever regularity scale you wish, roughly speaking).
Thank for the answer, this issue with $u(0)$ is problematic for me because as you said, even if assume compatibility between $f$ and $u_0$, $u$ is defined a.e in time. In classical Sobolev theory, we would encode the boundary conditions within the Sobolev itself (and use Lax-Milgram for ex). Can we do something similar with time conditions ie "encode" the initial condition in the Bochner space itself?
No, you're misunderstanding the problem, I think. You shouldn't view the condition $u(0)=u_0$ as an "initial/boundary" condition being part of the problem. The issue here is not to find a suitable weak formulation encoding the "initial condition", but that the whole problem is overdetermined (unless you impose an additional compatibility condition on $u_0,f$). Think of the smooth case where $f$ is continuous up to the boundary $t=0$: you rproblem is clearly ill-posed, unless $u_0=K f(0)$. So either you need to assume something more on $u_0,f$, or the problem does not make sense.
I think you misunderstood my comment. Suppose that the compatibility condition holds for starters. Im asking: is ir possible to encode such an initial condition in the Bocnher space? I mean: think about the time independent Laplace equation with the Dirichlet boundary condition. In that case, we look for weak solutions in the space $H^1_0$, which encodes the Dirichlet conditon. Can we do the same in a Bochner space ie to have "contain" the initial condition?
I'm pretty sure I didn't misunderstand your comment! Usual weak formulations encode simultaneously a PDE and a boundary conditions because integration by parts gives rise to an extra term for regular solutions, which you then force to cancel. In the absence of time derivatives in the PDE no such terms arise, and therefore you cannot encode initial condition in this way. The only thing you can reasonably do is force the initial condition by requiring time-continuity, e.g. $u\in C([0,T);L^2)$ or whatever. But then again you need an extra compatibility condition.
I suggest migrating to Math.SE.
Okay I see now! If we are given $f$ then $u(0)$ is determined by $f$ completely and is completely useless. Therefore temporal regularity comes mainly from this $f$ we are given. It makes complete sense now: we should look at $f$ in some appropriate Bochner space to add meaningful temporal regularity.Thank you!
Exactly. This is what I tried to explain in my thrid point in my answer aboe: whatever regularity you need/want for $u$ can only come from the same regularity on $f$.
|
2025-03-21T14:48:30.987953
| 2020-05-16T18:33:32 |
360524
|
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|
Stack Exchange
|
How does Conway's proposed compromise for constructing the real numbers in ONAG actually work?
I have also asked this question on Math Stack Exchange (link).
In On Numbers and Games, after discussing the inclusion of the real numbers in the surreal numbers, No, Conway discusses the merits of his construction versus historical ones. While constructing the reals via Dedekind cuts is straightforward, it has the clunky feature that multiplication $xy$ is defined in four cases based on the signs of $x$ and $y$. This means that proving the associativity of multiplication requires examining eight cases. The benefit of the surreal multiplication definition, where all numbers are defined as cuts of (previously constructed) numbers, is that multiplication is cleanly defined "genetically" in terms of its left and right options, without need for case splitting. However, using the surreal numbers to construct the real numbers has its own aesthetic drawbacks, namely being unnecessarily counter-intuitive, and giving special treatment to the dyadic rationals.
Conway then proposes the following compromise:
There is another way out. If we adopt a classical approach as far as the construction of $\mathbb Q$, and the define the reals as Dedekind sections of $\mathbb Q$ with the definitions of addition and multiplication given in this book, then all formal laws have 1-line proofs and there is no case splitting.
My question is, how does Conway's suggestion work, in detail? From what I can surmise, Conway is proposing defining real numbers $x$ as $x=\{x^L|x^R\}$, where $x^L$ and $x^R$ range over rational numbers, so $x^L<x^R$ for all such options, and at most one rational number does not appear as $x^L$ or $x^R$. Then the product of two real numbers would be defined as
$$
xy=\{x^Ly+xy^L-x^Ly^L,x^Ry+xy^R-x^Ry^R\mid x^Ly+xy^R-x^Ly^R,x^Ry+xy^L-x^Ry^L\}
$$
However, how does one define the product $x^Ly$? This is the product of a real number with a rational number, and rational numbers are to be defined "classically" without left and right options, so the above definition does not suffice. One could define, for real $x$ and rational $q$,
$$
xq=qx:=\begin{cases}\{x^Lq\mid x^Rq\} & q>0 \\ \{x^Rq\mid x^Lq\} & q <0 \\ 0 &q=0 \end{cases}
$$
but this would cause the case-splitting issue that Conway was trying to resolve.
I will note that the same issue arise when defining the sum of real numbers, but it can be circumvented. The surreal definition of addition is
$$
x+y=\{x+y^L,x^L+y\mid x+y^R,x^R+y\},
$$
which in Conway's compromise would run into the same problem where $x+y^L$, the sum of a real number and rational number, needs to be defined separately. However, this can be modified to work, if you instead define
$$
x+y=\{x^L+y^L\mid x^R+y^R\},
$$
and indeed this is exactly how addition of Dedekind cuts is defined. I wonder if there is a similar fix for multiplication of cuts.
I already commented that on MSE but the fix for addition does not yield a group law.
Does it not work to first identify the rational number $x$ with the real number ${x^L\mid x^R}$?
@TimothyChow I think this is what one would want to avoid: the reliance on dyadic numbers and perhaps also induction.
@TimothyChow I think the definition of multiplication would then be circular. To define the product of a real and a rational, you would first have to define the product of a real and a rational.
Sorry for not being clear. I was thinking: (1) Define $\mathbb Q$ classically. This way you get all the non-dyadic stuff. (2) Define $\mathbb R$ as you said, letting $x = {x^L \mid x^R}$ where $x^L$ and $x^R$ range over $\mathbb Q$. (3) Prove as a theorem that $\mathbb Q$ is isomorphic to a certain subfield of $\mathbb R$. This is what I meant by "identifying" a rational number with a certain real number. (4) Use this identification to "uniformize" the definition of $\mathbb R$ so that you don't have to split into cases any more.
I know Conway doesn't state the steps (3) and (4) explicitly, but he may have taken them for granted. If this works, it's still nicer than the usual Dedekind cut construction (my reference is "baby Rudin") because you don't have to treat multiplication of negative numbers specially. Looking again at ONAG, I think that Conway is most concerned about case-splitting according to signs, so the extra steps (3) and (4) may be a small price to pay.
I just noticed something. We can't prove that $\mathbb Q$ is a subfield until we have the field operations, but perhaps we can show isomorphism as an ordered set. Not sure if this is enough to push everything through though...
|
2025-03-21T14:48:30.988262
| 2020-05-16T19:11:25 |
360528
|
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|
Stack Exchange
|
Universal model category as a $\text{sSet}$-enriched co-completion
It's a standard fact that given a small category $\mathcal{C},$ the category of pre-sheaves $\text{Psh}(\mathcal{C})$ is the free co-completion of it.
I'm sure this can be done not only for $\text{Set}$-enriched categories but for general $\mathcal{V}$-enriched categories, with the appropriate notions of $\mathcal{V}$-enriched colimit, and functor preserving the enrichment, and I just found it in section 4.4. of Kelly's Basic Concepts of Enriched Category Theory.
So now the question is: can one prove proposition 2.3 in this paper about simplicial pre-sheaves on $\mathcal{C}$ being the universal model category on $\mathcal{C}$ by just doing the $\mathcal{V}$-enriched co-completion with $\mathcal{V}=\text{sSet}$?
Naive comment: did you try to consider the category of presheaves valued in $\mathcal{V}$ ?
@PhilippeGaucher Yes, of course it will be that. But there are details to be added to show the universal property in the general case, and I did not want to rediscover the wheel.
Given $C$ a small category (eventually, a small simplicial category) I denote by $UC$ the projective model structure on the category of simplicial presheaves on $C$ as in the paper. Using the kind of argument you have in mind we obtain the following theorem:
Theorem: If $M$ is a simplicial model category, then there is an equivalence of categories between:
(Simplicial) Functors $C \to M$ taking values in the full subcategory of cofibrant objects.
Simplicial left Quillen functor $UC \to M$.
In one direction, the equivalence is simply given by restricting to the Yoneda embedding $ C \to UC$ as representable are cofibrant in the projective model structure, this forces the composite functor $C \to UC \to M$ to take values in cofibrant objects. In the converse direction, one takes the unique simplicial left adjoint functor $UC \to M$ and check, using the axiom of simplicial model category for $M$ that this is a left Quillen functor.
However, this is not what the paper you mention proves.
There, they start from a model category $M$ that is not assumed to be a simplicial model category, and a functor $C \to M$ not assumed to takes values in cofibrant objects. And construct a left Quillen functor $UC \to M$ by considering (and choosing) a cofibrant simplicial resolution of the functor $C \to M$ they started from. In particular, the "uniqueness" of the left Quillen functor obtained this way, is only up to homotopy (to be more precise, up to a contractible space of choices).
One abstract way to understand the relation between the two is as follows:
Given $M$ a combinatorial left proper model category, there is a Quillen equivalent simplicial model structure on the category $sM$ on the category of simplicial objects of $M$, (this is explained in the paper "Replacing model categories with simplicial one" by Dugger)
The evaluation at $[0]$ gives a left Quillen equivalence $sM \to M$
One way to understand the non-simplicial theorem is that if you start from $C \to M$, you can see it as a functor $C \to sM$ taking values in constant simplicial objects, then take a levelwise cofibrant replacement to obtain a functor $C \to sM$ taking value in cofibrant object, apply the "simplicial theorem" to get a Quillen functor $UC \to sM$ and finally, post compose with Quillen functor $sM \to M$ that evaluate at $[0]$.
Now for the model structure on $sM$ to exist we need $M$ to be combinatorial and left proper, if you are willing to work with a left semi-model structure instead it is enough to assume that $M$ is an accessible model category (no properness assumption).
But in some sense the central observation of the paper you quote, is that, even if the model structure on $sM$ cannot be constructed, the overall construction make sense with no assumption $M$ (other than being a model category, I guess they also need functorial factorization, I do not remember).
Thanks for the explanation. I am wondering if from the theorem about simplicial model categories one can then derive also the result in the paper about general model categories?
@IvanDiLiberti this the paper mentioned in the question we are talking about ?
@IvanDiLiberti I linked to that paper in my question.
@giuseppe : I've added a comment on the relation between the two at the end. Does it helps ? At some point, even if it would be not true that you cannot use the result you mention to prove the proposition. This is really the best one can do: To apply the theorem you have in mind you need a simplicial model category, and I do not know any other way to make one appear, than this construction.
@giuseppe I think the fact you can’t recover the result straight from the idea of the free cocompletion is just an indication of the depth of the result. The purely simplicial theorem would have been considerably less significant a result.
|
2025-03-21T14:48:30.988718
| 2020-05-16T19:12:18 |
360529
|
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|
Stack Exchange
|
Condition on a function to have a Fourier transform in $L^{2-\varepsilon}$
It is known that in general the Fourier transform of $L^p(\mathbb{R})$ functions for $p>2$ are not even function. However, for regular enough functions, the regularitytransfers into decay for $\hat f$ and with sufficient decay one obtains $\hat f \in L^{2-\varepsilon}$ for some sufficiently small $\varepsilon >0$.
Is there a condition, which does not require regularity (neither $H^s$ or $\mathcal{C}^{\alpha}$ regularity) for functions $f \in L^1 \cap L^{\infty}$ to ensure $\hat f \in L^{2-\varepsilon}$ for some small $\varepsilon >0$?
For example, take $f=1_{[0,1]} \in L^1\cap L^{\infty}$, then $\hat f (\xi)=\frac{\sin \xi}{\xi} \in L^{1+\varepsilon}$. Is there a general class of function with such properties?
Sidenote: Since $\widehat{\widehat{f}} = \check{f} := x\mapsto f(-x)$, it is equivalent to ask for a characterisation of the image of $L^{2-\epsilon}$ under FT. Since the FT maps $L^1$ into $C_0\subseteq L^\infty$ and $L^2$ into $L^2$, it maps $L^{2-\epsilon}$ into $L^q$ with $q=\frac{2-\epsilon}{1-\epsilon}$ by the Riesz-Thorin theorem.
Therefore $f\in L^q$ is at least a necessary condition for $\widehat{f}$ being in $L^{2-\epsilon}$.
@JohannesHahn this is a good observation, I will edit my question and try to phrase more carefully what I am asking. Typically, we can assume that $f \in L^1 \cap L^{\infty}$. Then we have $\hat f \in L^2 \cap L^{\infty}$ at least, which does not help in proving further integrability properties such as $f \in L^{2-\varepsilon}$.
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