I need to show that an ergodic automorphism $T$ of a compact abelian group is mixing of all orders (via looking at what at what it does to the dual basis)
I can do it for the first order (i.e just mixing). In general I understand I need to show why we can't have $\chi_0(x) \chi_1(T^n_1 x) \chi_2(T^{n_1 +n_2})...\chi_k(T^{n_1 +n_2..+n_k}) = 1$ as $n_i \to \inf$
Let $(P,\le)$ be a poset and $S,T\subseteq P$ such that the following holds, $$\forall t\forall s(t\in T\land s\in S\to s\le t)$$does this type of relation between $S$ and $T$ has a standard name?
The below code checks all $\Bbb{Z}$-module homomorphisms (determined by assigning $a = f(1)$) and finds the ones (for each modulus $m$) that will work to factorize any integer $0 \lt x \lt m$.
The reason this works is because for powers and prime numbers you don't need an efficient integer fac...
I'm not joking, you can factor an integer in as little as a few lines of code as long as $m$ and $a$ are chosen for your given environment's integer size. Otherwise precomputation is necessary. Still though, it's an existence proof
Suppose I have a measurable function $f$ in two real variables $x,y$. I want to check the convergence of the integral of $f$ over R^2. Under which conditions can I do this by checking the convergence of the individual integrals? I.e., if $\int f(x,y) dx$ converges (for fixed $y$), and $\int f(x,y) dy$ converges (for fixed $x$), does the convergence of the double integral of $f$ over R^2 follow? Does this require Fubini?
I mean, Fubini-Tonelli already gives you equivalence, so yes. I doubt that there is a general condition which allows you to deduce convergence of the iterated integral just from the inner integral (also note that the inner integral doesn't need to converge everywhere). Of course, there are other methods to see whether an integral converges: either finding an integrable dominating function or seeing whether you can break your function down into something continuous on compact subsets should take care of most non-pathological functions.
@MikeMiller My earlier map was wrong. I'll write down the correct map. I'll work finite-stage, so let $G_{n, m}$ be the Grassmannian of $n$-planes in $\Bbb R^m$ and $\gamma_{n, m}$ be the tautological $n$-plane bundle over this, $\gamma_{n, m}^\perp$ orthocomplement in $G_{n, m} \times \Bbb R^m$.
Let $\Bbb R^m \to \gamma_{n, m}^\perp$ be a proper map; make it transverse to the zero section and take preimage to get a submanifold of $\Bbb R^m$ of codimension $\text{codim}_{\gamma_{n, m}^\perp}(0) = m - n$, so of dimension $n$. The map was the same as a map $S^m \to \text{Th}(\gamma_{n, m}^\perp)$ preserving $\infty$ by one point compactifying. This gives a map $\Omega^m \text{Th}(\gamma_{n, m}^\perp) \to \coprod_{\dim M = n} \text{Emb}(M, \Bbb R^m)/\text{Diff}(M)$.
I would like to say you can take the limit as $m \to \infty$ to get a map $\Omega^\infty \text{Th}(\gamma_n^\perp) \to \coprod_{\dim M = n} B\text{Diff}(M)$.
I will also mention that Balarka Sen and Mike Miller are often also in the main chatroom. But perhaps now it's reasonable to wait first whether you get access to that room.
This is Exercise II.6 of Mac Lane and Moerdijk's, "Sheaves in Geometry and Logic [. . .]". According to the first few pages of this Approach0 search, it is new to MSE.
The Details:
The functors $\Lambda$ and $\Gamma$ are discussed in the following:
Just what is Mac Lane & Moerdijk's $\Lambda$...
@Thorgott My question was under the assumption that both single integrals converge everywhere. But I understand that estimating one inner integral, and then calculating the outer integral and applying Fubini-Tonelli, might be more promising in general.
@MikeMiller There are natural inclusions $G_{n, m} \subseteq G_{n, m+1}$. Clearly, $\gamma_{n, m+1}|G_{n, m} = \gamma_{n, m}$ so $\gamma_{n, m+1}^\perp|G_{n, m+1} = \gamma_{n, m}^\perp \oplus \varepsilon^1$, so $\text{Th}(\gamma_{n, m+1}^\perp|G_{n, m}) = \text{Th}(\gamma_{n, m}^\perp) \wedge S^1 = \Sigma \text{Th}(\gamma_{n, m}^\perp)$, so we have an inclusion $\Sigma \text{Th}(\gamma_{n, m}^\perp) \to \text{Th}(\gamma_{n, m+1}^\perp)$.
There's a natural map $X \to \Omega \Sigma X$ adjoint to the identity map $\Sigma X \to \Sigma X$, using which we get a map $\Omega^m \text{Th}(\gamma_{n, m}^\perp) \to \Omega^{m+1} \text{Th}(\gamma_{n, m+1}^\perp)$.
Take colimit over these to get $\Omega^\infty \text{Th}(\gamma_n^\perp)$
Hello. This is perhaps a strange question, but I'm looking for two mathematical quantities that can take the place of A and B in the statement 'A is smaller than B'. It would be good if A and B were non-trivial quantities (e.g. not 1 and 2), and even better if they were quantities such that it is an open question in mathematics whether A really is smaller than B. If anyone can give a good pair of quantities, I'd be much obliged.
Neural networks are commonly used for classification tasks, in fact from this post it seems like that's where they shine brightest.
When we want to classify using NNs, we just take the last layer to take values in $[0,1]$; typically, by taking the last layer to be the sigmoid function $x \maps...
Given a map $f : X \to \Omega^n S^n$ from a compact manifold $X$, this gives rise to a proper map $F : X \times \Bbb R^n \to \Bbb R^n$ which we can make transverse to $0$ and pullback to get a submanifold $F^{-1}(0) = M$ of $X \times \Bbb R^n$ with normal bundle $\nu = (dF)^*(T_0 \Bbb R^n)$ which is rank $n$ trivial, so we have an isomorphism $TM \oplus \varepsilon^n \cong TX \oplus \varepsilon^n$.
This is a stable isomorphism $TM \to TX$ which covers the projection $M \to X$ I think
Should be a bit more precise. It's an isomorphism $TM \oplus \nu \cong T(X \times \Bbb R^n)|_M \cong \pi^* TX \oplus \varepsilon^n$, where $\pi : M \to X$ is restriction of the projection $X \times \Bbb R^n \to X$.
So that's the stable isomorphism $TM \to \pi^* TX$ (aka isomorphism $TM \oplus \varepsilon^n \to \pi^* TX \oplus \varepsilon^n$) of bundles over $M$ so a stable isomorphism $TM \to TX$ covering $\pi : M \to X$.
Let's try conversely. $M, X$ be equidimensional compact manifolds. Suppose $(F, f) : (TM, M) \to (TX, X)$ is a map of pairs such that $F : TM \to TX$ is a stable isomorphism covering $f : M \to X$.
$f$ factors as an embedding and a projection $M \to X \times \Bbb R^n \to X$ for some large $n$ so we get an isomorphism $TM \oplus \nu \cong f^* TX \oplus \varepsilon^n$, so an isomorphism $TM \oplus \nu \to TX \oplus \varepsilon^n$ covering $f$. Identify $TM \oplus \nu$ with $TX \oplus \varepsilon^n$ using the stable isomorphism $F$ so that we have an isomorphism $TX \oplus \varepsilon^n \to TX \oplus \varepsilon^n$ of bundles over $X$.
Hm, getting a little confused
We have two isomorphisms (1) $TM \oplus \nu \cong (df)^* TX \oplus \varepsilon^n$ and (2) $F : TM \oplus \varepsilon^n \to TX \oplus \varepsilon^n$.
OK, so two isomorphisms $TM \oplus \nu \to TX \oplus \varepsilon^n$ and $TM \oplus \varepsilon^n \to TX \oplus \varepsilon^n$ covering $f : M \to X$. Certainly this means $\nu \cong \varepsilon^n$ as bundles over $M$
Ok, so the map is probably given by collapsing everything outside a little neighborhood of $M$ in $X \times \Bbb R^n$, which gives $\Sigma^n X \to \text{Th}(\nu) \to S^n$ where the last map is induced from the proper diffeomorphism $\nu \to M \times \Bbb R^n$ composed with $M \times \Bbb R^n \to \Bbb R^n$
So that's $X \to \Omega^n S^n$
That should mean $\Omega^\infty S^\infty$ classifies all stable isomorphisms $TM \to TX$ (aka isomorphism $TM \oplus \varepsilon^n_M \to TX \oplus \varepsilon^n_X$) covering $M \to X$ upto concordance, i.e., stable isomorphisms $(F_1, f_1), (F_2, f_2) : (TM_0, M_0), (TM_1, M_1) \to (TX, X)$ are concordant if there is a stable isomorphism $(F, f) : (TW, W) \to (TX, X)$ such that $W$ is a manifold with boundary $\partial W = M_0 \sqcup M_1$ and $(F, f)$ restricts to $(F_i, f_i)$ on $M_i$)
@solisoc this isn't quite the kind of example you wanted, but you can frame stuff like "Are there any odd perfect numbers" in a similar way. Let A be the set of even perfect numbers and B be the set of perfect numbers. One certainly has that A is a subset of B (if it's an even perfect number, then it's perfect). What's open is whether there's odd perfect numbers. If there aren't, then actually A=B. So that's analogous to "we know A isn't bigger than B, but we don't know if B is bigger than A."
Another route would be via the Riemann hypothesis. There's a bunch of results of the following form: "Let X be the largest difference between the real part of a Riemann zeta zero and 1/2. Then X does not exceed Y."
for some choice of Y
The Riemann hypothesis is that this is true no matter what (positive) Y you pick
The US went from "don't worry, everything is fine, we don't need any particular measure" to being the third country in the world by number of cases as soon as they started testing for them (also looks well on its way to overtake Italy and China in the next few days)
it is all about money if you think about it, they are aware of the dangerous situation but choose not to act and lie to people just so they keep them working . the ones who can afford it has been taking safe measurements from the start .
Trump is a business man so nothing he said is for the best of the people , just his pocket and the pockets of his surroundings
@TedShifrin Too late in many places at this point, testing en masse would lead to greater spread (at least in CA, NY, IL). Probably already the case everywhere, but obviously we will never know.
The below code checks all $\Bbb{Z}$-module homomorphisms (determined by assigning $a = f(1)$) and finds the ones (for each modulus $m$) that will work to factorize any integer $0 \lt x \lt m$.
The reason this works is because for powers and prime numbers you don't need an efficient integer fac...
This is Exercise II.6 of Mac Lane and Moerdijk's, "Sheaves in Geometry and Logic [. . .]". According to the first few pages of this Approach0 search, it is new to MSE.
The Details:
The functors $\Lambda$ and $\Gamma$ are discussed in the following:
Just what is Mac Lane & Moerdijk's $\Lambda$...
Neural networks are commonly used for classification tasks, in fact from this post it seems like that's where they shine brightest.
When we want to classify using NNs, we just take the last layer to take values in $[0,1]$; typically, by taking the last layer to be the sigmoid function $x \maps...
