One way I see is the following.
To start with, $\pi_{4}(S^{3}) \simeq \pi_{4}(S^{2})$ because of the existence of the Hopf fibre sequence $S^{1} \rightarrow S^{3} \rightarrow S^{2}$ and moreover the isomorphism is given by composition with the Hopf map. Hence, as was observed in the comments, yo...
Hi everyone, assume that a statement $\phi$ written in first-order logic implies AC in ZF. Can we always find a model of ZF+not $\phi$? Does this question make sense?
I just want to read the basics with a knowledge of only up to the isomorphism theorems for groups, without getting confused with mentions of compact, continuous etc.
So the basics of general linear groups, unitary groups, orthogonal groups, Lorentz group etc.
Hi, there is a question which I can not to solve it , please can someone help me ? a) prove that h : #R^2 → S^1 × S^1# that defined as : #h(t, s) = (f(t), f(s))# is a Local depomorphism. b) prove that if #L := {(x, y) ∈ R^2| y = ax + b}# is a Straight on the plane איקמ #h|L# is immersion.
@LeakyNun Yeah alright, this is simple. B is the homotopy limit of C -> D <- pt. A is the homotopy limit of B -> C <- pt. Superimpose these diagrams to get that A is the homotopy limit of pt -> D <- pt.
Hi, there is a question which I can not to solve it , please can someone help me ? a) prove that h : #R^2 → S^1 × S^1# that defined as : #h(t, s) = (f(t), f(s))# is a Local depomorphism. b) prove that if #L := {(x, y) ∈ R^2| y = ax + b}# is a Straight on the plane, then #h|L# is immersion.
Could anybody take a look at my question on conditionally sampling from two distributions? https://math.stackexchange.com/q/3304535/47771 This should be a quite natural problem in applications, but I don't know what I need to google to read about it.
@LeakyNun I didn't specifically mean UCT is a special case of AHSS, rather I meant that conceptually it seems to do something more powerful than what UCT does. UCT relates (co)homology with arbitrary $G$ coefficients to (co)homology with $\mathbb{Z}$ coefficients, whereas AHSS relates a (co)homology theory with ordinary homology (by relating the (co)homology groups from one theory with the ordinary (co)homology groups),
and for example UCT does this as well since $H(_; G)$ is an ordinary homology theory for any abelian group $G$. Then again I may be wrong since I'm just learning about these things
@Perturbative They are not really related. Let's plug in $H^*(-;G)$ as the extraordinary cohomology theory in the AHSS. Call it $h(-)$ for convenience. Then AHSS promises us a spectral sequence starting from $H^*(X;h(pt))$ to $h(X)$. We start from singular cohomology with coefficient groups equal to $h(pt)$. But $h(pt) = G$ in degree zero and is otherwise zero. So you are starting from $H^*(X;G)$ and going to $H^*(X;G)$.
You don't pass through $\Bbb Z$-cohomology on the way. The AHSS provides no new information for ordinary cohomology theories.
The AHSS is in some sense a vast generalization of cellular cohomology. What I told you there is the $E^2$ page of a spectral sequence. The $E^1$ page (of the most common derivation of this SS for a CW complex $X$) is $C^*_{cell}(X;h(pt))$. The differential is the cellular differential. When $h$ is an ordinary cohomology theory the story ends after this page.
But in general if $X^k$ is the $k$-skeleton of a CW complex and $f: S^k \to X^k$ is the attaching map for a $(k+1)$-cell, the induced map $h(S^k) \to h(X^k)$ depends on a lot more than just the top cells of $X^k$; it depends also on how they were attached to lower cells. This is in stark contrast to ordinary cohomology (when you just need to know the degree to each top cell, and from there cook up the cellular chain complex)
And that's where the higher differentials come from
Suppose I've got a polytope M in an m-dimensional space, and I project it down to some polytope N in n-dimensional space. If I pick one of the vertices of N, is there anything I can say in general about the dimension of its preimage in M? (i'm assuming that dim(M)=m and dim(N)=n as well)
What does it mean exactly to integrate a function $f\colon G\to\Bbb C$ against the (left) Haar measure on $G$? ($G$ is a locally compact topological group)
(I'm confused by the codomain being $\Bbb C$ instead of $\Bbb R$)
> For brevity, we have chosen not to state or prove such additional results in this article since, thanks to the a priori estimates yielded by Prop. 9.2, their statements and proofs are similar to the ones given in [50, Theorem 2].
Sure, and that seems strange to me as I said. I know the journal has a good reputation (I had seen the name before even though it is way outside my own field), but for any up-and-coming journal any sentence like that would have been a huge red flag
I'm checking that a thing those notes defined really is a *-algebra, but I'm doing some stupid mistake in checking that $(ab)^\ast=b^\ast a^\ast$ because it's not working
I have a locally compact group $G$ with it's left Haar measure $\mu$ and its modular function $\Delta\colon G\to\Bbb R^+$. I look at $C_c(G)$, the space of complex valued functions on $G$ with compact support, and I want to turn this vector space of functions into a *-algebra
If you fix $g\in G$ then $\mu_g$ defined by $\mu_g(E)=\mu(Eg)$ is also a left invariant measure, so it must differ from $\mu$ by a constant, $\Delta(g)$ is the constant such that $\mu_g=\Delta(g)\mu$
$\Delta$ is actually a (continuous) group homomorphism $G\to\Bbb R^+$
The multiplication is convolution, $(ab)(g)=\int_G a(h)b(h^{-1}g)\mathrm{d}\mu(h)$, while the involution is defined by $a^\ast(g)=\Delta(g)^{-1}\overline{a(g^{-1})}$
Suppose I have n-by-m complex matrices $A,B$. Then the Hilbert-Schmidt inner product is just $\langle A,B\rangle = \mathrm{Tr}(A^* B)$ where $*$ is hermitian conjugate
why know that there are as many conjugacy classes as there are representations, but is there are nice correspondence? The general answer I think is that we don't know
There is another way to go about the one for $S_n$ that also works out nicely (and gives the same answer which is a bit surprising). But that one does not work out as nicely for the other Coxeter groups
the size of the conjugacy classes sum to the order of the group. the sum of the squares of the dimension of irreps (or the matrix factors) sum to the order of the group. So it seems unlikely that there is a relation between two kinds of quantities which are related like that
The KL basis allows for a way that directly associates 2-sided ideals to partitions in such a way that the ideal gives the same irrep as one would associate to the partition via Specht modules
> Not wishing the Germans to know that he was an expert in hydrodynamics, since he feared that if they found out he would be forced to undertake war work for them, Leray claimed to be a topologist. He worked only on topological problems for the years he was held captive in the camp.
Although he had undertaken some topological work it was not easy for Leray to work on the topic without reading topological literature. He was able to obtain some papers through Hopf who was at this time in Zürich but much of Leray's work was done independently of the developments which had taken place in the s…
