I was worried about computations like $H_n(X,A;\Bbb Q)$ coming up in the exam and now that I learned that this is just $H_n(X,A) \otimes \Bbb Q$ I want to milk this fact
But algebraic numbers, real numbers, rational numbers, integers are probably the only realistic examples
the only reason I even listed R as having relevancy is cause the singular cohomology with coefficients in R of a smooth manifold is canonically isomorphic to the de Rham cohomology
Now that I think of it, I think I saw the first fact already. It's basically just saying that the non torsion parts and the torsion parts are 0, right? Do I need a finiteness condition for that
For a finitely generated group, the result actually holds in each degree individually. If $G$ is finitely generated abelian and $G\otimes\mathbb{Q}=0$ and $G\otimes\mathbb{Z}_p=0$ for all primes $p$, then $G=0$. For we can write $G=\mathbb{Z}^r\oplus\bigoplus\mathbb{Z}_{p_i^{k_i}}$. Tensoring with $\mathbb{Q}$ kills torsion and gives $G\otimes\mathbb{Q}=\mathbb{Q}^r=0$, whence $r=0$, so $G=\bigoplus\mathbb{Z}_{p_i^{k_i}}$. But now tensoring with $\mathbb{Z}_p$ kills $\mathbb{Z}_{q^k}$ for any prime $q\neq p$, so $G\otimes\mathbb{Z}_p=\bigoplus\mathbb{Z}_{p^{k_i}}=0$ (my indices suck, but w/…
but I think for a not f.g. group, the result only remains true if you have the hypothesis in all degrees simultaneously and it's a homological argument with the Tor functor instead
wait no, I'm being stupid in some way, shape or form, but I'm not sure which
a module always injects into the product of its localizations
but the stronger form is not how it's stated in Hatcher, so something is fishy
ah, I figured the issue
Hatcher wants to state an iff. Vanishing of Q- and Z_p-homology does indeed imply vanishing of Z-homology in each degree individually. The converse obviously fails (see projective space), so if you want an iff, you have to make a statement in all degrees simultaneously.
They start with "Elements of Geometry and Topology" which is basically topology I, then do Topologie I + II where they handle homology & cohomology in detail, and only then they continue with Algebraische Topologie I + II
2+2 is 4, yet I'm currently unable to prove that H^2\otimes H^2->H^4 is surjective
We just did singular homology, then cellular homology, then cohomology and now characteristic classes. The lecture is pretty sparse in some aspects and we have skipped some things like e.g. Poincaré duality. But I've additionally supplied myself by reading a good amount of what's in ch.2/3 of Hatcher along the way.
(but it turns out I will have to know a lot about higher homotopy groups of spheres to understand the computation of the rational cobordism ring, so that's a temporary bummer)
With density matrix $\rho=\sum_{a,b=0}^1\rho_{a,b}|a\rangle\langle b|$ and it's transpose $\rho^T=\sum_{a,b=0}^1\rho_{a,b}|b\rangle\langle a|$. How to confirm that $\rho^T$ is positive and trace preserving.
The only thing I could observe was the given matrix is *skew-symmetric*, and the determinant is $(a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2})^2.$However I could not proceed any further than that. The rank is ofcourse = the number of non zero rows in the row echelon form of the matrix, but taking that route seems to be too tedious , and I believe the purpose of giving a skew symmetric matrix was …
@satan29 The rank of a matrix $M$ is the largest integer $n$ such that an $n\times n$ submatrix of $M$ is invertible (resp has maximal rank, resp has nonzero determinant)
so here "determinants of the third order" refer to the determinants of the $3\times 3$ submatrices
Is the integral $$\int_{0}^{2\pi}\int_{0}^{2\pi}\left[\cos(nx + y)\right]^p\text{ d}x\text{ d}y$$ a well-known value, where $p$ is an arbitrary integer, and $n \geq 1$ is an integer?
It appears that if $p$ is odd, the integral is $0$. I've not figured out what the even pattern is yet.
Sorry, more specifically... so you get $(e^{i(nx+y)} + e^{-i(nx+y)})^p$ and then you rewrite that as $(2\cos(i(nx + y)) + \sin(i(nx+y)) + \sin(-i(nx+y)))^p$, so at the end, using $\sin(x) = -\sin(x)$, this appears to be $2^p\cos^p(i(nx+y))$ if I did it right
Okay, so let's use the binomial theorem... you would get $$(e^{i(nx+y)} + e^{-i(nx+y)})^p = \sum_{k=0}^{p}\binom{p}{k}e^{ik(nx+y)}e^{-i(p-k)(nx+y)}$$ and then we want to integrate this with respect to $x$ and then with respect to $y$...
@Astyx Right, so it looks like in the summation, I get $\binom{p}{k}e^{in(2k-p)x}e^{in(2k-p)y}$... so I assume we then convert these to the complex $\cos + i \sin$ forms
Well, let's see. Let $t = n - k$, then since $k$ goes from $0$ to $p$, we have that $t$ goes from $n$ to $n-p$... so we would end up with $\sum_{t=n-p}^{n}\binom{n-p}{n}$ with the exponential in the summation... I have a feeling this is not what you intended
Oh, k. So let's let $t = p - k$. Then we have that $k$ goes from $0$ to $p$, so that $t$ goes from $p$ to $0$, and we have $k = p - t$ so that $2k = 2p - 2t$ and $2k - p = p - 2t$, hence we have$$\sum_{t=0}^{p}\binom{p}{p-t}e^{i(p-2t)(nx+y)}$$ assuming I did the binomial coefficient right
@Astyx So yes, I'd agree they're the same sum, but how does this help me translate to trig functions? I suppose... Oh, I see what's going on here. Pull out a negative from $(p-2t)$...
So then you basically get... what is this... $$2(e^{i(nx+y)} + e^{-i(nx+y)})^p = \sum_{k=0}^{p}\binom{p}{k}\left[e^{i(2k-p)(nx+y)} + e^{-i(2k-p)(nx+y)} \right]$$
So for the double integral of this over $[0, 2\pi] \times [0, 2\pi]$, I assume we're probably going to want to use the addition identity on the right-hand side
@Astyx So let's say I wanted to integrate something like $(e^{i(nx_1 + y_1)})^p(e^{i(mx_2 + y_2)})^q$ over the same domain in 4-space (i.e., $[0, 2\pi]$ each variable). Would this approach make sense?
Let's say we got $C_f := k[x,y]/(f)$ where $f$ an irreducible poly in $k[x,y]$, and $k$ alg. closed. Say we want factor the ideal $(x-a)$ explicitly as a product of prime ideals (I think this was building up to the fact that we can only do that when our noetherian domain of dimension 1 is a Dedekind domain).
We write $f(a,y) := c \Pi_{i=1}^s (y-b_i)^{e_i}$ (with $b_i \neq b_j$ whenever $i \neq j$) so that $f(x,y) = (x-a)g(x,y) + c \Pi_{i=1}^s (y-b_i)^{e_i}$ with $g(x,y) \in k[x,y]$ and $M_i := (x-a , y-b_i)$ would be the maximal ideals of $C_f$ that contain $(x-a)$.
Naively, I would think that this would result in $$\dfrac{1}{inp}(e^{2\pi i} - 1) \cdot \dfrac{1}{ip}(e^{2\pi i} - 1) \cdot \dfrac{1}{imq}(e^{2\pi i} - 1) \cdot \dfrac{1}{iq}(e^{2 \pi i} - 1)$$ but of course, you're talking to someone who's never taken a course on complex variables
@TedShifrin Hi, here's a random observation. Suppose $X$ is a stratified space with two strata, and you chip off a little piece out of the bottom stratum
Say the normal link at any point of the bottom stratum is $A$, then chipping out a little neighborhood of a closed disk on the bottom stratum is tantamount to chipping off $(A \times D^n) \cup_{A \times \partial D^n} (C(A) \times \partial D^n)$
This one isn't a Fourier problem, thankfully. But if this is true... this must mean that complex random variables to some powers, when multiplied, have moments equal to $0$ if we assume $x_1, y_1, x_2, y_2$ are uniformly distributed in $[0, 2\pi]$...
This post comes out from another question in the comment box of this answer.
Let $G$ be a free group with countable-infinitely many generators acting on $\Bbb R^2$ properly, and $H\subsetneqq G$ be an infinite cyclic subgroup of $G$. Let $p:\Bbb R^2\to \Bbb R^2/H$ be the orbit map, i.e., a coveri...
So I guess if you chip off a little closed disk $D = D(0; R) \subset \Bbb C$ (and all the normal junk around it) from the singular stratum, you cut out $D \times C(S^1 \sqcup S^1)$ basically, which leaves off a boundary that is two copies $S^1 \times D^2 \times \{i\}$ with a double cone filling in every pair $S^1 \times \{x\} \times \{0, 1\}$ for all $x \in \partial D^2$.
Yeah, the topological type (not upto homotopy) is (piece of stratum) x cone(link)
Aha, I see what's going on. It's like, initially you had the constant family of a pair of hyperplanes in $\Bbb C^2$ which kept intersecting at $0$ transversely. Now you do some kind of stratified surgery at the origin such that in a little neighborhood it looks like the family which is a pair of transverse hyperplanes in $\Bbb C^2$, then became the family $xy = 1$ (by the homotopy $xy = t$, $0 \leq t \leq 1$), and then back to transversely intersecting hyperplanes again
(Remembered this picture from elsewhere, in a different context)
Your example and the Milnor fiber comment is very interesting because that means my chipping operation is hopefully algebraically meaningful. I am trying to do something which only happens upto chipping, and I was wondering how much you lose.
So the differential geometry question is to compare the integral of curvature on a neighborhood in the smoothing with the integral of curvature on a neighborhood of the singular thing. There is a defect, which you can interpret in terms of Gauss maps.
Yeah, I'm not enough of a topologist to intuit your chipping :)
With density matrix $\rho=\sum_{a,b=0}^1\rho_{a,b}|a\rangle\langle b|$ and it's transpose $\rho^T=\sum_{a,b=0}^1\rho_{a,b}|b\rangle\langle a|$. How to confirm that $\rho^T$ is positive and trace preserving.
