The real answer is: if the induced $K$-module structure on $H$ is trivial and the factor system associated to the sequence $1 \to H \to G \to K \to 1$ is trivial in $H^2(K,H)$ :^)
If a smooth function $f$ between 2 manifolds is a local diffeomorphism when restricted to some open set, is also $df$ locally an isomorphism between tangent spaces?
if i have the state $\frac{\sqrt{2}}{2}|H\rangle_1|H\rangle_2\pm e^{i\phi}|V\rangle_1|V\rangle_2$ how can i represent it as a vector with $|H\rangle, |V\rangle$ as my basis?
One of the hints on my problem set says "Hint: work in local coordinates so that the function is of the form $f: \mathbb{R}^k \to \mathbb{R}$ and one can then think of $df: \mathbb{R}^k \to \mathbb{R}^k$" is this a clear typo and it should be $df: \mathbb{R}^k \to \mathbb{R}$? Or are we somehow padding things with 0s?
@heather I guess the proper reply is that your question didn't make a lot of sense: The way you wrote your state it is already expressed in terms of the $H,V$ basis.
if the wavefunction of your system is of the factorizable form $|\psi\rangle_1 \otimes |\phi\rangle_2$, then that means that the first system is in state $\psi$ and your second is in state $\phi$.
@Mathei: To TeX this is too much work right now, but I mean that $1\to H\to G\to K\to 1$ splits with $\beta\colon K\to G$ is equivalent to its splitting with $\alpha\colon G\to H$.
ah. so if you write $|H\rangle = \binom{1}{0}$ and $|V\rangle = \binom{0}{1}$, you want to know what $|H\rangle_1\otimes |H\rangle_2$ looks like (and similarly for the other)
@heather Uh...for that you'd just take the coefficients in front of the kets, no? That is, $(\frac{\sqrt{2}}{2},0,0,\exp(\mathrm{i}\phi))$ if you order your basis as $|H\rangle_1 |H\rangle_2,|H\rangle_1 |V\rangle_2,|V\rangle_1 |H\rangle_2,|V\rangle_1 |V\rangle_2$.
@Ted You're thinking about the module case, that's not true if we're working with non-abelian groups. Consider $1 \to A_3 \to S_3 \to \Bbb{Z}/(2) \to 1$
one thing you can eyeball from this is that there's four such products that you can form, and these span the vector space. so you can describe any pair of photons as a superposition of factorizable states
@TedShifrin I can assure you it's really just the standard tensor product of vector spaces, but physicists are terrible at explaining it and some even tend to use the signs $\otimes,\oplus,\times$ seemingly at random at times to denote either of the operations ;)
@ACuriousMind so let's say that before that i perform an operation that produces that state, i have a single photon, 45 degrees from both horizontal and vertical, that is, i think $\frac{1}{\sqrt{2}}|H\rangle+\frac{1}{\sqrt{2}}|V\rangle$
would i just represent that as $\begin{bmatrix}\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{bmatrix}$? or could i represent it as $\begin{bmatrix}\frac{1}{\sqrt{2}}\\ 0\\ 0\\ \frac{1}{\sqrt{2}}\end{bmatrix}$?
Well @TedShifrin as I proved the other day, $S^n, n > 1$ is simply connected, but $S^n \times S^1$ is not, so that extra circle can cause all sorts of problems.....
@TedShifrin $O(n)/SO(n) \cong \Bbb Z_2 \cong \langle -I_n \rangle$ We want their intersection to be $\langle I_n \rangle$, i.e. $-I_n \notin SO(n)$, i.e. $\det(-I_n) = (-1)^n \ne 1$, i.e. $n$ is odd.
@heather The first. For the latter you'd need to actually have the state of a second particle/qubit/whatever written in your state, but what you wrote is just the state of one object.
I mean, $\otimes$ is used for a general monoidal products all the time, and any category with finite products is monoidal with respect to the product bifunctor, so there's nothing inherently wrong with $SU(2) \otimes SU(3)$ :P
I was having a discussion with a physicists who was talking about generators of a lie group and meant generators of the lie algebra. I was thinking about actual generators of the group, so I was really confused how an uncountable group could have a finite set of generators
You shold see the way we try to teach that stuff to grad students with on a few weeks of group theory, no algebra or differential topology or anything. It isnt pretty.
I think it started when I was studying a project on the death penalty in middle school and 625 was the number of executions past a certain point or something
and then I started feeling like i'd see it everywhere
@TedShifrin OH I finally got it! You mean for every $x \in \mathbb{R}^k$ we associate to it the differential at that point $df_x \in L(\mathbb{R}^k, \mathbb{R})$.
But then how is that differ from informal reasoning or philosophy. Under the context of informal mathematics, what actions and reasoning will be constitute as "mathematical"?
for there are deterministic things that simply follow no reason nor (much) pattern such as chaotic processes
and there are things that are simply unpredictable and nonrandom
and of course at the very far end of the spectrum, there are unknown unknowns, those we don't even know how to discuss because we simply don't know anyting about them, and we don't even know that we don't know (they are nonexistent in our views and reasoning)
Every space can be turned into a separable one adding a single point so there are plenty of examples of separable compact spaces that are not metrizable
