Hey guys - Hope you can answer a quick question. Does it make sense to consider the tensor product of two Hilbert spaces with different scalar fields? Most Litterature considers both fields either real or complex, but as you can see in my latest question i seem to extend the procedure to the mixed case.
Sorry. So yes I was going on the right line of reasoning. So first of all my claim is that it suffices to prove this for transposition. The reason it suffices to prove it for transposition is that given $g \in S_n$ it is a product of transposition denote them by $g_1 ... g_n$. Therefore we have $det(p(g_1 ... g_n))$ since $p$ is a homomorphism we have $p(g_1 ... g_n) = p(g_1) ...p(g_n)$.
determinant is multiplicative so we have $det(p(g)) = det(p(g_1)) ... det(p(g_n))$.
Given a transposition $\lambda \in S_n$ we have it switches coloumns of identity matrix which is the same as multiplying 1 by (-1). Therefore $det(p(g)) = (-1)^n$
okay lets us do it. Degree 1 presentation is the same as maps $p : G \rightarrow F^{\times}$ and similiarly degree 1 presentation of $G/[G,G]$ is same as $\phi : G/[G,G] \rightarrow F^{\times}$.
Given $p \in Hom(G,F^{\times})$ we construct $\phi_p : G/[G,G]$ as follows. For simplicity denote $[G,G] = G`$. $\phi_p(gG`) \mapsto p(g)$. We have to show this is well defined. First we prove that $[G,G] \subset Ker(p)$.
yeah $\theta$ is arbitrarily. So we have constructed one way that is we have constructed given $p \in Hom(G,F^{\times})$ we constructed $\phi_p \in Hom(M,F^{\times})$ where $M = G/G`$.
now we need other direction.
Given $\theta \in Hom(M,F^{\times})$ we need to construct $\sigma \in Hom(G,F^{\times})$.
We did group theory built up from nothing, then Galois theory, then a massive thing at the end with module theory (structure theorem, jordan canonical form, etc) and homological algebra
next semester I will try something different because I always tend to burn myself up by the end of the semester. Next semester I will wake up early like 6 am go to gym and study. I will go gym no matter how busy I am and also wake up early 6 am no matter what.
My top. professor did Ph.D. with this guy: https://en.wikipedia.org/wiki/Benedict_Gross Pretty famous for his work with elliptic curves and Birch/Swinnerton Dyer
Just need to confrm a problem I've done. If a group has order 1925 and more than one sylow 5-subgroup what is the number of sylow 5-subgroups? I got that there are 11 sylow 5-subgroups. we may write the order as 5^2 * 7 * 11. Hence, we must have that the number is a divisor of 77 and that the number = 1 mod 5 by Sylow's Third Theorem. Is this correct?
what's that really obvious result about set cardinalities that's equivalent to the AoC? That a countable union of non empty sets something, or a cartesian product of something is something?
you've shown degree 1 reps correspond to nth roots of unity. should mention the converse as well.
also if the original problem is for "all reps" then you should probably mention maschke's theorem, and mention the fact that irreps of abelian groups are degree 1
Let's consider $\Bbb Z/3\Bbb Z$. Let $\alpha,\beta$ be representations with $\alpha(k)=\omega^k$ and $\beta(k)=\omega^{-k}$. What would the isomorphism $\Bbb C\to\Bbb C$ be?
First of all, it needs to be a linear map, so it's multiplication by some $c$.
Write $G=\prod_{i=1}^k \Bbb Z/n_k\Bbb Z$ and use $e_i$ for the usual coordinate tuples. The homomorphisms $G\to S^1$ are in bijection with ordered tuples $(\zeta_1,\cdots,\zeta_k)$ where each $\zeta_i$ is an $n_i$th root of unity
