Recent content by fishturtle1

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    Challenge Math Challenge - May 2021

    Proof: We have \begin{align*} \tau(h) \circ \text{Sym}(\varphi) &= \tau(h)\circ \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\ &= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ \tau(g) \circ \varphi \circ \rho(g^{-1}) \\ &= \frac{1}{\vert G \vert}...
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    Challenge Math Challenge - May 2021

    For ##\varphi \in \text{Hom}_{\mathbb{K}} ((\rho, V), (\tau, W))##, we have ##\tau(h) \text{Sym}(\varphi) = \text{Sym}(\varphi) \rho(h)## Proof: We have \begin{align*} \tau(h) \circ \text{Sym}(\varphi) &= \tau(h) \circ \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ...
  3. F

    Challenge Math Challenge - May 2021

    I think the calculation is \begin{align*} \text{Sym}(\varphi) &= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\ &= \frac{1}{\vert G \vert} \sum_{g \in G} \varphi \\ &= \frac{1}{\vert G \vert} \cdot \vert G \vert \varphi \\ &= \varphi \\ \end{align*} Edit...
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    Challenge Math Challenge - May 2021

    that clears things up, thank you!
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    Challenge Math Challenge - May 2021

    One other question, what is ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##? I'm pretty sure ##\text{Hom}_{\mathbb{K}}(V, W)## is the set of all ##\mathbb{K}##-linear maps from ##V## to ##W##. And we can make ##V## into a ##\mathbb{K}G## module by defining ##g \cdot v = \rho(g)v## I think?? So...
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    Challenge Math Challenge - May 2021

    I think I get it now, I had to look up the definition of GL(V). So, ##\rho(g^{-1})## is an automorphism of ##V## and ##\tau(g)## is an automorphism of ##W## and ##\varphi## is a k linear map from ##V## to ##W##.
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    Challenge Math Challenge - May 2021

    Sorry for the dumb question but do the ##\circ##'s in 4) mean function composition or matrix multiplication? As I understand it, ##\tau(g)## and ##\rho(g^{-1})## are matrices in ##GL(W)## and ##GL(V)## resp. ?
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    Challenge Math Challenge - January 2021

    Thank you for your time and feedback; and sorry for confusion! Here are two things I would add to make post #20 clearer: Lemma: Let ##G## be a group and ##H## a subgroup. If ##[G: H] = n##, then there exists a homomorphism ##f : G \rightarrow S_n## with ##\ker f \le H##. Proof: Let ##X## be...
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    Finding the dimension of a subspace

    We know ##y : V \rightarrow \mathbb{R}## is a linear transformation. By Rank-Nullity theorem we have ##\dim V = rank(y) + null(y)##. We note that ##null(y) = \dim W##. If ##rank(y) = 0##, then ##\dim V = \dim W## and so ##\dim W = n##. If ##rank(y) \neq 0##, then ##rank(y) = 1##. And so ##\dim...
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    Finding the dimension of a subspace

    Thank you for your replies. Yes, I think we can use Rank-Nullity theorem, I will try it. And yeah, I'm happy to use a more standard notation if that's better; I was trying to mimic the textbook.
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    Finding the dimension of a subspace

    I am stuck on finding the dimension of the subspace. Here's what I have so far. Proof: Let ##W = \lbrace x \in V : [x, y] = 0\rbrace##. We see ##[0, y] = 0##, so ##W## is non empty. Let ##u, v \in W## and ##\alpha, \beta## be scalars. Then ##[\alpha u + \beta v, y] = \alpha [u, y] + \beta [v...
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    Showing two groups are equal/Pontryagin duality

    We want to show ##\vert (H^\perp)^\perp \vert = \vert H \vert##. We have ##\vert (H^\perp)^\perp \vert = [\widehat{\widehat{G}} : H^\perp] = \frac{\vert\widehat{\widehat{G}}\vert }{\vert H^\perp\vert} = \frac{\vert G \vert}{[G: H]} = \vert H \vert##
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