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4

If $g$ acts by scalar multiplication, then its image in $\text{GL}(W_{\lambda})$ is central. This need not imply that $g$ itself is central; for example, $G$ could be a product of two simple groups, $W_{\lambda}$ could be an irrep of one of them, and $g$ could live in the other. But it's true if $W_{\lambda}$ is a faithful representation.


2

Suppose $\sigma: H\to\mathrm{GL}(W)$ is a complex irreducible representation of $H$. Call the trace $\chi$. Then character theory says that $\sum_{h\in H}\chi(h)$ equals $|H|$ if $\sigma$ is trivial and $0$ otherwise. Additionally, the sum $\sum_{h\in H}\sigma(h)$ is an interwtining operator, and Schur's lemma says the intertwining operators of a complex ...


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The nonzero vector $(x_1,x_2,\ldots,x_n)$ of $X$ generates an irreducible submodule isomorphic to $W_1$ if and only if there are $FG$-isomorphisms $\phi_i:W_1 \to W_i$ with $\phi_i(x_1) = x_i$ for all $i$. Otherwise it generates a reducible submodule isomorphic to a number of copies of $W_1$. The number will be equal to the maximum number of the $x_i$ that ...


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$\sum_{1\le i,j\le d_\rho} |\hat{f}(\rho)_{i,j}|^2 = ||\hat{f}(\rho)||^2$ is just the definition of the Frobenius norm of the matrix $\hat{f}(\rho)$.


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When we study the character table of a group, say $S_3$, what vector space are we looking at? When we are studying character table, we are not looking at a vector space.. We are looking at a collection of vector spaces.. Dimension of each of these vector spaces is related to the order of the group (in the case of finite groups).. To each irreducible ...


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It's important that on the first page the paper states that, without loss of generality, all of their irreducible representations are unitary, so that $\rho(g)^\dagger=\rho(g)^{-1}$ for all $g\in G$. By the way, the beginning of section $3$ is in error. The sum $\sum_{h\in H}\rho(h)$ is not a projection, because it is not normalized correctly. Instead, ...


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Let $G$ be the group in question. In the first case, you are decomposing the $G$-representation $\rho_a \otimes \rho_b$ into $G$-representations. In the second case, $\rho_a(g_1) \otimes \rho_b(g_2)$ is actually a value of the $(G \times G)$-representation $\rho_a \otimes \rho_b$, so it will decompose into $(G \times G)$-representations ...


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Since $\Gamma_{0,1}$ is contained in $\Lambda^2V\subset V\otimes V$, you see that $\Gamma_{a,b}\subset Sym^a V\otimes Sym^b \Gamma_{0,1}\subset \otimes^a V\otimes \otimes^b(\otimes^2 V)\subset\otimes^{a+2b}V$.


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This follows from the following form of Frobenius reciprocity: for every $k[G]$-module $V$ and every $k[H]$-module $U$, there is an isomorphism of groups $$ {\rm Hom}_G(V,{\rm Ind}_{G/H}U) \cong {\rm Hom}_H({\rm Res}_{G/H}V,U).$$ So if the right hand side is non-trivial, then so is the left hand side. But if $V$ is a simple module (i.e. the representation is ...


1

Let $S$ be a real $6$-dimensional vector space with basis $1,i,j,k,r,s$. We can define the multiplication on this basis as we want and then extend bilinearly to $S$. This yields "sextonians", as you wish, with $i^2=j^2=k^2=-1, r^2 =s^2=1$. The question is which identities we would require form this bilinear product. We know that there is no real normed ...


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Indeed the argument used is not quite clear as it stands, because context is missing. Apparently something is already known about uniqueness of decomposition of characters into irreducibles. What is known here is that we have a $\Bbb R G$-module that is reducible, and whose character is $\chi+\overline\chi$; its character must break up as the sum of at least ...


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Here we realize $D_8$ in terms of its usual generators and relations: $$D_8 = \langle a, b : a^4 = b^2 = (ab)^2 = 1 \rangle .$$ Note that by using the relations, we can write any element of $D_8$ as $a^l b^m$. The conditions $\phi(a)^4 = 1$ and $\phi(b)^2 = 1$, which imply $\phi(a) = i^r$ for some $r$ and $\phi(b) = (-1)^s$ for some $s$ are consequences of ...


1

What you call "exponentiation" means conjugacy: $\phi$ is a homomorphism $G\rightarrow Aut(V)$ for some vector space $V$; an (invertible) matrix is an automorphism of $V$ (once you choose a basis), and $\phi^A$ is the representation $x\mapsto A\phi(x)A^{-1}$ (or replace $A$ with $A^{-1}$ depending on the author's conventions). In other words, it's just the ...


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Sorry to resurrect such an old post... The matrix you wrote is not in $\text{SO}(5)$, as it not an orthogonal matrix. Only for simply connected Lie groups can a representation of the Lie algebra be lifted to a representation of the Lie group. $\text{SO}(5)$ is not simply connected. So not every representation of its Lie algebra $(\text{B}_2)$ can be lifted ...



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