# Tag Info

7

We claim that given the conditions set forth in the problem, $\text{Ind}_{Z(G)}^G \rho$ is always not irreducible. Set $Z = Z(G)$. Let $W$ be an irreducible $Z$-representation. We would like to compute$$\langle\text{Ind}\,W, \text{Ind}\,W\rangle_G = \langle W, \text{Res}\,\text{Ind}\,W\rangle_H$$by Frobenius reciprocity. Since $Z$ is abelian, we know $W$ is ...

3

No, in general this won't be irreducible: for example, if $G$ is a non-trivial finite group with $Z(G)$ trivial, then $\text{Ind}_{Z(G)}^G\text{triv}$ is the regular $G$-representation on ${\mathbb C}[G]$, which contains the proper non-trivial submodule ${\mathbb C}\cdot\sum_g g$.

3

Let $\rho: H \to \text{GL}(W)$ be a faithful representation, and let $V = \text{Ind}^{G}_{H} \, W$, given by $\hat{\rho}: G \to \text{GL}(V)$. Suppose there is a non-identity $g\in G$ such that $g\in \text{ker} \, \hat{\rho}$. Choose $S$ a set of coset representatives. By Mackey's formula, we see that $$\chi_{V} (g) = [G:H] (\text{dim} \, W) = \sum_{gsH = ... 3 As an alternative to Sameer's answer, you could argue as follows: If V is the H-module corresponding to \rho, then \text{Ind}_H^G \rho = {\mathbb C}[G]\otimes_{{\mathbb C}[H]} V, which as a vector space decomposes as the direct sum of g_i\otimes V for a choice \{g_i\}_i of H-right-coset representatives in G, among which we choose e as the ... 3 If A is a finite dimensional algebra, then every simple A-module is finite dimensional, since it is a quotient of the regular module (if S is a simple left module, then for any 0\neq s\in S, a\mapsto as is a non-zero homomorphism A\to S, which is surjective since S is simple). 2 As far as I can see, your reasoning is fine, and you don't even need an action of G on V^B: It suffices to pick some v\in V^B and to consider the map G\to V given by g\mapsto g.v. Since v\in V^B, this map factors through G/B\to V, and your arguments show that it must be constant. 2 There is a proof and it's pretty quick. Below, \mathbb{F} is used to denote the trivial representation of the trivial group. Let R be the regular representation of a group G. Then, R = \mathrm{Ind}_{1}^{G}(\mathbb{F}). Let V be an irreducible representation of G. Then, by Frobenius reciprocity$$\dim \mathrm{Hom}_{G}(R, V) = \dim ...

2

The statement that any subrepresentation of $V \oplus V$ is of the form $\{0\} \oplus V$ or $V \oplus \{0\}$ is false; a counterexample is given by $\{(v, v)\text{ }|\text{ }v \in V\}$. Write $V = \bigoplus_i U_i$ with $U_i$ irreducible. We claim there exists $Y \subset V$ with $V = X \oplus Y$. Proceed by contradiction, and choose $X \subset V$ of maximal ...

2

Here is a portion of the answer. All of these calculations are coming from here. In terms of that website, we think of $SU(3)$ as $A_2$. Then, the $10$ dim rep has heighest weight $(3,0)$ in their notation (and the $\overline{10}$ has heighest weight $(0,3)$). The $27$ d rep is $(2,2)$. The notation the website uses is $X[3,0]$, $X[0,3]$, or $X[2,2]$ ...

2

I assume by $\mathbb Z_k$ you mean $\mathbb Z/k\mathbb Z$, the cyclic group of order $k$. (Most number theorists don't like this notation.) In this case $\mathbb Z/k\mathbb Z$ is isomorphic to the group of $k$-th roots of unity under multpilication. Because it is abelian, all irreducible representations are 1-dimensional, i.e., linear characters. For ...

2

Write $G = S_n$. Recall that $\mathbb{C}^n = P \oplus \text{span}(1, 1, \dots, 1)$. Let $\phi: \mathbb{C}^n \to P$ be the projection and $e_1, \dots, e_n$ the standard basis for $\mathbb{C}^n$. $($Therefore, $\phi$ is given explicitly as $\phi(x_1, \dots, x_n) = (x_1 - \overline{x}, \dots, x_n - \overline{x})$ where $\overline{x} = (\sum x_i)/n$.$)$ Let $f_i ... 1 Consider first of all a nonzero irreducible subrepresentation$W \subset V^{\oplus n}$. Each projection map$\pi_j: W \to V$is either zero or an isomorphism, by Schur's Lemma. Choose$j_0$so that$\pi_{j_0}$is an isomorphism. For every$1 \le k \le n$, the map$$\pi_k \circ \pi_{j_0}^{-1}: V \to V$$is, by Schur's Lemma, a scalar; call this scalar ... 1 Let$H = (e)$, the trivial group in$G$, and let$W$denote the trivial representation of$H$. Then$V = \text{Ind}_H^G W$is the regular representation of$G$. If$U$is any irreducible representation of$G$, we know by Frobenius Reciprocity that$$\langle V, U\rangle_G = \langle \text{Ind}_H^G W, U\rangle_G = \langle W, \text{Res}_H^G U\rangle_H.$$Since$H$... 1 Try span of$[1\;0]^t$, i.e. $$V = \left\{ \begin{bmatrix}a\\0\end{bmatrix} \mid a\in\mathbb Z/p\mathbb Z \right\}.$$ You can simply compute everything, but in fact there is a direct way to see that there can not be a complementary representation and that hence Maschke's theorem must fail. If$V$had a complement, then the matrix$M$would be ... 1 By using a stronger result, I will give a short and general answer. Let$\chi$be a the corresponding character of$H$and$\chi^G$be induced character of$G$. Lemma:$Ker(\chi^G)=Core_G(Ker(\chi))=\bigcap (Ker\chi)^g$Thus, If$Ker(\chi)=e$then clearly$Ker(\chi^G)=e$. Hence,$\chi^G\$ is faitful.

Only top voted, non community-wiki answers of a minimum length are eligible