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15

Okay, we're going to have to use some heavy artillery to start off, but I can't think of another way to begin. Suppose $\rho: G\to \text{GL}_{2} (\mathbb{C})$ is nontrivial. Observe that since $G$ is simple and the representation is nontrivial, we must have $\text{ker} \, \rho =\text{ker}\, \chi = (e)$ (where $\chi$ is the character of this representation). ...

9

Using a lot less machinery: Such a representation must land in $SL_2(\mathbb C)$, and finite subgroups of $SL_2(\mathbb C)$ are completely classified (they are either cyclic, dihedral, tetrahedral, octahedral or isocahedral). None of these groups are simple, unless cyclic of prime order. One can prove this classification by letting a finite subgroup $G$ of ...

4

The group algebra is a functor $\mathsf{Grp} \to \mathsf{Alg}_K$ which is left adjoint to the "group of units" functor $\mathsf{Alg}_K \to \mathsf{Grp}$, $A \mapsto A^\times$. It is defined for all unital $K$-algebras. No non-trivial ring has the property that its underlying semigroup is a group, because $0$ is not invertible. If $V$ is some $K$-module, ...

3

The condition you verified is indeed enough, this is due to complete reducibility for compact Lie groups. Roughly, suppose you have a subrepresentation then by complete reducibility it must have a complementary subrepresentation. Now you can define a linear map that is the identity on one component and zero on the other, this will be a non-scalar map that ...

3

Character Theory of Finite Groups - I. Martin Isaacs

3

The fact that there are more than $n/p^2$ irreducible characters whose squares sum to $|G|$ implies that at least one nontrivial character $\chi$ has degree less than $p$. Since $G$ is simple, $\chi$ is faithful, so it restricts to a faithful character $\chi_P$ of a Sylow $p$-subgroup $P$ of $G$. The degrees of the irreducible constituents of $\chi_P$ all ...

3

Harmonic analysis means different things to different people, but to me it's describing the decomposition into irreducibles of the regular representation of a group $G$ on the space $L^2(G)$ of square-integrable $f:G\rightarrow\Bbb{C}$ via some appropriate "Plancherel theorem''. This may be something you're familiar with, but I'll say it anyway: the idea is ...

3

1. It suffices to show that the action of ${SU}(n)$ on a single basis polynomial ${x_1}^{a_1} {x_2}^{a_2} \dots {x_n}^{a_n}$ $($where $a_1 + \dots + a_n = m$$) is continuous. That is, let A = (a_{ij}) \in {SU}(n) be a matrix, then we need to show that the change of variables$$(x_1, x_2, \dots, x_n) \mapsto \left(\sum a_{1i} x_i, \sum a_{2i} x_i, \dots, ... 2 Here's how I would view the problem; I don't actually use Mackey's formula, but, in this special case, Frobenius reciprocity leads to the same thing. Inducing the trivial character of$H$up to$G$produces the permutation character of$G$acting on a transitive$G$-set with$H$as one of the point-stabilizers, i.e., the$G$-set$G/H$. The inner product you ... 2 Hint: if$\chi \in Irr (G)$, and$\phi$is an irreducible constituent of$\chi_H$(so$[\chi_H,\phi] \geq 1$), then by Frobenius Reciprocity$\phi^G(1)=\phi(1)[G:H] \geq \cdots + \chi(1) + \cdots \geq n$. 2 It seems OP's question (v1) is essentially asking: Given a Lie group$G$, with a Lie group representation$R:G\to GL(V,\mathbb{F})$, with corresponding Lie algebra$L={\rm Lie}(G)$, and with a corresponding Lie algebra representation$\rho :L\to {\rm End}(V,\mathbb{F})$, would$$\tag{1} \forall g\in G, x\in L:~~\rho({\rm ... 1 It's almost the same as Jyrki's comment, but you don't need to argue with dimensions: If$V$is an${\mathfrak s}{\mathfrak l}_2({\mathbb C})$-module for which you have found some direct sum decomposition$V = \bigoplus\limits_{n\in{\mathbb Z}} V^{\prime}_n$into${\mathbb C}$-subspaces such that$V^{\prime}_n$consists of$h$-eigenvectors of eigenvalue$n$, ... 1 I.e. the common eigenspace (for Eigenvalue 1) of the generators: For this, you simply calculate the nullspace bases of the generators: null:=List(GeneratorsOfGroup(g),x->NullspaceMat(x-x^0)); (Instead of GeneratorsOfGroup(g) you could use mats) For getting the intersection it is easiest to go to vector spaces: ... 1 Just some observations (not a complete answer): First, if$p^3\nmid |G|$, then the Sylow-$p$'s have order$p$or$p^2$and are thus abelian. So assume$p^3\mid |G|$. If I recall correctly, you can write$|G|$as a sum of$n$squares, where$n$is the number of conjugacy classes in$G$. We are given that$n>\dfrac{|G|}{p^2}$, so$|G|\$ is the sum of over ...

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