Tag Info

Hot answers tagged

3

In representation theory people often use the word representation as a synonym for module, and for good reason. A representation of a Lie algebra is the same thing as a module for its universal enveloping algebra, a representation of a quiver is the same as a module over its path algebra, etc. One important example is that a representation of a finite ...


3

For a given group $G$ and field $F$, the category of group representations of $G$ (in the sense of group homomorphisms from $G$ into $GL(n,F)$) is equivalent to the category of finite dimensional modules over the group ring $F[G]$. Similarly, the category of $F$-algebra representations of an algebra $A$ (in the sense of ring homomorphisms from $A$ into ...


2

We can argue as follows. Let $n\geq 3$ and $V$ be an $FS_n$-module over a field $F$ with $\dim V = n$ and basis $\mathcal B = \{e_1,\ldots,e_n\},$ just as described in the question. Suppose that $0\neq a\in V$ spans a $1$-dimensional $FS_n$-submodule of $V,$ i.e. $\dim FS_na = 1.$ For any $f\in V,$ let's denote by $f_j$ the coefficient of $e_j$ when ...


2

Since $$ \langle \pi(a)\zeta, \zeta\rangle = \langle \pi(a)\eta, \eta\rangle \quad\forall a\in A \qquad(\ast) $$ Define $u : \pi(A)\zeta \to \pi(A)\eta$ by $$ u(\pi(a)\zeta) := \pi(a)\eta $$ Check that $u$ is a well-defined isometry (Use Equation $(\ast)$ with $a^{\ast}a$). Now $u$ extends to an isometry $$ u : H\to H $$ which must be a unitary (Why?). Now ...


1

The image is an angle preserving depiction of a certain Euclidian space together with some of its elements: The Euclidean space in question is the real part of the dual ${\mathfrak h}^{\ast}$ of the Cartan subalgebra ${\mathfrak h}$ of ${\mathfrak s}{\mathfrak l}_3({\mathbb C})$ of diagonal matrices, equipped with the restriction of the dual of the ...


1

One can write the formula more simply as so: $$\sum_{V~{\rm irr}} (\dim V)\chi_V(g)=\begin{cases}|G| & g=e \\ 0 & \rm otherwise \end{cases} $$ One can replace $g$ with $h^{-1}g$ and $g=e$ with $g=h$ in order to get the original version. This equality comes from computing the trace of the linear map $x\mapsto gx$ on the regular representation $\Bbb ...


1

When $n=2$, the braid group $B_n$ is just the free group on one generator $\sigma$, and the representation is determined by the matrix via which $\sigma$ acts on $\mathbf{C}^2$; for $z=-1$ this is $$\sigma \mapsto \left( \begin{matrix} 2 & -1 \\ 1 & 0 \end{matrix} \right). $$ This matrix is conjugate to $$\left( \begin{matrix} 1 & 1 \\ 0 & 1 ...


1

According to Wikipedia, a Lie algebra is called semisimple if it is a direct sum of simple Lie algebras (and a Lie algebra is called simple if its adjoint representation is irreducible, i.e., if it has no proper ideal). The decomposition $L=L_1\oplus \cdots \oplus L_n$ is unique up to permutations of the summands. Indeed, the simple ideals $L_i$ are ...


1

Since $G$ is abelian, the irreducible representations (over $\Bbb C$) have to be 1-dimensional. The possible actions are governed by the following: Note that the group action is determined by the action of any two generating elements. Since the representation is 1-dimensional, any element $g$'s action is just multiplication by some scalar $\lambda_g \in ...


1

I don't have a complete answer, but I think I can add some information. In the spirit of you last comment, the way toweards characterizing the cone of simple tensors is looking at maximal linear subspaces contained in it. Each line in $V$ determins a linear subspace of dimension $dim(W)$ contained in the cone and vice versa. These two families are exactly ...


1

Exercise: let $G$ be a simple group and fix a field $K$. Then every nontrivial linear representation of minimal dimension is faithful and irreducible unless $G$ is cyclic of order $p$ and $p=0$ in $K$. In particular, if the simple group $G$ admits a nontrivial linear representation over $K$ (e.g., $G$ is finite), then it admits a faithful irreducible ...



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