Search Results

This is a comment, mostly on terminology that I think caused some confusion, not an answer, but I don't have enough "influence" to post this as a comment. For real reductive groups, the Jacquet funct …

You can use Atlas of Lie groups software to compute representations.

Your two representations of $GL_n$ are not isomorphic, because one of them contains $\Lambda_1$ with multiplicity 2 and the other with multiplicity 1, $\Lambda_1^*$ and $\Lambda_1$ being non-isomorphi …

It's more common to talk about the GT basis of $\mathfrak{gl}_n$-modules. In the case of the adjoint representation, the first row of the GT scheme is $(1,0,\ldots,0,-1)$ and each subsequent row satis …

The answer is affirmative. It is enough to provide a way to uniquely determine the highest weight $\lambda$ a Verma module $M(\lambda)$ from the composition factors of $M(\lambda)$. Every composition …

I have some trouble understanding notation (what is $K$?), so here is a very general answer. The case of ${\rm SL}(2,\Bbb{R})$ was studied in the old work of Repka and, if I am not mistaken, describ …

There is a fruitful way of thinking about the algebra $\mathbb{C}[M_{n,m}]^{{O_n}\times O_m}$ that originates from the $(GL_n, GL_m)$-duality. Namely, $$ \mathbb{C}[M_{n,m}]=\bigoplus_{\lambda}V_ …

The "trivial" case is when the simple highest weight module is a generalized Verma module (i.e. the module induced from a character of a parabolic). The multiplicity is given by the variant of Kostant …

Endow your algebra with the symmetric bilinear form making $\{e_i\}$ an orthornormal basis (the Gram matrix is the identity matrix). Since $c_{ij}^k=([e_i,e_j],e_k)$, the cyclicity condition is equiva …

The answer to the general question is "no": If $\mathfrak{g}$ is solvable, by Lie's theorem its commutant $\mathfrak{g}^{\prime}=[\mathfrak{g},\mathfrak{g}]$ is represented by strictly upper triang …

Based on comments from Eric Rowell and José Figueroa-O'Farrill, I assume that your $V$ is a half-spinor representation of $Spin_{10}$. The key issue is that your hypothetical decomposition of $S(V)$ i …

Note: This was written up concurrently with David's answer, but wasn't proofread and didn't get past the captcha stage due to technical problems. It is more common to consider the representation of …

A representation $\rho$ over a field $K$ is called absolutely irreducible if for any algebraic field extension $L/K$, the representation $\rho\otimes_K L$ obtained by extension of scalars is irreducib …

I assume that $\phi$ is an automorphism of $G.$ Note that if $\phi$ is inner then trivially $\rho$ and $\rho\circ\phi$ are equivalent, thus the answer depends only on the image of $\phi$ in the outer …

If the module $M$ is finite-dimensional then the answer seems to be affirmative. Let $Soc_I(M)$ be the maximal semisimple $I$-submodule of $M,$ which is non-zero by the finite-dimensio-nality of $M. …