Let $G$ be an affine, connected, reductive group and $X$ a $G$-module. Choose a maximal torus $T\subseteq G$, a Borel $B\subseteq G$ containing $T$ and let $U$ be the unipotent radical of $B$. Denote by the character group of $T$. Let $\Lambda\subseteq\mathbb{X}$ be the set of dominant weights of $G$ with respect to these choices. We can decompose the graded coordinate ring $\mathbb{C}[X]=\bigoplus_{\lambda\in\Lambda} V_{(\lambda)}$ into its isotypic components $V_{(\lambda)}$ of weight $\lambda$. Let
\[ \Lambda_X=\{ \lambda\in\Lambda \mid V_{(\lambda)}\ne\{0\}\} \]
be the set of weights that occur in $\mathbb{C}[X]$.
Let $V_{(\lambda)}\cong V_\lambda^{\oplus n_\lambda}$, where $V_\lambda$ is the irreducible module of highest weight $\lambda$. Each $V_\lambda$ has a highest weight vector which is unique up to scaling - let $f_{\lambda 1}, \ldots, f_{\lambda n_\lambda}\in V_{(\lambda)}$ be linearly independent highest weight vectors.
Claim. *For each $1\le k\le n_\lambda$, if the function $f:=f_{\lambda k}\in\mathbb{C}[X]$ is reducible, then there exist weights $\lambda_1,\ldots,\lambda_r\in\Lambda_X$ such that $\lambda$ is an $\mathbb N$-linear combination of the $\lambda_i$.*
Indeed, about a year ago this statement was completely unclear to me. However, it's actually not that hard to see and I felt like sharing my proof. (more…)
For the group $G=\mathrm{GL}_n(\mathbb{K})$ of invertible matrices over any field, there is the well-known determinant character $\det:G\to\mathbb{K}^\times$ which, well, maps any matrix to its determinant. It has the property that
\[ [G,G]=\mathrm{SL}_n(\mathbb{K})=\{ g\in G \mid \det(g)=1 \}, \]
where $[G,G]$ denotes the derived subgroup of $G$. In more geometric terms, $[G,G]$ is the vanishing of the regular function $\det-1$ on $G$. Because I find it rather cute, I will show that you can find a similar function for all linear, reductive groups.
(more…)