For $G=\mathbb G_m$, write $k[G]=k[t,t^{-1}]$. Since
$$
((a,b)t^r)(x)=(a^{-1}xb)^r=a^{-r}b^rt^r,
$$
the <weight space>[weight-space decomposition] is
$$
k[G]=\bigoplus_{r\in\mathbb Z}k_{(-r,r)}
$$
as a $G\times G$-representation.

If $V=\bigoplus_rV_r$ is the weight decomposition of a finite-dimensional $G$-module, the matrix-coefficient argument again gives $\operatorname{Hom}_G(V,k[G])\cong V^*$. A map from $k[G]$ to $V$ is independently determined by the image in $V_r$ of the basis vector $t^r$; only finitely many $V_r$ are nonzero. Thus
$$
\operatorname{Hom}_G(k[G],V)
\cong\bigoplus_rV_r
\cong V
$$
as vector spaces.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 167 / 1 / iii / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-167/1/iii/b/solution