# If $\chi$ a character of a torus $T$ in algebraic group $G$, and $E$ affords $\chi$, is $kGe_U\otimes_{k[T]}E\simeq kGe_Ue_\chi$?

Suppose $G=GL_n(\mathbb{F}_q)$, with $T$ and $U$ the standard maximal torus and unipotent radical. Assume that the characteristic is such that $|G|$ and $|T|$, $|U|$ are invertible in the field to avoid dividing by $0$.

If $\chi$ is an irreducible character of $T$, afforded by some representation $E$, is it true that $$k[G/U]\otimes_{k[T]}E \simeq kGe_U\otimes_{k[T]}E\simeq kGe_Ue_\chi$$ as $k[G]$-modules, where $e_\chi$ is the isotypic projection associated to $\chi$? The first isomorphism is okay I believe since $k[G/U]\simeq kGe_U$, where $e_U=|U|^{-1}\sum_{u\in U}u$ is the idempotent corresponding to the trivial representation, but I am unsure of the second. Here $k[G/U]$ is the $k$-vector space with basis the quotient set $G/U$, viewed as a $(k[G],k[T])$-bimodule via left and right multiplication.

• So by $k[G/U]$ you mean the coordinate algebra? I am a little confused by parts of this since once you consider $G$ as an algebraic group, usually you want to only consider representations over the same field as you have the group defined over (or an extension field), since otherwise the geometry of the group can not really get to play its part. – Tobias Kildetoft Apr 19 '17 at 6:21
• @TobiasKildetoft Sorry for the confusion, I just mean the group algebra of $G/U$ viewed as a $(k[G],k[T])$-bimodule via left and right multiplication. The $k[G]$-module $k[G/U]\otimes_{k[T]}E$ should be the module induced by parabolic induction through the Borel $B=TU$ with Levi $T$. – Camilla Vaernes Apr 19 '17 at 6:31
• But $G/U$ is not a group. So you just mean the vector space with that basis and with those actions? Unfortunately, I know practically nothing about cross-characteristic representations for finite groups of Lie type (which seem to be the groups you really want here, rather than algebraic groups). – Tobias Kildetoft Apr 19 '17 at 7:20
• @TobiasKildetoft Ah you're absolutely right, that's what I meant. My motivation here is looking at the fixed points $\mathbf{G}^F$ for $\mathbf{G}=GL_n(\overline{\mathbb{F}}_q)$. I'll try to edit it. – Camilla Vaernes Apr 19 '17 at 7:31
• Right, as I said, I unfortunately know practically nothing about the cross-characteristic case, being mainly familiar with the theory in the defining characteristic where on can exploit the overlying algebraic group in many nice ways. – Tobias Kildetoft Apr 19 '17 at 7:35

You can choose the representation affording the character $\chi$ just to be $k[T]e_\chi$. Note that multiplication gives an isomorphism $k[G] e_U \otimes_{k[T]} k[T] \cong k[G] e_U$ as left $k[G]$-modules which maps the submodule $k[G] e_U \otimes_{k[T]} k[T] e_\chi$ onto $k[G]e_U e_\chi$.