Let $G = GL_n$. By algebraic PeterWeyl theorem, we have $$ \mathbb{C}[G] = \bigoplus_{\lambda} V_{\lambda} \otimes V_{\lambda}^*, $$ where $\lambda$'s are dominant weights. Let $U^$ be the unipotent subgroup of $G$ consisting of all unipotent lower triangular matrices in $G$. Let $U^$ act on $G$ by left multiplication. Then we have $$ \mathbb{C}[G/U^] = \bigoplus_{\lambda} V_{\lambda}^*. $$ Let $B$ be the Borel subgroup of $G$ consisting of all upper triangular matrices in $G$. Do we have $$ \mathbb{C}[B] = \bigoplus_{\lambda \in \mathfrak{h}^*} V_{\lambda}^*? $$ Here $\lambda \in \mathfrak{h}^*$ can be nondominant. Thank you very much.

3$\begingroup$ Why should $\mathbb C[B]$ be a $G$representation? And what would $V_\lambda$ mean if $\lambda$ isn't dominant? As yet, this question is nonsensical. It is true that $B$ embeds as a dense set in $G/U^$, so $\mathbb C[B]$ contains $\mathbb C[G/U^]$. Maybe you should say what you mean in the $G=SL_2$ case. $\endgroup$– Allen KnutsonSep 6, 2015 at 3:02

2$\begingroup$ What does $V_\lambda^*$ for $\lambda$ not dominant mean? $\endgroup$– Ben Webster ♦Sep 6, 2015 at 3:02

$\begingroup$ I have the vague memory there is a statement like this involving Verma modules? Trying to guess what it could be: Maybe, as a $B$rep, $\mathbb{C}[B] = \bigoplus_{\lambda \in \mathfrak{h}^{\ast}} M_{\lambda}^{\ast}$, where $M_{\lambda}$ is the Verma module. And then $\mathbb{C}[G/U^{1}] \cap M_{\lambda}^{\ast}$ is $V_{\lambda}^{\ast}$ for $\lambda$ dominant and $0$ otherwise? @BenWebster you are the person whom I usually ask stuff like this. Is anything like my memory right? $\endgroup$– David E SpeyerSep 6, 2015 at 14:05
2 Answers
To expand on David's comment: let $M_\lambda$ be the lowest weight Verma for $\lambda$ an integral weight, and consider $x\in M^*_{\lambda}$, we have a function on the Borel given by $f_x(b)=\langle \ell, b^{1} x\rangle=\langle b\ell, x\rangle$ for $\ell$ the lowest weight vector in $M_\lambda$. This gives a map $\oplus M^*_\lambda \to \mathbb{C}[B]$. If we factor $b=nt$, then we see $f_x(b)=\lambda(t)f_x(n)$. The function $f_x(n)$ is independent of $\lambda$ (identifying all the lowest weight Verma modules as $\mathfrak{n}$modules), so we can think of $x$ as lying in $U(\mathfrak{n})^*\cong \mathbb C[N]$ (here I'm taking restricted dual). Since $\mathbb C[B]\cong \mathbb C[T]\otimes \mathbb C[N]$, we see that this gives that the map is injective and surjective. This isomorphism is precisely set up so that the functions on $G/U_$ are those which come from $x$ in a finite dimensional subrepresentation of $M_\lambda^*$.
EDIT: And if you're wondering what that f.d. sub is, $M_\lambda$ has by assumption a lowest weight vector of weight $\lambda$, and so $M^*_{\lambda}$ has a highest weight vector of weight $\lambda$. Thus, it contains $V_{\lambda}$ if $\lambda$ is antidominant, and no f.d. sub otherwise.

$\begingroup$ Thanks Ben! And, just for the record, the maximal finite dimensional subrep of $M_{\lambda}^{\ast}$ is $V_{\lambda}^{\ast}$ if $\lambda$ is dominant, and $0$ otherwise, right? $\endgroup$ Sep 6, 2015 at 18:32

$\begingroup$ @DavidSpeyer Those aren't quite the conventions I used, but up to some minus signs, yes. $\endgroup$– Ben Webster ♦Sep 6, 2015 at 18:39

$\begingroup$ So $\lambda$ runs through all possible weights and the isomorphism $\bigoplus_{h\in S} M_\lambda^* \to \mathbb{C}[B]$ is an isomorphism of $B$modules? How do you make the connection to $G/U_{}$? $\endgroup$ Sep 6, 2015 at 20:21

$\begingroup$ @VítTuček The description in the original posting. This wasn't explicitly said, but if you follow the argument through, the isomorphism $\mathbb C[G/U_]\cong \oplus_{\lambda}V_{\lambda}^*$ is given by the functions $f_x$ for $x$ in $V_{\lambda}$. $\endgroup$– Ben Webster ♦Sep 7, 2015 at 12:39
This type of question is considered in the paper Longest weight vectors and excellent filtrations by Wilberd van der Kallen (Math. Z. 201, 1931 (1989); MR). van der Kallen works over an arbitrary algebraically closed field $k$, and states his results for the Borel subgroup $B$ of a connected simplyconnected semisimple algebraic group $G$ defined over $k$ (e.g., $G = SL_n$). In this context, he shows that the coordinate algebra $k[B]$ admits a filtration as a $B \times B$module with sections of the form $P(\lambda) \otimes Q(\lambda)$ for $\lambda$ an integral weight, where $P(\lambda)$ is a "dual Joseph module," and $Q(\lambda)$ is "minimal relative Schubert module."
I am not wellversed on all of the terminology and conventions in van der Kallen's paper, so will leave it to the reader to consult van der Kallen's paper for the precise definitions of the relevant modules appearing in the filtration. I'm also not sure how this description fits with Ben's description when $k = \mathbb{C}$.

2$\begingroup$ The paper was about longest weight vectors, not just lowest weight vectors. $\endgroup$ Sep 7, 2015 at 7:23

$\begingroup$ The P(−λ) are the $V(λ)^*$ of Jianrong Li, so there must be multiplicities. It must already be visible for $SL_2$ that there are multiplicities so that the map $\bigoplus_{h\in S} M_\lambda^* \to \mathbb{C}[B]$ is not surjective. $\endgroup$ Sep 7, 2015 at 11:24

$\begingroup$ @Wilberd van der Kallen, thank you very much. In $SL_2$ case, how can we show that there is multiplicity of $V^*_{\lambda}$ in $\mathbb{C}[B]$? Do we have $\mathbb{C}[B] = \bigoplus_{\lambda} M_{\lambda}^*$, where $M_{\lambda}$ is a Verma module with highest weight $\lambda$? $\endgroup$ Sep 9, 2015 at 9:37

$\begingroup$ One must use the action on $\mathbb C[B]$ from the left and from the right. Filter by requiring that weights are no further than $d$ from zero for both actions. That is, look at maximal submodules (for the double action) with that property. This gives a canonical filtration which one should make explicit in terms of the coordinates. You should find multiplicities for $d$ less than 5. $\endgroup$ Sep 9, 2015 at 9:45

1$\begingroup$ @ L Spice: The article is moreorless available online: gdz.sub.unigoettingen.de/dms/load/img/?PID=GDZPPN002436086 $\endgroup$ Dec 11, 2016 at 15:31