Let $G$ be a connected semisimple simply connected group over an algebraically connected field. Let $w_0$ be the longest element in the Weyl group $W$ and $s_i$ be a simple reflection in $W$. Choose a Borel subgroup $B$ in $G$. It is known that the closure of the Bruhat cell $Bw_0s_iB$ in $G$ is defined by a single function (unique up to a scalar), say $f\in k[G]$. Is it true that under the action of $B\times B$, $f$ has the weight $(\chi_i,w_0\chi_i)$, where $\chi_i$ is the fundamental weight corresponding to $s_i$, why?
2 Answers
The function $f$ that cuts out the closure of this Bruhat cell is known  it's a generalised minor.
Let $\omega$ be a fundamental weight. Let $v$ be a highest weight vector in $L(\omega)$ and $w\in L(\omega)^\ast$ be dual to a lowest weight vector. Then $f(g)=\langle w,gv\rangle$ is the desired function.

$\begingroup$ >Thanks for your answer, could you be more specific about why this $f$ defines that Bruhat cell? $\endgroup$ May 4, 2023 at 9:36

$\begingroup$ What does it mean to be dual to a lowest weight vector? That it pairs non$0$ly with a lowestweight vector, and $0$ly with all other weight vectors? Is that different from just being a highestweight vector in $L(\omega)^*$? \\ Also, my computation gave a weight $(w_0\omega, \omega)$ (or at least a multiple of that), and it seems that yours does too, not $(\omega, w_0\omega)$. Is that correct? $\endgroup$– LSpiceMay 4, 2023 at 13:15

1$\begingroup$ @AllenLee, re, since this function transforms by the weight $(w_0\omega, \omega)$ under $B\times B$, it suffices to check what elements of the Weyl group have a representative on which $f$ doesn't vanish. For $n \in N_G(T)(k)$, we have that $f(n)$ is non$0$ if and only if $n\omega  w_0\omega$ equals $0$, i.e., if and only if $\omega$ is fixed by $n^{1}w_0$. It's not obvious to me that's equivalent to $s_i \not\le n^{1}w_0$, but, if so, then that's equivalent to $s_i \not\le w_0 n$, and so to $n \not\le w_0 s_i$. $\endgroup$– LSpiceMay 4, 2023 at 13:29
$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}$Let $k$ be the ground field. Choose a representative $n_0$ for $w_0$ in $N_G(T)(k)$.
I get an answer that is unsatisfactory in two ways. First, I have to replace $\chi_i$ by a positive multiple $\chi$, and I cannot yet identify the multiplier. (It seems likely that, if $\chi$ is a nontrivial multiple $N\chi_i$ of $\chi_i$, then a function $F$ on $B w_0 B$ with weight $(w_0\chi_i, \chi_i)$ such that $F(n_0)^N$ equals $f(n_0)$ extends to $G$; but I do not yet see for sure why this is so.) Second, I get that the weight is $(w_0\chi, \chi)$, not $(\chi, w_0\chi)$. Probably, I am just messing up something in the normalisation.
Let $\omega$ be the weight of $f$ for $B \times B$, and put $\chi = \omega(1, \cdot)$. Evaluation at $n_0$ shows that $\omega = (w_0\chi, \chi)$, so it suffices to show that $\chi$ is a positive multiple of $\chi_i$.
For each $j$, write $U_j$ for the root group associated to $\alpha_j$. Computing in $\SL_2$ shows that there are a representative $n_j$ in $N_G(T)(k)$ of $s_j$, and functions $u_\ell, u_r : \GL_1 \to U_j$, such that $\lim_{t \to 0} u_\ell(t)n_j\alpha_j^\vee(t)u_r(t) = 1$.
Per the comments, here are the deails. We have that $\operatorname C_G\bigl(\ker(\alpha_j)^\circ_\text{smooth}\bigr)$ is a Levi subgroup of $G$, so its derived subgroup $G_j$ is a rank$1$, simply connected group, hence is $\SL_2$. We may choose the isomorphism so that it carries $\alpha_j^\vee(t)$ to $\operatorname{diag}(t, t^{1})$, and $U_j$ to $\left\{\begin{pmatrix} 1 & * \\ & 1 \end{pmatrix}\right\}$. Write $n_j$, $u_\ell(t)$, and $u_r(t)$ for the preimages in $G$ of $\begin{pmatrix} & 1 \\ 1 \end{pmatrix}$, $\begin{pmatrix} 1 & (1 + t)t^{1} \\ & 1 \end{pmatrix}$, and $\begin{pmatrix} 1 & (1  t)t^{1} \\ & 1 \end{pmatrix}$, respectively. Then, if I have not messed things up, we have that $u_\ell(t)n_j\alpha_j^\vee(t)u_r(t)$ is the preimage of $\begin{pmatrix} 1 + t & t \\ t & 1  t \end{pmatrix}$.
With these definitions $g_j(t) \mathrel{:=} n_0 n_j^{1} u_\ell(t)\alpha_j^\vee(t)^{1}n_j u_r(t)$ lies in $B w_0 B$, so that $f(g_j(t))$ is non$0$, for all $t$, and $\lim_{t \to 0} g_j(t) = n_0 n_j^{1}$ belongs to $B w_0 s_j B$, so that $\lim_{t \to 0} f(g_j(t))$ is $0$ exactly if $j = i$.
We have that $$ g_j(t) = U_\ell(t)^{1}g_j(1)u_r(1)^{1}\alpha_j^\vee(t)u_r(t), $$ where $U_\ell(t) = n_0 n_j^{1}u_\ell(1)u_\ell(t)^{1}n_j n_0^{1}$, so $$ f(g_j(t)) = \omega(U_\ell(t), u_r(1)^{1}\alpha_j^\vee(t)u_r(t))f(g_j(1)) = \chi(\alpha_j^\vee(t))f(g_j(1)) = t^{\langle\chi, \alpha_j^\vee\rangle}f(g_j(1)), $$ for all $t$. Since $\lim_{t \to 0} f(g_j(t))$ exists, we have that $\langle\chi, \alpha_j^\vee\rangle$ is nonnegative; and $\langle\chi, \alpha_j^\vee\rangle$ is positive if and only if $\lim_{t \to 0} f(g_j(t)) = 0$, which happens if and only if $j = i$.

$\begingroup$ Thanks for your answer, could you please elaborate how to get $u_{\mathcal{l}}$ and $u_r$? $\endgroup$ May 4, 2023 at 15:05
