# Weight of the defining function of a Bruhat cell in a simply connected semisimple group

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?

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.

• >Thanks for your answer, could you be more specific about why this $f$ defines that Bruhat cell? Commented May 4, 2023 at 9:36
• What does it mean to be dual to a lowest weight vector? That it pairs non-$0$-ly with a lowest-weight vector, and $0$-ly with all other weight vectors? Is that different from just being a highest-weight 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? Commented May 4, 2023 at 13:15
• @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 $n \not\le w_0 s_i$. Commented May 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 non-trivial 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 pre-images 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 pre-image 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 non-negative; 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 $$j = i$$.

• Thanks for your answer, could you please elaborate how to get $u_{\mathcal{l}}$ and $u_r$? Commented May 4, 2023 at 15:05
• @AllenLee, re, I have edited in some details. Commented May 4, 2023 at 17:08