# Union of all Schubert divisors coming from a hyperplane

Let $$k$$ be a field with $$\operatorname{char}(k)=0$$. Let $$G$$ be a connected quasi-split reductive algebraic group definied over $$k$$, with maximally split maximal torus $$T$$, Borel $$B \supset T$$ and the Weyl group $$W=N_G(T)/T$$ with $$B$$-simple reflections $$S$$. For $$I \subset S$$, we define $$W_I \subset W$$ to be the subgroup of $$W$$ generated by the elements of $$I$$. Then $$P:=BW_IB$$ is a parabolic subgroup of $$G$$ containing $$B$$. Furthermore let $$W^I:=\{w \in W \mid l(ws)>l(w) \space \forall s\in I\}$$ be the set of minimal coset representatives for $$W/W_I$$, which has an element of maximal length $$w_0^P$$. Denote by $$X(w_0^P)=Bw_0^PP/P$$ the big cell in $$G/P$$. Then $$Z:=G/P-X(w_0)$$ is of codimension 1 and the union of all Schubert divisors in $$G/P$$.

If $$k$$ is algebraically closed then Billey and Laakshmibai state in "Singular Loci of Schubert Varieties", p.44 (MSN), that $$Z=G/P \cap H$$, where $$H \subset \operatorname{Proj}(H^0(G/P,L(\lambda)))$$ is a hyperplane defined by an extremal weight vector of $$H^0(G/P,L(\lambda))$$. Here $$L(\lambda)$$ is a very ample line bundle on $$G/P$$ associated to a dominant weight $$\lambda$$ s.t $$P=P_\lambda$$.

Does this also hold for $$k$$ not algebraically closed? I was wondering if the union of all Schubert divisors in $$G/P$$ coming from a linear homogeneous polynomial.

• Let $\Gamma={\rm Gal}(\bar k/k)$. I assume that $T$ and $B$ are defined over $k$. Then $\Gamma$ naturally acts on the generating set $S=S(G,T,B)$. The answer to your question seems to depend on whether the subset $I\subset S$ is $\Gamma$-stable. – Mikhail Borovoi 2 days ago
• If $I$ is not $\Gamma$-stable, then $P=P_I$ is not defined over $k$, and so I would expect that your set $Z=Z_I$ is not defined over $k$, and therefore, it is not the set of zeros of a polynomial defined over $k$, so the answer is No. – Mikhail Borovoi 2 days ago
• If $I$ is $\Gamma$-stable, then maybe Yes. Please edit your question. The last sentence is grammatically incorrect, and I don't understand what you are asking. – Mikhail Borovoi 2 days ago
• What does it mean $P=P_\lambda$ ? – Mikhail Borovoi 2 days ago
• @MikhailBorovoi, maybe it is the notation for the parabolic subgroup corresponding to the non-negative (or maybe the non-positive?) eigenspaces for the adjoint action of the cocharacter $\lambda$? – LSpice 2 days ago