# (Why) are Deligne-Lusztig varieties nonempty?

Background: Let $$G$$ be a reductive $$\mathbb F_q$$-group and let $$X$$ be the variety of Borel subgroups of $$G$$. By the Bruhat decomposition, the $$G$$-orbits in the space $$X\times X$$ (with diagonal action) are indexed by elements $$w$$ of the universal Weyl group $$W$$ of $$G$$. Let $$\cal O(w)$$ be the orbit corresponding to $$w$$. The Deligne-Lusztig variety $$X(w)$$ is defined to be the intersection of $$\cal O(w)$$ with the graph of the Frobenius endomorphism of $$X$$.

Deligne and Lusztig argue that $$X(w)$$ is of pure dimension $$\ell(w)$$, the length of $$w$$, by a transversality argument. My question is about this argument. I assume the argument to be standard, since Deligne and Lusztig devote only a sentence to it, but there's one point I'm having trouble with.

Two subvarieties of an ambient variety are transverse at a point $$x$$ if the sum of their tangent spaces at $$x$$ generates the ambient tangent space. When $$x$$ is a point of transversality, there is a formula for the codimension of the intersection of the varieties (near $$x$$): it is the sum of the codimensions of the subvarieties.

Since Frobenius has derivative zero, any intersection point of $$\cal O(w)$$ and the graph of Frobenius is a point of transversality. It follows (by the previous paragraph and the formula for the dimension of $$\cal O(w)$$) that if $$\cal O(w)$$ and the graph of Frobenius intersect, then their intersection, the Deligne-Lusztig variety $$X(w)$$, has pure dimension $$\ell(w)$$.

Question: (Why) does $$\cal O(w)$$ intersect the graph of Frobenius at all? In other words, (why) is the Deligne-Lusztig variety $$X(w)$$ non-empty?

Additional thoughts: It's easy to see that $$X(1) = X(\mathbb F_q)\neq\varnothing$$. Using the dimension estimate, you can also show that $$X(w_0)\neq\varnothing$$, where $$w_0$$ is the longest element of $$W$$: indeed, $$X(w_0)$$ is a dense subvariety of $$X$$. However, I don't see how to conclude from formal properties of the Bruhat stratification and the dimension estimate that the intersection must be nonempty.

$$\newcommand\FF{\overline{\mathbb F}}\DeclareMathOperator\Fr{Fr}$$The Lang (or Lang–Steinberg) map $$G(\FF) \to G(\FF)$$ given by $$g \mapsto g^{-1}\Fr(g)$$ is surjective. In particular, there is some $$g \in G(\FF)$$ such that $$g^{-1}\Fr(g)$$ represents $$w$$. If $$B$$ is a rational Borel containing the reference torus, then $$g B_{\FF} g^{-1}$$ and $$\Fr(g B_{\FF} g^{-1})$$ are in relative position $$w$$.
• @DavidSchwein, by the way, surjectivity of the LS map is used all over; but one of my favourite applications of a different but related principle (Lemma 7.3) is used in the proof of Theorem 7.2 of Steinberg - Endomorphisms of linear algebraic groups, which states that every surjective endomorphism of a linear algebraic group $G$ over an algebraically closed field preserves a Borel subgroup of $G$. Commented Aug 21, 2022 at 19:15