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.


1 Answer 1


$\newcommand\FF{\overline{\mathbb F}}\DeclareMathOperator\Fr{Fr}$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 $w$.

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    $\begingroup$ Thanks Loren! You are always answering my questions :D $\endgroup$ Aug 20, 2022 at 22:54
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    $\begingroup$ @DavidSchwein, no problem! My understanding of DL varieties is very small, so I'm glad to find an occasional question I can answer. $\endgroup$
    – LSpice
    Aug 21, 2022 at 0:07
  • $\begingroup$ @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 $G$. $\endgroup$
    – LSpice
    Aug 21, 2022 at 19:15
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    $\begingroup$ That's a very nice result. I suppose that once you know Theorem 7.2, you can begin to understand these endomorphisms through their effect on pinnings and so on. $\endgroup$ Aug 22, 2022 at 12:53
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    $\begingroup$ Ah, I see. So one particular point here is that conjugation by a unipotent element preserves a Borel, but not necessarily a maximal torus. $\endgroup$ Aug 22, 2022 at 13:25

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