**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.