# Asymptotics of regular semisimple elements in finite groups of lie type

Let $$G$$ be a reductive algebraic group defined over $$\mathbb{F}_{q}$$. Let $$G(\mathbb{F_{q}})$$ denote the finite group of fixed points under the Frobenius map $$F$$. Now, we know that the set of regular semisimple elements is dense in $$G$$.

Let $$RSS(G(\mathbb{F}_{q}))$$ denote the set of regular semisimple elements in $$G(\mathbb{F}_{q})$$. I want to know, what can one say about the following

$$\lim_{q\to \infty} \frac{|RSS(G(\mathbb{F}_{q}))|}{|G(\mathbb{F}_{q})|}$$. I think that this answer probably is 1. This is what happens when we look at the group $$GL_{n}(k)$$. The following question probably asks something similar for $$SL_{n}(k)$$:Regular semisimple elements in $SL(n,q)$

There is a comment on this question, which says this result follow from "Deligne's proof of Weil Conjectures". I am not much aware of this. I hope to get some kind of answer to my question, or reference where this question has been studied in details.

Thank you!!

• Sorry to be pedantic, but doesn’t one just need the Lang-Weil estimate (aka the Weil bound —- or actually any nontrivial bound —- for curves), not the full Weil conjectures? [[On second thought, doesn’t one just need the bound #|V(\F_q)| <<_V q^{\dim{V}}, which follows by e.g. Noether normalization? Probably I’m overlooking something...]] Commented Jun 5, 2019 at 17:17

At least in the split semisimple case, this has been worked out explicitly, e.g., by Jouve, Kowalski and Zywina (lemma 4.5 in "Splitting fields of characteristic polynomials of random elements in arithmetic groups"), using structure theory instead of the Weil conjecture (which is also certainly applicable). I don't know how/if this proof extends to the non-split case.

This seems to have nothing to do with algebraic groups. I suppose that you mean that $$G$$ is a connected reductive group. So you have an irreducible algebraically variety $$G$$ and an open subvariety $$X={\rm RSS}(G)\subset G$$, both defined over a finite field $$k={\Bbb{F}}_q$$. You wish to prove that $$\lim_{n\to\infty} \frac{\#X(\Bbb F_{q^n})}{\#G(\Bbb F_{q^n})}=1.$$ Let $$r$$ denote the dimension of $$G$$. The idea is that both $$G$$ and $$X$$ have approximately $$(q^n)^r$$ $$\Bbb F_{q^n}$$-points.

Write $$G=W\smallsetminus Y$$, $$X=W\smallsetminus Z$$, where $$W$$ is an irreducible projective variety of dimension $$r$$, and $$Y$$ and $$Z$$ are projective subvarieties of $$W$$, possibly reducible, of dimensions $$r_Y\le r-1$$ and $$r_Z\le r-1$$.

Using the Lang-Weil estimate (see these lectures or the original paper), we obtain that $$\lim_{n\to\infty} \frac{\#W(\Bbb F_{q^n})}{(q^n)^r}=1,$$ and $$\lim_{n\to\infty} \frac{\#Y(\Bbb F_{q^n})}{(q^n)^{r-1}}\le m_Y,$$ where $$m_Y$$ is the number of irreducible components of $$Y$$ of dimension $$r-1$$. It follows that $$\lim_{n\to\infty} \frac{\#G(\Bbb F_{q^n})}{(q^n)^r}=\lim_{n\to\infty} \frac{\#W(\Bbb F_{q^n})}{(q^n)^r}-\lim_{n\to\infty} \frac{\#Y(\Bbb F_{q^n})}{(q^n)^r}=1-0=1.$$ Similarly we obtain that $$\lim_{n\to\infty} \frac{\#X(\Bbb F_{q^n})}{(q^n)^r}=1.$$ Thus $$\lim_{n\to\infty} \frac{\#X(\Bbb F_{q^n})}{\#G(\Bbb F_{q^n})}=1,$$ as required.

• @Kiddo: Do you need more details? Commented Jun 6, 2019 at 0:56
• this might be a very stupid question still I am asking. Isn't the OP asking what's the limit when $q\to \infty$. In your answer the $q$ remains fixed all the time. Probably there is something obvious that I don't understand.
– Riju
Commented Jun 6, 2019 at 4:47
• @Riju: $q$ is fixed, but $q^n\to\infty$. I denote by $\#G(\Bbb F_{q^n})$ the number that OP denotes by $|G(\Bbb F_q)|$. Commented Jun 6, 2019 at 15:11