Let $G$ be a simple linear algebraic group, and $F$ a Frobenius map, i.e. some power of $F$ is the standard Frobenius map which raises matrix entries to the $q$-th power. Then $G^F$ is a group of Lie type over a field of $q$ elements and I note that all simple groups of Lie type can be obtained this way.

(I am using `Frobenius map' in the sense of Carter as I want to include Ree and Suzuki groups.)

I believe the following to be true and would like help with a reference:

If $q$ is large enough, then every torus in $G^F$ contains a regular semisimple element.

I would, moreover, like a precise definition of `large enough.' The statement and the definition of large enough can be derived from the following statement (although I'm less confident that this one is correct!):

If a torus $T$ in $G^F$ is non-degenerate, then $T$ contains a regular semisimple element.

(The statement now follows because if $q-1$ is greater than the height of the maximum root of $G$, then every torus of $G^F$ is non-degenerate. This result can be derived from the proof of Prop. 3.6.6 of Carter's "Finite Groups of Lie type")

If anyone can help with a proof/reference for either of the italicised statements I would be very grateful!


3 Answers 3


This question is interesting as it forces one to look closely at rational points of $F$-stable tori in $G$.

If $T$ is such torus then $F$ acts the lattice of cocharacters $X_*(T)$ thus inducing an automorphism of the set of coroots. Let $\sigma$ be this automorphism and suppose it has order $m$ (this is all we need to know about it). Let $\epsilon$ be a generator of the multiplicative group of the field in $q^m$ elements. Then for every coroot $\gamma^\vee$ we have that $\gamma^\vee(\epsilon)^F=(\sigma\gamma^\vee)(\epsilon^q)$. Now set $H_\gamma:=\prod_{i=0}^{m-1}(\sigma^i\gamma^\vee)(\epsilon^{q^i})$. Then $H_\gamma^F=H_\gamma$ by construction. Let $\alpha_1,\ldots, \alpha_l$ be a basis of simple roots in $X(T)$ and $a_1,\ldots, a_l\in\mathbb{Z}$. Define $H(a_1,\ldots, a_l):=H_{\alpha_1}^{a_1}\cdots H_{\alpha_l}^{a_l}$, an $F$-stable element of $T$.

We want fo find $a_i$ such that $H(a_1,\ldots, a_l)$ is regular (subject to some condition on $q$ to be determined). Let $a=\sum_{i-1}^la_i\alpha_i^\vee$, an element of $X_*(T)$, and let $e_\gamma\in g_\gamma$, a root vector in $g=Lie(G)$. Set $\tau=\sigma^{-1}$. Then $(Ad\ H(a_1,\ldots, a_l))\cdot e_\gamma= \epsilon^{M(a)}e_\gamma$ where $M(a)=\langle\gamma,\sum_{i=0}^{m-1}q^i\sigma^i(a)\rangle= \sum_{i=0}^{m-1}q^i\langle\tau^i\gamma,a\rangle$. Now $\epsilon$ is a primitive root of $1$ of order $q^m-1$. Suppose we found $a$ such that $\langle\beta,a\rangle\ne 0$ and $-(q-1)< \langle\beta,a\rangle< q-1$ for all roots $\beta$. Then it follows from the uniqueness of $q$-adic expansions that $-(q^m-1)< M(a)< q^m-1$ and $M(a)\ne 0$, so that $H(a_1,\ldots, a_l)$ is regular as wanted. The condition on $a$ will depend on $q$, of course, but NOT on the choice of $T$, and it is quite easy to make it explicit.

Perhaps I should also mention that in the notation of Prop. 3.6.6 in Carter's book the condition $q\ge|\alpha|+1$ for all $\alpha$ is not always tha same as $q-1>ht(\alpha)$ for all positive roots $\alpha$ (especially when there are roots of different lengths).

  • $\begingroup$ Thank you for this Alexander. I need some time to digest what you have written, but it looks like you've answered the question exactly as I wanted... And good point about the distinction between $|\alpha|$ and $ht(\alpha)$ in the use of Prop. 3.6.6 - my mistake. $\endgroup$
    – Nick Gill
    Sep 21, 2012 at 12:18

This line of auestioning is natural but needs more careful formulations to deal with the subtle things that go on for finite groups of Lie type. Presumably "maximal tori" of the finite group are meant to be the groups of rational points of maximal tori of the ambient algebraic group stable under $F$. But working in this set-up gets tricky, depending for instance on how big the field is and how close the tori are to being split over the field of definition; working with twisted groups or groups of Suzuki/Ree type usually requires further refinements.

It's probably easiest to answer affirmatively the broad question of whether such finite "tori" contain regular elements (in the algebraic group sense) when $q$ is large enough; but getting an optimal lower bound on $q$ might depend on the isogeny type, etc. The picture is fairly clear from the work of Cartter's student Derizotis explained briefly in Carter's section 3.8 at the end of Chapter 3. But here the algebraic group is simply connected and "simple" in the sense of algebraic groups, so the simple adjoint groups might need extra discussion. For example, in the simply connected case one has Steinberg's uniform count of semisimple classes in the finite group (recovered by Deriziotis): $q^\ell$ with $\ell$ the rank of $G$.

Consider the case of a simply connected group split over $\mathbb{F}_p$. Here the conjugacy classes under $G^F$ of $F$-stable maximal tori of $G$ are parametrized by classes of the Weyl group $W$. These correspond in the alcove pictures of Deriziotis to small alcoves surrounding various "special" points: a square in type $B_2$ (where $W$ has five classes and types of tori ranging from split to anisotropic have multiplicities 1, 1, 2, 2, 2). Each small alcove corresponds to a semisimple class in $G^F$, with regular ones corresponding to interior fixed points under $F$. To ensure regular elements occur in all types of finite tori, you want $q$ big enough so at least one special point lies far inside the big alcove and its small alcoves all have neighboring alcoves inside the big alcove. Easy to visualize, harder to compute in general.

Note especially the comment by Carter at the bottom of page 105, concerning the occurrence of regular elements in a given semisimple class of the finite group. For example, in the picture of the Brauer complex for $B_2(5)$ on the next page, you can see 13 (out of 25) interior fixed points of $F$ which correspond to classes containing regular elements. But the corresponding tables in Srinivasan's 1968 Trans. Amer. Math. Soc. paper on characters of that finite group illustrate the fact that some of the maximal tori fail to have regular elements over such a small field even though 5 exceeds the Coxeter number here. (Regular elements are detected indirectly from centralizers.) The paper does have minor errors, but is mostly reliable.

[If I read this example correctly, it gives a negative answer to your question about nondegenerate tori of the algebraic group. But I haven't looked closely at that material.]

In any case, it's worth exploring a number of small rank groups to pinpoint what information is of most interest to you. The subject becomes quite intricate for arbitrary groups of Lie type, but the papers by Carter and Deriziotis are well worth looking at. Groups of type $G_2$ (for which an old paper by Chang and Ree computes characters and classes) are especially nice because there is only one isogeny class of groups to consider.

SUMMARY: The qualitative question (existence of regular elements in all finite tori for sufficiently large $q$) is probably best understood, without case-by-case study, in terms of the geometry of alcoves; but I don't think Deriziotis or others formalized this. The quantitative question (computing actual numbers of regular elements) requires case-by-case work. This comes in two flavors: (1) counting the total number of regular semisimple elements in $G^F$, as in the new preprint by Fulman-Guralnick here and papers they cite; (2) counting the number of regular semisimple elements in each type of finite torus (these being parametrized by $F$-conjugacy classes in $W$), as in older work of Fleischmann-Janiszczak in J. Algebra 155 (1993). In each case one looks for answers in the form of polynomials in $q$, which might be zero for some $q$ depending on $G^F$ and in (2) also on the type of torus. Apparently approach (1) leads to nicer and more applicable results.

  • $\begingroup$ Jim, thank you. Your clarification of what I mean by maximal tori of the finite group is spot on. I don't know what you mean by "this already gets tricky" though. Are you suggesting that I can't always define maximal tori in this way? Or that answering the posed question is tricky for this definition? I hope the latter! $\endgroup$
    – Nick Gill
    Sep 20, 2012 at 8:49
  • $\begingroup$ Jim's answer suggests that a "unified proof" of the first italicised statement WITH A PRESCRIBED DEFINITION OF `LARGE ENOUGH' is not known and I would need to go through each family of groups of Lie type individually... As for the non-prescribed form of the first statement: I'd be very happy to hear other contributions to complement Jim's hints for a unified proof. $\endgroup$
    – Nick Gill
    Sep 20, 2012 at 8:56
  • $\begingroup$ @Nick: I added some material, but the story gets long. More recent literature is referenced in my 2006 LMS Lecture Notes 326 (Cambridge Press). $\endgroup$ Sep 20, 2012 at 12:17
  • $\begingroup$ Thank you again Jim, I shall follow up the references you give. I should probably add that I've seen the statement "all tori contain regular elements for q large enough" in Section 4 of Fulman & Guralnick's paper on derangements, but they don't give a reference. $\endgroup$
    – Nick Gill
    Sep 20, 2012 at 13:38

This is an old question now but I had cause to look at it recently. I thought it was worthwhile pointing out that Carter's proof about the existence of nondegenerate maximal tori in Proposition 3.6.6 is something of a red herring. In fact, I claim that if $F$ is a Frobenius endomorphism (not generalised, so ruling out Suzuki and Ree groups) and $G^F$ is not of type ${}^2\mathrm{A}_{2n}$ then a maximally split $F$-stable maximal torus $T \leqslant G$ is non-degenerate if $q > 3$. The point about these assumptions is that I want any root to be orthogonal to any other root in its $F$-orbit. Note that one can also easily show the statement still holds for the groups $\mathrm{SU}_{2n+1}(q)$ by working in a matrix representation of the group. The proof below, however, will not work in this case.

Once one has this statement one can see that a non-degenerate maximal torus can have no rational regular elements, as pointed out by Jim.

To show that $T$ is non-degenerate we must show that for any root $\alpha \in \Phi \subseteq X(T)$ there exists an element $t \in T^F$ such that $\alpha(t) \neq 1$. We will denote by $\langle -,-\rangle : X(T) \times \check{X}(T)$ the usual perfect pairing between the character and cocharacter groups of $T$. Now as $F$ is a Frobenius endomorphism (not generalised) we have $F$ acts as $q\tau$ on the character group $X(T)$ respectively $q\check{\tau}$ on the cocharacter group $\check{X}(T)$. Here $\tau : X(T) \to X(T)$ and $\check{\tau} : \check{X}(T) \to \check{X}(T)$ are finite order automorphisms, of $\mathbb{Z}$-modules, such that $\tau(\Phi) = \Phi$ and $\check{\tau}(\check{\Phi}) = \check{\Phi}$.

Given $\alpha \in \Phi$ we denote by $k\geqslant 1$ the smallest integer such that $\check{\tau}^k(\check{\alpha}) = \check{\alpha}$. Given an element $\zeta \in \mathbb{F}_{q^k}^{\times}$ we then define an element $t_{\alpha}(\zeta) \in T^F$ by setting

\begin{equation*} t_{\alpha}(\zeta) = \check{\alpha}(\zeta)\cdot\check{\tau}(\check{\alpha})(\zeta^q) \cdots \check{\tau}^{k-1}(\check{\alpha})(\zeta^{q^{k-1}}) \end{equation*}

Note this is clearly fixed by $F$ because for any element $\check{\beta}(\lambda) \in T$ we have $F(\check{\beta}(\lambda)) = \check{\tau}(\beta)(\lambda^q)$.

Now we can fix a $W = N_G(T)/T$-invariant bilinear form $(-,-)$ on the real vector space $V = \mathbb{R} \otimes_{\mathbb{Z}} X(T)$ such that

\begin{equation*} \langle x, \check{\alpha}\rangle = \frac{2(x,\alpha)}{(\alpha,\alpha)} \end{equation*}

for all $x \in V$ and $\alpha \in \Phi$, see the proof of 7.1.8 in Springer's LAGs. By 10.3.2(iii) of Springer's LAGs, and our assumption that $G^F$ is not of type ${}^2\mathrm{A}_{2n}$, we have for any $1 < i < k$ that

\begin{equation*} 0 = (\tau^i(\alpha),\alpha) \Rightarrow 0 = \langle \tau^i(\alpha),\check{\alpha}\rangle = \langle \alpha,\check{\tau}^i(\check{\alpha})\rangle. \end{equation*}

Here I've just used the compatibility of the perfect pairing with $F$. Note this statement obviously fails in the case where $G^F$ is of type ${}^2\mathrm{A}_{2n}$ and also when the automorphism is a Suzuki/Ree automorphism. With this we see that

\begin{equation*} \alpha(t_{\alpha}(\zeta)) = \zeta^{\langle \alpha,\check{\alpha}\rangle}\zeta^{q\langle \alpha,\check{\tau}(\check{\alpha})\rangle}\cdots \zeta^{q^{k-1}\langle \alpha,\check{\tau}^{k-1}(\check{\alpha})\rangle} = \zeta^{\langle \alpha,\check{\alpha}\rangle} = \zeta^2. \end{equation*}

Hence, if there exists an element $\zeta$ such that $\zeta^2 \neq 1$ then $\alpha(t_{\alpha}(\zeta)) \neq 1$. In particular, if $q>3$ then the torus is non-degenerate.

In fact even when $q = 2$ one can have a non-degenerate maximal torus. By Proposition 3.6.7 of Carter’s book we must have $F$ is twisted. However one can easily check that $\mathrm{SU}_n(2)$, with $n \geqslant 3$, has a non-degenerate maximal torus.

  • $\begingroup$ Hi Jay, Thanks for you answer. I'll have to think some more to understand what you've written. I had a quick look at Carter's Proposition 3.6.1 (in Finite Groups of Lie type), and it seems to me that maximally split maximal tori may be non-degenerate for $q=3$ too (at least for some simple groups $G$ and some endomorphisms $F$). Does this seem reasonable? $\endgroup$
    – Nick Gill
    Apr 21, 2016 at 8:50
  • $\begingroup$ Yes, certainly! If $q=3$ and $F$ is split then for any root $\alpha$ if there exists a root $\beta$ with $\langle \alpha,\check{\beta}\rangle = -1$ then we have $\alpha(\check{\beta}(-1)) = (-1)^{\langle \alpha,\check{\beta}\rangle} = -1 \neq 1$ so the maximal torus is certainly non-degenerate. I didn't check the details but there should always be a root with this property. $\endgroup$
    – Jay Taylor
    Apr 21, 2016 at 9:04
  • $\begingroup$ The special unitary groups $\mathrm{SU}_n(q)$, with $n \geqslant 3$, always have a non-degenerate maximally split maximal torus. This is easy to see from a matrix representation. This is because all bar one of the elements are contained in the field $\mathbb{F}_{q^2}$, so one has enough wiggle room. However, when $q=2$ then one can check that the maximally split torus in ${}^3\mathrm{D}_4(2)$, ${}^2\mathrm{D}_n(2)$ and ${}^2\mathrm{E}_6(2)$ are all degenerate, even though they're non-trivial. $\endgroup$
    – Jay Taylor
    Apr 21, 2016 at 9:06
  • $\begingroup$ Great, thanks for this, very helpful. $\endgroup$
    – Nick Gill
    Apr 21, 2016 at 9:46
  • $\begingroup$ "Once one has this statement one can see that a non-degenerate maximal torus can have no rational regular elements, as pointed out by Jim." Certainly there are some maximal tori in some groups with rational regular elements (in which case the torus is non-degenerate). I guess that this sentence is maybe the opposite of what you mean? $\endgroup$
    – LSpice
    Jun 4, 2016 at 21:26

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