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Take an algebraic group $G$ defined over a finite field $K$. Suppose its Lie algebra $\mathfrak{g}$ is simple. It should follow that $G$ is almost-simple. (By this I mean not that $G(K)$ is simple -- I've heard there is a tricky though in some sense inessential counterexample -- but rather that $G$ has no normal algebraic subgroups of dimension greater than $0$ and less than $ \text{dim}\, G(K)$.)

Does anybody know of a concise and straightforward proof of this fact? One can't proceed as for $K=\mathbb{R}$, since $\exp$ is not defined (except on nilpotent subalgebras of $\mathfrak{g}$). I feel this should be simpler than studying the simplicity of $G(K)$. Is this so?

IMPORTANT EDIT: I just realized that I meant to ask the converse of this question. If $G$ is almost-simple in the sense of algebraic groups, how do you prove that $\mathfrak{g}$ is simple? I agree that the direction stated above is very easy.

Yet another important edit: there are exceptions to the converse! See the answers below. They all seem to come from non-trivial centers due to small characteristic. What I really need is the following:

Let $G$ be an almost-simple linear algebraic group. Let $V$ be a subvariety of $G$ going through the origin. Assume $0< dim(V)< dim(G)$. Let $\mathfrak{v}$ be the tangent space to $V$ at the origin. Prove that there is no ideal $\mathfrak{w}$ of $\mathfrak{g}$ containing $\mathfrak{v}$. (Or, what is the same: prove that the conjugates of $g \mathfrak{v} g^{-1}$ of $\mathfrak{v}$ span $\mathfrak{g}$.)

Yet another edit: Aha - not even that is true.

Well, can we at least give a quick proof, without case-work, of the following? Given a classical almost-simple group $G$ (defined as a subgroup of $\SL_n$ by means of equations with integer coefficients), show that there is a constant $c$ such that, for every field $K$ of characteristic $>c$, the Lie algebra $\mathfrak{g}$ of $G(K)$ is simple.

Surely there should be a reduction to the case over $\mathbb{C}$, or a proof from scratch?

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    $\begingroup$ $\mathrm{exp}$ is not defined even on nilpotent subalgebras in the case of small characteristic $p$, because then the denominator $n!$ is congruent to zero modulo $p$.... $\endgroup$ – Mikhail Borovoi Feb 16 '16 at 18:39
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    $\begingroup$ You can look for references in the book by Platonov and Rapinchuk, Section 7.2. $\endgroup$ – Mikhail Borovoi Feb 16 '16 at 18:58
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    $\begingroup$ You can find proofs in the book by Gunter Malle and Donna Testerman, Section 24.2. $\endgroup$ – Mikhail Borovoi Feb 16 '16 at 20:35
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See Bourbaki, Groupes et algèbres de Lie, IV.2.7, Corollary of Theorem 5 (due to Tits). We assume that $K$ has at least four elements and that the Lie algebra is absolutely simple. We assume also that $G$ is quasi-split, which is the case when $K$ is an algebraically closed field or a finite field. We construct a $BN$-pair taking for $B$ a Borel subgroup. Since the root system is irreducible, the Coxeter system is irreducible. If we denote by $G(K)^+$ the (normal) subgroup of $G(K)$ generated by the unipotent elements, Corollary of Theorem 5 says that $G(K)^+/(G(K)^+\cap Z(K))$ is either a nonabelian simple group or reduces to one element. This proves that then the group $G(K)^+$ has no nontrivial noncentral normal subgroups. When $G$ is quasi-split and simply connected, $G(K)^+=G(K)$ (Steinberg). Thus $G(K)$ has no nontrivial noncentral normal subgroups, which seems to answer the question over an algebraically closed field and over a finite field.

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  • $\begingroup$ Thanks, but, unfortunately, I really meant to ask for the converse. See below. Do you have an idea? $\endgroup$ – H A Helfgott Feb 17 '16 at 23:53
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This is relatively elementary, in the sense of not depending on detailed study of BN-pair structure and root system. It goes back to Chevalley's early work on linear algebraic groups and their Lie algebras. The books called Linear Algebraic Groups, by Borel, me, and Springer, treat this kind of question in similar ways. In my book, for example, see Chapter III, $\S9$ and $\S10$, especially Cor. 10.4A. By comparing the Lie algebra of $G$ with the tangent space at the identity element, one sees right away that $G$ and Lie$(G) = \mathfrak{g}$ have the same dimension. After a little more work, one sees that a closed normal subgroup $H$ of $G$ has as Lie algebra an ideal $\mathfrak{h}$ in $\mathfrak{g}$.

In particular, if $\mathfrak{g}$ is simple (usually required by the definition to be nonabelian), then $G$ can't have a proper nontrivial closed connected normal subgroup. In other words, $G$ is simple in the sense of algebraic groups. Simplicity (in some sense) of the group of rational points over a non-algebraically closed field such as a finite field would then require more careful study.

None of this depends on the nature of the field of definition or its characteristic, but when that field is finite you get a well-studied finite group such as $\mathrm{SL}_n(\mathbb{F}_q)$ which is close to being simple for most $q$ but of course may have a nontrivial center.

[The more difficult direction is to start with a simple algebraic group and then study its Lie algebra. The most comprehensive results here were obtained by G.D.M. Hogeweij in his old Utrecht thesis (partly published in Indag. Math.). But there was quite a bit of work done previously in special cases. One sees for instance that the exceptional algebraic group $E_8$ has a simple Lie algebra over any algebraically closed field.]

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    $\begingroup$ This is so simple! $\endgroup$ – Mikhail Borovoi Feb 17 '16 at 6:16
  • $\begingroup$ Thank you very much for this - unfortunately I just realized that what I meant to ask is what you call the more difficult direction. Is it possible to give a conceptual proof of the statement in that direction (that is: the Lie algebra of a group that is simple in the sense of algebraic groups is simple)? It need not cover twisted groups and the like - only algebraic subgroups of $\text{SL}_n$ need be considered - but I would like to avoid case-work as much as possible. $\endgroup$ – H A Helfgott Feb 17 '16 at 23:52
  • $\begingroup$ @HAHelfgott: This is not true in characteristics 2 and 3, see the two papers by Hogeweij. $\endgroup$ – Mikhail Borovoi Feb 18 '16 at 6:57
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    $\begingroup$ @HA Helfgott: Even when $p>3$, there is still a problem with special linear groups $G=\mathrm{SL}_n$ in case $p|n$, since then the Lie algebra of $G$ has a one-dimensional center consisting of scalar matrices. $\endgroup$ – Jim Humphreys Feb 18 '16 at 16:46
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    $\begingroup$ @HAHelfgott: Yes, it still occurs! Consider the scalar matrix $\mathrm{diag}(1,1,1)$, in characteristic 3 it has trace 0, hence it is contained in the Lie algebra $\mathfrak{g}=\mathfrak{sl}(3,K)$, and it is clearly central. Therefore, $\mathfrak{g}$ is not simple. $\endgroup$ – Mikhail Borovoi Feb 19 '16 at 8:54
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Your final request, even if augmented to require the locally closed subvariety $V$ to be smooth, is false over every field $k$ of characteristic 2 and 3, with counterexamples of any rank $n \ge 1$ over any $k$ of characteristic 2, with $V \subset G$ of very large dimension (unbounded as $n$ grows). So avoiding finite fields of bounded size is insufficient in characteristics 2 and 3. A source of counterexamples is certain "well-known" exceptional (non-central) isogenies through which Frobenius non-trivially factors, and these isogenies only exist in such characteristics.

I will describe counterexamples with $G = {\rm{Spin}}_{2n+1}$ for any $n \ge 1$, and a variant of the same ideas (but carried out by more conceptual methods based on the structure theory of semisimple groups rather than based on manipulations in linear algebra) gives counterexamples for ${\rm{Sp}}_{2n}$ ($n \ge 1$) and ${\rm{F}}_4$ over any field of characteristic 2 and for G$_2$ over any field of characteristic 3; see section 7.1 of the book "Pseudo-reductive groups" for a unified development of the relevant general principles (but again, the phenomenon at hand has been widely known for many decades).

Let $V = k^{2n+1}$ and $$q = x_0^2 + x_1 x_2 + \dots + x_{2n-1} x_{2n}.$$ The associated symmetric bilinear form $B_q(v,v') = q(v+v') - q(v)-q(v')$ has defect space $V^{\perp} = \{v \in V\,|\,B_q(v,\cdot) = 0\}$ that is a line since ${\rm{char}}(k)=2$ (this line is $k e_0$). On $V/V^{\perp}$, the induced bilinear form $\overline{B}_q$ is symplectic since ${\rm{char}}(k)=2$. Thus, we get a natural composite homomorphism $$f: G = {\rm{Spin}}_{2n+1} \rightarrow {\rm{SO}}_{2n+1} \rightarrow {\rm{Sp}}(\overline{B}_q) = {\rm{Sp}}_{2n}.$$

Let $T$ be a split maximal $k$-torus in $G$, and let $\Delta$ be the base of $\Phi(G, T)$ associated to a choice of positive system of roots $\Phi^+$. The multiplication map $\prod_{a \in \Delta} \mathbf{G}_m \rightarrow T$ defined by $(\lambda_a) \mapsto \prod a^{\vee}(\lambda_a)$ is an isomorphism since $G$ is simply connected. In particular, ${\rm{Lie}}(T)$ is the direct sum of the "coroot lines" ${\rm{Lie}}(a^{\vee}(\mathbf{G}_m)) = {\rm{Lie}}(a^{\vee})(\partial_t|_{t=1})$.

The kernel $\mathfrak{n} := \ker {\rm{Lie}}(f) = {\rm{Lie}}(\ker f)$ is an ${\rm{Ad}}_G$-stable subspace of $\mathfrak{g}$, and it is nonzero and proper. Explicitly, it is the direct sum of the root lines for the short roots and the coroot lines ${\rm{Lie}}(a^{\vee}(\mathbf{G}_m))$ for the coroots associated to the short roots in $\Delta$. (In fact, $\mathfrak{n}$ is the unique minimal non-central ${\rm{Ad}}_G$-stable subspace of $\mathfrak{g}$.) Hence, $\mathfrak{n} = {\rm{T}}_e(V)$ for the smooth locally closed subvariety $V \subset G$ through the identity $e$ given as the closed subvariety of the open cell associated to $\Phi^+$ by forming the direct product (embedding via multiplication into $G$) of the $T$-root groups for the short roots and the direct factors $a^{\vee}(\mathbf{G}_m)$ for short $a \in \Delta$.

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  • $\begingroup$ Aha, interesting - thanks! Now I'm at a loss, however. Say you have a proper, irreducible subvariety $V$ of $G$ ($\dim V>0$) such that its tangent space $\mathfrak{v}$ at the origin is invariant under $Ad_G$. Does it follow that $g V g^{-1} = V$ for all $g$ in $G$? Is that even possible, given that $G$ is simple? It certainly isn't when the stabilizer $Stab(V) = \{h\in G: h V = V\}$ is non-trivial, since then the stabilizer would be a non-trivial, normal, proper subgroup of $G$. $\endgroup$ – H A Helfgott Feb 22 '16 at 10:34
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    $\begingroup$ @HAHelfgott: In the examples above $V$ is not normalized by $G$. Pick short $a\in\Phi^+$, so $V=U^{-}\times T\times U^{+}$ where $U^{-}$ and $U^{+}$ are direct products in any order of short roots in $\Phi^{\pm}$ (consider unipotent radicals of Borels of $G_{<}$ as in Prop. 7.1.7(1) in "Pseudo-reductive groups"). The reflection $r_a$ preserves $\Phi^{+} - \{a\}$, so conjugation on $V$ by a representative $n_a\in N_G(T)(k)$ of $r_a$ only swaps where $U_{\pm a}$ appear in the direct product. But $U_{-a}TU_a \ne U_aTU_{-a}$ by ${\rm{SL}}_2$-considerations, so $n_aVn_a^{-1}\ne V$. $\endgroup$ – nfdc23 Feb 23 '16 at 1:27
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    $\begingroup$ Interesting. Is $G$ still $\Spin_{2n+1}$ here? Just for concreteness - can you come up with a "small" enough example that you can write this in terms of matrices? I'd really like to see what is going on here. Also, see the new question at the end of the original post. Thanks! $\endgroup$ – H A Helfgott Feb 25 '16 at 15:57
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    $\begingroup$ @HAHelfgott: Yes, $G$ is Spin$_{2n+1}$ or Sp$_{2n}$ (or F$_4$ or G$_2$). Personally, I find descriptions in terms of root groups a much more illuminating to "see" what is going on than to calculate with matrices (which is more concrete but not necessarily instructive). For $n=2$ this is an exceptional endomorphism of Sp$_5$, so you can write down $V$ in that case using the standard root groups. (Also, in my preceding comment I should have chosen $\Phi^+$ relative to which $a$ is a simple root, by the way.) $\endgroup$ – nfdc23 Feb 25 '16 at 16:36
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    $\begingroup$ I don't see how such a relationship helps us to make progress on ruling out the breakdown of simplicity of Lie algebras, so I hadn't considered things like that to be of interest. To make progress on such questions of simplicity one must use structure theory via roots and weights (as has been thoroughly worked out in the literature on modular Lie algebras). That explains conceptually why one should only consider $p \ge 5$ and $p \nmid (n+1)$ for type A$_n$, and happily those are the only obstructions: see Theorem 18.3 of homepages.warwick.ac.uk/~masdf/modular/ln_11mar.pdf $\endgroup$ – nfdc23 Feb 28 '16 at 2:39

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