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Let $G$ be a semi-simple algebraic group over a field $K$, I am considering a question about whether there exists a finite set of semi-simple $K$-subgroups, say $H_1,...,H_r$, such that for any semi-simple $K$-subgroup $H\subset G$, $H$ is conjugate to one of the $H_i$ by an element in $G(K)$?

I know the answer is "no" for $K=\mathbb{Q}$. In fact, for any quaternion algebra $\mathcal{D}$ over $\mathbb{Q}$, the group $\mathrm{SL}(\mathcal{D})$ can be embedded in $\mathrm{SL}_4(\mathbb{Q})$. As $\mathcal{D}$ varies, we got infinitely many semi-simple subgroups $\mathrm{SL}(\mathcal{D})$ of $\mathrm{SL}_4(\mathbb{Q})$ that are not isomorphic to each other, therefore not conjugate to each other.

Now I would like to ask what happens if $K=\mathbb{R}$, I think the answer is yes, but how to prove it? And what about $K=\mathbb{Q}_p$ or some other fields?

Thanks very much!

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    $\begingroup$ A positive answer in the case when $K$ is algebraically closed of characteristic zero implies a positive answer for real and $p$-adic fields, by the finiteness results of Borel-Serre. For the algebraically closed case, I guess it's well-known (I guess it's known that $G$ has finitely many maximal connected subgroups up to conjugacy - there are finitely many parabolic ones, and the others are reductive; if the latters are finitely many, then the finiteness of semisimple subgroups up to conjugacy follows by induction) $\endgroup$
    – YCor
    Commented Apr 28, 2016 at 15:16
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    $\begingroup$ The algebraically closed case seems to be well-known, and actually Dynkin provides a recipe (see books.google.fr/…). On Google I also found algorithms about the real case. But a direct proof a the finiteness result would be interesting (without going into this combinatorial cuisine) $\endgroup$
    – YCor
    Commented Apr 28, 2016 at 16:01
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    $\begingroup$ @YCor: A modern exposition of Dynkin's results on maximal subalgebras of semisimple Lie algebras is given in Chapter 6 of the survey: books.google.fr/books/about/… $\endgroup$
    – Mikhail Borovoi
    Commented Apr 28, 2016 at 21:45

2 Answers 2

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For fields of characteristic zero one can argue as follows: Assume first that $K$ is algebraically closed. Since semisimple subgroups of $G$ correspond bijectively to semisimple subalgebras of $\mathfrak g={\rm Lie}\,G$ it suffices to consider the same problem in $\mathfrak g$. The advantage is that the subalgebras of $\mathfrak g$ form the points of a variety $SA(\mathfrak g)$ (a closed subvariety of the Grassmannian) on which $G$ acts. Now we use that semisimple subalgebras are rigid, i.e., if two are "nearby" they are conjugated. The proof uses the following idea: the tangent space of $SA(\mathfrak g)$ in some $\mathfrak h$ can be interpreted as the space of cocycles while the tangent space to the orbit as coboundaries. Vanishing of some cohomology group (that's where semisimplicity is used) yields equality of both tangent spaces. I remember a survey of Kraft on this topic. This means now that the orbits of semisimple algebras are open in $SA(\mathfrak g)$. Clearly, there are only finitely many of them. That $G$ is semisimple is immaterial. So we get: any linear algebraic group contains only finitely many conjugacy classes of semisimple subgroups.

If $K$ is not algebraically closed then the same result would follow if $(G/H)(K)$ decomposes into finitely many $G(K)$-orbits for any semisimple $H$. This is known for local fields by Borel-Serre.

In positive characteristic, the finiteness result above is wrong: there are continuous families of pairwise non conjugate semisimple subgroups. But it may be still true that there are only finitely many classes of maximal subgroups. This probably follows from works of Seitz et at.

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    $\begingroup$ I was aware of the cohomological argument, but I don't see how it works. This argument (at least, the one I have in mind, using $H^1(H,\mathfrak{g})=0$) shows that two close semisimple subalgebras are conjugated under a linear automorphism of $\mathfrak{g}$. But you need to have them conjugated by some element of $G$ (or, essentially equivalently, by some Lie algebra automorphism of $\mathfrak{g}$). $\endgroup$
    – YCor
    Commented Apr 28, 2016 at 21:09
  • $\begingroup$ Very nice answer. @YCor, if $G$ where semisimple (as asked originally) then it was of finite index in $\text{Aut}(\mathfrak{g})$ and you'd be happy. For the general case (char 0) you can always conjugate $H$ into your favorite Levy and work there (all Levy's are conjugated). $\endgroup$
    – Uri Bader
    Commented Apr 29, 2016 at 7:21
  • $\begingroup$ Friedrich, do you have in mind a simple to understand example for your statement in char$>0$? I would like to see it please. $\endgroup$
    – Uri Bader
    Commented Apr 29, 2016 at 7:25
  • $\begingroup$ @YCor: The tangent space to the Grassmannian is ${\rm Hom}(\mathfrak h,\mathfrak g/\mathfrak h)$. So the cohomology group to look at should be $H^1(\mathfrak h,\mathfrak g/\mathfrak h)$. $\endgroup$
    – Friedrich Knop
    Commented Apr 29, 2016 at 8:01
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    $\begingroup$ @user89334: No concrete example but one can get one as follows: Start with an irreducible faithful representation $U$ of $H$ with $\dim H^1(H,K)\ge2$. Then each $u\in H^1$ gives rise to an extension $U(u)$ of $U$ by $K$. Any two of them are isomorphic if and only if their parameters $u$ differ by a scalar. Identify $U(u)$ with $K^n$. Then most images of $H$ in $GL(U(u))=GL(n)$ will not be conjugate. $\endgroup$
    – Friedrich Knop
    Commented Apr 29, 2016 at 8:13
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In case it is useful to the OP, here is a reference:

Richardson, R. A rigidity theorem for subalgebras of Lie and associative algebras. Illinois J. Math. 11 1967 92–110.

Proposition 12.1 seems to be the relevant result.

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