# Are the automorphism groups of simple symmetric cones algebraic groups?

This question arises when I tried to understand Chapter 2 of the celebrated book "Smooth compactification of locally symmetric varieties" by Ash–Mumford–Rapoport–Tai.

The setting is as follows: consider a finite dimensional real vector space $$V$$ and a symmetric cone $$C$$ in $$V$$ (symmetric=open, self-dual and homogeneous). Then they proceed to consider the automorphism group $$G$$ of $$(V,C)$$, namely the group of linear automorphisms of $$V$$ that preserve $$C$$.

It is fairly easy to show that $$G$$ is real reductive, in the sense that $$G\subseteq \mathrm{GL}(V)$$ is closed in the Euclidean topology and is stable under conjugate transpose.

But when they considered the boundary components of $$C$$, they implicitly use that $$G$$ is an algebraic group (or more precisely, there is an algebraic group $$\mathcal{G}$$ over $$\mathbb{R}$$ such that $$G=\mathcal{G}(\mathbb{R})^+$$). For example, this is used on Page 54 to guarantee the representability of $$\mathrm{Norm}(C_0)$$. I cannot really figure out a proof of the algebraicity of $$G$$. By some easy manipulation, we can readily reduce the problem to simple symmetric cones.

So my question is: is the automorphism group of a simple symmetric cone an algebraic group?

In the classical case, this seems easy and follows from the explicit computations in Faraut–Korányi's book. But what about the semi-classical case and the exceptional case?

• Any linear reductive Lie group is algebraic. One way so see this is to use the classification. The other is to use reductivity on the action of G on the polynomials to find that the Zariski closure of G has the same dimension as G.
– user473423
Commented Aug 29, 2022 at 15:52
• @Echo it depends on the exact meaning of "algebraic". For instance, there is no real algebraic group $G$ such that $G(\mathbf{R})$ is isomorphic as a Lie group to $\mathrm{PSL}_2(\mathbf{R})$.
– YCor
Commented Aug 29, 2022 at 15:59
• @Echo I think both approaches only work if we know that G is algebraic a priori. For a real reductive group defined as a a matrix group closed under conjugate transpose, I do not really think there is a full classification (Maybe I'm wrong?). In the book jmilne.org/math/CourseNotes/LAG.pdf P160, Milne claimed what you claim. But the reference he referred to is totally irrelevant. So if what you claim is correct, could you please point out a reference? btw, I also noticed that there is a related discussion mathoverflow.net/questions/28849/… Commented Aug 29, 2022 at 18:47
• @YCor You mean the real points of the group scheme $PSL_2$? No probably you mean $SL_2({\mathbb R})/\pm 1$, which is the connected component. Yes indeed, I interpreted the question as a question on the connected component. My thinking was that you complexify the group and then you are in the classification of complex algebraic groups which essentially coincides with the classification of complex Lie algebras.
– user473423
Commented Aug 30, 2022 at 6:20
• @Echo yes of course, I don't mean the $\mathbf{R}$-points of $\mathrm{PSL}_2$ (one should never refer to $\mathrm{PSL}_n$ as a group scheme because of these ambiguities and because it's the same as the group scheme $\mathrm{PGL}_n$ which doesn't prompt these issues).
– YCor
Commented Aug 30, 2022 at 8:49

I think I figure a proof after discussing with Yu Zhao.

First of all, by Koecher-Vinberg theorem, we can put a structure of Euclidean Jordan algebra on $$V$$ so that the closure $$\bar{C}$$ of $$C$$ is the set of squares in this Jordan algebra. To make a distinction, we denote the Jordan algebra as $$A$$ (the underlying vector space of $$A$$ is $$V$$).

Now the square map $$A\rightarrow A$$ is clearly polynomial, so the image $$\bar{C}$$ is semi-algebraic by Tarski-Seidenberg theorem.

Next we consider the map $$GL(V)\times A\rightarrow A$$. The inverse image of $$A-\bar{C}$$ is semi-algebraic, so is its intersection with $$GL(V)\times \bar{C}$$. The image of this intersection under the projection $$GL(V)\times A\rightarrow GL(V)$$ is semi-algebraic, again by Tarski-Seidenberg theorem. But the complement of the image is nothing but $$Aut(\bar{C},V)$$, which is equal to $$Aut(C,V)$$ as by the convexity of $$C$$, the interior of $$\bar{C}$$ is $$C$$.

Now let $$G'$$ be the Zariski closure of $$Aut(C,V)$$ in $$GL(V)$$. But the Zariski closure of $$Aut(C,V)$$ in $$End(V)$$ has the same dimension as $$Aut(C,V)$$, since $$Aut(C,V)$$ is semi-algebraic. Similarly, the Zariski closure of $$Aut(C,V)$$ in $$End(V)$$ has the same dimension as Zariski closure of $$Aut(C,V)$$ in $$GL(V)$$. It follows that $$G'$$ and $$G$$ have the same dimension. Now the Lie algebra consideration shows that $$G^+=G'^+$$, which is exactly what we expected.

Claim: A connected Lie subgroup of $$GL_n({\mathbb R})$$, which ist stable under $$\theta(x)=x^{-t}$$, is the 1-component of an algebraic group.

For this let $$\mathfrak g$$ be the Lie algebra and let $$\mathfrak z$$ be its center, which is the Lie algebra of the center of $$G$$. The center is stable under $$\theta$$ as $$\theta$$ is a homomorphism. Let $$X\in \mathfrak z$$, then $$\theta(X)=-X^t\in\mathfrak z$$, hence $$[X,X^t]=0$$, or $$XX^t=X^tX$$, i.e., $$X$$ is normal, hence digonalisable over $$\mathbb C$$ and so is every $$X$$ in the center, which therefore is simultaneously diagonalisable, hence a torus, hence algebraic. Let $$\mathfrak h$$ be the semisimple Lie algebra withe $$\mathfrak{g}=\mathfrak{z}\oplus\mathfrak{h}$$. Then $$\mathfrak h=\mathfrak{k}\oplus\mathfrak{p}$$, where the Killing form is positive on $$\mathfrak p$$ and negative on $$\mathfrak k$$. In $$gl_n(\mathbb{C})$$ the compact form of $$\mathfrak h$$ ist $$\mathfrak{h}^c=\mathfrak k\oplus i\mathfrak p$$. As ist Killing form is strictly negative, it exponentiates to a compact subgroup of $$GL_n(\mathbb{C})$$. As compact connected Lie groups are algebraic, the group $$G$$ is a product of two algebraic groups, hence algebraic.