Are the automorphism groups of simple symmetric cones algebraic groups?

Question

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?

Answer 1

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.

Answer 2

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.