1
$\begingroup$

There is any relation between semisimple lie algebras and symmetric cones? I'm saying this because the classification of the euclidean Jordan algebras, dual by Koecher-Vinberg theorem to homogeneous self-dual convex cones, is very similar to the well-known classification for semisimple Lie algebras in terms of root systems (although not identical). Maybe a way of phrasing differently the idea might be what is the connection between symmetric spaces and symmetric cones? I'm trying to proof semisimplicity of a Lie algebra whose generated Lie group is the group of automorphisms of a cone, that's the interest behind my question.

$\endgroup$
4
  • 1
    $\begingroup$ Keyword: Weyl chamber $\endgroup$
    – YCor
    Aug 14, 2022 at 12:42
  • $\begingroup$ Please explain more, don't see clearly what does it have to do with homogeneous self-dual cones. $\endgroup$ Aug 15, 2022 at 8:29
  • 1
    $\begingroup$ You wrote "between semisimple lie algebras and convex cones". Please edit to clarify if you're exclusively interested in relations with homogeneous self-dual cones. $\endgroup$
    – YCor
    Aug 15, 2022 at 8:35
  • 1
    $\begingroup$ You're right. Thanks. $\endgroup$ Aug 16, 2022 at 10:17

1 Answer 1

4
$\begingroup$

I'm not super familiar with the theory of Jordan algebras or self-dual homogeneous cones so I might be barking up the wrong tree with this answer but here goes. I don't think the cones I'm about to describe are convex but it is a link between semisimple Lie algebras, Jordan algebras and homogeneous cones so it might be helpful.

Consider a semisimple Lie algebra and $\mathfrak{g}$ take a pair of complementary parabolic subalgebras $\mathfrak{p}$, $\mathfrak{q}$. Complementary meaning that $\mathfrak{p} \oplus \mathfrak{q}^\perp = \mathfrak{g}$ where $\mathfrak{q}^\perp$ is the nilradical of $ \mathfrak{q}$. Assume moreover the nilradicals are abelian so that $\mathfrak{p}^\perp \oplus (\mathfrak{p} \cap \mathfrak{q}) \oplus \mathfrak{q}^\perp$ is a Lie algebra grading $\mathfrak{g}_{-1}\oplus\mathfrak{g}_{0}\oplus\mathfrak{g}_{1}$.

If you add in a choice of Cartan involution $\tau$ which swaps this grading around ($\tau\mathfrak{g}_{1} = \mathfrak{g}_{-1}$, $\tau\mathfrak{g}_{0} = \mathfrak{g}_{0}$) we can make a triple product on $\mathfrak{g}_{1}$: $$ B(X,Y,Z) := \frac{1}{2}[[\tau Y, X],Z] \in \mathfrak{g}_{1}.$$

In general this is a Jordan triple system but if we add the requirement that $\mathfrak{p}$, $\mathfrak{q}$ are in the same conjugacy class this becomes a Jordan algebra. Then you get a cone by considering the quadratic $P(X)(Y) := B(X,Y,X) = \frac{1}{2}[[\tau Y, X],X]$ and looking at the elements where $P(X)$ is invertible (or maximum rank in the non-self-dual case). Indeed you can decompose $\mathfrak{g}_{1}$ into orbits based on the rank of $P(X)$.

Then the classification of these possibilities is precisely the classification of parabolic subalgebras whose nilradicals are abelian and which are conjugate to their complementary parabolics. It turns out that there are 1 of these each in the complex Lie algebras of type $A_n$, $B_n$, $C_n$, $D_n$ ($n$ odd), $E_7$ and 3 in $D_n$ ($n$ even). You also get some real versions of this but not in every real form and the complex orbits (on $\mathfrak{g}_{1}$) may split into several ones each. Effectively we are looking at the classification of self-dual symmetric R-spaces and we can view these cone structures as prehomogeneous cones (i.e. open dense orbits) in their tangent space (here $\mathfrak{g}_{1} \cong \mathfrak{g}/\mathfrak{p} \cong T_\mathfrak{p}M$ and $\mathfrak{g}_0$ is the Levi subalgebra which acts on $\mathfrak{g}_{1}$).

More detail can be found in Gindikin and Kaneyuki's "On the automorphism group of the generalised conformal structure of a symmetric R-space".

$\endgroup$
3
  • $\begingroup$ Interesting answer! I am used to the terminology "opposite" for parabolic subalgebras where you use "complementary". Maybe just different specialities, though! $\endgroup$
    – LSpice
    Aug 18, 2022 at 21:59
  • 2
    $\begingroup$ I tend to use "opposite" to refer to a specific complementary parabolic subalgebra with respect to some CSA/root system but that might just be a distinction in my head that no-one else uses. The rationale being that it feels like there should only be one "opposite" thing but "complementary" feels less unique. $\endgroup$
    – Callum
    Aug 18, 2022 at 22:53
  • 2
    $\begingroup$ Re$\newcommand\m{\mathfrak}$, fair enough! I use "opposite with respect to $\m t$" (or "… to $\m m$", where $\m m = \m p \cap \m q$ is the common Levi) in that sense, and "opposite" to mean "opposite with respect to some $\m t$" (which is then the same as "complementary"). For me, complementary risks confusion since, as you point out, it's not $\m q$ but $\m q^\perp$ that's a complement to $\m p$ in the "usual" sense. But, whatever background one is coming from, you certainly clearly define what you mean, so no confusion can result. $\endgroup$
    – LSpice
    Aug 19, 2022 at 2:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.