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Do all simple Lie algebras have just one quadratic Casimir invariant as the Harish-Chandra isomorphism seems to imply or are there counter-examples?

bonif
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  • The quadratic Casimir element is defined by $\Omega =\sum_i X_iX^i \in U(L)$. So it is a single invariant, depending on the invariant form $B$. If $L$ is simple, $B$ is a multiple of the Killing form. – Dietrich Burde Sep 27 '23 at 19:23
  • For non simple semisimple Lie algebras there are counter-examples where the quadratic Casimir is not a single invariant. I want to know if there are such counterexamples for simple ones. – bonif Sep 27 '23 at 19:24
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    Up to the scaling $B(x,y)=\lambda \kappa(x,y)$ with the Killing form $\kappa$, $\Omega$ is unique for simple Lie algebras. – Dietrich Burde Sep 27 '23 at 19:27
  • How about the simple $so(3, 1)$ algebra? – bonif Sep 27 '23 at 19:33
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    Killing Form is unique up to scaling for absolutely simple Lie algebras, which by definition are the ones that stay simple after complexification. – Torsten Schoeneberg Sep 27 '23 at 19:59
  • Are there any other known examples of simple not absolutely simple lie algebra? – bonif Sep 27 '23 at 20:08
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    @bonif: Those are exactly the simple complex Lie algebras but viewed (via scalar restriction) as real Lie algebras. E.g. if consider the three-dimensional complex $\mathfrak{sl}_2(\mathbb C)$ as a six-dimensional real Lie algebra, it is actually isomorphic to $\mathfrak{so}(3,1)$. This example stands out a bit, as all other ones have no isomorphism to other classical Lie algebras. Cf. e.g. https://math.stackexchange.com/q/4158558/96384, https://math.stackexchange.com/q/3931489/96384, https://math.stackexchange.com/q/3068134/96384, https://math.stackexchange.com/q/3931433/96384. – Torsten Schoeneberg Sep 28 '23 at 05:42
  • Well that's not true then: What you mean (that the space of quadratic invariants has dimension $2$ instead of $1$) holds for any (excluding exceptional small $n$) $SL_n(\mathbb C)$ or $SO_n(\mathbb C)$ or $Sp_n(\mathbb C)$ considered as real Lie groups. – Torsten Schoeneberg Sep 29 '23 at 01:12
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    That is not a real distinction as $SO(3,1)$ is a complex group that we have "realified". Just because you started thinking of it from the point of view of real groups doesn't change the fact that it admits a complex structure. It is isomorphic to $PSL(2,\mathbb{C}) \times \mathbb{Z}_2$, I believe (basically, up to connected components and covers, it is $SL(2,\mathbb{C})$). – Callum Sep 29 '23 at 13:06
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    But as I am saying, it is a realified complex group. Its connected component is $PSL(2,\mathbb{C})$ viewed as a real group. – Callum Sep 29 '23 at 17:54
  • @Callum Thanks I see it now. So upon exponentiation of the Lie algebras the subalgebra $su(2)$ survives as subgroup in $SO^+(3,1)$ but not $sl(2,R)$ unlike in $SL(2,C)$ viewed as real , right? – bonif Oct 02 '23 at 10:31
  • @bonif I don't really know what you mean by that. Any subalgebra generates a subgroup (by the subgroups-subalgebras theorem). In this case, a $\mathfrak{su}(2)$ subalgebra will generate a subgroup isomorphic to $SU(2)$ or $SO(3)$ (I think probably only the latter) and a $\mathfrak{sl}(2,\mathbb{R})$ subalgebra will generate a subgroup of the form $SL(2,\mathbb{R})$, $PSL(2,\mathbb{R})$ or a cover of these. – Callum Oct 02 '23 at 12:06
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    @bonif Firstly $\mathfrak{so}(3,1)= \mathfrak{su}(2) \oplus\mathfrak{sl}(2,\mathbb{R})$ is not a valid decomposition, at least not as a Lie algebra direct sum. Secondly, the Lorentz group certainly does have such subgroups. They are the known as the Bianchi VIII type subgroups in the classification – Callum Oct 02 '23 at 15:51

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