5
$\begingroup$

I know that for a complex Lie algebra, Schur's lemma can be used to show that any invariant bilinear form on a simple Lie algebra is a scalar multiple of the killing form, but Schur's lemma does not hold for $\mathbb{R}$, so I guess this isn't true for real simple Lie algebras. Does someone know a example? I can't seem to find one.

$\endgroup$
1
  • 3
    $\begingroup$ Well to have any chance to find such an example, one has to look at simple Lie algebras $\mathfrak g$ over $\mathbb R$ where the centroid $K := End_{ad_{\mathfrak g}}(\mathfrak g)$ is bigger than $\mathbb R$. The simplest example for this would be $\mathfrak{sl}_2(\mathbb C)$ but viewed as a six-dimensional real Lie algebra. Have you tried this one? Concretely, I would try the Killing form $\kappa(\cdot, \cdot)$ on it versus $\kappa(i\cdot, \cdot)$. $\endgroup$ Dec 2, 2020 at 22:26

2 Answers 2

5
$\begingroup$

Actually for a Lie algebra $\mathfrak{g}$ over a field $K$ and a field extension $K\subset L$, we have (by a straightforward linear algebra argument), with hopefully self-explanatory notation $$\dim_K \mathrm{InvBil}_K(\mathfrak{g})=\dim_L \mathrm{InvBil}_L(\mathfrak{g}\otimes_KL).$$

Consequence: for $\mathfrak{g}$ semisimple and $K$ of characteristic zero, $\dim_K \mathrm{InvBil}_K(\mathfrak{g})$ is equal to the number of simple factors of the "complexification" $\mathfrak{g}\otimes_K\bar{K}$.

In particular, $\dim_K \mathrm{InvBil}_K(\mathfrak{g})=1$ if and only if $\mathfrak{g}$ is absolutely simple.

Hence, to get counterexamples, it is enough to exhibit simple Lie algebras that are not absolutely simple. If $L$ is a finite extension of $K$ of degree $\ge 2$ and $n\ge 2$, $\mathfrak{sl}_n(L)$, viewed as Lie algebra over $K$, is such a Lie algebra.

$\endgroup$
2
$\begingroup$

Let $\mathfrak{g}\subset\mathfrak{gl}_n(\mathbb{C})$ be any complex Lie subalgebra such that $\mathfrak{g}_\mathbb{R}$, the “realification” of $\mathfrak{g}$ (just $\mathfrak{g}$ viewed as a real Lie algebra) has nondegenerate Killing form. In particular, $\mathfrak{g}\neq 0$. One case in which this property holds is whenever $\mathfrak{g}_\mathbb{R}$ is simple: see this post. (Also we have that $\mathfrak{g}_\mathbb{R}$ is a real simple Lie algebra iff $\mathfrak{g}$ is a simple complex Lie algebra, as they explain here.) A concrete example of this case may be the simple complex Lie algebra $\mathfrak{g}=\mathfrak{sl}_2(\mathbb{C})$.

Under the hypothesis “$\kappa_{\mathfrak{g}_\mathbb{R}}$ is nondegenerate” it happens that $\kappa_{\mathfrak{g}_\mathbb{R}}(\cdot,\cdot)$ and $\kappa_{\mathfrak{g}_\mathbb{R}}(i\cdot,\cdot)$ will be a counterexample (this idea was suggested in a comment by Torsten Schoeneberg). Specifically, we claim that these two are real-valued, real-bilinear, symmetric and invariant forms. On this case, then there is no real $\lambda$ such that $\lambda\kappa_{\mathfrak{g}_\mathbb{R}}(x,y)=\kappa_{\mathfrak{g}_\mathbb{R}}(ix,y)$ for all for $x,y\in\mathfrak{g}_\mathbb{R}$. Indeed, if such a $\lambda$ existed, then $0=\kappa_{\mathfrak{g}_\mathbb{R}}((i-\lambda)x,y)$, and by non-degeneracy of $\kappa_{\mathfrak{g}_\mathbb{R}}$ (that's our key hypothesis), it would follow that $(i-\lambda)x=0$, and since $\mathfrak{g}\neq 0$, that $\lambda=i$. But we were assuming $\lambda$ was real.

It is clear that $\kappa_{\mathfrak{g}_\mathbb{R}}(\cdot,\cdot)$ satisfies the required properties. It is also clear that $\kappa_{\mathfrak{g}_\mathbb{R}}(i\cdot,\cdot)$ is real-valued and real-bilinear. The proof of symmetry and invariance of $\kappa_{\mathfrak{g}_\mathbb{R}}(i\cdot,\cdot)$ is the typical given for a Killing form. Plus one observation: for $x\in\mathfrak{g}_\mathbb{R}$, the operators $\operatorname{ad}_{\mathfrak{g}_\mathbb{R}}(x)$ and “multiplication by $i$” (which is a real-linear endomorphism on the realification of any complex vector space) commute, due to the fact that the Lie bracket on $\mathfrak{g}_\mathbb{R}$ is just the commutator of matrices in $\mathfrak{gl}_n(\mathbb{C})$.

$\endgroup$
1
  • 1
    $\begingroup$ Everything you say is correct, I just want to point out that as soon as a Lie algebra (over any field with more than two elements) has two or more simple factors, the Killing form already is no longer the unique invariant bilinear form up to scaling, because one can "scale differently" on each of the factors. So we might as well restrict our attention to the simple case from the beginning, as OP does. $\endgroup$ May 13 at 17:05

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.