# Killing form of orthogonal Lie algebras

I have a question about the uniqueness, up to scale factor, of the Killing form on a Lie algebra $$\mathfrak{g}$$. I know that it is defined as $$B(X,Y)=\operatorname{tr}(\operatorname{ad}_X\circ \operatorname{ad}_Y),$$ for every $$X,Y\in\mathfrak{g}$$. It is known that, if $$\mathfrak{g}$$ is a simple Lie algebra, any bilinear, symmetric non-degenerate quadratic form which is $$\operatorname{ad}$$-invariant equals $$B$$, up to multiplications for a constant. Now, we know that the Lie algebra $$\mathfrak{o}(n)$$ of the orthogonal group $$O(n)$$ is simple if $$n\neq 4$$ and $$\mathfrak{o}(4)\cong\mathfrak{o}(3)\oplus\mathfrak{o}(3),$$ i.e. $$\mathfrak{o}(4)$$ is semisimple.

In the case $$n=4$$, what can we say about the Killing form? Is it the unique quadratic form as above, up to multiplications for a constant, although $$\mathfrak{o}(4)$$ is not simple?

• The Killing form is a unique multiple of the trace form for all orthogonal Lie algebras, yes. But there is no longer only a unique non degenerate ad-invariant quadratic form on $\mathfrak{so}(4)$. – Dietrich Burde Aug 11 at 15:12

Let $$\mathfrak g$$ be a semisimple Lie algebra with $$\mathfrak g \simeq \bigoplus_{i=1}^n \mathfrak g_i$$ and the $$\mathfrak g_i$$ simple.
Check that for any ad-invariant bilinear form on this, the $$\mathfrak g_i$$ are pairwise orthogonal (I used that simple Lie algebras are perfect for this, maybe there's an easier proof). So for such a form to be non-degenerate, it has to restrict to something non-degenerate on each $$\mathfrak g_i$$. Hence the restriction to each $$\mathfrak g_i$$ is a scaled version of the respective Killing form.
However, you can scale with different constants $$c_i \neq 0$$ on each summand!
• Perfect, thank you! So, in particular, this implies that there is no unique $\operatorname{Ad}(O(4))$-invariant inner product on $\mathfrak{o(4)}$, doesn't it? – Lukath Aug 11 at 21:29