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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?

glS
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Lukath
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    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 '20 at 15:12
  • What you claim to be true for all simple Lie algebras is actually only true for absolutely simple Lie algebras (which, however, $\mathfrak{o}(3)$ is). Cf. https://math.stackexchange.com/q/3931433/96384 – Torsten Schoeneberg Jan 11 '21 at 04:07

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Let $\mathfrak g$ be a semisimple Lie algebra with $\mathfrak g \simeq \bigoplus_{i=1}^n \mathfrak g_i$ and the $\mathfrak g_i$ absolutely 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!

Added: I just saw that Dietrich Burde had already answered a near-duplicate of this some years ago here.

Torsten Schoeneberg
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    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 '20 at 21:29
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    Well yes, and it's not even unique up to scaling. It is unique up to (possibly different) scaling on each of those two summands though. – Torsten Schoeneberg Aug 11 '20 at 22:55