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?
Best Answer
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.