Let $G \subset GL(n,\mathbb{R})$ be a finite-dimensional matrix Lie group. I am using the inner product $$\langle \xi, \eta \rangle := \frac{1}{2}tr(\xi^T\eta)$$ for $\xi, \eta \in \mathfrak{g}$. And of course the induced Lie bracket is $$[\xi,\eta]=\xi\eta-\eta\xi$$ where $\xi\eta$ is matrix multiplication. This is the common layout for matrix Lie groups laid out in Hall's Lie Groups, Lie Algebras, and Representations.

For $G=SO(3)$, we know that $\langle \xi, [\xi, \eta]\rangle = 0$ every single time. I understand the Lie bracket is a kind of a generalization of the cross product, so I was wondering if this holds for arbitrary real matrix Lie groups with this particular inner product?

  • $\begingroup$ a 3 by 3 skew symmetric matrix is defined by three (real) numbers which make a vector...., and the bracket is the traditional cross product of the two vectors, put back as skew symmetric. $\endgroup$
    – Will Jagy
    Dec 13, 2022 at 4:23
  • $\begingroup$ en.wikipedia.org/wiki/Skew-symmetric_matrix#Cross_product $\endgroup$
    – Will Jagy
    Dec 13, 2022 at 4:25
  • $\begingroup$ @WillJagy Of course, funny how I missed that. Still, I am curious if this holds for all matrix Lie groups equipped with this particular metric. Even better, is there an inner product where it always holds? $\endgroup$ Dec 13, 2022 at 5:54
  • 2
    $\begingroup$ In general that inner product won't be well behaved with respect to the bracket but something like $\langle\xi,\eta\rangle := \mathrm{tr}(\xi\eta)$ will be. If the Lie abgebra is simple this will be in fact a scale of the Killing form $\mathrm{tr}(\mathrm{ad}_\xi \circ \mathrm{ad}_\eta)$. These "trace forms" are invariant in the sense that $\langle[\xi,\psi],\eta\rangle = \langle\xi,[\psi,\eta]\rangle $. Thus we can see $ \langle\xi,[\xi,\eta]\rangle = \langle[\xi,\xi],\eta\rangle = 0$. Note this holds for any invariant inner product. $\endgroup$
    – Callum
    Dec 13, 2022 at 12:18

1 Answer 1



$\xi =\pmatrix{1&0\\1&1}, \eta = \pmatrix{2&-1\\0&1}$, $G= GL_2(\mathbb R)$ or any subgroup thereof which contains those elements.

As Callum notes in a comment, if instead you had taken (any scalar multiple of) the usual trace form $\langle x, y \rangle := tr(xy)$, the property would hold.

FYI, if $\mathfrak g$ is any Lie algebra over a field $K$, and $V$ a representation of $\mathfrak g$, there is the important concept of an invariant bilinear form $B: V\times V\rightarrow K$, which is one such that


for all $X \in \mathfrak g$, $v,w \in V$. For the adjoint representation $V= \mathfrak g$, this simplifies to

$$B([v,X], w) = B(v,[X,w])$$

which of course entails $B(X, [X,w])=0$ for all $X,w \in \mathfrak g$. It is well-known that the usual trace form mentioned above is invariant on all of $\mathfrak{gl}_n(K)$. The counterexample shows that your form is not.

  • $\begingroup$ Interesting how this is the same metric I had, but with the transpose removed. I will read up on this killing form. Thank you for the answer. $\endgroup$ Dec 13, 2022 at 18:47
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    $\begingroup$ @SpencerKraisler The reason for that is that $\mathfrak{so}(n)$ is precisely the set of matrices with $A^T = -A$ so your form is $-\frac{1}{2}\mathrm{tr}(\xi\eta)$. This choice ensures that it is positive definite, a requirement for an inner product. I should have made this more clear in my other comment but the trace form is not positive definite so is a bilinear form but not technically an inner product. It can be negative definite (as it is in this example) and then we say the Lie algebra is compact and we can make an invariant inner product by multiplying by $-1$. $\endgroup$
    – Callum
    Dec 14, 2022 at 0:04

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