# Ad/ad- invariant inner products on a Lie Algebra?

This might be a very trivial question for those of you well versed in Lie algebras, and if so I apologise... Any help is very much appreciated!

Let $G$ be a simply connected Lie group with Lie algebra $\mathfrak{g}$. If I have an $Ad_G$-invariant metric on $G$, it is my understanding that this gives rise to an $ad_{g}$-invariant inner product on $\mathfrak{g}$, let's call this $\langle . , . \rangle$. Is this inner product also invariant under $Ad_G$, so that

$\langle [w,u] , v \rangle+ \langle u , [w,v] \rangle = 0 \quad$ for $u,v,w\in \mathfrak{g}$

and

$\langle [w,u] , [w,v] \rangle = \langle u , v \rangle \quad$?

Or have I missunderstood something? It almost seems too simple to be true...

• An "$Ad_G$-invariant metric on $G$" doesn't make sense, since $Ad_G$ is a $G$-action on $\mathfrak{g}$, not on $G$; the corresponding action on $G$ is simply called conjugation. – YCor Nov 3 '17 at 22:05
• For any connected Lie group (simply connected is unnecessary here) there is a bijection between the set of left-invariant symmetric 2-tensors on $G$ and the set of quadratic forms on $\mathfrak{g}$ (mapping $\mu$ to its value on the tangent space at 1). This restricts to a bijection between bi-invariant symmetric 2-tensors on $G$ with $ad_g$-invariant quadratic forms on $\mathfrak{g}$. Also Riemannian metrics correspond to definite positive quadratic forms in this bijection. – YCor Nov 3 '17 at 22:06

The first identity is indeed coming from an ad-invariant metric, see the paper by G. Ovando Lie algebras with ad-invariant metric for a survey, with some low-dimensional examples. There are also several related questions here at this site:

Which Lie groups have Lie algebras admitting an Ad-invariant inner product?

There is a one-to-one correspondence between (a) left-invariant metrics on a connected simply connected Lie group $G$ and (b) Ad-invariant scalar products on the Lie algebra $\rm{Lie}(G)$.

Edit: The second identity, for all $w$, would imply $\langle u,v\rangle=0$, by taking $w=0$.

• Yes, I saw that, but I don't really understand if there are inner products such that both the statements above are true..? – Lou Nov 3 '17 at 19:44
• You can try a low-dimensional example, say, $\mathfrak{sl}(2)$, with the Killing form. The second identity cannot be true for all $w$. Take $w=0$. However, for the Heisenberg Lie algebra, the Killing form is identically zero, so it is true. – Dietrich Burde Nov 3 '17 at 19:45
• That indeed seems to very obviously contradict my second statement. I however thought this followed from the statement just below theorem 2.2 on page 3 in arxiv.org/pdf/1603.06528.pdf so now I'm confused... – Lou Nov 3 '17 at 19:50
• Just below Theorem $2.2$ it says "For every representation of a compact topological group" etc., so nothing about ad-invariant inner products of Lie algebras. – Dietrich Burde Nov 3 '17 at 19:56
• Gosh, you're completely right. I took the representation to be the adjoint and tried to derive my 'identity', which obviously doesn't work. Thank you so much! – Lou Nov 3 '17 at 20:03