The Jacobi identity of a Lie algebra says that $ad: \mathfrak g \to End(\mathfrak g)$ is a derivation. I am a bit emberassed but what is the easieast way to see that for every $X \in \mathfrak g$, $ad_X: \mathfrak g \to \mathfrak g, Y \mapsto [X,Y]$ is a Lie homomorphism, i.e., $[ad_X Y, ad_X Z] = ad_X([Y,Z])$ ?

Edit: The identity in the above line is wrong. It is $ad_X$ a derivation (endomorphism satisfying the product rule), not a Lie homomorphism (respecting the Lie bracket). I was a bit confused here, sorry.

  • 1
    $\begingroup$ Apply the Jacobi identity that you yourself mentioned (after writing out what everything means). $\endgroup$ – Tobias Kildetoft Jun 25 '15 at 19:47

Continuing with @TobiasKildetoft's comment: I really think one should think that the Jacobi identity is exactly the assertion that "ad" is a Lie algebra homomorphism.

So, for example, it is not a good idea to try to present the literal Jacobi-identity formula in a "symmetric" form (as some sources do), which would necessarily obscure the assertion that "ad" is a Lie algebra hom.

It is also unfortunate that few sources seem to let on that this is what the Jacobi identity is asserting!

So, don't think of the Jacobi identity as a formula that is required to hold, but as the property that "ad" is a Lie algebra hom, ... and then you can recover the formula whenever you want.

A (slightly) more interesting related question is about why the Lie algebras of classical matrix groups do satisfy the Jacobi identity! :) That is, we really should check the property rather than accidentally taking it for granted. Of course, then the structural sense of "ad" as derivative of "Ad", which in these linear cases is "conjugation", explain it, when one spells things out.

  • $\begingroup$ I totally agree with you and I am trying to build a better understanding of these notions. That is, I really want to understand what the real meaning is and your answer is very helpful because you stress whats really important. I just wonder why it is so easy to write down that $ad: \mathfrak g \to End(\mathfrak g)$ is a homomorphism, while i am a bit confused about how to see that $ad_X$ is a homomorphism. $\endgroup$ – Mekanik Jun 25 '15 at 20:15
  • $\begingroup$ Oh, I think I had a very basic misunderstanding: $ad_X$ is a linear morphism, it is NOT a Lie algebra homomorphism. $\endgroup$ – Mekanik Jun 25 '15 at 20:24

$\mathrm{ad}_X$ is not a Lie Homomorphism, but $\mathrm{ad}$ is. We can define a map $\mathrm{Ad}:G\to\mathrm{Aut}(G):=\lambda x.\lambda y.\ xyx^{-1}$, whose differential is then $\mathrm{ad}:TG\to T\mathrm{Aut}(G):=\lambda X.\lambda Y.\ [X,Y]$. The homomorphism property on this map is then precisely the Jacobi identity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.