2

This follows from a similar question before(Adjoint map is a Lie homomorphism), but I have a question other than the statement itself.

I am trying to understand the proof of Proposition 3.8 in Brian Hall's book Lie Groups, Lie Algebras, and Representations.

The proof shows that $ad_{[X, Y]}(Z) = [[X, Y], Z] = [X, [Y, Z]] - [Y, [X, Z]] = [ad_X, ad_Y](Z)$ where the second equality follows from the Jacobi identity.

My question is about the equation $[ad_X, ad_Y](Z) = [X, [Y, Z]] - [Y, [X, Z]]$. I recognized that this follows from the property that $[,]=−$ in a subspace $\mathfrak{g}$ of an associative algebra where $XY - YX \in \mathfrak{g}$ for all $X, Y \in \mathfrak{g}$, but I'm not sure whether this is the only possible Lie bracket on $End(\mathfrak{g})$. I don't think this uniqueness property has been mentioned anywhere before this proposition in Hall's book, but I am guessing it is the unique one.

May you tell me whether my guess is correct and explain why?

mlcv2022
  • 663
  • 3
    The Lie bracket described is the natural one to consider but of course we can define any valid Lie bracket we like. For example setting every bracket equal to 0 makes it an abelian Lie algebra – Callum Jul 02 '22 at 07:54
  • ${\rm End}(\mathfrak{g})$ is a vector space. As you know from the classification of, say, low-dimensional Lie algebras, there are in general several different Lie algebra structures on a given vector space. – Dietrich Burde Jul 02 '22 at 08:45

1 Answers1

3

It's not unique. Proposition 3.8 is stating that the specific map $ad:\mathfrak{g} \rightarrow End\left( \mathfrak{g} \right) $ is a Lie algebra homomorphism. While when we talk about Lie bracket in Lie algebra $End(\mathfrak{g} )$, there's no need to introduce $ad$, as long as the Lie bracket you defined satisfy some properties.

Even if for linear operator defined just as $ad_X$, we can find example to show the Lie bracket is not unique. One way to define Lie bracket(a map $\mathfrak{g} \times \mathfrak{g} \rightarrow \mathfrak{g} $) is $\left( ad_X,ad_Y \right) \longmapsto ad_Xad_Y-ad_Yad_X$ while another is $\left( ad_X,ad_Y \right) \longmapsto ad_Yad_X-ad_Xad_Y $.

narip
  • 67
  • I am a bit confused now. Is it that when we talk about the Lie algebra $End(\mathfrak{g})$ we always define its Lie bracket as $[X, Y] = XY - YX$? Otherwise I can't see $,=[,[,]]−[,[,]]$ – mlcv2022 Jul 02 '22 at 07:26
  • 1
    @mlcv2022 Let's just drop $ad$, and talk about define lie bracket in lie algebra $\mathfrak{g} $. We can define it as $\left[ g_1,g_2 \right] =g_1g_2-g_2g_1$ or $\left[ g_1,g_2 \right] =g_2g_1-g_1g_2$ while all satisfy the definition of lie bracket. This is what I want to convey. – narip Jul 02 '22 at 07:42
  • 1
    @mlcv2022 $\left[ ad_X,ad_Y \right] \equiv ad_Yad_X-ad_Xad_Y$ is a legal lie bracket, agree? But if we define this way, yeah, it seems that $ad$ is not lie algebra homomorphism. But for lie algebra homomorphism, I think exist one definition of lie bracket make the linear map to be lie algebra homomorphism is enough? – narip Jul 02 '22 at 07:50
  • Thank you, that clarifies a lot! – mlcv2022 Jul 02 '22 at 07:57