Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ be a Lie group and $\mathfrak{g}$ its Lie algebra.

The adjoint representation of the Lie algebra $\mathfrak{g}$ is defined as:

$$ \text{ad: } \mathfrak{g} \rightarrow \text{End}(\mathfrak{g}), X \mapsto [X,\cdot] $$

Now, it holds true that

$$ \text{ker ad} = \mathfrak{z}(\mathfrak{g}) = \{X \in \mathfrak{g} : [X,Y] = 0 \quad\forall\; Y \in \mathfrak{g}\}.$$

On the other hand, the definition of the kernel of this homomorphism is (at least in my mind)

$$ \text{ker ad} = \{ X \in \mathfrak{g} : [X,\cdot] = \text{id}, \text{ i.e. } [X,Y] = Y \quad \forall \;Y \in \mathfrak{g} \}, $$

since the group identity in the endomorphism group is the identity-map.

Evidently, the two sets are not the same, but where is my mistake?

share|improve this question
1  
This is a homomorphism of Lie algebras, not groups. The kernel is as a linear map. i.e. the preimage of zero. –  KotelKanim Aug 8 '12 at 17:17
add comment

1 Answer 1

up vote 2 down vote accepted

As KotelKanim mentionned in the comments, the adjoint homomorphism is a Lie algebra homomorphism, and as such, the kernel is the preimage of the zero vector. The second "definition" of the kernel you gave is what you would expect if Lie algebras were groups - which is not the case. So that much is settled.

Let me just point out two facts:

  • The fact that the adjoint homomorphism really is a Lie algebra homomorphism (and not merely a linear map) is because of the Jacobi identity. In my opinion, this is the strongest motivation for this axiom of a Lie algebra.
  • (Here, I assume the Lie algebra is finite-dimensional.) What you have shown is that, any Lie algebra with trivial center admits a faithful representation. However, much more is true. Indeed, Ado's theorem states that any Lie algebra over a field of characteristic zero admits a faithful representation. Later, the restriction on the characteristic was removed, and so we get the following (and perhaps surprising) result: Every Lie algebra admits a faithful representation. (For more information, see e.g. this Wikipedia article).
share|improve this answer
    
Ado's theorem only works for finite dimensional Lie algebras. I'm not sure about the generalization you mention. –  Jason DeVito Aug 8 '12 at 18:39
    
@JasonDeVito I'm sorry, I was assuming finite dimension. I will add this. –  M Turgeon Aug 8 '12 at 18:51
    
@JasonDeVito The generalisation to positive characteristic was proven by Iwasawa, see for example Jacobson's Lie algebras. –  M Turgeon Aug 8 '12 at 18:55
1  
@Jason: it is still true that every Lie algebra admits a faithful representation (not necessarily finite-dimensional). This is a corollary of Poincaré-Birkhoff-Witt: see math.stackexchange.com/questions/3031/… . –  Qiaochu Yuan Aug 8 '12 at 20:31
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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