Kernel of adjoint of Lie algebra

Question

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?

Answer 1

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).