📂Representation Theory

Lie Algebra Isomorphism

Definition¹

Let $\mathfrak{g}$ and $\mathfrak{h}$ be Lie algebras. If the linear map $\phi : \mathfrak{g} \to \mathfrak{h}$ satisfies the following, it is called a Lie algebra homomorphism.

$$ \phi([X, Y]_{\mathfrak{g}}) = [\phi(X), \phi(Y)]_{\mathfrak{h}}, \qquad \forall X, Y \in \mathfrak{g} $$

If $\phi$ is bijective, it is called a Lie algebra isomorphism. If $\mathfrak{h} = \mathfrak{g}$, then $\phi$ is called a Lie algebra automorphism.

Remark

Since a Lie algebra is equipped with a binary operation $[\cdot, \cdot]$, it is natural to consider maps that preserve this operation.

Theorem

Let $\mathfrak{g}$ and $\mathfrak{h}$ be Lie algebras. If $\phi : \mathfrak{g} \to \mathfrak{h}$ is a Lie algebra homomorphism, then the kernel of $\phi$ is an ideal of $\mathfrak{g}$.

Proof

Ideal

A subalgebra $\mathfrak{h}$ of $\mathfrak{g}$ is called an ideal of $\mathfrak{g}$ if it satisfies the following.

$$ [X, H] \in \mathfrak{h} \quad \forall X \in \mathfrak{g}, \forall H \in \mathfrak{h} $$

The kernel of $\phi$ is as follows.

$$ \ker \phi = \left\{ X \in \mathfrak{g} : \phi(X) = 0 \right \} $$

For $H \in \ker \phi$, showing that $\phi([X, H]) = 0$ completes the proof.

$$ \phi([X, H]) = [\phi(X), \phi(H)] = [\phi(X), 0] = 0 $$

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