Basic Concepts of Lie Algebra

Lie Algebra

Definition 1. A Lie algebra is a vector space $L$ over a field $K$ with a Lie bracket
$$
\begin{aligned}
\left[ \cdot,\cdot \right] &: L \times L \to L\\\
(x,y) &\mapsto [x,y],
\end{aligned}
$$
such that

(i) $(x,y)\mapsto [x,y]$ is linear in $x$ and in $y$;

(ii) $[x,x]=0$ for all $x\in L$;

(iii) (Jacobi identity)
$$
[[x,y],z] + [[y,z],x] + [[z,x],y] = 0 \quad \text{for all $x,y,z\in L$ .}
$$
​

Proposition 1. (Anti-symmetry) $[x,y] = -[y,x]$ for all $x,y\in L$.

Proof: For all $x,y\in L$,
$$
\begin{aligned}
0 &= [x+y,x+y] \\\
&= [x,x] + [x,y] + [y,x] + [y,y] \\\
&= [x,y] + [y,x].
\end{aligned}
$$

Remark. Assuming (i) and (iii), then the above proposition is equivalent to (ii).

Subalgebra

Notation. If $H,K$ are subspaces of the Lie algebra $L$, then $[H,K]$ is the subspace spanned by $[x,y]$ with $x\in H,y\in K$.

By Proposition 1, we easily obtain the following proposition.

Proposition 2. $[H,K]=[K,H]$ for subspaces $H,K$ of $L$.

Definition 2. A subspace $H$ of a Lie algebra $L$ is a subalgebra of $L$ if
$$
[H,H]\subseteq H.
$$

Commutators

Example 1. (Commutators)

$A$ is an associative algebra over $K$, with a product
$$
(x,y)\mapsto xy.
$$
Then $A$ becomes a Lie algebra if equipped with
$$
[x,y]=xy-yx.
$$
Such an associative algebra $A$ with the above Lie bracket is denoted by $[A]$.

Proof: It is easy to show (i) and (ii), now we only show the Jacobi identity.

Indeed, using the definition of the bracket, we compute
$$
[[x,y],z] = (xy-yx)z - z(xy-yx) = xyz - yxz - zxy + zyx.
$$

$$
[[y,z],x] = (yz-zy)x - x(yz-zy) = yzx - zyx - xyz + xzy.
$$

$$
[[z,x],y] = (zx-xz)y - y(zx-xz) = zxy - xzy - yzx + yxz.
$$

Adding these three expressions, we obtain
$$
[[x,y],z]+[[y,z],x]+[[z,x],y]=(xyz - yxz - zxy + zyx)+(yzx - zyx - xyz + xzy)+(zxy - xzy - yzx + yxz)=0.
$$

Ideal

Definition 3. A subset $I$ of $L$ is an ideal of $L$ if $I$ is a subspace of $L$ and
$$
[L,I]\subseteq I.
$$

Note that
$$
[L,I]=[I,L],
$$
so all ideals here are two-sided.

Proposition 3.
(i) $H\cap K$ is a subalgebra of $L$ if $H$ and $K$ are subalgebras of $L$.

(ii) $H\cap K$ is an ideal of $L$ if $H$ and $K$ are ideals of $L$.

(iii) If $H$ is an ideal of $L$ and $K$ is a subalgebra of $L$, then $H+K$ is a subalgebra of $L$.

(iv) If $H,K$ are ideals of $L$, then $H+K$ is an ideal of $L$.

Quotient Algebra (Factor Algebra)

Definition 4. Let $I$ be an ideal of $L$. The quotient space $L/I$ consists of cosets $I + x$ for $x \in L$, where
$$
I + x = \{ y + x \mid y \in I \}.
$$

Proposition 4. $L/I$ becomes a Lie algebra if equipped with $[\cdot, \cdot]$ given by
$$
[I + x, I + y] = I + [x, y].
$$
Proof : We only show it is well-defined, (i) and (ii) is trivial and the Jacobi identity follows directly from the Jacobi identity of $L$. If $I+x=I+x^\prime$ and $I+y=I+y^\prime$, we need to show
$$
[I+x,I+y]=[I+x^\prime,I+y^\prime],
$$
which it equivalent to
$$
I+[x,y]=I+[x^\prime,y^\prime],
$$
which is also equivalent to
$$
[x,y]-[x^\prime,y^\prime]\in I.
$$
In fact,
$$
[x,y]-[x^\prime,y^\prime]=[x,y]-[x^\prime,y]+[x^\prime,y]-[x^\prime,y^\prime]=[x-x^\prime,y]+[x^\prime,y-y^\prime].
$$
Since $I$ is an ideal, we get the desired result. $\square$

Homomorphisms

Definition 5. Let $L_1, L_2$ be Lie algebras. A map $\varphi: L_1 \to L_2$ is a homomorphism of Lie algebras if $\varphi$ is linear and
$$
\varphi([x, y]) = [\varphi(x), \varphi(y)] \quad \text{for} \quad x, y \in L_1.
$$

If $\varphi$ is in addition bijective, then $\varphi$ is an isomorphism of Lie algebras. In this case, we say $L_1$ and $L_2$ are isomorphic.

Theorem 1. Let $\varphi: L_1 \to L_2$ be a homomorphism of Lie algebras, then

(1) $\text{Im}(\varphi)$ is a subalgebra of $L_2$,

(2) $\ker(\varphi)$ is an ideal of $L_1$,

(3) $L_1 / \ker \varphi$ is isomorphic to $\text{Im} \varphi$.

Proof : (1) and (2) are trivial, we only show (3). Define a map
$$
\begin{aligned}
\Psi: L_1 / \ker \varphi &\to \text{Im} \varphi\\\
x + \ker \varphi&\mapsto \varphi(x).
\end{aligned}
$$

Since
$$
\varphi(x) = \varphi(y) \Longleftrightarrow \varphi(x - y) = 0 \Longleftrightarrow x - y \in \ker \varphi\iff x + \ker \varphi = y + \ker \varphi,
$$

we know $\Psi$ is well-defined and bijective.

Moreover, for $x, y \in L_1$,
$$
\Psi([x+\ker \varphi, y+\ker\varphi])=\Phi([x,y]+\ker\varphi)=\varphi([x,y])
$$
and
$$
[\Psi(x+\ker\varphi),\Psi(y+\ker\varphi)]=[\varphi(x),\varphi(y)].
$$
Hence by $\varphi$ is a homomorphism, we get
$$
\Psi([x+\ker \varphi, y+\ker\varphi])=[\Psi(x+\ker\varphi),\Psi(y+\ker\varphi)],
$$
which means $\Psi$ is also an isomorphism of Lie algebras. $\square$

Proposition 5. Let $I$ be an ideal of $L$ and $H$ be a subalgebra of $L$. Then

(1) $I$ is an ideal of $I+H$;

(2) $I \cap H$ is an ideal of $H$;

(3) $(I+H)/I$ is isomorphic to $H/(I \cap H)$.

Proof. (1) and (2) are routine. For (3), define
$$
\begin{aligned}
\theta : H &\to (I+H)/I\\\
x &\mapsto I+x.
\end{aligned}
$$

Then $\theta$ is linear and is a surjective homomorphism of Lie algebras.

Now
$$
x \in \ker \theta
\iff \theta(x)=I+0
\iff I+x=I
\iff x \in I.
$$

Since also $x \in H$, this means
$$
\ker \theta = I \cap H.
$$

By the previous theorem, we obtain
$$
H/(I \cap H) \cong (I+H)/I.\quad \square
$$

Reference

This series of articles on Lie algebras is based on Carter R., Lie Algebras of Finite and Affine Type (Cambridge University Press, 2005), as well as the class notes for Introduction to Lie Algebras by Chenwei Ruan at BIMSA.

The cover image of this article was taken on the Hooker Valley Track, Aoraki / Mount Cook, New Zealand.