Study Notes on Lie Algebras, Algebraic Groups, and Lie Groups (J.S. Milne): Some Sources of Lie Algebras

Lie Algebras and Lawvere Theories

1. Definition of a Lie Algebra

Let $k$$k$ be a field. A Lie algebra is a $k$$k$-vector space $\mathfrak{g}$$\mathfrak{g}$ equipped with a bilinear map $[-,-]:\mathfrak{g}\times \mathfrak{g}\to \mathfrak{g}$$[-,-] : \mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}$, called the Lie bracket, satisfying:

1. Alternativity: $[x,x]=0$$[x,x] = 0$ for all $x\in \mathfrak{g}$$x \in \mathfrak{g}$ (equivalent to antisymmetry $[x,y]=-[y,x]$$[x,y] = -[y,x]$ if $\mathrm{char}k\ne 2$$\operatorname{char}k \neq 2$);

2. Jacobi identity: $[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0$$[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0$ for all $x,y,z\in \mathfrak{g}$$x,y,z \in \mathfrak{g}$.

Given an associative algebra $A$$A$, the commutator $[x,y]:=xy-yx$$[x,y] := xy - yx$ turns $A$$A$ into a Lie algebra, denoted $[A]$$[A]$.

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2. Lawvere Theory of Lie Algebras

A Lawvere theory is a small category with objects $0,1,2,\dots$$0,1,2,\dots$ such that each $n$$n$ is the $n$$n$-fold product $1^n$$1^n$.
The Lawvere theory of Lie algebras $\mathcal{L}$$\mathcal{L}$ is generated by:

• a binary operation $\mu :2\to 1$$\mu : 2 \to 1$ (the Lie bracket $[-,-]$$[-,-]$);

• and $k$$k$-linear operations (unary operations $r:1\to 1$$r : 1 \to 1$ for each $r\in k$$r \in k$, satisfying linearity axioms).

The axioms are encoded by commutative diagrams in $\mathcal{L}$$\mathcal{L}$:

• Antisymmetry: $\mu \circ ⟨{\pi }_1,{\pi }_2⟩+\mu \circ ⟨{\pi }_2,{\pi }_1⟩=0$$\mu \circ \langle \pi_1,\pi_2 \rangle \;+\; \mu \circ \langle \pi_2,\pi_1 \rangle = 0$ as morphisms $2\to 1$$2 \to 1$.

• Jacobi identity:

  $$\mu \circ ⟨\mu \circ ⟨{\pi }_1,{\pi }_2⟩,{\pi }_3⟩+\mu \circ ⟨\mu \circ ⟨{\pi }_2,{\pi }_3⟩,{\pi }_1⟩+\mu \circ ⟨\mu \circ ⟨{\pi }_3,{\pi }_1⟩,{\pi }_2⟩=0$$

  as a morphism $3\to 1$$3 \to 1$.

• Linearity: e.g., $\lambda \cdot (\mu (x,y))=\mu (\lambda x,y)$$\lambda \cdot (\mu(x,y)) = \mu(\lambda x, y)$, etc.

Models: A finite-product-preserving functor $\mathfrak{g}:\mathcal{L}\to \mathbf{Set}$$\mathfrak{g} : \mathcal{L} \to \mathbf{Set}$ is exactly a Lie algebra (the underlying set is $\mathfrak{g}(1)$$\mathfrak{g}(1)$, the bracket is $\mathfrak{g}(\mu )$$\mathfrak{g}(\mu)$). The category of models $\mathrm{Mod}(\mathcal{L})$$\mathrm{Mod}(\mathcal{L})$ is isomorphic to the usual category ${\mathrm{Lie}}_k$$\mathrm{Lie}_k$ of Lie algebras over $k$$k$.

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3. Adjointness and Subalgebra Constructions

Let $k$$k$ be a field and $A$$A$ an associative $k$$k$-algebra.
Let $\ast :A\to A$$*: A \to A$ be a linear involution (i.e., $(a^{\ast }{)}^{\ast }=a$$(a^*)^* = a$ and $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$$(ab)^* = b^*a^*$).
Denote by $[A]$$[A]$ the Lie algebra with bracket $[x,y]=xy-yx$$[x,y] = xy - yx$.

Lemma. $\alpha :=-(-{)}^{\ast }:A\to A$$\alpha := -(-)^*: A \to A$ is a Lie algebra homomorphism.

Proof.

$$\alpha ([x,y])=-(xy-yx{)}^{\ast }=(yx-xy{)}^{\ast }=x^{\ast }{y}^{\ast }-y^{\ast }{x}^{\ast }=(-x^{\ast })(-y^{\ast })-(-y^{\ast })(-x^{\ast })=[\alpha (x),\alpha (y)].$$

$◻$$\square$

Corollary. $\mathsf{Eq}(\alpha ,{\mathrm{id}}_A)=\{x\in A:-x^{\ast }=x\}=\{x\in A:x+x^{\ast }=0\}$$\mathsf{Eq}(\alpha,\mathrm{id}_A) = \{x \in A : -x^*=x\} = \{x \in A : x+x^*=0\}$ is a Lie subalgebra of $[A]$$[A]$.

Remark. For any Lawvere theory $\mathbb{T}$$\mathbb{T}$, the category $\mathrm{Mod}(\mathbb{T})$$\mathrm{Mod}(\mathbb{T})$ is complete and cocomplete.
In particular, $\mathrm{Mod}(\mathcal{L})={\mathrm{Lie}}_k$$\mathrm{Mod}(\mathcal{L}) = \mathrm{Lie}_k$ is complete and cocomplete.

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4. Classical Lie Subalgebras via Bilinear Forms

Let $V$$V$ be a finite-dimensional vector space over $k$$k$, and let $\beta :V\times V\to k$$\beta: V \times V \to k$ be a non-degenerate $k$$k$-bilinear form.

Define $\ast :\mathrm{End}(V)\to \mathrm{End}(V)$$*: \mathrm{End}(V) \to \mathrm{End}(V)$ by $\beta (\alpha x,y)=\beta (x,{\alpha }^{\ast }y)$$\beta(\alpha x,y) = \beta(x,\alpha^* y)$ for $\alpha \in \mathrm{End}(V)$$\alpha \in \mathrm{End}(V)$ and $x,y\in V$$x,y \in V$.

Then

$$\beta ((a+b)x,y)=\beta (ax,y)+\beta (bx,y)=\beta (x,a^{\ast }y)+\beta (x,b^{\ast }y)=\beta (x,(a^{\ast }+b^{\ast })y)=\beta (x,(a+b{)}^{\ast }y),$$

hence $(a+b{)}^{\ast }=a^{\ast }+b^{\ast }$$(a+b)^* = a^* + b^*$. Similarly, $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$$(ab)^* = b^*a^*$.

If $\beta$$\beta$ is symmetric or skew-symmetric (i.e., $\beta (x,y)=\beta (y,x)$$\beta(x,y) = \beta(y,x)$ or $\beta (x,y)=-\beta (y,x)$$\beta(x,y) = -\beta(y,x)$), then

$$\beta (ax,y)=\beta (x,a^{\ast }y)=\pm \beta (a^{\ast }y,x)=\pm \beta (y,a^{\ast \ast }x)=\beta (a^{\ast \ast }x,y)⟹a=a^{\ast \ast },$$

so $\ast$$*$ is an involution.

Now consider $\mathsf{Eq}(\alpha ,{\mathrm{id}}_{\mathrm{End}(V)})$$\mathsf{Eq}(\alpha,\mathrm{id}_{\mathrm{End}(V)})$. The condition $a+a^{\ast }=0$$a+a^*=0$ is equivalent to $\beta (ax,y)+\beta (x,ay)=0$$\beta(ax,y) + \beta(x,ay) = 0$ because

$$\beta ((a+a^{\ast })x,y)=0=\beta (ax,y)+\beta (a^{\ast }x,y)=\beta (ax,y)+\beta (x,ay).$$

Thus we obtain the Lie subalgebra

$$\{a\in \mathfrak{gl}(V):\beta (ax,y)+\beta (x,ay)=0\}.$$

Examples:

• Orthogonal Lie algebra ${\mathfrak{o}}_n=\{A\in M_n(k):A+A^t=0\}$$\mathfrak{o}_n = \{A \in M_n(k) : A+A^t=0\}$ (take $\beta$$\beta$ the standard dot product).

• Symplectic Lie algebra: let $\begin{array}{c}J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)\end{array}$$J = \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix}$ and $\beta (x,y):=x^{t}Jy$$\beta(x,y) := x^t J y$. Then

  $${\mathfrak{sp}}_{2n}=\{A\in M_{2n}(k):(Ax{)}^{t}Jy+x^{t}JAy=0\}=\{A\in M_{2n}(k):x^t(A^{t}J+JA)y=0\},$$

  i.e.,

  $${\mathfrak{sp}}_{2n}=\{A\in M_{2n}(k):A^{t}J+JA=0\}.$$

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5. Trace and the Special Linear Lie Algebra

Lemma. $\mathrm{Tr}:M_n(k)\to k$$\mathrm{Tr}: M_n(k) \to k$ is a Lie algebra homomorphism.

Proof. $\mathrm{Tr}([x,y])=0=[\mathrm{Tr}(x),\mathrm{Tr}(y)]$$\mathrm{Tr}([x,y]) = 0 = [\mathrm{Tr}(x), \mathrm{Tr}(y)]$ (the bracket on the right is the trivial bracket in $k$$k$). $◻$$\square$

Corollary. $\mathfrak{sl}(n)=\{A\in M_n(k):\mathrm{Tr}(A)=0\}$$\mathfrak{sl}(n) = \{A \in M_n(k) : \mathrm{Tr}(A)=0\}$ is a Lie subalgebra of ${\mathfrak{gl}}_n(k)$$\mathfrak{gl}_n(k)$.