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

Let $k$$k$ be a field and $A$$A$ be an algebra. Let $A\to A$$A \to A$ be a linear involution. Let $[A]$$[A]$ denote the Lie algebra given by $[x,y]=xy-yx$$[x,y] = xy - yx$.

Lemma. $\alpha :=-(-{)}^{\ast }:V\to V$$\alpha := -(-)^*: V \to V$ 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)]$$\alpha([x,y]) = -(xy-yx)^* = (yx-xy)^* = x^*y^* - y^*x^* = (-x^*)(-y^*) - (-y^*)(-x^*) = [\alpha(x),\alpha(y)]$. $◻$$\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. Let $\mathbb{T}$$\mathbb{T}$ be a Lawvere theory; then $\mathsf{Mod}(\mathbb{T})$$\mathsf{Mod}(\mathbb{T})$ is complete and cocomplete.

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

Define $\ast :\mathrm{End}(V)\to \mathrm{End}(V)$$*: \mathrm{End}(V) \to \mathrm{End}(V)$ via $\beta (\alpha x,y)=\beta (x,{\alpha }^{\ast }y)$$\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)$$\beta(x,y) = -\beta(y,x)$), then $\beta (ax,y)=\beta (x,a^{\ast }y)=\pm \beta (a^{\ast }y,x)=\pm \beta (y,a^{\ast \ast }x)=\beta (a^{\ast \ast }x,y)⟹a=a^{\ast \ast }$$\beta(ax,y) = \beta(x,a^* y) = \pm\beta(a^*y,x) = \pm\beta(y,a^{**}x) = \beta(a^{**}x,y) \implies a = a^{**}$. Hence, $\ast$$*$ forms an involution.

Now let us consider $\mathsf{Eq}(\alpha ,{\mathrm{id}}_{\mathrm{End}(V)})$$\mathsf{Eq}(\alpha,\mathrm{id}_{\mathrm{End}(V)})$. $a+a^{\ast }=0⟺\beta (ax,y)+\beta (x,ay)=0$$a+a^*=0 \iff \beta(ax,y) + \beta(x,ay) = 0$.

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

Hence, we obtain the Lie subalgebra:

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

Here are some examples of Lie subalgebras of this form:

${\mathfrak{o}}_n=\{A\in M_n(k):A+A^t=0\}$$\mathfrak{o}_n = \{A \in M_n(k) : A+A^t=0\}$.

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

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

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)]$$\mathrm{Tr}([x,y]) = 0 = [\mathrm{Tr}(x), \mathrm{Tr}(y)]$. $◻$$\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.

The following example shows that $\mathrm{Lie}(k)$$\mathrm{Lie}(k)$ is not a normal category; that is, a monomorphism is not always an equalizer.

Consider the algebra of upper triangular matrices ${\mathfrak{b}}_n$$\mathfrak{b}_n$. This forms a Lie subalgebra because upper triangular matrices form a subalgebra of $M_n(k)$$M_n(k)$.

Furthermore, $[-]:{\mathrm{Alg}}_k\to \mathrm{Lie}(k)$$[-]: \mathrm{Alg}_k \to \mathrm{Lie}(k)$ is a right adjoint functor; hence, it preserves limits, and therefore, it preserves monomorphisms.