Computing Lie Algebras with Dual Numbers

The Lie Algebra Functor of an Affine Algebraic GroupThe Functoriality on MorphismsHow to Compute Matrix ExamplesThe General Linear GroupThe Special Linear GroupThe Multiplicative GroupThe Additive GroupThe Diagonal TorusThe Borel Subgroup of Upper Triangular MatricesThe Unipotent Upper Triangular GroupThe Orthogonal GroupThe Orthogonal Group of a General Bilinear FormThe Symplectic GroupThe Finite Group Scheme $\mu_m$The Differential of the Determinant Is the TraceSummary

 

The Lie Algebra Functor of an Affine Algebraic Group

Let $k$$k$ be a field, and let ${\mathbf{CAlg}}_k$$\mathbf{CAlg}_k$ denote the category of commutative $k$$k$-algebras.

An affine algebraic group over $k$$k$ may be viewed as a representable group-valued functor

$$G:{\mathbf{CAlg}}_k\to \mathbf{Grp}.$$

For every $k$$k$-algebra $R$$R$, define the algebra of dual numbers over $R$$R$ by

$$R[\epsilon ]:=R[x]/(x^2).$$

There is a natural projection

$${\pi }_R:R[\epsilon ]\to R,\epsilon \mapsto 0.$$

Applying the group functor $G$$G$, we get a group homomorphism

$$G(R[\epsilon ])\stackrel{G({\pi }_R)}{\to }G(R).$$

The Lie algebra functor of $G$$G$ is defined by

$$\mathrm{Lie}(G)(R):=\ker (G(R[\epsilon ])\to G(R)).$$

Thus $\mathrm{Lie}(G)(R)$$\operatorname{Lie}(G)(R)$ consists of those $R[\epsilon ]$$R[\varepsilon]$-points of $G$$G$ which reduce to the identity element of $G(R)$$G(R)$.

Informally, its elements are infinitesimal elements of the form

$$1+\epsilon X.$$

For matrix groups this notation is literal.

The Functoriality on Morphisms

Let

$$\phi :G\to H$$

be a morphism of affine algebraic groups. In functorial language, this means that for every $R$$R$ we have a group homomorphism

$${\phi }_R:G(R)\to H(R),$$

natural in $R$$R$.

For every $R$$R$, there is a commutative square

$$\begin{array}{ccc}G(R[\epsilon ])& ⟶& G(R)\\ \downarrow {\phi }_{R[\epsilon ]}& & \downarrow {\phi }_R\\ H(R[\epsilon ])& ⟶& H(R).\end{array}$$

Therefore, if

$$X\in \ker (G(R[\epsilon ])\to G(R)),$$

then

$${\phi }_{R[\epsilon ]}(X)\in \ker (H(R[\epsilon ])\to H(R)).$$

Hence $\phi$$\varphi$ induces a natural map

$$\mathrm{Lie}(\phi {)}_R:\mathrm{Lie}(G)(R)\to \mathrm{Lie}(H)(R),$$

given simply by restriction:

$$\mathrm{Lie}(\phi {)}_R={\phi }_{R[\epsilon ]}{|}_{\mathrm{Lie}(G)(R)}.$$

So the Lie algebra functor on morphisms is literally obtained by restricting the original morphism to the infinitesimal kernel.

Equivalently, if $X$$X$ is written as an infinitesimal element $1+\epsilon X$$1+\varepsilon X$, then

$$\phi (1+\epsilon X)=1+\epsilon \mathrm{Lie}(\phi )(X).$$

This is the functor-of-points version of the differential at the identity.

How to Compute Matrix Examples

Let

$$G\subseteq {\mathrm{GL}}_n$$

be a matrix algebraic group. Then

$$\mathrm{Lie}(G)(R)=\ker (G(R[\epsilon ])\to G(R)).$$

An element in this kernel must be of the form

$$I+\epsilon X,X\in M_n(R).$$

So the computation is always:

1. Write an infinitesimal element as $I+\epsilon X$$I+\varepsilon X$.

2. Substitute it into the defining equations of $G$$G$.

3. Keep only first-order terms in $\epsilon$$\varepsilon$.

4. The resulting linear equations define $\mathrm{Lie}(G)$$\operatorname{Lie}(G)$.

This is the sense in which the Lie algebra is the linearization of the algebraic group at the identity.

The General Linear Group

For

$$G={\mathrm{GL}}_n,$$

we have

$${\mathrm{GL}}_n(R[\epsilon ])=\{A\in M_n(R[\epsilon ])\mid A\text{ is invertible}\}.$$

An element reducing to $I$$I$ is exactly

$$I+\epsilon X,X\in M_n(R).$$

Its inverse is

$$(I+\epsilon X{)}^{-1}=I-\epsilon X.$$

Therefore there are no further conditions on $X$$X$, and

$$\mathrm{Lie}({\mathrm{GL}}_n)(R)=M_n(R).$$

Thus

$$\mathrm{Lie}({\mathrm{GL}}_n)={\mathfrak{gl}}_n.$$

The Lie bracket is the usual matrix commutator:

$$[X,Y]=XY-YX.$$

The Special Linear Group

The special linear group is defined by

$${\mathrm{SL}}_n(R)=\{A\in {\mathrm{GL}}_n(R)\mid \det \left(A\right)=1\}.$$

Take an infinitesimal element

$$I+\epsilon X.$$

We use the standard first-order identity

$$\det (I+\epsilon X)=1+\epsilon \mathrm{tr}(X).$$

Therefore

$$I+\epsilon X\in {\mathrm{SL}}_n(R[\epsilon ])$$

if and only if

$$1+\epsilon \mathrm{tr}(X)=1.$$

Hence

$$\mathrm{tr}(X)=0.$$

So

$$\mathrm{Lie}({\mathrm{SL}}_n)(R)=\{X\in M_n(R)\mid \mathrm{tr}(X)=0\}.$$

Thus

$$\mathrm{Lie}({\mathrm{SL}}_n)={\mathfrak{sl}}_n.$$

The Multiplicative Group

The multiplicative group is

$${\mathbb{G}}_m(R)=R^{\times }.$$

Then

$$\mathrm{Lie}({\mathbb{G}}_m)(R)=\ker ((R[\epsilon ]{)}^{\times }\to R^{\times }).$$

An invertible element reducing to $1$$1$ is exactly

$$1+\epsilon a,a\in R.$$

Multiplication gives

$$(1+\epsilon a)(1+\epsilon b)=1+\epsilon (a+b).$$

Therefore

$$\mathrm{Lie}({\mathbb{G}}_m)(R)\cong R.$$

So

$$\mathrm{Lie}({\mathbb{G}}_m)\cong k$$

as a one-dimensional abelian Lie algebra.

The Additive Group

The additive group is

$${\mathbb{G}}_a(R)=R$$

with addition as the group law.

Then

$${\mathbb{G}}_a(R[\epsilon ])=R[\epsilon ],$$

and the kernel of

$$R[\epsilon ]\to R$$

is

$$\epsilon R.$$

Thus

$$\mathrm{Lie}({\mathbb{G}}_a)(R)\cong R.$$

So

$$\mathrm{Lie}({\mathbb{G}}_a)\cong k.$$

Again, this is a one-dimensional abelian Lie algebra.

The Diagonal Torus

Let

$$T_n={\mathbb{G}}_m^n$$

be the diagonal torus inside ${\mathrm{GL}}_n$$\operatorname{GL}_n$.

Its $R$$R$-points are diagonal matrices

$$\mathrm{diag}(t_1,\dots ,t_n),t_i\in R^{\times }.$$

An infinitesimal element is

$$\mathrm{diag}(1+\epsilon a_1,\dots ,1+\epsilon a_n).$$

Equivalently,

$$I+\epsilon \mathrm{diag}(a_1,\dots ,a_n).$$

Therefore

$$\mathrm{Lie}(T_n)(R)=\{\mathrm{diag}(a_1,\dots ,a_n)\mid a_i\in R\}.$$

So $\mathrm{Lie}(T_n)$$\operatorname{Lie}(T_n)$ is the Lie algebra of diagonal matrices. It is abelian.

The Borel Subgroup of Upper Triangular Matrices

Let $B_n\subseteq {\mathrm{GL}}_n$$B_n\subseteq\operatorname{GL}_n$ be the subgroup of invertible upper triangular matrices.

An infinitesimal element has the form

$$I+\epsilon X.$$

It is upper triangular if and only if $X$$X$ is upper triangular.

Therefore

$$\mathrm{Lie}(B_n)(R)=\{X\in M_n(R)\mid X_{ij}=0\text{ for }i>j\}.$$

So

$$\mathrm{Lie}(B_n)={\mathfrak{b}}_n,$$

the Lie algebra of upper triangular matrices.

The Unipotent Upper Triangular Group

Let $U_n\subseteq B_n$$U_n\subseteq B_n$ be the subgroup of upper triangular matrices with all diagonal entries equal to $1$$1$.

An infinitesimal element is

$$I+\epsilon X.$$

The condition that it is upper triangular says that $X$$X$ is upper triangular. The condition that its diagonal entries are equal to $1$$1$ says

$$1+\epsilon X_{ii}=1,$$

hence

$$X_{ii}=0.$$

Therefore

$$\mathrm{Lie}(U_n)(R)=\{X\in M_n(R)\mid X_{ij}=0\text{ for }i\ge j\}.$$

So $\mathrm{Lie}(U_n)$$\operatorname{Lie}(U_n)$ is the Lie algebra of strictly upper triangular matrices.

The Orthogonal Group

Assume first that $\mathrm{char}k\ne 2$$\operatorname{char}k\neq 2$. Let $O_n$$O_n$ be the orthogonal group for the standard symmetric form:

$$O_n(R)=\{A\in {\mathrm{GL}}_n(R)\mid A^{T}A=I\}.$$

Take

$$A=I+\epsilon X.$$

Then

$$A^{T}A=(I+\epsilon X{)}^T(I+\epsilon X).$$

Expanding and using ${\epsilon }^2=0$$\varepsilon^2=0$, we get

$$(I+\epsilon X^T)(I+\epsilon X)=I+\epsilon (X^T+X).$$

The condition $A^{T}A=I$$A^TA=I$ therefore becomes

$$X^T+X=0.$$

Hence

$$\mathrm{Lie}(O_n)(R)=\{X\in M_n(R)\mid X^T+X=0\}.$$

So

$$\mathrm{Lie}(O_n)={\mathfrak{o}}_n.$$

For the special orthogonal group $S{O}_n$$SO_n$, one also imposes $\det \left(A\right)=1$$\det(A)=1$. But when $\mathrm{char}k\ne 2$$\operatorname{char}k\neq 2$, the condition

$$X^T+X=0$$

already implies

$$\mathrm{tr}(X)=0.$$

Hence

$$\mathrm{Lie}(S{O}_n)={\mathfrak{so}}_n=\{X\in M_n(k)\mid X^T+X=0\}.$$

In characteristic $2$$2$, one has to be more careful, since symmetric and alternating behavior degenerate.

The Orthogonal Group of a General Bilinear Form

More generally, let $Q$$Q$ be the matrix of a nondegenerate bilinear form. Define

$$O(Q)(R)=\{A\in {\mathrm{GL}}_n(R)\mid A^{T}QA=Q\}.$$

Take

$$A=I+\epsilon X.$$

Then

$$A^{T}QA=(I+\epsilon X{)}^{T}Q(I+\epsilon X).$$

Expanding gives

$$Q+\epsilon (X^{T}Q+QX).$$

So the defining condition becomes

$$X^{T}Q+QX=0.$$

Therefore

$$\mathrm{Lie}(O(Q))(R)=\{X\in M_n(R)\mid X^{T}Q+QX=0\}.$$

This is the Lie algebra preserving the bilinear form $Q$$Q$ infinitesimally.

The Symplectic Group

Let

$$\begin{array}{c}J=\left(\begin{array}{cc}0& I_n\\ -I_n& 0\end{array}\right).\end{array}$$

The symplectic group is

$${\mathrm{Sp}}_{2n}(R)=\{A\in {\mathrm{GL}}_{2n}(R)\mid A^{T}JA=J\}.$$

Take

$$A=I+\epsilon X.$$

Then

$$A^{T}JA=(I+\epsilon X{)}^{T}J(I+\epsilon X)=J+\epsilon (X^{T}J+JX).$$

Therefore the infinitesimal condition is

$$X^{T}J+JX=0.$$

Hence

$$\mathrm{Lie}({\mathrm{Sp}}_{2n})(R)=\{X\in M_{2n}(R)\mid X^{T}J+JX=0\}.$$

If

$$\begin{array}{c}X=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right),\end{array}$$

then this condition is equivalent to

$$D=-A^T,B^T=B,C^T=C.$$

Thus

$$\begin{array}{c}{\mathfrak{sp}}_{2n}(R)=\left\{\left(\begin{array}{cc}A& B\\ C& -A^T\end{array}\right)\right|B^T=B,C^T=C\}.\end{array}$$

The Finite Group Scheme ${\mu }_m$$\mu_m$

The group scheme ${\mu }_m$$\mu_m$ is defined by

$${\mu }_m(R)=\{t\in R^{\times }\mid t^m=1\}.$$

An infinitesimal element has the form

$$1+\epsilon a.$$

The defining equation gives

$$(1+\epsilon a{)}^m=1.$$

Since ${\epsilon }^2=0$$\varepsilon^2=0$, we have

$$(1+\epsilon a{)}^m=1+m\epsilon a.$$

Thus the condition is

$$ma=0.$$

Therefore

$$\mathrm{Lie}({\mu }_m)(R)=\{a\in R\mid ma=0\}.$$

If $k$$k$ is a field and $\mathrm{char}k\nmid m$$\operatorname{char}k\nmid m$, then

$$\mathrm{Lie}({\mu }_m)=0.$$

But if $\mathrm{char}k=p$$\operatorname{char}k=p$ and $p\mid m$$p\mid m$, then

$$\mathrm{Lie}({\mu }_m)\cong k.$$

In particular, in characteristic $p$$p$,

$$\mathrm{Lie}({\mu }_p)\cong k.$$

This is a useful warning: even a finite group scheme can have a nonzero Lie algebra if it is non-reduced.

The Differential of the Determinant Is the Trace

Now consider the determinant morphism

$$\det :{\mathrm{GL}}_n\to {\mathbb{G}}_m.$$

By functoriality of the Lie construction, it induces

$$\mathrm{Lie}(\det ):\mathrm{Lie}({\mathrm{GL}}_n)\to \mathrm{Lie}({\mathbb{G}}_m).$$

Using the previous computations, this is a map

$${\mathfrak{gl}}_n\to k.$$

By definition, it is obtained by restricting determinant to the infinitesimal kernel. Thus we compute

$$\det (I+\epsilon X)=1+\epsilon \mathrm{tr}(X).$$

Since

$$\mathrm{Lie}({\mathbb{G}}_m)(R)=\{1+\epsilon a\mid a\in R\}\cong R,$$

we get

$$\mathrm{Lie}(\det )(X)=\mathrm{tr}(X).$$

Therefore

$$\boxed{{\displaystyle d(\det {)}_I=\mathrm{tr}.}}$$

This gives a conceptual definition of trace:

$$\mathrm{tr}=\mathrm{Lie}(\det ).$$

In words, trace is the infinitesimal part of determinant at the identity.

The usual coordinate formula

$$\mathrm{tr}(X)=X_{11}+\cdots +X_{nn}$$

is therefore not the most fundamental definition. It is the matrix expression of the differential of the determinant character.

Summary

The Lie algebra functor of an affine algebraic group is defined by

$$\mathrm{Lie}(G)(R)=\ker (G(R[\epsilon ])\to G(R)).$$

For a morphism

$$\phi :G\to H,$$

the induced Lie algebra map is simply the restriction

$$\mathrm{Lie}(\phi {)}_R={\phi }_{R[\epsilon ]}{|}_{\mathrm{Lie}(G)(R)}.$$

For matrix groups, this gives a simple computational method:

$$I+\epsilon X$$

is substituted into the defining equations of the group, and the coefficient of $\epsilon$$\varepsilon$ gives the defining equations of the Lie algebra.

Thus:

$$\mathrm{Lie}({\mathrm{GL}}_n)={\mathfrak{gl}}_n,$$

$$\mathrm{Lie}({\mathrm{SL}}_n)={\mathfrak{sl}}_n,$$

$$\mathrm{Lie}(T_n)=\text{diagonal matrices},$$

$$\mathrm{Lie}(B_n)=\text{upper triangular matrices},$$

$$\mathrm{Lie}(U_n)=\text{strictly upper triangular matrices},$$

$$\mathrm{Lie}(O(Q))=\{X\mid X^{T}Q+QX=0\},$$

$$\mathrm{Lie}({\mathrm{Sp}}_{2n})=\{X\mid X^{T}J+JX=0\},$$

and

$$\mathrm{Lie}(\det )=\mathrm{tr}.$$

The slogan is:

$$\boxed{{\displaystyle \text{The Lie algebra is the first-order part of the algebraic group at the identity.}}}$$

Or even more functorially:

$$\boxed{{\displaystyle \mathrm{Lie}=\ker (G(-[\epsilon ])\to G(-)).}}$$