Smooth Functor of Points II: Lie Algebras as Infinitesimal Kernels

Lie groups as group-valued functorsThe Lie algebra as the dual-number kernelExample: $GL_n$The Lie bracket from a second-order commutatorThe matrix-group calculationWhy the bracket satisfies the Lie algebra axiomsParallel with algebraic groupsThree technical cautionsSummary

In the first part, we saw that for a smooth manifold $M$$M$, its tangent bundle can be recovered as

$$TM\cong M(D),D=\mathbb{R}[\epsilon ]/({\epsilon }^2),$$

where

$$M(A)={\mathrm{Hom}}_{C^{\mathrm{\infty }}\mathrm{Ring}}(C^{\mathrm{\infty }}(M),A).$$

Thus tangent vectors are dual-number points.

We now apply this to Lie groups.

The result is the smooth analogue of the algebraic-geometric definition of the Lie algebra of an algebraic group:

$$\boxed{{\displaystyle \mathrm{Lie}(G)=\ker (G(D)\to G(\mathbb{R})).}}$$

Here $G$$G$ is a Lie group, interpreted as a group-valued functor on $C^{\mathrm{\infty }}$$C^\infty$-rings.

The bracket also has a functorial origin: it is the second-order term in the group commutator of two first-order infinitesimal elements.

Lie groups as group-valued functors

Let $G$$G$ be a Lie group. In categorical terms, $G$$G$ is a group object in $\mathbf{Man}$$\mathbf{Man}$. Thus there are smooth maps

$$m:G\times G\to G,$$

$$e:\ast \to G,$$

$$i:G\to G,$$

for multiplication, identity, and inverse.

Applying $C^{\mathrm{\infty }}(-)$$C^\infty(-)$ gives morphisms

$$m^{\ast }:C^{\mathrm{\infty }}(G)\to C^{\mathrm{\infty }}(G\times G),$$

$$e^{\ast }:C^{\mathrm{\infty }}(G)\to C^{\mathrm{\infty }}(\ast )=\mathbb{R},$$

$$i^{\ast }:C^{\mathrm{\infty }}(G)\to C^{\mathrm{\infty }}(G).$$

One important point must be kept explicit:

$$C^{\mathrm{\infty }}(G\times G)$$

is generally not the ordinary tensor product

$$C^{\mathrm{\infty }}(G){\otimes }_{\mathbb{R}}{C}^{\mathrm{\infty }}(G).$$

The correct statement is that in the category of $C^{\mathrm{\infty }}$$C^\infty$-rings,

$$C^{\mathrm{\infty }}(G\times G)\cong C^{\mathrm{\infty }}(G){\otimes }_{\mathrm{\infty }}{C}^{\mathrm{\infty }}(G),$$

where ${\otimes }_{\mathrm{\infty }}$$\otimes_\infty$ denotes the coproduct of $C^{\mathrm{\infty }}$$C^\infty$-rings.

Therefore $C^{\mathrm{\infty }}(G)$$C^\infty(G)$ is not best viewed as an ordinary Hopf algebra. It is better viewed as a cogroup object in $C^{\mathrm{\infty }}$$C^\infty$-rings, or informally as a Hopf $C^{\mathrm{\infty }}$$C^\infty$-ring.

For every $C^{\mathrm{\infty }}$$C^\infty$-ring $A$$A$, define

$$G(A)={\mathrm{Hom}}_{{\mathbf{C}}^{\mathrm{\infty }}\mathbf{Ring}}(C^{\mathrm{\infty }}(G),A).$$

We now define the group structure on $G(A)$$G(A)$.

Let

$$\alpha ,\beta \in G(A).$$

That is,

$$\alpha ,\beta :C^{\mathrm{\infty }}(G)\to A.$$

By the coproduct property, they induce a unique morphism

$$\alpha {\otimes }_{\mathrm{\infty }}\beta :C^{\mathrm{\infty }}(G\times G)\to A.$$

Define

$$\alpha \cdot \beta =(\alpha {\otimes }_{\mathrm{\infty }}\beta )\circ m^{\ast }.$$

The identity element is induced by

$$C^{\mathrm{\infty }}(G)\stackrel{e^{\ast }}{\to }\mathbb{R}\to A.$$

The inverse of $\alpha$$\alpha$ is

$${\alpha }^{-1}=\alpha \circ i^{\ast }.$$

The group axioms follow from the group-object axioms for $G$$G$ in $\mathbf{Man}$$\mathbf{Man}$. Since $C^{\mathrm{\infty }}(-)$$C^\infty(-)$ is contravariant and fully faithful on manifolds, the commutative diagrams defining associativity, identity, and inverse dualize into the required diagrams of $C^{\mathrm{\infty }}$$C^\infty$-rings, and hence into group laws after applying ${\mathrm{Hom}}_{C^{\mathrm{\infty }}\mathrm{Ring}}(-,A)$$\operatorname{Hom}_{C^\infty\mathrm{Ring}}(-,A)$.

Thus a Lie group defines a group-valued functor

$$G(-):{\mathbf{C}}^{\mathrm{\infty }}\mathbf{Ring}\to \mathbf{Grp}.$$

This is the smooth functor of points of the Lie group.

The Lie algebra as the dual-number kernel

Take

$$D=\mathbb{R}[\epsilon ]/({\epsilon }^2).$$

The projection

$$p:D\to \mathbb{R}$$

induces a group homomorphism

$$G(D)\to G(\mathbb{R}).$$

Since

$$G(\mathbb{R})\cong G$$

as an ordinary group of real points, it has an identity element $e$$e$.

Define

$$\mathfrak{g}=\mathrm{Lie}(G)=\ker (G(D)\to G(\mathbb{R})).$$

Thus an element of $\mathfrak{g}$$\mathfrak g$ is a $D$$D$-point of $G$$G$ whose underlying real point is the identity.

Equivalently, an element $X\in \mathfrak{g}$$X\in\mathfrak g$ is a $C^{\mathrm{\infty }}$$C^\infty$-ring morphism

$${\varphi }_X:C^{\mathrm{\infty }}(G)\to D$$

such that

$$p\circ {\varphi }_X=e^{\ast }.$$

Therefore it is uniquely of the form

$${\varphi }_X(f)=f(e)+\epsilon X(f),$$

where

$$X:C^{\mathrm{\infty }}(G)\to \mathbb{R}$$

is a derivation at the identity:

$$X(fg)=f(e)X(g)+g(e)X(f).$$

Hence

$$\mathfrak{g}\cong {\mathrm{Der}}_e(C^{\mathrm{\infty }}(G),\mathbb{R}).$$

This is the classical tangent space $T_{e}G$$T_eG$, but it has now been obtained as a functorial infinitesimal kernel:

$$\boxed{{\displaystyle \mathrm{Lie}(G)=\ker (G(\mathbb{R}[\epsilon ]/{\epsilon }^2)\to G(\mathbb{R})).}}$$

Example: $G{L}_n$$GL_n$

For

$$G=G{L}_n,$$

one has

$$G(D)=G{L}_n(D).$$

An element of $G(D)$$G(D)$ lying over the identity matrix is of the form

$$I+\epsilon X,$$

where

$$X\in M_n(\mathbb{R}).$$

Indeed,

$$(I+\epsilon X)(I-\epsilon X)=I-{\epsilon }^2{X}^2=I.$$

Thus

$$\mathrm{Lie}(G{L}_n)\cong M_n(\mathbb{R})={\mathfrak{gl}}_n(\mathbb{R}).$$

This recovers the usual Lie algebra of $G{L}_n$$GL_n$, but it does so without first invoking curves, velocities, or left-invariant vector fields.

The Lie bracket from a second-order commutator

The bracket can also be defined through the functor of points.

Let

$$D_{12}=\mathbb{R}[{\epsilon }_1,{\epsilon }_2]/({\epsilon }_1^2,{\epsilon }_2^2).$$

This is again a $C^{\mathrm{\infty }}$$C^\infty$-ring. Its $C^{\mathrm{\infty }}$$C^\infty$-structure is given by Taylor expansion truncated by the relations

$${\epsilon }_1^2={\epsilon }_2^2=0.$$

There are two morphisms

$$D\to D_{12},\epsilon \mapsto {\epsilon }_1,$$

and

$$D\to D_{12},\epsilon \mapsto {\epsilon }_2.$$

Given

$$X,Y\in \mathfrak{g}\subset G(D),$$

let $X_1$$X_1$ denote the image of $X$$X$ in $G(D_{12})$$G(D_{12})$ along $\epsilon \mapsto {\epsilon }_1$$\varepsilon\mapsto\varepsilon_1$, and let $Y_2$$Y_2$ denote the image of $Y$$Y$ along $\epsilon \mapsto {\epsilon }_2$$\varepsilon\mapsto\varepsilon_2$.

Consider the group commutator

$$X_1{Y}_2{X}_1^{-1}{Y}_2^{-1}\in G(D_{12}).$$

Let

$$I=({\epsilon }_1{\epsilon }_2)\subset D_{12}.$$

Then

$$I^2=0.$$

Moreover,

$$D_{12}/I=\mathbb{R}[{\epsilon }_1,{\epsilon }_2]/({\epsilon }_1^2,{\epsilon }_2^2,{\epsilon }_1{\epsilon }_2).$$

In this quotient, the mixed term ${\epsilon }_1{\epsilon }_2$$\varepsilon_1\varepsilon_2$ vanishes. Hence the two first-order infinitesimal elements commute to first order, and the commutator becomes the identity in $G(D_{12}/I)$$G(D_{12}/I)$.

Therefore

$$X_1{Y}_2{X}_1^{-1}{Y}_2^{-1}\in \ker (G(D_{12})\to G(D_{12}/I)).$$

This kernel is canonically identified with

$$\mathfrak{g}{\otimes }_{\mathbb{R}}I.$$

Indeed, since $I^2=0$$I^2=0$, an element of this kernel is a square-zero infinitesimal displacement of the identity, hence is determined by an identity-based derivation with values in $I$$I$.

Because $I$$I$ is generated by ${\epsilon }_1{\epsilon }_2$$\varepsilon_1\varepsilon_2$, there is a unique element

$$[X,Y]\in \mathfrak{g}$$

such that

$$X_1{Y}_2{X}_1^{-1}{Y}_2^{-1}=1+{\epsilon }_1{\epsilon }_2[X,Y].$$

This defines the Lie bracket.

The notation on the right means that the corresponding $C^{\mathrm{\infty }}$$C^\infty$-ring morphism is

$$f\mapsto f(e)+{\epsilon }_1{\epsilon }_2[X,Y](f).$$

Thus

$$\boxed{{\displaystyle [X,Y]\text{ is the second-order term in the commutator of two first-order infinitesimal elements.}}}$$

The matrix-group calculation

For

$$G=G{L}_n,$$

an element of the identity fiber over $D$$D$ is of the form

$$I+\epsilon X,X\in {\mathfrak{gl}}_n(\mathbb{R}).$$

Now take

$$X_1=I+{\epsilon }_{1}X,Y_2=I+{\epsilon }_{2}Y.$$

Since

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

and

$$(I+{\epsilon }_{2}Y{)}^{-1}=I-{\epsilon }_{2}Y,$$

we compute

$$X_1{Y}_2{X}_1^{-1}{Y}_2^{-1}=(I+{\epsilon }_{1}X)(I+{\epsilon }_{2}Y)(I-{\epsilon }_{1}X)(I-{\epsilon }_{2}Y).$$

Expanding and using

$${\epsilon }_1^2={\epsilon }_2^2=0,$$

we get

$$X_1{Y}_2{X}_1^{-1}{Y}_2^{-1}=I+{\epsilon }_1{\epsilon }_2(XY-YX).$$

Therefore

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

So the functor-of-points definition recovers the usual matrix commutator.

Why the bracket satisfies the Lie algebra axioms

The bracket is bilinear because addition in the identity fiber over dual numbers comes from multiplication of infinitesimal group elements:

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

Thus the second-order commutator term is linear in each variable.

Antisymmetry comes from the group identity

$$[b,a]=[a,b{]}^{-1}$$

for the commutator

$$[a,b]=ab{a}^{-1}{b}^{-1}.$$

Since

$$(1+{\epsilon }_1{\epsilon }_{2}Z{)}^{-1}=1-{\epsilon }_1{\epsilon }_{2}Z,$$

one gets

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

The Jacobi identity comes from a group-commutator identity at third order.

Let

$$D_{123}=\mathbb{R}[{\epsilon }_1,{\epsilon }_2,{\epsilon }_3]/({\epsilon }_1^2,{\epsilon }_2^2,{\epsilon }_3^2).$$

Put

$$X,Y,Z\in \mathfrak{g}$$

in the three infinitesimal directions ${\epsilon }_1,{\epsilon }_2,{\epsilon }_3$$\varepsilon_1,\varepsilon_2,\varepsilon_3$.

The commutator of first-order terms produces second-order terms such as

$${\epsilon }_1{\epsilon }_2[X,Y].$$

Commuting such a second-order term with a first-order term produces a third-order term such as

$${\epsilon }_1{\epsilon }_2{\epsilon }_3[[X,Y],Z].$$

The group-theoretic Hall-Witt identity implies that the product of the three corresponding third-order commutators is trivial. Extracting the coefficient of

$${\epsilon }_1{\epsilon }_2{\epsilon }_3$$

gives

$$[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0.$$

Thus the bracket satisfies Jacobi.

In this sense, the Lie algebra axioms are infinitesimal shadows of group identities.

Parallel with algebraic groups

For an affine algebraic group

$$G=\mathrm{Spec}A$$

over a field $k$$k$, the functor of points is

$$G(R)={\mathrm{Hom}}_{k\mathrm{Alg}}(A,R).$$

The Lie algebra is

$$\mathrm{Lie}(G)=\ker (G(k[\epsilon ]/{\epsilon }^2)\to G(k)).$$

The bracket again comes from the second-order group commutator.

For Lie groups, the same pattern holds after replacing ordinary commutative algebras by $C^{\mathrm{\infty }}$$C^\infty$-rings:

$${\mathbf{CAlg}}_k⇝{\mathbf{C}}^{\mathrm{\infty }}\mathbf{Ring}.$$

The dictionary is:

$$\mathcal{O}(X)\leftrightarrow C^{\mathrm{\infty }}(M),$$

$${\mathrm{Hom}}_{k\mathrm{Alg}}(\mathcal{O}(X),R)\leftrightarrow {\mathrm{Hom}}_{C^{\mathrm{\infty }}\mathrm{Ring}}(C^{\mathrm{\infty }}(M),A),$$

$$k[\epsilon ]/{\epsilon }^2\leftrightarrow \mathbb{R}[\epsilon ]/{\epsilon }^2\text{ as a }{C}^{\mathrm{\infty }}\text{-ring}.$$

Thus the shared principle is:

$$\boxed{{\displaystyle \text{The Lie algebra of a group object is its identity fiber over the dual numbers.}}}$$

And the shared source of the bracket is:

$$\boxed{{\displaystyle \text{The Lie bracket is the second-order term of the group commutator.}}}$$

Three technical cautions

First, one must use $C^{\mathrm{\infty }}$$C^\infty$-ring morphisms, not ordinary $\mathbb{R}$$\mathbb R$-algebra morphisms.

The expression

$${\mathrm{Hom}}_{\mathbb{R}\mathrm{Alg}}(C^{\mathrm{\infty }}(M),A)$$

does not encode the smooth structure correctly. It ignores the operations associated to arbitrary smooth functions.

The correct expression is

$${\mathrm{Hom}}_{{\mathbf{C}}^{\mathrm{\infty }}\mathbf{Ring}}(C^{\mathrm{\infty }}(M),A).$$

Second, one should not write

$$C^{\mathrm{\infty }}(M\times N)=C^{\mathrm{\infty }}(M){\otimes }_{\mathbb{R}}{C}^{\mathrm{\infty }}(N)$$

as an ordinary tensor product identity.

The correct statement is

$$C^{\mathrm{\infty }}(M\times N)\cong C^{\mathrm{\infty }}(M){\otimes }_{\mathrm{\infty }}{C}^{\mathrm{\infty }}(N),$$

where ${\otimes }_{\mathrm{\infty }}$$\otimes_\infty$ is the coproduct in the category of $C^{\mathrm{\infty }}$$C^\infty$-rings.

Third, ordinary manifolds are not enough as test objects.

If one only evaluates at $\mathbb{R}$$\mathbb R$, one sees ordinary points:

$$M(\mathbb{R})\cong M.$$

To see tangent vectors, one must allow infinitesimal test objects such as

$$\mathbb{R}[\epsilon ]/{\epsilon }^2.$$

This is why the functor of points should be defined on $C^{\mathrm{\infty }}$$C^\infty$-rings, or at least on a suitable subcategory containing Weil algebras.

Summary

The functorial definition of the Lie algebra of a Lie group is fully rigorous.

A smooth manifold $M$$M$ defines a $C^{\mathrm{\infty }}$$C^\infty$-ring

$$C^{\mathrm{\infty }}(M).$$

Its functor of points is

$$M(A)={\mathrm{Hom}}_{{\mathbf{C}}^{\mathrm{\infty }}\mathbf{Ring}}(C^{\mathrm{\infty }}(M),A).$$

The dual numbers

$$D=\mathbb{R}[\epsilon ]/({\epsilon }^2)$$

are a $C^{\mathrm{\infty }}$$C^\infty$-ring. Therefore

$$M(D)$$

is the set of first-order infinitesimal points of $M$$M$.

For a point $x\in M$$x\in M$,

$$T_{x}M={(M(D)\to M(\mathbb{R}))}^{-1}(x).$$

For a Lie group $G$$G$,

$$\mathrm{Lie}(G)=\ker (G(D)\to G(\mathbb{R})).$$

The bracket is defined by the second-order commutator:

$$X_1{Y}_2{X}_1^{-1}{Y}_2^{-1}=1+{\epsilon }_1{\epsilon }_2[X,Y].$$

This recovers the usual matrix formula

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

for $G{L}_n$$GL_n$.

Thus the analytic tangent-space definition can be replaced by a functorial infinitesimal definition. A Lie algebra is not first a space of velocities of curves; it is the first-order identity fiber of a group-valued smooth functor.

The conceptual chain is:

$$\text{Lie group}⇝\text{group-valued }{C}^{\mathrm{\infty }}\text{-functor}⇝\text{dual-number identity fiber}⇝\text{Lie algebra}.$$

This is the smooth analogue of the algebraic-group construction.