The Lie Algebra of a Non-representable Aut

 

When Aut Is No Longer an Algebraic GroupInternal automorphisms in a cartesian enrichmentIf the units are not representedThe affine finite-dimensional caseInfinite dimension: what breaksThe automorphism group functorLie of a group functorThe linear algebra of dual numbersLinearizing structure preservationAssociative algebrasLie algebrasCoalgebrasHopf algebrasPROP-algebrasThe Lie bracket

 

When Aut Is No Longer an Algebraic Group

The finite-dimensional slogan is familiar:

$$\mathrm{Der}(A)=\mathrm{Lie}(\mathrm{Aut}(A)).$$

But this slogan hides a choice of language.

In finite dimension, $\mathrm{Aut}(A)$$\operatorname{Aut}(A)$ is often an affine group scheme, so it is natural to take its Lie algebra in the usual algebraic-geometric sense. In infinite dimension, this representability usually fails. The right object to keep is not necessarily an algebraic group, but the automorphism group functor.

The point of this note is that the definition of derivations does not really require $\mathrm{Aut}$$\operatorname{Aut}$ to be representable. It only requires $\mathrm{Aut}$$\operatorname{Aut}$ to be a group functor.

Before specializing to vector spaces, it is useful to isolate the general enriched-categorical construction.

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Internal automorphisms in a cartesian enrichment

Let $\mathcal{V}$$\mathcal V$ be a cartesian monoidal category, and let $\mathcal{C}$$\mathcal C$ be a $\mathcal{V}$$\mathcal V$-enriched category.

For objects $X,Y\in \mathcal{C}$$X,Y\in\mathcal C$, write

$${\underset{―}{\mathrm{Hom}}}_{\mathcal{C}}(X,Y)\in \mathcal{V}$$

for the enriched Hom-object.

Composition is a morphism in $\mathcal{V}$$\mathcal V$:

$${\circ }_{X,Y,Z}:\underset{―}{\mathrm{Hom}}(Y,Z)\times \underset{―}{\mathrm{Hom}}(X,Y)⟶\underset{―}{\mathrm{Hom}}(X,Z),$$

and the identity of $X$$X$ is a morphism

$$e_X:1_{\mathcal{V}}⟶\underset{―}{\mathrm{Hom}}(X,X).$$

Therefore

$$\underset{―}{\mathrm{End}}(X):=\underset{―}{\mathrm{Hom}}(X,X)$$

is a monoid object in $\mathcal{V}$$\mathcal V$.

This is the first construction. Cartesian enrichment gives internal endomorphism monoids.

To get automorphisms, we must take the units of this monoid.

There is a small point here which is worth making explicit. In a general category $\mathcal{V}$$\mathcal V$, we do not literally have elements of

$$\underset{―}{\mathrm{End}}(X).$$

So the phrase “invertible elements” is only internal language. The actual definition must be given by morphisms.

Let

$$M:=\underset{―}{\mathrm{End}}(X)$$

with multiplication and unit

$$\mu :M\times M\to M,\eta :1\to M.$$

Assume $\mathcal{V}$$\mathcal V$ has finite limits. Define the object of units $M^{\times }$$M^\times$ by the equalizer

$$M^{\times }⟶M\times M\rightrightarrows M\times M.$$

The first map is

$$M\times M\stackrel{(\mu ,\mu \circ \tau )}{\to }M\times M,$$

where

$$\tau :M\times M\to M\times M$$

is the symmetry. The second map is the constant unit map

$$M\times M⟶1\stackrel{(\eta ,\eta )}{\to }M\times M.$$

Thus $M^{\times }$$M^\times$ is the universal object of pairs whose two products are both the unit.

In internal language, one may write

$$M^{\times }=\{(a,b)\mid ab=1,ba=1\}.$$

But this is only shorthand. The real definition is the equalizer above.

Let

$$i:M^{\times }\to M\times M$$

be the equalizer map, and write its two components as

$$u:M^{\times }\to M,v:M^{\times }\to M.$$

The defining equalities are

$$\mu \circ (u,v)=\eta \circ {!}_{M^{\times }},$$

and

$$\mu \circ (v,u)=\eta \circ {!}_{M^{\times }}.$$

So $u$$u$ is the universal unit and $v$$v$ is its universal inverse.

The object $M^{\times }$$M^\times$ is naturally a group object in $\mathcal{V}$$\mathcal V$.

The unit is induced by

$$(\eta ,\eta ):1\to M\times M.$$

The inverse is induced by the symmetry

$$M\times M\to M\times M,$$

which swaps the two components.

The multiplication

$$M^{\times }\times M^{\times }\to M^{\times }$$

is the unique morphism classified by the pair

$$\mu \circ (u{\pi }_1,u{\pi }_2):M^{\times }\times M^{\times }\to M$$

and

$$\mu \circ (v{\pi }_2,v{\pi }_1):M^{\times }\times M^{\times }\to M.$$

In internal language this is

$$(a,b)(a^{\prime },b^{\prime })=(a{a}^{\prime },b^{\prime }{b}^{\prime }).$$

The second coordinate appears in the reverse order because the inverse of a product is the product of the inverses in the opposite order.

Applying this to

$$M={\underset{―}{\mathrm{End}}}_{\mathcal{C}}(X),$$

we define

$${\underset{―}{\mathrm{Aut}}}_{\mathcal{C}}(X):={\underset{―}{\mathrm{End}}}_{\mathcal{C}}(X{)}^{\times }.$$

Thus:

$$\boxed{{\displaystyle \text{Cartesian enrichment gives internal End as a monoid object.}}}$$

If the base has finite limits, then:

$$\boxed{{\displaystyle \text{internal Aut is the group object of units of internal End.}}}$$

This is the categorical source of automorphism objects.

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If the units are not represented

Even when one does not have the object $M^{\times }$$M^\times$ internally, there is still a functor of units.

For each test object $T\in \mathcal{V}$$T\in\mathcal V$, the set

$${\mathrm{Hom}}_{\mathcal{V}}(T,M)$$

is an ordinary monoid. Hence we can take its group of units:

$$T⟼{\mathrm{Hom}}_{\mathcal{V}}(T,M{)}^{\times }.$$

This gives a group-valued functor

$$M^{\times }:{\mathcal{V}}^{\mathrm{op}}\to \mathbf{Grp}.$$

If the internal object of units exists, it represents this functor. If it does not, the functor still makes sense.

This distinction is exactly what becomes important in infinite-dimensional algebra.

Representability is extra structure. The group functor is the more stable object.

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The affine finite-dimensional case

Now take

$$\mathcal{V}={\mathbf{Aff}}_k$$

with its cartesian monoidal structure.

For finite-dimensional vector spaces $X,Y$$X,Y$, the Hom-object is the affine scheme representing the functor

$$R⟼{\mathrm{Hom}}_R(X{\otimes }_{k}R,Y{\otimes }_{k}R).$$

If

$$\dim X=m,\dim Y=n,$$

then

$$\underset{―}{\mathrm{Hom}}(X,Y)\cong {\mathbb{A}}_k^{nm}.$$

Composition is matrix multiplication, hence polynomial, so it is a morphism of affine schemes.

Thus finite-dimensional vector spaces form a category enriched in affine schemes.

For a finite-dimensional $V$$V$, we have

$$\underset{―}{\mathrm{End}}(V)\cong {\mathbb{A}}_k^{n^2}.$$

The automorphism object is the group of units:

$$\underset{―}{\mathrm{Aut}}(V)=\mathrm{GL}(V).$$

This is not just an arbitrary open subscheme of $\mathrm{End}(V)$$\operatorname{End}(V)$. It is the principal open subset

$$D(\det )\subseteq {\mathbb{A}}_k^{n^2}.$$

Therefore

$${\mathrm{GL}}_n=\mathrm{Spec}k[x_{ij},t]/(t\det -1)$$

is affine.

Now suppose $V$$V$ carries an algebraic structure $\alpha$$\alpha$, for example a multiplication

$$m:V\otimes V\to V.$$

An element $g\in \mathrm{GL}(V)$$g\in\operatorname{GL}(V)$ preserves $m$$m$ precisely when

$$g\circ m=m\circ (g\otimes g).$$

In coordinates, these are polynomial equations. Hence the structure-preserving automorphisms form a closed subgroup scheme of $\mathrm{GL}(V)$$\operatorname{GL}(V)$.

This is the finite-dimensional reason why automorphism groups of algebraic structures are often affine group schemes.

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Infinite dimension: what breaks

Now let

$$V=\underset{i\in I}{⨁}k{e}_i$$

with $I$$I$ infinite.

For a commutative $k$$k$-algebra $R$$R$, the actual scalar extension of $V$$V$ is

$$V_R:=V{\otimes }_{k}R\cong R^{(I)},$$

the finitely supported $I$$I$-tuples in $R$$R$.

So the natural functor of points of the infinite-dimensional affine space is

$$F(R)=V{\otimes }_{k}R\cong R^{(I)}.$$

This functor is generally not represented by an affine scheme.

Indeed, if

$$F(R)={\mathrm{Hom}}_{k\text{-Alg}}(A,R)$$

were represented by some affine scheme $\mathrm{Spec}A$$\operatorname{Spec}A$, then $F$$F$ would preserve products, since representable functors preserve limits. In particular,

$$F\left(\prod _{n}k\right)\cong \prod _{n}F(k)$$

would have to hold.

But here the left-hand side is

$${\left(\prod _{n}k\right)}^{(I)}.$$

This consists of finitely supported functions

$$I\to \prod _{n}k.$$

Equivalently, it consists of sequences of vectors in $k^{(I)}$$k^{(I)}$ whose supports are contained in one common finite subset of $I$$I$.

The right-hand side is

$$\prod _n{k}^{(I)}.$$

This consists of sequences of finitely supported vectors, but the finite support is allowed to depend on $n$$n$.

These are not the same when $I$$I$ is infinite. For instance, choose distinct indices

$$i_1,i_2,i_3,\cdots \in I.$$

The sequence

$$(e_{i_1},e_{i_2},e_{i_3},\dots )$$

belongs to

$$\prod _n{k}^{(I)},$$

but it does not have uniformly finite support. Hence it does not come from

$${\left(\prod _{n}k\right)}^{(I)}.$$

Conceptually, the obstruction is simple:

$$\underset{i\in I}{⨁}$$

does not commute with

$$\prod _n.$$

Thus the actual functor

$$R\mapsto V{\otimes }_{k}R$$

is ind-like. It is built from finite-dimensional affine spaces:

$$V=\underset{W\subset V,\dim W<\mathrm{\infty }}{\underrightarrow{\lim }}W.$$

But it is generally not an ordinary affine scheme.

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The automorphism group functor

Let $V$$V$ now be a possibly infinite-dimensional $k$$k$-vector space equipped with some algebraic structure $\alpha$$\alpha$.

For each commutative $k$$k$-algebra $R$$R$, set

$$V_R:=V{\otimes }_{k}R.$$

The structure $\alpha$$\alpha$ extends to a structure ${\alpha }_R$$\alpha_R$ on $V_R$$V_R$.

Define

$${\mathrm{Aut}}_{\alpha }(R):=\{g:V_R\to V_R|g\text{ is an }R\text{-linear automorphism preserving }{\alpha }_R\}.$$

This is a group-valued functor

$${\mathrm{Aut}}_{\alpha }:k\text{-}\mathbf{CAlg}\to \mathbf{Grp}.$$

In finite dimension, this functor is often represented by an affine group scheme.

In infinite dimension, it usually is not.

But the functor still exists. This is the object one should keep.

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Lie of a group functor

For any group functor

$$G:k\text{-}\mathbf{CAlg}\to \mathbf{Grp},$$

define its infinitesimal kernel over $R$$R$ by

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

In particular,

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

This definition uses only the functor of points. It does not require $G$$G$ to be represented by a scheme.

For a structured vector space $(V,\alpha )$$(V,\alpha)$, we therefore set

$${\mathrm{Der}}_{\alpha }(V):=\mathrm{Lie}({\mathrm{Aut}}_{\alpha })(k).$$

The next point is to see why elements of this kernel have the familiar form

$$\mathrm{id}+\epsilon D.$$

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The linear algebra of dual numbers

Let $R$$R$ be a commutative $k$$k$-algebra.

As an $R$$R$-module,

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

Since

$$V{\otimes }_k-$$

preserves direct sums, we get

$$V{\otimes }_{k}R[\epsilon ]/{\epsilon }^2\cong (V{\otimes }_{k}R)\oplus \epsilon (V{\otimes }_{k}R).$$

That is,

$$V_{R[\epsilon ]}\cong V_R\oplus \epsilon V_R.$$

Thus every element of $V_{R[\epsilon ]}$$V_{R[\varepsilon]}$ can be written uniquely as

$$x+\epsilon y,x,y\in V_R.$$

The reduction map

$$R[\epsilon ]/{\epsilon }^2\to R,\epsilon \mapsto 0$$

induces

$$V_{R[\epsilon ]}\to V_R,x+\epsilon y\mapsto x.$$

Now let

$$g:V_{R[\epsilon ]}\to V_{R[\epsilon ]}$$

be an $R[\epsilon ]$$R[\varepsilon]$-linear automorphism which reduces to the identity on $V_R$$V_R$.

Then, for $x\in V_R$$x\in V_R$, we must have

$$g(x)\equiv x(\mathrm{mod}\epsilon ).$$

Hence there is a unique $R$$R$-linear map

$$D:V_R\to V_R$$

such that

$$g(x)=x+\epsilon D(x).$$

Since $g$$g$ is $R[\epsilon ]$$R[\varepsilon]$-linear, this determines $g$$g$ on all of $V_{R[\epsilon ]}$$V_{R[\varepsilon]}$:

$$g(x+\epsilon y)=g(x)+\epsilon g(y)=x+\epsilon D(x)+\epsilon y.$$

Thus

$$g(x+\epsilon y)=x+\epsilon (y+D(x)).$$

In other words,

$$g=\mathrm{id}+\epsilon D.$$

Conversely, every $R$$R$-linear map

$$D:V_R\to V_R$$

defines an $R[\epsilon ]$$R[\varepsilon]$-linear automorphism

$$\mathrm{id}+\epsilon D:V_{R[\epsilon ]}\to V_{R[\epsilon ]},$$

with inverse

$$\mathrm{id}-\epsilon D.$$

So invertibility imposes no condition on $D$$D$.

The real condition is that

$$\mathrm{id}+\epsilon D$$

must preserve the structure ${\alpha }_{R[\epsilon ]}$$\alpha_{R[\varepsilon]}$.

That condition is what linearizes into the derivation rule.

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Linearizing structure preservation

Suppose the structure contains an operation

$$\theta :V^{\otimes p}\to V^{\otimes q}.$$

An automorphism $g$$g$ preserves $\theta$$\theta$ if

$$g^{\otimes q}\circ \theta =\theta \circ g^{\otimes p}.$$

Substitute

$$g=\mathrm{id}+\epsilon D.$$

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

$$(\mathrm{id}+\epsilon D{)}^{\otimes q}={\mathrm{id}}^{\otimes q}+\epsilon \sum _{a=1}^q{\mathrm{id}}^{\otimes (a-1)}\otimes D\otimes {\mathrm{id}}^{\otimes (q-a)}.$$

Similarly,

$$(\mathrm{id}+\epsilon D{)}^{\otimes p}={\mathrm{id}}^{\otimes p}+\epsilon \sum _{b=1}^p{\mathrm{id}}^{\otimes (b-1)}\otimes D\otimes {\mathrm{id}}^{\otimes (p-b)}.$$

Equating the coefficients of $\epsilon$$\varepsilon$ gives

$$\sum _{a=1}^q({\mathrm{id}}^{\otimes (a-1)}\otimes D\otimes {\mathrm{id}}^{\otimes (q-a)})\circ \theta =\sum _{b=1}^p\theta \circ ({\mathrm{id}}^{\otimes (b-1)}\otimes D\otimes {\mathrm{id}}^{\otimes (p-b)}).$$

This is the general Leibniz rule.

So a derivation is not an arbitrary formula. It is the first-order linearization of a structure-preserving equation.

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Associative algebras

For an associative algebra, the multiplication is

$$m:A\otimes A\to A.$$

Here $p=2$$p=2$ and $q=1$$q=1$. The general formula gives

$$D\circ m=m\circ (D\otimes \mathrm{id})+m\circ (\mathrm{id}\otimes D).$$

Equivalently,

$$D(xy)=D(x)y+xD(y).$$

This is the usual Leibniz rule.

If the algebra is unital, preservation of the unit also gives

$$D(1)=0.$$

In fact this also follows from the Leibniz rule applied to $1\cdot 1$$1\cdot 1$.

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Lie algebras

For a Lie algebra, the bracket is

$$[-,-]:\mathfrak{g}\otimes \mathfrak{g}\to \mathfrak{g}.$$

Again $p=2$$p=2$ and $q=1$$q=1$. The formula gives

$$D[x,y]=[Dx,y]+[x,Dy].$$

Thus Lie algebra derivations are precisely infinitesimal automorphisms of the Lie bracket.

No finite-dimensionality is needed.

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Coalgebras

For a coalgebra, the comultiplication is

$$\mathrm{\Delta }:C\to C\otimes C.$$

Here $p=1$$p=1$ and $q=2$$q=2$. The formula gives

$$(D\otimes \mathrm{id}+\mathrm{id}\otimes D)\circ \mathrm{\Delta }=\mathrm{\Delta }\circ D.$$

This is the coderivation rule.

If the coalgebra is counital, one also has

$$ϵ\circ D=0.$$

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Hopf algebras

For a Hopf algebra $H$$H$, an infinitesimal automorphism should preserve both the algebra and coalgebra structures. Thus $D:H\to H$$D:H\to H$ should be both a derivation and a coderivation:

$$D(xy)=D(x)y+xD(y),$$

and

$$\mathrm{\Delta }D=(D\otimes \mathrm{id}+\mathrm{id}\otimes D)\mathrm{\Delta }.$$

The antipode condition does not need to be imposed separately.

The reason is that the antipode is the convolution inverse of the identity map. It is uniquely determined by the bialgebra structure. Therefore a bialgebra automorphism automatically preserves the antipode.

The same is true infinitesimally. A map $D$$D$ which is both a derivation and a coderivation automatically satisfies

$$D\circ S=S\circ D.$$

So the Hopf case is essentially the simultaneous algebra and coalgebra case.

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PROP-algebras

The same discussion applies to PROP-algebras.

Let $\mathcal{P}$$\mathcal P$ be a PROP, and let $V$$V$ be a $\mathcal{P}$$\mathcal P$-algebra. For every operation

$$\theta \in \mathcal{P}(p,q),$$

there is a structure map

$${\theta }_V:V^{\otimes p}\to V^{\otimes q}.$$

The automorphism group functor is defined by

$${\mathrm{Aut}}_{\mathcal{P}}(V)(R)=\{g:V_R\to V_R|g^{\otimes q}\circ ({\theta }_V{)}_R=({\theta }_V{)}_R\circ g^{\otimes p}\text{ for all }\theta \in \mathcal{P}(p,q)\}.$$

Its infinitesimal elements are maps

$$\mathrm{id}+\epsilon D$$

where $D:V_R\to V_R$$D:V_R\to V_R$ satisfies, for every $\theta \in \mathcal{P}(p,q)$$\theta\in\mathcal P(p,q)$,

$$\sum _{a=1}^q({\mathrm{id}}^{\otimes (a-1)}\otimes D\otimes {\mathrm{id}}^{\otimes (q-a)})\circ {\theta }_V=\sum _{b=1}^p{\theta }_V\circ ({\mathrm{id}}^{\otimes (b-1)}\otimes D\otimes {\mathrm{id}}^{\otimes (p-b)}).$$

Thus derivations of associative algebras, Lie algebras, coderivations of coalgebras, and biderivations of bialgebras are all instances of the same linearized PROP equation.

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The Lie bracket

Dual numbers give the tangent space. The bracket comes from second-order infinitesimal commutators.

Take

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

Let

$$g_1=\mathrm{id}+{\epsilon }_1{D}_1,g_2=\mathrm{id}+{\epsilon }_2{D}_2.$$

Then

$$g_1^{-1}=\mathrm{id}-{\epsilon }_1{D}_1,g_2^{-1}=\mathrm{id}-{\epsilon }_2{D}_2.$$

A direct expansion gives

$$g_1{g}_2{g}_1^{-1}{g}_2^{-1}=\mathrm{id}+{\epsilon }_1{\epsilon }_2(D_1{D}_2-D_2{D}_1).$$

Therefore the Lie bracket is

$$[D_1,D_2]=D_1{D}_2-D_2{D}_1.$$

In the automorphism functor, the group commutator stays inside the group. Therefore the commutator of two infinitesimal structure-preserving maps is again structure-preserving.

This is why the derivations form a Lie algebra.