Group-like Elements as Points: Monoidal Adjunctions from Coalgebras to Hopf Algebras

Group-like Elements as Points: From Coalgebras to GroupsGroup-like elements as coalgebraic pointsFrom sets to coalgebrasThe basic adjunctionRecovering the original setThe group-like part of a coalgebraThe monoidal structureFrom monoids to bialgebrasHopf algebras in a cartesian worldFrom groups to Hopf algebrasThe group algebra adjunctionGroups sit inside Hopf algebrasThe conceptual picture

 

Group-like Elements as Points: From Coalgebras to Groups

There is a small construction in coalgebra theory that becomes much more interesting once it is viewed categorically.

Let $k$$k$ be a field and let ${\mathrm{Coalg}}_k$$\operatorname{Coalg}_k$ denote the category of counital coassociative $k$$k$-coalgebras. For a coalgebra $C$$C$, recall that an element $g\in C$$g\in C$ is called group-like if

$$\mathrm{\Delta }(g)=g\otimes g,\epsilon (g)=1.$$

Write

$$G(C)=\{g\in C\mid \mathrm{\Delta }(g)=g\otimes g,\epsilon (g)=1\}.$$

At first sight, $G(C)$$G(C)$ is simply a distinguished subset of the underlying vector space of $C$$C$. But the defining equations suggest something more structural. They say precisely that $g$$g$ behaves like the unique basis element of the one-dimensional coalgebra $k$$k$.

This turns group-like elements into points, the group-like functor into a representable functor, and eventually leads to an adjunction between groups and Hopf algebras.

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Group-like elements as coalgebraic points

Regard $k$$k$ itself as a coalgebra by

$${\mathrm{\Delta }}_k(1)=1\otimes 1,{\epsilon }_k(1)=1.$$

A linear map

$$f:k\to C$$

is completely determined by the vector

$$g=f(1).$$

For $f$$f$ to be a coalgebra morphism, we must have

$${\mathrm{\Delta }}_{C}f=(f\otimes f){\mathrm{\Delta }}_k$$

and

$${\epsilon }_{C}f={\epsilon }_k.$$

Evaluating at $1$$1$ gives

$${\mathrm{\Delta }}_C(g)=g\otimes g,{\epsilon }_C(g)=1.$$

Thus

$$g\in G(C)$$

if and only if the corresponding map

$$k\to C$$

is a coalgebra morphism.

Therefore there is a natural bijection

$$G(C)\cong {\mathrm{Coalg}}_k(k,C),$$

and hence

$$G\cong {\mathrm{Coalg}}_k(k,-).$$

So $G$$G$ is representable.

This gives a useful reinterpretation of the usual definition. A group-like element is not merely a vector satisfying two equations. It is a coalgebraic point

$$k⟶C.$$

The equations are simply what the statement “this map is a coalgebra morphism” looks like after choosing the element $1\in k$$1\in k$.

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From sets to coalgebras

There is a natural construction in the opposite direction.

Given a set $X$$X$, let

$$k[X]=\underset{x\in X}{⨁}kx$$

be the free vector space on $X$$X$. Declare each basis vector to be group-like:

$$\mathrm{\Delta }(x)=x\otimes x,\epsilon (x)=1.$$

Extending linearly gives a coalgebra structure on $k[X]$$k[X]$.

Indeed,

$$(\mathrm{\Delta }\otimes 1)\mathrm{\Delta }(x)=x\otimes x\otimes x=(1\otimes \mathrm{\Delta })\mathrm{\Delta }(x),$$

and the counit identities are immediate.

A function

$$u:X\to Y$$

extends linearly to

$$k[u]:k[X]\to k[Y].$$

Since basis elements are sent to basis elements, and hence to group-like elements, this is automatically a coalgebra morphism.

Thus we obtain a functor

$$k[-]:\mathrm{Set}\to {\mathrm{Coalg}}_k.$$

It is the ordinary free vector space functor, but equipped with the coalgebra structure for which the distinguished basis consists entirely of group-like elements.

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The basic adjunction

Suppose that

$$F:k[X]\to C$$

is a coalgebra morphism.

Since every $x\in X$$x\in X$ is group-like in $k[X]$$k[X]$, its image $F(x)$$F(x)$ must be group-like in $C$$C$. Hence restriction to the basis gives a function

$$X\to G(C).$$

Conversely, given any function

$$f:X\to G(C),$$

the universal property of the free vector space produces a unique linear extension

$$\tilde{f}:k[X]\to C.$$

Since every $f(x)$$f(x)$ is group-like,

$$\mathrm{\Delta }(\tilde{f}(x))=f(x)\otimes f(x)=(\tilde{f}\otimes \tilde{f})\mathrm{\Delta }(x),$$

and

$$\epsilon (\tilde{f}(x))=1.$$

Thus $\tilde{f}$$\widetilde f$ is a coalgebra morphism.

We obtain a natural bijection

$${\mathrm{Coalg}}_k(k[X],C)\cong \mathrm{Set}(X,G(C)).$$

Therefore

$$k[-]⊣G.$$

This already gives a satisfying interpretation of the two functors:

$$X⟼\text{the coalgebra freely spanned by group-like points},$$

while

$$C⟼\text{the set of coalgebraic points of }C.$$

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Recovering the original set

The unit of this adjunction is particularly simple.

Every element $x\in X$$x\in X$ is group-like in $k[X]$$k[X]$, so there is a canonical map

$$X\to G(k[X]).$$

In fact this is a bijection.

Let

$$g=\sum _x{a}_{x}x$$

be group-like. Then

$$\mathrm{\Delta }(g)=\sum _x{a}_{x}x\otimes x,$$

whereas

$$g\otimes g=\sum _{x,y}{a}_x{a}_{y}x\otimes y.$$

Comparing coefficients gives

$$a_x{a}_y=0(x\ne y)$$

and

$$a_x=a_x^2.$$

Since $k$$k$ is a field, at most one coefficient is nonzero, and every nonzero coefficient satisfying $a_x^2=a_x$$a_x^2=a_x$ must be $1$$1$. The condition

$$\epsilon (g)=1$$

then shows that exactly one coefficient is $1$$1$.

Hence every group-like element is one of the original basis vectors:

$$G(k[X])\cong X.$$

Thus the unit

$$X\to G(k[X])$$

is an isomorphism, and consequently

$$k[-]:\mathrm{Set}\to {\mathrm{Coalg}}_k$$

is fully faithful.

So $\mathrm{Set}$$\operatorname{Set}$ sits inside ${\mathrm{Coalg}}_k$$\operatorname{Coalg}_k$ as the full subcategory of coalgebras of the form $k[X]$$k[X]$.

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The group-like part of a coalgebra

The counit of the adjunction at $C$$C$ is

$$k[G(C)]\to C.$$

If $[g]$$[g]$ denotes the formal basis vector corresponding to $g\in G(C)$$g\in G(C)$, the map is simply

$$[g]⟼g.$$

Distinct group-like elements of a coalgebra are linearly independent. Hence this map is injective.

We may therefore regard

$$k[G(C)]$$

as a canonical subcoalgebra of $C$$C$: the subcoalgebra spanned by all group-like elements.

Since $k[-]$$k[-]$ is fully faithful and right adjoint to $G$$G$ on its essential image, the coalgebras of the form $k[X]$$k[X]$ form a coreflective full subcategory of ${\mathrm{Coalg}}_k$$\operatorname{Coalg}_k$.

The coreflection of $C$$C$ is precisely

$$k[G(C)]⟶C.$$

Thus $G$$G$ does not merely extract a set of distinguished vectors. Together with $k[-]$$k[-]$, it extracts a canonical coalgebraic part of $C$$C$.

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The monoidal structure

There is another feature of this adjunction that turns out to be crucial.

The category of sets is cartesian monoidal:

$$(\mathrm{Set},\times ,\ast ),$$

while ${\mathrm{Coalg}}_k$$\operatorname{Coalg}_k$ is symmetric monoidal under the usual tensor product of coalgebras:

$$({\mathrm{Coalg}}_k,\otimes ,k).$$

For sets $X$$X$ and $Y$$Y$ there is a natural isomorphism

$$k[X\times Y]\cong k[X]\otimes k[Y],$$

given on basis elements by

$$(x,y)⟼x\otimes y.$$

Likewise,

$$k[\ast ]\cong k.$$

These maps respect the coalgebra structures, so

$$k[-]:(\mathrm{Set},\times ,\ast )⟶({\mathrm{Coalg}}_k,\otimes ,k)$$

is strong symmetric monoidal.

The group-like functor has the corresponding property.

Given coalgebras $C$$C$ and $D$$D$, their tensor product has comultiplication

$${\mathrm{\Delta }}_{C\otimes D}=(1\otimes \tau \otimes 1)({\mathrm{\Delta }}_C\otimes {\mathrm{\Delta }}_D),$$

so

$$\mathrm{\Delta }(c\otimes d)=\sum (c_{(1)}\otimes d_{(1)})\otimes (c_{(2)}\otimes d_{(2)}).$$

If $c$$c$ and $d$$d$ are group-like, then

$$\mathrm{\Delta }(c\otimes d)=(c\otimes d)\otimes (c\otimes d),$$

and

$$\epsilon (c\otimes d)=1.$$

Thus there is a natural map

$$G(C)\times G(D)\to G(C\otimes D),(c,d)\mapsto c\otimes d.$$

It is in fact a bijection.

Indeed, if $z\in G(C\otimes D)$$z\in G(C\otimes D)$, define

$$c=(1\otimes {\epsilon }_D)(z),d=({\epsilon }_C\otimes 1)(z).$$

Both $c$$c$ and $d$$d$ are group-like, and applying the counit identities to

$$\mathrm{\Delta }(z)=z\otimes z$$

gives

$$z=c\otimes d.$$

Hence

$$G(C\otimes D)\cong G(C)\times G(D),$$

and clearly

$$G(k)\cong \ast .$$

Therefore

$$G:({\mathrm{Coalg}}_k,\otimes ,k)⟶(\mathrm{Set},\times ,\ast )$$

is also strong symmetric monoidal.

So the adjunction

$$k[-]⊣G$$

is not merely an adjunction of ordinary categories. It is compatible with the monoidal structures on both sides.

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From monoids to bialgebras

Once the monoidal structure is visible, bialgebras appear almost automatically.

A monoid object in

$$(\mathrm{Set},\times )$$

is simply an ordinary monoid.

A monoid object in

$$({\mathrm{Coalg}}_k,\otimes )$$

is a coalgebra $B$$B$ equipped with coalgebra morphisms

$$m:B\otimes B\to B,\eta :k\to B,$$

satisfying associativity and unitality.

But saying that $m$$m$ and $\eta$$\eta$ are coalgebra morphisms is exactly the usual compatibility condition between the algebra and coalgebra structures of a bialgebra.

Thus

$$\mathrm{Mon}({\mathrm{Coalg}}_k)\simeq {\mathrm{Bialg}}_k.$$

The strong monoidal functors above therefore send monoids to monoids.

For a monoid $M$$M$, the coalgebra $k[M]$$k[M]$ inherits the ordinary monoid algebra multiplication and becomes a bialgebra.

Conversely, if $B$$B$ is a bialgebra, then $G(B)$$G(B)$ inherits a monoid structure. Categorically, multiplication is obtained as

$$G(B)\times G(B)\cong G(B\otimes B)\stackrel{G(m)}{\to }G(B).$$

On elements this is simply

$$(g,h)⟼gh.$$

The usual fact that the product of two group-like elements is group-like is therefore not an isolated computation. It is a consequence of the strong monoidality of $G$$G$.

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Hopf algebras in a cartesian world

The Hopf case reveals an even more striking principle.

A Hopf algebra can be described by a PROP whose generators include multiplication, unit, comultiplication, counit and antipode, together with the usual algebra, coalgebra, compatibility and antipode relations.

Now interpret this PROP not in ${\mathrm{Vect}}_k$$\operatorname{Vect}_k$, but in an arbitrary cartesian monoidal category

$$(\mathcal{C},\times ,1).$$

Every object $X$$X$ of a cartesian monoidal category carries a canonical comonoid structure:

$${\mathrm{\Delta }}_X=⟨1_X,1_X⟩:X\to X\times X$$

and

$${\epsilon }_X:X\to 1.$$

Moreover, this comonoid structure is forced by the cartesian product: there is no additional choice.

The comultiplication part of the Hopf structure therefore collapses to the diagonal.

What remains is a multiplication

$$m:X\times X\to X,$$

a unit

$$e:1\to X,$$

and an antipode

$$S:X\to X.$$

The antipode identities become

$$m(S\times 1){\mathrm{\Delta }}_X=e{\epsilon }_X$$

and

$$m(1\times S){\mathrm{\Delta }}_X=e{\epsilon }_X.$$

Since ${\mathrm{\Delta }}_X$$\Delta_X$ is the diagonal, these say internally that

$$S(x)x=e,xS(x)=e.$$

They are precisely the inverse axioms.

Thus a model of the Hopf PROP in a cartesian monoidal category is nothing other than a group object.

In particular,

$$P_{\mathrm{Hopf}}\text{-}\mathrm{Mod}(\mathrm{Set})\simeq \mathrm{Grp}.$$

So the familiar fact that group-like elements of a Hopf algebra form a group is already encoded at the level of the PROP.

When the Hopf theory is transported from the tensor world of coalgebras to the cartesian world of sets, it becomes ordinary group theory.

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From groups to Hopf algebras

The left adjoint now has an immediate interpretation on groups.

Given a group $\mathrm{\Gamma }$$\Gamma$, the coalgebra $k[\mathrm{\Gamma }]$$k[\Gamma]$ already has

$$\mathrm{\Delta }(g)=g\otimes g,\epsilon (g)=1.$$

The group multiplication extends linearly to the usual multiplication on the group algebra, and inversion extends linearly to

$$S(g)=g^{-1}.$$

Thus $k[\mathrm{\Gamma }]$$k[\Gamma]$ becomes a Hopf algebra.

So the functor $k[-]$$k[-]$ restricts to

$$k[-]:\mathrm{Grp}\to {\mathrm{HopfAlg}}_k.$$

In the other direction, if $H$$H$ is a Hopf algebra, then $G(H)$$G(H)$ is a group.

Its multiplication is inherited from $H$$H$, its identity is $1_H$$1_H$, and the inverse of a group-like element $g$$g$ is

$$S(g).$$

Indeed,

$$S(g)g=gS(g)=1.$$

Hence there is a functor

$$G:{\mathrm{HopfAlg}}_k\to \mathrm{Grp}.$$

This is not merely a pair of related constructions. The original adjunction survives at the Hopf level.

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The group algebra adjunction

Let $\mathrm{\Gamma }$$\Gamma$ be a group and $H$$H$ a Hopf algebra.

Suppose first that

$$F:k[\mathrm{\Gamma }]\to H$$

is a Hopf algebra morphism.

Every $g\in \mathrm{\Gamma }$$g\in\Gamma$ is group-like in $k[\mathrm{\Gamma }]$$k[\Gamma]$, so $F(g)$$F(g)$ is group-like in $H$$H$. Since $F$$F$ preserves multiplication,

$$F(gh)=F(g)F(h).$$

Thus restriction gives a group homomorphism

$$\mathrm{\Gamma }\to G(H).$$

Conversely, suppose

$$f:\mathrm{\Gamma }\to G(H)$$

is a group homomorphism.

It extends uniquely by linearity to

$$\tilde{f}:k[\mathrm{\Gamma }]\to H.$$

Since $f$$f$ is multiplicative,

$$\tilde{f}(gh)=f(gh)=f(g)f(h),$$

so $\tilde{f}$$\widetilde f$ is an algebra morphism.

Since every $f(g)$$f(g)$ is group-like,

$$\mathrm{\Delta }(f(g))=f(g)\otimes f(g),$$

and

$$\epsilon (f(g))=1,$$

so $\tilde{f}$$\widetilde f$ is also a coalgebra morphism.

Finally,

$$S(f(g))=f(g{)}^{-1}=f(g^{-1}),$$

so the antipodes are compatible.

Thus $\tilde{f}$$\widetilde f$ is a Hopf algebra morphism.

These two constructions are inverse and natural, giving

$${\mathrm{HopfAlg}}_k(k[\mathrm{\Gamma }],H)\cong \mathrm{Grp}(\mathrm{\Gamma },G(H)).$$

Therefore

$$k[-]:\mathrm{Grp}\rightleftarrows {\mathrm{HopfAlg}}_k:G$$

is an adjunction.

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Groups sit inside Hopf algebras

As before,

$$G(k[\mathrm{\Gamma }])\cong \mathrm{\Gamma }.$$

Hence the unit

$$\mathrm{\Gamma }\to G(k[\mathrm{\Gamma }])$$

is an isomorphism.

It follows that the group algebra functor

$$k[-]:\mathrm{Grp}\to {\mathrm{HopfAlg}}_k$$

is fully faithful.

Thus groups form a full subcategory of Hopf algebras through the group algebra construction.

For any Hopf algebra $H$$H$, the counit of the adjunction is

$$k[G(H)]\to H.$$

Since distinct group-like elements are linearly independent, this map is injective. Its image is the Hopf subalgebra spanned by the group-like elements of $H$$H$.

So every Hopf algebra contains a canonical group algebra:

$$k[G(H)]\hookrightarrow H.$$

This is the largest part of $H$$H$ that is built purely from group-like points.

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The conceptual picture

The construction begins with a very elementary-looking definition:

$$\mathrm{\Delta }(g)=g\otimes g,\epsilon (g)=1.$$

But these equations can be reorganized into a sequence of categorical observations.

First,

$$G(C)\cong {\mathrm{Coalg}}_k(k,C),$$

so group-like elements are represented points.

Then

$$k[-]⊣G$$

relates sets to coalgebras.

Both functors are strong symmetric monoidal:

$$k[X\times Y]\cong k[X]\otimes k[Y],$$

and

$$G(C\otimes D)\cong G(C)\times G(D).$$

Consequently, algebraic structure can be transported across the adjunction.

Monoids become bialgebras, while the Hopf PROP, when interpreted in a cartesian monoidal category, becomes the theory of group objects.

This produces the second adjunction

$$k[-]:\mathrm{Grp}\rightleftarrows {\mathrm{HopfAlg}}_k:G.$$

The familiar statement

the group-like elements of a Hopf algebra form a group

is therefore only the visible shadow of a more structural fact.

The group-like functor sends the Hopf theory from the tensor world of coalgebras into the cartesian world of sets. In the cartesian world, comultiplication becomes the diagonal, the counit becomes the terminal map, and the antipode becomes inversion.

What remains is precisely a group.

So the word group-like is not merely suggestive terminology. Categorically, these points become genuine group elements once the Hopf structure is viewed through the appropriate monoidal functor.