Representable Functors, Group-Like Elements, and Primitives in Hopf Algebras

Primitive Elements: The Lie Algebra Hidden Inside a Hopf AlgebraSome Immediate Properties of Primitive ElementsPrimitive Elements Form a Lie AlgebraPrimitive Elements Define a FunctorThe Universal Enveloping Algebra Is Its Left AdjointRepresentability and the Monoidal Origin of Group-Like and Primitive ElementsGroup-Like Elements Are RepresentablePrimitive Elements Are RepresentableA General Principle Behind the Two RepresentationsStrong Monoidality and the Group-Like MultiplicationPrimitive Elements and the Additive GroupThe Two Strong Monoidal Functors

 

Primitive Elements: The Lie Algebra Hidden Inside a Hopf Algebra

A Hopf algebra carries both multiplication and comultiplication:

$$m:H\otimes H\to H,\mathrm{\Delta }:H\to H\otimes H.$$

It is precisely the interaction between these two structures that allows both groups and Lie algebras to arise naturally inside a Hopf algebra.

We are already familiar with the group-like elements:

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

They form a group under multiplication.

In parallel, an element $x\in H$$x\in H$ is called primitive if

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

We write

$$P(H)=\mathrm{Prim}(H)=\{x\in H\mid \mathrm{\Delta }(x)=x\otimes 1+1\otimes x\}.$$

At first sight, this merely looks like a way of selecting a special class of elements from a Hopf algebra. In fact, the primitive elements automatically form a Lie algebra, and this construction defines a functor

$$P:{\mathbf{Hopf}}_k⟶{\mathbf{Lie}}_k.$$

More importantly, it is the right adjoint of the universal enveloping algebra functor:

$$U⊣P.$$

In characteristic zero, this adjunction has a particularly strong property:

$$P(U(\mathfrak{g}))=\mathfrak{g}.$$

Thus, if we first pass from a Lie algebra to its universal enveloping algebra and then extract the primitive elements, no new primitive directions appear.

The goal of this article is to explain this functor and prove this statement.

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Some Immediate Properties of Primitive Elements

Let $x\in P(H)$$x\in P(H)$. Starting from

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

apply $\epsilon \otimes \mathrm{id}$$\varepsilon\otimes\operatorname{id}$. We obtain

$$x=\epsilon (x)1+x,$$

and hence

$$\epsilon (x)=0.$$

Using the antipode identity

$$m(S\otimes \mathrm{id})\mathrm{\Delta }=\eta \epsilon ,$$

we similarly obtain

$$S(x)+x=0,$$

so that

$$S(x)=-x.$$

Thus a primitive element satisfies

$$\mathrm{\Delta }(x)=x\otimes 1+1\otimes x,\epsilon (x)=0,S(x)=-x.$$

This should be compared with a group-like element:

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

The group-like condition is multiplicative, while the primitive condition is linear.

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Primitive Elements Form a Lie Algebra

Every associative algebra $A$$A$ becomes a Lie algebra under the commutator bracket

$$[x,y]=xy-yx.$$

Now let $x,y\in P(H)$$x,y\in P(H)$. Since $\mathrm{\Delta }$$\Delta$ is an algebra homomorphism,

$$\begin{array}{rl}\mathrm{\Delta }([x,y])& =\mathrm{\Delta }(xy-yx)\\ & =\mathrm{\Delta }(x)\mathrm{\Delta }(y)-\mathrm{\Delta }(y)\mathrm{\Delta }(x).\end{array}$$

Substituting

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

we get

$$\begin{array}{rl}\mathrm{\Delta }([x,y])& =(x\otimes 1+1\otimes x)(y\otimes 1+1\otimes y)\\ & -(y\otimes 1+1\otimes y)(x\otimes 1+1\otimes x).\end{array}$$

After expanding, the mixed terms cancel, leaving

$$\mathrm{\Delta }([x,y])=[x,y]\otimes 1+1\otimes [x,y].$$

Therefore

$$[x,y]\in P(H).$$

Hence $P(H)$$P(H)$ is a Lie subalgebra of the commutator Lie algebra $H_{\mathrm{Lie}}$$H_{\mathrm{Lie}}$:

$$P(H)\subseteq H_{\mathrm{Lie}}.$$

Notice that no cocommutativity assumption on $H$$H$ is needed.

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Primitive Elements Define a Functor

Let

$$f:H\to K$$

be a morphism of Hopf algebras.

If $x\in P(H)$$x\in P(H)$, then

$$\begin{array}{rl}{\mathrm{\Delta }}_K(f(x))& =(f\otimes f){\mathrm{\Delta }}_H(x)\\ & =(f\otimes f)(x\otimes 1+1\otimes x)\\ & =f(x)\otimes 1+1\otimes f(x).\end{array}$$

Therefore

$$f(x)\in P(K).$$

Since $f$$f$ is also an algebra homomorphism,

$$f([x,y])=[f(x),f(y)].$$

Thus restriction gives a Lie algebra homomorphism

$$P(f):P(H)\to P(K).$$

Hence we obtain a functor

$$P:{\mathbf{Hopf}}_k\to {\mathbf{Lie}}_k.$$

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The Universal Enveloping Algebra Is Its Left Adjoint

Let $\mathfrak{g}$$\mathfrak g$ be a Lie algebra. Its universal enveloping algebra $U(\mathfrak{g})$$U(\mathfrak g)$ carries a canonical Hopf algebra structure determined on $x\in \mathfrak{g}$$x\in\mathfrak g$ by

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

$$\epsilon (x)=0,$$

and

$$S(x)=-x.$$

Therefore the canonical map

$$i:\mathfrak{g}\to U(\mathfrak{g})$$

actually lands in the primitive elements:

$$\mathfrak{g}⟶P(U(\mathfrak{g})).$$

More importantly, there is a natural bijection

$${\mathrm{Hom}}_{\mathbf{Hopf}}(U(\mathfrak{g}),H)\cong {\mathrm{Hom}}_{\mathbf{Lie}}(\mathfrak{g},P(H)).$$

Thus

$$U⊣P.$$

Let us see why.

Suppose we are given a Lie algebra homomorphism

$$f:\mathfrak{g}\to P(H).$$

Forgetting the coalgebra structure for the moment, the universal property of $U(\mathfrak{g})$$U(\mathfrak g)$ gives a unique algebra homomorphism

$$\tilde{f}:U(\mathfrak{g})\to H.$$

Since $f(x)$$f(x)$ is primitive,

$${\mathrm{\Delta }}_H(f(x))=f(x)\otimes 1+1\otimes f(x).$$

Hence for every $x\in \mathfrak{g}$$x\in\mathfrak g$,

$${\mathrm{\Delta }}_H\tilde{f}(x)=(\tilde{f}\otimes \tilde{f}){\mathrm{\Delta }}_{U(\mathfrak{g})}(x).$$

Both sides are algebra homomorphisms from $U(\mathfrak{g})$$U(\mathfrak g)$ to $H\otimes H$$H\otimes H$, and they agree on the generators $\mathfrak{g}$$\mathfrak g$. Therefore they agree everywhere.

The counit and antipode are handled similarly.

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

Equivalently:

The primitive condition is exactly the extra condition needed for a Lie algebra map $\mathfrak{g}\to H_{\mathrm{Lie}}$$\mathfrak g\to H_{\mathrm{Lie}}$ to extend to a Hopf algebra map $U(\mathfrak{g})\to H$$U(\mathfrak g)\to H$.

Representability and the Monoidal Origin of Group-Like and Primitive Elements

The group-like and primitive-element constructions admit another categorical interpretation. Both are representable, and both interact strongly with tensor products.

These two facts describe different aspects of the construction:

• representability tells us that a group-like or primitive element can be regarded as a morphism from a universal probe;

• strong monoidality explains how the multiplication of a Hopf algebra induces algebraic structure on the collection of such elements.

It is useful to examine these two ideas separately.

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Group-Like Elements Are Representable

For a Hopf algebra $H$$H$, let

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

The elements of $G(H)$$G(H)$ form a group under the multiplication of $H$$H$.

Consider the infinite cyclic group $\mathbb{Z}$$\mathbb Z$. For every group $\mathrm{\Gamma }$$\Gamma$, choosing a group homomorphism

$$\mathbb{Z}\to \mathrm{\Gamma }$$

is equivalent to choosing the image of $1\in \mathbb{Z}$$1\in\mathbb Z$, hence to choosing an arbitrary element of $\mathrm{\Gamma }$$\Gamma$. Therefore

$$|\mathrm{\Gamma }|\cong {\mathrm{Hom}}_{\mathbf{Grp}}(\mathbb{Z},\mathrm{\Gamma }).$$

Now use the adjunction

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

We obtain

$$\begin{array}{rl}|G(H)|& \cong {\mathrm{Hom}}_{\mathbf{Grp}}(\mathbb{Z},G(H))\\ & \cong {\mathrm{Hom}}_{{\mathbf{HopfAlg}}_k}(k[\mathbb{Z}],H).\end{array}$$

Thus the underlying-set-valued group-like functor is represented by

$$k[\mathbb{Z}].$$

In other words,

$$|G(H)|\cong {\mathrm{Hom}}_{{\mathbf{HopfAlg}}_k}(k[\mathbb{Z}],H).$$

If $z$$z$ denotes the canonical generator of $k[\mathbb{Z}]$$k[\mathbb Z]$, then

$$\mathrm{\Delta }(z)=z\otimes z,\epsilon (z)=1,S(z)=z^{-1}.$$

A Hopf algebra morphism

$$f:k[\mathbb{Z}]\to H$$

is completely determined by $f(z)$$f(z)$, and the Hopf compatibility forces $f(z)$$f(z)$ to be group-like.

Thus $k[\mathbb{Z}]$$k[\mathbb Z]$ is the universal Hopf algebra containing one group-like element.

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Primitive Elements Are Representable

Now consider

$$P(H)=\{x\in H\mid \mathrm{\Delta }(x)=x\otimes 1+1\otimes x\}.$$

Let $\mathfrak{a}=kx$$\mathfrak a=kx$ be the one-dimensional abelian Lie algebra.

For every Lie algebra $\mathfrak{g}$$\mathfrak g$, a Lie algebra morphism

$$\mathfrak{a}\to \mathfrak{g}$$

is completely determined by the image of $x$$x$, and this image may be any element of $\mathfrak{g}$$\mathfrak g$. Hence

$$|\mathfrak{g}|\cong {\mathrm{Hom}}_{{\mathbf{Lie}}_k}(\mathfrak{a},\mathfrak{g}).$$

Using the adjunction

$$U:{\mathbf{Lie}}_k\rightleftarrows {\mathbf{HopfAlg}}_k:P,$$

we obtain

$$\begin{array}{rl}|P(H)|& \cong {\mathrm{Hom}}_{{\mathbf{Lie}}_k}(\mathfrak{a},P(H))\\ & \cong {\mathrm{Hom}}_{{\mathbf{HopfAlg}}_k}(U(\mathfrak{a}),H).\end{array}$$

Since $\mathfrak{a}$$\mathfrak a$ is one-dimensional and abelian,

$$U(\mathfrak{a})\cong k[t],$$

with

$$\mathrm{\Delta }(t)=t\otimes 1+1\otimes t,\epsilon (t)=0,S(t)=-t.$$

Therefore

$$|P(H)|\cong {\mathrm{Hom}}_{{\mathbf{HopfAlg}}_k}(k[t],H).$$

A Hopf algebra morphism

$$f:k[t]\to H$$

is determined by $f(t)$$f(t)$, and Hopf compatibility says precisely that

$$\mathrm{\Delta }(f(t))=f(t)\otimes 1+1\otimes f(t).$$

Thus $k[t]$$k[t]$ is the universal Hopf algebra containing one primitive element.

We therefore obtain the parallel picture

$$\begin{array}{rl}\text{group-like element}& ⟷k[\mathbb{Z}]\to H,\\ \text{primitive element}& ⟷k[t]\to H.\end{array}$$

The representing objects themselves reflect the two defining equations:

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

and

$$\mathrm{\Delta }(t)=t\otimes 1+1\otimes t.$$

The first is multiplicative, while the second is its infinitesimal, additive analogue.

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A General Principle Behind the Two Representations

Both calculations are instances of the same elementary categorical fact.

Suppose

$$L:\mathcal{C}\rightleftarrows \mathcal{D}:R$$

is an adjunction, and suppose a Set-valued functor on $\mathcal{C}$$\mathcal C$ is represented by an object $A$$A$:

$$F(X)\cong {\mathrm{Hom}}_{\mathcal{C}}(A,X).$$

Then

$$F(R(Y))\cong {\mathrm{Hom}}_{\mathcal{D}}(L(A),Y).$$

Thus representability is transported across the adjunction.

For group-like elements, the generic element of a group is represented by

$$\mathbb{Z},$$

and the left adjoint sends it to

$$k[\mathbb{Z}].$$

For primitive elements, the generic element of a Lie algebra is represented by the one-dimensional abelian Lie algebra

$$\mathfrak{a},$$

and the left adjoint sends it to

$$U(\mathfrak{a})=k[t].$$

Hence the representing Hopf algebras are not accidental. They are the images of the generic one-generator objects under the corresponding left adjoints.

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Strong Monoidality and the Group-Like Multiplication

Representability tells us what an individual group-like element is. To understand why group-like elements form a group, it is useful to move one categorical level down and regard a Hopf algebra as a monoid object in coalgebras equipped with an antipode.

Let

$$G_0:({\mathbf{Coalg}}_k,\otimes ,k)⟶(\mathbf{Set},\times ,\ast )$$

be the group-like-element functor.

It is strong symmetric monoidal:

$$G_0(C\otimes D)\cong G_0(C)\times G_0(D).$$

Under this isomorphism,

$$(g,h)⟼g\otimes h.$$

Indeed, if $g$$g$ and $h$$h$ are group-like, then

$$\mathrm{\Delta }(g\otimes h)=(g\otimes h)\otimes (g\otimes h).$$

Now a bialgebra $H$$H$ is precisely a monoid object in the monoidal category of coalgebras. Its multiplication and unit are coalgebra morphisms

$$m:H\otimes H\to H,\eta :k\to H.$$

A strong monoidal functor sends monoid objects to monoid objects. Therefore $G_0$$G_0$ sends the multiplication of $H$$H$ to

$$G_0(H)\times G_0(H)\cong G_0(H\otimes H)\stackrel{G_0(m)}{\to }{G}_0(H).$$

Explicitly,

$$(g,h)⟼g\otimes h⟼gh.$$

Thus the multiplication on group-like elements is not an additional construction. It is the Hopf multiplication transported through the strong monoidal functor $G_0$$G_0$.

The strong monoidal argument first gives a monoid. When $H$$H$ is a Hopf algebra, the antipode satisfies

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

for every group-like element $g$$g$, so this monoid is in fact a group.

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Primitive Elements and the Additive Group

There is a completely parallel construction for primitive elements, but with an important change in the target monoidal structure.

For a coaugmented coalgebra $C$$C$, let

$$P_0(C)=\{x\in C\mid \mathrm{\Delta }(x)=x\otimes 1+1\otimes x\}.$$

Then

$$P_0:({\mathbf{Coalg}}_k^{\mathrm{coaug}},\otimes ,k)⟶({\mathbf{Vect}}_k,\oplus ,0)$$

is strong symmetric monoidal:

$$P_0(C\otimes D)\cong P_0(C)\oplus P_0(D).$$

The comparison map is

$$(x,y)⟼x\otimes 1+1\otimes y.$$

Now take a Hopf algebra $H$$H$. Its multiplication

$$m:H\otimes H\to H$$

is a morphism of coaugmented coalgebras. Applying $P_0$$P_0$ gives

$$P_0(H\otimes H)\stackrel{P_0(m)}{\to }{P}_0(H).$$

Using strong monoidality, this becomes

$$P(H)\oplus P(H)⟶P(H).$$

What operation is this?

Starting with $(x,y)$$(x,y)$,

$$(x,y)⟼x\otimes 1+1\otimes y,$$

and then applying multiplication gives

$$m(x\otimes 1+1\otimes y)=x+y.$$

Therefore the multiplication transported through $P_0$$P_0$ is

$$(x,y)⟼x+y.$$

The unit becomes

$$0,$$

and the antipode acts on primitive elements by

$$S(x)=-x.$$

Thus the group structure produced on primitive elements is precisely their additive group structure:

$$(P(H),+,0,-).$$

If we further compose with the underlying-set functor

$$U:({\mathbf{Vect}}_k,\oplus ,0)⟶(\mathbf{Set},\times ,\ast ),$$

then

$$U\circ P_0$$

is again strong symmetric monoidal, and it sends a Hopf algebra to the ordinary additive group underlying its primitive vector space.

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The Two Strong Monoidal Functors

We can now place the two constructions side by side:

$$G_0:(\mathbf{Coalg},\otimes )⟶(\mathbf{Set},\times ),$$

with

$$G_0(C\otimes D)\cong G_0(C)\times G_0(D),$$

and

$$P_0:({\mathbf{Coalg}}^{\mathrm{coaug}},\otimes )⟶(\mathbf{Vect},\oplus ),$$

with

$$P_0(C\otimes D)\cong P_0(C)\oplus P_0(D).$$

For the same Hopf multiplication

$$m:H\otimes H\to H,$$

the first functor produces

$$(g,h)⟼gh,$$

while the second produces

$$(x,y)⟼x+y.$$

Thus

$$\begin{array}{rl}G(H)& \text{ sees the multiplicative group carried by group-like points},\\ P(H)& \text{ sees the additive group carried by infinitesimal points}.\end{array}$$

This is the categorical form of the passage from a group to its infinitesimal linearization.

There is, however, one further piece of structure on $P(H)$$P(H)$.

The additive group law on $P(H)$$P(H)$ is explained entirely by the strong monoidality of $P_0$$P_0$. The Lie bracket

$$[x,y]=xy-yx$$

is additional information. It comes from the noncommutativity of the multiplication of $H$$H$, together with the fact that primitive elements are closed under commutators.

Hence the two structures on $P(H)$$P(H)$ have conceptually different origins:

$$\begin{array}{rl}\text{strong monoidality}& ⇝x+y,\\ \text{noncommutative multiplication}& ⇝[x,y].\end{array}$$

This distinction mirrors ordinary Lie theory. The differential of group multiplication at the identity is

$$d{m}_{(e,e)}:\mathfrak{g}\oplus \mathfrak{g}\to \mathfrak{g},$$

and is simply

$$(X,Y)⟼X+Y.$$

The Lie bracket is not this first derivative; it records a higher-order failure of commutativity.

In the Hopf-algebraic picture, group-like elements retain the multiplicative structure itself, while primitive elements retain its infinitesimal additive structure, with the commutator supplying the additional Lie bracket.