Invertible Representations: Picard Groups, Hopf Characters, and Group-Like Elements

Picard Groups, Hopf Characters, and Invertible RepresentationsInvertible objects as loopsWhy the inverse is the dualThe representable functor of Hopf charactersTensor product realizes the group lawReconstructing the invertible part

 

Picard Groups, Hopf Characters, and Invertible Representations

What structure is carried by the invertible objects of a monoidal category?

Taking their isomorphism classes gives a group—the Picard group. For representations of a Hopf algebra, this group is the value at the ground field of a representable group-valued functor. Taking finite direct sums of these invertible representations then recovers a tensor category of graded vector spaces.

We follow these three steps: extract a group, identify its functorial meaning, and reconstruct the tensor category generated by its elements.

Invertible objects as loops

Let $(\mathcal{C},\otimes ,\mathbf{1})$$(\mathcal C,\otimes,\mathbf1)$ be a monoidal category. An object $X$$X$ is tensor-invertible if there exists $Y$$Y$ such that

$$X\otimes Y\cong \mathbf{1},Y\otimes X\cong \mathbf{1}.$$

Assuming their isomorphism classes form a set, these classes constitute the Picard group

$$\mathrm{Pic}(\mathcal{C})=\{[X]:X\text{ is tensor-invertible}\},$$

with multiplication

$$[X][Y]=[X\otimes Y].$$

There is a useful geometric interpretation. Regard $\mathcal{C}$$\mathcal C$ as a bicategory $B\mathcal{C}$$B\mathcal C$ with one object $\bullet$$\bullet$. Objects of $\mathcal{C}$$\mathcal C$ become $1$$1$-morphisms $\bullet \to \bullet$$\bullet\to\bullet$, morphisms become $2$$2$-morphisms, and tensor product becomes composition.

A tensor-invertible object is therefore a self-equivalence of $\bullet$$\bullet$.

Retain only these invertible objects and their isomorphisms. The resulting monoidal groupoid ${\mathcal{C}}^{\times }$$\mathcal C^\times$ is a 2-group. In its delooping, we can picture an object as a loop and an isomorphism as a homotopy between loops. Tensor product composes loops, and a tensor inverse reverses one up to homotopy.

Thus

$${\pi }_1(B{\mathcal{C}}^{\times })\cong \mathrm{Pic}(\mathcal{C}).$$

Here the homotopy groups are those of the associated classifying space.

If $\mathcal{C}$$\mathcal C$ is braided, the braiding gives

$$X\otimes Y\cong Y\otimes X,$$

so the Picard group is abelian. Without a braiding, it need not be.

Why the inverse is the dual

Duality belongs to the same bicategorical picture. The evaluation and coevaluation maps

$$\mathrm{ev}:X^{\ast }\otimes X\to \mathbf{1},\mathrm{coev}:\mathbf{1}\to X\otimes X^{\ast }$$

are the counit and unit of an adjunction, and their snake identities are the triangle identities.

Thus dual objects correspond to adjoint $1$$1$-morphisms.

An equivalence in a bicategory can be equipped with an adjoint equivalence structure. Consequently, a tensor inverse can be equipped with duality data. Uniqueness of duals then gives

$$[X{]}^{-1}=[X^{\ast }].$$

Equivalently, duality gives an adjunction

$$-\otimes X⊣-\otimes X^{\ast }.$$

If $Y$$Y$ is a tensor inverse of $X$$X$, then $-\otimes Y$$-\otimes Y$ is a quasi-inverse, hence also a right adjoint, of $-\otimes X$$-\otimes X$. Uniqueness of right adjoints yields $Y\cong X^{\ast }$$Y\cong X^*$.

The distinction is that duality only requires evaluation and coevaluation to satisfy the snake identities. For an invertible object, these maps are isomorphisms.

The representable functor of Hopf characters

Let $H$$H$ be a Hopf algebra over a field $k$$k$, with coproduct $\mathrm{\Delta }$$\Delta$, counit ${ϵ}_H$$\epsilon_H$, and antipode $S$$S$. Consider its finite-dimensional left representations.

If $V$$V$ is tensor-invertible, then for some $W$$W$,

$$V\otimes W\cong k.$$

Taking dimensions gives

$${\dim }_k(V){\dim }_k(W)=1.$$

Therefore every invertible representation is one-dimensional.

A one-dimensional representation is specified by a unital algebra homomorphism

$$\chi :H⟶k.$$

Allowing the scalars to vary leads to the functor

$${\mathcal{G}}_H(-):{\mathrm{CAlg}}_k⟶\mathbf{Set},A⟼{\mathrm{Alg}}_k(H,A).$$

Its representability follows immediately from the universal property of abelianization. Every map from $H$$H$ to a commutative algebra factors uniquely through

$$H_{\mathrm{ab}}=H/⟨ab-ba:a,b\in H⟩.$$

Hence there is a natural isomorphism of functors

$${\mathcal{G}}_H(-)\cong {\mathrm{Alg}}_k(H_{\mathrm{ab}},-).$$

Thus ${\mathcal{G}}_H(-)$$\mathcal G_H(-)$ is represented by the commutative algebra $H_{\mathrm{ab}}$$H_{\mathrm{ab}}$.

The Hopf structure descends to $H_{\mathrm{ab}}$$H_{\mathrm{ab}}$.

The induced Hopf structure equips ${\mathcal{G}}_H(-)$$\mathcal G_H(-)$ with natural group operations. On $A$$A$-points, multiplication is convolution:

$$(\chi \ast \psi )(h)=m_A((\chi \otimes \psi )(\mathrm{\Delta }h))=\sum \chi (h_{(1)})\psi (h_{(2)}).$$

Its identity is ${\eta }_A\circ {ϵ}_H$$\eta_A\circ\epsilon_H$, and its inverse operation is

$${\chi }^{-1}=\chi \circ S.$$

These operations commute with postcomposition along algebra homomorphisms $A\to B$$A\to B$. Therefore

$${\mathcal{G}}_H(-):{\mathrm{CAlg}}_k⟶\mathbf{Grp}$$

is a group-valued functor whose underlying set-valued functor is representable: an affine group scheme, expressed through its functor of points.

The commutativity of the representing algebra does not imply that the group-valued functor takes values in abelian groups. Commutativity of the group law corresponds to cocommutativity of the coproduct on $H_{\mathrm{ab}}$$H_{\mathrm{ab}}$.

Tensor product realizes the group law

Write $k_{\chi }$$k_\chi$ for the one-dimensional representation defined by $\chi :H\to k$$\chi:H\to k$. On a tensor product,

$$h\cdot (1\otimes 1)=\sum \chi (h_{(1)})\psi (h_{(2)})(1\otimes 1).$$

The coefficient is exactly $(\chi \ast \psi )(h)$$(\chi*\psi)(h)$. Therefore

$$k_{\chi }\otimes k_{\psi }\cong k_{\chi \ast \psi }.$$

The identity element of ${\mathcal{G}}_H(k)$$\mathcal G_H(k)$ gives the trivial representation. The dual action

$$(h\cdot f)(v)=f(S(h)\cdot v)$$

gives

$$(k_{\chi }{)}^{\ast }\cong k_{\chi \circ S}.$$

Thus every one-dimensional representation is invertible, with inverse its dual. Since two one-dimensional representations are isomorphic precisely when their characters agree, we obtain

$$\mathrm{Pic}({\mathrm{Rep}}_k^{\mathrm{fd}}(H))\cong {\mathcal{G}}_H(k).$$

Abelianization establishes representability; tensor product identifies the group of $k$$k$-points with the Picard group.

Reconstructing the invertible part

Having identified the group of invertible representations, what tensor category do these representations generate?

Write

$$\mathrm{\Gamma }=\mathrm{Pic}({\mathrm{Rep}}_k^{\mathrm{fd}}(H))\cong {\mathcal{G}}_H(k).$$

Take finite direct sums of the one-dimensional representations $k_{\chi }$$k_\chi$. An object in this subcategory can be written as

$$V=\underset{\chi \in \mathrm{\Gamma }}{⨁}{V}_{\chi }\otimes k_{\chi },$$

where the $V_{\chi }$$V_\chi$ are multiplicity spaces and the total dimension is finite.

Writing the group operation in $\mathrm{\Gamma }$$\Gamma$ multiplicatively, tensor product follows the rule

$$(V\otimes W{)}_{\eta }=\underset{\chi \psi =\eta }{⨁}{V}_{\chi }\otimes W_{\psi }.$$

This is exactly the tensor product of $\mathrm{\Gamma }$$\Gamma$-graded vector spaces. Hence

$$⟨\text{invertible representations}{⟩}_{\oplus }{\simeq }_{\otimes }{\mathrm{Vect}}_{\mathrm{\Gamma }}^{\mathrm{fd}}.$$

Here we take finite direct sums of invertible representations, not an additional closure under extensions. The tensor identifications are coherent: on the standard one-dimensional models, they are given by $1\otimes 1\mapsto 1$$1\otimes1\mapsto1$.

There is a Hopf-algebraic explanation of this reconstruction.

Let $H^{\circ }$$H^\circ$ denote the finite dual of $H$$H$. Every character $\chi :H\to k$$\chi:H\to k$ lies in $H^{\circ }$$H^\circ$, because its kernel is an ideal of codimension one. Multiplicativity becomes

$$\chi (ab)=\chi (a)\chi (b)⟺{\mathrm{\Delta }}_{H^{\circ }}(\chi )=\chi \otimes \chi ,$$

and preservation of the unit becomes

$${ϵ}_{H^{\circ }}(\chi )=1.$$

Thus the characters are precisely the group-like elements of $H^{\circ }$$H^\circ$, and their multiplication is convolution:

$$\mathrm{\Gamma }\cong G(H^{\circ }).$$

Distinct group-like elements are linearly independent. Their span therefore gives a Hopf algebra embedding

$$k[\mathrm{\Gamma }]\hookrightarrow H^{\circ }.$$

Finally, a $\mathrm{\Gamma }$$\Gamma$-grading is exactly a right $k[\mathrm{\Gamma }]$$k[\Gamma]$-coaction: for $v\in V_g$$v\in V_g$, set

$$\rho (v)=v\otimes g.$$

The tensor product of coactions multiplies the labels:

$$\rho (v\otimes w)=(v\otimes w)\otimes gh(v\in V_g,w\in W_h).$$

Consequently,

$$⟨\text{invertible representations}{⟩}_{\oplus }{\simeq }_{\otimes }{\mathrm{Vect}}_{\mathrm{\Gamma }}^{\mathrm{fd}}{\simeq }_{\otimes }{\mathrm{Comod}}^{\mathrm{fd}}(k[\mathrm{\Gamma }]).$$

The Picard group records the multiplication of the invertible representations. Taking their finite direct sums turns that group multiplication into a linear tensor category.