Taken from An Invitation to Quantum Groups and Duality by Timmermann.

Let A be a quantum group with a comultiplication \Delta. We make the following definitions. A corepresentation of A on a complex vector space V is a linear map \chi:V\rightarrow V\otimes A that dualises representations with the coassociativity and counit properties:

(I\otimes \Delta)\circ \chi=(\chi\otimes I)\circ \chi, and

(I \otimes\varepsilon)\circ\chi=I.

Now we wish to dualise the terms invariantirreducible, unitary, intertwiner, equivalent, matrix elements. As we want a theory dual to group representation we won’t use Timmermann’s definitions at face value but instead construct them from their group representation counterparts. This might prove difficult. In attempting to dualise group representation theory as quantum group corepresentation theory is ‘everything’ dualised?  Our first term here provides a problem. Do we need an invariant corepresentation or a ‘coinvariant representation’?


An invariant subspace of a group representation \Phi:(V,G) is a subspace W\subset V such that

\Phi(w,g)\in W for all w\in W and g\in G.

This means that for the family of linear maps \{\rho(g):g\in G\}W is stable subspace. How this is dualised is important in how we may hope to write a corepresentation as a direct sum of irreducible corepresentations so we need the right definition. Timmermann calls W\subset V invariant if \chi(W)\subset W\otimes A. If we could view the co-representation as a family of endomorphisms on V then we might be able to write down a definition of co-invariant or maybe say that Timmermann’s definition is what we need.

As an example of what we might need to do let \Phi:F(G)\times G\rightarrow G be the regular action of a group and let W\subset be the subspace of constant functions. This set is invariant. Thinking about this yields a definition of co-invariant. 

A subspace W\subset V is co-invariant for \chi if W\otimes A\subset \chi(W).

Timmermann’s definition makes good sense. One is confused between invariant co-representation and co-invariant co-representation. I am guessing that Timmermann’s definition will allow us to do what we want.


We call a representation irreducible if it has no non-trivial subspaces. We can do the exactly the same for co-representations.


No problem here we want \langle \chi(x),\chi(y)\rangle_{V\otimes A}=\langle x,y\rangle where we might take the Haar inner product on A\langle a,b\rangle=\eta(a^\ast b), where \eta is the Haar state on AI forgot this.

Intertwiners & Equivalent Corepresentations

In the group representation picture, an intertwiner of \Phi_1 and \Phi_2 is a bijective linear map which makes the following diagram commute.


We can find an appropriate definition of an intertwiner of two corepresentations \chi_1 and \chi_2 of a quantum group A by reversing arrows:


Two representations \Phi_1 and \Phi_2 if there exists an intertwiner for them. The intertwiner by definition is bijective. Two corepresentations can now be said to be equivalent if there they have an intertwiner.

Schur’s Lemma

Suppose that \chi_1:U\rightarrow U\otimes A and \chi_2:V\rightarrow V\otimes A are two irreducible corepresentations of a quantum group and let f:U\rightarrow V an intertwiner. 

  • If \chi_1 and \chi_2 are not equivalent then f\equiv 0
  •  If U is complex and g:U\rightarrow U is an intertwiner of U with itself then g=\lambda\,I

First we show that the kernel and image of an intertwiner are invariant subspaces. The kernel is invariant by the following diagram:


Similarly the image is invariant by the following diagram:


Hence by irreducibility the kernel and image are either trivial or the whole space. The rest of the theorem follows through from the classical argument \bullet

Now we want to work towards some orthogonality relations.

From this point we follow Timmermann. Timmermann works with compact algebraic quantum groups of which finite quantum groups are.

Theorem 3.2.1

Let \chi be a corepresentation of (A,\Delta) over a vector space V

  1. If (A,\Delta) is a algebraic compact quantum group, then \chi is equivalent to a unitary corepresentation.
  2. If \chi is unitary and W\subset V is an invariant subspace, then the complement W^\perp is also invariant.
  3. Every element v\in V is contained in some finite-dimensional invariant subspace of V. In particular, V has finite dimension if \chi is irreducible.
  4. If for every finite-dimensional invariant subspace W\subset V, the restriction \chi_{\left|W\right.} is equivalent to a unitary corepresentation, in particular, if (A,\Delta) is an algebraic compact quantum group, then \chi is equivalent to a direct sum of finite-dimensional irreducible unitary corepresentations.


1. Choose some Hermitian inner product \langle\cdot|\cdot\rangle_V on V. Since the Haar stare of (A,\Delta) is positive and faithful (proof in here), (a,b)\mapsto \eta(a^\ast b) defines a Hermitian inner product on A. By a standard argument  \langle v\otimes a|w\otimes b\rangle=\langle w|v\rangle_V \eta(a^*b) defines a Hermitian product on V\otimes A. As \chi is injective, we can define a second Hermitian inner product \langle\cdot|\cdot\rangle'_V on V by the formula

\displaystyle \langle v|w\rangle'_V:=\langle \chi(v),\chi(u)\rangle=\sum\langle v_{(0)}|w_{(0)}\rangle_V\cdot \eta\left(v_{(1)}^\ast w_{(1)}\right) for all v,\,w\in V,

and \chi(v) expressed in Sweedler’s notation. An associated A-valued product

\displaystyle\langle\chi(v)|\chi(w)\rangle_A'=\sum\langle v_{(0)}|w_{(0)}\rangle'_V v_{(1)}^*w_{(1)}=\sum \langle v_{(0)}|w_{(0)}\rangle_V\eta(v_{(1)}^\ast w_{(1)})v_{(2)}^\ast w_{(2)}

for all v,\,w\in V. Timmermann goes on to show that \chi is unitary with respect the inner product

2. This is a standard argument according to Timmermann.

3. Denote by \pi:A'\rightarrow \text{Hom}(V) the representation

\pi(f)v=(I_V\otimes f)(\chi(v)).

Note that \text{Hom}(U,V) is the space of intertwiners from U to V.

Since \pi(\varepsilon)=I_V, by definition, the subspace \pi(A')v\subset V contains v. Evidently, \pi(A')v is invariant for \pi, so by a previous proposition of Timmermann (3.1.7 (vi)) also for \chi. If \chi(v)=\sum_i v_i\otimes a_i, where v_i\in V and a_i\in A, then \pi(A')v is contained in the linear span of the v_i. Therefore \pi(A')v has finite dimension.

4. Follows from 3. & 4. and Zorn’s Lemma.

Proposition 3.2.3

Two irreducible unitary co-representations are equivalent if and only if they admit a unitary intertwiner.

Schur’s Orthogonality Relations

The matrix elements of irreducible co-representations satisfy an analogue of Schur’s orthogonality relations known from the representation theory of compact groups. It is precisely this kind of a result that we need to write a quantised Diaconis Upper Bound Lemma.


Let (A,\Delta) be a Hopf *-algebra with normalised integral h, let \chi_V and \chi_W be co-representations on finite-dimensional vector spaces V and W, respectively, and let R\in\text{Hom}(V,W). Denote by X and Y the co-representation operators corresponding to \chi_V and \chi_W, respectively, and define S,\,T\in\text{Hom}(V,W) by 

S:=(I\otimes h)(Y^{-1}(R\otimes 1)X) and T:=(I\otimes h)(Y(R\otimes 1)X^{-1}).

Then S,\,T\in \text{Hom}(\chi_V,\chi_W). If R\in\text{Hom}(\chi_V,\chi_W), then S=T=R \bullet

Proposition 3.2.6

Let (A,\Delta) be a Hopf *-algebra with a normalised integral h, and let \chi_V and \chi_W be inequivalent irreducible co-representations of (A,\Delta) on vector spaces V and W, respectively. Then for all s\in\mathcal{C}(\chi_V) and b\in\mathcal{C}(\chi_W),


If \chi_V and \chi_W are unitary, then h(b^\star a)=0=h(ba^\star) for all a\in\mathcal{C}(\chi_V) and b\in\mathcal{C}(\chi_W).

Proof : As the co-representations are irreducible we can take the vector spaces V and W to be finite dimensional. Let X and Y be the corepresentation operators defined by

X(v\otimes 1_A)=\chi_V(v) and Y(w\otimes 1_A)=\chi_W(w).

Let a be the matrix element of X given by

a=\left(\langle v_2|\otimes I\right)X\left(|v_1\rangle\otimes I\right)

and b be the matrix element of Y given by

b=\left(\langle w_2|\otimes I\right)Y\left(|w_1\rangle \otimes I\right).

…I am completely lost now!