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In order to further advance, with some finer results, we need to integrate over [math]G[/math]. In the classical case the existence of such an integration is well-known, as follows:

Proposition

Any commutative Woronowicz algebra, [math]A=C(G)[/math] with [math]G\subset U_N[/math], has a unique faithful positive unital linear form [math]\int_G:A\to\mathbb C[/math] satisfying

[[math]] \int_Gf(xy)dx=\int_Gf(yx)dx=\int_Gf(x)dx [[/math]]

called Haar integration. This Haar integration functional can be constructed by starting with any faithful positive unital form [math]\varphi\in A^*[/math], and taking the Cesàro limit

[[math]] \int_G=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\varphi^{*k} [[/math]]

where the convolution operation for linear forms is given by [math]\phi*\psi=(\phi\otimes\psi)\Delta[/math].

Show Proof

This is the existence theorem for the Haar measure of [math]G[/math], in functional analytic formulation. Observe first that the invariance conditions in the statement read:

[[math]] d(xy)=d(yx)=dx\quad,\quad\forall y\in G [[/math]]

Thus, we are looking indeed for the integration with respect to the Haar measure on [math]G[/math]. Now recall that this Haar measure exists, is unique, and can be constructed by starting with any probability measure [math]\mu[/math], and performing the following Cesàro limit:

[[math]] dx=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^nd\mu^{*k}(x) [[/math]]

In functional analysis terms, this corresponds precisely to the second assertion.

■

The above statement and proof, which are quite brief, are of course more of a reminder, with all the technical details missing. However, we will reprove all this later on, as a particular case of a general Haar integration existence result, in the general Woronowicz algebra setting. In general now, let us start with a definition, as follows:

Definition

Given an arbitrary Woronowicz algebra [math]A=C(G)[/math], any positive unital tracial state [math]\int_G:A\to\mathbb C[/math] subject to the invariance conditions

[[math]] \left(\int_G\otimes id\right)\Delta=\left(id\otimes\int_G\right)\Delta=\int_G(.)1 [[/math]]

is called Haar integration over [math]G[/math].

As a first observation, in the commutative case, this notion agrees with the one in Proposition 3.13. To be more precise, Proposition 3.13 tells us that any commutative Woronowicz algebra has a Haar integration in the above sense, which is unique, and which can be constructed by performing the Cesàro limiting procedure there.

Before getting into the general case, let us discuss the group dual case. Here things are quite elementary, and we have the following result:

Proposition

Given a discrete group [math]\Gamma= \lt g_1,\ldots,g_N \gt [/math], the Woronowicz algebra [math]A=C^*(\Gamma)[/math] has a Haar functional, given on the standard generators [math]g\in\Gamma[/math] by:

[[math]] \int_{\widehat{\Gamma}}g=\delta_{g,1} [[/math]]

This functional is faithful on the image on [math]C^*(\Gamma)[/math] in the regular representation. Also, in the abelian case, we obtain in this way the counit of [math]C(\widehat{\Gamma})[/math].

Show Proof

Consider indeed the left regular representation [math]\pi:C^*(\Gamma)\to B(l^2(\Gamma))[/math], given by [math]\pi(g)(h)=gh[/math], that we already met in chapter 1. By composing it with the functional [math]T\to \lt T1,1 \gt [/math], the functional [math]\int_{\widehat{\Gamma}}[/math] that we obtain is given by:

[[math]] \int_{\widehat{\Gamma}}g= \lt g1,1 \gt =\delta_{g,1} [[/math]]

But this gives all the assertions in the statement, namely the existence, traciality, left and right invariance properties, and faithfulness on the reduced algebra. As for the last assertion, this is clear from the Pontrjagin duality isomorphism.

■

With a bit of functional analysis knowledge, one can improve the above result, with a proof of the fact that the Haar integration is unique, and appears via a Cesàro limiting procedure, as in Proposition 3.13. We will do this directly, in the general case.

In order to discuss now the general case, that of the arbitrary Woronowicz algebras, let us define the convolution operation for linear forms by:

[[math]] \phi*\psi=(\phi\otimes\psi)\Delta [[/math]]

We have then the following technical result, from Woronowicz's paper :

Proposition

Given an arbitrary unital linear form [math]\varphi\in A^*[/math], the limit

[[math]] \int_\varphi a=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\varphi^{*k}(a) [[/math]]

exists, and for a coefficient of a corepresentation [math]a=(\tau\otimes id)v[/math], we have

[[math]] \int_\varphi a=\tau(P) [[/math]]

where [math]P[/math] is the orthogonal projection onto the [math]1[/math]-eigenspace of [math](id\otimes\varphi)v[/math].

Show Proof

By linearity, it is enough to prove the first assertion for elements of the following type, where [math]v[/math] is a Peter-Weyl corepresentation, and [math]\tau[/math] is a linear form:

[[math]] a=(\tau\otimes id)v [[/math]]

Thus we are led into the second assertion, and more precisely we can have the whole result proved if we can establish the following formula, with [math]a=(\tau\otimes id)v[/math]:

[[math]] \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\varphi^{*k}(a)=\tau(P) [[/math]]

In order to prove this latter formula, observe that we have:

[[math]] \varphi^{*k}(a) =(\tau\otimes\varphi^{*k})v =\tau((id\otimes\varphi^{*k})v) [[/math]]

Consider now the following matrix, which is a usual complex matrix:

[[math]] M=(id\otimes\varphi)v [[/math]]

In terms of this matrix, we have the following formula:

[[math]] \begin{eqnarray*} ((id\otimes\varphi^{*k})v)_{i_0i_{k+1}} &=&\sum_{i_1\ldots i_k}M_{i_0i_1}\ldots M_{i_ki_{k+1}}\\ &=&(M^k)_{i_0i_{k+1}} \end{eqnarray*} [[/math]]

Thus for any [math]k\in\mathbb N[/math] we have the following formula:

[[math]] (id\otimes\varphi^{*k})v=M^k [[/math]]

It follows that our Cesàro limit is given by the following formula:

[[math]] \begin{eqnarray*} \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\varphi^{*k}(a) &=&\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\tau(M^k)\\ &=&\tau\left(\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^nM^k\right) \end{eqnarray*} [[/math]]

Now since [math]v[/math] is unitary we have [math]||v||=1[/math], and we conclude that we have:

[[math]] ||M||\leq1 [[/math]]

Thus, by standard calculus, the above Cesàro limit on the right exists, and equals the orthogonal projection onto the [math]1[/math]-eigenspace of [math]M[/math]:

[[math]] \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^nM^k=P [[/math]]

Thus our initial Cesàro limit converges as well, to [math]\tau(P)[/math], as desired.

■

When [math]\varphi[/math] is faithful, we have the following finer result, also from Woronowicz :

Proposition

Given a faithful unital linear form [math]\varphi\in A^*[/math], the limit

[[math]] \int_\varphi a=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\varphi^{*k}(a) [[/math]]

exists, and is independent of [math]\varphi[/math], given on coefficients of corepresentations by

[[math]] \left(id\otimes\int_\varphi\right)v=P [[/math]]

where [math]P[/math] is the orthogonal projection onto [math]Fix(v)=\{\xi\in\mathbb C^n|v\xi=\xi\}[/math].

Show Proof

In view of Proposition 3.16, it remains to prove that when [math]\varphi[/math] is faithful, the [math]1[/math]-eigenspace of [math]M=(id\otimes\varphi)v[/math] equals [math]Fix(v)[/math].

“[math]\supset[/math]” This is clear, and for any [math]\varphi[/math], because we have:

[[math]] v\xi=\xi\implies M\xi=\xi [[/math]]

“[math]\subset[/math]” Here we must prove that, when [math]\varphi[/math] is faithful, we have:

[[math]] M\xi=\xi\implies v\xi=\xi [[/math]]

For this purpose, we use a positivity trick. Consider the following element:

[[math]] a=\sum_i\left(\sum_jv_{ij}\xi_j-\xi_i\right)\left(\sum_kv_{ik}\xi_k-\xi_i\right)^* [[/math]]

We want to prove that we have [math]a=0[/math]. Since [math]v[/math] is biunitary, we have:

[[math]] \begin{eqnarray*} a &=&\sum_i\left(\sum_j\left(v_{ij}\xi_j-\frac{1}{N}\xi_i\right)\right)\left(\sum_k\left(v_{ik}^*\bar{\xi}_k-\frac{1}{N}\bar{\xi}_i\right)\right)\\ &=&\sum_{ijk}v_{ij}v_{ik}^*\xi_j\bar{\xi}_k-\frac{1}{N}v_{ij}\xi_j\bar{\xi}_i-\frac{1}{N}v_{ik}^*\xi_i\bar{\xi}_k+\frac{1}{N^2}\xi_i\bar{\xi}_i\\ &=&\sum_j|\xi_j|^2-\sum_{ij}v_{ij}\xi_j\bar{\xi}_i-\sum_{ik}v_{ik}^*\xi_i\bar{\xi}_k+\sum_i|\xi_i|^2\\ &=&||\xi||^2- \lt v\xi,\xi \gt -\overline{ \lt v\xi,\xi \gt }+||\xi||^2\\ &=&2(||\xi||^2-Re( \lt v\xi,\xi \gt )) \end{eqnarray*} [[/math]]

By using now our assumption [math]M\xi=\xi[/math], we obtain from this:

[[math]] \begin{eqnarray*} \varphi(a) &=&2\varphi(||\xi||^2-Re( \lt v\xi,\xi \gt ))\\ &=&2(||\xi||^2-Re( \lt M\xi,\xi \gt ))\\ &=&2(||\xi||^2-||\xi||^2)\\ &=&0 \end{eqnarray*} [[/math]]

Thus [math]a=0[/math], and by positivity we obtain [math]v\xi=\xi[/math], as desired.

■

We can now formulate the general Haar measure result, due to Woronowicz :

Theorem

Any Woronowicz algebra has a unique Haar integration, which can be constructed by starting with any faithful positive unital state [math]\varphi\in A^*[/math], and setting

[[math]] \int_G=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\varphi^{*k} [[/math]]

where [math]\phi*\psi=(\phi\otimes\psi)\Delta[/math]. Moreover, for any corepresentation [math]v[/math] we have

[[math]] \left(id\otimes\int_G\right)v=P [[/math]]

where [math]P[/math] is the orthogonal projection onto [math]Fix(v)=\{\xi\in\mathbb C^n|v\xi=\xi\}[/math].

Show Proof

Let us first go back to the general context of Proposition 3.16 above. Since convolving one more time with [math]\varphi[/math] will not change the Cesàro limit appearing there, the functional [math]\int_\varphi\in A^*[/math] constructed there has the following invariance property:

[[math]] \int_\varphi*\varphi=\varphi*\int_\varphi=\int_\varphi [[/math]]

In the case where [math]\varphi[/math] is assumed to be faithful, as in Proposition 3.17 above, our claim is that we have the following formula, valid this time for any [math]\psi\in A^*[/math]:

[[math]] \int_\varphi*\psi=\psi*\int_\varphi=\psi(1)\int_\varphi [[/math]]

It is enough to prove this formula on a coefficient of a corepresentation, [math]a=(\tau\otimes id)v[/math]. In order to do so, consider the following matrices:

[[math]] P=\left(id\otimes\int_\varphi\right)v\quad,\quad Q=(id\otimes\psi)v [[/math]]

In terms of these matrices, we have:

[[math]] \left(\int_\varphi*\psi\right)a =\left(\tau\otimes\int_\varphi\otimes\psi\right)(v_{12}v_{13}) =\tau(PQ) [[/math]]

Similarly, we have the following computation:

[[math]] \left(\psi*\int_\varphi\right)a =\left(\tau\otimes\psi\otimes\int_\varphi\right)(v_{12}v_{13}) =\tau(QP) [[/math]]

Finally, regarding the term on the right, this is given by:

[[math]] \psi(1)\int_\varphi a=\psi(1)\tau(P) [[/math]]

Thus, our claim is equivalent to the following equality:

[[math]] PQ=QP=\psi(1)P [[/math]]

But this latter equality follows from the fact, coming from Proposition 3.17 above, that [math]P=(id\otimes\int_\varphi)v[/math] equals the orthogonal projection onto [math]Fix(v)[/math]. Thus, we have proved our claim. Now observe that our formula can be written as:

[[math]] \psi\left(\int_\varphi\otimes id\right)\Delta=\psi\left(id\otimes\int_\varphi\right)\Delta=\psi\int_\varphi(.)1 [[/math]]

This formula being true for any [math]\psi\in A^*[/math], we can simply delete [math]\psi[/math], and we conclude that the invariance formula in Definition 3.14 holds indeed, with [math]\int_G=\int_\varphi[/math]. Finally, assuming that we have two invariant integrals [math]\int_G,\int_G'[/math], we have:

[[math]] \begin{eqnarray*} \left(\int_G\otimes\int_G'\right)\Delta &=&\left(\int_G'\otimes\int_G\right)\Delta\\ &=&\int_G(.)1\\ &=&\int_G'(.)1 \end{eqnarray*} [[/math]]

Thus we have [math]\int_G=\int_G'[/math], and this finishes the proof.

■

As a first observation, in the case of the classical groups, and of the group duals, we recover the various Haar measure results mentioned before. As another illustration, for the basic product operations, we have the following result, due to Wang :

Proposition

We have the following results:

• For a product [math]G\times H[/math], we have [math]\int_{G\times H}=\int_G\otimes\int_H[/math].
• For a dual free product [math]G\,\hat{*}\,H[/math], we have [math]\int_{G\,\hat{*}\,H}=\int_G*\int_H[/math].
• For a quotient [math]G\to H[/math], we have [math]\int_H=\left(\int_G\right)_{|C(H)}[/math].
• For a projective version [math]G\to PG[/math], we have [math]\int_{PG}=\left(\int_G\right)_{|C(PG)}[/math].

Show Proof

These formulae all follow from the invariance property, as follows:

(1) Here the tensor product form [math]\int_G\otimes\int_H[/math] satisfies the left and right invariance properties of the Haar functional [math]\int_{G\times H}[/math], and so by uniqueness, it is equal to it.

(2) Here the situation is similar, with the free product of linear forms being defined with some inspiration from the discrete group case, where [math]\int_{\widehat{\Gamma}}g=\delta_{g,1}[/math].

(3) Here the restriction [math]\left(\int_G\right)_{|C(H)}[/math] satisfies by definition the required left and right invariance properties, so once again we can conclude by uniqueness.

(4) Here we simply have a particular case of (3) above.

■

In practice, the last assertion in Theorem 3.18 is the most useful one. By applying it to the Peter-Weyl corepresentations, we obtain the following alternative statement:

Theorem

The Haar integration of a Woronowicz algebra is given, on the coefficients of the Peter-Weyl corepresentations, by the Weingarten formula

[[math]] \int_Gu_{i_1j_1}^{e_1}\ldots u_{i_kj_k}^{e_k}=\sum_{\pi,\sigma\in D_k}\delta_\pi(i)\delta_\sigma(j)W_k(\pi,\sigma) [[/math]]

valid for any colored integer [math]k=e_1\ldots e_k[/math] and any multi-indices [math]i,j[/math], where:

• [math]D_k[/math] is a linear basis of [math]Fix(u^{\otimes k})[/math].
• [math]\delta_\pi(i)= \lt \pi,e_{i_1}\otimes\ldots\otimes e_{i_k} \gt [/math].
• [math]W_k=G_k^{-1}[/math], with [math]G_k(\pi,\sigma)= \lt \pi,\sigma \gt [/math].

Show Proof

As a first observation, the above formula computes indeed the Haar integral, because the coefficients of the Peter-Weyl corepresentations span a dense subalgebra:

[[math]] A=\overline{span\left(u_{i_1j_1}^{e_1}\ldots u_{i_kj_k}^{e_k}\Big| e,i,j,k\right)} [[/math]]

Regarding now the proof, we know from Theorem 3.18 that the integrals in the statement form altogether the orthogonal projection [math]P[/math] onto the following space:

[[math]] Fix(u^{\otimes k})=span(D_k) [[/math]]

Consider now the following linear map:

[[math]] E(x)=\sum_{\pi\in D_k} \lt x,\pi \gt \pi [[/math]]

By a standard linear algebra computation, it follows that we have [math]P=WE[/math], where [math]W[/math] is the inverse on [math]span(D_k)[/math] of the restriction of [math]E[/math]. But this restriction is the linear map given by [math]G_k[/math], and so [math]W[/math] is the linear map given by [math]W_k[/math], and this gives the result.

■

We will be back to the above two Haar measure theorems, which are both fundamental, with versions, illustrations and applications, on several occasions, later on.

General references

Banica, Teo (2024). "Introduction to quantum groups". arXiv:1909.08152 [math.CO].