Flat matrix models for quantum permutation groups

Introduction

The quantum permutation group $S_{N}^{+}$ was introduced by Wang in . Of particular interest are the quantum subgroups $\mathcal{G}\subset S_{N}^{+}$ appearing from random matrix representations $\pi:C(S_{N}^{+})\to M_{N}(C(X))$ via the Hopf image construction . One key problem is the computation of the law of the main character of $\mathcal{G}$. See , , .

A number of general algebraic and analytic tools for dealing with such questions have been developed , , , , . However, at the level of concrete examples, only two types of models $\pi:C(S_{N}^{+})\to M_{N}(C(X))$ have been succesfully investigated, so far. The first example, coming from the Pauli matrices, was investigated in . The second example, coming from deformed Fourier matrices, was investigated in .

Our purpose here is to advance on such questions:

The Pauli matrix construction and the deformed Fourier matrix one are both of type $\pi:C(S_{N}^{+})\to C(U_{B},\mathcal{L}(B))$, with $B$ being a finite dimensional $C^{*}$-algebra. We will investigate here the case where $B=C^{*}_{\sigma}(G)$ is a cocycle twist of a finite group algebra, which generalizes the Pauli matrix construction. Our main result will be the computation of the law of the main character.

We will present as well a “universal” construction, inspired from the Sinkhorn algorithm , . This algorithm starts with a $N\times N$ matrix having positive entries and produces, via successive averagings over rows/columns, a bistochastic matrix. We will find here an adaptation of this algorithm to Wang’s magic unitaries , which conjecturally produces an inner faithful representation of $C(S_{N}^{+})$.

There are of course many questions raised by the present work. Regarding the generalized Pauli matrix construction, our results, and also , , suggest that the associated quantum group should be a twist of $PU_{n}$. Also, this construction still remains to be unified with the deformed Fourier matrix one. Regarding the Sinkhorn type models, here our computer simulations suggest that we should get a free Poisson law , , but so far, we have no convincing abstract methods in order to approach this question.

The paper is organized as follows: 1-2 contain preliminaries and generalities, in 3-4 we study the generalized Pauli models, and in 5-6 we study the Sinkhorn type models.

Acknowledgements. The present work was started at the Fields Institute conference “Quantum groups and Quantum information theory”, Herstmonceux 2015, and we would like to thank the organizers for the invitation. IN received financial support from the ANR grants RMTQIT ANR-12-IS01-0001-01 and StoQ ANR-14-CE25-0003-01.

Quantum permutations

We are interested in what follows in the quantum permutation group $S_{N}^{+}$, and in the random matrix representations of the associated Hopf algebra $C(S_{N}^{+})$.

Our starting point is the following notion, coming from Wang’s paper :

A magic unitary is a square matrix over a $C^{*}$-algebra, $u\in M_{N}(A)$, whose entries are projections, summing up to $1$ on each row and each column.

At $N=2$ these matrices are as follows, with $p$ being a projection:

At $N=3$ it is known from that the entries of $u$ must commute as well. At $N\geq 4$ the entries of $u$ no longer automatically commute. Indeed, we have here the following example, with $p,q\in B(H)$ being non-commuting projections:

The following key definition is due to Wang :

$C(S_{N}^{+})$ is the universal $C^{*}$-algebra generated by the entries of a $N\times N$ magic unitary matrix $w=(w_{ij})$, with the morphisms defined by

as comultiplication, counit and antipode.

This algebra satisfies Woronowicz’ axioms in , , and the underlying space $S_{N}^{+}$ is therefore a compact quantum group, called quantum permutation group.

Observe that any magic unitary $u\in M_{N}(A)$ produces a representation $\pi:C(S_{N}^{+})\to A$, given by $\pi(w_{ij})=u_{ij}$. In particular, we have a representation as follows:

The corresponding embedding $S_{N}\subset S_{N}^{+}$ is an isomorphism at $N=2,3$, but not at $N\geq 4$, where $S_{N}^{+}$ is infinite. Moreover, it is known that we have $S_{4}^{+}\simeq SO_{3}^{-1}$, and that any $S_{N}^{+}$ with $N\geq 4$ has the same fusion semiring as $SO_{3}$. See , .

Our claim now is that, given a magic unitary $u\in M_{N}(A)$, we can associate to it a certain quantum permutation group $\mathcal{G}\subset S_{N}^{+}$. In order to perform this construction, we use the notions of Hopf image and inner faithfulness, from :

The Hopf image of a $C^{*}$-algebra representation $\pi:C(\mathcal{G})\to A$ is the smallest Hopf $C^{*}$-algebra quotient $C(\mathcal{G})\to C(\mathcal{G}^{\prime})$ producing a factorization as follows:

The representation $\pi$ is called inner faithful when $\mathcal{G}=\mathcal{G}^{\prime}$.

Here $\mathcal{G}$ can be any compact quantum group, in the sense of , .

As a basic example, when $\mathcal{G}=\widehat{\Gamma}$ is a group dual, $\pi:C^{*}(\Gamma)\to A$ must come from a unitary group representation $\Gamma\to U_{A}$, and the minimal factorization is the one obtained by taking the image, $\Gamma\to\Gamma^{\prime}\subset U_{A}$. Thus $\pi$ is inner faithful when $\Gamma\subset U_{A}$.

Now back to our above claim, we can now formulate:

Associated to any magic unitary $u\in M_{N}(A)$ is the smallest quantum permutation group $\mathcal{G}\subset S_{N}^{+}$ producing a factorization

of the representation $\pi:C(S_{N}^{+})\to A$ given by $w_{ij}\to u_{ij}$.

At the level of examples, let us recall that a Latin square is a matrix $L\in M_{N}(1,\ldots,N)$ having the property that each of its rows and columns is a permutation of $1,\ldots,N$. For instance, associated to any finite group $\mathcal{G}$ is the Latin square $(L_{\mathcal{G}})_{ij}=ij^{-1}$, with $i,j,ij^{-1}\in\mathcal{G}$ being regarded as elements of $\{1,\ldots,N\}$, where $N=|\mathcal{G}|$.

With these conventions, we have the following result:

If $u\in M_{N}(A)$ comes from a Latin square $L\in M_{N}(1,\ldots,N)$, in the sense that $u_{ij}=p_{L_{ij}}$, with $p_{1},\ldots,p_{N}\in A$ being projections summing up to $1$, then:

$\mathcal{G}\subset S_{N}^{+}$ is the subgroup of $S_{N}$ generated by the rows of $L$.

In particular, when $L=L_{\mathcal{G}}$, we obtain the group $\mathcal{G}$ itself.

In addition, this is the only case where $\mathcal{G}$ is classical.

These results are well-known, the proof being as follows:

(1) This comes from the fact that we have a factorization $\pi:C(S_{N}^{+})\to C(\mathcal{G})\subset A$.

(2) This follows from (1), because the rows of $L_{\mathcal{G}}$ generate the group $\mathcal{G}$ itself.

(3) This follows by using the Gelfand theorem. For details here, see . ∎

Cocyclic models

We are interested in what follows in representations of type $\pi:C(S_{N}^{+})\to M_{N}(C(X))$, and in the computation of their Hopf images. As a motivation, it is known that the existence of an inner faithful representation of type $\pi:C(\mathcal{G})\to M_{N}(C(X))$ implies that $L^{\infty}(\mathcal{G})$ has the Connes embedding property. For a discussion here, see , , .

The key example of a magic unitary matrix $u\in M_{N}(A)$ over a random matrix algebra, $A=M_{N}(C(X))$, appears at $N=4$, in connection with the Pauli matrices:

which commutes with canonical integration maps, and is faithful.

The point now is that the combinatorics of the variables $x\to Proj(g_{i}xg_{j}^{*})$ can be shown to be the same as the Weingarten combinatorics of the variables $w_{ij}\in C(S_{4}^{+})$. This gives the integration assertion, and the faithfulness assertion follows from it. See . ∎

At $N\geq 5$ now, since the dual of $S_{N}^{+}$ is not amenable, we cannot have a faithful representation $\pi:C(S_{N}^{+})\to M_{N}(C(X))$. Our purpose will be find such a representation which is inner faithful, or at least which is “as inner faithful” as possible.

With these conventions, we can formulate:

A magic unitary $u\in M_{N}(\mathcal{L}(B))$ is called:

Flat, if each $u_{ij}\in\mathcal{L}(B)$ is a rank $1$ projection.

Split, if $u_{ij}=Proj(e_{i}f_{j}^{*})$, for certain sets $\{e_{i}\},\{f_{i}\}\subset U_{B}$.

Fully split, if $u_{ij}=Proj(g_{i}xg_{j}^{*})$, with $\{g_{i}\}\subset U_{B}$, and $x\in U_{B}$.

If $G$ is an abelian group, $|G|=N$, then $u_{ij}=Proj((F_{G})_{i-j})$ is fully split.

Let us clarify now the relation with Theorem 1.5. We first have:

The split magic unitaries which produce Latin squares are those of the form $u_{ij}=Proj(g_{i}g_{j}^{*})$, with $\{g_{1},\ldots,g_{N}\}\subset U_{B}$ being pairwise orthogonal, and forming a group $G\subset PU_{B}$. For such a magic unitary, the associated Latin square is $L_{G}$.

Assume indeed that $u_{ij}=Proj(e_{i}f_{j}^{*})$ produces a Latin square.

(1) Our first claim is that we can assume $e_{1}=f_{1}=1$. Indeed, given $x,y\in U_{B}$ the matrix $u_{ij}^{\prime}=Proj(xe_{i}f_{j}^{*}y)$ is still magic, and in the case where $u$ comes from a Latin square, $u_{ij}=Proj(\xi_{L_{ij}})$, we have $u_{ij}^{\prime}=Proj(\xi_{L_{ij}}^{\prime})$ with $\xi^{\prime}_{ab}=x\xi_{ab}y$, and so $u^{\prime}$ comes from $L$ as well. Thus, by taking $x=e_{1}^{*},y=f_{1}$, we can assume $e_{1}=f_{1}=1$.

(2) Our second claim is that we can assume $u_{ij}=Proj(e_{i}e_{j}^{*})$. Indeed, since $u$ is magic, the first row of vectors $\{1,f_{2}^{*},\ldots,f_{N}^{*}\}\subset PU_{B}$ must appear as a permutation of the first column of vectors $\{1,e_{2},\ldots,e_{N}\}\subset PU_{B}$. Thus, up to a permutation of the columns, and a rescaling of the columns by elements in $Z(U_{B})$, we can assume $f_{i}=e_{i}$, and we obtain $u_{ij}=Proj(e_{i}e_{j}^{*})$. Observe that this permutation/rescaling of the columns won’t change the fact that the associated Latin square $L$ comes or not from a group.

(3) Let us construct now $G$. The Latin square condition shows that for any $i,j$ there is a unique $k$ such that $e_{i}e_{j}=e_{k}$ inside $PU_{B}$, and our claim is that the operation $(i,j)\to k$ gives a group structure on the set of indices. Indeed, all the group axioms are clear from definitions, and we obtain in this way a subgroup $G\subset PU_{B}$, having order $N$.

(4) With $G$ being constructed as above, we have $u_{ij}=Proj(e_{i}e_{j}^{*})=Proj(e_{ij^{-1}})$. Thus we have $u_{ij}^{\prime}=Proj(\xi_{L_{ij}})$ with $\xi_{k}=e_{k}$ and $L_{ij}=ij^{-1}$, and we are done. ∎

In order to further process the above result, we will need:

The algebra $C^{*}(G)$, with multiplication $g\cdot h=\sigma(g,h)gh$, is denoted $C^{*}_{\sigma}(G)$.

Observe that $g\cdot h=\sigma(g,h)gh$ is associative, and that we have $g\cdot 1=1\cdot g=g$, due to the 2-cocycle condition. Thus $C^{*}_{\sigma}(G)$ is an associative algebra with unit 1. In fact, $C^{*}_{\sigma}(G)$ is a $C^{*}$-algebra, with the involution making the canonical generators $g\in C^{*}_{\sigma}(G)$ unitaries. The canonical trace on $C^{*}_{\sigma}(G)$ coincides then with that of $C^{*}(G)$.

With this notion in hand, we can now formulate:

The split magic unitaries which produce Latin squares are precisely those of the form $u_{ij}=Proj(g_{i}g_{j}^{*})$, with $\{g_{1},\ldots,g_{N}\}$ being the standard basis of a twisted group algebra $C^{*}_{\sigma}(G)$. In this case, the associated Latin square is $L_{G}$.

It follows from definitions that $\sigma$ is a 2-cocycle, and our claim now is that we have $B=C^{*}_{\sigma}(G)$. Indeed, this is clear when $\sigma=1$, because by linear independence we can define a linear space isomorphism $B\simeq C^{*}(G)$, which follows to be a $C^{*}$-algebra isomorphism. In the general case, where $\sigma$ is arbitrary, the proof is similar. ∎

At the level of examples now, we can use the following construction:

The map $\sigma((i,a),(j,b))=<i,b>$ is a $2$-cocycle on $G=H\times\widehat{H}$.

These results are all well-known, the proof being as follows:

The fact that $\varphi$ is multiplicative follows from:

Recall now that the involution of $C^{*}_{\sigma}(G)$ is the one making the canonical generators $g\in C^{*}_{\sigma}(G)$ unitaries. Since we have $g_{ia}g_{-i,-a}=<-i,a>g_{00}=<-i,a>$, it follows that we have $g_{ia}^{*}=<i,a>g_{-i,-a}$, and the involutivity check goes as follows:

In order to prove the bijectivity of $\varphi$, consider the following linear map:

It is routine to check that $\varphi,\psi$ are inverse to each other, and this finishes the proof.

But these matrices are proportional, by factors $1,i,1,i$, to the Pauli matrices. ∎

We have now all the needed ingredients for generalizing the Pauli matrix construction. The result here, conceptually motivated by Proposition 2.6 above, is as follows:

where $\{g_{1},\ldots,g_{N}\}\subset U_{B}$ is the standard basis of the algebra $B=C^{*}_{\sigma}(G)$. Moreover:

As an example, we can use $G=H\times\widehat{H}$, with $\sigma((i,a),(j,b))=<i,b>$.

When the cocycle is trivial, we obtain the Fourier matrix representation.

The first assertion follows from Proposition 2.6 and its proof, (1) and (2) follow from Proposition 2.7, and (3) follows from Proposition 2.3. ∎

Laws of characters

Let $\pi:C(S_{N}^{+})\to C(\mathcal{G})\to M_{K}(C(X))$ be a Hopf image factorization, mapping $w_{ij}\to v_{ij}\to u_{ij}$, and let $\chi=\sum_{i}v_{ii}$.

$\int_{\mathcal{G}}=\lim_{k\to\infty}\frac{1}{k}\sum_{r=1}^{k}\int_{\mathcal{G}}^{r}$, with $\int_{\mathcal{G}}^{r}=(tr\circ\pi)^{*r}$, where $\phi*\psi=(\phi\otimes\psi)\Delta$.

$\int_{\mathcal{G}}^{r}v_{i_{1}j_{1}}\ldots v_{i_{p}j_{p}}=(T_{p}^{r})_{i_{1}\ldots i_{p},j_{1}\ldots j_{p}}$, where $(T_{p})_{i_{1}\ldots i_{p},j_{1}\ldots j_{p}}=tr(u_{i_{1}j_{1}}\ldots u_{i_{p}j_{p}})$.

The moments of $\chi$ with respect to $\int_{\mathcal{G}}^{r}$ are the numbers $c_{p}^{r}=Tr(T_{p}^{r})$.

The first assertion, which is the key one, was proved in in the case $X=\{.\}$, and then in in the general, parametric case. The second assertion is elementary, and the third one follows from it, by summing over indices $i_{k}=j_{k}$. ∎

As a main consequence, if we denote by $\mu,\mu^{r}$ the laws of the main character $\chi$ with respect to the Haar functional $\int_{\mathcal{G}}$, and with its truncated version $\int_{\mathcal{G}}^{r}=(tr\circ\pi)^{*r}$, we have a convergence in moments $\mu^{r}\to\mu$. Following now , we have:

For a representation coming from a split matrix, $u_{ij}=Proj(e_{i}f_{j}^{*})$, the truncated measure $\mu^{r}$ is the law of the Gram matrix of the vectors

with respect to the normalized trace of the $N^{r}\times N^{r}$ matrices.

According to Proposition 3.1 (3), the moments of $\mu^{r}$ are given by:

In the case of a split magic unitary, $u_{ij}=Proj(e_{i}f_{j}^{*})$, since the vectors $e_{i}f_{j}^{*}$ are all of norm 1, with respect to the canonical scalar product, we therefore obtain:

Now by changing the order of the terms in the product, this gives:

In terms of the vectors $\xi_{i_{1}\ldots i_{r}}=e_{i_{1}}f_{i_{2}}^{*}\otimes\ldots\otimes e_{i_{r}}f_{i_{1}}^{*}$ in the statement, and then of their Gram matrix $G_{i_{1}\ldots i_{r},j_{1}\ldots j_{r}}=<\xi_{i_{1}\ldots i_{r}},\xi_{j_{1}\ldots j_{r}}>$, we obtain the following formula:

But this gives the formula in the statement, and we are done. ∎

In the fully split case now, we have the following result:

For a representation coming from a fully split matrix, $u_{ij}=Proj(g_{i}xg_{j}^{*})$, the truncated measure $\mu^{r}$ is the law of the Gram matrix of the vectors

with respect to the usual integration over $M_{N^{r}}(C(U_{B}^{r}))$.

The idea is that the computations in the proof of Proposition 3.2 apply, with $e_{i}=g_{i}x$ and $f_{i}=g_{i}$, and with an integral $\int_{U_{B}^{r}}$ added. To be more precise, we can start with the same formula as there, stating that the moments of $\mu^{r}$ are given by:

In the case of a fully split matrix, $u_{ij}=Proj(g_{i}xg_{j}^{*})$, since the vectors $g_{i}xg_{j}^{*}$ are all of norm 1, we therefore obtain:

Now by changing the order of the terms in the product, this gives:

In terms of the vectors $\xi_{i_{1}\ldots i_{r}}^{x_{1}\ldots x_{r}}=g_{i_{1}}x_{1}g_{i_{2}}^{*}\otimes\ldots\ldots\otimes g_{i_{r}}x_{r}g_{i_{1}}^{*}$ in the statement, and then of their Gram matrix $G_{i_{1}\ldots i_{r},j_{1}\ldots j_{r}}^{x_{1}\ldots x_{r}}=<\xi_{i_{1}\ldots i_{r}}^{x_{1}\ldots x_{r}},\xi_{j_{1}\ldots j_{r}}^{x_{1}\ldots x_{r}}>$, we therefore obtain:

But this gives the formula in the statement, and we are done. ∎

Cocyclic abelian models

Let us go back now to the cocyclic abelian models, from Theorem 2.8 (1) above. We will explicitely compute the law of the main character, for these models.

By appplying the general formula in Theorem 3.3, we first have:

with respect to the usual integration over $M_{n^{2r}}(C(U_{n}^{r}))$.

We use the general formula found in Theorem 3.3 above. The Gram matrix that we are interested in, having now double indices, is given by:

In the case of a cocyclic abelian model, as in the statement, we can use for computations the isomorphism found in the proof of Proposition 2.7, namely:

With this identification made, the scalar products can be computed as follows:

Thus the Gram matrix that we are interested in is given by:

But this gives the formula in the statement, and we are done. ∎

The point now is that the Gram matrix in Proposition 4.1 is circulant, and so is diagonal in Fourier transform. By diagonalizing it, we obtain the following result:

As already mentioned, the idea will be that of applying a discrete Fourier transform. With $F_{ij}=\frac{1}{\sqrt{n}}<i,j>$, having as inverse $\bar{F}_{ij}=\frac{1}{\sqrt{n}}<-i,j>$, we have:

We can rewrite this formula in the following way:

By summing over $a_{i},b_{i}$, we must have $c_{i}=d_{i}$ and $s_{i-1}=c_{i}+p_{i}$. By changing the indices of summation, $i_{x}\to i_{x}-p_{x}$ and $j_{x}\to j_{x}-p_{x}$, we obtain:

We conclude that $\mu^{r}$ is the law of the following diagonal random matrix:

Now observe that we have $<k,i>x_{i,j+c}=(A_{k}xB_{c})_{ij}$, where $A_{k}:e_{i}\to<k,i>e_{i}$ and $B_{c}:e_{i}\to e_{i+c}$. In addition, we have $B_{c}A_{k}=W_{kc}$, and this gives:

Thus, we have obtained the formula in the statement. ∎

By making now some final manipulations, of probabilistic nature, everything simplifies in the formula in Proposition 4.2, and we obtain the following result:

We use the formula in Proposition 4.2 above. Observe first that the matrices $W_{kc}:e_{i}\to<k,i>e_{i+c}$ appearing there, called Weyl matrices, satisfy:

This is indeed already known from the cocyclic picture, and can be checked as well directly. Consider now the following group, obtained by tensoring such matrices:

With these notions in hand, Proposition 4.2 tells us that $\mu^{r}$ appears as average over the above Weyl group $W$ of the laws of the following variables:

The point now is that the random Weyl matrices $W_{k_{i}c_{i}}$ can be “absorbed” into the Haar distributed unitaries $x_{i}$, and we obtain that $\mu^{r}$ is the law of the following variable:

Now since the product $x_{1}\ldots x_{r}\in U_{n}$ is Haar distributed when the individual variables $x_{1}\in U_{n},\ldots,x_{r}\in U_{n}$ are each Haar distributed, this gives the result.

Summarizing, we have obtained Diaconis-Shahshahani variables . The asymptotics can be investigated by using the Weingarten formula, and are well-known, see , . Note also that by , the moments of the variable $|Tr(x)|^{2}$ are:

From a quantum group viewpoint, Theorem 4.3 suggests that the underlying quantum group should be a twist of $PU_{n}$. There is actually more evidence pointing towards this, coming from , . We intend to investigate these facts in some future work.

Universal models

$X_{N}$, the space of all $N\times N$ flat magic unitaries $u=(u_{ij})$.

$K_{N}$, the space of all $N\times N$ magic bases $\xi=(\xi_{ij})$.

With these notations, the abstract model spaces $X_{N},K_{N}$ that we are interested in, and some related spaces, are as follows:

We have inclusions and surjections as follows,

where $X_{N},Y_{N}$ consist of bistochastic/stochastic matrices, and $K_{N}$ is the lift of $X_{N}$.

In order to get some insight into the structure of $X_{N},K_{N}$, we use inspiration from the Sinkhorn algorithm , . This algorithm starts with a $N\times N$ matrix having positive entries and produces, via successive averagings over rows/columns, a bistochastic matrix. In our situation, we would like to have an “averaging” map $Y_{N}\to Y_{N}$, whose infinite iteration lands in the model space $X_{N}$. Equivalently, we would like to have an “averaging” map $U_{N}^{N}\to U_{N}^{N}$, whose infinite iteration lands in $K_{N}$.

In order to construct such averaging maps, we use the orthogonalization procedure coming from the polar decomposition. First, we have the following result:

We have orthogonalization maps as follows,

where $\alpha(x)_{i}=Pol([(x_{i})_{j}]_{ij})$, and $\beta(p)=(P^{-1/2}p_{i}P^{-1/2})_{i}$, with $P=\sum_{i}p_{i}$.

We can now compute the projections $p_{i}^{\prime}=Proj(x_{i}^{\prime})$. Indeed, the coefficients of these projections are given by $(p_{i}^{\prime})_{kl}=\bar{U}_{ik}U_{il}$ with $U=MP^{-1/2}$, and we obtain, as desired:

An alternative proof uses the fact that the elements $p_{i}^{\prime}=P^{-1/2}p_{i}P^{-1/2}$ are self-adjoint, and sum up to 1. The fact that these elements are indeed idempotents can be checked directly, via $p_{i}P^{-1}p_{i}=p_{i}$, because this equality holds on $\ker p_{i}$, and also on $x_{i}$. ∎

As an illustration, here is how the orthogonalization works at $N=2$:

At $N=2$ the orthogonalization procedure for $(Proj(x),Proj(y))$ amounts in considering the vectors $(x\pm y)/\sqrt{2}$, and then rotating by $45^{\circ}$.

By performing a rotation, we can restrict attention to the case $x=(\cos t,\sin t)$ and $y=(\cos t,-\sin t)$, with $t\in(0,\pi/2)$. Here the computations are as follows:

Thus the orthogonalization procedure replaces $(Proj(x),Proj(y))$ by the orthogonal projections on the vectors $(\frac{1}{\sqrt{2}}(1,1),\frac{1}{\sqrt{2}}(-1,1))$, and this gives the result. ∎

With these preliminaries in hand, let us discuss now the version that we need of the Sinkhorn algorithm. The orthogonalization procedure is as follows:

The orthogonalization maps $\alpha,\beta$ induce maps as follows,

which are the transposition maps on $K_{N},X_{N}$, and which are projections at $N=2$.

It follows from definitions that $\Phi(x)$ is obtained by putting the components of $x=(x_{i})$ in a row, then picking the $j$-th column vectors of each $x_{i}$, calling $M_{j}$ this matrix, then taking the polar part $x_{j}^{\prime}=Pol(M_{j})$, and finally setting $\Phi(x)=x^{\prime}$. Thus:

Thus, the first assertion is clear, and the second assertion is clear too. ∎

At $N=3$ now, the algorithm doesn’t stop any longer after 1 step. We obtain, after an infinite iteration, one of the 2 possible magic matrices coming from Latin squares.

Our first claim is that the algorithm converges, as follows:

The maps $\Phi,\Psi$ increase the volume,

and respectively land, after an infinite number of steps, in $K_{N}/X_{N}$.

As a main application of the above conjecture, the infinite iteration $(\Phi^{2})^{\infty}:U_{N}^{N}\to K_{N}$ would provide us with an integration on $K_{N}$, and hence on the quotient space $K_{N}\to X_{N}$ as well, by taking the push-forward measures, coming from the Haar measure on $U_{N}^{N}$.

In relation now with the matrix model problematics, we have:

The universal $N\times N$ flat matrix representation

is faithful at $N=4$, and is inner faithful at any $N\geq 5$.

Regarding the $N\geq 5$ conjecture, the problem here is that of proving that the truncated moments $c_{p}^{r}$ in Proposition 3.1 converge with $r\to\infty$ to the Catalan numbers.

Linear algebra

Our purpose here is to advance towards a unification of the two conjectures formulated in section 5 above. The point indeed is that when trying to approach Conjecture 5.7 with the probabilistic tools coming from Proposition 3.1, the estimates that are needed seem to be related to those required for approaching Conjecture 5.6.

We first have the following definition, inspired from Proposition 3.1:

The first few values of these matrices, at $p=1,2,3$, are as follows:

The interest in these matrices, in connection with Conjecture 5.7, comes from:

For the universal model, the matrices $T_{p}$ in Proposition 3.1 are

where $dx$ is the measure on the model space $K_{N}$ coming from Conjecture 5.6.

This is a trivial statement, because by definition of $T_{p}$, we have:

Thus the formula in the statement holds indeed. ∎

These vectors appear in the representation theory of $S_{N}^{+}$. See .

At $p=1$, we obtain the 1-eigenvector of $T_{1}^{x}=(1/N)_{ij}$:

At $p=2$ now, the two vectors constructed above are as follows:

In general, we have the following result:

If $\{x_{ij}\}_{i}$ are pairwise orthogonal then $(T_{p}^{x})^{*}\xi_{||\ldots|}=\xi_{||\ldots|}$ and $T_{p}^{x}\xi_{\sqcap\hskip-1.42262pt\sqcap\ldots\sqcap}=\xi_{\sqcap\hskip-1.42262pt\sqcap\ldots\sqcap}$.

If $\{x_{ij}\}_{j}$ are pairwise orthogonal then $T_{p}^{x}\xi_{||\ldots|}=\xi_{||\ldots|}$ and $(T_{p}^{x})^{*}\xi_{\sqcap\hskip-1.42262pt\sqcap\ldots\sqcap}=\xi_{\sqcap\hskip-1.42262pt\sqcap\ldots\sqcap}$.

If $\{x_{ij}\}_{i}$ or $\{x_{ij}\}_{j}$ are pairwise orthogonal then $<T_{p}^{x}\xi_{||\ldots|},\xi_{||\ldots|}>=N^{p}$.

We have $<T_{p}^{x}\xi_{\sqcap\hskip-1.42262pt\sqcap\ldots\sqcap},\xi_{\sqcap\hskip-1.42262pt\sqcap\ldots\sqcap}>=N$, without assumptions on $x$.

It is elementary to see that we have $(T_{p}^{x})^{*}=T_{p}^{x^{*}}$, and so it is enough to establish the assertions in (1,2) regarding the eigenvalues of $T_{p}^{x}$. The proof goes as follows:

(1) Assuming that $\{x_{ij}\}_{i}$ are pairwise orthogonal, we have indeed:

(2) Assuming now that $\{x_{ij}\}_{j}$ are pairwise orthogonal, we have indeed:

Here we have used, $p$ times via a recurrence, the fact that given an orthonormal basis $\{e_{k}\}$ we have $\sum_{k}<x,e_{k}><e_{k},y>=<x,y>$, for any two vectors $x,y$.

(3) The scalar product in the statement is given by:

When $\{x_{ij}\}_{j}$ are pairwise orthogonal, by using (2) we obtain $N^{p}$, as claimed. Since $(T_{p}^{x})^{*}=T_{p}^{x^{*}}$, the result follows to hold when $\{x_{ij}\}_{i}$ are pairwise orthogonal too.

(4) We have the following computation, valid for any $x$:

But this proves the last assertion, and we are done. ∎

The above computations suggest the following definition:

Observe that, according to the formula of $T_{p}^{x}$, we have:

We have the following statement, supported by computer calculations:

For any $x\in U_{N}^{N}$, and any $p\geq 2$, we have

with equality iff $x\in K_{N}$, in which case $F_{p}(x)=1$.

By a compacity argument, this would prove that our Sinkhorn type algorithm converges. Thus, we have here a first step towards unifying Conjecture 5.6 and Conjecture 5.7.

Let us restrict now attention to the case $p=2$. Here we have:

At $N=2$, by writing the inequality in Conjecture 6.5 in terms of the orthogonal projections $P,Q,R,S$ on the vectors $x_{ij}$, we are led to the following statement:

where $P^{\prime},Q^{\prime},R^{\prime},S^{\prime}$ are the following orthogonal projections

with all the inverses taken in the sense of Moore-Penrose.

We only know how to prove a special case of the statement above:

We can write $P,R$ by using the Halmos normal form :

By using the condition (3) in the statement, we can replace the first term in the direct sums above by . Now by using the fact that $H,W$ commute, we have:

We therefore have the following estimate:

Thus we have obtained the desired inequality. ∎

References