guide:C541bc3902

[math] \newcommand{\mathds}{\mathbb}[/math]

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As already mentioned, we will be interested in what follows in the case where the deformation matrix [math]Q[/math] is generic. Our genericity assumptions are as follows:

Definition

We use the following notions:

• We call [math]p_1,\ldots,p_m\in\mathbb T[/math] root independent if for any [math]r_1,\ldots, r_m\in\mathbb Z[/math] we have:
  [[math]] p_1^{r_1}\ldots p_m^{r_m}=1\implies r_1=\ldots=r_m=0 [[/math]]
• A matrix [math]Q\in M_{G\times H}(\mathbb T)[/math], taken to be dephased,
  [[math]] Q_{0c}=Q_{i0}=1 [[/math]]
  is called generic if the elements [math]Q_{ic}[/math], with [math]i,c\neq0[/math], are root independent.

In what follows we will do the computation for such matrices. Our main result will show that the associated quantum group does not depend in fact of the matrix. In order to do the computation, we will need the following technical result:

Proposition

Assume that [math]Q\in M_{G\times H}(\mathbb T)[/math] is generic, and set:

[[math]] \theta_{ic}^{ke}=\frac{Q_{i,e-c}Q_{i-k,e}}{Q_{ie}Q_{i-k,e-c}} [[/math]]

For every [math]k \in G[/math], we have a representation [math]\pi^k:\Gamma_{G,H}\to U_{|H|}[/math] given by:

[[math]] \pi^k(c^{(i)})\epsilon_e=\theta_{ic}^{ke}\epsilon_{e-c} [[/math]]

The family of representations [math](\pi^k)_{k \in G}[/math] is projectively faithful, in the sense that if for some [math]t \in \Gamma_{G,H}[/math] we have that [math]\pi^k(t)[/math] is a scalar matrix for any [math]k[/math], then [math]t=1[/math].

Show Proof

The representations [math]\pi^k[/math] arise as above. With [math]\Gamma_{G,H}=T\rtimes H[/math], as in the proof of Proposition 16.11, we see that for [math]t\in\Gamma_{G,H}[/math] such that [math]\pi^k(t)[/math] is a scalar matrix for any [math]k[/math], then [math]t\in T[/math], since the elements of [math]T[/math] are the only ones having their image by [math]\pi^k[/math] formed by diagonal matrices. Now write [math]t[/math] as follows, with the generators of [math]T[/math] being as in the proof of Proposition 16.11, and with [math]R_{ic}\in\mathbb Z[/math] being certain integers:

[[math]] t=\prod_{i \not=0, c\not=0} ((-c)^{(0)}(c)^{(i)})^{R_{ic}} [[/math]]

Consider now the following quantities:

[[math]] \begin{eqnarray*} A(k,e) &=&\prod_{i\neq0}\prod_{c\neq0}(\theta_{ic}^{ke}(\theta_{0c}^{ke})^{^{-1}})^{R_{ic}}\\ &=&\prod_{i\neq0}\prod_{c\neq0} (\theta_{ic}^{ke})^{R_{ic}}(\theta_{0c}^{ke})^{-R_{ic}}\\ &=&\prod_{i\neq0}\prod_{c\neq0}(\theta_{ic}^{ke})^{R_{ic}} \cdot\prod_{c\neq0}(\theta_{0c}^{ke})^{-\sum_{i\neq0}R_{ic}}\\ &=&\prod_{j\neq0}\prod_{c\neq0} (\theta_{jc}^{ke})^{R_{jc}} \cdot\prod_{c\neq0}\prod_{j\neq0}(\theta_{jc}^{ke})^{\sum_{i\neq0}R_{ic}}\\ &=&\prod_{j\neq0}\prod_{c\neq0}(\theta_{jc}^{ke})^{R_{jc}+\sum_{i\neq0}R_{ic}} \end{eqnarray*} [[/math]]

We have then the following formula, valid for any [math]k,e[/math]:

[[math]] \pi^k(t)(\epsilon_e)= A(k,e)\epsilon_e [[/math]]

Our assumption is that for any [math]k[/math], and for any [math]e,f[/math], we have:

[[math]] A(k,e)=A(k,f) [[/math]]

By using now the root independence of the elements [math]Q_{ic}[/math], with [math]i,c\neq0[/math], we see that this implies [math]R_{ic}=0[/math] for any [math]i,c[/math], and this proves our assertion.

■

We will need as well the following technical result:

Proposition

Consider a surjective Hopf algebra map

[[math]] \pi:C^*(\Gamma)\rtimes C(H)\to L [[/math]]

such that [math]\pi_{|C(H)}[/math] is injective, and such that for [math]r\in\Gamma[/math] and [math]f\in C(H)[/math], we have:

[[math]] \pi(r\otimes 1)=\pi(1\otimes f)\implies r=1 [[/math]]

Then [math]\pi[/math] is an isomorphism.

Show Proof

We use here various Hopf algebra tools. Consider the following algebra:

[[math]] A=C^*(\Gamma)\rtimes C(H) [[/math]]

In order to prove the result, we start with the following standard Hopf algebra exact sequence, where [math]i(f)=1\otimes f[/math], and where [math]p=\varepsilon\otimes 1[/math]:

[[math]] \mathbb C\to C(H)\overset{i}\to A \overset{p}\to C^*(\Gamma)\to \mathbb C [[/math]]

Since [math]\pi\circ i[/math] is injective, and the Hopf subalgebra [math]\pi\circ i(C(H))[/math] is central in [math]L[/math], we can form the following quotient Hopf algebra:

[[math]] \overline{L} = L/ (\pi\circ i(C(H))^+L [[/math]]

We obtain in this way another exact sequence, as follows:

[[math]] \mathbb C\longrightarrow C(H)\overset{\pi\circ i}\longrightarrow L\overset{q}\longrightarrow\overline{L}\longrightarrow\mathbb C [[/math]]

Note that this sequence is indeed exact, e.g. by centrality. Thus, we get the following diagram with exact rows, with the Hopf algebra map on the right being surjective:

[[math]] \xymatrix@R=50pt@C=50pt {\mathbb C\ar[r]&C(H)\ar@2@{-}[d]\ar[r]^i&A\ar[d]^\pi\ar[r]^p&C^*(\Gamma)\ar[r]\ar[d]&\mathbb C\\ \mathbb C\ar[r]&C(H)\ar[r]^{\pi\circ i}&L\ar[r]^q&\overline{L}\ar[r]&\mathbb C} [[/math]]

Since a quotient of a group algebra is still a group algebra, we get a commutative diagram with exact rows as follows:

[[math]] \xymatrix@R=50pt@C=50pt {\mathbb C\ar[r]&C(H)\ar@2@{-}[d]\ar[r]^i&A\ar[d]^\pi\ar[r]^p&C^*(\Gamma)\ar[r]\ar[d]&\mathbb C\\ \mathbb C\ar[r]&C(H)\ar[r]^{\pi\circ i}&L\ar[r]^{q'}&C^*(\overline{\Gamma})\ar[r]&\mathbb C} [[/math]]

Here the map on the right is induced by a surjective group morphism, as follows:

[[math]] u:\Gamma\to\overline{\Gamma}\quad,\quad g\to\overline{g} [[/math]]

By the five lemma, which is something very classical in algebra, we just have to show that [math]u[/math] is injective. So, let [math]g \in \Gamma[/math] be such that [math]u(g)=1[/math]. We have then:

[[math]] q' \pi(g \otimes 1) =u p(g\otimes 1) =u(g) =\overline{g} =1 [[/math]]

For [math]g\in\Gamma[/math], let us set:

[[math]] _gA=\left\{a \in A \ \Big| \ p(a_1) \otimes a_2= g \otimes a\right\} [[/math]]

[[math]] _{\overline{g}}L= \left\{l \in L \ \Big| \ q'(l_1) \otimes l_2= \overline{g} \otimes l\right\} [[/math]]

The commutativity of the square on the right ensures that we have:

[[math]] \pi(_gA) \subset {_{\overline{g}}L} [[/math]]

Then with the previous [math]g[/math], we have, by exactness of the sequence:

[[math]] \pi(g \otimes 1) \in {_{\overline{1}}L} = \pi i (C(H)) [[/math]]

Thus, for some [math]f \in C(H)[/math], we must have:

[[math]] \pi(g \otimes 1)= \pi(1 \otimes f) [[/math]]

We conclude by our assumption that [math]g=1[/math].

■

We have now all the needed ingredients for proving a main result, as follows:

Theorem

When [math]Q[/math] is generic, the minimal factorization for [math]\pi_Q[/math] is

[[math]] \xymatrix@R=45pt@C=40pt {C(S_{G\times H}^+)\ar[rr]^{\pi_Q}\ar[rd]&&M_{G\times H}(\mathbb C)\\&C^*(\Gamma_{G,H})\rtimes C(G)\ar[ur]_\pi&} [[/math]]

where on the bottom

[[math]] \Gamma_{G,H}\simeq\mathbb Z^{(|G|-1)(|H|-1)}\rtimes H [[/math]]

is the discrete group constructed above.

Show Proof

We want to apply Proposition 16.13 to the following morphism, arising from the factorization in Theorem 16.9, where [math]L[/math] denotes the Hopf image of [math]\pi_Q[/math]:

[[math]] \theta : C^*(\Gamma_{G,H})\rtimes C(G)\to L [[/math]]

To be more precise, this morphism produces the following commutative diagram:

[[math]] \xymatrix@R=40pt@C=40pt {C(S_{G\times H}^+) \ar[rr]^{\pi_Q} \ar[dr]_{} \ar@/_/[ddr]_{}& & M_{G\times H}(\mathbb C) \\ & L \ar[ur]_{} & \\ & C^*(\Gamma_{G,H})\rtimes C(G) \ar@{-- \gt }[u]_\theta \ar@/_/[uur]_{\pi}& } [[/math]]

The first observation is that the injectivity assumption on [math]C(G)[/math] holds by construction, and that for [math]f \in C(G)[/math], the matrix [math]\pi(f)[/math] is “block scalar”, the blocks corresponding to the indices [math]k[/math] in the basis [math]\varepsilon_{ke}[/math] in the basis from Proposition 16.13. Now for [math]r \in \Gamma_{G,H}[/math] with [math]\theta(r\otimes 1)=\theta(1 \otimes f)[/math] for some [math]f \in C(G)[/math], we see, using the commutative diagram, that we will have that [math]\pi(r \otimes 1)[/math] is block scalar. By Proposition 16.11, the family of representations [math](\pi^k)[/math] of [math]\Gamma_{G,H}[/math], corresponding to the blocks [math]k[/math], is projectively faithful, so [math]r=1[/math]. We can apply indeed Proposition 16.13, and we are done.

■

Summarizing, we have computed the quantum permutation groups associated to the Di\c t\u a deformations of the tensor products of Fourier matrices, in the case where the deformation matrix [math]Q[/math] is generic. For some further computations, in the case where the deformation matrix [math]Q[/math] is no longer generic, we refer to and follow-up papers.

General references

Banica, Teo (2024). "Invitation to Hadamard matrices". arXiv:1910.06911 [math.CO].