10b. Quantum groups

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At the group level now, the situation is much more rigid, and becomes quite interesting. We have the following result from , to start with:

Theorem

The following inclusion of compact groups is maximal,

[[math]] \mathbb TO_N\subset U_N [[/math]]

in the sense that there is no intermediate compact group in between.

Show Proof

In order to prove this result, consider as well the following group:

[[math]] \mathbb TSO_N=\left\{wU\Big| w\in\mathbb T,U\in SO_N\right\} [[/math]]

Observe that we have [math]\mathbb TSO_N=\mathbb TO_N[/math] if [math]N[/math] is odd. If [math]N[/math] is even the group [math]\mathbb TO_N[/math] has two connected components, with [math]\mathbb TSO_N[/math] being the component containing the identity. Let us denote by [math]\mathfrak{so}_N,\mathfrak u_N[/math] the Lie algebras of [math]SO_N,U_N[/math]. It is well-known that [math]\mathfrak u_N[/math] consists of the matrices [math]M\in M_N(\mathbb C)[/math] satisfying [math]M^*=-M[/math], and that:

[[math]] \mathfrak{so}_N=\mathfrak u_N\cap M_N(\mathbb R) [[/math]]

Also, it is easy to see that the Lie algebra of [math]\mathbb TSO_N[/math] is [math]\mathfrak{so}_N\oplus i\mathbb R[/math].

\underline{Step 1}. Our first claim is that if [math]N\geq 2[/math], the adjoint representation of [math]SO_N[/math] on the space of real symmetric matrices of trace zero is irreducible.

Let indeed [math]X \in M_N(\mathbb R)[/math] be symmetric with trace zero. We must prove that the following space consists of all the real symmetric matrices of trace zero:

[[math]] V=span\left\{UXU^t\Big|U \in SO_N\right\} [[/math]]

We first prove that [math]V[/math] contains all the diagonal matrices of trace zero. Since we may diagonalize [math]X[/math] by conjugating with an element of [math]SO_N[/math], our space [math]V[/math] contains a nonzero diagonal matrix of trace zero. Consider such a matrix:

[[math]] D=\begin{pmatrix} d_1\\ &\ddots\\ &&d_N \end{pmatrix} [[/math]]

We can conjugate this matrix by the following matrix:

[[math]] \begin{pmatrix} 0&-1&0\\ 1&0&0\\ 0&0&I_{N-2} \end{pmatrix}\in SO_N [[/math]]

We conclude that our space [math]V[/math] contains as well the following matrix:

[[math]] D'=\begin{pmatrix} d_2\\ &d_1\\ &&d_3\\ &&&\ddots\\ &&&&d_N \end{pmatrix} [[/math]]

More generally, we see that for any [math]1\leq i,j\leq N[/math] the diagonal matrix obtained from [math]D[/math] by interchanging [math]d_i[/math] and [math]d_j[/math] lies in [math]V[/math]. Now since [math]S_N[/math] is generated by transpositions, it follows that [math]V[/math] contains any diagonal matrix obtained by permuting the entries of [math]D[/math].

But it is well-known that this representation of [math]S_N[/math] on the diagonal matrices of trace zero is irreducible, and hence [math]V[/math] contains all such diagonal matrices, as claimed.

In order to conclude now, assume that [math]Y[/math] is an arbitrary real symmetric matrix of trace zero. We can find then an element [math]U\in SO_N[/math] such that [math]UYU^t[/math] is a diagonal matrix of trace zero. But we then have [math]UYU^t \in V[/math], and hence also [math]Y\in V[/math], as desired.

\underline{Step 2}. Our claim is that the inclusion [math]\mathbb TSO_N\subset U_N[/math] is maximal in the category of connected compact groups.

Let indeed [math]G[/math] be a connected compact group satisfying:

[[math]] \mathbb TSO_N\subset G\subset U_N [[/math]]

Then [math]G[/math] is a Lie group. Let [math]\mathfrak g[/math] denote its Lie algebra, which satisfies:

[[math]] \mathfrak{so}_N\oplus i\mathbb R\subset\mathfrak g\subset\mathfrak u_N [[/math]]

Let [math]ad_{G}[/math] be the action of [math]G[/math] on [math]\mathfrak g[/math] obtained by differentiating the adjoint action of [math]G[/math] on itself. This action turns [math]\mathfrak g[/math] into a [math]G[/math]-module. Since [math]SO_N \subset G[/math], [math]\mathfrak g[/math] is also a [math]SO_N[/math]-module. Now if [math]G\neq\mathbb TSO_N[/math], then since [math]G[/math] is connected we must have:

[[math]] \mathfrak{so}_N\oplus i\mathbb{R}\neq\mathfrak g [[/math]]

It follows from the real vector space structure of the Lie algebras [math]\mathfrak u_N[/math] and [math]\mathfrak{so}_N[/math] that there exists a nonzero symmetric real matrix of trace zero [math]X[/math] such that:

[[math]] iX\in\mathfrak g [[/math]]

We know that the space of symmetric real matrices of trace zero is an irreducible representation of [math]SO_N[/math] under the adjoint action. Thus [math]\mathfrak g[/math] must contain all such [math]X[/math], and hence [math]\mathfrak g=\mathfrak u_N[/math]. But since [math]U_N[/math] is connected, it follows that [math]G=U_N[/math].

\underline{Step 3}. Let us compute now the commutant of [math]SO_N[/math] in [math] M_N(\mathbb C)[/math]. Our first claim is that at [math]N=2[/math], this commutant is as follows:

[[math]] SO_2' =\left\{\begin{pmatrix} \alpha&\beta\\ -\beta&\alpha \end{pmatrix}\Big|\alpha,\beta\in\mathbb C\right\} [[/math]]

As for the case [math]N\geq3[/math], our claim here is that this commutant is as follows:

[[math]] SO_N'=\left\{\alpha I_N\Big|\alpha\in\mathbb C\right\} [[/math]]

Indeed, at [math]N=2[/math], the above formula is clear. At [math]N\geq 3[/math] now, an element in [math]X\in SO_N'[/math] commutes with any diagonal matrix having exactly [math]N-2[/math] entries equal to [math]1[/math] and two entries equal to [math]-1[/math]. Hence [math]X[/math] is diagonal. Now since [math]X[/math] commutes with any even permutation matrix, and we have assumed [math]N\geq 3[/math], it commutes in particular with the permutation matrix associated with the cycle [math](i,j,k)[/math] for any [math]1 \lt i \lt j \lt k[/math], and hence all the entries of [math]X[/math] are the same. We conclude that [math]X[/math] is a scalar matrix, as claimed.

\underline{Step 4}. Our claim now is that the set of matrices with nonzero trace is dense in [math]SO_N[/math].

At [math]N=2[/math] this is clear, since the set of elements in [math]SO_2[/math] having a given trace is finite. So assume [math]N \gt 2[/math], and consider a matrix as follows:

[[math]] T\in SO_N\simeq SO(\mathbb R^N)\quad,\quad Tr(T)=0 [[/math]]

Let [math]E\subset\mathbb R^N[/math] be a 2-dimensional subspace preserved by [math]T[/math], such that:

[[math]] T_{|E} \in SO(E) [[/math]]

Let [math]\varepsilon \gt 0[/math] and let [math]S_\varepsilon \in SO(E)[/math] satisfying the following condition:

[[math]] ||T_{|E}-S_\varepsilon|| \lt \varepsilon [[/math]]

Moreover, in the [math]N=2[/math] case, we can assume that [math]T[/math] satisfies as well:

[[math]] Tr(T_{|E})\neq Tr(S_\varepsilon) [[/math]]

Now define [math]T_\varepsilon\in SO(\mathbb R^N)=SO_N[/math] by the following formulae:

[[math]] T_{\varepsilon|E}=S_\varepsilon\quad,\quad T_{\varepsilon|E^\perp}=T_{|E^\perp} [[/math]]

It is clear that we have the following estimate:

[[math]] ||T-T_\varepsilon|| \leq ||T_{|E}-S_\varepsilon|| \lt \varepsilon [[/math]]

Also, we have the following estimate, which proves our claim:

[[math]] Tr(T_\varepsilon)=Tr(S_\varepsilon)+Tr(T_{|E^\perp})\neq0 [[/math]]

\underline{Step 5}. Our claim now is that [math]\mathbb TO_N[/math] is the normalizer of [math]\mathbb TSO_N[/math] in [math]U_N[/math], i.e. is the subgroup of [math]U_N[/math] consisting of the unitaries [math]U[/math] for which, for all [math]X\in\mathbb TSO_N[/math]:

[[math]] U^{-1}XU \in\mathbb TSO_N [[/math]]

Indeed, [math]\mathbb TO_N[/math] normalizes [math]\mathbb TSO_N[/math], so we must prove that if [math]U\in U_N[/math] normalizes [math]\mathbb TSO_N[/math] then [math]U\in\mathbb TO_N[/math]. First note that [math]U[/math] normalizes [math]SO_N[/math], because if [math]X \in SO_N[/math] then:

[[math]] U^{-1}XU \in\mathbb TSO_N [[/math]]

Thus we have a formula as follows, for some [math]\lambda\in\mathbb T[/math] and [math]Y\in SO_N[/math]:

[[math]] U^{-1}XU=\lambda Y [[/math]]

If [math]Tr(X)\neq0[/math], we have [math]\lambda\in\mathbb R[/math] and hence:

[[math]] \lambda Y=U^{-1}XU \in SO_N [[/math]]

The set of matrices having nonzero trace being dense in [math]SO_N[/math], we conclude that [math]U^{-1}XU \in SO_N[/math] for all [math]X\in SO_N[/math]. Thus, we have:

[[math]] \begin{eqnarray*} X \in SO_N &\implies&(UXU^{-1})^t(UXU^{-1})=I_N\\ &\implies&X^tU^tUX= U^tU\\ &\implies&U^tU \in SO_N' \end{eqnarray*} [[/math]]

It follows that at [math]N\geq 3[/math] we have [math]U^tU=\alpha I_N[/math], with [math]\alpha \in \mathbb T[/math], since [math]U[/math] is unitary. Hence we have [math]U=\alpha^{1/2}(\alpha^{-1/2}U)[/math] with:

[[math]] \alpha^{-1/2}U\in O_N\quad,\quad U\in\mathbb TO_N [[/math]]

If [math]N=2[/math], [math](U^tU)^t=U^tU[/math] gives again [math]U^tU=\alpha I_2[/math], and we conclude as before.

\underline{Step 6}. Our claim is that the inclusion [math]\mathbb TO_N\subset U_N[/math] is maximal.

Assume indeed that [math]\mathbb TO_N\subset G\subset U_N[/math] is a compact group such that [math]G\neq U_N[/math]. It is a well-known fact that the connected component of the identity in [math]G[/math] is a normal subgroup, denoted [math]G_0[/math]. Since we have [math]\mathbb TSO_N\subset G_0 \subset U_N[/math], we must have:

[[math]] G_0=\mathbb TSO_N [[/math]]

But since [math]G_0[/math] is normal in [math]G[/math], the group [math]G[/math] normalizes [math]\mathbb TSO_N[/math], and hence [math]G\subset\mathbb TO_N[/math], which finishes the proof.

■

Along the same lines, still following , we have as well the following result:

Theorem

The following inclusion of compact groups is maximal,

[[math]] PO_N\subset PU_N [[/math]]

in the sense that there is no intermediate compact group in between.

Show Proof

This follows from Theorem 10.3. Indeed, assuming [math]PO_N\subset G \subset PU_N[/math], the preimage of this subgroup under the quotient map [math]U_N\to PU_N[/math] would be then a proper intermediate subgroup of [math]\mathbb TO_N\subset U_N[/math], which is a contradiction.

■

Finally, still following , we have as well the following result:

Theorem

The following inclusion of compact quantum groups is maximal,

[[math]] O_N\subset O_N^* [[/math]]

in the sense that there is no intermediate compact quantum group in between.

Show Proof

Consider indeed a sequence of surjective Hopf [math]*[/math]-algebra maps as follows, whose composition is the canonical surjection:

[[math]] C(O_N^*)\overset{f}\longrightarrow A\overset{g}\longrightarrow C(O_N) [[/math]]

This produces a diagram of Hopf algebra maps with pre-exact rows, as follows:

[[math]] \xymatrix@R=45pt@C=33pt{ \mathbb C\ar[r]&C(PO_N^*)\ar[d]^{f_|}\ar[r]&C(O_N^*)\ar[d]^f\ar[r]&C(\mathbb Z_2)\ar[r]\ar@{=}[d]&\mathbb C\\ \mathbb C\ar[r]&PA\ar[d]^{g_|}\ar[r]&A\ar[d]^g\ar[r]&C(\mathbb Z_2)\ar[r]\ar@{=}[d]&\mathbb C\\ \mathbb C\ar[r]&PC(O_N)\ar[r]&C(O_N)\ar[r]&C(\mathbb Z_2)\ar[r]&\mathbb C} [[/math]]

Consider now the following composition, with the isomorphism on the left being something well-known, coming from , as explained in chapter 9:

[[math]] C(PU_N)\simeq C(PO_N^*)\overset{f_|}\longrightarrow PA\overset{g_|}\longrightarrow PC(O_N)\simeq C(PO_N) [[/math]]

This induces, at the group level, the folowing embedding:

[[math]] PO_N\subset PU_N [[/math]]

Thus [math]f_|[/math] or [math]g_|[/math] is an isomorphism. If [math]f_|[/math] is an isomorphism we get a commutative diagram of Hopf algebra morphisms with pre-exact rows, as follows:

[[math]] \xymatrix@R=45pt@C=33pt{ \mathbb C\ar[r]&C(PO_N^*)\ar@{=}[d]\ar[r]&C(O_N^*)\ar[d]^f\ar[r]&C(\mathbb Z_2)\ar[r]\ar@{=}[d]&\mathbb C\\ \mathbb C\ar[r]&C(PO_N^*)\ar[r]&A\ar[r]&C(\mathbb Z_2)\ar[r]&\mathbb C} [[/math]]

Then [math]f[/math] is an isomorphism. Similarly if [math]g_|[/math] is an isomorphism, then [math]g[/math] is an isomorphism, and this gives the result. See .

■

In connection now with our question, which is that of classifying the intermediate groups [math]O_N\subset G\subset U_N[/math], the above results lead to a dichotomy, coming from:

[[math]] PG\in\{PO_N,PU_N\} [[/math]]

In the lack of a classification result here, which is surely well-known, here are some basic examples of such intermediate groups, which are all well-known:

Proposition

We have compact groups [math]O_N\subset G\subset U_N[/math] as follows:

• The following groups, depending on a parameter [math]r\in\mathbb N\cup\{\infty\}[/math],
  [[math]] \mathbb Z_rO_N=\left\{wU\Big|w\in\mathbb Z_r,U\in O_N\right\} [[/math]]
  whose projective versions equal [math]PO_N[/math], and the biggest of which is the group [math]\mathbb TO_N[/math], which appears as affine lift of [math]PO_N[/math].
• The following groups, depending on a parameter [math]d\in 2\mathbb N\cup\{\infty\}[/math],
  [[math]] U_N^d=\left\{U\in U_N\Big|\det U\in\mathbb Z_d\right\} [[/math]]
  interpolating between [math]U_N^2[/math] and [math]U_N^\infty=U_N[/math], whose projective versions equal [math]PU_N[/math].

Show Proof

All the assertions are elementary, the idea being as follows:

(1) We have indeed compact groups [math]\mathbb Z_rO_N[/math] with [math]r\in\mathbb N\cup\{\infty\}[/math] as in the statement, whose projective versions are given by:

[[math]] P\mathbb Z_rO_N=PO_N [[/math]]

At [math]r=\infty[/math] we obtain the group [math]\mathbb TO_N[/math], and the fact that this group appears as the affine lift of [math]PO_N[/math] follows exactly as in the sphere case, as in the proof of Theorem 10.1.

(2) As a first observation, the following formula, with [math]d\in\mathbb N\cup\{\infty\}[/math], defines indeed a closed subgroup [math]U_N^d\subset U_N[/math]:

[[math]] U_N^d=\left\{U\in U_N\Big|\det U\in\mathbb Z_d\right\} [[/math]]

In the case where [math]d[/math] is even, this subgroup contains the orthogonal group [math]O_N[/math]. As for the last assertion, namely [math]PU_N^d=PU_N[/math], this follows either be suitably rescaling the unitary matrices, or by applying the result in Theorem 10.3.

■

The above result suggests that the solutions of [math]O_N\subset G\subset U_N[/math] should come from [math]O_N,U_N[/math], by succesively applying the following constructions:

[[math]] G\to\mathbb Z_rG\quad,\quad G\to G\cap U_N^d [[/math]]

These operations do not exactly commute, but normally we should be led in this way to a 2-parameter series, unifying the two 1-parameter series from (1,2). However, some other groups like [math]\mathbb Z_NSO_N[/math] work too, so all this is probably a bit more complicated.

In what follows we will be mostly interested in the group [math]\mathbb TO_N[/math], which fits with the spheres and tori that we already have, in view of our axiomatization purposes. This particular group [math]\mathbb TO_N[/math], and the whole series [math]\mathbb Z_rO_N[/math] with [math]r\in\mathbb N\cup\{\infty\}[/math] that it is part of, is known to be easy, the precise result, from Tarrago-Weber , being as follows:

Theorem

We have the following results:

• [math]\mathbb TO_N[/math] is easy, the corresponding category [math]\bar{P}_2\subset P_2[/math] consisting of the pairings having the property that when flattening, we have the following global formula:
  [[math]] \#\circ=\#\bullet [[/math]]
• [math]\mathbb Z_rO_N[/math] is easy, the corresponding category [math]P_2^r\subset P_2[/math] consisting of the pairings having the property that when flattening, we have the following global formula:
  [[math]] \#\circ=\#\bullet(r) [[/math]]

Show Proof

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

(1) If we denote the standard corepresentation by [math]u=zv[/math], with [math]z\in\mathbb T[/math] and with [math]v=\bar{v}[/math], then in order to have [math]Hom(u^{\otimes k},u^{\otimes l})\neq\emptyset[/math], the [math]z[/math] variabes must cancel, and in the case where they cancel, we obtain the same Hom-space as for [math]O_N[/math].

Now since the cancelling property for the [math]z[/math] variables corresponds precisely to the fact that [math]k,l[/math] must have the same numbers of [math]\circ[/math] symbols minus [math]\bullet[/math] symbols, the associated Tannakian category must come from the category of pairings [math]\bar{P}_2\subset P_2[/math], as claimed.

(2) This is something that we already know at [math]r=1,\infty[/math], where the group in question is [math]O_N,\mathbb TO_N[/math]. The proof in general is similar, by writing [math]u=zv[/math] as above.

■

Quite remarkably, the above result has the following converse, also from :

Theorem

The proper intermediate easy compact groups

[[math]] O_N\subset G\subset U_N [[/math]]

are precisely the groups [math]\mathbb Z_rO_N[/math], with [math]r\in\{2,3,\ldots,\infty\}[/math].

Show Proof

According to our conventions for the easy quantum groups, which apply of course to the classical case, we must compute the following intermediate categories:

[[math]] \mathcal P_2\subset D\subset P_2 [[/math]]

So, assume that we have such a category, [math]D\neq\mathcal P_2[/math], and pick an element [math]\pi\in D-\mathcal P_2[/math], assumed to be flat. We can modify [math]\pi[/math], by performing the following operations:

(1) First, we can compose with the basic crossing, in order to assume that [math]\pi[/math] is a partition of type [math]\cap\ldots\ldots\cap[/math], consisting of consecutive semicircles. Our assumption [math]\pi\notin\mathcal P_2[/math] means that at least one semicircle is colored black, or white.

(2) Second, we can use the basic mixed-colored semicircles, and cap with them all the mixed-colored semicircles. Thus, we can assume that [math]\pi[/math] is a nonzero partition of type [math]\cap\ldots\ldots\cap[/math], consisting of consecutive black or white semicircles.

(3) Third, we can rotate, as to assume that [math]\pi[/math] is a partition consisting of an upper row of white semicircles, [math]\cup\ldots\ldots\cup[/math], and a lower row of white semicircles, [math]\cap\ldots\ldots\cap[/math]. Our assumption [math]\pi\notin\mathcal P_2[/math] means that this latter partition is nonzero.

For [math]a,b\in\mathbb N[/math] consider the partition consisting of an upper row of [math]a[/math] white semicircles, and a lower row of [math]b[/math] white semicircles, and set:

[[math]] \mathcal C=\left\{\pi_{ab}\Big|a,b\in\mathbb N\right\}\cap D [[/math]]

According to the above, we have [math]\pi\in \lt \mathcal C \gt [/math]. The point now is that we have:

(1) There exists [math]r\in\mathbb N\cup\{\infty\}[/math] such that [math]\mathcal C[/math] equals the following set:

[[math]] \mathcal C_r=\left\{\pi_{ab}\Big|a=b(r)\right\} [[/math]]

This is indeed standard, by using the categorical axioms.

(2) We have the following formula, with [math]P_2^r[/math] being as above:

[[math]] \lt \mathcal C_r \gt =P_2^r [[/math]]

This is standard as well, by doing some diagrammatic work.

With these results in hand, the conclusion now follows. Indeed, with [math]r\in\mathbb N\cup\{\infty\}[/math] being as above, we know from the beginning of the proof that any [math]\pi\in D[/math] satisfies:

[[math]] \pi \in \lt \mathcal C \gt = \lt \mathcal C_r \gt =P_2^r [[/math]]

We conclude from this that we have an inclusion as follows:

[[math]] D\subset P_2^r [[/math]]

Conversely, we have as well the following inclusion:

[[math]] P_2^r = \lt \mathcal C_r \gt = \lt \mathcal C \gt \subset \lt D \gt =D [[/math]]

Thus we have [math]D=P_2^r[/math], and this finishes the proof. See .

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As a conclusion, [math]\mathbb TO_N[/math] is indeed the “privileged” unitary group that we were looking for, with the remark that its arithmetic versions [math]\mathbb Z_rO_N[/math] are interesting as well.

It remains now to discuss the reflection group case. Here the problem is that of classifying the intermediate compact groups [math]H_N\subset G\subset K_N[/math], but the situation is more complicated than in the continuous group case, with the 2-parameter series there being now replaced by a 3-parameter series. Instead of getting into this quite technical subject, let us just formulate a basic result, explaining what the 3 parameters are:

Proposition

We have compact groups [math]H_N\subset G\subset K_N[/math] as follows:

• The groups [math]\mathbb Z_rH_N[/math], with [math]r\in\mathbb N\cup\{\infty\}[/math].
• The groups [math]H_N^s=\mathbb Z_s\wr S_N[/math], with [math]s\in 2\mathbb N[/math].
• The groups [math]H_N^{sd}=H_N^s\cap U_N^d[/math], with [math]d|s[/math] and [math]s\in 2\mathbb N[/math].

Show Proof

The various constructions in the statement produce indeed closed subgroups [math]G\subset K_N[/math], and the condition [math]H_N\subset G[/math] is clearly satisfied as well.

■

The same discussion as in the continuous case applies, the idea being that the constructions [math]G\to\mathbb Z_rG[/math] and [math]G\to G\cap H_N^{sd}[/math] can be combined, and that all this leads in principle to a 3-parameter series. All this is, however, quite technical. Fortunately, exactly as in the continuous case, a solution to these classification problems comes from the notion of easiness. We have indeed the following result, coming from , :

Theorem

The following groups are easy:

• [math]\mathbb Z_rH_N[/math], the corresponding category [math]P_{even}^r\subset P_{even}[/math] consisting of the partitions having the property that when flattening, we have the following global formula:
  [[math]] \#\circ=\#\bullet(r) [[/math]]
• [math]H_N^s=\mathbb Z_s\wr S_N[/math], the corresponding category [math]P_{even}^{(s)}\subset P_{even}[/math] consisting of the partitions having the property that we have the following formula, in each block:
  [[math]] \#\circ=\#\bullet(s) [[/math]]

In addition, the easy solutions of [math]H_N\subset G\subset K_N[/math] appear by combining these examples.

Show Proof

All this is well-known, the idea being as follows:

(1) The computation here is similar to the one in the proof of Theorem 10.7, by writing the fundamental representation [math]u=zv[/math] as there.

(2) This is something very standard and fundamental, known since the paper , and which follows from a long, routine computation, perfomed there.

As for the last assertion, things here are quite technical, and for the precise statement and proof of the classification result, we refer here to paper .

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Summarizing, the situation here is more complicated than in the continuous group case. However, in what regards the “standard” solution, this is definitely [math]\mathbb TH_N[/math].

General references

Banica, Teo (2024). "Affine noncommutative geometry". arXiv:2012.10973 [math.QA].