12d. Maximality questions

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Let us go back now to the standard cube, and to edge problems, but without the easiness assumption, this time. An interesting family of questions here is that of proving that the easy solutions of various edge problems are in fact the only ones, even in the non-easy case. We will see that these are several results and conjectures here.

We have the following result from , to start with:

Theorem

The following inclusions are maximal:

• [math]\mathbb TO_N\subset U_N[/math].
• [math]PO_N\subset PU_N[/math].

Show Proof

In order to prove these results, consider as well the group [math]\mathbb TSO_N[/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 as well 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}. Our claim is that the commutant of [math]SO_N[/math] in [math] M_N(\mathbb C)[/math] is as follows:

• [math]SO_2' =\left\{\begin{pmatrix} \alpha&\beta\\ -\beta&\alpha \end{pmatrix}\Big|\alpha,\beta\in\mathbb C\right\}[/math].
• If [math]N\geq 3[/math], [math]SO_N'=\{\alpha I_N|\alpha\in\mathbb C\}[/math].

Indeed, at [math]N=2[/math] this is a direct computation. 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 a diagonal matrix. Now since [math]X[/math] commutes with any even permutation matrix and [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 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 let:

[[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] with [math]||T_{|E}-S_\varepsilon|| \lt \varepsilon[/math], and with [math]Tr(T_{|E}) \not= Tr(S_\varepsilon)[/math], in the [math]N=2[/math] case. Now define [math]T_\varepsilon\in SO(\mathbb R^N)=SO_N[/math] by:

[[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 computation:

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

Thus, we have proved our claim.

\underline{Step 5}. Our claim 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]]

It is clear that [math]\mathbb TO_N[/math] normalizes [math]\mathbb TSO_N[/math], so we must show 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]. Indeed if [math]X \in SO_N[/math] then:

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

Thus [math]U^{-1}XU= \lambda Y[/math] with [math]\lambda\in\mathbb T[/math], [math]Y\in SO_N[/math]. If [math]Tr(X)\neq0[/math], we have [math]\lambda\in\mathbb R[/math] and so:

[[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 that [math]U^tU=\alpha I_2[/math], and we conclude as in the previous case.

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

Suppose 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].

\underline{Step 7}. Our claim is that the inclusion [math]PO_N\subset PU_N[/math] is maximal in the category of compact groups.

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

■

In connection now with the “edge question” of classifying the intermediate groups [math]O_N\subset G\subset U_N[/math], the above result leads to a dichotomy, coming from:

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

Here are some basic examples of such intermediate groups:

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_r=O_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, and well-known.

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The above results suggest 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 constructions [math]G\to\mathbb Z_rG[/math] and [math]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) above. However, some other groups like [math]\mathbb Z_NSO_N[/math] work too, so all this is probably a bit more complicated.

We have as well the following related result, also from :

Theorem

The inclusion of compact quantum groups

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

is maximal in the category of compact quantum groups.

Show Proof

The idea is that this follows from the result regarding [math]PO_N\subset PU_N[/math], by taking affine lifts, and using algebraic techniques. 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=40pt@C=40pt{ \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 , that we will explain in chapter 16 below:

[[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 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=40pt@C=40pt{ \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. For further details on all this, we refer to .

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Summarizing, we are reaching to the conclusion formulated in the beginning of this chapter, namely that some of the easy solutions of the easy edge problems for the standard cube stay unique, even when lifting the easiness assumption.

In relation with these questions, we have as well the well-known and popular question of proving that the quantum group inclusion [math]S_N\subset S_N^+[/math] is maximal, in the sense that there is no intermediate quantum group, as follows:

[[math]] S_N\subset G\subset S_N^+ [[/math]]

As evidence for this latter conjecture, the inclusions [math]S_4\subset S_4^+[/math] and [math]S_5\subset S_5^+[/math] can be both shown to be maximal, by using advanced quantum algebra techniques, including some recent planar algebra work. However, there is no good idea so far in order to deal with the general case. We refer to , and related papers for a discussion here.

Finally, for our discussion to be complete, let us discuss as well twisting results. For this purpose, let us go back to the standard cube, namely:

[[math]] \xymatrix@R=20pt@C=20pt{ &K_N^+\ar[rr]&&U_N^+\\ H_N^+\ar[rr]\ar[ur]&&O_N^+\ar[ur]\\ &K_N\ar[rr]\ar[uu]&&U_N\ar[uu]\\ H_N\ar[uu]\ar[ur]\ar[rr]&&O_N\ar[uu]\ar[ur] } [[/math]]

According to the general Schur-Weyl twisting method from chapter 7, all these quantum groups can be twisted. In addition, the continuous twists were computed in chapter 7, and the discrete objects were shown in chapter 10 to be equal to their own twists. Thus, we are led to the following conclusion, in relation with the standard cube:

Theorem

The Schur-Weyl twists of the main quantum groups are

[[math]] \xymatrix@R=20pt@C=20pt{ &K_N^+\ar[rr]&&U_N^+\\ H_N^+\ar[rr]\ar[ur]&&O_N^+\ar[ur]\\ &K_N\ar[rr]\ar[uu]&&\bar{U}_N\ar[uu]\\ H_N\ar[uu]\ar[ur]\ar[rr]&&\bar{O}_N\ar[uu]\ar[ur] } [[/math]]

and we will call this diagram “twisted standard cube”.

Show Proof

This follows indeed from the above discussion.

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This construction raises the perspective of finding the twisted versions of the above classification results. Following , in the uniform case, the result here is as follows:

Theorem

The classical and free uniform twisted easy quantum groups are

[[math]] \xymatrix@R=7pt@C=7pt{ &&K_N^+\ar[rr]&&K_N^{++}\ar[rr]&&\ U_N^+\ \\ &H_N^{s+}\ar[ur]&&&&\\ H_N^+\ar[rrrr]\ar[ur]&&&&O_N^+\ar[uurr]\\ \\ &&K_N\ar[rrrr]\ar@.[uuuu]&&&&\ \bar{U}_N\ \ar@.[uuuu]\\ &H_N^s\ar[ur]&&&&\\ H_N\ar@.[uuuu]\ar[ur]\ar[rrrr]&&&&\bar{O}_N\ar@.[uuuu]\ar[uurr] \\ } [[/math]]

where [math]H_s=\mathbb Z_s\wr S_N[/math], [math]H_N^{s+}=\mathbb Z_s\wr_*S_N^+[/math] with [math]s=4,6,8\ldots[/math]\,, and where [math]K_N^+=\widetilde{K_N^+}[/math].

Show Proof

This follows indeed from Theorem 12.1 above, dealing with the untwisted case, and from the above discussion, regarding the twists.

■

We can merge the above result with the untwisted result, and we are led to the following statement, also from :

Theorem

The uniform classical/twisted and free quantum groups are

[[math]] \xymatrix@R=2mm@C=12mm{ &&U_N,\bar{U}_N\ar@/^/[drr]\\ K_N\ar[rr]\ar@/^/[urr]&&K_N^+\ar[r]&K_N^{++}\ar[r]&U_N^+\\ \\ H_N^s\ar[rr]\ar[uu]&&H_N^{s+}\ar[uu]\\ \\ H_N\ar[rr]\ar[uu]\ar@/_/[drr]&&H_N^+\ar[rr]\ar[uu]&&O_N^+\ar[uuuu]\\ &&O_N,\bar{O}_N\ar@/_/[urr]} [[/math]]

where [math]H_N^s=\mathbb Z_s\wr S_N[/math], [math]H_N^{s+}=\mathbb Z_s\wr_*S_N^+[/math], with [math]s\in\{2,4,\ldots,\infty\}[/math], and [math]K_N^{++}=\widetilde{K}_N^+[/math].

Show Proof

This is a slight extension of Theorem 12.1, the idea being as follows:

(1) All the above quantum groups are quizzy, and the uniformity condition is clear as well, for each of the quantum groups under consideration. Finally, all these quantum groups are either classical/twisted or free.

(2) In order to prove now the converse, in view of our Schur-Weyl twisting method, which only needs a category of partitions as input, it is enough to deal with the [math]q=1[/math] case. So, consider a uniform category of partitions [math]D\subset P_{even}[/math]. We must prove that in the classical/free cases, the solutions are:

[[math]] \xymatrix@R=2mm@C=11mm{ &&\mathcal P_2\ar@/_/[dll]\\ \mathcal P_{even}\ar[dd]&&\mathcal{NC}_{even}\ar[ll]\ar[dd]&\mathcal{NC}_{even}^-\ar[l]&\mathcal{NC}_2\ar[l]\ar@/_/[ull]\ar[dddd]\\ \\ P_{even}^s\ar[dd]&&NC_{even}^s\ar[ll]\ar[dd]\\ \\ P_{even}&&NC_{even}\ar[ll]&&NC_2\ar@/^/[dll]\ar[ll]\\ &&P_2\ar@/^/[ull]} [[/math]]

To be more precise, in the classical case, where [math]\backslash\hskip-2.1mm/\in D[/math], we must prove that the only solutions are the categories [math]P_2,\mathcal P_2,P_{even}^s[/math], and that in the free case, where [math]D\subset NC_{even}[/math], we must prove that the only solutions are the categories [math]NC_2,\mathcal{NC}_2,\mathcal{NC}_{even}^-,NC_{even}^s[/math].

(3) We jointly investigate these two problems. Let [math]B[/math] be the set of all possible labelled blocks in [math]D[/math], having no upper legs. Observe that [math]B[/math] is stable under the switching of colors operation, [math]\circ\leftrightarrow\bullet[/math]. We have two possible situations, as follows:

\underline{Case 1}. The set [math]B[/math] consists of pairings only. Here the pairings in question can be either all labelled pairings, namely [math]\circ-\circ[/math], [math]\circ-\bullet[/math], [math]\bullet-\circ[/math], [math]\bullet-\bullet[/math], or just the matching ones, namely [math]\circ-\bullet[/math], [math]\bullet-\circ[/math], and we obtain here the categories [math]P_2,\mathcal P_2[/math] in the classical case, and the categories [math]NC_2,\mathcal{NC}_2[/math] in the free case.

\underline{Case 2}. [math]B[/math] has at least one block of size [math]\geq 4[/math]. In this case we can let [math]s\in\{2,4,\ldots,\infty\}[/math] to be the length of the smallest [math]\circ\ldots\circ[/math] block, and we obtain in this way the category [math]P_{even}^s[/math] in the classical case, and the categories [math]\mathcal{NC}_{even}^-,NC_{even}^s[/math] in the free case.

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Finally, we have as well the following classification result:

Theorem

The easy quantum groups [math]H_N\subset G\subset O_N^+[/math] and their twists are

[[math]] \xymatrix@R=7mm@C=20mm{ &O_N\ar[r]&O_N^*\ar[rd]\\ H_N\ar[r]\ar[ur]\ar[rd]&H_N^\Gamma\ar[r]&H_N^{\diamond k}\ar[r]&O_N^+\\ &\bar{O}_N\ar[r]&\bar{O}_N^*\ar[ru]} [[/math]]

and the set formed by these quantum groups is stable by intersections.

Show Proof

There are several things to be proved here, the idea being as follows:

(1) According to the various classification and twisting results presented so far in this book, and to some straightforward extensions of them, the easy quantum groups [math]H_N\subset G\subset O_N^+[/math] and their twists are the quantum groups in the above diagram.

(2) Regarding now the intersection assertion, some straightforward computations show that we have the following intersection diagram:

[[math]] \xymatrix@R=7mm@C=20mm{ &O_N\ar[r]&O_N^*\ar[rd]\\ H_N\ar[r]\ar[ur]\ar[rd]&H_N^*\ar[r]\ar[ur]\ar[dr]&H_N^+\ar[r]&O_N^+\\ &\bar{O}_N\ar[r]&\bar{O}_N^*\ar[ru]} [[/math]]

But with this diagram in hand, the assertion follows. Indeed, the intersections between the quantum groups [math]O_N^\times[/math] are their twists are all on this diagram, and hence on the diagram in the statement as well. Regarding now the intersections of an easy quantum group [math]H_N\subset G\subset H_N^+[/math] with the twists [math]\bar{O}_N,\bar{O}_N^*[/math], we can use again the above diagram. Indeed, from [math]H_N^+\cap\bar{O}_N^*=H_N^*[/math] we deduce that both [math]K=G\cap\bar{O}_N,K'=G\cap\bar{O}_N^*[/math] appear as intermediate easy quantum groups [math]H_N\subset K^\times\subset H_N^*[/math], and we are done.

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For more details on the above, we refer to and related papers, for the most dealing with noncommutative geometry. And with the remark that noncommutative geometry has something to do with all this because groups and quantum groups are not everything, let's not forget about spheres, tori and other homogeneous spaces and manifolds, which are equally in need of structure and classification results, often in relation with structure and classification results for some related groups or quantum groups.

In relation with this, we have already talked a bit about noncommutative geometry at the end of chapter 11, and we will be back to this in chapter 15 below.

As a conclusion now to all this, the classification of the compact quantum groups is a very interesting topic, and the general idea is that of stating and proving the results in the easy case first, and then trying to lift the easiness assumption. But this is of course just a general idea, and we will in chapters 13-14 below that many things can be done in a completely different way, using maximal tori, without reference to easiness.

General references

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