13b. Monomial spheres

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

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Let us discuss now an alternative take on these questions, following , based on the notion of “monomiality”, which applies to the spheres, which are not easy. Looking back at the definition of the spheres that we have, and at the precise relations between the coordinates, we are led into the following notion:

Definition

A monomial sphere is a subset [math]S\subset S^{N-1}_{\mathbb C,+}[/math] obtained via relations

[[math]] x_{i_1}^{e_1}\ldots x_{i_k}^{e_k}=x_{i_{\sigma(1)}}^{f_1}\ldots x_{i_{\sigma(k)}}^{f_k}\quad,\quad\forall (i_1,\ldots,i_k)\in\{1,\ldots,N\}^k [[/math]]

with [math]\sigma\in S_k[/math] being certain permutations, and with [math]e_r,f_r\in\{1,*\}[/math] being certain exponents.

This definition is quite broad, and we have for instance as example the sphere [math]S^{N-1}_{\mathbb C,\times}[/math] coming from the relations [math]ab^*c=cb^*a[/math], corresponding to the following diagram:

[[math]] \xymatrix@R=10mm@C=5mm{\circ\ar@{-}[drr]&\bullet\ar@{-}[d]&\circ\ar@{-}[dll]\\\circ&\bullet&\circ} [[/math]]

This latter sphere is actually a quite interesting object, coming from the projective space considerations in , . However, while being monomial, this sphere does not exactly fit with our noncommutative geometry considerations here.

To be more precise, according to the work in , , this sphere is part of a triple [math](S^{N-1}_{\mathbb C,\times},\mathbb T_N^\times,U_N^\times)[/math], satisfying a simplified set of noncommutative geometry axioms. However, according to the work by Mang-Weber , the quantum group [math]U_N^\times[/math] has no reflection group counterpart [math]K_N^\times[/math]. Thus, this sphere does not exactly fit with our axiomatics here.

In view of these difficulties, we will restrict now the attention to the real case. Let us first recall, from the various classification results established in chapter 6, that we have the following fundamental result, dealing with the real case:

Theorem

There are exactly [math]3[/math] real easy geometries, namely

[[math]] \mathbb R^N\subset\mathbb R^N_*\subset\mathbb R^N_+ [[/math]]

coming from the following categories of pairings [math]D[/math],

[[math]] P_2\supset P_2^*\supset NC_2 [[/math]]

whose associated spheres are as follows,

[[math]] S^{N-1}_\mathbb R\subset S^{N-1}_{\mathbb R,*}\subset S^{N-1}_{\mathbb R,+} [[/math]]

and whose tori, unitary and reflection groups are given by similar formulae.

Show Proof

This is something that we know from chapter 6, coming from the fact that [math]G=O_N^*[/math] is the unique intermediate easy quantum group [math]O_N\subset G\subset O_N^+[/math].

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Let us focus now on the spheres, and try to better understand their “easiness” property, with results getting beyond what has been done above, in the general easy context. That is, our objects of interest in what follows will be the 3 real spheres, namely:

[[math]] S^{N-1}_\mathbb R\subset S^{N-1}_{\mathbb R,*}\subset S^{N-1}_{\mathbb R,+} [[/math]]

Our purpose in what follows we will be that of proving that these spheres are the only monomial ones. Following , in order to best talk about monomiality, in the present real case, it is convenient to introduce the following group:

[[math]] S_\infty=\bigcup_{k\geq0}S_k [[/math]]

To be more precise, this group appears by definition as an inductive limit, with the inclusions [math]S_k\subset S_{k+1}[/math] that we use being given by:

[[math]] \sigma\in S_k\implies\sigma(k+1)=k+1 [[/math]]

In terms of [math]S_\infty[/math], the definition of the monomial spheres reformulates as follows:

Proposition

The monomial spheres are the algebraic manifolds [math]S\subset S^{N-1}_{\mathbb R,+}[/math] obtained via relations of type

[[math]] x_{i_1}\ldots x_{i_k}=x_{i_{\sigma(1)}}\ldots x_{i_{\sigma(k)}},\ \forall (i_1,\ldots,i_k)\in\{1,\ldots,N\}^k [[/math]]

associated to certain elements [math]\sigma\in S_\infty[/math], where [math]k\in\mathbb N[/math] is such that [math]\sigma\in S_k[/math].

Show Proof

We must prove that the relations [math]x_{i_1}\ldots x_{i_k}=x_{i_{\sigma(1)}}\ldots x_{i_{\sigma(k)}}[/math] are left unchanged when replacing [math]k\to k+1[/math]. But this follows from [math]\sum_ix_i^2=1[/math], because:

[[math]] \begin{eqnarray*} &&x_{i_1}\ldots x_{i_k}x_{i_{k+1}}=x_{i_{\sigma(1)}}\ldots x_{i_{\sigma(k)}}x_{i_{k+1}}\\ &\implies&x_{i_1}\ldots x_{i_k}x_{i_{k+1}}^2=x_{i_{\sigma(1)}}\ldots x_{i_{\sigma(k)}}x_{i_{k+1}}^2\\ &\implies&\sum_{i_{k+1}}x_{i_1}\ldots x_{i_k}x_{i_{k+1}}^2=\sum_{i_{k+1}}x_{i_{\sigma(1)}}\ldots x_{i_{\sigma(k)}}x_{i_{k+1}}^2\\ &\implies&x_{i_1}\ldots x_{i_k}=x_{i_{\sigma(1)}}\ldots x_{i_{\sigma(k)}} \end{eqnarray*} [[/math]]

Thus we can indeed “simplify at right”, and this gives the result.

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As already mentioned, following , our goal in what follows will be that of proving that the 3 main spheres are the only monomial ones. In order to prove this result, we will use group theory methods. We call a subgroup [math]G\subset S_\infty[/math] filtered when it is stable under concatenation, in the sense that when writing [math]G=(G_k)[/math] with [math]G_k\subset S_k[/math], we have:

[[math]] \sigma\in G_k,\pi\in G_l\implies \sigma\pi\in G_{k+l} [[/math]]

With this convention, we have the following result:

Theorem

The monomial spheres are the subsets [math]S_G\subset S^{N-1}_{\mathbb R,+}[/math] given by

[[math]] C(S_G)=C(S^{N-1}_{\mathbb R,+})\Big/\Big \lt x_{i_1}\ldots x_{i_k}=x_{i_{\sigma(1)}}\ldots x_{i_{\sigma(k)}},\forall (i_1,\ldots,i_k)\in\{1,\ldots,N\}^k,\forall\sigma\in G_k\Big \gt [[/math]]

where [math]G=(G_k)[/math] is a filtered subgroup of [math]S_\infty=(S_k)[/math].

Show Proof

We know from Proposition 13.8 that the construction in the statement produces a monomial sphere. Conversely, given a monomial sphere [math]S\subset S^{N-1}_{\mathbb R,+}[/math], let us set:

[[math]] G_k=\left\{\sigma\in S_k\Big|x_{i_1}\ldots x_{i_k}=x_{i_{\sigma(1)}}\ldots x_{i_{\sigma(k)}},\forall (i_1,\ldots,i_k)\in\{1,\ldots,N\}^k\right\} [[/math]]

With [math]G=(G_k)[/math] we have then [math]S=S_G[/math]. Thus, it remains to prove that [math]G[/math] is a filtered group. But since the relations [math]x_{i_1}\ldots x_{i_k}=x_{i_{\sigma(1)}}\ldots x_{i_{\sigma(k)}}[/math] can be composed and reversed, each [math]G_k[/math] follows to be stable under composition and inversion, and is therefore a group. Also, since the relations [math]x_{i_1}\ldots x_{i_k}=x_{i_{\sigma(1)}}\ldots x_{i_{\sigma(k)}}[/math] can be concatenated as well, our group [math]G=(G_k)[/math] is stable under concatenation, and we are done.

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At the level of examples, according to our definitions, the simplest filtered groups, namely [math]\{1\}\subset S_\infty[/math], produce the simplest real spheres, namely:

[[math]] S^{N-1}_{\mathbb R,+}\supset S^{N-1}_\mathbb R [[/math]]

In order to discuss now the half-classical case, we need to introduce and study a certain privileged intermediate filtered group [math]\{1\}\subset S_\infty^*\subset S_\infty[/math], which will eventually produce the intermediate sphere [math]S^{N-1}_{\mathbb R,+}\supset S^{N-1}_{\mathbb R,*}\supset S^{N-1}_\mathbb R[/math]. This can be done as follows:

Proposition

Let [math]S_\infty^*\subset S_\infty[/math] be the set of permutations having the property that when labelling cyclically the legs as follows

[[math]] \bullet\circ\bullet\circ\ldots [[/math]]

each string joins a black leg to a white leg.

• [math]S_\infty^*[/math] is a filtered subgroup of [math]S_\infty[/math], generated by the half-classical crossing.
• We have [math]S_{2k}^*\simeq S_k\times S_k[/math], and [math]S^*_{2k+1}\simeq S_k\times S_{k+1}[/math], for any [math]k\in\mathbb N[/math].

Show Proof

The fact that [math]S_\infty^*[/math] is indeed a subgroup of [math]S_\infty[/math], which is filtered, is clear. Observe now that the half-classical crossing has the “black-to-white” joining property:

[[math]] \xymatrix@R=10mm@C=5mm{\circ\ar@{-}[drr]&\bullet\ar@{.}[d]&\circ\ar@{-}[dll]\\\bullet&\circ&\bullet} [[/math]]

Thus this crossing belongs to [math]S_3^*[/math], and it is routine to check that the filtered subgroup of [math]S_\infty[/math] generated by it is the whole [math]S_\infty^*[/math]. Regarding now the last assertion, observe first that the filtered subgroups [math]S_3^*,S_4^*[/math] consist of the following permutations:

[[math]] \xymatrix@R=10mm@C=5mm{\circ\ar@{-}[d]&\bullet\ar@{.}[d]&\circ\ar@{-}[d]\\\bullet&\circ&\bullet}\qquad\qquad \item[a]ymatrix@R=10mm@C=5mm{\circ\ar@{-}[drr]&\bullet\ar@{.}[d]&\circ\ar@{-}[dll]\\\bullet&\circ&\bullet}\qquad\qquad \item[a]ymatrix@R=10mm@C=3mm{\circ\ar@{-}[d]&\bullet\ar@{.}[d]&\circ\ar@{-}[d]&\bullet\ar@{.}[d]\\\bullet&\circ&\bullet&\circ} [[/math]]

[[math]] \xymatrix@R=10mm@C=3mm{\circ\ar@{-}[drr]&\bullet\ar@{.}[d]&\circ\ar@{-}[dll]&\bullet\ar@{.}[d]\\\bullet&\circ&\bullet&\circ}\qquad\qquad \item[a]ymatrix@R=10mm@C=3mm{\circ\ar@{-}[drr]&\bullet\ar@{.}[drr]&\circ\ar@{-}[dll]&\bullet\ar@{.}[dll]\\\bullet&\circ&\bullet&\circ}\qquad\qquad \item[a]ymatrix@R=10mm@C=3mm{\circ\ar@{-}[d]&\bullet\ar@{.}[drr]&\circ\ar@{-}[d]&\bullet\ar@{.}[dll]\\\bullet&\circ&\bullet&\circ} [[/math]]

Thus we have [math]S_3^*=S_1\times S_2[/math] and [math]S_4^*=S_2\times S_2[/math], with the first component coming from dotted permutations, and with the second component coming from the solid line permutations. The same argument works in general, and gives the last assertion.

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Now back to the main 3 real spheres, the result is as follows:

Proposition

The basic monomial real spheres, namely

[[math]] S^{N-1}_\mathbb R\subset S^{N-1}_{\mathbb R,*}\subset S^{N-1}_{\mathbb R,+} [[/math]]

come respectively from the filtered groups [math]S_\infty\supset S_\infty^*\supset\{1\}[/math].

Show Proof

This is clear by definition in the classical and in the free cases. In the half-liberated case, the result follows from Proposition 13.10 (1).

■

Now back to the general case, with the idea in mind of proving the uniqueness of the above spheres, consider a monomial sphere [math]S_G\subset S^{N-1}_{\mathbb R,+}[/math], with the filtered group [math]G\subset S_\infty[/math] taken to be maximal, as in the proof of Theorem 13.9. We have the following result:

Proposition

The filtered group [math]G\subset S_\infty[/math] associated to a monomial sphere [math]S\subset S^{N-1}_{\mathbb R,+}[/math] is stable under the following operations, on the corresponding diagrams:

• Removing outer strings.
• Removing neighboring strings.

Show Proof

Both these results follow by using the quadratic condition:

(1) Regarding the outer strings, by summing over [math]a[/math], we have:

[[math]] \begin{eqnarray*} Xa=Ya &\implies&Xa^2=Ya^2\\ &\implies&X=Y \end{eqnarray*} [[/math]]

We have as well the following computation:

[[math]] \begin{eqnarray*} aX=aY &\implies&a^2X=a^2Y\\ &\implies&X=Y \end{eqnarray*} [[/math]]

(2) Regarding the neighboring strings, once again by summing over [math]a[/math], we have:

[[math]] \begin{eqnarray*} XabY=ZabT &\implies&Xa^2Y=Za^2T\\ &\implies&XY=ZT \end{eqnarray*} [[/math]]

We have as well the following computation:

[[math]] \begin{eqnarray*} XabY=ZbaT &\implies&Xa^2Y=Za^2T\\ &\implies&XY=ZT \end{eqnarray*} [[/math]]

Thus [math]G=(G_k)[/math] has both the properties in the statement.

■

We can now state and prove a main result, from , as follows:

Theorem

There is only one intermediate monomial sphere

[[math]] S^{N-1}_\mathbb R\subset S\subset S^{N-1}_{\mathbb R,+} [[/math]]

namely the half-classical real sphere [math]S^{N-1}_{\mathbb R,*}[/math].

Show Proof

We will prove that the only filtered groups [math]G\subset S_\infty[/math] satisfying the conditions in Proposition 13.12 are those correspoding to our 3 spheres, namely:

[[math]] \{1\}\subset S_\infty^*\subset S_\infty [[/math]]

In order to do so, consider such a filtered group [math]G\subset S_\infty[/math]. We assume this group to be non-trivial, [math]G\neq\{1\}[/math], and we want to prove that we have [math]G=S_\infty^*[/math] or [math]G=S_\infty[/math].

\underline{Step 1}. Our first claim is that [math]G[/math] contains a 3-cycle. Assume indeed that two permutations [math]\pi,\sigma\in S_\infty[/math] have support overlapping on exactly one point, say:

[[math]] supp(\pi)\cap supp(\sigma)=\{i\} [[/math]]

The point is then that the commutator [math]\sigma^{-1}\pi^{-1}\sigma\pi[/math] is a 3-cycle, namely:

[[math]] (i,\sigma^{-1}(i),\pi^{-1}(i)) [[/math]]

Indeed the computation of the commutator goes as follows:

[[math]] \xymatrix@R=7mm@C=5mm{\pi\\ \sigma\\ \pi^{-1}\\ \sigma^{-1}}\qquad \item[a]ymatrix@R=6mm@C=5mm{\\ \\ =}\qquad \item[a]ymatrix@R=5mm@C=5mm{ \circ&\circ\ar@{-}[drr]&\circ&\bullet\ar@{-}[dl]&\circ\ar@{.}[d]&\circ\ar@{-}[d]&\circ\ar@{.}[d]\\ \circ\ar@{.}[d]&\circ\ar@{.}[d]&\circ\ar@{-}[d]&\bullet\ar@{-}[dr]&\circ&\circ\ar@{-}[dll]&\circ\\ \circ&\circ&\circ\ar@{-}[dr]&\bullet\ar@{-}[dll]&\circ\ar@{-}[d]&\circ\ar@{.}[d]&\circ\ar@{.}[d]\\ \circ\ar@{.}[d]&\circ\ar@{-}[d]&\circ\ar@{.}[d]&\bullet\ar@{-}[drr]&\circ\ar@{-}[dl]&\circ&\circ\\ \circ&\circ&\circ&\bullet&\circ&\circ&\circ } [[/math]]

Now let us pick a non-trivial element [math]\tau\in G[/math]. By removing outer strings at right and at left we obtain permutations [math]\tau'\in G_k,\tau''\in G_s[/math] having a non-trivial action on their right/left leg, and the trick applies, with:

[[math]] \pi=\tau'\otimes id_{s-1}\quad,\quad \sigma=id_{k-1}\otimes\tau'' [[/math]]

Thus, [math]G[/math] contains a 3-cycle, as claimed.

\underline{Step 2}. Our second claim is [math]G[/math] must contain one of the following permutations:

[[math]] \xymatrix@R=10mm@C=2mm{ \circ\ar@{-}[dr]&\circ\ar@{-}[dr]&\circ\ar@{-}[dll]\\ \circ&\circ&\circ}\qquad\qquad \item[a]ymatrix@R=10mm@C=2mm{ \circ\ar@{-}[drr]&\circ\ar@{.}[d]&\circ\ar@{-}[dr]&\circ\ar@{-}[dlll]\\ \circ&\circ&\circ&\circ} [[/math]]

[[math]] \xymatrix@R=10mm@C=2mm{ \circ\ar@{-}[dr]&\circ\ar@{-}[drr]&\circ\ar@{.}[d]&\circ\ar@{-}[dlll]\\ \circ&\circ&\circ&\circ}\qquad\qquad \item[a]ymatrix@R=10mm@C=2mm{ \circ\ar@{-}[drr]&\circ\ar@{.}[d]&\circ\ar@{-}[drr]&\circ\ar@{.}[d]&\circ\ar@{-}[dllll]\\ \circ&\circ&\circ&\circ&\circ} [[/math]]

Indeed, consider the 3-cycle that we just constructed. By removing all outer strings, and then all pairs of adjacent vertical strings, we are left with these permutations.

\underline{Step 3}. Our claim now is that we must have [math]S_\infty^*\subset G[/math]. Indeed, let us pick one of the permutations that we just constructed, and apply to it our various diagrammatic rules. From the first permutation we can obtain the basic crossing, as follows:

[[math]] \xymatrix@R=5mm@C=5mm{ \circ\ar@{-}[d]&\circ\ar@{-}[dr]&\circ\ar@{-}[dr]&\circ\ar@{-}[dll]\\ \circ\ar@{-}[dr]&\circ\ar@{-}[dr]&\circ\ar@{-}[dll]&\circ\ar@{-}[d]\\ \circ&\circ&\circ&\circ} \qquad \item[a]ymatrix@R=5mm@C=5mm{ \\ \to\\}\qquad \item[a]ymatrix@R=6mm@C=5mm{ \circ\ar@{-}[ddr]\ar@/^/@{.}[r]&\circ\ar@{-}[ddl]&\circ\ar@{-}[ddr]&\circ\ar@{-}[ddl]\\ \\ \circ\ar@/_/@{.}[r]&\circ&\circ&\circ} \qquad \item[a]ymatrix@R=5mm@C=5mm{ \\ \to\\}\qquad \item[a]ymatrix@R=6mm@C=5mm{ \circ\ar@{-}[ddr]&\circ\ar@{-}[ddl]\\ \\ \circ&\circ} [[/math]]

Also, by removing a suitable [math]\slash\hskip-2.1mm\backslash[/math] shaped configuration, which is represented by dotted lines in the diagrams below, we can obtain the basic crossing from the second and third permutation, and the half-liberated crossing from the fourth permutation:

[[math]] \xymatrix@R=10mm@C=2mm{ \circ\ar@{.}[drr]&\circ\ar@{.}[d]&\circ\ar@{-}[dr]&\circ\ar@{-}[dlll]\\ \circ&\circ&\circ&\circ}\qquad\qquad \item[a]ymatrix@R=10mm@C=2mm{ \circ\ar@{-}[dr]&\circ\ar@{.}[drr]&\circ\ar@{.}[d]&\circ\ar@{-}[dlll]\\ \circ&\circ&\circ&\circ}\qquad\qquad \item[a]ymatrix@R=10mm@C=2mm{ \circ\ar@{.}[drr]&\circ\ar@{.}[d]&\circ\ar@{-}[drr]&\circ\ar@{-}[d]&\circ\ar@{-}[dllll]\\ \circ&\circ&\circ&\circ&\circ} [[/math]]

Thus, in all cases we have a basic or half-liberated crossing, and so, as desired:

[[math]] S_\infty^*\subset G [[/math]]

\underline{Step 4}. Our last claim, which will finish the proof, is that there is no proper intermediate subgroup as follows:

[[math]] S_\infty^*\subset G\subset S_\infty [[/math]]

In order to prove this, observe that [math]S_\infty^*\subset S_\infty[/math] is the subgroup of parity-preserving permutations, in the sense that “[math]i[/math] even [math]\implies[/math] [math]\sigma(i)[/math] even”.

Now let us pick an element [math]\sigma\in S_k-S_k^*[/math], with [math]k\in\mathbb N[/math]. We must prove that the group [math]G= \lt S_\infty^*,\sigma \gt [/math] equals the whole [math]S_\infty[/math]. In order to do so, we use the fact that [math]\sigma[/math] is not parity preserving. Thus, we can find [math]i[/math] even such that [math]\sigma(i)[/math] is odd. In addition, up to passing to [math]\sigma|[/math], we can assume that [math]\sigma(k)=k[/math], and then, up to passing one more time to [math]\sigma|[/math], we can further assume that [math]k[/math] is even. Since both [math]i,k[/math] are even we have:

[[math]] (i,k)\in S_k^* [[/math]]

We conclude that the following element belongs to [math]G[/math]:

[[math]] \sigma(i,k)\sigma^{-1}=(\sigma(i),k) [[/math]]

But, since [math]\sigma(i)[/math] is odd, by deleting an appropriate number of vertical strings, [math](\sigma(i),k)[/math] reduces to the basic crossing [math](1,2)[/math]. Thus [math]G=S_\infty[/math], and we are done.

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As already mentioned in the above, the story is not over with this kind of result, because the complex case still remains to be worked out.

General references

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