11a. Ad-hoc twists

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

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We have seen so far that the abstract noncommutative geometries, taken in a “spherical” sense, with coordinates bounded by [math]||x_i||\leq1[/math], can be axiomatized with the help of quadruplets [math](S,T,U,K)[/math]. There are 9 main such geometries, as follows:

[[math]] \xymatrix@R=35pt@C=38pt{ \mathbb R^N_+\ar[r]&\mathbb T\mathbb R^N_+\ar[r]&\mathbb C^N_+\\ \mathbb R^N_*\ar[u]\ar[r]&\mathbb T\mathbb R^N_*\ar[u]\ar[r]&\mathbb C^N_*\ar[u]\\ \mathbb R^N\ar[u]\ar[r]&\mathbb T\mathbb R^N\ar[u]\ar[r]&\mathbb C^N\ar[u] } [[/math]]

An important question, that we would like to investigate now, concerns the twisting of these geometries, by suitably replacing commutation with anticommutation:

[[math]] ab=ba\quad\to\quad ab=\pm ba [[/math]]

[[math]] abc=cba\quad\to\quad abc=\pm cba [[/math]]

We will see that this is possible, and that we have twisted geometries, as follows:

[[math]] \xymatrix@R=35pt@C=36pt{ \mathbb R^N_+\ar[r]&\mathbb T\mathbb R^N_+\ar[r]&\mathbb C^N_+\\ \bar{\mathbb R}^N_*\ar[u]\ar[r]&\mathbb T\bar{\mathbb R}^N_*\ar[u]\ar[r]&\bar{\mathbb C}^N_*\ar[u]\\ \bar{\mathbb R}^N\ar[u]\ar[r]&\mathbb T\bar{\mathbb R}^N\ar[u]\ar[r]&\bar{\mathbb C}^N\ar[u] } [[/math]]

Here the bars stand, as before in this book, for anticommutation twists. However, all this is quite tricky, and before starting, a few general comments:

(1) First of all, our work is motivated by the general commutation/anticommutation duality from quantum mechanics, in a general sense. If there is one thing to be learned from basic quantum mechanics, and let us recommend here again our favorite books, namely Dirac , Feynman , Griffiths , von Neumann , Weinberg , Weyl , this is the fact that there is no commutation without anticommutation.

(2) Mathematically, and in relation with what we have been doing so far here, we have already met [math]q=-1[/math] twists on several occasions, and notably in relation with the computation of the quantum isometry groups [math]G^+(T)[/math] of our tori [math]T[/math], with one of our 7 noncommutative geometry axioms stating that we must have [math]K=G^+(T)\cap K_N^+[/math]. And the point is that [math]G^+(T)[/math], quite surpringly, often happens to be a [math]q=-1[/math] twist.

(3) And there are countless other reasons, both mathematical and physical, to look at anticommutation and [math]q=-1[/math] twists. If you are a bit familiar for instance with Drinfeld-Jimbo , , you probably know that many geometric objects can be deformed with the help of a parameter [math]q\in\mathbb C[/math], the interesting case being [math]q\in\mathbb T[/math], and more specifically the case where [math]q[/math] is a root of unity. And aren't [math]q=\pm1[/math] the simplest roots of unity.

(4) Summarizing, we have motivations. However, when getting to work, several surprises are waiting for us. First if the fact that the Drinfeld-Jimbo deformation procedure , produces non-semisimple objects at roots of unity [math]q\neq1[/math], and in particular at [math]q=-1[/math]. Thus, this very popular theory is useless for us, not to say wrong in our opinion, and we must come up with new definitions for everything, at [math]q=-1[/math].

(5) Fortunately, this is possible, with the correct objects at [math]q=-1[/math] having emerged, in a somewhat discreet way, not to contradict much the popular belief, and get sent to the Inquisition or something, in a number of technical papers, all peer-reviewed and published, on quantum groups and noncommutative geometry, all over the late 00s and 10s, starting with which launched everything, with the correct twist of [math]O_N[/math].

(6) And so, getting back now to the question of twisting the 9 geometries that we have, this is definitely possible, thanks to all this underground, while ironically public, [math]q=-1[/math] knowledge accumulated over the years, and we will explain this, in this chapter. With the technical remark that the twisted geometries do not exactly satisfy our axioms from chapter 4, but are not far from them either, and we will comment on this.

(7) Finally, this chapter will be a modest introduction to all this. The geometries of [math]\bar{\mathbb R}^N,\bar{\mathbb C}^N[/math] for instance are potentially as wide as those of [math]\mathbb R^N,\mathbb C^N[/math], and with many classical techniques applying well, and there is certainly room for writing a book on this topic, “twisted geometry”. Let me also mention that, in the lack of such a book, you can always ask my colleague Bichon about such things, he's the one who knows.

In order to get started now, the best is to deform first the simplest objects that we have, namely the quantum spheres. This can be done as follows:

Theorem

We have quantum spheres as follows, obtained via the twisted commutation relations [math]ab=\pm ba[/math], and twisted half-commutation relations [math]abc=\pm cba[/math],

[[math]] \xymatrix@R=13mm@C=10mm{ S^{N-1}_{\mathbb R,+}\ar[r]&\mathbb TS^{N-1}_{\mathbb R,+}\ar[r]&S^{N-1}_{\mathbb C,+}\\ \bar{S}^{N-1}_{\mathbb R,*}\ar[r]\ar[u]&\mathbb T\bar{S}^{N-1}_{\mathbb R,*}\ar[r]\ar[u]&\bar{S}^{N-1}_{\mathbb C,*}\ar[u]\\ \bar{S}^{N-1}_\mathbb R\ar[r]\ar[u]&\mathbb T\bar{S}^{N-1}_\mathbb R\ar[r]\ar[u]&\bar{S}^{N-1}_\mathbb C\ar[u]} [[/math]]

with the precise signs being as follows:

• The signs on the bottom correspond to anticommutation of distinct coordinates, and their adjoints. That is, with [math]z_i=x_i,x_i^*[/math] and [math]\varepsilon_{ij}=1-\delta_{ij}[/math], the formula is:
  [[math]] z_iz_j=(-1)^{\varepsilon_{ij}}z_jz_i [[/math]]
• The signs in the middle come from functoriality, as for the spheres in the middle to contain those on the bottom. That is, the formula is:
  [[math]] z_iz_jz_k=(-1)^{\varepsilon_{ij}+\varepsilon_{jk}+\varepsilon_{ik}}z_kz_jz_i [[/math]]

Show Proof

This is something elementary, from , the idea being as follows:

(1) Here there is nothing to prove, because we can define the spheres on the bottom by the following formulae, with [math]z_i=x_i,x_i^*[/math] and [math]\varepsilon_{ij}=1-\delta_{ij}[/math] being as above:

[[math]] C(\bar{S}^{N-1}_\mathbb R)=C(S^{N-1}_{\mathbb R,+})\Big/\Big \lt x_ix_j=(-1)^{\varepsilon_{ij}}x_jx_i\Big \gt [[/math]]

[[math]] C(\bar{S}^{N-1}_\mathbb C)=C(S^{N-1}_{\mathbb C,+})\Big/\Big \lt z_iz_j=(-1)^{\varepsilon_{ij}}z_jz_i\Big \gt [[/math]]

(2) Here our claim is that, if we want to construct half-classical twisted spheres, via relations of type [math]abc=\pm cba[/math] between the coordinates [math]x_i[/math] and their adjoints [math]x_i^*[/math], as for these spheres to contain the twisted spheres constructed in (1), the only possible choice for these relations is as follows, with [math]z_i=x_i,x_i^*[/math] and [math]\varepsilon_{ij}=1-\delta_{ij}[/math] being as above:

[[math]] z_iz_jz_k=(-1)^{\varepsilon_{ij}+\varepsilon_{jk}+\varepsilon_{ik}}z_kz_jz_i [[/math]]

But this is something clear, coming from the following computation, inside of the quotient algebras corresponding to the twisted spheres constructed in (1) above:

[[math]] \begin{eqnarray*} z_iz_jz_k &=&(-1)^{\varepsilon_{ij}}z_jz_iz_k\\ &=&(-1)^{\varepsilon_{ij}+\varepsilon_{ik}}z_jz_kz_i\\ &=&(-1)^{\varepsilon_{ij}+\varepsilon_{jk}+\varepsilon_{ik}}z_kz_jz_i \end{eqnarray*} [[/math]]

Thus, we are led to the conclusion in the statement, the spheres being given by:

[[math]] C(\bar{S}^{N-1}_{\mathbb R,*})=C(S^{N-1}_{\mathbb R,+})\Big/\Big \lt x_ix_jx_k=(-1)^{\varepsilon_{ij}+\varepsilon_{jk}+\varepsilon_{ik}}x_kx_jx_i\Big \gt [[/math]]

[[math]] C(\bar{S}^{N-1}_{\mathbb C,*})=C(S^{N-1}_{\mathbb C,+})\Big/\Big \lt z_iz_jz_k=(-1)^{\varepsilon_{ij}+\varepsilon_{jk}+\varepsilon_{ik}}z_kz_jz_i\Big \gt [[/math]]

Thus, we have constructed our spheres, and embeddings, as desired.

■

Let us twist now the unitary quantum groups [math]U[/math]. We would like these to act on the corresponding spheres, [math]U\curvearrowright S[/math]. Thus, we would like to have morphisms, as follows:

[[math]] \Phi(x_i)=\sum_jx_j\otimes u_{ji} [[/math]]

Now with [math]z_i=x_i,x_i^*[/math] being as before, and with [math]v_{ij}=u_{ij},u_{ij}^*[/math] constructed accordingly, the above formula and its adjoint tell us that we must have:

[[math]] \Phi(z_i)=\sum_jz_j\otimes v_{ji} [[/math]]

Thus the variables [math]Z_i=\sum_jz_j\otimes v_{ji}[/math] on the right must satisfy the twisted commutation or half-commutation relations in Theorem 11.1, and this will lead us to the correct twisted commutation or half-commutation relations to be satisfied by the variables [math]v_{ij}[/math]. In practice now, let us first discuss the twisting of [math]O_N,U_N[/math]. Following in the orthogonal case, and then in the unitary case, the result here is as follows:

Theorem

We have twisted orthogonal and unitary groups, as follows,

[[math]] \xymatrix@R=15mm@C=15mm{ O_N^+\ar[r]&U_N^+\\ \bar{O}_N\ar[r]\ar[u]&\bar{U}_N\ar[u]} [[/math]]

defined via the following relations, with the convention [math]\alpha=a,a^*[/math] and [math]\beta=b,b^*[/math]:

[[math]] \alpha\beta=\begin{cases} -\beta\alpha&{\rm for}\ a,b\in\{u_{ij}\}\ {\rm distinct,\ on\ the\ same\ row\ or\ column\ of\ }u\\ \beta\alpha&{\rm otherwise} \end{cases} [[/math]]

These quantum groups act on the corresponding twisted real and complex spheres.

Show Proof

Let us first discuss the construction of the quantum group [math]\bar{O}_N[/math]. We must prove that the algebra [math]C(\bar{O}_N)[/math] obtained from [math]C(O_N^+)[/math] via the relations in the statement has a comultiplication [math]\Delta[/math], a counit [math]\varepsilon[/math], and an antipode [math]S[/math]. Regarding [math]\Delta[/math], let us set:

[[math]] U_{ij}=\sum_ku_{ik}\otimes u_{kj} [[/math]]

For [math]j\neq k[/math] we have the following computation:

[[math]] \begin{eqnarray*} U_{ij}U_{ik} &=&\sum_{s\neq t}u_{is}u_{it}\otimes u_{sj}u_{tk}+\sum_su_{is}u_{is}\otimes u_{sj}u_{sk}\\ &=&\sum_{s\neq t}-u_{it}u_{is}\otimes u_{tk}u_{sj}+\sum_su_{is}u_{is}\otimes(-u_{sk}u_{sj})\\ &=&-U_{ik}U_{ij} \end{eqnarray*} [[/math]]

Also, for [math]i\neq k,j\neq l[/math] we have the following computation:

[[math]] \begin{eqnarray*} U_{ij}U_{kl} &=&\sum_{s\neq t}u_{is}u_{kt}\otimes u_{sj}u_{tl}+\sum_su_{is}u_{ks}\otimes u_{sj}u_{sl}\\ &=&\sum_{s\neq t}u_{kt}u_{is}\otimes u_{tl}u_{sj}+\sum_s(-u_{ks}u_{is})\otimes(-u_{sl}u_{sj})\\ &=&U_{kl}U_{ij} \end{eqnarray*} [[/math]]

Thus, we can define a comultiplication map for [math]C(\bar{O}_N)[/math], by setting:

[[math]] \Delta(u_{ij})=U_{ij} [[/math]]

Regarding now the counit [math]\varepsilon[/math] and the antipode [math]S[/math], things are clear here, by using the same method, and with no computations needed, the formulae to be satisfied being trivially satisfied. We conclude that [math]\bar{O}_N[/math] is a compact quantum group, and the proof for [math]\bar{U}_N[/math] is similar, by adding [math]*[/math] exponents everywhere in the above computations.

Finally, the last assertion is clear too, by doing some elementary computations, of the same type as above, and with the remark that the converse holds too, in the sense that if we want a quantum group [math]U\subset U_N^+[/math] to be defined by relations of type [math]ab=\pm ba[/math], and to have an action [math]U\curvearrowright S[/math] on the corresponding twisted sphere, we are led to the relations in the statement. We refer to for further details on all this.

■

In order to discuss now the half-classical case, given three coordinates [math]a,b,c\in\{u_{ij}\}[/math], let us set [math]span(a,b,c)=(r,c)[/math], where [math]r,c\in\{1,2,3\}[/math] are the number of rows and columns spanned by [math]a,b,c[/math]. In other words, if we write [math]a=u_{ij},b=u_{kl},c=u_{pq}[/math] then [math]r=\#\{i,k,p\}[/math] and [math]l=\#\{j,l,q\}[/math]. With this convention, we have the following result:

Theorem

We have intermediate quantum groups as follows,

[[math]] \xymatrix@R=12mm@C=12mm{ O_N^+\ar[r]&\mathbb TO_N^+\ar[r]&U_N^+\\ \bar{O}_N^*\ar[r]\ar[u]&\mathbb T\bar{O}_N^*\ar[r]\ar[u]&\bar{U}_N^*\ar[u]\\ \bar{O}_N\ar[r]\ar[u]&\mathbb T\bar{O}_N\ar[r]\ar[u]&\bar{U}_N\ar[u]} [[/math]]

defined via the following relations, with [math]\alpha=a,a^*[/math], [math]\beta=b,b^*[/math] and [math]\gamma=c,c^*[/math],

[[math]] \alpha\beta\gamma=\begin{cases} -\gamma\beta\alpha&{\rm for}\ a,b,c\in\{u_{ij}\}\ {\rm with}\ span(a,b,c)=(\leq 2,3)\ {\rm or}\ (3,\leq 2)\\ \gamma\beta\alpha&{\rm otherwise} \end{cases} [[/math]]

which act on the corresponding twisted half-classical real and complex spheres.

Show Proof

We use the same method as for Theorem 11.2, but with the combinatorics being now more complicated. Observe first that the rules for the various commutation and anticommutation signs in the statement can be summarized as follows:

[[math]] \begin{matrix} r\backslash c&1&2&3\\ 1&+&+&-\\ 2&+&+&-\\ 3&-&-&+ \end{matrix} [[/math]]

Let us first prove the result for [math]\bar{O}_N^*[/math]. We must construct here morphisms [math]\Delta,\varepsilon,S[/math], and the proof, similar to the proof of Theorem 11.2, goes as follows:

(1) We first construct [math]\Delta[/math]. For this purpose, we must prove that [math]U_{ij}=\sum_ku_{ik}\otimes u_{kj}[/math] satisfy the relations in the statement. We have the following computation:

[[math]] \begin{eqnarray*} U_{ia}U_{jb}U_{kc} &=&\sum_{xyz}u_{ix}u_{jy}u_{kz}\otimes u_{xa}u_{yb}u_{zc}\\ &=&\sum_{xyz}\pm u_{kz}u_{jy}u_{ix}\otimes\pm u_{zc}u_{yb}u_{xa}\\ &=&\pm U_{kc}U_{jb}U_{ia} \end{eqnarray*} [[/math]]

We must show that, when examining the precise two [math]\pm[/math] signs in the middle formula, their product produces the correct [math]\pm[/math] sign at the end. But the point is that both these signs depend only on [math]s=span(x,y,z)[/math], and for [math]s=1,2,3[/math] respectively, we have:

-- For a [math](3,1)[/math] span we obtain [math]+-[/math], [math]+-[/math], [math]-+[/math], so a product [math]-[/math] as needed.

-- For a [math](2,1)[/math] span we obtain [math]++[/math], [math]++[/math], [math]--[/math], so a product [math]+[/math] as needed.

-- For a [math](3,3)[/math] span we obtain [math]--[/math], [math]--[/math], [math]++[/math], so a product [math]+[/math] as needed.

-- For a [math](3,2)[/math] span we obtain [math]+-[/math], [math]+-[/math], [math]-+[/math], so a product [math]-[/math] as needed.

-- For a [math](2,2)[/math] span we obtain [math]++[/math], [math]++[/math], [math]--[/math], so a product [math]+[/math] as needed.

Together with the fact that our problem is invariant under [math](r,c)\to(c,r)[/math], and with the fact that for a [math](1,1)[/math] span there is nothing to prove, this finishes the proof for [math]\Delta[/math].

(2) The construction of the counit, via the formula [math]\varepsilon(u_{ij})=\delta_{ij}[/math], requires the Kronecker symbols [math]\delta_{ij}[/math] to commute/anticommute according to the above table. Equivalently, we must prove that the situation [math]\delta_{ij}\delta_{kl}\delta_{pq}=1[/math] can appear only in a case where the above table indicates “+”. But this is clear, because [math]\delta_{ij}\delta_{kl}\delta_{pq}=1[/math] implies [math]r=c[/math].

(3) Finally, the construction of the antipode, via the formula [math]S(u_{ij})=u_{ji}[/math], is clear too, because this requires the choice of our [math]\pm[/math] signs to be invariant under transposition, and this is true, the above table being symmetric.

We conclude that [math]\bar{O}_N^*[/math] is indeed a compact quantum group, and the proof for [math]\bar{U}_N^*[/math] is similar, by adding [math]*[/math] exponents everywhere in the above. Finally, the last assertion is clear too, exactly as in the proof of Theorem 11.2. We refer to for details.

■

The above results can be summarized as follows:

Theorem

We have quantum groups as follows, obtained via the twisted commutation relations [math]ab=\pm ba[/math], and twisted half-commutation relations [math]abc=\pm cba[/math],

[[math]] \xymatrix@R=12mm@C=12mm{ O_N^+\ar[r]&\mathbb TO_N^+\ar[r]&U_N^+\\ \bar{O}_N^*\ar[r]\ar[u]&\mathbb T\bar{O}_N^*\ar[r]\ar[u]&\bar{U}_N^*\ar[u]\\ \bar{O}_N\ar[r]\ar[u]&\mathbb T\bar{O}_N\ar[r]\ar[u]&\bar{U}_N\ar[u]} [[/math]]

with the various signs coming as follows:

• The signs for [math]\bar{O}_N[/math] correspond to anticommutation of distinct entries on rows and columns, and commutation otherwise, with this coming from [math]\bar{O}_N\curvearrowright\bar{S}^{N-1}_\mathbb R[/math].
• The signs for [math]\bar{O}_N^*,\bar{U}_N,\bar{U}_N^*[/math] come as well from the signs for [math]\bar{S}^{N-1}_\mathbb R[/math], either via the requirement [math]\bar{O}_N\subset U[/math], or via the requirement [math]U\curvearrowright S[/math].

Show Proof

This is a summary of Theorem 11.2 and Theorem 11.3, and their proofs.

■

Moving ahead now, and back to our geometric program, we have twisted the spheres and unitary groups [math]S,U[/math], and we are left with twisting the tori and reflection groups [math]T,K[/math]. But these are “discrete” objects, which can only be rigid, so let us formulate:

Definition

The twists of the basic quantum tori and reflection groups,

[[math]] \xymatrix@R=11mm@C=11mm{ T_N^+\ar[r]&\mathbb TT_N^+\ar[r]&\mathbb T_N^+\\ T_N^*\ar[r]\ar[u]&\mathbb TT_N^*\ar[r]\ar[u]&\mathbb T_N^*\ar[u]\\ T_N\ar[r]\ar[u]&\mathbb TT_N\ar[r]\ar[u]&\mathbb T_N\ar[u]} \qquad\quad\qquad \item[a]ymatrix@R=11mm@C=11mm{ H_N^+\ar[r]&\mathbb TH_N^+\ar[r]&K_N^+\\ H_N^*\ar[r]\ar[u]&\mathbb TH_N^*\ar[r]\ar[u]&K_N^*\ar[u]\\ H_N\ar[r]\ar[u]&\mathbb TH_N\ar[r]\ar[u]&K_N\ar[u]} [[/math]]

are by definition these tori and reflection groups themselves.

With this definition in hand, we are done with our twisting program for the triples [math](S,T,U,K)[/math], and we have now candidates [math]\bar{\mathbb R}^N[/math], [math]\bar{\mathbb C}^N[/math] and [math]\bar{\mathbb R}^N_*[/math], [math]\bar{\mathbb C}^N_*[/math] for new noncommutative geometries, to be checked from our axiomatic viewpoint, and then developed.

General references

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