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In short, how do we get the formula for the NC torus? I find the equations in many places (including here) but I still have no idea for how this comes from the torus. If my understanding is correct, then you take the torus and look at its algebra of (smooth or just continuous?) functions. Then make this noncommutative and that is your noncommutative torus. There are different ways to do this which lead to different anti-symmetric matrices $\theta$ which yield different tori. But how do I do this in practice? And is there any way to go backwards: Is there a way to take the information $UV = e^{2\pi i \theta} VU$ and do some math and say, "hey this is a torus in some sense!"?

When looking at noncommutative $\mathbb{R}^4$ for example, it is easy to explain what it has to do with $\mathbb{R}^4$ and how to get it. Is there no other way to simply explain this for the torus? Perhaps because the algebra of functions on the torus is much more complicated than that of $\mathbb{R}^4$?

I am also curious if there is some way to realize $\mathbb{T}^n_{\theta}$ as a quotient of $\mathbb{R}^4_{\theta}$ by some NC lattice (if such a thing exists). That might provide a nice construction of the NC torus.

A reference would be great. I find a lot of references to papers or books that talk about the NC torus a good way into the text and then it often invokes a lot of background. Maybe it really does take that much set up? Any help would be greatly appreciated. Thank you.

*Note: I put this on MO by mistake earlier, but I feel it's better suited here. I think I deleted the MO one...

user46348
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Recall that any $f \in C^\infty(\mathbb{T}^n)$ can be uniquely written as a convergent Fourier series $$ f = \sum_{\mathbb{k} \in \mathbb{Z}^n} f_{\mathbb{k}} U_{\mathbb{k}}, \quad f_{\mathbb{k}} := \int_{\mathbb{T}^n} e^{-2\pi i \langle \mathbb{k},t \rangle}f(t)\,dt, $$ where for each $\mathbb{k} \in \mathbb{Z}^n$, $$ \forall t \in \mathbb{T}^n, \quad U_{\mathbb{k}}(t) := e^{2\pi i \langle \mathbb{k},t\rangle}. $$ Recall, moreover, that the usual pointwise multiplication of functions corresponds to the convolution product of Fourier series: $$ \forall f, \; g \in C^\infty(\mathbb{T}^n), \quad fg = \sum_{\mathbb{k} \in \mathbb{Z}} \left(\sum_{\mathbb{k}^\prime \in \mathbb{Z}^n} f_{\mathbb{k}-\mathbb{k}^\prime}g_{\mathbb{k}^\prime} \right)U_{\mathbb{k}}. $$ In other words, $C^\infty(\mathbb{T}^n)$ is generated by unitaries $\{U_{\mathbb{k}}\}_{\mathbb{k} \in \mathbb{Z}^n}$ satisfying the relations $$ \forall \mathbb{k}, \; \mathbb{k}^\prime \in \mathbb{Z}^n, \quad U_{\mathbb{k}}U_{\mathbb{k}^\prime} = U_{\mathbb{k}+\mathbb{k}^\prime}, $$ which implies, in turn, that $$ \forall \mathbb{k}, \; \mathbb{k}^\prime \in \mathbb{Z}^n, \quad U_{\mathbb{k}^\prime}U_{\mathbb{k}} = U_{\mathbb{k}}U_{\mathbb{k}^\prime}. $$

Now, suppose that $\Theta : \mathbb{Z}^n \times \mathbb{Z}^n \to \mathbb{T} := \mathbb{R}/\mathbb{Z}$ is a (normalised) $2$-cocycle, i.e., that it satisfies

  1. For all $\mathbb{k}\in\mathbb{Z}^n$, $\Theta(\mathbb{0},\mathbb{k}) = \Theta(\mathbb{k},\mathbb{0}) = 0$ (normalisation).
  2. For all $\mathbb{p}$, $\mathbb{q}$, $\mathbb{r} \in \mathbb{Z}^n$, $\Theta(\mathbb{p},\mathbb{q}+\mathbb{r}) + \Theta(\mathbb{q},\mathbb{r}) = \Theta(\mathbb{p},\mathbb{q}) + \Theta(\mathbb{p}+\mathbb{q},\mathbb{r})$ ($2$-cocycle condition).

Then you can replace the usual convolution of Fourier series by a deformed convolution of Fourier series to obtain the noncommutative $n$-torus $C^\infty(\mathbb{T}^n_\Theta) := (C^\infty(\mathbb{T}^n),\star_\Theta)$: $$ \forall f,g \in C^\infty(\mathbb{T}^n), \quad f \star_\Theta g := \sum_{\mathbb{k} \in \mathbb{Z}} \left(\sum_{\mathbb{k}^\prime \in \mathbb{Z}^n} e^{-2\pi i \Theta(\mathbb{k}-\mathbb{k}^\prime,\mathbb{k}^\prime)}f_{\mathbb{k}-\mathbb{k}^\prime}g_{\mathbb{k}^\prime} \right)U_{\mathbb{k}}. $$ In terms of our unitary generators $\{U_{\mathbb{k}}\}_{\mathbb{k} \in \mathbb{Z}^n}$, this reduces to defining $$ \forall \mathbb{k}, \; \mathbb{k}^\prime \in \mathbb{Z}^n, \quad U_{\mathbb{k}} \star_\Theta U_{\mathbb{k}^\prime} := e^{-2\pi i \Theta(\mathbb{k},\mathbb{k}^\prime)}U_{\mathbb{k}+\mathbb{k}^\prime}, $$ which, in turn, implies the commutation relations $$ \forall \mathbb{k}, \; \mathbb{k}^\prime \in \mathbb{Z}^n, \quad U_{\mathbb{k}^\prime} \star_\Theta U_{\mathbb{k}} = e^{2\pi i \theta(\mathbb{k},\mathbb{k}^\prime)}U_{\mathbb{k}} \star_\Theta U_{\mathbb{k}^\prime}, $$ where $\theta : \mathbb{Z}^n \times \mathbb{Z}^n \to \mathbb{T}$ is defined by $$ \forall \mathbb{k}, \; \mathbb{k}^\prime \in \mathbb{Z}^n, \quad \theta(\mathbb{k},\mathbb{k}^\prime) := \Theta(\mathbb{k},\mathbb{k}^\prime) - \Theta(\mathbb{k}^\prime,\mathbb{k}). $$ As it turns out, $\theta$ is an alternating bicharacter, i.e.,

  1. For all $\mathbb{k} \in \mathbb{Z}^n$, $\theta(\mathbb{k},\mathbb{k}) = 0$ (alternating).
  2. For all $\mathbb{k} \in \mathbb{Z}^n$, $\mathbb{k}^\prime \mapsto \theta(\mathbb{k},\mathbb{k}^\prime)$ and $\mathbb{k}^\prime \mapsto \theta(\mathbb{k}^\prime,\mathbb{k})$ both define homomorphisms $\mathbb{Z}^n \to \mathbb{T}$ (bicharacter).

Conversely, every alternating bicharacter $\theta : \mathbb{Z}^n \to \mathbb{Z}^n$ can be induced in this way from a (normalised) $2$-cocycle $\Theta : \mathbb{Z}^n \to \mathbb{Z}^n$.

Now, what happens when $\Theta$ and $\Theta^\prime$ both induce the same alternating bicharacter $\theta$? It's an old theorem of Kleppner's that this happens if and only if $\Theta$ and $\Theta^\prime$ are cohomologous, i.e., there exists some $T : \mathbb{Z}^n \to \mathbb{T}$ such that $$ \forall \mathbb{k}, \; \mathbb{k}^\prime \in \mathbb{Z}^n, \quad \Theta^\prime(\mathbb{k},\mathbb{k}^\prime) = \Theta(\mathbb{k},\mathbb{k}^\prime) + T(\mathbb{k}) + T(\mathbb{k}^\prime) - T(\mathbb{k}+\mathbb{k}^\prime) ; $$ in other words, $\Theta^\prime - \Theta = dT$. In that case, we can define an explicit $\mathbb{T}^n$-equivariant isomorphism $\Psi_T : C^\infty(\mathbb{T}^n_\Theta) \to C^\infty(\mathbb{T}^n_{\Theta^\prime})$ by $$ \forall \mathbb{k} \in \mathbb{Z}^n, \quad \Psi_T(U_{\mathbb{k}}) = e^{2\pi i T(\mathbb{k})} U_{\mathbb{k}}. $$ Hence, up to equivariant isomorphism, for any alternating bicharacter $\theta$ (or equivalently, by Kleppner, for any class $\theta \in H^2(\mathbb{Z}^n,\mathbb{T})$, where $H^2(\mathbb{Z}^n,\mathbb{T})$ is the second group cohomology of $\mathbb{Z}^n$ with coefficients in $\mathbb{T}$), we can define the noncommutative $n$-torus $C^\infty(\mathbb{T}^n_\theta)$ as the algebra generated by unitaries $\{U_{\mathbb{k}}\}_{\mathbb{k} \in \mathbb{Z}^n}$ satisfying the commutation relations $$ \forall \mathbb{k}, \; \mathbb{k}^\prime \in \mathbb{Z}^n, \quad U_{\mathbb{k}^\prime} U_{\mathbb{k}} = e^{2\pi i \theta(\mathbb{k},\mathbb{k}^\prime)}U_{\mathbb{k}} U_{\mathbb{k}^\prime}; $$ to construct it, we just take $C^\infty(\mathbb{T}^n_\theta) := C^\infty(\mathbb{T}^n_\Theta)$ for any $2$-cocycle $\Theta$ inducing $\theta$.

Let me be more explicit about what appears in the literature. We have an isomorphism $$ \{\text{alternating bicharacters $\theta : \mathbb{Z}^n \times \mathbb{Z}^n \to \mathbb{T}$}\} \cong \mathbb{T}^{n(n-1)/2} $$ given by $$ \theta \mapsto (\theta(e_i,e_j))_{1 \leq i < j \leq n}, $$ where $\{e_i\}_{i=1}^n$ is the standard ordered basis for $\mathbb{R}^n$; conversely, the alteranting bicharacter $\theta$ corresponding to $(\theta_{ij})_{1\leq i < j \leq n} \in \mathbb{T}^{n(n-1)/2}$ is given by $$ \forall \mathbb{k}, \; \mathbb{k}^\prime \in \mathbb{Z}^n, \quad \theta(\mathbb{k},\mathbb{k}^\prime) = \sum_{1 \leq i < j \leq n} \theta_{ij}(k_i k^\prime_j - k_j k^\prime_i). $$ Then there are three conventions that tend to appear for constructing $2$-cocycles $\Theta : \mathbb{Z}^n \times \mathbb{Z}^n \to \mathbb{T}$ inducing $\theta$:

  1. For all $\mathbb{k}$, $\mathbb{k}^\prime \in \mathbb{Z}^n$, set $\Theta(\mathbb{k},\mathbb{k}^\prime) := \sum_{1 \leq i < j \leq n} \theta_{ij} k_i k^\prime_j$.
  2. For all $\mathbb{k}$, $\mathbb{k}^\prime \in \mathbb{Z}^n$, set $\Theta(\mathbb{k},\mathbb{k}^\prime) := -\sum_{1 \leq i < j \leq n} \theta_{ij} k_j k^\prime_i$.
  3. For each $1 \leq i < j \leq n$, choose $\tfrac{\theta_{ij}}{2}$ such that $2 \tfrac{\theta_{ij}}{2} = \theta_{ij}$. Hence, for all $\mathbb{k}$, $\mathbb{k}^\prime \in \mathbb{Z}^n$, set $\Theta(\mathbb{k},\mathbb{k}^\prime) := \sum_{1\leq i < j \leq n}\tfrac{\theta_{ij}}{2}(k_i k^\prime_j - k_j k^\prime_i)$.

The first two conventions give you non-alternating bicharacters, which complicates a fair bit of the algebra, but turn out to be far more robust from a group-cohomological perspective; in particular, they give honest-to-goodness splittings of the short exact sequence $$ 0 \to \ker(\Theta \mapsto \theta) \to \{\text{$2$-cocycles}\} \xrightarrow{\Theta \mapsto \theta} \{\text{alternating bicharacters}\} \to 0. $$ The third convention gives you alternating bicharacters, which simplifies a fair bit of the algebra, but all the non-uniqueness can give you headaches when dealing with certain technicalities.

  • Hi Branimir. Is it true that any alternating bi-character over $ \mathbb{Z}^{n} $ is equivalent to a non-degenerate one? – Transcendental Oct 16 '16 at 12:00
  • Given that you're talking about equivalence, I presume you have in mind a non-zero $2$-cocycle $\Theta$ that happens to be an alternating bicharacter itself. By non-degeneracy of $\Theta$, do you mean that $\Theta$ itself is non-degenerate as a bicharacter or that the canonically associated alternating bicharacter $\theta$ is non-degenerate as a bicharacter (i.e., that the cohomology class $[\Theta]$ is non-degenerate)? – Branimir Ćaćić Oct 16 '16 at 14:33
  • Hi Branimir. If I recall correctly, every $ 2 $-cocycle $ \Theta $ is co-homologous to a bi-character, which may not necessarily be the canonically associated alternating bi-character $ \theta $. In fact, I believe that in general, $ \theta $ is co-homologous to $ \Theta^{2} $ instead of $ \Theta $. We may thus restrict our attention to $ 2 $-cocycles that are bi-characters. Let me then modify my question as follows: Is any bi-character co-homologous to an alternating bi-character that is non-degenerate? – Transcendental Oct 16 '16 at 20:35
  • The answer is no. Let $\alpha \in \mathbb{R}/\mathbb{Z}$ be irrational and consider the bicharacter $\Theta$ on $\mathbb{Z}^3$ defined by $\Theta(\mathbb{x},\mathbb{y}) = \alpha x_1 y_2$, so that the corresponding alternating bicharacter $\theta$ encoding the cohomology class $[\Theta]$ of $\Theta$ is given by $\theta(\mathbb{x},\mathbb{y}) = \alpha(x_1y_2 - x_2y_1)$. – Branimir Ćaćić Oct 16 '16 at 20:51
  • Then any alternating bicharacter $\Theta^\prime$ cohomologous to $\Theta$ must necessarily take the form $\Theta^\prime(\mathbb{x},\mathbb{y}) = \beta(x_1y_2-x_2y_1)$ for $\beta \in \mathbb{R}/\mathbb{Z}$ with $2\beta = \alpha$, and such $\Theta^\prime$ is clearly degenerate (with $\mathbb{Z}e_3$ in its kernel). – Branimir Ćaćić Oct 16 '16 at 20:51
  • One minor point, Branimir. I don’t think that it’s quite right to define an $ n $-dimensional non-commutative torus as a universal $ C^{\ast} $-algebra generated by a sequence $ (U_{\mathbf{p}}){\mathbf{p} \in \mathbb{Z}^{n}} $ of unitary elements such that $ U{\mathbf{q}} U_{\mathbf{p}} = e^{2 \pi i \theta(\mathbf{p},\mathbf{q})} U_{\mathbf{p}} U_{\mathbf{q}} $ for all $ \mathbf{p},\mathbf{q} \in \mathbb{Z}^{n} $, where $ \theta $ is an alternating bi-character on $ \mathbb{Z}^{n} $. These commutation relations say nothing about the universal $ C^{\ast} $-algebra being finitely generated. – Transcendental Oct 19 '16 at 20:48
  • Defining universal $C^$-algebras is always a messy business. In practice I'd much rather deal with explicit constructions via strict deformation quantisation or even with Olesen–Pedersen–Takesaki's dynamical characterisation. But you're right, I'm sweeping something under the rug: I'm clumsily trying to say that a noncommutative $n$-torus as a twisted group $C^$-algebra of $\mathbb{Z}^n$ (with basically the same issues with completions as in the untwisted case). – Branimir Ćaćić Oct 19 '16 at 22:18