The general context for the question - is belief that quantization of compact symplectic manifolds can be endowed with the structure of $C^*$-algebra (see MO230695). The particular question is about the most simple example - sphere $S^2$ with the standard symplectic form.

I think that corresponding $C^*$-algebra can be defined explicitly by generators and relations. However the difficulty is that not all the relations define the $C^*$-algebra (see MO151809) , so I wonder whether the relations below work ?

QUESTION: do these relations define $C^*$-algebra:

three self-adjoint generators: 

$x_1, x_2 , x_3: x_i^*=x_i$

and four relations, three relations says that $x_i$ generate $so(3)$ Lie algebra: 

$$ [x_1, x_2] = 2 i x_3, ~~ [x_2, x_3] = 2 i x_1 , ~~ [x_3, x_1] = 2 i x_2, $$

the fourth relation is sphere condition of radius R: $$ x_1^2 + x_2^2 + x_3^2 = R^2 - 1 .$$

I guess the norms of these elements should be equal to $R$. Is it correct ? We have irrep of dim=R and can calculate everything there, may be it is helpful ?

MOTIVATION: I need to explain why the relations above correspond to quantization (deformation quantization) of the standard $S^2$ with the standard symplectic form (scaled by $R$).

High level words are the following: we are representing $S^2$ as a coadjoint orbit for Lie algebra $so(3)=\mathbb{R}^3$ (i.e. consider $ \sum x_i^2 = R^2$), quantization of corresponding linear bracket on $\mathbb{R}^3=so(3)$ is universal enveloping algebra $U(so(3))$, use "quantization commute with reduction" go back to quantization of $S^2$ from quantization of $\mathbb{R}^3$ as a factor algebra by the ideal generated by the central element $ \sum x_i^2 - R^2 + 1$. (Similar one can deal with flag manifolds and su(n),so(n) ).

In more simple words the following is done: we consider $S^2 \subset \mathbb{R}^3, \sum x_i^2 = R^2$. Instead of quantizing $S^2$ directly, we wanna to quantize $\mathbb{R}^3$ with very simple Poisson bracket $\{x_1 , x_2 \} = 2x_3 , ...$. Why ? Because the quantization of any linear Poisson bracket is very simple just substitute brackets by commutators: $ [ x_1 , x_2 ] = 2i h x_3,...$. (For simplicity I put $h=1$). What we must guarantee is that the standard symplectic structure on $S^2$ and the above Poisson structure on $\mathbb{R}^3$ are somehow related to each other. This is indeed true: the point is that $\sum x_i^2$ is Poisson CENTRAL ELEMENT (i.e. it Poisson commutes with any $x_i$), and hence the Poisson structure on $\mathbb{R}^3$ can be pushed down to the factor algebra $Fun(\mathbb{R}^3)/ (\sum x_i^2 = R^2) = Fun(S^2)$. So we get Poisson structure on $S^2$, actually it is the standard symplectic structure on $S^2$ - because it is rotation invariant and the only invariant structure is the standard one. The final step is go to quantization. We are in situation when we have a Poisson manifold $\mathbb{R}^3$ and submanifold $S^2$ defined by the Poisson central element constraint $\sum x_i^2 = R^2$, it is natural to believe that "quantization commute with reduction" and we can quantize $\mathbb{R}^3$ and then go back to quantization of $S^2$ just by factorization of quantized algebra by something like $\sum x_i^2 - R^2 +-... $ . Indeed that can be done. There are some subtle points here and use of Duflo map is necessary (that is why we get $R^2-1$, not just $R^2$ in quantum case), as was conjectured in our paper and proved by Cattaneo. So we arrive to the generators and relations in our question.


Remark 1: There are many "quantum spheres" (e.g. Podles q-sphere), but to the best of my understanding non of them are related to the question above. (See e.g. The Garden of Quantum Spheres, MO10581 ).

Remark 2: We have discussed this example with quantization of $S^2$ from very various points of view (geometric quantization, deformation quantization, reducations) in our paper and all them agree correctly with each other, so I am pretty sure that should work, but I am far from $C^*$-star algebra theory, that is why I am asking here.

Remark 3: As mentioned in MO230695 it is natural to believe that quantized algebra has unique irreducible representation of dimension calculated by the index like formula, under condition that symplectic form satisfies integrality condition. That indeed can be seen in the example above. Indeed: generators $x_i$ form $so(3)$ Lie algebra so irreps are numerated by natural numbers by their dimensions, additionally we have constraint $\sum x_i^2 = R^2-1$, hence we see that the only irrep which satisfies that constraint is $R$-dimensional irrep - i.e. it is indeed unique.

Remark 4: Why we consider quantum constraint $\sum x_i^2 = R^2 - 1$, but not naively expected $\sum x_i^2 = R^2$ ? It is subtle point. It is not related to existence of $C^*$-algebra, so we may just use $R^2$ in the question. Just this is related to subtle effect that with such normalization we indeed will get quantization of sphere of radius $R$. It is related to Sommerfeld's "+ 1/2 correction", half-forms and Duflo map, see our paper, (our conjecture has been proved by Cattaneo, so $R^2-1$, not $R^2$ is not a conjecture, but a theorem in deformation quantization framework).

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    $\begingroup$ When you ask "do these relations define C*-algebra", are you asking for existence or uniqueness? For existence, all you need is to find a triple of complex matrices satisfying these relations, and that is not specifically a C*-algebra question. For instance, the usual generators of $so(3)$ work with $R = 3$ I think. BTW the norm of the $x_i$ cannot equal $R$ because we have $x_i^2 \leq (R^2 - 1)I$ and hence $\|x_i\| \leq \sqrt{R^2 - 1}$. $\endgroup$ – Nik Weaver Feb 13 '16 at 21:36
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    $\begingroup$ Uniqueness is a separate issue. It would not be surprising. You would need to show that any C*-algebra generated by operators satisfying these relations is simple, which might be done using a group action on the C*-algebra. $\endgroup$ – Nik Weaver Feb 13 '16 at 21:37
  • $\begingroup$ @NikWeaver Thank you for your comments ! With the norm it is good catch... I somehow sometimes forget about this $R^2-1$, not $R^2$. But now it becomes more strange - for usual commutative S^2 norms of $x_i$ equal to $R$, now in quantum case we see that norms $<sqrt(R^2-1)$ - not a nice number ... $\endgroup$ – Alexander Chervov Feb 14 '16 at 9:12
  • $\begingroup$ @NikWeaver Concerning existence... Well taking usual generators of so(3) it is clearly NOT what I want. I want size of the algebra to be the same as in commutative case, not the finite-dimensional algebra. And morever finite-dim irrep exists for integer $R$... but it is natural to hope for existence of algebra for all $R$, while your argument does not work for non-integer $R$. It is analogous with quantum torus, the algebra exists for all $q$, but for $q^n=1$ we have finite-dim irrep (unique). $\endgroup$ – Alexander Chervov Feb 14 '16 at 9:17
  • $\begingroup$ @NikWeaver The analogy with q-torus is quite complete: $q^n=1$ and $R\in Z$ both corresponds to the same condition -- symplectic form represents integral cohomology class. $\endgroup$ – Alexander Chervov Feb 14 '16 at 9:19

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