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A spectral triple consist from $(A,H,D)$ where $A$ is some unital $*$-subalgebra of $B(H)$ and $D$ is unbounded operator with compact resolvent such that for all $a\in A$ the commutator $[D,a]$ is bounded. The motivating example is $C^{\infty}(M)$ where $M$ is smooth compact manifold. However the definition is such that it seems to me that every $*$-subalgebra $B \subset A$ is also good (i.e. $(B,H,D)$ will be again a spectral triple).

What is the relevance of such construction of taking the subalgebra of $A$?

Maybe my question is not rigorous however I do not expect the precise answer, any comments, motivation, background and examples are welcome.

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    $\begingroup$ The closure of $A$ in the operator norm on $B(H)$ will be a $C^\ast$-algebra, which you can think of as the $C^\ast$-algebra of continuous scalar-valued functions on the underlying noncommutative topological space of your spectral triple, whilst the closure of $A$ in the norm $$\|a\|_D := \|a\|_{B(H)} + \|[D,a]\|_{B(H)}$$ will be a Banach algebra, dense and closed under the holomorphic functional calculus in the operator norm closure of $A$ in any reasonable example, which you can think of as the Banach algebra of all continuously differentiable scalar-valued functions on … $\endgroup$ – Branimir Ćaćić May 15 '16 at 0:49
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    $\begingroup$ …the noncommutative manifold (morally) defined by your spectral triple. In the case where $(A,H,D) = (C^\infty(M),L^2(M,E),D)$ is an honest-to-goodness commutative spectral triple, then this is literally true. Speaking of which, if $(C^\infty(M),L^2(M,E),D)$ is a commutative spectral triple, and $B$ is just some unital $\star$-subalgebra of $C^\infty(M)$, then $(B,L^2(M,E),D)$ is still a spectral triple, but will almost never be a commutative spectral triple. $\endgroup$ – Branimir Ćaćić May 15 '16 at 0:52
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    $\begingroup$ Thank you, so at least in the commutative example there is no point in taking some (proper) subalgebra of the initial algebra $A$. However in the noncommutative case (I mean, the situation where $A$ is noncommutative) how do we know that the underlying algebra $A$ for which we succeed to define the spectral triple $(A,H,D)$ is the correct one? Since every subalgebra $B$ of $A$ will again lead to some spectral triple. $\endgroup$ – truebaran May 15 '16 at 1:18
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    $\begingroup$ The choice will be fairly clear in examples. In the case of noncommutative tori, there's an obvious algebra of noncommutative Fourier polynomials (which you can enlarge, if you prefer, into an algebra of rapidly decaying noncommutative Fourier series, the usual smooth algebra for noncommutative tori), whilst in the case of quantum groups, there's usually some sort of algebra of "matrix coefficients" that's at the heart of all algebraic computations. A basic sanity check, though, is what you want as your $C^\ast$-completion, and hence as your underlying noncommutative topological space. $\endgroup$ – Branimir Ćaćić May 15 '16 at 1:21

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