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.

the correct one? Since every subalgebra $B$ of $A$ will again lead to some spectral triple. $\endgroup$ – truebaran May 15 '16 at 1:18