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Apr
4
comment Most natural equivalence between $C^*$-algebras in NCG
One more crucial fact: the crossed product $C_0(X) \rtimes_r G$ of the $C^\ast$-algebra of a locally compact Hausdorff space by a free and proper action of a locally compact topological group is, in general, only Morita equivalent to the $C^\ast$-algebra of the quotient space $C_0(X/G)$. Indeed, $C_0(X) \rtimes_r G$ is necessarily noncommutative the moment $G$ is non-Abelian.
Apr
2
answered Local form of Dirac operator
Mar
20
comment How to compute $\sum_{x \in \mathbb{Z}^n} e^{-x^TMx}$ efficiently
In your case, $S_M = \Theta\left(0 \vert \tfrac{1}{\pi} i M\right)$, where $\tfrac{1}{\pi} i M$ is indeed complex symmetric with strictly positive definite imaginary part.
Jan
29
comment An example of an Azumaya algebra that isn't free over its centre
In the context of operator algebras, the situation that immediately comes to mind is the algebra $\Gamma(X,\operatorname{End}(E))$ of (continuous) global sections of the endormorphism bundle $\operatorname{End}(E)$ of a vector bundle $E$ on a compact space $X$, in the case that $\operatorname{End}(E)$ is itself a non-trivial vector bundle. This isn't quite what you're looking for, but I suspect it's not too far off, at least morally.
Sep
6
comment Formula for the distance in noncommutative geometry
On the other hand, by definition, since $D$ is a Dirac-type operator, $D^2$ is of Laplace-type, so that $D^2 = -g^{ij}\partial_i\partial_j + \text{lower order terms}$ locally, and hence you can read off $\sigma(D^2)(\omega) = g(\omega,\omega)$ for any $1$-form $\omega$. As a result, $c := \sigma(D)$ really does satisfy the Clifford relation.
Sep
6
comment Formula for the distance in noncommutative geometry
Observe that the principal symbol $\sigma(D)$ of the first-order differential operator $D$ can be recovered as $\sigma(D)(df) := i[D,f]$ for $f \in C^\infty(M)$. On the one hand, by the basic theory of (pseudo)differential operators, the principal symbol $\sigma(D^2)$ of $D^2$ satisfies $\sigma(D^2)(\omega) = \sigma(D)(\omega)^2$ for any $1$-form $\omega$. [...]
Aug
26
comment An extension of $K$-theory to topological $^*$-algebras
You can forget about involutions entirely, since the $K$-theory of a $C^\ast$-algebra as a $C^\ast$-algebra (i.e., in terms of projections and unitaries) is identical to its $K$-theory as a Banach algebra (i.e., in terms of idempotents and invertible matrices).
Aug
15
comment Formula for the distance in noncommutative geometry
This certainly takes care of the problem at the base point $x$, but you're still left, in general, with non-differentiability of the distance function on the cut locus of $x$, no?
Aug
15
comment Formula for the distance in noncommutative geometry
@NikWeaver Thank you for the correction. I had a silly idée fixe about approximating in the Lipschitz seminorm, so I just didn't see that it was enough to bound it from above.
Aug
15
revised Formula for the distance in noncommutative geometry
Corrected major error in answer to Question 1, thanks to Nik Weaver's comment.
Aug
15
awarded  Yearling
Aug
14
awarded  Revival
Aug
14
revised Formula for the distance in noncommutative geometry
deleted 472 characters in body
Aug
14
comment Realisation of the noncommutative torus as a universal $ C^{*} $-algebra
+1 I'm very late to the party, but this is an excellent answer.
Aug
14
answered Formula for the distance in noncommutative geometry
May
16
answered Name for construction on two vector bundles
Apr
26
comment A survey for various $K$-homology theories and their relationship
A possible first step towards this might be found in Hilsum's notion of bordism of unbounded KK-cycles---nothing to do with cobordism of bounded cycles---which has recently been developed further by Deeley--Goffeng--Mesland: arxiv.org/abs/1503.07398
Mar
22
awarded  Announcer
Feb
27
awarded  Announcer
Jan
4
comment Correspondences as generalized morphism between $C^*$-algebras
See, for instance, mathoverflow.net/questions/82871/… and similar discussions on MathOverflow and Math.SE. I suppose the point is that naive $\ast$-homomorphisms $A \to B$ aren't necessarily all that natural in the nonunital case, when seen through the guiding lens of Gelfand–Naimark.