90
votes

Accepted

### What is a foliation and why should I care?

Without any disrespect, let me say that I find it incredible that someone naturally cares about non-commutative geometry but needs convincing about actual geometry (this just goes to highlight that ...

21
votes

### What is a foliation and why should I care?

Here is another reason about why people care about foliations: If you care about dynamical systems, you should care about foliations.
For instance, if you have a nowhere singular vector field on a ...

20
votes

### What is a foliation and why should I care?

Probably there are many reasons why people care about foliations, but for someone coming from operator algebras one of the main reasons is the connection to von Neumann algebra theory. In brief, every ...

20
votes

### Quantum corrections to geometry

Physicist chiming in; "quantum corrections of a geometry" represents the vague idea that gravity should be quantum.
Classically, gravity is a geometric theory: "reality" is modelled as a (Lorentzian) ...

15
votes

Accepted

### What is the significance of the Jiang Su algebra in classification of C$^*$ -algebras?

Diego's answer in the comments above is related to why we would expect any classifiable $C^\ast$-algebra to satisfy $A\cong A\otimes \mathcal Z$:
Since $\mathcal Z$ is separable, nuclear, unital, ...

14
votes

Accepted

### Vanishing of Hochschild homology of a category

This precise question was phrased as the vanishing conjecture in Hochschild homology and semiorthogonal decompositions. But we now know that there exist so called (quasi)phantom categories, which give ...

14
votes

Accepted

### Why does Riesz's Representation Theorem apply in quantum mechanics?

Okay, there is a lot of confusion in this question.
First, I'm not sure why you say ``it is common to begin the discussion with embedding $A$'' into $B(H)$. The point of the C${}^*$-algebra approach ...

13
votes

Accepted

### Reference for Connes Bourbaki membership or otherwise

Connes was a member, according to M. Mashaal "Bourbaki: A secret society of mathematicians", AMS 2006 (translated from the French by A. Pierrehumbert). It says so on page 18; see the link. But, as ...

Community wiki

13
votes

Accepted

### Making sense of "every non-commutative algebra has its own internal time evolution (aka a one-parameter group)"?

Given any von Neumann algebra $M$, we can define its noncommutative $\def\L{{\cal L}} \L^p$-spaces $\L^p(M)$ for any $\def\C{{\bf C}} p∈\C$ such that $\Re p≥0$.
Here I use the notation $\L^p:={\rm L}^{...

12
votes

Accepted

### A non nuclear $C^*$ algebra $A$ for which the algebraic tensor product $A\otimes A$ admits a unique $C^*$ norm

Pisier https://arxiv.org/abs/1908.02705 very recently constructed a non-nuclear $C^\ast$-algebra $A$ with the weak expectation property (WEP) and the local lifting property (LLP). By a celebrated ...

12
votes

### Why does Riesz's Representation Theorem apply in quantum mechanics?

$\DeclareMathOperator\Ann{Ann}\DeclareMathOperator\Tr{Tr}$My answer is somewhat complementary to Nik Weaver's, and admitedly more focused on Question 2 since I have nothing more to add to the latter ...

12
votes

### Reference for the Swan-Serre theorem as a monoidal equivalence

Recall that the equivalence in question is given by taking global sections of a vector bundle, $V\mapsto \Gamma(X,V)$.
Now if you have a section $s$ of $V$ and a section $t$ of $W$, this gives you a ...

11
votes

Accepted

### Most natural equivalence between $C^*$-algebras in NCG

Here I list some facts that may be useful for building your intuition:
1. Two commutative Morita equivalent $C^*$-algebra are in fact $*$-isomorphic.
2 If $A$ is $C^*$-algebra and you take $B=M_n(A)$ ...

11
votes

Accepted

### Is a "smooth" finite-dimensional algebra separable modulo its radical?

Let $K$ be an algebraic closure of $k$.
The following lemma must surely be well-known, but I haven't found an explicit reference, so I'll include a proof at the end of this post.
Lemma. If $S$ is ...

11
votes

Accepted

### $*$-algebras, completions, and $K$-theory

Any infinite discrete group $\Gamma$ with Kazhdan's property (T) gives an example. Since it is not amenable, the full and reduced C*-algebras (which are both completions of the group algebra) do not ...

11
votes

### Quantum corrections to geometry

I presume this refers to quantum corrections to the metric tensor. These are expected to be important in general relativity when the curvature of space-time approaches the Planck scale $\sqrt{G\hbar/c^...

10
votes

Accepted

### Can $C^*$-algebra of continuous functions on $R^n$ ($S^n$) be characterized alternatively?

Yes, it can be defined as the univeral commutative $C^*$-algebra with unit, generated by $n+1$ self adjoint elements $x_1,...,x_{n+1}$ subject to the relation $x_1^2+...+x_{n+1}^2=1$. Here universal ...

10
votes

Accepted

### $K$-theory and surjective norm-decreasing $*$-homomorphisms between $C^*$-algebras

Yes to both.$\newcommand{\Cst}{{\rm C}^*}$
The standard example for the first is: take a discrete group $G$ and let $A$ be its full $\Cst$-algebra, $B$ its reduced $\Cst$-algebra. There is a ...

9
votes

### $C^{*}$-correspondences viewed as generalized endomorphisms

Expanding my comment as per suggestion above. All this is explained in more detail in the reference I gave though.
A correspondence from $A$ to $B$ is a Hilbert $B$-module on which $A$ acts non-...

9
votes

Accepted

### K theory long exact sequence

Regarding the first question:
If $X$ is quasi-compact quasi-separated and $U \to X$ is a quasi-compact open immersion, then Thomason-Trobaugh showed that there is a "proto-localization sequence", i.e....

9
votes

### What is a foliation and why should I care?

Here's why I care about foliations. It is always interesting when a structure can be expressed in terms of simpler structures. For instance a torus is the union of circles making it into a cartesian ...

9
votes

Accepted

### Isomorphism in cyclic cohomology vs isomorphism in Hochschild cohomology

This was something that used to puzzle me when I was first learning this stuff. Refreshing my memory just now, I think that the trick is to use the fact that the Connes–Tsygan sequence has some ...

9
votes

### Baum Connes conjecture and Atiyah-Singer index theorem

I find it overwhelmingly difficult to do this question proper justice. So instead I give a highly condensed answer and refer to Connes NCG, Section II.10 for more details (as well as all the ...

9
votes

### preliminary reading recommendation before embarking on Connes non commutative geometry book?

To understand everything in Connes' book you would need expertise in many different fields. My advice would be to browse it and see if anything attracts your interest. Then you can read up on the ...

9
votes

### On differential equation $Z'=Z^2-Z$ on a $C^*$ algebra

The solution $Z(t)$ of your differential equation with $Z(0) = Z_0$ satisfies
$$ Z(t) (e^t + (1-e^t) Z_0) = Z_0 $$
In order for this to be periodic with period $p$, you'd need
$(1-e^p) Z_0 (1-Z_0) = ...

9
votes

Accepted

### tangent bundle on noncommutative manifold

Noncommutative Riemannian (spin) geometry via spectral triples is grounded in an approach to noncommutative differential calculus that privileges the cotangent bundle over the tangent bundle: given a ...

8
votes

### Can $C^*$-algebra of continuous functions on $R^n$ ($S^n$) be characterized alternatively?

Podles defined quantum 2-spheres in such a universal way. In a special case they restrict to $C(S^2)$. The description in this case is: the universal unital C*-algebra generated by operators $A$ and $...

8
votes

Accepted

### Path algebras are formally smooth

Assuming standard results on lifting idempotents, it's not hard to check that a path algebra $kQ$ satisfies the lifting property that Ginzburg uses to define formal smoothness in Definition 19.1.1.
...

8
votes

### $K$-theory and surjective norm-decreasing $*$-homomorphisms between $C^*$-algebras

There are even commutative counterexamples. Let $A = C[0,2]$ and let $A_0$ be the $*$-subalgebra of all polynomials in $x$. Then let $p: C[0,2] \to C[0,1]$ be the restriction map.
(My first example ...

8
votes

Accepted

### Noncommutative torus as a von Neumann algebra

No. It's irreducible. The element $U$ generates the maximal abelian subalgebra $L^\infty({\mathbb T})$ and hence one computes the commutant:
$$\{U,V\}'=\{U\}'\cap\{V\}'=L^\infty({\mathbb T})\cap\{V\}'=...

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