86
votes
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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 ...
- 3,967
24
votes
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Mysterious quotes (at least for me)
Here is a guess about the remark of Orlov. Suppose that one wants to define a good notion of noncommutative scheme, given that an affine noncommutative scheme is an associative algebra. Trying to ...
- 5,559
21
votes
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If the direct image of f preserves coherent sheaves on noetherian schemes, how to show f is proper?
Well, as already explained in the comment by Vivek Shende, the answer is no: there are non-proper morphisms with push-forward that preserves coherence. I've thought a bit on this the last couple of ...
- 4,639
19
votes
Accepted
Why is "naive" definition of non-commutative spectrum bad?
I have also wondered about this question and recently came across some papers that seem to answer it.
First of all, the paper
Manuel L. Reyes, Obstructing extensions of the functor Spec to ...
- 5,559
19
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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 ...
- 37.2k
19
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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 ...
- 346
19
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
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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, ...
- 2,021
14
votes
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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 ...
- 37.2k
13
votes
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Realisation of the noncommutative torus as a universal $ C^{*} $-algebra
According to what I have seen in the literature so far, the standard procedure consists of two main steps:
Prove the existence of a universal $ C^{*} $-algebra $ A_{\theta} $ generated by two ...
- 1,366
13
votes
Mysterious quotes (at least for me)
Alain Connes: "a noncommutative algebra creates its own intrinsic time".
First of all, as Yemon Choi commented, this quote of Alain Connes is a slogan, not a theorem.
"Most of the NC algebras ...
- 22.9k
13
votes
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Relationship between Hochschild cohomology and Drinfeld centers
Classically, Hochschild cohomology is an invariant defined for associative algebras while the Drinfeld centre is an invariant defined for monoidal categories. The latter is a categorification of the ...
- 5,559
13
votes
Accepted
A survey for various $K$-homology theories and their relationship
I would say that there are really only two definitions of K-homology commonly used in the literature (apart from the naive definition via the Bott spectrum): "analytic K-homology" and "geometric K-...
- 27.2k
13
votes
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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 ...
13
votes
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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 ...
- 1,436
12
votes
Accepted
Non commutative topological manifolds
Theorem Let $A$ be a unital ring and $I_1,\dots,I_n \subset A$ be 2-sided commutative ideals such that $A=I_1+\dots + I_n$. Then, $A$ is commutative.
Proof: If $A=I_1+\dots+I_n$, then $1 = x_1+\dots+...
- 24.4k
12
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Relationship between Hochschild cohomology and Drinfeld centers
The key formula is, as you note, $End_{A\text{-mod-}A} (A)$. We can then interpret this same formula in lots of settings.
In the category of sets, $A$ is an ordinary algebra, and $End_{A\text{-mod-}...
- 26.5k
12
votes
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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 ...
- 2,021
12
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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 ...
- 5,834
12
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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}^{...
- 29.5k
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 ...
- 8,130
11
votes
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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 $B=M_n(A)$ ...
- 8,660
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 ...
- 191
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^...
- 147k
10
votes
If the direct image of f preserves coherent sheaves on noetherian schemes, how to show f is proper?
If $f:X\to Y$ is separated of finite type between noetherian schemes and $f_*$ preserves coherence, then $f$ is proper. Here is a proof that follows the geometric idea (given in the comments of Piotr ...
- 6,199
10
votes
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Commutative spectral triples
From the perspective of the Gelfand–Naimark theorem, the heart of the reconstruction theorem is the following statement, Theorem 11.4 in Connes's paper:
Let $\mathcal{A}$ be a commutative unital ...
- 2,145
10
votes
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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 ...
- 8,660
10
votes
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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 ...
- 29.7k
10
votes
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$K$-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 ...
- 23.9k
9
votes
Smooth Affine algebras are Calabi-Yau
If $X$ is smooth projective and $D = \cup D_i \subset X$ is an ample divisor so that $Y = X \setminus D$ is affine, then there is an exact sequence
$$
\oplus {\mathbb Z}D_i \to Pic X \to Pic Y \to 0.
$...
- 31.5k
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