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7 votes
2 answers
1k views

Is there a notion of point in noncommutative geometry?

It is not clear to me whether there is a general notion of point in NCG. I have heard (more through physics) that the notion of a point becomes meaningless or ill-defined in noncommutative spaces, but ...
Esmond's user avatar
  • 136
4 votes
1 answer
270 views

Kaplansky inverse element theorem on group C-star algebra

In a class talking about $C^*$ algebra and (higher) index theory, I heard a theorem (related to Kaplansky, proved?), that is Suppose $\Gamma$ is a group (admitting Haar measure if necessary) while $\...
YOTAL's user avatar
  • 193
1 vote
0 answers
86 views

Inner product on Standard form of von Neumann algebra

Let $(M, H, H_+,J)$ be a standard form of a von Neumann algebra $M$ acting on a complex Hilbert space $H$ endowed with a self-dual cone $H_+$. Is it true that $$\langle x,yz\rangle=\langle zx,y\rangle$...
Guest's user avatar
  • 131
1 vote
2 answers
126 views

Upper triangular $2\times2$-matrices over a Baer *-ring

Let $A$ be a Baer $*$-ring. Let us denote $B$ by the space of all upper triangular matrices $\left(\begin{array}{cc} a_1& a_2 \\ 0 & a_4 \end{array}\right)$ where $a_i$'s are in $A$. Is $B$ ...
ABB's user avatar
  • 4,058
3 votes
1 answer
293 views

Is the square diagram of index and exponential maps in $K$-theory of $C^*$-algebras anti-commutative?

Assume we have a $3\times 3$ grid with rows and columns being short exact sequences of $C^*$-algebras. This gives a grid of 6-term exact sequences: 3 "horizontal" sequences and 3 "vertical" sequences,...
Fiktor's user avatar
  • 1,284
2 votes
1 answer
464 views

Computing noncommutative geometries

I find myself needing to construct some noncommutative geometries. I want to take various (algeba-) geometric objects and look at their noncommutative analogs. Is there a constructive way to do this? ...
user46348's user avatar
  • 161
14 votes
2 answers
549 views

$\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $\ell^1(G)$ is a flat (right) $\mathbb{Z}G$-module?

Let $G$ be a countable discrete group (not necessarily abelian), and suppose the group ring $\mathbb{Z}G$ is a left-Noetherian ring, for example, when $G$ is a polycyclic-by-finite group. Denote the ...
Jiang's user avatar
  • 1,528
21 votes
0 answers
869 views

Noncommutative arithmetic mean geometric mean inequality and symmetric polynomials

While analyzing convergence speed of stochastic-gradient methods for convex optimization problems, Recht et al (2011) posed a tantalizing conjecture. It seems quite tricky, so after having struggled a ...
Suvrit's user avatar
  • 28.6k
43 votes
8 answers
3k views

How to quantify noncommutativity?

If I have two operators or finite-dimensional matrices $A$ and $B$, how can I quantify the amount to which they commute or don't commute? (I would consider it a big plus if it is computable easily for ...
Jiahao Chen's user avatar
  • 1,890
40 votes
9 answers
10k views

Simplest examples of rings that are not isomorphic to their opposites

What are the simplest examples of rings that are not isomorphic to their opposite rings? Is there a science to constructing them? The only simple example known to me: In Jacobson's Basic Algebra (...
Amritanshu Prasad's user avatar
4 votes
2 answers
286 views

unitary reduction of $q$-normal matrices

The unitary reduction of normal matrices is a well-known fact: if $A\in M_n(\mathbb C)$ commutes with its Hermitian adjoint $A^*$, then there exists a unitary $U\in\mathbb U_n$ and a diagonal matrix $...
Denis Serre's user avatar
  • 52.3k
6 votes
1 answer
319 views

Does there exist any massive proper $C^*$-subalgebra?

Definition 1: Suppose $B$ is a $C^* $-algebra. $A$ is massive $C^* $-subalgebra of $B$ iff 1. $A$ is a subalgebra of $B$; 2. for each irreducible representation $\pi$ of $B$ representation $\pi|_A$ is ...
Fiktor's user avatar
  • 1,284
3 votes
2 answers
1k views

Spectral decomposition for an arbitrary linear combination of position and momentum operators

Suppose we have the Hilbert space L2(Rn) and we have n operators Qi and n operators Pi defined in the usual way by: Qi ψ(q1,q2,...,qn) = qi ψ(q1,q2,...,qn) Pi ψ(q1,q2,...,qn) = -i $\frac{...
StevenJ's user avatar
  • 195
14 votes
9 answers
2k views

Examples of noncommutative analogs outside operator algebras?

Theo's question made me wonder if there are other "noncommutative analogs" outside of operator algebras. Some noncommutative analogs from operator algebras include: A $C^\ast$-algebra is a ...
Dave Penneys's user avatar
  • 5,425
6 votes
3 answers
324 views

Inverses in convolution algebras

Let $G$ be a locally compact totally disconnected group, and to make life easy let's suppose its Haar measure is bi-invariant. Let $C_c(G)$ be the space of locally constant complex functions on $G$ ...
D. Savitt's user avatar
  • 2,713
14 votes
2 answers
899 views

Do torsion-free groups give projectionless group ($C^\ast$) algebras?

One of the reasons I study von Neumann algebras is that they always have plenty of projections. There are many projectionless $C^\ast$-algebras ($0$ and possibly $1$ are the only projections), but the ...
Dave Penneys's user avatar
  • 5,425
9 votes
1 answer
509 views

Maximal localizations of von Neumann algebras

Suppose M is a von Neumann algebra. Denote by L its maximal noncommutative localization, i.e., the Ore localization with respect to the set of all left and right regular elements, i.e., elements whose ...
Dmitri Pavlov's user avatar
9 votes
7 answers
2k views

Hochschild/cyclic homology of von Neumann algebras: useless?

Hochschild homology gives invariants of (unital) $k$-algebras for $k$ a unital, commutative ring. If we let our algebra $A$ be the group ring $k[G]$ for $G$ a finite group, we get group homology. ...
Dave Penneys's user avatar
  • 5,425
13 votes
2 answers
723 views

Ideals in Factors

One can easily prove that factors have no nontrivial ultraweakly closed 2-sided ideals as these are equivalent to nontrivial central projections. One can also show type $I_n$, type $II_1$, and type $...
Dave Penneys's user avatar
  • 5,425