62

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 there is a wide variety of ways of thinking in mathematics). I would need convincing the other way around (e.g. How are C* algebras relevant in foliation theory ...

dg.differential-geometry oa.operator-algebras differential-topology noncommutative-geometry foliations

47

I'm can't claim to have studied the relevant history in a lot of detail, but count me a skeptic of Landsman's claim. Let's take this little paper and the companion that it cites as a test case, which I hope we can all agree is "real physics". The authors are clearly well versed in the calculus of variations and the representation theory of Lie groups. ...

fa.functional-analysis mp.mathematical-physics oa.operator-algebras ho.history-overview quantum-mechanics

36

Ozawa, Schechtman, and I finally wrote up what we know on this question. The estimate is that for every $\epsilon > 0$ there is a constant $C_\epsilon$ so that for every $n$, $\lambda(n)\le C_\epsilon n^{\epsilon}$. The paper can be downloaded from the arXiv.

26

(1) No. The Calkin algebra contains an uncountable family of mutually orthogonal nonzero projections.
(2) Yes. Embed $B(H)$ into $B(H \otimes H)$ by the map $A \mapsto A \otimes I$, then pass to $Q(H\otimes H) \cong Q(H)$.
(3) No. The Calkin algebra is simple.

26

This reminds me the following anecdote. K. Friedrichs once met Heisenberg on a conference. He thanked Heisenberg for creation of quantum mechanics which benefited mathematics so much, and added:
"But mathematicians gave much in return."
Heisenberg: "What?"
Friedrichs: "For example, von Neumann explained the difference between symmetric and self-adjoint ...

fa.functional-analysis mp.mathematical-physics oa.operator-algebras ho.history-overview quantum-mechanics

25

Yes. One can take $B$ to be the direct sum of all separable C*-algebras. A more interesting answer would be $B={\mathcal Q}(\ell_2)\otimes C^{\ast}(F_\infty)$. For an explanation, let me start with another question of you: It is an open problem
whether $B = {\mathcal Q}(\ell_2)$, the Calkin algebra, suffices or not.
It's written in my textbook with Nate ...

23

I'm a little puzzled by the tone of the original question. My personal view is that operator algebras are intrinsically interesting, and if there are good applications to other fields, so much the better ... I think this is a pretty common attitude, probably among people in most areas of pure math.
Anyway, I am not the most qualified to describe some of ...

22

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 define the spectrum of an associative algebra leads to various problems (c.f. this answer), so a different approach is needed. On the other hand there are ...

22

I'd like to try to give a more comprehensive answer.
In the elementary formulation of quantum mechanics, pure states are represented by unit vectors in a complex Hilbert space $H$ and observables are represented by unbounded self-adjoint operators on $H$. The expected value of a measurement of the observable $A$ in the state $v$ is $\langle Av,v\rangle$. We ...

20

The von Neumann algebra $M$ generated by $\mathcal O_\infty$ is all of $B(\mathcal F(H))$.
Indeed, if $a$ belongs to its commutant, let me prove that $a$ is a multiple of the identity. First since for all $v \in H$, $s_v^* (a \Omega)= a (s_v^* \Omega)=0$, we have that $a \Omega=\lambda \Omega$ for some $\lambda \in \mathbb C$. Then for every $\xi \in \...

answered Aug 30 '12 at 11:55

Mikael de la Salle

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20

Edit: as pointed out in the comments, the following answers the question for unital C*-algebras presented in terms of generators and relations. When I say C*-algebra, I really mean unital C*-algebra.
It may depend on what exactly you mean by "concrete", but I highly doubt that there is a general solution to this; finding a concrete realization of a ...

19

The answer is yes, but I don't know where it is written. If $p$ is not central, then $pAp^\perp\neq\{0\}$ and one can take $x\in pAp^\perp$ such that $0<\|x\|<1/2$. Then, $$q:=\left(\begin{array}{cc} \frac{1+\sqrt{1-4xx^*}}{2} & x\\ x^*& \frac{1-\sqrt{1-4x^*x}}{2}\end{array}\right)\quad{\rm in}\quad \left(\begin{array}{cc} pAp & pAp^\perp \\...

19

If $C^*(G)$ is isomorphic to $C^*_r(G)$, then $C^*_r(G)$ has a 1-dimensional representation, i.e. $G$ has a 1-dimensional representation $\chi$ weakly contained in the regular representation $\lambda_G$. Then $1_G=\chi\otimes\overline{\chi}$ is weakly contained in $\lambda_G\otimes\overline{\lambda_G}\simeq \infty.\lambda_G$, hence $G$ is amenable.

19

Answering the question in the body of the original post, which seems to be more restricted than the implicit question in the title of the post....
The answer is YES. See
L. Terrell Gardner, On isomorphisms of $C^\ast$-algebras.
Amer. J. Math. 87 (1965) 384–396.
MathReview
Roughly speaking, the proof works by considering the ${\rm C}^*$-...

18

I would recommend you to look at Jones' survey paper from 1986, entitled A New Knot Polynomial and Von Neumann Algebras. It is very readable. Let me try to make a brief summary, though.
The basic object you start with is a $II_1$ factor. This is a von Neumann algebra $M$ with trivial center $Z(M)\simeq \mathbb{C}$, possessing a faithful trace $\tau : M \...

18

Here's an argument showing that in the ${\rm II}_1$ case the flip automorphism is never inner.
Let $M$ be a type ${\rm II}_1$ factor and $\tau$ its trace, so that $M \subset L^2(M, \tau)$, and $M$ acts standardly on $L^2(M, \tau)$. Suppose that the flip automorphism is implemented by a unitary $U \in \mathcal U(M \overline \otimes M)$. I'll reach a ...

18

Since free C*-algebras don't exist, we can't give a concrete description of all relations that are allowed. Instead, we need to give axioms that determine what collections of n-tuples $(a_1,\dots,a_n)$ in $A$ are allowed, where $A$ varies over all C*-algebras. It is important that the elements not "know about" the ambient C*-algebra, so ``a is in a ...

18

A direct and precise way to arrive at your answer is to appeal to the theory of Fourier transforms of distributions. If I define the Fourier transform as
$$F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{ikx}\,dx,$$
then the equation $f(x+1)+f(x)=g(x)$ becomes upon Fourier transformation
$$e^{-ik}F(k)+F(k)=G(K)\Rightarrow F(k)=\frac{G(k)}{1+e^{-ik}}.$$...

answered Jun 24 '19 at 14:57

Carlo Beenakker

99.8k99 gold badges239239 silver badges349349 bronze badges

18

As jjcale mentions in a comment, the index of a Fredholm operator is very important in physics. One way to define the Chern number of a topological insulator is in terms of the index of a Fredholm operator, as explained in [1].
There is also the concept of an index of a pair of projections. This is seen a lot recently in physics papers, for example in [2]. ...

fa.functional-analysis mp.mathematical-physics oa.operator-algebras ho.history-overview quantum-mechanics

17

There are $2^c$ mutually non-equivalent irreducible representations. Since $\ell_\infty(N)$ has $2^c$ many pure states (there are $2^c$ many ultrafilters on $N$), any $C^*$-algebra containing $\ell_\infty(N)$ has at least as much pure states. Since $R$ has $c$ many unitary elements, there are $2^c$ many mutually non-equivalent pure states. This is a very old ...

17

Here's what Dan Voiculescu himself gave as motivation:
Around 1982, I realized that the right way to look at certain operator
algebra problems was by imitating some basic probability theory. More
precisely, in noncommutative probability theory a new kind of
independence can be defined by replacing tensor products with free
products and this can ...

answered Jun 27 '13 at 17:57

Carlo Beenakker

99.8k99 gold badges239239 silver badges349349 bronze badges

17

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 foliation of a smooth manifold has an associated von Neumann algebra and interesting properties of the von Neumann algebra are reflected in geometric ...

dg.differential-geometry oa.operator-algebras differential-topology noncommutative-geometry foliations

17

Yes. I will show that any two positive elements of $A$ commute. Since every element is a linear combination of positive elements, this suffices.
Say $a$ and $b$ are positive. Then $a^{1/2}ba^{1/2} \in A_{sa}$, so by hypothesis $ba^{1/2}a^{1/2} = ba \in A_{sa}$. That is, $ba = (ba)^* = a^*b^* = ab$. QED

16

Connes' critique was recently analyzed by Kanovei, Katz, and Mormann in this article in Foundations of Science (see also arXiv 1211.0244). Here is the abstract:
We examine some of Connes' criticisms of Robinson's infinitesimals starting in 1995. Connes sought to exploit the Solovay model S as ammunition against non-standard analysis, but the model tends ...

16

The relationship between operator algebras and braids is fairly straightforward to explain, and is nicely written up in many places (e.g. in Kauffman's Knots and Physics). Jones studied representations of the braid group $B_n$ into the Temperley-Lieb algebra $TL_n$. The existence of such a representation is not so surprising (the following explanation is ...

answered May 24 '12 at 7:52

Daniel Moskovich

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16

C*-algebras don't see the ISP. The operators $T\in B(H)$ and $T\oplus T\in B(H\oplus H)$
generate isomorphic C*-algebras, but the latter clearly has non-trivial invariant subspaces.
To have both operators in the same Hilbert space, pick isometries $v_1,v_2\in B(H)$ with orthogonal ranges that add up to $H$. Then
$$
T\mapsto v_1Tv_1^*+v_2Tv_2^*
$$
is an ...

16

The "canonical" example of a map that is $k$-positive but not $(k+1)$-positive is the map defined by
$$
\Phi_k(X) = k\cdot\mathrm{Tr}(X)I_n - X.
$$
Above, $n$ denotes the size of $X$ (i.e., $X \in M_n$) and $I_n$ is the $n \times n$ identity matrix. This map was introduced in "J. Tomiyama. On the geometry of positive maps in matrix algebras II. Linear ...

answered Jul 15 '15 at 16:57

Nathaniel Johnston

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15

I'm a bit wary of resurrecting such an old question, but given that the precise content of the reconstruction theorem doesn't seem to be terribly well disseminated, please permit me to cross-post from math.SE and then make some extra comments:
"To be absolutely clear about the state of the art, Connes's theorem actually tells you the following:
A unital ...

15

In Theorem 4.6 of their paper
http://www.univie.ac.at/nuhag-php/bibtex/open_files/deha85_CanniereHaagerup.pdf
de Canni`ere and Haagerup construct an explicit sequence of finitely supported functions on the free group $\mathbb{F}_N$ ($N\geq 2$), defining positive multipliers of the reduced C*-algebra $C^*_r(\mathbb{F}_N)$, and such that the corresponding ...

15

This paper contains a proposed solution to the problem. The acknowledgements suggest that MO might have facilitated the solution, should it be correct. (I'm speculating that Gil Kalai found out about Nik Weaver's formulation of the problem from this MO question.)

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