22
votes
Accepted
Is the opposite category of commutative von Neumann algebras a topos?
The opposite category of commutative von Neumann algebras is not a topos
because categorical products with a fixed object do not always preserve small colimits.
See Theorem 6.4 in Andre Kornell's ...
- 36.3k
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}^{...
- 36.3k
12
votes
Separable von Neumann algebra
This is proven for instance here, Corollary 1.3.17 p. 26.
I am far from being an expert, however it seems to me that the main ingredient in the proof is the fact that if a von Neumann algebra is ...
- 65.1k
12
votes
In which sense the GNS-construction is a functor?
Although this response is a bit late, perhaps this perspective may help nonetheless. It only addresses the question about functoriality of the GNS construction.
The GNS construction is not quite a ...
- 121
12
votes
Accepted
When is $\inf_{n\geq0}x^n\neq0$?
Basically, Borel functional calculus translates pointwise convergence of functions into convergence in the strong operator topology. Since the functions $(x^n)$ converge to the function $\mathbb{1}_{\{...
- 4,950
12
votes
Accepted
Actions of locally compact groups on the hyperfinite $II_1$ factor
The answer to (1) (for second countable groups) is yes: every locally compact second countable group $G$ admits a countinuous, faithful outer action on the hyperfinite $II_1$ factor. This is ...
- 1,134
12
votes
Accepted
Can one associate a "nice" topos to a von Neumann algebra?
(I'm going to be a bit informal to be able to go to the point relatively directly, but if you want more details on some specific aspect. I can try to add them)
Toposes are closely related to ...
- 40.2k
12
votes
Simplicity of group $C^\ast$-algebra implies fullness of group-von Neumann algebra?
No, whenever $\Gamma$ is an infinite direct product of C$^*$-simple groups, we obtain a counterexample. For instance, taking $\Gamma = \mathbb{F}_2^{(\mathbb{N})}$ to be the direct sum of infinitely ...
- 4,011
11
votes
Accepted
Is a C*-algebra with an isomorphic predual a von Neumann algebra?
Via my colleague Garth Dales, some observations which answer your question in the negative, even in the abelian case:$\newcommand{\N}{{\mathbb N}}$
We know that $K$ is hyper-Stonean iff $C(K)$ is ...
- 25.5k
11
votes
Accepted
The double dual of the unitization of a $C^*$-algebra
Believe it or not, these are $*$-isomorphisms as C${}^*$-algebras. If $J$ is a closed two-sided ideal of $B$ then $J^{**}$ is a weak* closed two-sided ideal of $B^{**}$, and every weak*-closed two-...
- 42.1k
11
votes
Accepted
Tensor product of a von Neumann algebra and $L_\infty $
If $\mathbb F_I$ denotes the free group on $I$ generators with $\lvert I \rvert > 1$, then $L^\infty(0, 1) \overline \otimes L \mathbb F_I$ is not isomorphic to a von Neumann subalgebra of $L \...
- 4,594
11
votes
Accepted
Impact of annihilators in C*-algebras
An AW${}^*$-algebra is a C${}^*$-algebra which satisfies this condition for both right and left annihilators. So every AW${}^*$-algebra has your property, and any C${}^*$ algebra that is isomorphic to ...
- 42.1k
11
votes
Accepted
Positive cone in Haagerup L²-space: how much information does it contain?
For countably decomposable von Neumann algebras, the pair $(L^2(M),L^2_+(M))$ characterizes $M$ up to Jordan $\ast$-isomorphism by the result of Section 3 from [1]. Moreover, a Jordan $\ast$-...
- 1,027
10
votes
Accepted
Von Neumann Algebra isomorphism extension
No. This will almost never be true (subalgebras of the compacts are the only cases I can think of where it could work). The easiest example is probably $C[0,1].$ It has an injective homomorphism to $...
- 2,679
10
votes
Accepted
A small corner w.r.t. a masa in a von Neumann algebra
This is not true. Here's an example for which it fails:
Let $\{ r_n \}_{n \in \mathbb N}$ be an enumeration of the rationals. Construct an orthogonal set $\left\{ f_{j, k}^{l, m} \right\}_{j, k, l, m ...
- 4,594
10
votes
What does it mean for a category to admit direct integrals?
This is a question that I have been working on recently together with Robert Furber and Bas Westerbaan. Let me sketch what we know so far, starting with the case of (infinite) direct sums, treated in ...
- 5,795
10
votes
Accepted
When a normal functional is restricted to a vn Neumann sub-algebra
No, such a property does not hold. For instance, you could take $H = \mathbb{C}^2$ and $M \cong \mathbb{C} \oplus \mathbb{C}$ the subalgebra of diagonal matrices in $B(H)$. Denoting by $E : M_2(\...
- 4,011
10
votes
Accepted
Normal linear functionals on bicommutants of C*-algebras
Doesn't this imply that all normal functionals on $\pi_u(A)''$ are linear combinations of vector states, hence (SOT) continuous?
It sounds like you understand the proof, but are suspicious of this ...
- 42.1k
10
votes
Sigma-weakly dense *-subalgebra of von Neumann algebra has increasing net of positive elements convergent to the identity
In his comment, Narutaka Ozawa already gave a positive answer in the abelian case. The following argument gives a positive answer in the separable case, or more generally when $M$ is countably ...
- 4,011
9
votes
Normal $*$-homomorphism
No, this already fails in the abelian case. Take $M = L^\infty[0,1]$ with the trace coming from integration against Lebesgue measure. Let $\pi$ take $f(x)$ to $f(x^2)$, i.e., composition with the ...
- 42.1k
9
votes
Accepted
Is $L(\mathbb{Z}*\mathbb{Z}_{2})$ a free group factor?
This should just be a comment- but for some reason I couldn't add a comment.
It seems to me that using Corollary 5.3 of this paper by Dykema, we indeed get a positive answer to your question.
...
- 456
9
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\}'=...
- 9,631
9
votes
Accepted
When a $C^*$-algebra is an ideal in its second dual?
Warning: the following is just what I found from some work on MathSciNet, following Yemon's hint in the comments. It's not meant to be accurate historical notes.
A "dual" $C^\ast$-algebra ...
- 18.5k
9
votes
Accepted
Polar decomposition in abstract von Neumann algebra
I would say that the polar decomposition of $m \in M$ is the unique pair $(v,a)$ of elements in $M$ satisfying the following (algebraic) properties.
$m = va$.
$v$ is a partial isometry and $a$ is ...
- 4,011
9
votes
Accepted
Defining the abstract tensor product of W*-algebras via a universal property
You can see this is false by taking $N = \mathbb{C}$. Then, given a von Neumann algebra $M$, you are asking for a von Neumann algebra $\widetilde{M}$ and a weak* dense embedding $\iota: M \to \...
- 42.1k
9
votes
Accepted
von Neumann algebra of canonical commutation relations
The C$^*$-algebra generated by the exponentials $e^{isQ}$ and $e^{itP}$, $s,t \in \mathbb{R}$, is the CCR algebra. The von Neumann algebra they generate in this representation is all of $B(L^2(\mathbb{...
- 42.1k
9
votes
Accepted
von Neumann subalgebra having separable predual
Not necessarily. In fact, even a single operator $x$ can generate a nonseparable vNa. For example, in the sequence space $l^2([0, 1])$, the diagonal operator given by $x(e_t) = te_t$ for all $t \in [0,...
- 1,262
8
votes
Accepted
How the modular theory of von Neumann algebras, deal with generating C*-algebras?
All groups algebra are trivial examples because for them the modular time evolution is (can be chosen) trivial: the modular time evolution attached to a state is trivial if and only if the state is a ...
- 40.2k
8
votes
Accepted
In which sense the GNS-construction is a functor?
If you want $\tilde{\varphi}$ to be normal then this is false. But first let me point out that there is a sense in which the GNS construction is a functor. Note that $\varphi$ induces an isometric ...
- 42.1k
8
votes
Accepted
Why are ultraweak *-homomorphisms the `right' morphisms for von Neumann algebras (and say, not ultrastrong)?
A $*$-homomorphism between two von Neumann algebras is weak* to weak* continuous if and only if it is ultrastrong to ultrastrong continuous. See Proposition III.2.2.2 of Blackadar's book (which, ...
- 42.1k
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