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 ...

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

6,26311 gold badge1919 silver badges3232 bronze badges

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 ...

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 ...

15

I suggest that you look into the paper Quantum Collections by Andre Kornell, available here on the arXiv. The abstract reads:
We develop the viewpoint that the opposite of the category of $W^\ast$-algebras and unital normal $\ast$-homomorphisms is analogous to the category of sets and functions. For each pair of $W^\ast$-algebras, we construct their free ...

14

What about $A = A_1 \oplus A_2$ and $M = (A_1\otimes 1) \oplus (A_2 \otimes B(H_2))$?
It seems like your condition just says that $A\otimes 1 \subseteq M \subseteq A \otimes B(H_2)$, did you leave something out?
Edit: in case $A$ is a factor, the answer is yes. Ge and Kadison, On tensor products of von Neumann algebras, Invent. Math. 123 (1996), 453-466.

13

(i) Obviously, a vN subalgebra $N$ of a B-complemented vN algebra $M$ is also B-complemented if there exists a conditional expectation from $M$ onto $N$. (ii) Haagerup and Pisier's proof actually says if $N$ is a B-complemented vN algebra which contains a (possibly non-unital) copy of $M_2(N)$ with a conditional expectation, then $N$ is CB-complemented and ...

13

The Weyl algebra construction can be done abstractly for any real vector space (even infinite-dimensional) endowed with an antisymmetric bilinear form, thanks to B. Blackadar's universal C*-algebra construction using generators and relations ("Shape theory for C∗-algebras", Math. Scand. 56 (1985) 249–275). However, since you asked for a concrete ...

answered Aug 16 '13 at 1:11

Pedro Lauridsen Ribeiro

4,56911 gold badge2020 silver badges5050 bronze badges

12

This is not an answer to the original question, but rather a summary and conclusion of the discussion that Dmitri and I had in the comments to his accepted answer. At the outset, I should make two caveats: (1) I am not a set theorist. (2) In addition to the conversations with Dmitri, my answer is heavily influenced by conversations with (indeed, based on ...

answered Mar 17 '12 at 5:17

Theo Johnson-Freyd

44.4k88 gold badges8989 silver badges272272 bronze badges

12

Pick distinct complex numbers $\lambda_1,\ldots,\lambda_k$ and consider the element
$$
X:=\Bigg(\,\underbrace{\begin{smallmatrix}
\lambda_1&1&0&0&0\\
0&\lambda_1&1&0&0\\
0&0&\ddots&\ddots&0\\
0&0&0&\lambda_1&1\\
0&0&0&0&\lambda_1\\
\end{smallmatrix}}_{n_1}\,\Bigg)
\oplus
\Bigg(\,\...

answered Nov 23 '13 at 20:42

André Henriques

37.3k55 gold badges9393 silver badges227227 bronze badges

12

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 create their own intrinsic time" would be a bit more correct, more precisely:
Theorem: a von Neumann algebra of type $\rm III$ creates its own intrinsic time up to ...

answered Nov 7 '14 at 23:11

Sebastien Palcoux

18.9k44 gold badges5151 silver badges128128 bronze badges

12

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 $\mathbb{1}_{\{1\}}$, $y=\inf x^n$ is equal to the projection onto the eigenspace of $x$ corresponding to the eigenvalue $1$, i.e. it is non-zero iff $1$ is an eigenvalue of $...

11

Here is a counterexample. I don't see any easy way to augment the question to something more natural.
Let $Q$ be a $w$-rigid II$_1$ factor with trivial fundamental group, e.g., $Q = L( \mathbb Z^2 \rtimes SL_2(\mathbb Z) )$. Let $\mathcal S \subset \mathbb R_+^*$ be a non-trival subgroup. Set $M = *_{s \in \mathcal S} Q^s$, and $N = Q * M$. (We may take $M$...

11

Thierry Coquand and Bas Spitters have several papers on this topic, see Integrals and Valuations, and Metric Boolean Algebras and Constructive Measure Theory.

answered Sep 10 '13 at 11:09

Carlo Beenakker

100k99 gold badges239239 silver badges350350 bronze badges

11

First of all, one should mention that not every triple (X,B,μ) (i.e., what is often called a measure space)
satisfies the property that its C*-algebra of bounded functions is a von Neumann algebra (= W*-algebra) or that the map L^∞→(L_1)* is an isomorphism.
One has to impose additional conditions to ensure that this is true.
If we do not assume such a ...

answered Jan 11 '14 at 11:38

Dmitri Pavlov

18.7k44 gold badges3939 silver badges100100 bronze badges

11

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 infinite dimensional, then it contains
an infinite family of non-zero pairwise orthogonal projections (see Proposition 1.3.16).

answered Mar 15 '16 at 14:21

Francesco Polizzi

52.9k33 gold badges141141 silver badges227227 bronze badges

10

This is not true. Popa showed in "Orthogonal pairs of ∗-subalgebras in finite von Neumann algebras" (1983), that if $F$ is a free group with arbitrary cardinality than any abelian von Neumann subalgebra of the group von Neumann algebra $LF$ must have separable predual.
Edit: This doesn't even hold when $M$ is abelian since $\ell^\infty(\mathbb R)$ has no ...

10

Yes, it has property $(\Gamma)$. This follows from Stalder's proof of inner amenability plus a fact that the semigroup $\langle T_m, T_n \rangle$ admits an approximately invariant subsets having proportional measures. Here, $T_m$ is the $m$-times map on $[0,1)$, $T_m x = mx \mod 1$. As Stalder proves, $\sum_{1\le i,j \le n} z^{m^i n^j}$ is approximately ...

10

Here is an adaptation of the standard proof that $G$ is amenable if $LG$ is injective. (I believe for instance that it is contained in the book of Brown and Ozawa).
Suppose $p \in LG$ is a non-zero central projection such that $p LG$ is injective. Thus, there exists a conditional expectation $E: \mathcal B(p \ell^2 G) \to p LG$. If we view $\ell^\infty G ...

10

The map is continuous, see the argument below. This also means that the group isomorphism is a homeomorphism.
Let $(u_n)_n$ be a sequence in $N$ that converges to $1$ in strong operator topology. Then we have to show that $\varphi(u_n^\ast\cdot u_n)$ converges to $\varphi$ for all $\varphi\in A_\ast$.
We can assume that $\varphi(x)=\langle\xi,x\eta\rangle$ ...

10

Given any two tracial von Neumann algebras $(N_1, \tau_1)$ and $(N_2, \tau_2)$ the $L^2$ space of the free product $(N_1 * N_2, \tau)$ canonically decomposes as
$$
L^2(N_1 * N_2, \tau) = \mathbb C \oplus_{n \in \mathbb N} \bigoplus_{i_1 \not= i_2, i_2 \not= i_3, \ldots, i_{n - 1} \not= i_n } \overline {\otimes}_{k = 1}^n L^2_0(N_{i_k}, \tau_{i_k}),
$$
where ...

10

Since a von Neumann algebra is injective if and only if its commutant is injective, this is the same as finding injective von Neumann subalgebras of $VN(\Gamma)' \cong VN(\Gamma)$. Maximal injective subalgebras of $VN(\Gamma)$ have been studied quite a bit. In particular, they always exist and for $\Gamma$ a free group, the von Neumann subalgebra generated ...

10

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 $\ell^\infty(\mathbb{N})$--by point evaluation at the rationals--that generates $\ell^\infty(\mathbb{N})$. It also sits inside $L^\infty[0,1].$ But $\ell^\...

10

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 $\left\{ f_{j, k}^{l, m} \right\}_{j, k, l, m \in \mathbb N} \subset L^2 (\mathbb R)$ such that each $f_{j, k}^{l, m}$ is valued in $\{ -1, 0, 1 \}$, has only finitely many discontinuity points, and ...

10

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 attributed to Blattner, and is stated explicitly in Proposition 1 of the following article:
R. J. Plymen. "Automorphic group representations: The hyperfinite $II_1$ ...

9

The standard reference for such matters is Guichardet's paper
Sur la catégorie des algèbres de von Neumann.
Bulletin des Sciences mathématiques 90 (1966), 41–64.
PDF file: http://math.berkeley.edu/~pavlov/scans/guichardet.pdf
I don't think separability (or the more general property of σ-finiteness) is important.
Requiring it is more of a tradition than a ...

answered Jan 23 '12 at 12:00

Dmitri Pavlov

18.7k44 gold badges3939 silver badges100100 bronze badges

9

Steve Vickers’ A monad of valuation locales presents a strong monad on the category of locales, a localic analogue of the Giry monad. It is commutative, i.e. product valuations exist and a Fubini Theorem holds. Concrete representations are given for the tensor product of lattices and for the modular monoid. Vickers combines domain theoretic measure theory ...

9

The result holds in a C*-algebra. The sketch of the proof I know, which was given to me as a series of exercises in a class on C*-algebras, is as follows. First one constructs isomorphisms from each C*-algebra on the following list to the next one (below all C*-algebras are unital, $\mathbb{Z}_2$ denotes the cyclic group of order $2$, and $M_2$ denotes the C*...

answered Dec 11 '13 at 2:23

Qiaochu Yuan

94.3k2929 gold badges336336 silver badges623623 bronze badges

9

I doubt there is a good general answer to the question. If $\pi$ is the GNS representation associated to a state $\omega$ and $\omega$ is invariant under $\phi$ then the automorphism of $\pi(A)$ is unitarily implemented since $$\langle \pi(\phi(x)),\pi(\phi(y))\rangle = \omega(\phi(y)^*\phi(x)) = \omega(\phi(y^*x)) = \omega(y^*x) = \langle \pi(x),\pi(y)\...

Only top voted, non community-wiki answers of a minimum length are eligible

#### Related Tags

von-neumann-algebras × 433oa.operator-algebras × 352

fa.functional-analysis × 102

c-star-algebras × 75

subfactors × 56

operator-theory × 34

reference-request × 32

gr.group-theory × 31

measure-theory × 17

noncommutative-geometry × 15

qa.quantum-algebra × 13

planar-algebras × 12

gn.general-topology × 9

ra.rings-and-algebras × 9

rt.representation-theory × 8

ergodic-theory × 7

linear-algebra × 5

mp.mathematical-physics × 5

banach-spaces × 5

noncommutative-algebra × 5

lattice-theory × 5

group-algebras × 5

operator-spaces × 5

harmonic-analysis × 4

group-cohomology × 4