I've listened to many interviews and lectures of Alain Connes, in which he says something which goes roughly as follows

"Every non-commutative algebra has its own time (evolution of), by which I mean a one-parameter group."

I find this statement somewhat mysterious and intriguing at the same time.

Question. What is the precise statement of this result and how to can this proven explicitly even for a simple scenario like, say, the von Neumann algebra $M_2(\mathbb C)$ of $2 \times 2$ complex matrices ?

Disclaimer. I have essentially no knowledge of non-comutative geometry, etc. I'd appreciate a very simple construction / proof; nothing too fancy. Thanks.

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    $\begingroup$ I have seen this quote before, with varying levels of precision from Connes, and it is best thought of as being in the tradition of Francophone epigrams, or "poetically true" rather than "literally true" $\endgroup$
    – Yemon Choi
    Mar 14 at 19:40
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    $\begingroup$ In particular, the given claim applied to $M_2({\bf C})$ just yields a time evolution that is trivial, i.e. staying the same for all time. What Connes is referring to is, AFAICT from various readings, the so-called "modular flow" of a von Neumann algebra. The technical details are outlined at en.wikipedia.org/wiki/… - I'm aware that this doesn't meet your request of a simple account but it might provide some of the right words to search for or ask about. $\endgroup$
    – Yemon Choi
    Mar 14 at 19:45
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    $\begingroup$ @DmitriPavlov I was aware that this can be done, although not of the details that you provide in your impressive answer; but I stand by my first comment in light of the fact that not every noncommutative algebra over C is a von Neumann algebra, as Connes is clearly well-aware. $\endgroup$
    – Yemon Choi
    Mar 15 at 0:34
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    $\begingroup$ @dohmatob All von Neumann algebras with a faithful normal tracial state will have trivial modular flow. In particular this is true for all finite-dimensional von Neumann algebras (= finite sums of finite matrix algebras). Hence why I find the slogan of Connes that you have quoted somewhat exasperating (and this is even before we get into non-selfadjoint operator algebras, or Banach star-algebras that are not Cstar algebras, etc). As Dmitri's answer indicates at the end, you need to go to the setting of Type III von Neumann algebras to get something nontrivial. $\endgroup$
    – Yemon Choi
    Mar 15 at 0:36
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    $\begingroup$ Sometimes I think that people whose only exposure to the NC world is through a vN background or by reading the sweeping statements from NCG, could do with reading and working through Howie's Fundamentals of Semigroup Theory, or reading the classical literature on nest algebras $\endgroup$
    – Yemon Choi
    Mar 19 at 15:06

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 $\L^p:={\rm L}^{1/p}$, where the right side is the usual notation from measure theory and real analysis. (This notation is explained in more detail in another answer.)

We have $\L^0(M)≅M$, $\L^1(M)≅M_*$, and $\L^{1/2}(M)$ is the standard form of $M$ due to Haagerup. The spaces $\L^p(M)$ for all $p∈\C$ form a $\C$-graded *-algebra, where the involution is a $\C$-antilinear map $\L^p(M)→\L^{\bar p}(M)$. The multiplication is well-defined because of Hölder's inequality, which in this context says that we have a map $$\L^p(M)⊗_\C\L^q(M)→\L^{p+q}(M).$$

In fact, we can do better: the induced map $$\L^p(M)⊗_{\L^0(M)}\L^q(M)→\L^{p+q}(M)$$ is an isomorphism for any $p,q∈\C$ such that $\Re p≥0$, $\Re q≥0$. Here the tensor product on the left side is purely algebraic, it happens to be automatically complete.

In particular, the bimodules $\L^p(M)$ are invertible for any $\def\I{{\bf I}} p∈\I=\{z∈\C\mid \Re z=0\}$, with the inverse being the bimodule $\L^{-p}(M)=\L^{\bar p}(M)$.

Thus, we have a morphism of 2-groups $$\def\VNA{{\sf VNA}} \L(M)\colon\I→\VNA^⨯_M,$$ where $\VNA$ denotes the bicategory of von Neumann algebras, W*-bimodules, and (bounded) intertwiners, $\VNA^⨯$ denotes its maximal 2-subgroupoid, i.e., we only take invertible bimodules (alias Morita equivalences) and invertible intertwiners, and $\VNA^⨯_M$ denotes the 2-subgroupoid of $\VNA^⨯$ on a single object, namely, $M$, whose endomorphisms form a monoidal groupoid, i.e., a 2-group.

The bicategory $\VNA^⨯$ is analogous to the bicategory of Lie groupoids, Morita equivalences (given by invertible bibundles), and invertible morphisms of bibundles. In particular, 1-morphisms in $\VNA^⨯$ can be seen as invertible maps of corresponding noncommutative measurable spaces in the same way as Morita equivalences of Lie groupoids can be seen as invertible maps of stacks corresponding to these Lie groupoids.

Thus, the canonical homomorphism of 2-groups $\L(M)\colon\I→\VNA^⨯_M$ equips every von Neumann algebra $M$ with a canonical action of the Lie group $\I=i{\bf R}$ of imaginary numbers. This is precisely the Tomita–Takesaki modular flow, expressed in a canonical manner without arbitrary choices.

If $M$ is a type III von Neumann algebra, this action is not isomorphic to the trivial action.

This can be most easily seen as follows. The trivial action sends all $t∈\I$ to the $M$-$M$-bimodule $M$, with the obvious trivial coherence data. If α is an isomorphism from the trivial action to the action $\L(M)$, then for each $t∈\I$ we have an isomorphism of $M$-$M$-bimodules $$α_t\colon M→\L^t(M).$$ Such an isomorphism is uniquely characterized by the element $α_t(1)∈\L^t(M)$, which must be a central element of support 1. Now the relations imposed by the definition of a homomorphism of 2-groups tell us that $$α_s(1)α_t(1)=α_{s+t}(1)∈\L^{s+t}(M)$$ and $α_0(1)=1∈\L^0(M)=M$. Thus, $$t↦α_t(M)∈\L^t(M)$$ is a one-parameter group of unitary elements of $\L(M)$. Such one-parameter groups are in bijection with faithful semifinite normal weights $μ$ on $M$: given $μ$, we set $$α_t(1)=μ^t∈\L^t(M).$$ The elements $α_t(1)$ are central in $\L^t(M)$ if and only if $μ$ is a trace. (If $μ$ is finite, then $μ(ab-ba)=0$ can be rewritten as $(μa-aμ)(b)=0$, i.e., $aμ=μa$ for all $a∈M$. That is to say, $α_1(1)=μ^1=μ∈\L^1(M)$ is a central element.)

Thus, the action $\L(M)$ is isomorphic to the trivial action if and only if $M$ admits a faithful semifinite normal trace, i.e., $M$ is a semifinite von Neumann algebra, equivalently, a direct integral of type I and type II factors.

Hence, if $M$ is a type III algebra, the action is nontrivial.

This approach also makes it clear the extent to which the traditional modular automorphism groups depend on the choice of a faithful semifinite normal weight. Given $t\in\I$ and a faithful semifinite normal weight $μ$, the traditional Tomita–Takesaki modular automorphism is $$σ_μ^t\colon M→M,\qquad x↦σ_μ^t(x)=μ^t x μ^{-t}.$$

We have a canonical homomorphism of 2-groups $$\def\Aut{\mathop{\rm Aut}} T\colon \Aut(M) → \VNA^⨯_M$$ that sends $σ∈\Aut(M)$ to the $M$-$M$-bimodule $T(σ)$. This is indeed a homomorphism because we have canonical isomorphisms $T(σ)⊗_M T(σ')≅T(σσ')$ that satisfy the relevant coherence conditions.

Now consider the $M$-$M$-bimodule $M_{μ,t}=T(σ_μ^t)$, where the right action is twisted by $σ_μ^t$: $a⋅m⋅b=amσ_μ^t(b)=amμ^t b μ^{-t}$. We have a canonical isomorphism of $M$-$M$-bimodules $$ρ_{μ,t}\colon M_{μ,t}→\L^t(M), \qquad m↦mμ^t.$$

Furthermore, the isomorphisms $ρ_{μ,t}$ are compatible with tensor products: $$ρ_{μ,s}⊗_M ρ_{μ,t}≅ρ_{μ,s+t}.$$

This amounts to saying that for any faithful semifinite normal weight $μ$, the composition of the homomorphism of groups $$σ_μ\colon \I → \Aut(M)$$ with the homomorphism of 2-groups $$T\colon\Aut(M) → \VNA^⨯_M$$ that twists the right action is canonically isomorphic (via the isomorphism $ρ_μ$) to the homomorphism of 2-groups $$\L(M)\colon \I → \VNA^⨯_M.$$

Observe that both $σ_μ$ and the isomorphism $$ρ_μ\colon T∘σ_μ → \L(M)$$ depend on the choice of a faithful semifinite normal weight $μ$, but the homomorphism $\L(M)$ does not. Thus, $\L(M)$ expresses the modular automorphism group in a coordinate-free way.

  • $\begingroup$ Awesome! Is there some book to read more about this stuff? Specifically I would like to know whether there are some other invertible bimodules and where to find them, if yes. Also, many thanks for lots of fantastic material on your home page. $\endgroup$ Mar 15 at 8:15
  • $\begingroup$ @მამუკაჯიბლაძე: Invertible bimodules (i.e., Morita equivalences) of von Neumann algebras have been studied a lot. Any such bimodule can be obtained as a composition of an amplification (an M-M⊗B(H)-bimodule M⊗H, for a Hilbert space H) and a reduction (an M-pMp-bimodule Mp, for a projection p∈M with central support 1). Morita equivalent von Neumann algebras have isomorphic centers, and Morita equivalences respect direct integral decompositions of von Neumann algebras, so it suffices to study Morita equivalences of factors. $\endgroup$ Mar 15 at 16:14
  • $\begingroup$ @მამუკაჯიბლაძე: All type I factors are Morita equivalent. If Morita equivalent type III factors act on a separable Hilbert space, they are isomorphic. Any type II_∞ factor is Morita equivalent to a type II_1 factor. Finally, for a type II_1 factor M any Morita equivalent type II_1 factor has the form End(P), where P is a finitely generated module over M, characterized by its dimension dim(P)∈(0,∞); we have P=M^n⊕pMp, where n is the integer part of dim(P) and p is a projection in M of the trace dim(P)−n. $\endgroup$ Mar 15 at 16:17
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    $\begingroup$ @მამუკაჯიბლაძე: The canonical reference is Marc Rieffel's paper Morita equivalence for C*-algebras and W*-algebras. $\endgroup$ Mar 15 at 16:19
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    $\begingroup$ @მამუკაჯიბლაძე: I would say that the Picard 2-group of a von Neumann algebra categorifies the unitary group of a Hilbert space. Unlike the unitary group, it is equipped with a canonical 1-parameter subgroup. I don't think there is much literature on Picard 2-groups of von Neumann algebras, since operator algebraists rarely talk about categories. However, André Henriques's papers probably say something about Picard 2-groups. André is actually here on MathOverflow, we could try to ask him. $\endgroup$ Mar 15 at 18:11

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