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.



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 $\L^p:={\rm L}^{1/p}$, where the right side is the usual notation from measure theory and real analysis.
(This notation is explained in more detail in [another answer](https://mathoverflow.net/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical/20820#20820).)


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](https://arxiv.org/abs/1309.7856):
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.


