Arveson index and curvature Can someone explain me what is the intuitive idea behind Arveson Index and curvature of $E_0$ semigroups. I was reading the standard paper of Arveson, but is lost and yet to get intuition about it. An index is generally invariant under certain operations. Waht are the actions for which this index (and curvature) are invariant? Advanced thanks for any help suggestion. 
 A: The intuition behind Arveson's index is that it's invariant under "small" perturbations of the generator. But this doesn't really make sense mathematically, so the formal definition is that it's invariant under cocycle conjugacy - details are given below.

Arveson's index is an invariant in the following sense:


*

*If $\alpha$ and $\beta$ are E$_0$-semigroups which are conjugate, i.e.
$$\beta_t(x)=U\alpha_t(U^\ast xU)U^\ast \qquad (t\geq0,~x\in B(H))$$
for some unitary $U$ then they have the same index.

*If there exists a strongly continuous family of unitaries $U_t$ with $U_s\alpha_s(U_t)=U_{s+t}$ (such a family is called a cocycle for $\alpha$) such that
$$\beta_t(x)=U_t\alpha_t(x)U_t^\ast \qquad (t\geq0,~ x\in B(H))  $$
then $\alpha$ and $\beta$ have the same index.
I claim that item 2. somehow corresponds to a "small" perturbation of the generator. Note that the generator of $\alpha$ will be an (unbounded) $\ast$-derivation $\delta$. Pick a self-adjoint operator $d\in B(H)$ and consider the inner perturbation $\delta'=\delta+i[d,\cdot]$ and let $\beta$ be the semigroup generated by $\delta'$. Then there exists a norm-continuous cocycle $(U_t)_{t\geq0}$ such that $\beta_t(x)=U_t\alpha_t(x)U_t^\ast$ for all $x\in B(H)$. It is determined by the differential equation $$ \frac{dU_t}{dt}=iU_t\alpha_t(d). $$ Thus you should think of item 2. as the index being invariant under small perturbations of the generator. I recomend you look at Arveson's book Noncommutative dynamics and E-semigroups for an exposition of all of this. There are also a bunch of really good expository articles kept alive on his old homepage http://math.berkeley.edu/~arveson/ . I recomend studying E$_0$-semigroups before, or at least along side, CP semigroups.
From the paper you're studying it is immediate that two unitarily equivalent CP semigroups will have the same generator, so the same index. A second question you might ask is "can I perturb the generator without changing the index?" and for that I suggest you look at his definition of rank. You should immediately spot exactly what perturbations you can make, i.e. in what sense the index is invariant.

For a more historical perspective of why that's a correct definition of the index consider semigroups of operators on a Hilbert space.
Pick a Fredholm operator $X$ and consider the semigroup $(X^n)_{n\in\mathbb{N}}$. Then the Fredholm index of $X$ is an invariant for the semigroup in the following two senses:


*

*Another semigroup $(Y_n)\_{n\in\mathbb{N}}$ which is unitarily equivalent to  $(X^n)_{n\in\mathbb{N}}$ will have the same index.

*If we perturb the ''generator'' $X$ by a compact operator then the new semigroup also has the same index 


For a C$_0$-semigroup $S=(S_r)_{r\in\mathbb{R}}$ the situation is more complicated, however for a semigroup of isometries the Stone generator is a maximal symmetric operator and its deficiency index (see http://en.wikipedia.org/wiki/Extensions_of_symmetric_operators ) is an invariant for the semigroup in the following sense:


*

*A semigroup of isometries $T=(T_r)_{r\geq0}$ unitarily equivalent to $S$ will have the same index.

*If we perturb the generator of $S$ by a bounded self-adjoint operator then the index will remain the same.


Thus the deficiency index could be thought of as a sort of continuous version of Fredholm index. Starting from here, Powers introduced a "noncommutative Fredholm index" in the paper
 An index for semigroups of $\ast$-endomorphisms on $B(H)$ and $II_1$ factors. The definition in the continuous case involved detailed analysis of the generator of the semigroup, but Arveson showed that for $B(H)$ it was equivalent to his definition of index, which has an algebraic flavour. Thus, by construction, Arveson's index is an invariant for an E$_0$-semigroup in pretty much the same way as Fredholm index and deficiency index are invariants for semigroups of operators on Hilbert space.
