# KMS, Gibbs & zeta function

Time to wrap up this series on the Bost-Connes algebra. Here's what we have learned so far : the convolution product on double cosets

$\begin{bmatrix} 1 & \mathbb{Z} \\ 0 & 1 \end{bmatrix} \backslash \begin{bmatrix} 1 & \mathbb{Q} \\ 0 & \mathbb{Q}_{> 0} \end{bmatrix} / \begin{bmatrix} 1 & \mathbb{Z} \\ 0 & 1 \end{bmatrix}$

is a noncommutative algebra, the Bost-Connes Hecke algebra $\mathcal{H}$, which is a bi-chrystalline graded algebra (somewhat weaker than 'strongly graded') with part of degree one the group-algebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$. Further, $\mathcal{H}$ has a natural one-parameter family of algebra automorphisms $\sigma_t$ defined by $\sigma_t(X_n) = n^{it}X_n$ and $\sigma_t(Y_{\lambda})=Y_{\lambda}$.

For any algebra $A$ together with a one-parameter family of automorphisms $\sigma_t$ one is interested in KMS-states or Kubo-Martin-Schwinger states with parameter $\beta$, $KMS_{\beta}$ (this parameter is often called the 'invers temperature' of the system) as these are suitable equilibria states. Recall that a state is a special linear functional $\phi$ on $A$ (in particular it must have norm one) and it belongs to $KMS_{\beta}$ if the following commutation relation holds for all elements $a,b \in A$

$\phi(a \sigma_{i\beta}(b)) = \phi(b a)$

Let us work out the special case when $A$ is the matrix-algebra $M_n(\mathbb{C})$. To begin, all algebra-automorphisms are inner in this case, so any one-parameter family of automorphisms is of the form

$\sigma_t(a) = e^{itH} a e^{-itH}$

where $e^{itH}$ is the matrix-exponential of the nxn matrix $H$. For any parameter $\beta$ we claim that the linear functional

$\phi(a) = \frac{1}{tr(e^{-\beta H})} tr(a e^{-\beta H})$

is a KMS-state.Indeed, we have for all matrices $a,b \in M_n(\mathbb{C})$ that

$\phi(a \sigma_{i \beta}(b)) = \frac{1}{tr(e^{-\beta H})} tr(a e^{- \beta H} b e^{\beta H} e^{- \beta H})$

$= \frac{1}{tr(e^{-\beta H})} tr(a e^{-\beta H} b) = \frac{1}{tr(e^{-\beta H})} tr(ba e^{-\beta H}) = \phi(ba)$

(the next to last equality follows from cyclic-invariance of the trace map).

These states are usually called Gibbs states and the normalization factor $\frac{1}{tr(e^{-\beta H})}$ (needed because a state must have norm one) is called the partition function of the system. Gibbs states have arisen from the study of ideal gases and the place to read up on all of this are the first two chapters of the second volume of "Operator algebras and quantum statistical mechanics" by Ola Bratelli and Derek Robinson.

This gives us a method to construct KMS-states for an arbitrary algebra $A$ with one-parameter automorphisms $\sigma_t$ : take a simple n-dimensional representation $\pi~:~A \mapsto M_n(\mathbb{C})$, find the matrix $H$ determining the image of the automorphisms $\pi(\sigma_t)$ and take the Gibbs states as defined before.

Let us return now to the Bost-Connes algebra $\mathcal{H}$. We don't know any finite dimensional simple representations of $\mathcal{H}$ but, sure enough, have plenty of graded simple representations. By the usual strongly-graded-yoga they should correspond to simple finite dimensional representations of the part of degree one $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$ (all of them being one-dimensional and corresponding to characters of $\mathbb{Q}/\mathbb{Z}$).

Hence, for any $u \in \mathcal{G} = \prod_p \hat{\mathbb{Z}}_p^{\ast}$ (details) we have a graded simple $\mathcal{H}$-representation $S_u = \oplus_{n \in \mathbb{N}_+} \mathbb{C} e_n$ with action defined by

$\begin{cases} \pi_u(X_n)(e_m) = e_{nm} \\ \pi_u(Y_{\lambda})(e_m) = e^{2\pi i n u . \lambda} e_m \end{cases}$

Here, $u.\lambda$ is computed using the 'chinese-remainder-identification' $\mathcal{A}/\mathcal{R} = \mathbb{Q}/\mathbb{Z}$ (details).

Even when the representations $S_u$ are not finite dimensional, we can mimic the above strategy : we should find a linear operator $H$ determining the images of the automorphisms $\pi_u(\sigma_t)$. We claim that the operator is defined by $H(e_n) = log(n) e_n$ for all $n \in \mathbb{N}_+$. That is, we claim that for elements $a \in \mathcal{H}$ we have

$\pi_u(\sigma_t(a)) = e^{itH} \pi_u(a) e^{-itH}$

So let us compute the action of both sides on $e_m$ when $a=X_n$. The left hand side gives $\pi_u(n^{it}X_n)(e_m) = n^{it} e_{mn}$ whereas the right-hand side becomes

$e^{itH}\pi_u(X_n) e^{-itH}(e_m) = e^{itH} \pi_u(X_n) m^{-it} e_m =$

$e^{itH} m^{-it} e_{mn} = (mn)^{it} m^{-it} e_{mn} = n^{it} e_{mn}$

proving the claim. For any parameter $\beta$ this then gives us a KMS-state for the Bost-Connes algebra by

$\phi_u(a) = \frac{1}{Tr(e^{-\beta H})} Tr(\pi_u(a) e^{-\beta H})$

Finally, let us calculate the normalization factor (or partition function) $\frac{1}{Tr(e^{-\beta H})}$. Because $e^{-\beta H}(e_n) = n^{-\beta} e_n$ we have for that the trace

$Tr(e^{-\beta H}) = \sum_{n \in \mathbb{N}_+} \frac{1}{n^{\beta}} = \zeta(\beta)$

is equal to the Riemann zeta-value $\zeta(\beta)$ (at least when $\beta > 1$).

Summarizing, we started with the definition of the Bost-Connes algebra $\mathcal{H}$, found a canonical one-parameter subgroup of algebra automorphisms $\sigma_t$ and computed that the natural equilibria states with respect to this 'time evolution' have as their partition function the Riemann zeta-function. Voila!

So in an informal way one can say that KMS states are like (normalized) twisted traces?

A small remark, the books by Bratteli are available at his website as (big) pdf files:

Also, a naive question I'd like to know your answer to. Imagine that you are talking about this to people who just finished a first course in algebraic geometry, and once you start talking about the groupring $$\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$$ someone comes with the following:

"Ok, we know that the real circle can be defined as the quotient $$\mathbb{R}/\mathbb{Z}$$, and that its coordinate ring is the Laurent polynomial ring $$\mathbb{R}[x,x^{-1}]$$. Why don't we consider that quotient to be something like a rational circle and study it using the ring $$\mathbb{Q}[x,x^{-1}]$$ instead of making all that mess with adeles and the so?"

(In case I wasn't clear, I am more intrested in the "pedagogical" part on how to answer these questions that in the math of it)

I'd probably start gently by asking them what kind of group G has real group algebra $$\mathbb{R} G = \mathbb{R}[x,x^{-1}]$$? Hopefully, they'll answer 'G must be infinite cyclic!' and then I'll ask them whether they believe that $$\mathbb{R}/\mathbb{Z}$$ is generated by one element? finitely many elements? countably many elements?
By that time they will understand that these group-algebras are strange beasts and have nothing to do with affine algebraic geometry.

The groupalgebra of $$\mathbb{Q}/\mathbb{Z}$$ is 'somewhat' nicer in that it is at least countably generated.

So even if there is no affine variety associated to the group-algebra I'll ask them what they think the 'points' of any geometrical object associated to it should be. Again, hopefully, they'll respond something like : the one-dimensional representations of the group.

Fine! so what are they? this may take some time to explain to them but eventually we'll agree that this is something like $$\hat{\mathbb{Z}}$$ which then may take us eventually to $$\prodp \mathbb{Z}_p$$ and then if there is still plenty of time and interest left to the adeles...