We take a brief detour to talk about some operations on power series that arise in calculations of probabilities in the grand canonical ensemble. We already saw an example of this when we computed the probability that all particles lie in a ball of radius $p^{-r}$ was $$\frac{1}{Z(t, \beta, \lambda)} \sum_N p^{-r\beta {N \choose 2} – rN} \frac{t^N}{N!} Z(N, \beta, \lambda).$$

The series arising in this probability is yet another generating function for $Z(N, \beta, \lambda)$. From physical consideration we can view the $p^{-r\beta {N \choose 2} – rN}$ appearing in the coefficient as arising from the additional interaction energy caused by forcing all of our particles into the same ball. But mathematically, we may view this simply as a transform on the ring of formal power series.

Let $\alpha = \sum a_n t^n \in \mathbb C[[t]]$ be a power series with coefficients in $\mathbb C$ (or another favorite field). Given indeterminant $Q$ define the $Q$-transform $$T(Q,\alpha) = \sum a_n Q^{{n \choose 2}} t^n.$$ We also define a convolution operator $\star_Q: \mathbb C[[t]] \times \mathbb C[[t]] \rightarrow \mathbb C[[t]]$ by $$\sum a_n t^n \star_Q \sum b_m t^m = \sum_{m,n} a_n b_m Q^{nm} t^{n+m}.$$ When $Q=1$ the $Q$-transform does nothing, and the associated convolution is just regular multiplication of series.

**Theorem:** For any $Q\neq 0$, $\mathbb C[[t]]$ is a ring under $+$ and $\star_Q$.

**Exercise**: Prove this.

**Proposition:** Let $\alpha, \beta \in \mathbb C[[t]]$, and $Q \neq 0$. Then, $T(Q, \alpha \beta) = T(Q, \alpha) \star_Q T(Q, \beta).$

We sometime say $\star_Q$ is the convolution that *linearizes* the transform, which I guess it does if you think multiplicatively.

**Proposition:** $T(Q, T(Q,(\alpha)) = T({Q^2},\alpha).$

### Notation for the $p$-adic Grand Canonical Ensemble

We set $Q = p^{-r \beta}$ and write $T_r := T(p^{-r \beta})$, and $\star_r$ for the associated convolution (this notation is a bit ambiguous, but from here forward we will basically be working in the grand canonical ensemble). The $p$-power theorem now says $Z(t, \beta, \lambda) = T_1 Z(t/p, \beta, \lambda)^p.$ More generally we have:

**Theorem:** Let $\lambda_r$ be Haar measure restricted to $p^r \mathbb Z_p$. Then,

- $Z(t, \beta, \lambda_r) = T_r Z(t/p^r, \beta, \lambda).$
- $Z(t, \beta, \lambda_{r}) = T_{1} Z(t/p, \beta, \lambda_{r+1})^{\star_r p}.$

*Example. *We have $$\begin{eqnarray} Z(t, \beta, \lambda) &=& T_1 Z(t/p, \beta, \lambda)^{\star_0 p} = T_1 (T_1 Z(t/p^2, \beta, \lambda_1)^{\star_1 p})^{\star_0 p} \\ &=& (T_2 Z(t/p^2, \beta, \lambda)^{\star_1 p})^{\star_0 p} = (Z(t/p^2, \beta, \lambda_2)^{\star_1 p})^{\star_0 p}. \end{eqnarray}.$$ We may continue this as far as we like.

For an event $A \in \Omega$, $\mathbb P(A)$ is a ratio of power series. For our operators $T_r$, let $T_r \mathbb P(A)$ denote $T_r$ operating on the numerator (but not the denominator).

**Lemma:** $T_r \mathbb P(A) = \mathbb P(p^r A).$ That is $$\begin{eqnarray}T_r \mathbb P(A) &=& \frac{1}{Z(t, \beta, \lambda)} \sum_{N=0}^{\infty} \frac{t^N}{N!} p^{-r\beta {N \choose 2} – rN} Z(N, \beta, \boldsymbol 1_{A_N} \lambda^N) \\ &=& \frac{1}{Z(t, \beta, \lambda)} \sum_{N=0}^{\infty} \frac{t^N}{N!} Z(N, \beta, \boldsymbol 1_{p^r A_N}, \lambda^N) = \mathbb P(p^r A). \end{eqnarray}$$

Now, given coset $j + p \mathbb Z_p$ define $\pi : j + p \mathbb Z_p \rightarrow \mathbb Z_p$ given by $\pi(j + p x) = x.$ The image of $\pi$ is the* inflation* of the coset to all of $\mathbb Z_p$. If $A \in \Omega(j + p \mathbb Z_p)$, then $\pi A$ is the event in $\Omega$ similar to $A$ except for scale. Note that $A = j + p \pi A$, and hence the following lemma.

**Lemma:** If $A \in \Omega(j + p \mathbb Z_p)$, then $\mathbb P(A) = T_1 \mathbb P(\pi A).$