he right hand side of is precisely $$\mu_1\left\|\sum_p\nu^p\right\|^2-n\sum_p\left\|\nu^p\right\|^2,$$ which by Lemma \[ineq-1\] is bounded above by $\mu_1n^2-n\left\|\mu\right\|^2$ with equality only where either $\rho^p=(1^{\mu_p})$ (case (ii)) or all $\rho^p$ are equal and $\mu=(t^{n/t})$ for some $t$ (case (i)).

Combining this with Lemma \[n-ineq\] we see that to obtain the maximum of the left hand side of we must also have $\rho^1\cup \cdots\cup\rho^r=\lambda$. In case (i) then, $\lambda$ is the union of $n/t$ copies of $\lambda_0$, the common value of $\rho^p$, and in case (ii), $\lambda=(1^n)$.

We first prove (ii). Using Lemma \[v-lambda\] we have $$-v(\lambda)=(2g-2+k) n(\lambda) +(g-1)n -\sum_{i=1}^kv(\lambda,\mu^i)=
   \frac{\delta}{n}n(\lambda)+(g -1)n
   +\frac1n\sum_{i=1}^k\left[\mu_1^in(\lambda)-nv(\lambda,\mu^i)\right].
\label{vlambda}$$ The terms $n(\lambda)$ and $\sum_{i=1}^n\left[\mu_1^in(\lambda)-nv(\lambda,\mu^i)\right]$ are all maximal at $\lambda=(1^n)$ (the last by Proposition \[opti\]). Hence $-v(\lambda)$ is also maximal at $(1^n)$, since $\delta\geq 0$. Now $n(\lambda)$ has a unique maximum at $(1^n)$ by Lemma \[n-ineq\], hence $-v(\lambda)$ reaches its maximum at other partitions if and only if $\delta=0$ and for each $i$ we have $\mu^i=(t_i^{n/t_i})$ for some positive integer $t_i\mid n$ (again by Proposition \[opti\]). In this case the maximum occurs only for $\lambda$ the union of $n/t$ copies of a partition $\lambda_0\in \calP_t$, where $t=\gcd t_i$. Now (ii) follows from Proposition \[affine-descrip\].

To prove (i) we use Lemma \[v-lambda\] and and find that $v((1^n))=-\Delta(\muhat)$ as claimed.

Let $\muhat=(\mu^1,\mu^2,\ldots,\mu^k)\in {\P_n}^k$ with $\delta(\muhat) \geq 0$. Suppose that $v(\lambda)$ is minimal. Then the coefficient of $q^{v(\lambda)}$ in $\calA_{\lambda \muhat}$ is $1$. \[lowest\]

We use the notation of the proof of Lemma \[v-fmla-lemma\]. Note that the coefficient of the lowest power of $q$ in $\calH_\lambda(\sqrt q,1/\sqrt q)
  \left(q^{-n(\lambda