By a theorem of Luck, $K^0 (BS_n) \simeq \mathbb{Z} \times \prod_p (\mathbb{Z}_{p}^\wedge)^{r(p,n)}$ where $r(p,n)$ is the number of partitions of $n$ into powers of $p$ (excluding the trivial partition $1+1 \cdots 1$). Wanting to compute the $r(p,n)$, you can repeat the procedure of making the generating function for all the $n$.

\begin{eqnarray*} \sum (r(p,n)+1)x^n &=& \prod_{j \geq 0} \frac{1}{1 - x^{p^j}}\\ &=& \exp \sum_{j=0}^\infty \sum_{k=1}^{\infty} \frac{x^{kp^j}}{k}\\ \end{eqnarray*}

Last night, I plotted a coarse approximation the quantity inside the exponential for $\mid x\mid <1$. You can start to see the divergences at all the roots of unity.

My question was whether these quantities have some sort of modular properties like $\eta (q)$. Or just estimates of $r(p,n)$ via this method. I am not well versed in this area so I don't know the difficulty of this question or whether this is addressed in the literature.

  • 2
    $\begingroup$ No, but the Mellin transform of $\log f(e^{-t})$ is $F(s)=\Gamma(s)\sum_{j\ge 0, k \ge 1} \frac{(kp^j)^{-s}}{k} = \Gamma(s)\frac{\zeta(s+1)}{1-p^{-s}}$ which has a functional equation. By comparison with $g(e^{-t}) = \log \prod_{n\ge 1} \frac{1}{1-e^{-nt}} = \sum_{k \ge 1} e^{-kt} \sigma_{-1}(k)$ its Mellin transform is $G(s) = \Gamma(s) \zeta(s)\zeta(s+1) = (2\pi)^{2s-1}(\frac{1}{s}-1)G(1-s)$. $\endgroup$ – reuns Jul 23 '17 at 1:40

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