the $SU(2)$ character, on $D_j$, for rotation through the angle $\theta$; i.e. $$\chi^j(\theta)={\sin((2j+1)\theta)\over\sin(\theta)}\no$$ Hence our explicit degeneracy formula for $0$-forms on $L(p)$ is $$\Gamma_n(0,p)={(n+1)\over p}\sum_{j=0}^{(p-1)}\exp[-2\pi i
j/p]\,{\sin(2\pi(n+1)j/p)\over\sin(2\pi j/p)}\no$$

We now have to find the analogous formula for the $1$-forms. The formula that results is $$\Gamma_n(1,p)={1\over p}\sum_{j=0}^{(p-1)}\exp[-2\pi i
j/p]\left\{n\chi^{(n+1)/2}(2\pi j/p)+(n+2)\chi^{(n-1)/2}(2\pi j/p)\right\}\no$$ or, more explicitly, $$\Gamma_n(1,p)={1\over p}\sum_{j=0}^{(p-1)}\exp[-2\pi i
j/p]\left\{n{\sin(2\pi(n+2)j/p)\over\sin(2\pi j/p)}+
(n+2){\sin(2\pi nj/p)\over\sin(2\pi j/p)}\right\}\no$$

To simplify the notation we introduce the $p$-averaged character’ $\charav{j}$ which we define by $$\charav{j}={1\over p}\sum_{j=0}^{(p-1)}\exp[-2\pi i
j/p]\,\chi^j(2 pi j/p)\no$$ Finally this gives us a concrete expression for $\tau(p,s)$, i.e. $$\eqalign{\tau(p,s)&=
\sum_n\left\{{2(n+1)\charav{n/2}\over \{n(n+2)\}^{2s}}
-{n\charav{(n+1)/2}+(n+2)\charav{(n-1)/2}\over
(n+1)^{2s}}\right\}\cr
 {}&=\tau_+(p,s)-\tau_-(p,s)\cr
\hbox{where}\quad&\cr
 \tau_+(p,s)&=\sum_n{2(n+1)\charav{n/2}\over \{n(n+2)\}^{s}},
\quad \tau_-(p,s)=\sum_n{n\charav{(n+1)/2}+(n+2)\charav{(n-1)/2}\over
(n+1)^{2s}}\cr}\no$$

To make further progress towards a computation of the determinants and torsion we need to be able to evaluate these $p$-averaged characters. This is a somewhat non-trivial combinatorial task but this task is eased if we use for $\chi^j(\theta)$, the alternative expression $$\chi^j(\theta)=\sum_{m=-j}^j\exp[2 i m\theta]\no$$ It is also necessary to divide $n$ up into its conjugacy classes mod $p$ by writing $$n=pk-j,\;k\in{\bf Z},\, j=0,1,\ldots,(p-1)\no$$ We eventually discover that $$\charav{(pk-j)/2}=
\cases{
       \cases{k&for $j=0,2,\ldots,(p-1)$\cr
              k&for $j=1$\cr
              (k-1)&for $j=3,5,\ldots,(p-2)$\cr}
             &if $p$ is odd\cr
           \strut&\null\cr
      \cases{0&for $j=0,2,\dots,(p-2)$ \cr
             2k&for $j=1$\cr
             (2k-1)&for $j=3,5,\ldots,(p-1)$\cr}
             &if $p$ is even\cr}  \no$$

We now lack only 