have $$\eqalign{\tau(p,s)&=\zeta_{\Delta_1^{F}}(s)-2\zeta_{\Delta_2^{F}}(s)+
3\zeta_{\Delta_3^{F}}(s)\cr
{}&=3\zeta_{\Delta_0^{F}}(s)-\zeta_{\Delta_1^{F}}(s),\quad
\hbox{using Poincar\'e duality}\cr}\no$$ Combining the standard decomposition $\Delta_p=(\dstar d+d\dstar)_p$, with the fact that $\ker\dstar\cap\ker d=\emptyset$, we further obtain the formula $$\tau(p,s)=2\zeta_{{\dstar d}_0}(s)-\zeta_{{\dstar d}_1}(s)\no$$ We now simplify our notation by labelling $$\tau_{+}(p,s)=2\zeta_{{\dstar d}_0}(s),\qquad
\tau_{-}(p,s)=\zeta_{{\dstar d}_1}(s)$$ For the individual zeta functions we denote the eigenvalues and their degeneracies by $\lambda_n(q,p)$ and $\Gamma_n(q,p)$ respectively giving the expressions $$\tau_{+}(p,s)=2\sum_n{\Gamma_n(0,p)\over
\lambda_n^s(0,p)},\qquad
\tau_{-}(p,s)=\sum_n{\Gamma_n(1,p)\over
\lambda_n^s(1,p)}\no$$ It remains to compute these eigenvalues and degeneracies. The former are actually independent of $p$ and are fairly easily calculated by the technique of starting with harmonic forms in ${\bf R}^{2n}$ and then restricting successively to $S^{2n+1}$ and $L(p)$. In any case they are given by $$\eqalign{\lambda_n(0,p)&=n(n+2),\;n=1,2,\dots\cr
           \lambda_n(1,p)&=(n+1)^2,\;n=1,2,\dots\cr}\no$$

To calculate the degeneracies is more difficult; we make use of the fact that $S^3$ is a group manifold and proceed as follows: Consider the Laplacians $\dstar d_q$ on $S^3$, and $\dstar d_q^{F}$ on $L(p)$ also, if $\lambda$ is an eigenvalue, denote the corresponding eigenspaces by $\Lambda_q(\lambda)$ and $\Lambda^{F}_q(\lambda)$ respectively. Let $$v(z)\in \Lambda_q(\lambda),\;\hbox{with }z\in S^3\subset {\bf C}^2,
\qquad \hbox{and }g\in {\bf  Z}_p,\;
\hbox{where }g\equiv\exp[2\pi i j/p],\;0\le j\le (p-1)\no$$ The element $g$ acts on $v(z)$ to give $g\cdot v(z)$ where $$\eqalign{g\cdot v(z)&=v(gz)\cr
gz&=(\exp[2\pi i j/p]z_1,\exp[2\pi i j/p]z_2)\cr}\no$$ The above definitions allow 