&   &   &   &   &   & 7 & 8 & 9 & 10 & \hspace{2cm} {\rm (A)} \\
M2   & 1 & 2 &   &   &   &   &   &   &   &    &
\end{array}$$ where $H_{F}^{(1)}(x_7, x_8, x_9, x_{10})$, $H_{F}^{(2)}(x_3, x_4, x_5, x_6)$, $H_{T}=H_{F}^{(1)} H_{F}^{(2)}$. (See (\[M552\]) below for the explicit solution in terms of the metric and antisymmetric tensor field.) This solution preserves $1/4$ of supersymmetry. The easiest way to check that one can take for the membrane harmonic function the product of the five-brane harmonic functions is to consider the field equation for the antisymmetric tensor field, $\partial_M (\sqrt{-g} F^{MNPQ})=0$[^9]. Substituting the metric associated to (A) according to the harmonic function rule, one obtains (for $N,P,Q$ in the membrane directions): \[HT\] (H\_F\^[(1)]{}(x’)\_x\^2 + H\_F\^[(2)]{}(x)\_[x’]{}\^2 )H\_T(x,x’)=0, where $x$ denotes coordinates $x_3,\cdots x_6$ and $x'$ denotes $x_7,\cdots x_{10}$. Notice that (\[HT\]) is the Laplace equation in the curved transverse space produced by the five-branes. Clearly, $H_T=H_F^{(1)} H_{F}^{(2)}$ is a solution. What is essential though for obtaining the anti-de Sitter product geometry is that the near-horizon limit of $H_T$ behaves as $1/r^2 r'^2$, so one could take a more general solution of (\[HT\]) with that property. For example, $H_T = 1+{Q_T \over 