=-Q_1Q_2/4(Q_1+ Q_2)$. The dilaton vanishes in the limit, $e^{-2\phi}=1$. The field strengths become \[fieldst\] H\_[01]{}=F\_[0]{}=-2A\_[0]{} ,  F\_[1’’c’]{}=-2Q\_1\_[’’c’]{} ,    F\_[1c]{}=-2Q\_2\_[c]{}, where $\e_{0\r}$, $\e_{\a,\b,\c}$ and $\e_{\a',\b',\c'}$ are volume forms on $adS_2$, $S^3$ and $S^3$, respectively. (We use the convention that the epsilon symbols with tangent space indices are constants.) The covariantly constant field strengths in (\[fieldst\]) support $adS_2$ and the two $S^3$’s, respectively. Each of the factors of the geometry together with the corresponding field strength represents a CFT, as discussed in the introduction.

(11,7)(0,0) (0,0)[(4,1)[$(adS_2 \xx E^2 \xx S^3 \xx S^3)_{1/4}$]{}]{} (3,1) (3,1)[(0,1)[5.1]{}]{} (3,1)[(1,0)[4.9]{}]{} (8,0.5)[(3,1)[$(adS_2 \xx E^5 \xx S^3)_{1/4}$]{}]{} (1,5.8)[(3,1)[$(adS_2 \xx E^5 \xx S^3)_{1/4}$]{}]{} (5,3)[(1,1)[$({\rm B})_{1/8}$]{}]{} (6,3.5)[(1,0)[1.9]{}]{} (5.5,4)[(0,1)[2]{}]{} (5.9,3.9)[(1,1)[2]{}]{} (5,5.8)[(1,1)[$(\rm b)_{1/8}$]{}]{} (8,3)[(1,1)[$(\rm b)_{1/8}$]{}]{} (8,5.8)[(1,1)[$({\cal M}^{10})_1$]{}]{} (6,.5)[$r$]{} (2.5,4)[$r'$]{}

The interpolation diagram for intersection (B) is shown in figure 4. It is now understood that (b) corresponds to the standard intersection $D0\perp D4\perp F1$ obtained from $M2\perp M2\perp M2$ by a dimensional reduction and two $T$-dualities. Since the reduction and $T$-dualities are along relative transverse directions, one still has the $adS$ near-horizon geometry and the same number of unbroken supersymmetries.

The next solution is found by adding instead of a wave a monopole to solution (A) and reducing to $D=10$, in the same 