supersymmetric states, holds in general for the two giant system. Looking at the labels, it is natural to interpret $O_A(b_0,b_1)$ as a state in which we deform only the larger threebrane. Recall from section 3, that deforming a single threebrane gives us a supersymmetric state so it seems natural for $O_A(b_0,b_1)$ to remain supersymmetric. Similarly, $O_F(b_0,b_1)$ can be interpreted as a state in which we deform only the smaller threebrane and a similar comment can be made. The fact that the combinations $O_B(b_0,b_1)+\sqrt{2}O_C(b_0,b_1)$ and $O_D(b_0,b_1)+\sqrt{2}O_E(b_0,b_1)$ are annihilated by $D$ implies that there are another two supersymmetric ways to deform the pair of threebranes. Finally, notice that if we set $O_B(b_0,b_1)-O_C(b_0,b_1)/\sqrt{2}\equiv O_{B-C}(b_0,b_1)$ and $O_D(b_0,b_1)-O_E(b_0,b_1)/\sqrt{2}\equiv O_{D-E}(b_0,b_1)$ we have $$\begin{aligned}
\nonumber
DO_{B-C}(b_0,b_1) = &&-2g_{YM}^2(N-b_0)\left[
O_{B-C}(b_0+1,b_1-2)-2O_{B-C}(b_0,b_1)+O_{B-C}(b_0-1,b_1+2)\right]\cr
DO_{D-E}(b_0,b_1) = &&-2g_{YM}^2(N-b_0)\left[
O_{D-E}(b_0+1,b_1-2)-2O_{D-E}(b_0,b_1)+O_{D-E}(b_0-1,b_1+2)\right]\end{aligned}$$ The right hand side again is a discretization of the second derivative. [*It is the Young diagram itself that is defining the lattice.*]{} After recalling that the number of boxes in each column sets the angular momentum and hence the radius[^4] of the corresponding threebrane, its clear that the radius of the giant graviton together with local physics in this radial direction has emerged.

Four Impurities
---------------

For the case of four impurities there 