the numbers above, take $n=3, k=4$. Then $$N = \dim  H^0\left({\mathbb{CP}}^3, O(4)\right) = {3+4 \choose 4} = 35$$ and a basis of $H^0({\mathbb{CP}}^3, O(4))$ is (in local coordinates) $$\label{eqn: basis elements}
    \{w_1, \ldots, w_{35}\} = \{ 1, z_1, z_2, z_3, z_1^2, \ldots, z_2z_3^3, z_3^4\}.$$ We choose a $G$ which is invariant under *every* permutation of the homogeneous coordinates $Z_i,\, i=0, \ldots, 3$ (where $z_i=\frac{Z_i}{Z_0}$). Taking into account these symmetries there are only five distinct basis elements: $$1,\ z_1,\ z_1^2,\ z_1z_2,\ z_1z_2z_3.$$ In the order the basis elements are listed in (\[eqn: basis elements\]) – first by degree then lexicographically – these are the 1st, 2nd, 5th, 6th, and 15th elements. In the notation used at the beginning of this section we pick diagonal entries of $G$: $a_i^{-1}$ in the row and column determined by the basis element $w_i$. Due to the symmetries we will have five parameters: $$G: a_1, a_2, a_5, a_6, a_{15}$$ The iterations of ${T_{\nu}}$ on these parameters – denote them by $a_{i,r}, r=0, 1, \ldots, \infty$ – will (after uniform scaling) tend toward the values $1, 4, 6, 12, 24$ for $i=1,2,5,6,15$ respectively. This can readily be checked by noting the Fubini-Study metric is the balanced metric 