# Positive definiteness of Fubini-Study metric

Define the Fubini-Study metric $$g_{i\overline{j}} = \frac{\delta_{i\overline{j}}(1+|\boldsymbol{z}|^2)-\overline{z}^jz^i}{(1+|\boldsymbol{z}|^2)^2}$$ for $i,j=1,\ldots,n$ and $z_i$ complex variables and $|\boldsymbol{z}|^2=\sum_{i=1}^n |z^i|^2.$

My GOAL is to show that, for every $k=1,\ldots,n,$ $$\det \left( g_{i\overline{j}} \right)_{1 \leq i,\overline{j}\leq k} = \frac{1+\sum_{i=k+1}^n |z^i|^2}{(1+|\boldsymbol{z}|^2)^{k+1}}.$$

• What exactly do you mean with an overbar on top of an index? After all, being an integer, the complex conjugate of the index equals the index itself. Jun 22, 2013 at 16:27
• it's a fairly common notation in theoretical physics to adorn the indices with the same notation as the variables. Less common here. It shouldn't be taken as literal conjugation. Just replace $\bar{j}$ with $j$ if it is troublesome. Jun 22, 2013 at 16:32
• right, sorry for the misunderstanding; James comment already explains.
– jj_p
Jun 22, 2013 at 16:38
• What does $z^2$ mean? If it means $\|(z_1,\ldots,z_n)^T\|^2$ for some vector norm, then your $(g_{ij})$ is simply a matrix of the form $aI+b\bar{z}z^T$, and its determinant follows from the Sherman-Morrison formula. Jun 22, 2013 at 16:52
• There's a typo in the denominator of your determinant formula. This is a place where a bit of theory is helpful. The associated Kähler form $\omega = \dfrac{i}{2\pi} \partial\bar\partial \log|z|^2$ is a positive $(1,1)$-form and its powers induce the natural volume forms on $k$-dimensional subspaces of $T_z\mathbb CP^n$ for every $k$. The $k$th power will have your $k$th determinant as the coefficient of $\left(\frac i{2\pi}\right)^k\,dz^1\wedge dz^{\bar 1}\wedge \dots \wedge dz^k\wedge dz^{\bar k}$. Jun 22, 2013 at 18:31

Let's write numerator matrix as $B_{i\bar{j}}=\delta_{ij}(1+|z|^2)-z_i\bar z_j$, where $|z|^2=\sum\limits_{i=1}^k|z_i|^2$. Notice that the first component $B^1_{i\bar j}={\delta_{ij}(1+|z|^2)}$ is a scalar multiply an identity matrix, and the second component $B^2_{i\bar j}=z_i\bar z_j$ is a symmetric matrix of rank one, and $B^1B^2=B^2B^1$, so there exists invertile matrix $P$, such that $PBP^{-1}= \left( \begin{array}{ccc} 1+|z|^2 \\ & \ddots & \\ & & 1+|z|^2 \end{array} \right)- \left( \begin{array}{cccc} |z|^2 & &0 &\\ & \ddots & & \\ 0& &0 \end{array} \right)= \left( \begin{array}{ccc} 1 \\ &1+|z|^2 & \\ &\ddots& \\ & & 1+|z|^2 \end{array} \right)$ so $\det(g_{i\bar j})=\frac{1}{(1+|z|^2)^{k+1}}$.
• OK: your formula agrees with mine for $k=n;$ you're left to prove or disprove that mine holds also for $k<n,$ where still $|\boldsymbol{z}|^2=\sum_{i=1}^n |z^i|^2.$