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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 $k=1,\ldots,n,$ $$ \det \left( g_{i\overline{j}} \right)_{1 \leq i,\overline{j}\leq k} = \frac{1+\sum_{i=k+1}^n |z^i|^2}{(1+|\boldsymbol{z}|^2)^{k+1}}.$$

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  • $\begingroup$ 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. $\endgroup$
    – celtschk
    Jun 22, 2013 at 16:27
  • $\begingroup$ 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. $\endgroup$
    – James S. Cook
    Jun 22, 2013 at 16:32
  • $\begingroup$ right, sorry for the misunderstanding; James comment already explains. $\endgroup$
    – jj_p
    Jun 22, 2013 at 16:38
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    $\begingroup$ 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. $\endgroup$
    – user1551
    Jun 22, 2013 at 16:52
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    $\begingroup$ 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 $\left(\frac i{2\pi}\right)^k\,dz^1\wedge dz^{\bar 1}\wedge \dots \wedge dz^k\wedge dz^{\bar k}$. $\endgroup$
    – Ted Shifrin
    Jun 22, 2013 at 18:31

1 Answer 1

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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 $\det(g_{i\bar j})=\frac{1}{(1+|z|^2)^{k+1}}$.

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  • $\begingroup$ 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 $|\boldsymbol{z}|^2=\sum_{i=1}^n |z^i|^2.$ $\endgroup$
    – jj_p
    Oct 4, 2013 at 9:04

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