I'm wondering if anyone could help me with calculating the Ricci tensor for the Grassmannian manifold. For Kähler manifolds we have: \begin{equation} R_{\mu \bar{\nu}} = -\partial_{\bar{\nu}} \partial_{\mu} \log \det g_{\mu\bar{\nu}} \end{equation}
For the Grassmannian, we are using the two-index notation \begin{equation} \Phi^{n\alpha} \quad,\qquad \text{where} \quad n=1\ldots N,\quad \alpha=1\ldots M \end{equation} And so, the Ricci tensor will have four indices, $R_{i\bar{j}\alpha\bar{\beta}}$.
The Kähler potential is given by: \begin{equation} K = \operatorname{Tr} \ln (\delta^{n\bar{m}} + \Phi^{n\gamma}\overline{\Phi}^{\bar{\gamma} \bar{m}}) \end{equation} Here we take the trace over the Latin indices, but we could also define $K = \operatorname{Tr} \ln (\delta^{\alpha\bar{\beta}} + \Phi^{n\alpha}\overline{\Phi}^{\bar{\beta}\bar{n}})$ and take the trace with respect to the Greek indices. The subscripts are reserved for lowering with the aid of the metric tensor. The latter is obtained as follows: \begin{gather} G_{i\bar{j} \alpha \bar{\beta}} = \dfrac{\partial}{\partial \Phi^{i\alpha}} \dfrac{\partial}{\partial \overline{\Phi}^{\bar{\beta}\bar{j}}} K = \operatorname{Tr} \left\{ \delta^{ni}\delta^{\alpha\bar{\beta}}(A^{-1})^{\bar{j} m} - \Phi^{n\beta} (A^{-1})^{\bar{j} i} \overline{\Phi}^{\bar{\alpha} \bar{l}} (A^{-1})^{\bar{l} m} \right\} \end{gather}
How should I proceed from here? I guess, the answer should be \begin{equation} R_{i\bar{j}\alpha\bar{\beta}} \propto (M+N)G_{i\bar{j}\alpha\bar{\beta}} \end{equation}