Here's a dumb question from a physicist. The problem comes from expanding the Kähler potential of the Grassmannian manifold in the vicinity of a certain point.
Consider two matrices, $\{A^i_j\}_{N\times N}$ and $\{\tilde{A}^\alpha_\beta\}_{M\times M}$, defined as: \begin{alignat}{9} \begin{alignedat}{9} A^i_j = \delta^i_j + \phi^i_\alpha\bar{\phi}^\alpha_j\\ \tilde{A}^\alpha_\beta = \delta^\alpha_\beta + \bar{\phi}^\alpha_i\phi^i_\beta \end{alignedat}\quad,\qquad \begin{alignedat}{9} i,j&=1\ldots N\\ \alpha,\beta&=1\ldots M \end{alignedat} \end{alignat} Here bar denotes Hermitean conjugation.
It's straightforward to show that $\operatorname{Tr{}} \ln(A) = \tilde{\operatorname{Tr{}}} \ln(\tilde{A}) \equiv K$, where the tilde over the trace reminds that the trace is taken over the greek indices. Indeed, since the logarithm is defined through a series, we have: \begin{equation} \operatorname{Tr{}} \ln(A) = \operatorname{Tr{}} f_k (\phi^i_\alpha\bar{\phi}^\alpha_j)^k = f_k \operatorname{Tr{}} (\phi^i_\alpha\bar{\phi}^\alpha_j)^k = f_k \tilde{\operatorname{Tr{}}} (\bar{\phi}^\alpha_i\phi^i_\beta)^k = \tilde{\operatorname{Tr{}}} f_k (\bar{\phi}^\alpha_i\phi^i_\beta)^k = \tilde{\operatorname{Tr{}}} \ln(\tilde{A}) \end{equation} Here $f_k$ are the coefficients in the Taylor expansion on the logarithm.
As I am constructing the small-$\phi$ expansion of the following quantity (the generalisation of the Fubini-Study metric): $$ G_{i\beta}^{j\alpha}= \dfrac{\partial}{\partial\phi^i_\alpha}\dfrac{\partial}{\partial\bar{\phi}_j^\beta} K $$ Depending on which definition of $K$ I'm using, I'm getting different results. The definition above renders: \begin{alignat}{9} {}^{(I)}G_{i\beta}^{j\alpha}&=\operatorname{Tr{}} (A^{-1}) \delta_i^j \delta^\alpha_\beta - \operatorname{Tr{}}(A^{-2}) \phi^j_\beta \bar{\phi}_i^\alpha \\ {}^{(II)}G_{i\beta}^{j\alpha}&=\tilde{\operatorname{Tr{}}} (\tilde{A}^{-1}) \delta_i^j \delta^\alpha_\beta - \tilde{\operatorname{Tr{}}}(\tilde{A}^{-2}) \phi^j_\beta \bar{\phi}_i^\alpha \end{alignat} OK, looks like the quantities $\operatorname{Tr{}} (A^{-1})$ and $\tilde{\operatorname{Tr{}}} (\tilde{A}^{-1})$ should coincide. However: \begin{alignat}{9} (A^{-1})^i_j &\approx \delta^i_j - \phi^i_\alpha \bar{\phi}^\alpha_j \\ (\tilde{A}^{-1})^\alpha_\beta &\approx \delta^\alpha_\beta - \bar{\phi}^\alpha_i \phi^i_\beta \end{alignat} Which gives for the traces: \begin{alignat}{9} \begin{alignedat}{9} \operatorname{Tr{}} (A^{-1}) &= N - \phi^i_\alpha \bar{\phi}^\alpha_i \\ \tilde{\operatorname{Tr{}}} (\tilde{A}^{-1}) &= M - \phi^i_\alpha \bar{\phi}^\alpha_i \end{alignedat} \quad,\qquad \begin{alignedat}{9} \operatorname{Tr{}} (A^{-2}) &= N - 2\phi^i_\alpha \bar{\phi}^\alpha_i \\ \tilde{\operatorname{Tr{}}} (\tilde{A}^{-2}) &= M - 2\phi^i_\alpha \bar{\phi}^\alpha_i \end{alignedat} \end{alignat} Clearly, the expansions ${}^{(I)}G_{i\beta}^{j\alpha}$ and ${}^{(II)}G_{i\beta}^{j\alpha}$ differ, and only match upon replacing $N$ with $M$.
What am I doing wrong?
I already discussed this with you, but given there are upvotes on the question I'll write up an answer. Consider a matrix as a function of a variable $A(u)$. Now how do derivatives act on the powers $A^n$ (which appear as terms in power series expansions) ? $$\partial_u A^n = A'A^{n-1}+A\,A'A^{n-2}+\dots+A^{n-1}A'$$ Now the derivative matrix $A'$ and $A$ don't necessarily commute, but if we take the derivative inside a trace there is no problem with the naive chain rule. $$\partial_u\,Tr(A^n)=n\,Tr(A^{n-1}A').$$ But if we take a second derivative we start to see problems because for instance $$Tr(A'A'A^{n-2})\neq Tr(A'AA'A^{n-3}).$$
In your case to be a little more specific where the manipulations fail I'll start calculating your $G$: $$Tr(\partial _{\bar{\phi}^\beta_j}(\ln A)^{i'}_{j'})=Tr((A^{-1})^{i'}_{k'}\partial _{\bar{\phi}^\beta_j}A^{k'}_{j'})=Tr((A^{-1})^{i'}_{k'}\phi^{k'}_\beta\delta^j_{j'})=(A^{-1})^j_{k'}\phi^{k'}_\beta$$ Now I was able to use the chain rule because it was in a trace, but now I am left with a sum over a specific row of the inverse matrix. If I take the next derivative, $$^{(I)}G^{j\alpha}_{i\beta}=(A^{-1})^j_i\delta^{\alpha}_{\beta}+\phi^{k'}_\beta\partial_{{\phi}^i_\alpha}(A^{-1})^j_{k'}.$$ The first term is a bit different from what you had, and the second I can't naively expand it as matrix elements of $A^{-2}$ in an obvious way. But if you expand it in terms of a power series you should be able to show it's equivalent to $^{(II)}G$, given that's how you demonstrated they were equivalent in the first place.