I am really struggling with this one: I have a two form $T=T_{i\overline{j}}dz_id\overline{z}_j$ and a Hermitian form $\omega=g_{i\overline{j}}dz_i\wedge d\overline{z_j}$. And I want to prove the formula $T\wedge \frac{\omega^{n-1}}{(n-1)!}=(g^{\overline{j}k}T_{\overline{k}j})\frac{\omega^n}{n!}$, with $g^{\overline{j}k}$ being the inverse matrix of $g_{k\overline{j}}$.

What have I tried: I tried calculating $n=1$ scenario, which reduces to $T=g^{\overline{1}1}T_{\overline{1}1}g_{\overline{1}1}dz_1d\overline{z_1}$.

However, when I tried to go to $n=2$, I have $LHS=(-T_{\overline{1}2}g_{2\overline{1}}-T_{2\overline{1}}g_{1\overline{2}}+T_{1\overline{1}}g_{2\overline{2}}+T_{2\overline{2}}g_{1\overline{1}})dz_1\wedge d\overline{z_1}\wedge dz_2\wedge d\overline{z_2}$ and $RHS=\frac{1}{2}(g^{1\overline{2}}T_{\overline{2}1}+g^{2\overline{1}}T_{\overline{1}2}+g^{1\overline{1}}T_{\overline{1}1}+g^{2\overline{2}}T_{\overline{2}2})det(g_{i\overline{j}})dz_1\wedge d\overline{z_1}\wedge dz_2\wedge d\overline{z_2}$, which I can't manage to fit together.

Now that I couldn't fit these together, I began to doubt whether the formula is true in general for Hermitian forms. Do we need the form $\omega$ to be Kahler?


  • $\begingroup$ First, you have a typo. It should be $\omega^{n-1}$. If this is just linear algebra, how can closedness of $\omega$ be remotely relevant? Your summation convention looks suspicious — ordinarily we sum lower $k\bar k$ and either $^k\,_k$ or $^{\bar k}\,_{\bar k}$. And my personal preference would be to use wedge products everywhere, since hermitian metrics are often written without the tensor product. $\endgroup$ Apr 8 at 16:53
  • $\begingroup$ You are right, it should be $\omega^{n-1}$. I guess in class we assumed $\omega$ to be kahler, and I think I have seen somewhere that under a proper change of basis Kahler forms have nice local representations. And I followed my prof in class on the summation convention. I guess this isn't the standard one because he saves the upper indices for the inverse matrix. $\endgroup$
    – kid111
    Apr 8 at 16:57
  • 1
    $\begingroup$ Oh, I wasn't quite following you. So have you used the formula for the inverse matrix? $\endgroup$ Apr 8 at 17:08
  • $\begingroup$ It might be wise ultimately to do the computation in a unitary coframe. $\endgroup$ Apr 8 at 17:13
  • $\begingroup$ I tried the inverse matrix formula just now, but I get LHS$=(T_{1\overline{2}}g^{1\overline{2}}+T_{2\overline{1}}g^{2\overline{1}}+T_{1\overline{1}}g^{1\overline{1}}+T_{2\overline{2}}g^{2\overline{2}})det(g_{i\overline{j}})dz_1\wedge d\overline{z_1}\wedge dz_2\wedge d\overline{z_2}$, which lacks a $\frac{1}{2}$ from the RHS? $\endgroup$
    – kid111
    Apr 8 at 17:27


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy