One formula for Hermitian form

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?

Help?

• 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. Apr 8 at 16:53
• 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. Apr 8 at 16:57
• Oh, I wasn't quite following you. So have you used the formula for the inverse matrix? Apr 8 at 17:08
• It might be wise ultimately to do the computation in a unitary coframe. Apr 8 at 17:13
• 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? Apr 8 at 17:27