The setting is:
- $X$ is a complex manifold;
- $E$ os a complex vector bundle over $X$;
- $s,t$ are sections of $E$;
- $\nabla$ is a hermitian connection on $E$.
Then we have a proposition saying $F_\nabla$, the curvature of $\nabla$, is a section of $[\Lambda^{1,1}]\otimes\operatorname{End}(E,h)$. The $[\Lambda^{1,1}]$ part is easy, once the other part has been established. And establishing that boils down to:
$$h(F_\nabla s,t)+h(s,F_\nabla t)=0.$$
To prove this, I tried computing the two summands and relating them to $d(dh(s,t))$ which I know to be zero. First of all:
$$d(dh(s,t))=d(h(\nabla s,t)+h(s,\nabla t))=dh(\nabla s,t)+dh(s,\nabla t),$$
first passage by hermitianity of $\nabla$, second passage by linearity of $d$ and definition of $dh$. I would expect those summands to be:
\begin{align*} dh(\nabla s,t)={}&h(F_\nabla s,t)+h(\nabla s,\nabla t), \\ dh(s,\nabla t)={}&h(\nabla s,\nabla t)+h(s,F_\nabla t). \end{align*}
But I also would like the $h(\nabla s,\nabla t)$ to cancel out, to then conclude what I wrote above and prove the proposition. So I think that i the first term, intuitively, we are passing a differential beyond a 1-form (when we jump the $\nabla s$ already present to put the $\nabla$ on the $t$), so maybe there is a minus:
$$dh(\nabla s,t)=h(F_\nabla s,t)-h(\nabla s,\nabla t).$$
With that and the above expression for $dh(s,\nabla t)$, I am done, if I prove those. So I try to. And I fail. Here is what I did. Locally (i.e. in a trivialization of $E$), we assume:
$$\nabla=d+A,$$
with $A\in\mathcal A^1(\operatorname{End}E)$. This means that, since we are in a trivialization and can therefore consider the constant sections $e_i$ as a basis, we can say:
$$s=\sum_is_i\otimes e_i,$$
and we then have:
$$ds=\sum_ids_i\otimes e_i.$$
As for $A$, we can say:
$$A=\sum\alpha_i\otimes T_i,$$
$\alpha_i$ being 1-forms on $X$ and $T_i$ being sections of $\operatorname{End}E$. Then:
$$As=\sum_i\alpha_i\otimes T_is.$$
Within these assumptions, I know that:
$$F_\nabla s=dA(s)+A\wedge A(s)=\sum_id\alpha_i\otimes T_is+\sum_{i,j}\alpha_i\wedge\alpha_j\otimes T_jT_is.$$
On simple tensors, $h$ behaves as follows:
$$h(\alpha\otimes s,\beta\otimes t)=h(s,t)\alpha\wedge\beta.$$
It is evident that $de_i=0$, whence $\nabla e_i=Ae_i=\sum_j\alpha_j\otimes T_je_i$. With all that, I proceed to the computation:
\begin{align*} dh(\nabla s,t)={}&d(h(ds+As,t))={} \\ {}={}&d\left(h\left(\sum_ids_i\otimes e_i+\sum_i\alpha_i\otimes T_is,t\right)\right)={} \\ {}={}&d\left(\sum_ih(e_i,t)ds_i+h(T_is,t)\alpha_i\right)={} \\ {}={}&\sum_idh(e_i,t)\wedge ds_i+\underbrace{\sum_ih(e_i,t)d^2s_i}_{\text{Vanishes since }d^2=0}+{} \\ &{}+\sum_idh(T_is,t)\wedge\alpha_i+\sum_ih(T_is,t)d\alpha_i={} \\ {}={}&\sum_i(h(\nabla e_i,t)+h(e_i,\nabla t))\wedge ds_i+{} \\ &{}+\sum_i(h(\nabla T_is,t)+h(T_is,\nabla t))\wedge\alpha_i+h\left(\sum_id\alpha_i\otimes T_is,t\right)={} \\ {}={}&\sum_{i,j}h(\alpha_j\otimes T_je_i,t)\wedge ds_i+{} \\ &{}+\sum_{i,j}h(e_i,dt_j\otimes e_j+\alpha_j\otimes T_jt)\wedge ds_i+{} \\ &{}+\sum_{i,j}h(d(T_is)_j\otimes e_j+\alpha_j\otimes T_jT_is,t)\wedge\alpha_i+{} \\ &{}+\sum_{i,j}h(T_is,dt_j\otimes e_j+\alpha_j\otimes T_jt)\wedge\alpha_i+h(dA(s),t)={} \\ {}={}&\sum_{i,j}h(T_je_i,t)\alpha_j\wedge ds_i+{} \\ &{}+\sum_{i,j}h(e_i,e_j)dt_j\wedge ds_i+\sum_{i,j}h(e_i,T_jt)\alpha_j\wedge ds_i+{} \\ &{}+\sum_{i,j}h(e_j,t)d(T_is)_j\wedge\alpha_i+\sum_{i,j}h(T_jT_is,t)\alpha_j\wedge\alpha_i+{} \\ &{}+\sum_{i,j}h(T_is,e_j)dt_j\wedge\alpha_i+\sum_{i,j}h(T_is,T_jt)\alpha_j\wedge\alpha_i+h(dA(s),t)={} \\ {}={}&\sum_{i,j}h(\underbrace{\alpha_j\wedge ds_i\otimes T_je_i}_{\text{These sum to }A\wedge ds},t)+{} \\ &{}-\sum_{i,j}h(\underbrace{ds_i\otimes e_i}_{\text{Summing to }ds},\overbrace{dt_j\otimes e_j}^{\text{Summing to }dt})-\sum_{i,j}h(ds_i\otimes e_i,\underbrace{\alpha_j\otimes T_jt}_{\text{Summing to }At})+{} \\ &{}-\sum_{i,j}h(\alpha_i\wedge d(T_is)_j\otimes e_j,t)-\sum_{i,j}h(\underbrace{\alpha_i\wedge\alpha_j\otimes T_jT_is}_{\text{These sum to }A\wedge A(s)},t)+{} \\ &{}-\sum_{i,j}h(\underbrace{\alpha_i\otimes T_is}_{\text{Summing to }As},\overbrace{dt_i\otimes e_j}^{\text{Summing to }dt})-\sum_{i,j}h(\alpha_i\otimes T_is,\underbrace{\alpha_j\otimes T_jt}_{\text{Summing to }At})+h(dA(s),t)={} \\ {}={}&h(A\wedge ds,t)-h(ds,dt)-h(ds,At)+{} \\ &{}-\sum_{i,j}h(\alpha_i\wedge T_i(ds)_j\otimes e_j,t)-h(A\wedge A(s),t)+{} \\ &{}-h(As,dt)-h(As,At)-h(dA(s),t)={} \\ {}={}&h(A\wedge ds-A\wedge A(s),t)-h(ds,dt+At)+{} \\ &{}-\sum_{i,j}h(\alpha_i\wedge ds_j\otimes T_ie_j,t)-h(As,dt+At)+h(dA(s),t)={} \\ {}={}&h(A\wedge ds-A\wedge A(s)-A\wedge ds+dA(s),t)-h(ds+As,dt+At)={} \\ {}={}&h(dA(s)-A\wedge As,t)-h(\nabla s,\nabla t). \end{align*}
Uh-oh. A sign is wrong. All signs appeared when I commuted wedges of 1-forms to have those forms in the right order. Being smart, I quickly obtain:
\begin{align*} dh(s,\nabla t)={}&d(h(s,\nabla t))=d\overline{h(\nabla t,s)}=\overline{d(h(\nabla t,s))}=\overline{dh(\nabla t,s)}={} \\ {}={}&\overline{h(dA(t)-A\wedge At,s)-h(\nabla t,\nabla s)}=\overline{h(dA(t)-A\wedge At,s)}-\overline{h(\nabla t,\nabla s)}={} \\ {}={}&h(s,dA(t)-A\wedge At)-h(\nabla s,\nabla t), \end{align*}
which gives another wrong sign. So where did I go wrong in the above?
Edit
I had meant to explain this above but I forgot. We may assume $T_i$ are represented by constant matrices, and denote their entries by $T^i_{jk}$ ($i$ for the $i$ in $T_i$, $jk$ for the entry indices in the matrix). The passage where I moved $d$ inside $T_i$ (i.e. $d(T_is)_j=T_i(ds)_j$) is thus justified:
\begin{align*} \alpha_i\wedge d(T_is)_j\otimes e_j={}&\alpha_i\wedge d\left(\sum_kT^i_{jk}s_ke_j\right)_j\otimes e_j=\alpha_i\wedge d\left(\sum_kT^i_{jk}s_k\right)\otimes e_j={} \\ {}={}&\alpha_i\wedge\left(\sum_kT^i_{jk}ds_k\right)\otimes e_j=\sum_k\alpha_i\wedge ds_k\otimes T^i_{jk}e_j, \end{align*}
now we remember we had a sum over $i,j$ outside, and sum over $j$ to get $T_ie_k$ on the right, and that would give us $A\wedge ds$.
I also just realized the extension is supposed to conjugate the form part of the second entry, i.e.:
$$h(\alpha\otimes s,\beta\otimes t)=h(s,t)\alpha\wedge\overline\beta.$$
I need that for the proof of the existence of the Chern connection to work. This only means there are a bunch of comnjugation bars missing from the above, but it does not harm the results, since the form parts were always taken out and then back in, and commuting wedges yields the same sign even if there is conjugation involved.
Now let me just fix a couple indices above and then post this edit.
I just redid the computations for $F_\nabla=dA+A\wedge A$, and realized the same sign problem was in there too, so I guess the first sign error was trusting memory to b right and mis-defining $A\wedge A(s)$. Define:
$$A\wedge A(s)=\sum_{i,j}\alpha_j\wedge\alpha_i\otimes T_jT_is,$$
and that sign fixes itself. Still, though now I correctly get:
$$dh(\nabla s,t)=h(F_\nabla s,t)-h(\nabla s,\nabla t),$$
I still get:
$$dh(s,\nabla t)=h(s,F_\nabla t)-h(\nabla s,\nabla t),$$
with the $h(\nabla s,\nabla t)$ not canceling out when I sum. Let me redo the "smart" computations with more care:
\begin{align*} dh(s,\nabla t)={}&d(h(s,\nabla t))=d\overline{h(\nabla t,s)}=\overline{d(h(\nabla t,s))}=\overline{dh(\nabla t,s)}={} \\ {}={}&\overline{h(dA(t)+A\wedge At,s)-h(\nabla t,\nabla s)}=\overline{h(dA(t)-A\wedge At,s)}-\overline{h(\nabla t,\nabla s)}={} \\ {}={}&\overline{\sum_ih(d\alpha_i\otimes T_it,s)}+\overline{\sum_{i,j}h(\alpha_i\wedge\alpha_j\otimes T_iT_jt,s)}+{} \\ &{}-\overline{\sum_{i,j}h(dt_i\otimes e_i+\alpha_i\otimes T_it,ds_j\otimes e_j+\alpha_j\otimes T_js)}={} \\ {}={}&\sum_i\overline{h(T_it,s)}\overline{d\alpha_i}+\sum_{i,j}\overline{h(T_iT_jt,s)}\overline{\alpha_i\wedge\alpha_j}+{} \\ &{}-\sum_{i,j}[\overline{h(e_i,e_j)}\overline{dt_i\wedge\overline{ds_j}}+\overline{h(e_i,T_js)}\overline{dt_i\wedge\overline{\alpha_j}}+{} \\ &{\hspace{1cm}}+\overline{h(T_it,e_j)}\overline{\alpha_i\wedge\overline{ds_j}}+\overline{h(T_it,T_js)}\overline{\alpha_i\wedge\overline{\alpha_j}}]={} \\ {}={}&\sum_ih(s,T_it)\overline{d\alpha_i}+\sum_{i,j}h(s,T_iT_jt)\overline{\alpha_i\wedge\alpha_j}+{} \\ &{}-\sum_{i,j}[h(e_j,e_i)\overline{dt_i}\wedge ds_j+h(T_js,e_i)\overline{dt_i}\wedge\alpha_j+{} \\ &{\hspace{1cm}}+h(e_j,T_it)\overline{\alpha_i}\wedge ds_j+h(T_js,T_it)\overline{\alpha_i}\wedge\alpha_j={} \\ {}={}&\sum_ih(s,d\alpha_i\otimes T_it)+\sum_{i,j}h(s,\alpha_i\wedge\alpha_j\otimes T_iT_jt)+{} \\ &{}+\sum_{i,j}[h(e_j,e_i)ds_j\wedge\overline{dt_i}+h(T_js,e_i)\alpha_j\wedge\overline{dt_i}+{} \\ &{\hspace{1cm}}+h(e_j,T_it)ds_j\wedge\overline{\alpha_i}+h(T_js,T_it)\alpha_j\wedge\overline{\alpha_i}={} \\ {}={}&h(s,dA(t))+h(s,A\wedge A(t))+{} \\ &{}+\sum_{i,j}[h(ds_j\otimes e_j,dt_i\otimes e_i)+h(\alpha_j\otimes T_js,dt_i\otimes e_i)+{} \\ &{\hspace{1cm}}+h(ds_j\otimes e_j,\alpha_i\otimes T_it)+h(\alpha_j\otimes T_js,\alpha_i\otimes T_it)={} \\ {}={}&h(s,dA(t)+A\wedge A(t))+h(ds,dt)+h(As,dt)+h(ds,At)+h(As,At)={} \\ {}={}&h(s,F_\nabla t)+h(\nabla s,dt)+h(\nabla s,At)={} \\ {}={}&h(s,F_\nabla t)+h(\nabla s,\nabla t). \end{align*}
This fixes the problem. The TL;DR version of the above computation mess is that, if $\alpha\in\mathcal A^k(X),\beta\in\mathcal A^h(X)$, then:
\begin{align*} \overline{h(\alpha\otimes s,\beta\otimes t)}={}&\overline{h(s,t)}\overline{\alpha\wedge\overline\beta}=h(t,s)\overline\alpha\wedge\beta=(-1)^{kh}h(t,s)\beta\overline\alpha=(-1)^{kh}h(\beta\otimes t,\alpha\otimes s), \end{align*}
otherwise known as:
$$h(\beta\otimes t,\alpha\otimes s)=(-1)^{kh}h(\alpha\otimes s,\beta\otimes t),$$
which generalizing to the fact that, if $s\in\mathcal A^k(E),t\in\mathcal A^h(E)$, then:
$$h(s,t)=(-1)^{kh}h(t,s),$$
and this was the second sign problem: commuting inside $h$ does not only conjugate, there is a sign factor.