Decomposition of hermitian form used in the definition of Griffiths/Nakano positivity Let $E$ be a hermitian holomorphic vector bundle over a complex manifold $X$. Then $\Theta(E)$, the curvature of $E$, is a section of $\bigwedge^{1,1}X\otimes\operatorname{End}(E)$. However, we have the following isomorphism: $$\bigwedge\nolimits^{\!1,1}X\otimes\operatorname{End}(E) \cong \left(T^{1,0}X\otimes E\otimes\overline{T^{1,0}X\otimes E}\right)^*.$$ Under this isomorphism, $\Theta(E)$ corresponds to a hermitian form $\theta_E$ on $T^{1,0}X\otimes E$. This is used to define Griffiths and Nakano positivity of $E$.
Now suppose $E$ and $F$ are two hermitian holomorphic vector bundles over $X$. The curvature satisfies $\Theta(E\otimes F) = \Theta(E)\otimes\operatorname{id}_F + \operatorname{id}_E\otimes\Theta(F)$. 

Is there a similar decomposition for $\theta_{E\otimes F}$ under the above isomorphism? If not, what about the case where $F$ is a line bundle?

I've been playing around with tensor products for a while now but I can't seem to get it straight.
 A: With your notations, the hermitian form $\theta_E$ on $T_X\otimes E$ defined by $\Theta_E$ is given in a somewhat more extrinsic way by
$$
\theta_E(v\otimes\sigma,v\otimes\sigma):=h(\Theta_E(v,\bar v)\cdot \sigma,\sigma), 
$$ 
where


*

*$h$ is the hermitian metric on $E$,

*$v\in T_X$, so that $\bar v\in \overline{T_X}$ (or, if you want, $v\in T^{1,0}_X$ so that $\bar v\in T^{0,1}_X$),

*$\sigma\in E$. 


Note that I gave the formula just for decomposable (rank one) tensors, but then you extend it to all tensors by sesquilinearity.
Now, let $(E,h_E)$ and $(F,h_F)$ be two holomorphic hermitian vector bundles. Take the product metric $h_{E\otimes F}=h_{E}\otimes h_F$ on their tensor product so that, as you said, the curvature of the corresponding Chern connection is given by
$$
\Theta(E\otimes F)=\Theta(E)\otimes\operatorname{Id}_F+\operatorname{Id}_E\otimes\Theta(F).
$$
Then, for $v\in T_X$, $\sigma\in E$ and $\tau\in F$, you get
$$
\begin{aligned}
\theta_{E\otimes F}(v\otimes\sigma\otimes\tau,v\otimes\sigma\otimes\tau) &=
h_{E\otimes F}(\Theta_{E\otimes F}(v,\bar v)\cdot(\sigma\otimes\tau),\sigma\otimes\tau) \\
&=h_{E\otimes F}((\Theta(E)(v,\bar v)\otimes\operatorname{Id}_F \\
&\qquad\qquad\qquad\qquad\qquad+\operatorname{Id}_E\otimes \Theta(F)(v,\bar v))\cdot(\sigma\otimes\tau),\sigma\otimes\tau) \\
&=h_{E\otimes F}((\Theta(E)(v,\bar v)\cdot\sigma)\otimes\tau+\sigma\otimes(\Theta(F)(v,\bar v)\cdot\tau),\sigma\otimes\tau) \\
&=h_{E\otimes F}((\Theta(E)(v,\bar v)\cdot\sigma)\otimes\tau,\sigma\otimes\tau) \\
&\qquad+h_{E\otimes F}(\sigma\otimes(\Theta(F)(v,\bar v)\cdot\tau),\sigma\otimes\tau) \\
&=h_E(\Theta(E)(v,\bar v)\cdot\sigma,\sigma)h_F(\tau,\tau) \\
&\qquad+h_E(\sigma,\sigma)h_F(\Theta(F)(v,\bar v)\cdot\tau,\tau) \\
&=\theta_E(v\otimes\sigma,v\otimes\sigma)h_F(\tau,\tau)+h_E(\sigma,\sigma)\theta_F(v\otimes\tau,v\otimes\tau).
\end{aligned}
$$
Thus, the decomposition you were looking for is perhaps the most natural one:
$$
\theta_{E\otimes F}=\theta_E\otimes h_F+h_E\otimes\theta_F.
$$
EDIT: Re-edited the answer coherently with this other answer.  
