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.



Decomposition of hermitian form used in the definition of Griffiths/Nakano positivity
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](https://mathoverflow.net/questions/134232/local-expression-involved-in-the-definition-of-positivity-of-vector-bundles/134569#134569) other answer. 


