I'm trying to prove the following:



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> Let $L$ be a positive holomorphic line bundle on a compact complex manifold $X$. For any hermitian holomorphic vector bundle $E$ on $X$, there is $k \in \mathbb{N}$ such that $E\otimes L^k$ is Nakano positive.
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> 
> 


I'm using the decomposition given by diverietti in [this answer](https://mathoverflow.net/a/133376/21564). Part of my problem is that (I think) I can find a sufficiently large choice of $k$ so that $E\otimes L^k$ is Nakano positive at a point, but I can't extend beyond that. If I try to prove that I can find such a $k$ locally, my bounds are no longer valid. Maybe I need to use the fact that $X$ is locally compact.



> 
> Does anyone have a reference (or proof) of the above fact?
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**Update:** I'm not actually sure that the statement I'm trying to prove is true, though I feel like it should be.



If $L$ is positive, $E\otimes L^k$ is Nakano positive for some $k$
This follows by the following general, quite elementary 


**Fact.** *Suppose you have a complex vector bundle $E\to X$ on a compact manifold $X$ and two hermitian form $h\_1$ and $h\_2$ on $E$, such that the first one is positive definite. Then, there exists a constant $C\_0>0$ such that for all $C\ge C\_0$ the hermitian form $C\,h\_1+h\_2$ is positive definite.*


*Proof.* Since both $h\_1$ and $h\_2$ are homogeneous, it suffices to check the statement on the unitary (with respect to $h\_1$) bundle $U(E,h\_1)\to X$, defined as the set of elements in $E$ of $h\_1$-norm equal to one. $U(E,h\_1)$ is obviously compact, since $X$ is. Let $m\_i$ be the minimum of $h\_i$ on $U(E,h\_1)$. By compactness, $m\_1>0$. We can also suppose $m\_2<0$, otherwise we are done by taking any $C\_0>0$. Now, take $C\_0$ to be any real number $>-m\_2/m\_1$. Then, for any $v\in E\setminus\{0\}$, you have
$$
\begin{aligned}
C\_0\,h\_1(v,v)+h\_2(v,v) & =||v||^2\_{h\_1}\bigl(C\_0\,h\_1(v/||v||\_{h\_1},v/||v||\_{h\_1})+h\_2(v/||v||\_{h\_1},v/||v||\_{h\_1})\bigr) \\
& \ge ||v||^2\_{h\_1}(C\_0\,m\_1+m\_2)>0. 
\end{aligned}
$$ 


Now, remark that $h\_E\otimes\theta\_{L}$ defines a positive definite hermitian form on $T\_X\otimes E\otimes L$ (this will be morally our $h\_1$) since by hypothesis $(L,h\_L)$ is positive, then apply the above fact with the other hermitian form defined by $\theta\_E\otimes h\_L$ (this will be morally our $h\_2$). Let $C\_0$ be the constant you get such that
$$
C\_0\,h\_E\otimes\theta\_{L}+\theta\_E\otimes h\_L
$$
is positive definite on $T\_X\otimes E\otimes L$ and $k$ any integer greater than $C\_0$. 


Next, by the decomposition you cited, we have
$$
\theta\_{E\otimes L^{\otimes k}}=\theta\_E\otimes h\_{L^{\otimes k}}+h\_E\otimes\theta\_{L^{\otimes k}}
$$ 


**Claim.** *The above expression defines a positive hermitian form on $T\_X\otimes E\otimes L^{\otimes k}$. Consequently $E\otimes L^{\otimes k}$ is Nakano positive.* 


*Proof.* Without loss of generality, we can suppose $k\ge 2$. Then, the claim follows from the following elementary identity (which will be proven subsequently):
$$
\theta\_{L^{\otimes k}}=k\,\theta\_L\otimes h\_{L^{\otimes(k-1)}}.
$$
Indeed, you now get
$$
\begin{aligned}
\theta\_{E\otimes L^{\otimes k}} &=
\theta\_E\otimes h\_{L^{\otimes k}}+h\_E\otimes\theta\_{L^{\otimes k}}\\
&= \theta\_E\otimes h\_L\otimes h\_{L^{\otimes(k-1)}}+k\,h\_E\otimes \theta\_L\otimes h\_{L^{\otimes(k-1)}}\\
&=(\theta\_E\otimes h\_L+k\,h\_E\otimes \theta\_L)\otimes h\_{L^{\otimes(k-1)}}\\
&\ge(\theta\_E\otimes h\_L+C\_0\,h\_E\otimes\theta\_{L})\otimes h\_{L^{\otimes(k-1)}},
\end{aligned}
$$
and the latter is clearly a positive definite hermitian form on $T\_X\otimes E\otimes L^{\otimes k}$.


Coming back to the identity, notice first that whenever $F$ is a line bundle, on $T\_X\otimes F$ we have
$$
\theta\_F=h\_F(\Theta\_F(\bullet,\overline\bullet)\,\bullet,\bullet)=\Theta\_F\otimes h\_F,
$$
since $\Theta\_F$ is just a $(1,1)$-form (seen as a hermitian operator on $T\_X$).
Thus, we obtain
$$
\begin{aligned}
\theta\_{L^{\otimes k}} &=\Theta\_{L^{\otimes k}}\otimes h\_{L^{\otimes k}}\\
& =k\,\Theta\_{L}\otimes h\_L\otimes h\_{L^{\otimes(k-1)}}\\
&=k\,\theta\_L\otimes h\_{L^{\otimes(k-1)}}.
\end{aligned}
$$


