If $L$ is positive, $E\otimes L^k$ is Nakano positive for some $k$

I'm trying to prove the following:

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.

I'm using the decomposition given by diverietti in this answer. 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 If I try to prove that I can find such a $k$ locally, my bounds are no longer valid.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?

Update: I'm not actually sure that the statement I'm trying to prove is true, though I feel like it should be.

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Simone's decomposition shows that you can make your bundle positive in a small neighborhood of a point (use the unit sphere fibration in $T_X$ to extend from the point to a small nbh). Now cover $X$ by finitely many such neighborhoods. –  Gunnar Magnusson Aug 11 '13 at 15:22

1 Answer

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}

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"by definition $m_1 > 0$" should maybe be by compactness? (Hola big brother!) –  Gunnar Magnusson Aug 11 '13 at 15:54
Yes, thanks!! I'll edit it! (Hola little brother, where are you?!) –  diverietti Aug 11 '13 at 16:23
Fantastic. I was looking at using the sphere bundle (as can be seen from my last MSE question), but I couldn't quite get it to work. I was trying to take the sphere bundles associated to $T_X$, $E$ and $L^{\otimes k}$ respectively, rather than the bundle $T_X\otimes E\otimes L^{\otimes k}$. This led to problems when dealing with tensors with rank greater than one. –  Michael Albanese Aug 11 '13 at 16:35
Glad that you are satisfied! –  diverietti Aug 11 '13 at 16:39
You say that $m_1 > 0$ by compactness. Compactness guarantees that $m_1$ exists; $m_1 > 0$ as $h_1$ is positive definite. Without compactness, you could try defining $m_1$ to be the infimum of $h_1$, but then we could only say $m_1 \geq 0$. –  Michael Albanese Aug 12 '13 at 2:33