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.

  • $\begingroup$ It is not clear to me which kind of decomposition you are looking for. The curvature of $E\otimes F$ gives an hermitian form on $T_X\otimes E\otimes F$, but this you know. How do you want to decompose this latter tensor product? $\endgroup$ – diverietti Jun 11 '13 at 9:51

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.

  • $\begingroup$ If I'm not mistaken, you can use this decomposition to show that for any hermitian holomorphic vector bundle $E$ on a compact complex manifold with a positive line bundle $L$ (so it is algebraic by the Kodaira Embedding Theorem), there is some $k_0 \in \mathbb{N}$ such that $E\otimes L^k$ is Griffiths positive for all $k \geq k_0$. $\endgroup$ – Michael Albanese Jun 12 '13 at 8:13
  • $\begingroup$ More than this, you can use it to show that it is Nakano positive! $\endgroup$ – diverietti Jun 12 '13 at 11:36
  • $\begingroup$ Of course. Is there a slick way to see this? At the moment I'm using Cauchy-Schwarz and an operator norm, then by compactness, all the relative terms are bounded. Finally, as $\theta_{L^k} = k\theta_L$, it is just a matter of choosing $k$ large enough. It is precisely how Serre's Theorem is proved in Huybrechts' Complex Geometry, except there is no mention of $\theta_E$, but rather the endomorphism $[i\Theta(E), \Lambda] : \bigwedge^{p,q}X\otimes E \to \bigwedge^{p,q}X\otimes E$. Do you know if there is some direct relationship between $\theta_E$ and $[i\Theta(E), \Lambda]$? $\endgroup$ – Michael Albanese Jun 13 '13 at 6:45
  • $\begingroup$ What do you mean? The former is an hermitian form, the latter an operator acting on forms... $\endgroup$ – diverietti Jun 13 '13 at 9:14
  • $\begingroup$ I suppose I was wondering about the map $\theta_E$ and the map $h_E([i\Theta(E), \Lambda]\cdot,\cdot)$, but the latter does not take a tangent vector as an input. $\endgroup$ – Michael Albanese Jun 16 '13 at 3:49

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