# Local expression involved in the definition of positivity of vector bundles

This is question follows on from this one.

In the linked question, the hermitian form $\theta_E$ on $T^{1,0}X\otimes E$ is defined globally as $$\theta_E(v\otimes\sigma,v\otimes\sigma):=h(i\Theta_E(v,\bar v)\cdot \sigma,\sigma).$$ Previous to seeing that answer, I had only seen a local expression for $\theta_E$ in the work of Demailly. Having never seen the above expression before, I consulted the references I had seen which contained the local expressions for $\theta_E$ and found two slightly different expressions.

In his online book Complex Algebraic and Differential Geometry (available here), Demailly write on page 338

$$i\Theta(E) = ic_{jk\lambda\mu}dz_{j}\wedge d\bar{z}_k\otimes e^{*}_{\lambda}\otimes e_{\mu}$$

where $\Theta(E)$ is the curvature of $E$. However, in his lecture notes $L^2$ vanishing theorems for positive line bundles and adjunction theory (available here), he writes on page 24

$$i\Theta(E) = c_{jk\lambda\mu}dz_{j}\wedge d\bar{z}_{k}\otimes e^{*}_{\lambda}\otimes e_{\mu}.$$

In both instances, he goes on to define a hermitian form by $$\zeta\otimes v \mapsto c_{jk\lambda\mu}\zeta^j\bar{\zeta}^kv^\lambda\bar{v}^\mu$$ which he calls $\theta_{E}$ in the book and $\tilde{\Theta}(E)$ in the lecture notes, but I'm sure they are intended to be the same as they are both used to define Griffiths and Nakano positivity in their respective documents.

If I have not made an error myself, I believe one of the two expressions has a typographical error. So my question is:

Which local expression is correct?

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My answer will consist mainly of a collection of trivial facts but which nonetheless often generate some confusion.

I begin by fixing some notation. Let $V$ be a complex vector space of complex dimension $n$, and $V^{\mathbb R}$ its underlying real vector space, together with the complex structure $J$ given by the multiplication by $i$ in $V$. Let $V^{\mathbb C}:=V\otimes_{\mathbb R}\mathbb C$ the complexification of $V$ and $J^{\mathbb C}=J\otimes\operatorname{Id}_{\mathbb C}$ the corresponding complexification of $J$. Finally, let $V^{1,0}\subset V^{\mathbb C}$ (resp. $V^{0,1}$) the eigenspace relative to the eigenvalue $i$ (resp. $-i$) of the operator $J^{\mathbb C}$. They are $\mathbb R$-isomorphic (and $\mathbb C$-antiisomorphic) via the conjugation. Fix the complex linear isomorphism \begin{aligned} \phi\colon & V\overset{\simeq}\to V^{1,0} \cr & v\mapsto \frac 12(v-iJv). \end{aligned} Now, suppose you have a symmetric sesquilinear form $h$ on $V$. Then, its real part $g=\Re h$ defines a symmetric bilinear form on $V^{\mathbb R}$ which is moreover $J$-invariant. Conversely, given a $J$-invariant symmetric bilinear form $g$ in $V^{\mathbb R}$ you can build a (unique) symmetric sesquilinear form $h$ on $V$ whose real part is $g$: just take $$h(\bullet,\bullet):=g(\bullet,\bullet)-ig(J\bullet,\bullet).$$ Next, given such a $J$-invariant $g$ (or, equivalently, such a $h$), consider its $\mathbb C$-bilinear extension $g^{\mathbb C}$ to $V^{\mathbb C}$. Since on $V^{\mathbb C}$ there exist a natural conjugation, we can define a symmetric sesquilinear form $H$ on $V^{\mathbb C}$ by setting $$H(\bullet,\bullet):=g^{\mathbb C}(\bullet,\overline\bullet).$$ If, by abuse of notation, we still call $H$ the restriction $H|_{V^{1,0}}$ of $H$ to $V^{1,0}$, it is straightforward to check that $$H(\phi(v),\phi(w))=\frac 12 h(v,w).$$ Of course, starting from a symmetric sesquilinear form $H$ on $V^{1,0}$ you can recover the corresponding $h$ and $g$.

We next pass to (minus) the imaginary part $\eta:=-\Im h$ of $h$. It is a skew-symmetric $2$-form on $V^{\mathbb R}$. Call $\omega$ its $\mathbb C$-bilinear extension to $V^{\mathbb C}$. It is straightforward to check that it is real, that is $\overline{\omega(\bullet,\bullet)}=\omega(\overline\bullet,\overline\bullet)$ and that it is of type $(1,1)$: $$\omega(\phi(v),\phi(w))=\omega\bigl(\overline{\phi(v)},\overline{\phi(w)}\bigr)=0,$$ for all $v,w\in V$. Conversely, given a real $(1,1)$-form $\omega$ on $V^{\mathbb C}$ you can recover $h$ (and thus $H$) simply by $$h(\bullet,\bullet)=2H\bigl(\phi(\bullet),\phi(\bullet)\bigr)=-2i\,\omega\bigl(\phi(\bullet),\overline{\phi(\bullet)}\bigr).\qquad(*)$$

All this said, let's pass now to curvature and vector bundles. The reason why usually one considers $i$ times the (Chern) curvature is because this makes the curvature a $(1,1)$-form with values in the hermitian operators acting on the hermitian vector bundle. That is, if $\langle\bullet,\bullet\rangle$ is the hermitian metric on the vector bundle $E$, then $$\langle i\Theta(E)\cdot\sigma,\tau\rangle=\langle\sigma,i\Theta(E)\cdot\tau\rangle,$$ the equality intended to be as an equality of $(1,1)$-forms on the complex manifold $X$. Now, if $\sigma=\tau$ and the curvature is contracted with the hermitian metric, you are left with a real $(1,1)$-form: indeed you have $$\overline{\langle i\Theta(E)\cdot\sigma,\sigma\rangle}=\langle\sigma,i\Theta(E)\cdot\sigma\rangle=\langle i\Theta(E)\cdot\sigma,\sigma\rangle.$$ By the previous discussion, you want to think about it as a symmetric sesquilinear form on $(1,0)$-vectors: this is often implicit in the formulae! By $(*)$, what you need is to multiply it by $-i$. So, the correct expression for $\theta_E(v\otimes\sigma,v\otimes\sigma)$, where $\sigma\in E$ and $v\in T^{1,0}_X$ is $$-i\langle i\Theta(E)(v,\overline v)\cdot\sigma,\sigma\rangle=\langle \Theta(E)(v,\overline v)\cdot\sigma,\sigma\rangle.$$ This also gives some precisions to this previous answer of mine (please note that, up to this factor of $-i$, my answer in question remains valid!).

In particular, the right local expression (whenever $E$ is endowed with a local orthonormal frame) is $$v\otimes\sigma\mapsto \sum_{j,k,\lambda,\mu}c_{jk\lambda\mu}v_j\overline v_k\sigma_\lambda\overline\sigma_\mu.$$
Note that this is a real number, since by the hermitian property of ($i$ times) the Chern curvature, in a local orthonormal frame for $E$ the curvature coefficients satisfy the hermitian relations $$\overline c_{jk\lambda\mu}=c_{kj\mu\lambda}.{}{}{}$$

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So does that mean in your previous answer you want $\theta_E(v\otimes\sigma, v\otimes\sigma) = h(\Theta(E)(v, \bar{v})\cdot\sigma, \sigma)$? Furthermore, just to be clear, the coefficients $c_{jk\lambda\mu}$ come from $i\Theta(E)$ not $\Theta(E)$, correct? –  Michael Albanese Jul 4 '13 at 4:22
Exactly. Moreover, the coefficients $c_{jk\lambda\mu}$ come from $\Theta(E)$. I mean that $\Theta(E)=\sum c_{jk\lambda\mu}dz_j\wedge d\bar z_k\otimes e_\lambda^*\otimes e_\mu$, and these $c_{jk\lambda\mu}$ are the ones which satisfy the hermitian relations above. –  diverietti Jul 5 '13 at 15:22