MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 2 down vote accepted

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 \begin{equation} h(\bullet,\bullet)=2H\bigl(\phi(\bullet),\phi(\bullet)\bigr)=-2i\,\omega\bigl(\phi(\bullet),\overline{\phi(\bullet)}\bigr).\qquad(*) \end{equation}

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 $$ \overline c_{jk\lambda\mu}=c_{kj\mu\lambda}.{}{}{} $$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.