research41 min studyLesson 948 in Mathematics

Chern curvature and positivity

Read positivity in complex tangent directions, distinguish line and vector-bundle notions, and integrate a correctly normalized Chern form.

Before you begin

On this page

The question for today

What you’ll understand

Test line-bundle curvature positivity using the chosen exponential-weight convention.
Distinguish Griffiths positivity from the stronger Nakano condition.
Compute a projective-line degree and identify the canonical twist in Kodaira vanishing.

Every Hermitian metric has positive squared norms, but its Chern curvature can be positive, negative, or indefinite. Curvature positivity measures second derivatives in complex directions, not positivity of the metric matrix itself. For line bundles the criterion is a positive real one-one form. For higher rank there are distinct tensor tests, and the direction of implication matters. We keep the curvature sign and two-pi normalization fixed while connecting local positivity to global degree and compact vanishing results.

A positive weight need not have positive curvature

For a line weight exponential minus phi, the curvature is partial bar-partial phi, and multiplying by i gives a real one-one form. Positive curvature means this form evaluates positively on every nonzero real vector and its J-rotation. Equivalently the complex Hessian of phi is positive definite. The scalar metric weight is always positive for real phi, even when that Hessian is negative. Reversing the sign of phi inverts the metric weight and reverses its curvature, as expected for a dual line.

Tensor powers multiply line curvature by the integer power, and tensor products add curvatures. Their normalized real Chern classes follow the same operations. Positivity is a condition in every nonzero tangent direction; a positive trace or one positive directional restriction cannot establish it. Semipositivity allows zero directions and must not be silently promoted to strict positivity. On a compact complex manifold a line bundle with a strictly positive curvature metric yields a Kähler form, but a general Kähler class need not be an integral Chern class.

H=e^{-\phi},\quad iF_L=i\partial\bar\partial\phi>0\iff(\phi_{j\bar k})>0,\qquad F_{L^{\otimes m}}=mF_L,\quad F_{L^*}=-F_L,\qquad c_1(L)_{\mathbb R}=\left[\frac{iF_L}{2\pi}\right]

The positivity sign refers to a real one-one form on complex tangent directions, not to the already-positive scalar H. The Chern representative is divided by two pi.↔ Scroll wide equations horizontally.

Vector positivity has different quantifiers

For a Hermitian holomorphic vector bundle, curvature defines a Hermitian form on the tensor product of tangent space and bundle fiber. Griffiths positivity tests that form on nonzero decomposable tensors consisting of one tangent vector and one fiber vector. Nakano positivity tests all nonzero tensors, including sums that do not factor. Consequently Nakano positivity implies Griffiths positivity, while the reverse implication is not generally valid. For rank one the distinction disappears because the fiber factor is one dimensional.

These definitions explain why a line-bundle argument cannot automatically be copied to every vector bundle. They also prevent a misleading duality shortcut: the simple line sign reversal does not by itself identify all higher-rank Nakano duality properties. The examples here remain scalar so that positivity can be checked directly from a Hessian. The vector definitions are stated to mark the boundary of that calculation, not to claim that computing a determinant or a few decomposable directions settles the stronger tensor condition.

\Theta_E\text{ Griffiths positive}:\ \Theta_E(\xi\otimes s,\xi\otimes s)>0;\qquad \Theta_E\text{ Nakano positive}:\ \Theta_E(u,u)>0\ (u\ne0);\qquad \text{Nakano}\Longrightarrow\text{Griffiths}

Theta here denotes the Hermitian curvature form associated with iF, not a new sign convention for F. In the first condition both xi and s are nonzero; in the second u ranges over the entire tangent-fiber tensor product.↔ Scroll wide equations horizontally.

Compact consequences need the correct bundle

Kodaira vanishing, in the form used here, says that for a positive holomorphic line L on a compact Kähler manifold, the higher sheaf cohomology of the canonical bundle tensored with L vanishes. The canonical factor is essential to this immediate statement; it is not a license to erase all higher cohomology of an arbitrary positive bundle with no twist. The result is invoked from the curvature identity and elliptic theory, rather than proved by the elementary degree integral below.

Kodaira embedding says that a positive holomorphic line bundle on a compact complex manifold has a sufficiently large tensor power whose sections define a projective embedding. Strict positivity and compactness matter. A local potential or a positive restriction along a single curve is far less information. Conversely, the standard positive metrics on projective hyperplane powers offer concrete normalizations: their curvature integrals over the projective line equal their integer degrees. This supplies a reliable sign check before applying more abstract positivity theorems.

Work it through

A degree-three positive line on the projective line

On the affine chart of the projective line use the metric $H=(1+|z|^2)^{-3}$ for $\mathcal O(3)$ . Compute its curvature, verify positivity, and integrate its normalized Chern form.

The potential is $\phi=3\log(1+|z|^2)$ . Differentiating gives $F_L=3(1+|z|^2)^{-2}dz\wedge d\bar z$ , so $iF_L$ is strictly positive.
With reciprocal coordinate w and frame change $e_w=z^3e_z$ , the weight becomes $(1+|w|^2)^{-3}$ . Thus the metric and curvature extend smoothly across infinity rather than being defined only on the affine chart.
Writing z in polar coordinates gives $\int iF_L=\int_0^{2\pi}\int_0^\infty 6r(1+r^2)^{-2}\,dr\,d\theta=6\pi$ . Dividing by $2\pi$ gives degree three. The dual has degree minus three and negative curvature.

The result

The metric has positive curvature in every complex direction, and the normalized integral recovers the integer degree with the correct dual sign.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Metric weights stay positive while curvature changes sign

Pause and Reset, set both $a=1,x=0.5$ , and predict the real-slice value one quarter and slope one. Regard the separately defined line potential as $\phi=a|z|^2$ , whose curvature coefficient is a. Reset and set both $a=-1,x=0.5$ , predicting value minus one quarter and slope minus one. The weight $e^{-\phi}$ remains positive in both cases, but curvature reverses sign. The lab plots a real slice, not bundle curvature. Edits retain time; Reset clears both controls and clock.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forPositivity of squared norms is automatic for an exponential weight; positivity of curvature is a second-derivative condition.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Nakano tests every tensor, so it includes the decomposable tensors used by Griffiths. The converse is not automatic, and neither determinant nor metric positivity supplies these curvature tests.

Your turn

Try it yourself

For a trivial line on complex two-space use $H=\exp(-2|z^1|^2+|z^2|^2)$ . Compute i times its Chern curvature, decide its positivity, and compare restriction to the first coordinate line with the whole-space test.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

First express H as exponential minus phi. The two diagonal complex-Hessian entries then give different signs.

Compare with the solution

Here $\phi=2|z^1|^2-|z^2|^2$ , so $iF=i(2\,dz^1\wedge d\bar z^1-dz^2\wedge d\bar z^2)$ . Its Hessian eigenvalues are two and minus one, making curvature indefinite, although H is everywhere a positive metric weight. Restriction to the line with second coordinate zero is positive, but the second coordinate direction is negative, so the line bundle metric is not positive over the whole complex two-space.

Keep the idea

Take it with you

Positive Chern curvature tests every complex tangent direction, not the sign of a metric weight.
Nakano positivity is stronger than Griffiths positivity, and they coincide for lines.
Normalized curvature integrals recover degrees; compact positivity theorems require their precise twists and hypotheses.

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