I'm confused by the following statement in complex differential geometry:

Let $E$ be a holomorphic vector bundle and let $E'$ be a holomorphic sub-bundle. Fix a unitary connection $\nabla$ on $E$ associated to some hermitian metric.

We can compare the curvature $F_{\nabla}$ on $E$ to the curvature of the connection $\nabla'$ on the sub-bundle obtained by orthogonal projection with respect to the metric.

The statement I am confused by is that $F_{\nabla'} \leq F_{\nabla}.$ I.e. "curvature decreases in holomorphic sub-bundles". This statement can be found in Griffiths and Harris, and in many other standard sources.

I am confused for the following reason:

Consider the inclusion of vector bundles on $\mathbb{C} P^1$:

$0 \to O(1) \to O(1) \oplus O(-1).$

Since curvature is proportional to the first Chern class, and $c_1(O(1))> c(O(1) \oplus O(-1))=0,$ this seems to contradict the above principle.

What am I misunderstanding?

Note: I thought initially that my confusion was due to some normalization, but since one can also consider $0 \to O(-1) \to O(1) \oplus O(-1),$ it seems like my confusion is due to something else.

  • $\begingroup$ The Chern class of a line bundle is (up to a factor) its curvature. But how does this works for non line bundles $E$ ? The curvature is an element of $\Gamma(\Omega^2\otimes \operatorname{End}(E))$. This does not define an element of $H^2$ in an obvious way. The first Chern class is obtained by taking the trace (and the other Chern classes by taking the coefficients of the characteristic polynomial). $\endgroup$ – Roland Mar 3 '18 at 13:38
  • 2
    $\begingroup$ You might want to think about (and understand) this post. $\endgroup$ – Ted Shifrin Mar 4 '18 at 0:37

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