7
$\begingroup$

Let $L$ be a holomorphic line bundle on a complex manifold $X$, and assume it is equipped with a singular hermitian metric $h$ with local weight $\varphi$. Then, one can show that the de Rham class of $\frac{i}{\pi}\partial \overline{\partial} \varphi$ coincides with the first Chern class $c_1(L)$ of the line bundle.

Is there a generalization of this result to higher-rank vector bundles?

For example, let $E$ be a holomorphic vector bundle of rank $r$ on a complex manifold $X$ and let $h$ be a singular hermitian metric on $E$. Can we describe the Chern clases $c_1(E),\ldots,c_r(E)$ in terms of $h$ and its curvature form?

$\endgroup$
7
$\begingroup$

Given a complex hermitian vector bundle $E$ of complex rank n over a smooth manifold $M$, a representative of each Chern class $ck(E)$ of $E$ are given as the coefficients of the characteristic polynomial of the curvature form $Ω$ of $E$.

$$\det \left(\frac {it\Omega}{2\pi} +I\right) = \sum_k c_k(E) t^k $$

Each Chern class $c_k$ is a real cohomology class

this expression for the Chern form expands as

$$ \sum_k c_k(E) t^k = \left[ I + i \frac{\mathrm{tr}(\Omega)}{2\pi} t + \frac{\mathrm{tr}(\Omega^2)-\mathrm{tr}(\Omega)^2}{8\pi^2} t^2 + i \frac{-2\mathrm{tr}(\Omega^3)+3\mathrm{tr}(\Omega^2)\mathrm{tr}(\Omega)-\mathrm{tr}(\Omega)^3}{48\pi^3} t^3 + \cdots \right]. $$

Let $S\in H^0(X,E)$ be a non-zero holomorphic section

Then the Poincaré-Lelong formula says that

$$dd^c\log ||S||=[Z_S]-\frac{\sqrt{-1}}{2\pi}\Theta(E)$$

where $\Theta$ is the Chern curvature. See http://www.mathunion.org/ICM/ICM1994.2/Main/icm1994.2.0817.0827.ocr.pdf

and http://arxiv.org/abs/alg-geom/9708003

In singular setting:

On a holomorphic line bundle a hermitian metric $h$ is just a scalar-valued function so that $\Theta=\bar\partial\partial \log h$ as long as $\log h\in L_{loc}^1(X)$ But for holomorphic vector bundles with rank $E ≥ 2 $ it is not clear what the appropriate notions of connection and curvature associated with $h$ are.

Let $h$ be a singular hermitian metric on a holomorphic vector bundle $E$ , and assume that $h$ is negatively curved in the sense of Griffiths. Then $\log \det h$ is a plurisubharmonic function and if $\det h\neq 0$ then $\log\det h\in L_{loc}^1(X)$ due to the proposition 1.1 of the nice paper http://link.springer.com/article/10.1007%2Fs11512-015-0212-4

$\endgroup$
5
$\begingroup$

As Hassan mentions, in the setting of singular metrics on vector bundles, the notion of curvature appears problematic, which is discussed in the paper of Raufi.

Still, as is also discussed in that paper, in the case that the singular metric is positively or negatively curved in the sense of Griffiths, one can give a reasonable meaning to $c_1(E,h)$ as a current by $c_1(E,h) := dd^c \log (\det h)$. This gives a way of defining the first Chern class without having defined a curvature matrix, but which coincides with the usual definition when $h$ is smooth.

Me, Raufi, Ruppenthal and Sera do a variant of this also for higher Chern classes, through local regularizations of the metric, under some additional assumption that the singularities of the metric are not too "big". This is so far not completely written up, but a discussion of it can be seen at https://www.birs.ca/events/2016/5-day-workshops/16w5080/videos/watch/201605020920-Larkang.html

The precise statement of our main result is discussed at around 37:40, and basically, in the positively curved case, we can define $c_k(E,h)$ if the set $L(\log (\det h^*))$ is contained in a variety of codimension $\geq k$, where $L(\log(\det h^*))$ is the complement of the set where $\log(\det h^*)$ is locally bounded.

By taking the "diagonal" metric $h = e^{-\varphi} Id$ on a trivial vector bundle, where $\varphi$ is a plurisubharmonic function, one sees that one should expect to need some condition on the singularities of $h$, since $(dd^c \varphi)^k$ does typically not have a natural meaning when $\varphi$ is plurisubharmonic and unbounded, and $k$ is too large.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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