# Terminology issue: meaning of 'ample class' ?

What is meant by an "ample class" in general? Motivation: In the document I am reading, the phrase in question is "fix an ample class $\alpha\in H^1(X,\Omega^1_X)$." I know what ampleness of a line bundle is. I have checked the only Wikipedia article that could be related (http://en.wikipedia.org/wiki/Ample_line_bundle). And I looked in Hartshorne.

Thanks.

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Charles' and Pete's answer are (almost) the same: First there is a map $\mathrm{dlog}\colon \mathcal{O}_X^\ast \rightarrow \Omega^1_X$ taking $f$ to $df/f$ (just to show that it also makes algebraic sense) which indeed induces a group homomorphism $H^1(X,\mathcal{O}_X^\ast)\rightarrow H^1(X,\Omega^1_X)$ giving one version of the Chern class. In the other version we have an exact sequence $0\rightarrow 2\pi i\mathbb Z\rightarrow \mathcal O_X\rightarrow \mathcal{O}_X^\ast\rightarrow0$ which gives a map $H^1(X,\Omega^1_X) \rightarrow H^2(X,2\pi i\mathbb Z)$. Combined with the inclusion $2\pi i\mathbb Z\subseteq\mathbb C$ and the projection on the $(1,1)$-part it gives the previous Chern class. Of course the sheaf $2\pi i\mathbb Z$ is isomorphic to $\mathbb Z$ but using the latter forces one to use the map $\mathbb Z \rightarrow \mathbb C$ taking $1$ to $2\pi i$. It is better to use the sheaf $2\pi i\mathbb Z$. One other reason for that is to keep track of complex conjugation. If $X$ comes from a real algebraic variety so that it has an antiholomorphic involution $\overline{(-)}$. Then we have $\overline{c_1(L)}=c_1(\overline L)$ when we let complex conjugation do what it should do on $2\pi i\mathbb Z$ (if one uses $\mathbb Z$ one has to throw in a sign). This is completely analogous to the case of étale cohomology where the first Chern class takes value in $H^2_{et}(X,\mathbb Z_\ell(1))$, where $\mathbb Z_\ell(1)$ is the inverse limit of $\{\mu_{\ell^n}\}$. Similarly the $n$'th Chern class lies most naturally in cohomology of $(2\pi i)^n\mathbb Z=(2\pi i\mathbb Z)^{\otimes n}$ resp. $\mathbb Z_\ell(n):=(\mathbb Z_\ell(1))^{\otimes n}$.

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Consider the exponential sequence of sheaves on $X$:

$0 \rightarrow \underline{\mathbb{Z}} \rightarrow \mathcal{O}_X \stackrel{\operatorname{exp}}{\rightarrow} \mathcal{O}_X^{\times} \rightarrow 0$.

The connecting map in sheaf cohomology gives a map

$c: H^1(X,\mathcal{O}_X^{\times}) \rightarrow H^2(X,\mathbb{Z})$.

In my experience, it is this map which is usually called the Chern class map. However, there is a natural map $H^2(X,\mathbb{Z}) \rightarrow H^2(X,\mathbb{C})$ and then we can use the Hodge decomposition $H^2(X,\mathbb{C}) = H^{0,2} \oplus H^{1,1} \oplus H^{2,0}$, where $H^{p,q} = H^q(X,\Omega^p)$. Under this decomposition the image of the Chern class map lands in $H^{1,1}$.

And then it is reasonable to say that the class $c(L) \in H^{1,1}$ is ample if $L$ is itself an ample line bundle.

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This is the first Chern class, usually denoted $c_1$. – Kevin H. Lin Apr 22 '10 at 3:17
Yeah, ok, nevermind. – Kevin H. Lin Apr 22 '10 at 3:53
I mean the map which appears in the universal coefficient theorem for cohomology...which is indeed defined by functoriality. – Pete L. Clark Apr 22 '10 at 4:31
By the way, with people like Ekedahl, Vistoli, BCnrd and VA lurking about, I feel a little silly answering algebraic geometry questions. [I am pretty good at algebraic geometry...for a number theorist.] Perhaps I should exercise more restraint. – Pete L. Clark Apr 22 '10 at 4:36
Dear Pete: I sure hope you will not restraint yourself (-:, because of the following: 1) nobody knows everything 2) experts may not answers all the questions and 3) it benefits everyone to see answers from a different perspective. Please keep on answering. – Hailong Dao Apr 22 '10 at 18:02

There is a map $d\log:H^1(X,\mathcal{O}_X^\times)\to H^1(X,\Omega_X^1)$ taking a line bundle to it's Chern class (at least, when everything is over $\mathbb{C}$, I believe this works). An ample class is then the Chern class of an ample line bundle.

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Did you mean $H^1(X,\mathcal O_X^*)$? – VA. Apr 22 '10 at 1:58
I did, yeah, apologies. – Charles Siegel Apr 22 '10 at 2:30