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If $X$ is a complex manifold, one can define the first Chern class of $L\in \textrm{Pic}\,X$ to be its image in $H^2(X,\textbf Z)$, by using the exponential sequence. So one can write something like $c_1(L)\in H^2(X,\textbf Z)$.

But if $X$ is a scheme (say of finite type) over any field, then I saw a definition of the first Chern class $c_1(L)$ just via its action on the Chow group of $X$, namely, on cycles it works as follows: for a $k$-dimensional subvariety $V\subset X$ one defines

\begin{equation} c_1(L)\cap [V]=[C], \end{equation}

where $L|_V\cong\mathscr O_V(C)$, and $[C]\in A_{k-1}X$ denotes the Weil divisor associated to the Cartier divisor $C\in\textrm{Div}\,V$ (the latter being defined up to linear equivalence). So then one shows that this descends to rational equivalence and we end up with a morphism $c_1(L)\cap -:A_kX\to A_{k-1}X$. So, my naive questions are:

$\textbf{1.}$ Where do Chern classes "live"? (I just saw them defined via their action on $A_\ast X$ so the only thing I can guess is that $c_1(L)\in \textrm{End}\,A_\ast X$ but does that make sense?)

$\textbf{2.}$ How to recover the complex definition by using the general one that I gave?

$\textbf{3.}$ Are there any references where to learn about Chern classes from the very beginning, possibly with the aid of concrete examples?

Thank you!

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2 Answers 2

up vote 13 down vote accepted

First let me lift the suspense: if $L$ is a line bundle and if your scheme $X$ has dimension $n$, then $c_1(L)\in A^1(X)=A_{n-1}(X)$, where $A(X)$ is the Chow group of $X$, graded by codimension (upper indices) or dimension (lower indices).

The definition is very simple: take a non-zero rational section $s\in \Gamma_{rat}(X,L)$.
Its divisor of zeros and poles $div(s)$ is a cycle of dimension $n-1$ and the rational equivalence class of that cycle is the requested first Chern class: $$c_1(L)=[div(s)]\in A_{n-1}(X)$$
If $X$ is smooth (or if you want to be more technical, just locally factorial) the first chern class yields an isomorphism $$c_1: Pic(X)\xrightarrow \cong A^1(X)\quad (*)$$

If $X$ is a smooth variety defined over $\mathbb C$, then $A(X)$ has the structure of a ring graded by codimension and there is a canonical morphism of graded rings $A^*(X)\to H^{2*}(X^{an},\mathbb Z)$, sending $c_1(L)\in A^1(X)$ to the analytically defined Chern class $c_1(L^{an})\in H^2(X^{an},\mathbb Z)$ obtained by the exponential sequence.
(More generally $A^i(X)$ is sent to $H^{2i}(X^{an},\mathbb Z)$: that's what the notation with the stars above means)

The canonical (but very difficult) reference is of course Fulton's Intersection Theory.
Edit
A more accessible resource is a projected book by Eisenbud and Harris , amusingly called 3264 & All That, a draft of which they generously put online.

Second Edit
As an answer to a question in atricolf's comment, note that the displayed isomorphism $(*)$ implies that in general $A^1(X)$ is very far from being isomorphic to $H^{2}(X^{an},\mathbb Z)$.
For an elliptic curve $X$, for example, $A^1(X)$ is isomorphic to $X\times \mathbb Z$, which has the cardinality $2^{\aleph_0}$ of the continuum, whereas $H^{2}(X^{an},\mathbb Z)$ is isomorphic to $ \mathbb Z$.

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I understand your definition, thank you! Unfortunately I don't know what the upper $^\ast$ stands for in your $H^{2\ast}(X^{an},\mathbb Z)$. So, if I understand well, in the complex case you send the class $c_1(L)$ (that you defined in general) to $c_1(L^{an})$ in $H^2$; but which one of the two shall I name "the first Chern class of $L$"? Are $A^\ast(X)$ and that $H^2$ isomorphic? As an aside, it is just Fulton's book (the end of the second chapter) that originated my question! –  Brenin Oct 11 '12 at 15:21
1  
The $*$ should just stand for any integer (the relevant values being those from 0 to $n$). –  Aaron Mazel-Gee Oct 11 '12 at 15:51
    
Dear atricolf, I have addressed your comment in two edits. As to the question of which is the genuine Chern class of a line bundle: both are useful. But the algebraic class is much finer, as the example of an elliptic curve shows:the algebraic Chern class determines the line bundle while the analytic Chern class certainly doesn't. –  Georges Elencwajg Oct 11 '12 at 17:32
    
Dear Georges, your answer continues to raise my curiosity. Now I see that, on an elliptic curve, cohomology does not "distinguish" points, as rational equivalence does. But could you please explain (or give a reference, that I didn't find) why $\textrm{Pic}\,X\cong X\times\mathbb Z$ and $H^2(X^{an},\mathbb Z)\cong\mathbb Z$? It looks as if the Neron-Severi group of $X$ were equal to this $H^2$, is it true? –  Brenin Oct 12 '12 at 8:53
    
Dear atricolf, any compact connected orientable real manifold $M$ of dimension $n$ has $H^{2}(M,\mathbb Z)=\mathbb Z$: this is a purely topological result. It is also classical that $X\xrightarrow \cong Pic^0(X):x\mapsto [x-0]$ is an isomorphism of groups and from there follows $Pic(X)=A^1(X) \cong X\times \mathbb Z$. And, yes, the Néron-severi group of the elliptic curve $X$ is $H^{2}(X^{an},\mathbb Z)$ –  Georges Elencwajg Oct 12 '12 at 10:50

Your conceptual questions have been nicely answered by Georges Elencwajg. For your third question, and to learn how all he says and more is developed from scratch, you may find very interesting the following freely available course notes, where the authors develop the whole machinery at an "introductory" level. The second one requires a previous course in algebraic geometry, but the first reference provides you exactly with the needed background before introducing Segre and Chern classes in general and intersection theory up to Hirzebruch-Riemann-Roch theorem:

To get an easier quick glimpse at all the topics covered by the mentioned master monograph by Fulton, look at his own overview:

Georges Elencwajg has recommended in several of his posts the future book by Eisenbud and Harris, but I have found all the links to be of an old 2010 version, I recommend anybody interested in the evolution of the book to get the latest version available, as it includes refinements, many more pictures and is more complete:

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Dear @Javier Álvarez, thank you very much for your answer! In fact, I stumbled into Gathmann's notes some weeks ago, and I find them very well written. Also, they are exactly at the right level for me. Instead, I used to read those by Vakil in parallel with Fulton's book... So thank you again for your nice advise! –  Brenin Jan 26 '13 at 20:10
    
@atricolf, if you already knew the references I guess my answer is not that useful, but thanks anyway! Take a look at the introductory very short book (~80p) by Fulton that I mentioned, not to be confused with his main monograph (>400p.), where he explains that Chern classes can be defined as operators satisfying the same properties as in singular homology, although one would like a fully satisfactory contravariant ("cohomology") theory of rational equivalence of cycles (there have been several attempts, not to mention the whole subject of motives...). –  Javier Álvarez Jan 26 '13 at 20:26
    
@atricolf: I have added a link to a more recent version of the wonderful draft book mentioned by Georges. It is filled with interesting things on intersections and applications to enumerative geometry. You may find enlightening how they introduce and develop Chern classes and how they give useful enumerative (virtual) information of neat classical examples. –  Javier Álvarez Jan 26 '13 at 20:41

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