# Where do Chern classes live? $c_1(L)\in \textrm{?}$

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

$$c_1(L)\cap [V]=[C],$$

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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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
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