# Chern classes are not numbers, are they?

Let $X$ be a smooth projective algebraic variety, say over $\mathbb C$. Let $E$ be a rank $r$ vector bundle on $X$. We can associate with $E$ its Chern classes $c_i(E)$. When I read "$c_i(E)$", the first thing I (automatically) do is to think where it lives. And it lives in the codimension $i$ part of the Chow ring of $X$, namely $$c_i(E)\in A^i(X).$$

If $r>n=\dim X$, I see we can identify $c_n(E)\in A^n(X)$ with an integer. But reading some papers, I got the impression that one can do the same for the other $c_i$'s as well. So my question is just about terminology: what does it mean, for a vector bundle $E$, to have even or odd $c_i$, if Chern classes live in $A^\ast(X)$?

• Nitpick: the restriction $r>n=\dim X$ is not necessary. A vector bundle $E$ has Chern classes $c_i(E)$ for all $i$; they just happen to be zero for $i\gt r$. – Georges Elencwajg Nov 20 '13 at 0:10
• It could be very helpful to have specific references. – Kevin Carlson Nov 20 '13 at 0:11
• Not a nitpick: it is a mistake to think that you can identify $A^n(X)$ with the integers. This is false already for curves since $A^1(X)=\operatorname{Pic}(X)$ – Georges Elencwajg Nov 20 '13 at 0:31
• @KevinCarlson: you are right, but I do not have one at hand right now. I was asking about an expression which I read and heard about several times, and I always asked myself "what do they mean?". Sorry about that. – Brenin Nov 20 '13 at 10:11

The only reasonable interpretation I can think of for "$c_i(E)$ is even" is that there is some class $a\in A^i(X)$ such that $c_i(E)=2a$.
However I have no interpretation for "$c_i(E)$ is odd", except the rather ridiculous one that it means "$c_i(E)$ is not even" ...

• I also think this is very reasonable. I will add an explicit quotation from a reference the next time I'll stumble upon a similar expression. – Brenin Nov 20 '13 at 10:13

In general, my guess would be that even or _odd refers to the parity of the degree. In situations where this is well-defined, this seems like the most reasonable explanation—though I don't know your references.

I don't have a very good intuition for the case $r\notin \{1,n\}$. But consider $A^1 (X)$, the group of divisors modulo rational equivalence. Two divisors can be equivalent only if they have the same degree, so the degree is well-defined for an element of $A^1 (X)$ and it makes sense to refer to its parity.

Again, knowing the specific references would be helpful. For all I know the general definition involved tensoring with $\mathbb{Z}/2$ or something.

• Dear User-33433, there is no reasonable notion of "degree" for a divisor, except in the very special case of divisors on a curve. ("Reasonable" meaning in particular compatible with linear equivalence) – Georges Elencwajg Nov 20 '13 at 0:02
• @GeorgesElencwajg I suppose I was imagining $X=\mathbb{P}^n$ (though if you tell me that there's no degree there, I will take that as my signal that it's time for me to start reviewing once again). – Slade Nov 20 '13 at 7:50
• Dear User-33433, let me reassure you: you are right that there is indeeed a notion of degree for Chow classes on $\mathbb P^n$ since $A^i(\mathbb P^n)=\mathbb Z$ for $0\leq i\leq n$. – Georges Elencwajg Nov 20 '13 at 9:56