# Characterization of Chern classes and Whitney product formula

Let E be a Complex vector bundle of rank r on M. Given $s_1, \cdots , s_r$ generic global sections, i can characterize the i-th Chern class as follows:

$C_i(E)= \eta_{V_i}$, and $V_i$ is the locus in M where $s_1, \cdots, s_i$ are not linearly indipendent

So if $s_1, \cdots s_i$ are everywhere linearly indipendent $C_i(E)=0$, in this way Chern classes measure the distance of a fiber bundle from being trivial.

Now i'm applying this characterization to the Whitney product formula on E,F vector bundles of rank $\geq 2$ on M:

$C(E \bigoplus F)=C(E)C(F)$, so, for example

$C_2(E\bigoplus F)=C_2(E)+C_2(F)+C_1(E)C_1(F)$

but this appeared strange to me, because $C_2(E\bigoplus F)=0$ if there are 2 not collinear sections on $E \bigoplus F$, while the right side of the equality is zero if there are 2 not collinear sections on E, two on F, and one section of E or F always non zero.

Then i thought that this could be because $C_2(E\bigoplus F)$ expresses a more "global" distance of $E\bigoplus F$ from being trivial, that is $E \bigoplus F$ is trivial if and only if both E and F are trivial and so $C_2(E\bigoplus F)$ is zero if both E and F have two always not collinear global sections.

Am i right or i misunderstood?

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$V_i$ should be the locus where $s_1$, $s_2$, ..., $s_{r-i+1}$ are not linearly independent. (You wrote $i$ in place of $r-i+1$ above.) I'm not sure whether this is the source of your confusion, or just a typo. – David Speyer Jan 31 '12 at 14:29
i don't agree with what you say, because $C_1(E) \in H^2(M)$ and you say that $C_1(E)$ is the Poincarè dual of $V_1$ of codimension r-1+1=r so $C_1(E) \in H^{2r}(M)$ – tigu Jan 31 '12 at 14:49
$C_1$ is in $H^2$, yes, which means that $V_1$ should be complex codimension $1$. And it is, using my formula. The locus where $r$ sections of a rank $r$ fail to be independent is a complex hypersurface because, in local coordinates, it is given by the vanishing of a single determinant. Remember than, when you have more vectors, it harder for them to be independent. – David Speyer Jan 31 '12 at 14:56
ok, now i understand. but with this characterization it suffeces to have $C_1(E)=0$ to say E is a trivial bundle because it means that E has r linearly indipendent sections? – tigu Jan 31 '12 at 16:10
No, but this there is a true statement like this. When you describe $V_i$ as the locus where some sections are dependent, there are a lot of subtleties in that description. It is only literally correct if there are "enough" holomorphic sections and if those sections are "generic enough". Otherwise, you have to start talking about keeping tracks of multiplicities and signs. (continued) – David Speyer Jan 31 '12 at 18:08