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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?

tigu
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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 E 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 E 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 E Speyer Jan 31 '12 at 18:08
  • The details of how you deal with the technicalities will be different depending on whether you are looking at algebraic/holomorphic sections (in this case, your best source is Fulton's "Intersection Theory") or whether you are looking at smooth sections (in this case, Milnor's "Characteristic Classes" has a good reputation, thought I haven't read it). I'll discuss a single example (continued). – David E Speyer Jan 31 '12 at 18:10
  • Look at $\mathbb{P}^n$ for $n \geq 2$, and let $V:=\mathcal{O}(1) \oplus \mathcal{O}(-1)$. The Chern class is $1-h^2$, where $h$ generates $H^2(\mathbb{P}^n)$. The point is that $\mathcal{O}(-1)$ has no nonzero holomorphic sections. So all the holomorphic sections of $V$ are sections of $\mathcal{O}(1)$. In particular, any two holomorphic sections of $V$ are proportional everywhere on $\mathbb{P}^n$. (continued) – David E Speyer Jan 31 '12 at 18:16
  • If we perturb the above sections so that they are not holomorphic then, I think, there will be are two curves $C$ and $C'$ on which they become dependent. One of these curve should be weighted positively and one negatively, and their degrees cancel in $H^2(\mathbb{P}^n)$.It would be fun to work out an example of this. – David E Speyer Jan 31 '12 at 18:17
  • Similarly, if you look at one holomorphic section of $V$, it will vanish along an entire hyperplane in $\mathbb{P}^n$. If you perturb this section to be smooth, then it will vanish along a codimension $2$ subvariety, with multiplicity $-1$. You might want to read http://math.stackexchange.com/questions/51193/geometric-motivation-for-negative-self-intersection/ for a further understanding of how negative numbers show up as intersection multiplicities. – David E Speyer Jan 31 '12 at 18:19
  • The true statement I allude to above should be something like "If $V$ has enough holomorphic sections, and $c_1(V)=0$, then $V$ is trivial." I don't know a precise statement of this. – David E Speyer Jan 31 '12 at 18:21

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