Comparing notions of degree of vector bundle
In this question, $X$ will be a smooth complex projective variety. This question will be about comparing two different ways of calculating the degree of a vector bundle on such an $X$.
I understand that, unless $X$ is a curve or $\mbox{Pic}(X)=\mathbb Z$, there is no well-defined notion of "degree" for vector bundles.
I would like to use $X=\mathbb{P}^2$ as an example. This is a variety with $\mbox{Pic}(X)=\mathbb Z$. A vector bundle $V$ on $\mathbb{P}^2$ has a well-defined degree, which we define to be the image of $\det V$ under the natural identification of $\mbox{Pic}(X)$ with $\mathbb Z$. For instance, if $V$ is the tangent bundle of $\mathbb{P}^2$, then $\det V=\mathcal O(3)$, and so $\deg V=3$.
I have also seen the following way of defining the degree, for bundles on projective varieties: since $X$ is projective, there is a map $\kappa$ from $X$ into a projective space of the same or higher dimension. Then we define a line bundle $\mathcal O_X(1)$ on $X$ by pulling back the $\mathcal O(1)$ from the projective space via $\kappa$. Then, the degree of a bundle $V$ on $X$ is a number which I have seen denoted either
$\deg(V):=c_1(\mathcal O_X(1))^{d-1}.c_1(V)$
or
$\deg(V):=c_1(\mathcal O_X(1))^{d-1}\cap c_1(V)$,
where $d=\dim(X)$.
Question:
In the case of $X=\mathbb P^2$, the map $\kappa$ is the identity, and we have $\deg(V)=c_1(\mathcal O(1))\cap c_1(V)$, where $\mathcal O(1)$ is just the hyperplane bundle of $\mathbb P^2$ itself.
When $V$ is the tangent bundle, how do I see that the number $c_1(\mathcal O(1))\cap c_1(V)$ is equal to $3$? My "instinct" is to think of $\mathcal O(1)$ as the line bundle of a divisor $D$, which in turn is determined by a hypersurface, specifically a projective line in $\mathbb P^2$, and then ask how many times that line is intersected by a "generic" section of $V$ (a vector field on $\mathbb P^2$, in this case, but I don't want this to depend too much on the fact that these sections are vector fields).
Is this the right way to think?
If it is, then two more questions:
(1) How can I see in a fairly intuitive way that this intersection number is $3$?
(2) Why would the intersection product of two Chern classes (elements of the cohomology ring $H^*(X,\mathbb Z)$) be equal to the intersection number of a divisor and a vector field (or of two divisors, more generally)? (I guess there is a cohomology $\leftrightarrow$ geometry philosophy that I need to understand better.)
There are a couple of different issues here which might best be addressed separately, so let me use the answerbox for more space. (I will try to give full details, so apologies if I tell you things you already know.)
For any vector bundle $V$ on any variety or complex manifold, we have $c_1(V)=c_1(\operatorname{det} V)$. This is a consequence of the the splitting principle.
So your correct claim that in your example $c_1 O(1) \cup c_1(V) = \operatorname{deg} ( \operatorname{det} V)=3$ is really just an equality involving the line bundle $K^* = \operatorname{det} V$ — you can completely ignore the fact that this line bundle is actually the determinant of some higher-rank bundle.
Now, why are the two sides the same? By your definition $\operatorname{deg} L= d$ means precisely that $L = O(1)^d$, since in this case $O(1)$ is the generator of $\operatorname{Pic} \mathbf P^2$. So here $\operatorname{deg} ( K^*)=3$ means that $K^* = O(3)$, hence $c_1(K^*)= 3 c_1 O(1)$. So the left-hand side of your equality becomes $3 c_1 O(1) \cup c_1 O(1)$. Since $c_1 O(1)$ is represented by a line in the plane, its cup-product with itself is indeed 1, so we get $3.1=3$ on the left-hand side too.
However: this is not generally true. The equality above hinged on the fact that the line bundle $O(1)$ generating $\operatorname{Pic} \mathbf P^2$ had $c_1(O(1) \cup c_1 (O(1)) = 1$. In general this will not be true. For example, suppose $X$ is a generic quartic surface in $\mathbf P^3$. Then its Picard group will be generated by a line bundle $L$, the restriction of $O(1)$ from $\mathbf P^3$ to the surface. Now $c_1(L)$ is represented by a hyperplane section of $X$, ie a quartic curve on $X$, and $c_1(L) \cup c_1(L) = 4$. So your equality would not hold on $X$: indeed the value of the left-hand side would be 4 times that of the right-hand side.
Finally let me address your question (2) in the body. You are wondering why the intersection product of Chern classes would equal the intersection number of divisors. I am not sure if I am interpreting the question correctly, but the vital point here is that, to a great extent, all the objects involved here — cohomology, characteristic classes, and intersection theory for divisors and algebraic cycles more generally — grew out of the attempt to formalise and generalise the same basic situation, namely counting the number of intersection points of two submanifolds of a compact manifold intersecting transversely. (I am not a historian of mathematics, so that statement should be liberally strewn with caveats, but the basic point is true.) Any form of intersection theory (cup product in topology, intersection numbers in algebraic geometry) is set up so that in this basic situation it returns the answer one naively expects.
To put it another way: in a definition such as
$$\deg(V):=c_1(\mathcal O_X(1))^{d-1}.c_1(V)$$
what one is really saying is
"intersect $d-1$ general hyperplanes in projective space and intersect the result with $X$ to get a smooth curve: then restrict $V$ to that curve and find its degree."
Here, the language of Chern classes and cup product allows one not to worry about any details like what "general" means, why this degree is independent of the choice of hyperplanes, and so on.