I am studying the Chern class using by some textbooks and lecture notes.
One day, I found an example of the first Chern class of $\mathbb{CP}^1$.
Let $\xi$ be a tautological line bundle of $\mathbb{CP}^1$ and $\eta$ be a trivial one. Then
$$
c_1(\xi)=-1,\ c_1(\eta)=0.
$$
I know the (first) Chern class of the trivial bundle equals to $0$.
How come $c_1(\xi)=-1$? ( $c(\xi)=1+c_1(\xi)$). Hence, does it means $c(\xi)=0$? )
And, what does $"-1"$ means? Is there some geometrical meaning?
Best Answer
Recall that $c_1(\xi)$ lives in $H^2(\mathbb{C}P^1)$, so when you write $c_1(\xi)=-1$, you actually mean $c_1(\xi)$ is the negative of the generator of $H^2(\mathbb{C}P^1)$. The first $1$ in $c(\xi)$, on the other hand, is the generator of $H^0(\mathbb{C}P^1)$.
As for the geometric reasoning, the top Chern class (highest non-zero Chern class) of a complex bundle is precisely the Euler class of the associated real bundle (forgetting the complex structure).
EDIT: Just a comment on why the first Chern class of the tautological bundle over $\mathbb{C}P^1$ is indeed negative of the generator: if you defined Chern classes axiomatically, this will actually be one of the axioms. The reason for this axiom is to ensure not all Chern classes are trivial. You can also get this result by looking at the construction of Chern classes.