# The first Chern class of projective line $\mathbb{CP}^1$

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?

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.

One of the good properties of Chern classes is that if $E$ is a complex vector bundle of rank $r$ on a manifold $X$ and $s$ is a global section of $E$ that intersects the zero-section transversely, then $c_r(E)$ is exactly the cohomology class in $H^{2r}(X,\mathbf Z)$ of the submanifold $Z(s)=\{x \in X \mid s(x)=0\}$.

As @fixedp says, the "1" here refers to the canonical generator of $H^2(\mathbf{CP}^1)$: namely, a point with its canonical orientation as a complex submanifold of $\mathbf{CP}^1$. So why is $c_1(\xi)$ the negative of this?

One way to see it is to consider the dual bundle $\xi^{\vee}$. One of the axioms of Chern classes is that $c_1(E)=-c_1(E^{\vee})$, so it will be enough to show that $c_1(\xi^\vee)=1$.

To show that, one proves that $\xi^\vee$ is the bundles whose holomorphic global sections are exactly the linear forms on $\mathbf{CP}^1$. An example of such a linear form is the homogeneous coordinate $x$: this has a single zero at the point $[0,1]$, so by the discussion in the first paragraph this shows $c_1(\xi^\vee)=1$ as required.