# Chern character of canonical line bundle over $\mathbb{CP}^1$

Let $$H \to \mathbb{CP}^1$$ be the canonical line bundle over $$\mathbb{CP}^1 =S^2$$.

Then from the text Vector Bundles and K-theory by Hatcher, given the chern character, $$ch$$, and first chern class $$c_1(H)$$, we get:

$$ch(H)=e^{c_1(H)}=1+c_1(H)$$

I don't understand where all the higher order $$c_1(H)$$ terms go when expanding the exponential. Could anyone explain this please?

• Note, incidentally, that "canonical bundle" to a topologist might mean the tautological bundle $\mathcal{O}(-1)$ (whose fibre at each point $p$ is the line represented by $p$), while to an algebraic geometer it means the (top exterior power of the) cotangent bundle, here $\mathcal{O}(-2)$. Nowadays, I believe the algebro-geometric meaning is more widely used, but it's prudent to specify which you mean. :) – Andrew D. Hwang May 8 '16 at 16:24

## 1 Answer

In the proof of Corollary 4.4, on page 101, Hatcher says that for the canonical line bundle on $\textbf{CP}^n$ and $c$ the first Chern class, the Chern character is given by $$1+ c +\frac{c^2}{2}+\frac{c^3}{3!}+\cdots + \frac{c^n}{n!}.$$ The reason why this sum stops at $n$ is because the cohomology groups of $\textbf{CP}^n$ are trivial past $H^n$ (and the Chern classes are elements in these groups). If we continue to apply the cup product to an element in $H^n$ (such as $c^n$) and an element in $H^1$ (such as $c$), then we should get an element of $H^{n+1}$, but that group is trivial. Hence we get zero, so all products $c^k$ for $k>n$ are zero, so they disappear in the expansion of the exponential.

Your question is when $n=1$, so everything vanishes for degree 2 and above.

• Is canonical line bundle the same as tautological line bundle? – wonderich Jul 9 '18 at 20:01
• @wonderich in my answer, yes. In general, depends on the context - see Andrew's comment to the question, or Section 1.1 of Hatcher's book, where he defines "canonical line bundle". – Jānis Lazovskis Jul 10 '18 at 19:50