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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?

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    $\begingroup$ 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. :) $\endgroup$ – Andrew D. Hwang May 8 '16 at 16:24
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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.

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  • $\begingroup$ Is canonical line bundle the same as tautological line bundle? $\endgroup$ – wonderich Jul 9 '18 at 20:01
  • $\begingroup$ @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". $\endgroup$ – Jānis Lazovskis Jul 10 '18 at 19:50

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