# Calculation of the first Chern class of the canonical line bundle over $\mathbb{CP}^n$

There are different ways of defining and thereafter calculating the Chern classes. Right now I'm studying from the lecture notes which introduce the first Chern class through the classifying spaces as follows:

The classifying bundle for $\mathbb{U}(1)$ is $\mathbb{S}^{\infty} \rightarrow \mathbb{CP}^{\infty}$. So the set of complex line bundles over $B$, which are $\mathbb{U}(1)$-bundles, is in the bijective correspondence with the maps from $B$ to $\mathbb{CP}^{\infty}$ taking up to homotopy. So for any bundle there exists a map induced on the cohomology rings from $\mathbb{Z}[t]$ to $H^{\bullet}(B)$. And the first Chern class is defined as the image of the generator $t$ in the latter map.

I have a canonical line bundle over $\mathbb{CP}^n$ and I want to calculate the first Chern class using classifying space theory. Do I have to follow the explicit construction of the map between $\mathbb{CP}^n$ and $\mathbb{CP}^{\infty}$ for this line bundle, because I couldn't do it, or there is another way using classifying spaces?

• Can "the guess" of the right map together with showing the commutativity of the corresponding diagram be one of the possible solutions to the problem above? – Mihail Jul 31 '17 at 21:02
• There are three complex line bundles on $\mathbb{CP}^n$ which have a serious claim to be called canonical , namely $\mathcal O_{\mathbb{CP}^n}(1), \mathcal O_{\mathbb{CP}^n}(-1)$ and $\mathcal O_{\mathbb{CP}^n}(-n-1)$. Algebraic geometers call these line bundles respectively the hyperplane bundle, the tautological bundle and the canonical bundle. – Georges Elencwajg Aug 1 '17 at 20:46
• @GeorgesElencwajg I'm sorry for the ambiguity. I wasn't thinking about the algebraic-geometrical context. Isn't the name 'canonical' preserved for the single line bundle in the smooth category? Do you advise to add the definition of the canonical line bundle in the question part? – Mihail Aug 1 '17 at 20:52
• Dear @Mihail, in the smooth category one wouldn't use canonical in the sense of $\mathcal O(-n-1)$ but it couldn't hurt to say that you mean the tautological line bundle $\mathcal O(-1)$, as is clear from your answer. – Georges Elencwajg Aug 1 '17 at 20:58

Moving from my own comment I have come to some idea which looks plausible to me. I would be grateful if you indicate mistakes or support.

The idea is to move the other way around from the given bijectivity between set of complex line bundles and $[\mathbb{CP}^n,\mathbb{CP}^{\infty}]$. The mentioned bijection is given by pull-backing the classifying bundle. So I wondered which bundle I will get if I pull-back it under the usual inclusion $\mathbb{CP}^n \hookrightarrow \mathbb{CP}^{\infty}$ and fortunately I get exactly the canonical line bundle! The total space of the pullbacked bundle is defined as $E=\{[l]\times x \in \mathbb{CP}^n\times S^{\infty}:x\text{ maps to } [x] \in \mathbb{CP}^k \text{ for } x\in S^{2k+1} \text{ under the usual quotient map and } [x]=[l] \}$

What can be said about $E$ ? It's second coordinate cannot exceed the sphere with dimension $2n+1$ because $[x]=[l]\in \mathbb{CP}^n = S^{2n+1}/ \sim$ so the set $E$ can be rephrased as $$\{[l]\times x \in \mathbb{CP}^n\times S^{2n+1}:x\text{ quotients to } [l] \in \mathbb{CP}^n \text{ for } x\in S^{2n+1}\}=$$ $$=\{[l]\times x \in \mathbb{CP}^n\times \mathbb{C}^{n+1}:x\text{ quotients to } [l] \in \mathbb{CP}^n \text{ for } x\in \mathbb{C}^{n+1}\}$$

But this is exactly the definition of the canonical line bundle!

The rest part of showing that the image of generator $t$ is again $t$ (actually $t+(t^{n+1})$) is rather a standart job, because the inducing map is inclusion.

Still can't say that I am pleased by this answer because it doesn't use constructive way but rather lucky guess.

• At least now I do know why it deserves a name 'canonical'. – Mihail Aug 1 '17 at 20:24