4

Let $L$ be a line bundle and $E$ a vector bundle of rank $r$, then how can we prove that $$c_1(L\otimes E)=rc_1(L)+c_1(E)?$$ Here $c_1$ means the first Chern class.

Michael Albanese
  • 104,029
  • 2
    The universal answer to such questions is " The splitting principle." –  Apr 07 '14 at 13:22
  • How do you define chern classes? A typical proof would involve the splitting principle (which says you may assume that E is a sum of $r$ line bundles) and begin with computing the chern class of a product of line bundles. – Tom Bachmann Apr 07 '14 at 13:23
  • can you exlain why $r$ must be behind of $c_1(L)$? –  Apr 07 '14 at 14:04
  • Dear Hasanjol: consider the case when $X=\mathbf{P}^1$, $E=\mathcal{O}\oplus\mathcal{O}$, and $L=\mathcal{O}(1)$. Do you see where the $r$ comes from in this case? –  Apr 07 '14 at 15:23
  • 2
    For line bundles $c_1(L \otimes E)=c_1(L) +c_1(E)$, so the formula for $E$ sum of line bundles is direct, each summand $c_1(L)$ being from one of summands of $L$. Splitting principle says that it is enough to check on such $L$. – evgeny May 17 '14 at 07:16

2 Answers2

5

Below I give two methods for establishing the identity you asked about, but these methods can also be used to establish the general identity $c_1(E_1\otimes E_2) = \operatorname{rank}(E_2)c_1(E_1) + \operatorname{rank}(E_1)c_1(E_2)$.

$1.$ Using the splitting principle.

Splitting Principle: Let $E \to Y$ be a rank $r$ complex vector bundle. There is a space, $X$ (namely the total space of the flag bundle of $E$), and a continuous map $f : X \to Y$ such that

  • $f^*E = L_1\oplus\dots\oplus L_r$ where each $L_i$ is a complex line bundle, and

  • $f^* : H^*(Y; \mathbb{Z}) \to H^*(X; \mathbb{Z})$ is injective.

As $c_1$ is additive with respect to direct sums and tensor products of line bundles, we see that

\begin{align*} f^*(c_1(L\otimes E)) &= c_1(f^*(L\otimes E))\\ &= c_1(f^*L\otimes f^*E)\\ &= c_1(f^*L\otimes(L_1\oplus\dots\oplus L_r))\\ &= c_1(f^*L\otimes L_1\oplus\dots\oplus f^*L\otimes L_r)\\ &= c_1(f^*L\otimes L_1) + \dots + c_1(f^*L\otimes L_r)\\ &= c_1(f^*L) + c_1(L_1) + \dots + c_1(f^*L) + c_1(L_r)\\ &= rc_1(f^*L) + c_1(L_1) + \dots + c_1(L_r)\\ &= rc_1(f^*L) + c_1(L_1\oplus\dots\oplus L_r)\\ &= rc_1(f^*L) + c_1(f^*E)\\ &= rf^*c_1(L) + f^*c_1(E)\\ &= f^*(rc_1(L) + c_1(E)). \end{align*}

As $f^*(c_1(L\otimes E)) = f^*(rc_1(L) + c_1(E))$ and $f^*$ is injective, we see that

$$c_1(L\otimes E) = rc_1(L) + c_1(E).$$

$2.$ Using the Chern character.

The Chern character is given by $\operatorname{ch}(E) = \operatorname{rank}(E) + c_1(E) + \dots$ where the dots indicate higher order terms, and has the property that $\operatorname{ch}(E_1\otimes E_2) = \operatorname{ch}(E_1)\operatorname{ch}(E_2)$.

So, by definition, we have

$$\operatorname{ch}(L\otimes E) = r + c_1(L\otimes E) + \dots$$

On the other hand, using the multiplicative property of the Chern character, we have

$$\operatorname{ch}(L\otimes E) = \operatorname{ch}(L)\operatorname{ch}(E) = (1 + c_1(L) + \dots)(r + c_1(E) + \dots) = r + (rc_1(L) + c_1(E)) + \dots$$

Comparing the degree two parts of both expressions, we see that $c_1(L\otimes E) = rc_1(L) + c_1(E)$.

Community
  • 1
Michael Albanese
  • 104,029
  • on proof with principal splitting, why are you change the whitney sum to addition of chern class? Is not it change to cup product of chern class? – Ramtin.VA Jun 16 '19 at 23:00
  • 1
    @Ramtin.VA The total Chern class satisfies $c(L_1\oplus L_2) = c(L_1)c(L_2)$, but if you compare the terms in degree 2 cohomology, you see that $c_1(L_1\oplus L_2) = c_1(L_1) + c_1(L_2)$. – Michael Albanese Sep 19 '19 at 21:22
  • May I ask you, could you prove the relation $c_2(V \otimes L)=c_2(V)+(r−1)c_1(V)c1(L)+{r \choose 2} c_1(L)^2$ using the Chern character approach? –  Dec 03 '21 at 11:08
  • @Mathias: Yes. I have included the proof here as an answer to your question. With enough work, you can use the Chern character approach to compute all the Chern classes of a tensor product of complex vector bundles. – Michael Albanese Jan 31 '22 at 05:46
2

In smooth case the first Chern class of bundle $B \rightarrow M$ is equal up to constant to the class in de-Rham cohomology of the trace of any curvature form $\Theta$: $$c_1(B)=\left[ \frac{2 \pi}{i} \operatorname{tr} \Theta \right] \in H_{dR}^2 (M),$$ here $\Theta$ is a differential 2-form with values in endomorphisms of $B$, so you can consider it as a matrix of usual 2-forms and take its trace, which will be usual 2-form.

Because you can suppose $$\Theta_{L \otimes E}=\Theta_E \otimes I_1+I_r \otimes\Theta_L,$$ your formula is very transparent.

You can find details in Griffiths, Harris "Principles of algebraic geometry-1", chapter 3.3, this formula is one page before Gauss-Bonnet formula.

evgeny
  • 3,931