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Let $X$ be a Riemann surface$^\dagger$ and consider a complex vector bundle $E$ over it. I know that the definition of the degree of the bundle $E$ is given by $$\text{deg} E = \frac{i}{2\pi} \int_X \text{trace}F_A$$ where $F_A$ is the curvature induced by an arbitrary connection $A$ on $E$.

I know that the degree depends solely on the topology of the bundle and not on the arbitrary connection $A$ used in the definition. I would like to know what are the degrees of related vector bundles such as:

  • The tensor product, $\text{deg} (E_1\otimes E_2)$.
  • The direct sum, $\text{deg} (E_1\oplus E_2)$.
  • Quotients, $\text{deg}(E/F)$ where $F\subset E$ is a subbundle.

I think I can prove that for direct sums and tensor products the degree is additive by using direct-sum and tensor-product connections, but I am at a loss at computing degree in the quotient bundle.

$^\dagger$: I restrict the question to Riemann surfaces in order for the degree to also be independent of the volume form of the base space $X$, although I know that the degree can be defined with respect to a given metric/volume/Kähler form in higher dimensions.

Michael Albanese
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topolosaurus
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    Are you aware that $\frac{i}{2\pi}\operatorname{trace}F_A$ is a representative for $c_1(E)$? – Michael Albanese Sep 02 '21 at 15:12
  • Yes, I understand that this is a representative of the Chern class. From this we can see that for direct sums $$c_1(E_1\oplus E_2)= c_0(E_1)\wedge c_1(E_2) + c_1(E_1)\wedge c_0(E_2) = c_1(E_1) + c_1(E_2)$$ and therefore the degree splits as sum of the degrees. – topolosaurus Sep 02 '21 at 15:17
  • Correct. Likewise, you just need to express $c_1(E\otimes F)$ and $c_1(E/F)$ in terms of $c_1(E)$ and $c_1(F)$ to answer your question. – Michael Albanese Sep 02 '21 at 15:19
  • @MichaelAlbanese I have not found much regarding the Chern class of products and quotients. I am trying to solve a problem that requires the computation of the degree of quotients of low-rank vector bundles. Are there any simplifications for say, line bundles or bundles over projective spaces? – topolosaurus Sep 02 '21 at 15:28

1 Answers1

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As I mentioned in the comments, $\frac{i}{2\pi}\operatorname{trace}F_A$ is a representative for $c_1(E)$ by Chern-Weil theory. Therefore

$$\deg(E) = \frac{i}{2\pi}\int_X\operatorname{trace}F_A = \int_Xc_1(E).$$

So formulae for the degree follow from formulae for the first Chern class.

For the tensor product, you can show that $c_1(E_1\otimes E_2) = \operatorname{rank}(E_1)c_1(E_2) + \operatorname{rank}(E_2)c_1(E_1)$ by using the splitting principle or the Chern character; see this answer for both methods in the special case where one of the vector bundles is a line bundle.

As for the other two cases, recall that the first Chern class is additive in short exact sequences. From the short exact sequence

$$0 \to E_1 \to E_1\oplus E_2 \to E_2,$$

we see that $c_1(E_1\oplus E_2) = c_1(E_1) + c_1(E_2)$. Likewise, from the short exact sequence

$$0 \to F \to E \to E/F \to 0,$$

we see that $c_1(E) = c_1(F) + c_1(E/F)$, so $c_1(E/F) = c_1(E) - c_1(F)$. Note that this can also be deduced from the first short exact sequence because Chern classes are topological and short exact sequences of continuous vector bundles always split.

Therefore

\begin{align*} \deg(E_1\otimes E_2) &= \operatorname{rank}(E_1)\deg(E_2) + \operatorname{rank}(E_2)\deg(E_1)\\ \deg(E_1\oplus E_2) &= \deg(E_1) + \deg(E_2)\\ \deg(E/F) &= \deg(E) - \deg(F). \end{align*}

Note that these identities are true on a compact Kähler manifold too.

Conjecture
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Michael Albanese
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