# Functorial properties of the degree in vector bundles

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.

• Are you aware that $\frac{i}{2\pi}\operatorname{trace}F_A$ is a representative for $c_1(E)$? Sep 2, 2021 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. Sep 2, 2021 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. Sep 2, 2021 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? Sep 2, 2021 at 15:28

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.