# Is a Chern class a property of a vector bundle or of a section of the bundle?

by Dwagg   Last Updated August 14, 2019 15:20 PM

Coming from a physics point of view, I learned from Green, Schwarz, Witten that (for instance) "the second Chern class of an $$SU(N)$$ gauge field" is an integral class of $$H^4(M,\mathbb R)$$ for a manifold $$M$$. In particular, if $$A = A^a_\mu(x) dx^\mu T^a$$ (where $$T^a$$ is Lie-algebra-valued and repeat indices are summed) is a particular $$SU(N)$$ gauge field (i.e. a section of the vector bundle $$E$$ of Lie-algebra valued one-forms over $$M$$), then we can form the field strength $$F = dA + A\wedge A$$, and the second Chern class is $$\text{Tr} F\wedge F/ 8\pi^2$$. So a Chern class is a property of a section of a vector bundle.

Learning Chern classes from Hatcher's book, I read that Chern classes assign to "each vector bundle $$E\to B$$ a class $$c^{2i} \in H^{2i}(B, \mathbb Z)$$". So the Chern class is a property of a vector bundle.

Is there some canonical choice of section (gauge field) made in the latter that is not made in the former, physics, definition? Any conceptual clarification would be helpful.

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#### Answers 1

One can prove that the Chern class is well-defined, independent of the choice of section. More precisely, $$\text{Tr} F \wedge F / 8 \pi^2$$ is not the Chern class itself, instead it is a closed form. One must prove that the cohomology class of that form is well-defined independent of the choice of $$A$$. That makes it a property of the vector bundle alone.