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.

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.

Lee Mosher
Lee Mosher
August 14, 2019 15:03 PM

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