I am a bit confused with the definition of the 1st (and 2nd by extension) Chern class in YM theories.

I understand that in general $c_i \in H^{2i}(M,\mathbb{Z})$ where $M$ is a smooth manifold. Then, for the curvature of a principal bundle being $F$ we know that $$ c_1 = \frac{i}{2\pi}\text{Tr}F $$ and $$ c_2 = \frac{i}{2\pi}\text{Tr}F \wedge F. $$ I see that $F$ is a 2-form and that $F \wedge F$ is a 4-form but since we consider the traces (both $F$ and $F \wedge F$ can be thought of as matrices, how can $c_1$ and $c_2$ be elements of $H^{2}$ and $H^{4}$ correspondingly? Are they not just numbers? Elements of $H^{2i}$ should be $2i$-forms, right?

Then, I know that the integral of $c_2$ over the manifold gives the instanton charge $k$. What quantity does the integral of $c_1$ give upon integration over $M$?

  • $\begingroup$ It is the trace over gauge indices. I.e. for U(1) gauge group you can drop the traces. $c_1$ cannot be integrated over $M$ (if $M$ is four-manifold), since you need a $4$-form to be able to integrate it. $\endgroup$ Commented Feb 22, 2016 at 18:45

1 Answer 1

  1. You wrote $$ c_2 = \mathrm{Tr}(F\wedge F)$$ which is a real-valued form, since $F$ is an $\mathrm{ad}$-valued form and the trace (which is supposed to be the adjoint trace in general) produces a number, so this is a real valued form.

  2. Elements of $H^{2n}(M,\mathbb{Z})$ are not forms. The deRham cohomology $H_\text{dR}^{2n}(M,\mathbb{R})$ of forms as the cohomology of the resolution of the constant sheaf $\mathbb{R}$ by differential $\mathbb{R}$-valued forms computes $H^{2n}(M,\mathbb{R})$, not $H^{2n}(M,\mathbb{Z})$. Note that $\mathbb{Z}$-valued forms don't make sense because $\mathbb{Z}$ is discrete.

  3. $H^{2n}(M,\mathbb{Z})$ embeds into $H_\text{dR}^{2n}(M,\mathbb{R})$ such that classes $[F]\in H^{2n}_\text{dR}(M,\mathbb{R})$ who have representants $F$ that integrate over every $2n$-cycle to integers are in the image of the canonical map $H^{2n}(M,\mathbb{Z})\to H^{2n}(M,\mathbb{R}) = H^{2n}_\text{dR}(M,\mathbb{R})$, cf. this math.SE question.

  4. Therefore, $c_2$ defines a class in integer cohomology only if it integrates to an integer (times some $4 \pi^2$) on the entire manifold. This integer is indeed the instanton charge $k$. Please note that $c_2$ itself is a form, which is a representant of a cohomology class, but not a class itself.

  5. You cannot integrate $c_1$ over the manifold, you can only integrate it over 2-cycles (more generally 2-chains, but that won't be an integer). I don't know of a generic intepretation for this integral because there seems to be no "physically natural" 2-cycle to choose in a more than two-dimensional manifold.

  • $\begingroup$ I am confused with the wording in comment 2. Can you please explain this a bit further? The constant sheaf $\mathbb{R}$ are real valued functions, right? $\endgroup$
    – Marion
    Commented Feb 22, 2016 at 19:28
  • $\begingroup$ @Marion It's a very short explanation of why the deRham cohomology computes the actual (singular) cohomology with real coefficients (otherwise known as the deRham theorem). What confuses you about it? (If you don't know how to compute cohomology by resolutions of sheaves, just ignore the part between "of forms" and "computes") $\endgroup$
    – ACuriousMind
    Commented Feb 22, 2016 at 19:30
  • $\begingroup$ Hi, yes I do not know how to compute cohomology by resolutions of sheaves. I have some understanding of de Rham cohomology because it is given in terms of forms. Also, in point 1 you got me confused as well on a second read. The trace produces a number, thus not a form. I do not know what an ad-valued form is. Finally, if elements of $H^{2n}(M,\mathbb{Z})$ are not forms what are they. So, what kind of number is this $c_1$? $\endgroup$
    – Marion
    Commented Feb 22, 2016 at 20:27
  • $\begingroup$ @Marion: $\mathrm{ad}$-valued means $F$ transforms in the adjoint representation of the gauge group, i.e. as an element of the Lie algebra. The trace is applied to the $\mathrm{ad}$-valued form $F$ pointwise, so that at every point, it produces a number out of the algebra element at that point. Elements of $H^{2n}$ are just cohomology classes. Cohomology of a space, in general, is not defined by forms, but by singular (co)homology. As I say, I don't know a physical significance of $c_1$. $\endgroup$
    – ACuriousMind
    Commented Feb 22, 2016 at 20:36
  • $\begingroup$ since you integrate $c_1$ over 2-cycles it seems like it corresponds to some kind of E/M flux, right? $\endgroup$
    – Marion
    Commented Feb 25, 2016 at 23:45

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