p-cycles and Fluxes

I would like to ask why the existence of a non-trivial p-cycle leads to a non-trivial flux. I would say that e.g. for a five-form $F_{(5)}$ field strength , the flux is: $$\int\limits_{\mathcal{C}^{5}}F_{(5)}$$ so in general: $$\int\limits_{\mathcal{C}^{p}}F_{(p)}$$

Is this correct? And why the geometry should be a p-cycle? Couldn't it be some other topology?

• This question could use a lot more context. What, exactly, do you mean by "non-trivial flux" (I would take $\int_{C^p} F_{(p)}\neq 0$ as the definition of "non-trivial flux")? What do you mean by "the geometry should be a p-cycle"? The geometry of what? Why are you talking about topology after that? Commented May 30, 2016 at 14:47
• You are right, thanks for the comment. First of all, yes, by non-trivial flux, I mean what you say. Secondly, you are correct, it is not well put; I mean why the compact manifold has to contain p-cycles in order to support fluxes. The topology has to do with the different choices of curves that one can use. But you are absolutely right, i could have stated it in a more proper way, thanks for the comment. Commented May 30, 2016 at 14:51

A p-cycle is a differential form that lives in $ker(\partial_p)$ for the differential $\partial_p$ (in grading $p$), and such a form is nontrivial if it is not in the image of $\partial_{p+1}$. Mathematically we can see this as a cycle that is not the boundary of anything, picture a circle around a torus that bounds no area on the torus. If one has a boundary we can have the Stokes' rule that $$\int_{\partial M}\omega = \int_M d\omega.$$ This is seen in Gauss' law. For a cocycle we then have a form $\omega \ne d\xi$, it is not the result of a coboundary, but where it is closed with $d\omega = 0$. Physically this means the field content of fields is not due to another field. This has some bearing of gauge invariance with ${\bf A}\rightarrow {\bf A} + {\bf d}\xi$ is such that $d^2\xi = 0$ gives gauge invariance. This is a topological form of a similar thing, and is seen in BRST quantization.
• First of all thank you for you comment Lawrence. I would say I understand your last two lines about the pure topological nature of the field. But in the end if $$d\omega=0$$ by itself (and not due to a boundary as you stated above) then I have: $$\int\limits_{\partial\mathcal{M}}\omega=0$$. This is a zero flux. So how can we achieve a non-zero one from the cocycle? Commented May 30, 2016 at 14:57
• It is more that for the $p$-form $\omega$ $\int_M \omega$ is non-zero, but where there does not exist a $p-1$-form $\chi$ such that $\omega = d\chi$ and $\int_M \omega = \int_{\partial M}\chi$. Commented May 30, 2016 at 16:51