Could somebody untangle following statement I found here:

the integer cohomology groups correspond to the quantization of the electric charge.

I know from pure mathematical side the meaning of cohomology groups, but not understand the translative part between physics and pure math at this point: Could somebody borrow some time to sketch how this identification/ correspondence is in detail established? ie how can the (presumably geometric) quantization procedure be "interpreted" / encoded purely in terms of certain cohomology group?

And furthermore, what is here the precise meaning of the notion of "a charge" from viewpoint of pure mathematical terminology? Up to now I thought that a charge in mathematical physics can be recognized as a quantity comming from the existence of global symmetry of the given system, so in simple terms certain intergral/ number witnessing the existence of such global symmetry.

Do they endow in the quoted sentence above certain "bulked" meaning to the term " charge" in order to making it to becomes the object subjected to quantization procedure?

  • $\begingroup$ Usually we prefer to think of magnetic charges in this topological way (see this answer of mine), but EM duality means you can apply the same idea to isolated electric charges. $\endgroup$
    – ACuriousMind
    Jul 5 at 15:49
  • $\begingroup$ @ACuriousMind: so essentially in this setting a charge is equivalent to an isomorphism class of a $U(1)$-principal bundle over $S^2$, which in turn uniquely determined by an isomorphism of a $U(1)$-principal bundle over $S^1$, which dictates how the overlaps of parts of the trivial cover are glued together, which in turn determined by a single number, the classical charge. That's funny from mathematical point of view, since one feels easily tempted to take it as generalized mathematical definition of a charge a isomorphy class of a principal G-bundle over arbitrary base space, where of $\endgroup$
    – user267839
    Jul 5 at 16:46
  • $\begingroup$ course the interesting ones are bundles over non contractable spaces. Thank you, thats a nice viewpoint! $\endgroup$
    – user267839
    Jul 5 at 16:47
  • $\begingroup$ Having this interpretation in our hand, how is the (geometric) quantization of this charge aka principal bundle related to certain cohomology groups? What I know, is that equivalence classes of G-principal bundles over nice enough base space X are classified by - well, how should the guy else be called ? - the classifying space BG via homotopy classes of pointed maps $[X, BG]$. This should be somehow related to equivariant cohomology groups... have still to think csrefully to make a precise statement. But after all it looks doable... $\endgroup$
    – user267839
    Jul 5 at 16:56
  • $\begingroup$ What I still not got, is how the "quantization" data is encoded there? That is an additional data together with the charges/ U(1) bundles. Above we roughly understood how the U(1) can be encoded in terms of cohomology groups, now the natural question is how it capteres the quantization data as "additional structure"? $\endgroup$
    – user267839
    Jul 5 at 17:00

1 Answer 1


In the following lecture notes you can find a description of your question. The relevant parts here are: chapter 1.3.4, , 7.1 and 8.

We treat the particle in an electric field as a Hamiltonian system in the following way:Let $(Q,g)$ be a Riemannian Manifold and construct $T^{*}Q$ as the cotangent bundle with the canonical symplectic structure $\omega_0$. Now let us assume we have an electric field, that is a closed 2-form $F$ on $Q$. Pulling back $F$ along the canonical projection $\pi:T^{*}Q \to Q$, we obtain a symplecitc form on $T^{*}Q$ of the following form, where $e$ is the electric charge: $$\omega=\omega_0+e\pi^{*} F.$$ It can be shown that $\omega$ is symplectic iff $F$ is closed. The proof can be found in the leccture notes. We now have obtained a Hamiltonian system $(T^{*}Q,\omega)$, which we want to quantize using the language of geometric quantization. In order to do this, we have to build a prequantum line bundle over this symplectic manifold, whose curvature is equal (up to some factors of i/h) to the symplectic form. In the lecture notes, you will find that for the system to be quantizable,a certain integrality condition needs to hold: the deRham cohomology class of the symplectic form needs to be in the image of the map $$i^2:H^2(T^{*}Q,\mathbb{Z}) \to H^2(T^{*}Q,\mathbb{C}).$$ Physically, this will correspond to the charge, $e$ being quantized, the condition for a quantization to exist will be a condition on $e$ from the symplectic form. We can now turn to the question whether the quantization was unique or not. So what we did mathematically during the geometric quantization procedure is that we have built a $U(1)$ bundle over $T^{*}Q$ for some Riemannian manifold Q, whose curvature was equal to the symplectic form $\omega$. The non-uniqueness, as discussed in the lecture notes chapter 8 is in one-to-one correspondence with $H^1(T^{*}Q,U(1))$.

So, I think, what they really mean, is the integrality condition, namely that in order for the hamiltonian system to be quantizable, the symplectic form needs to be entire, i.e. its derham class satisfies the condition given above, or in the notes chapter 8.

As to not confuse notions, here we quantized a charged particle moving in an electric field.


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