I am trying to understand the seminal paper "Topological gauge theories and group cohomology" by Dijkgraaf and Witten. They consider an oriented three-manifold $M$, compact Lie group $G$ and a $G$-bundle $E$ with a connection. In the case that $E$ is trivial they identify the connection on $E$ with a Lie algebra-valued one-form $A$. Then they choose an invariant bilinear form $\langle - , - \rangle$ on the Lie algebra and define the Chern-Simons action as $$ S(A) = \frac{k}{8 \pi^2} \int_M \mathrm{Tr} \left( A \wedge dA + \frac{2}{3} A \wedge A \wedge A \right), $$ where $k$ is an arbitrary integer. I think what is missing in this definition is normalization of $\mathrm{Tr}$ required to make this quantity gauge invariant modulo integers. According to my calculations, if $G$ is simply connected, then the correct condition is that $\mathrm{Tr}(h_{\alpha}^2) \in 2 \mathbb Z$ for every coroot $h_{\alpha}$. In particular this is satisfied for trace forms associated to finite-dimensional representations. Can someone confirm whether my conclusion in this case is correct?

Secondly, they proceed to the case in which $G$ is not simply connected. Then $E$ may be nontrivial. Their first definition of the Chern-Simons action in this case is as follows: take a four manifold $B$ whose boundary is $M$ (this is guaranteed to exist), extend $E$ and $A$ to a bundle with connection over $B$ (as far as I understand this may fail to exist, but at this point we assume existence) and put $$ S(A) = \frac{k}{8 \pi^2} \int_B \mathrm{Tr} \left( F \wedge F \right). $$ Then they write that "standard argument" shows that if $k$ is an integer, then $S(A)$ is independent, modulo $1$, of the choice of $B$ and the extension of $E$ and $A$ over $B$. They don't reproduce this standard argument, but let me write what I think they mean. Given some other $B'$ with appropriate extension we consider $B$ with its orientation reversed and glue the two together to obtain a closed four-manifold $X$. Then one probably needs to show that the two extensions of $(E,A)$ may be glued together to obtain a $G$-bundle with a connection over the whole $X$. I don't know if this is automatically true, but let's just assume this for now. Then the difference of two values of the Chern-Simons action is $$ S(A)_{B'} - S(A)_B = \frac{k}{8 \pi^2} \int_X \mathrm{Tr} \left( F \wedge F \right). $$ Now I think one would like to say that the right hand side is an integer. However, I'm not sure if this is true. For example for $G = \mathrm{U}(1)$ one can construct a line bundle over $X = \mathbb{CP}^2$ such that $ \int_X F \wedge F = 4 \pi^2,$ and in this case the right hand side of the formula above is $\frac{1}{2}$. Am I doing something wrong? Maybe some factor of $2$ is implicitly hidden in "$\mathrm{Tr}$" in the abelian case?

Further, I would like to remark that in the case that $G = \mathrm{SU}(n)$ and $\mathrm{Tr}$ - the trace form associated to the fundamental representation we have that $\frac{1}{8 \pi^2} \int_X \mathrm{Tr} (F \wedge F)$ is the integral of the second Chern class of $E$, which is known to be an integer. I don't know if similar statement can be made for other (semi)simple Lie groups.


1 Answer 1


Let me answer the second part of your question: In the Abelian case there is no trace at all. The term $$\frac{1}{8\pi^2} \int F\wedge F$$ is indeed the second Chern number of the $U(1)$ bundle. This is an integer only on Spin manifolds (which you can easily prove using the Atiyah-Singer index theorem). For non-Spin manifolds such as $\mathbb{CP}^2$ it can be a half-integer as you pointed out. This is just an indication that the theories with odd $k$ are not well defined on non-Spin manifolds. The theories with $k$ odd are sometimes called spin-Chern-Simons theories and require a choice of spin structure in addition to the $U(1)$ bundle, so they are only consistent on Spin manifolds. They are examples of spin-TQFTs.

Edit: The proof of integrality on spin manifolds directly follows from the AS index theorem which (forgetting about the metric) states that on a closed four-manifold

$$\text{ind} D = \frac{1}{8\pi^2} \int F\wedge F$$

$\text{ind} D$ just counts zero-modes of the Dirac operator, so it is an integer.

  • $\begingroup$ I am slightly confused with this description of the abelian case. Indeed, $\mathbb{CP}^2$ is not a spin manifold and according to the mathematical literature the second Chern number is even for spin manifolds. But here we need $X$ to be a spin manifold, while the original theory is formulated on $M$. One way to make sense of this is that we require that $M$ is a spin manifold and then we consider only such higher dimensional extensions $B$ which are spin manifolds. Then one would have to show that gluing of $B$ and $B'$ automatically yields a spin manifold. Is that what you have in mind? $\endgroup$
    – Blazej
    Commented Apr 4, 2020 at 7:51
  • $\begingroup$ Do you have a reference for that? There could be a difference in convention between the physics and the math literature, but I'm suspicious because if you take a look at my edit there is no reason why the index of the Dirac operator should be an even integer. Regarding the extension to X, indeed, in addition to the gauge bundle one needs to extend the spin structure. $\endgroup$
    – jpm
    Commented Apr 4, 2020 at 18:49
  • $\begingroup$ In general the differential form $c_1=\frac{F}{2\pi}$ has integer periods, hence so does its square, $c_1^2=\frac{F \wedge F}{4 \pi^2}$. There is no reason for the integral $\int_X c_1^2$ to be even if X is not a spin manifold, as shown by the example of the so-called tautological bundle over $\mathbb{CP}^2$. In this case the right hand side for the formula you write for the index is a half-integer. This is fine, because in the non-spin case the Dirac's operator is not defined. For spin manifolds square of every differential form with integer periods has even periods... $\endgroup$
    – Blazej
    Commented Apr 4, 2020 at 21:19
  • $\begingroup$ ... see for example en.m.wikipedia.org/wiki/Intersection_form_(4-manifold) . Thus the quantity you wrote down is indeed an integer (which clearly must be true for it to be the index). $\endgroup$
    – Blazej
    Commented Apr 4, 2020 at 21:24
  • 1
    $\begingroup$ I was saying that in the spin case $\int c_1^2$ (not $\int \frac{1}{2} c_1^2$) is an even integer. Thus $\int \frac{1}{2} c_1^2$ is an integer, not necessarily even. This is all in agreement with what you wrote. $\endgroup$
    – Blazej
    Commented Apr 6, 2020 at 8:02

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