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I have been reading Tong's notes on QHE and Gauge Theories, specifically the part about quantizing the Abelian U(1) Chern-Simons level at finite temperature in the presence of a monopole (These discussions begin on pages 150 and 391 in the documents I refer to, respectively). There, he talks about the proper way of performing the integral of the gauge variation of the CS term by defining the gauge field on patches and then gluing them together. Unfortunately, he never actually does it this way in the notes or refers to a text where this calculation is followed through. I can understand how to get the result he claims by integrating by parts, but I don't know how to justify the fact that you need to integrate by parts first. I'd like to learn how to be able to perform such calculations correctly, but don't know where to start. Here is my attempt.

Let us consider the manifold $M = S^1 \times S^2$, parametrized by coordinates $\tau \in [0,\beta]$ for the $S^1$, and $\theta \in [0, \pi]$ and $\varphi \in [0, 2\pi ]$ for the azimuthal and polar angles of the $S^2$, respectively, where $\beta$ is the inverse temperature and the size of the $S^1$. The gauge field configuration is such that

\begin{align} \frac{1}{2 \pi} \int_{S^2} F_{\theta \varphi} \,d^2x =1. \tag{1} \end{align}

In order to have such a configuration on the 2-sphere, we define the gauge field in two patches - $A^{(1)}$ and $A^{(2)}$. Let $\theta_1$ and $\theta_2$ be such that $0<\theta_1<\theta_2<\pi$. Then the non-zero components of the gauge field in the two patches are \begin{align} A^{(1)}_{\varphi}&= \frac{-1}{2} \frac{\cos \theta -1}{\sin \theta}, \text{ for } 0<\theta<\theta_2,\\ A^{(2)}_{\varphi}&= \frac{-1}{2} \frac{\cos \theta +1}{\sin \theta} \text{ for } \theta_1<\theta<\pi, \\ A^{(1)}_{\theta}&=A^{(2)}_{\theta}=A^{(1)}_0=A^{(2)}_0=0, \text{ everywhere}. \end{align}

The Chern-Simons action is \begin{equation} S_{CS} = \frac{k}{4\pi}\int_{S^1\times S^2} d^3x (A_0 F_{\theta \varphi} + A_{\theta} F_{ \varphi 0} +A_{\varphi} F_{0 \theta}). \end{equation}

Under a (non-single valued) gauge transformation \begin{equation} A_{\mu} \rightarrow A_{\mu} - \partial_{\mu} \alpha, \end{equation}

where $\alpha = \frac{\tau}{\beta}$, the action changes as \begin{equation} S_{CS} \rightarrow S_{CS} + \delta S_{CS}, \end{equation} where \begin{align} \delta S_{CS} = \frac{k}{4\pi} \int_{S^1\times S^2} d^3x \left(\frac{1}{\beta} F_{\theta \varphi} \right). \end{align} Clearly, here we have thrown out the terms $A_{\theta} F_{ \varphi 0}$ and $A_{\varphi} F_{0 \theta}$, since $\partial_{\theta} \frac{\tau}{\beta}=\partial_{\varphi} \frac{\tau}{\beta}=0$. This is something that I wasn't supposed to do, according to Tong. Proceeding with the $S^1$ integral, we're left with \begin{equation} \delta S_{CS} = \frac{k}{4\pi} \int_{S^2} F_{\theta \varphi} \, d^2x = \frac{k}{2}, \end{equation} where we have used equation (1). This result differs from the correct one by a factor of 2.


So there are several things that I'd like to ask.

  1. I don't see how the specific chart that I have chosen plays any role in the calculation. I found out about certain gluing rules (Čech Cohomology) on the intersection region (in this case $\theta_1 <\theta <\theta_2$), but I don't understand how to use them to properly perform the integral. How does this work in details?
  2. I don't understand how one can see that integration by parts is necessary. It seems to lead to contradictory results. For example \begin{align} \int_{S^2} A_{\varphi} F_{0 \theta} = \int_{S^2} A_{\varphi} (\partial_0 A_{\theta} - \partial_{\theta} A_0) = \int_{S^2} A_{\varphi} \partial_{\theta} A_0. \end{align} If $A_0 = a=const$, then \begin{equation} \int_{S^2} A_{\varphi} \partial_{\theta} A_0=0 = \int_{S^2} a \partial_{\theta} A_{\varphi} = a \int_{S^2} F_{\theta \varphi} = 2\pi a, \end{equation} Where we have neglected the boundary term in the integration by parts since $S^2$ has no boundary and used the explicit form of the charts to conclude that $F_{\theta \varphi} = \partial_{\theta} A_{\varphi}$. Is there anything that can signal to me from the get go that what I'm doing doesn't make sense?
  3. A reference that either goes through the details of how to perform such calculations would be invaluable. I have not been able to find anything that clarifies the above two questions.
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I haven't seen this clearly explained anywhere else, so this answer tries to be thorough. Contents:

  • Section 1 tries to define the $3$d Chern-Simons action using $3$d patches, so that the gauge field is well-defined within each patch, to see if this recovers the missing factor of $2$. It doesn't.

  • Section 2 explains why the definition using $3$d patches is ambiguous, which is presumably why Dijkgraaf and Witten didn't define it that way.

  • Section 3 explains how to define the $3$d Chern-Simons action properly using an integral over a $4$d manifold and then shows how to apply this definition to the configuration in the question.

  • Section 4 shows how the properly defined Chern-Simons action is affected by the gauge transform described in the question.

  • Section 5 shows how the properly defined Chern-Simons action can be written in terms of patches — using $4$d patches instead of $3$d patches — and shows that the "integration by parts" step that seems so mysterious in Tong's notes has a well-defined analogue in this context.

1. First try: defining the integral using $3$d patches

The gauge field needs to be defined patchwise, as shown in the question. This suggests that maybe we should define the Chern-Simons action patchwise, too. However, this doesn't explain the factor of $2$. This section shows the details.

Write the manifold $M=S^1\times S^2$ as $M=P\cup P'$, where $P$ and $P'$ cover the $\theta\leq \pi/2$ and $\theta\geq\pi/2$ hemispheres of the $S^2$, respectively, so that $P$ and $P'$ touch only along a $2$d interface (the Cartesian product of the equator of $S^2$ with the original $S^1$). Try defining $S_{CS}$ by $$ \newcommand{\pl}{\partial} S_{CS} =\frac{k}{4\pi} \left(\int_P A\,dA + \int_{P'} A'\,dA'\right) \tag{1} $$ where the gauge field one-form is $A$ in $P$ and $A'$ in $P'$, with $A'=-A$ in the intersection $P\cap P'$, as in the question. The orientations of $P$ and $P'$ are such that $\pl P$ and $\pl P'$ are oppositely-oriented versions of the $2$d interface $P\cap P'$. Under a gauge transformation $$ \delta A =d\alpha \hskip2cm \delta A' =d\alpha \tag{2} $$ the change in $S_{CS}$ is \begin{align} \delta S_{CS} &=\frac{k}{4\pi} \left(\int_P d\alpha\,dA + \int_{P'} d\alpha\,dA'\right) \\ &=\frac{k}{4\pi} \left(\int_{\pl P}A\,d\alpha + \int_{\pl P'} A'\,d\alpha\right) \\ &=\frac{k}{4\pi} \left(\int_{\pl P}A\,d\alpha + \int_{\pl P} A\,d\alpha\right) \\ &=\frac{k}{2\pi} \int_{\pl P}A\,d\alpha \\ &=\frac{k}{2\pi} \int_{S^1}d\alpha\int_\text{equator of $S^2$} A. \tag{3} \end{align} Now consider the equation numbered (1) in the question, which may be written in coordinate-free notation as $$ \frac{1}{2\pi}\int_{S^2} F_{S^2}=1. \tag{4} $$ This condition implies $$ \int_\text{equator of $S^2$} A=\pi, \tag{5} $$ so we end up with $$ \delta S_{CS} = \frac{k}{2}\int_{S^1}d\alpha. \tag{6} $$ This matches the result shown in the question, which is smaller than Tong's result by a factor of $2$. Apparently, defining the integral patchwise is not the key to answering the question. The next section shows that the definition is ambiguous, so we shouldn't be surprised that it fails to answer the question.

2. Why that definition is ambiguous

Dijkgraaf and Witten's paper is an authoritative reference on the definition of $3$d Chern-Simons theory, and they do not define the Chern-Simons action in terms of $3$d patches as I tried to do above. This section shows that the definition used above is ambiguous, which is presumably why Dijkgraaf and Witten didn't use it.

The gauge field is defined using $3$d patches whose intersections are open sets, so that transition functions can be defined. If we want to define the Chern-Simons action in terms of $3$d patches that touch only along $2$d interfaces, like I tried to do above, then we need to check that the value of the integral doesn't change when we move the interface around within the open sets where the patches overlap. If it does change, then the definition is ambiguous.

To check this, consider a pair of overlapping patches $Q$ and $Q'$, in which the gauge field is $A$ and $A'$ respectively. The intersection $O =Q\cap Q'$ is an open set in which both $A$ and $A'$ are defined. To avoid double-counting this in the integral, we can "trim" the patches $Q$ and $Q'$ down to smaller patches $P$ and $P'$ that touch only along a $2$d interface. Then we can try to define the integral by $$ \int_P A\,dA+\int_{P'}A'\,dA', \tag{a} $$ as I did above. But what happens if we move the interface to some other place within $Q\cap Q'$? Let $o\subset O$ denote the $3$d open set contained between two different choices for the $2$d interface. Then the difference between the two versions of the integral is $$ \int_o (A'\,dA'-A\,dA). \tag{b} $$ Since $A'$ and $A$ are both defined within $o$ and represent the same gauge-invariant field strength, we must have $dA'=dA$. This gives $$ \int_o (A'\,dA'-A\,dA) =\int_o (A'-A)\,dA. \tag{c} $$ Integrating by parts gives $$ \int_o (A'-A)\,dA=\int_{\pl o} A'\wedge A. \tag{c} $$ I don't see any reason why this last integral should be zero. This suggest that we cannot define the $3$d Chern-Simons action this way, because it would require making arbitrary choices about where to put the $2$d interfaces, and the value of the integral apparently depends on those choices.

3. Defining the action without patches

Dijkgraaf and Witten's paper and Tong's notes both mention that the Chern-Simons action for a $U(1)$ gauge field can be properly defined by treating the $3$d manifold as the boundary of a $4$d manifold. That's the definition I'll use in the rest of this answer. The definition is $$ \frac{k}{4\pi}\int_{M=\pl X} A\wedge dA := \frac{k}{4\pi}\int_{X} F\wedge F \tag{7} $$ where $F$ is the field-strength two-form and $X$ is a $4$d manifold with boundary $M=S^1\times S^2$. This definition is reasonable because $F\wedge F=d(A\,dA)$ locally. The advantage of the definition is that the two-form $F$ is defined globally, not just patchwise, so the right-hand side of (7) is unambiguous.

To apply this new definition to the configuration described in the question, define $X$ by taking the $S^1$ to be the boundary of a disk $D$, so that $X=D\times S^2$. Then we can write $$ F=F_D+F_{S^2}, \tag{8} $$ with $$ \frac{1}{2\pi}\int_{S^2} F_{S^2}=1. \tag{9} $$ (Equation (9) is the equation numbered (1) in the question.) To construct $F_D$, use polar coordinates $r,\alpha$ on the disk $D$ such that the boundary $\pl D=S^1$ is at $r=1$. (To relate this to Tong's notation, use $x^0=\alpha$ and $R=1$.) Then the two-form $$ F_D = d(r^2)\wedge d\alpha = 2 dr\wedge (r\,d\alpha) \tag{10} $$ is well-defined everywhere on $D$ (even though I wrote it using the periodic coordinate $\alpha$) and can be written as $$ F=dA_{S^1} \hskip1cm \text{with} \hskip1cm A_{S^1}=r^2\,d\alpha. $$ The two-form satisfies $$ \int_D F_D = \int_{D} dA_{S^1} = \int_{S^1} A_{S^1} = \int_{S^1} d\alpha = 2\pi. \tag{11} $$ Now, as Tong mentions, we can evaluate the action (7) like this: \begin{align} \frac{k}{4\pi}\int_{X} F\wedge F &= \frac{k}{2\pi}\int_{X} F_D\wedge F_{S^2} \\ &= \frac{k}{2\pi}\left(\int_{D} F_D\right)\left(\int_{S^2} F_{S^2}\right) = 2\pi k. \tag{12} \end{align} Notice the factor of $2$ that came from the two identical cross-terms when $F\wedge F$ is expanded using $F=F_D+F_{S^2}$.

4. The effect of the gauge transform

We can use this new definition of the action to show directly that the action changes by $2\pi$ times an integer under a transformation that has the form $A_{S^1}\to A_{S^1}+d\alpha$ on $S^1$. We can define this transformation on all of $D$ by $$ A_{S^1}\to A_{S^1}+r^2\,d\alpha = 2A_{S_1}. \tag{13} $$ This qualifies as a gauge transform on $S^1$ but not on $D$, so it doesn't necessarily leave $F_D$ invariant. In fact, the effect of this transformation on $F_D$ is $$ F_D\to 2F_D. \tag{14} $$ Now, using (12), we can see how the action transforms under (13): $$ \frac{k}{4\pi}\int_{X} F\wedge F \to 2\frac{k}{4\pi}\int_{X} F\wedge F. \tag{15} $$ In other words, under the transformation (13), which is a legitimate gauge transform over $M=S^1\times S^2$, the action changes by an amount equal to (12), which is $2\pi$ times an integer. This is the result that the question tried to derive — but got stuck on the ill-defined integration-by-parts step. The next section shows that the integration-by-parts trick has a well-defined version using the definition (7) of the action.

5. A well-defined version of the integration-by-parts step

Since $F\wedge F=d(A\,dA)$ locally, we should be able to write $$ \frac{k}{4\pi}\int_{X} F\wedge F = \frac{k}{4\pi}\sum_{Y\subset X}\int_{Y} d(A\wedge dA). \tag{16} $$ The sum is over a set of patches $Y$ that cover the $4$d manifold $X$ and that touch each other only along $3$d interfaces. Evaluating the left-hand side as in (12), we get \begin{align} \frac{k}{4\pi}\int_{X} F\wedge F &= \frac{k}{2\pi}\int_{X} F_D\wedge F_{S^2} \\ &= \frac{k}{2\pi}\left(\int_{D} F_D\right)\left(\int_{S^2} F_{S^2}\right) \\ &= \frac{k}{2\pi}\left(\int_{S^1} A_{S^1}\right)\left(\int_{S^2} dA_{S^2}\right) \tag{17} \end{align} (where the decomposition into patches is left implicit). In contrast, if we start by writing $F\wedge F=d(A\wedge dA)$ instead, then we get \begin{align} \frac{k}{4\pi}\int_{X} F\wedge F &= \frac{k}{4\pi}\sum_{Y\subset X}\int_{Y} d(A\wedge dA) \\ &= \frac{k}{4\pi}\sum_{Y\subset X}\int_{\pl Y} A\wedge dA \\ &= \frac{k}{4\pi}\sum_{Y\subset X}\int_{\pl Y} \Big(A_{S^1}\wedge dA_{S^2} +A_{S^2}\wedge dA_{S^1}\Big). \tag{18} \end{align} We can take each of these patches to have the form $$ Y = \text{(half of the disk $D$)}\times \text{(hemisphere of $S^2$)}. \tag{19} $$ Now, focus on the second term on the last line of (18), because this term corresponds to the terms that were equated to zero in the question. It would be zero if the integral were only over $S^1\times S^2$, as in the first section of this answer. But here the integral is over $\pl Y$, the $3$d boundary of a $4$d patch. This includes a part of the form $$ \text{(half of $S^1$)}\times S^2, \tag{20} $$ but it also includes $$ \text{(half of $D$)}\times \text{(equator of $S^2$)}, \tag{21} $$ and $dA_{S^1}$ is not zero on this part, so integrating by parts can give a non-zero result. The reason for this is that the boundary of a direct product space is $\partial (A \times B) = (\partial A \times \bar{B}) \bigcup ( \bar{A} \times \partial B)$. Integrating by parts gives \begin{align} \frac{k}{4\pi}\sum_{Y\subset X}\int_{\pl Y} &\Big(A_{S^1}\wedge dA_{S^2} +A_{S^2}\wedge dA_{S^1}\Big) \\ &= \frac{k}{2\pi}\sum_{Y\subset X}\int_{\pl Y} A_{S^1}\wedge dA_{S^2} \tag{22} \end{align} because the contributions from the adjacent parts of the different patches' boundaries cancel each other, so we see that (18) is consistent with (17), and both are consistent with Tong's result.

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  • $\begingroup$ I've deleted a number of obsolete comments and/or responses to them. $\endgroup$ – David Z Jun 29 at 5:02
  • $\begingroup$ Wow, this was an amazing answer. So am I to understand that all of the contradictions I was arriving at were due to the fact that the Chern-Simons action is not well-defined if it is thought of as a 3d action as opposed to the boundary of a 4d term? Also, I can't upvote this anymore, is it possible that I can still award a bounty? This has been very helpful and very thorough. $\endgroup$ – Stratiev Jun 29 at 10:28
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    $\begingroup$ @Stratiev You're right, I'm attributing the contradictions to an ambiguous definition of the action. I haven't seen that conclusion confirmed explicitly in the literature, but I haven't seen any evidence against it either. By the way, the text around equation (5.2) in Dijkgraaf and Witten's paper is where they say that the boundary-of-4d definition is a valid and sufficiently general definition for the $U(1)$ case. $\endgroup$ – Chiral Anomaly Jun 29 at 12:29
  • $\begingroup$ @Chiral Anomaly: Thank you for a very clear explanantion of something that I have always found opaque. $\endgroup$ – mike stone Jun 29 at 15:49

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