6
$\begingroup$

I am learning TQFT from compact Lie groups by Freed, Hopkins, Lurie, and Teleman: https://arxiv.org/abs/0905.0731 , and got stuck very hard even in the first section ($n = 1$), which was "trivial but included for completeness".

In particular, I have several unfamiliar terms while it discusses "1-dimensional pure gauge theory". Each term will require some explanations, if not too long. Any pointers to places where I can learn more about the terms will be highly appreciated.

1. 1-dimensional "pure" gauge theory

I have an impression that gauge theory is just bundle theory in math term. But how about the adjective "pure"?

2. The standard quantization procedure

Given a compact Lie group $G$, an abelian character $\lambda: G \to U(1)$, and a $G$-bundle with connection over the circle ($g$ being its holonomy), they define a 1d-TQFT by assigning to the circle the number $\lambda(g)$. Then "by the standard quantization procedure", they assign to the positively oriented point a subspace of $\mathbb{C}$, depending on if the abelian character is trivial or not.

I know little about geometric quantization - all I have read is J. Baez's informal introduction (http://www.math.ucr.edu/home/baez/quantization.html). But I have know idea how these two relate, and the article even claimed that this procedure relates to "the Gauss law in physics", making it even more mysterious for me..

3. Path integral over the groupoid $G//G$

Now the value assigned to the circle is just the dimension of the vector space assigned to the positively oriented point, and as above this value is either 0 or 1 depending on whether the character is nontrivial or not. The authors claimed that this may be understood as the result of the **path integral over the groupoid $G//G$ of connections on $S^1$ with respect to Haar measure:

$$ \frac{1}{|G|} \int_G \lambda(g)dg = 0 \mbox{ or } 1$$

.. This has nothing to do with what I think a path integral is: to me, a path integral is an integral over all path/section space with a suitable weight.


I hope I express my questions clear. If there's any confusion, please let me know. Thank you.

$\endgroup$
  • 2
    $\begingroup$ I think pure gauge theory is gauge theory without matter. $\endgroup$ – jjcale Oct 12 '19 at 17:01
  • $\begingroup$ Yes, that is the standard physics parlance. $\endgroup$ – Michael Engelhardt Oct 12 '19 at 17:53
  • $\begingroup$ What are matters formulated in mathematics, if fields are viewed as "sections of some bundles"? $\endgroup$ – Student Oct 14 '19 at 2:53
  • 1
    $\begingroup$ A matter field could be pretty much anything for which you can construct a covariant derivative using the gauge field. $\endgroup$ – Michael Engelhardt Oct 14 '19 at 16:42
5
$\begingroup$

A gauge theory in mathematical terms (as I understand it) is a field theory whose fields include a gauge field: a $G$-bundle with connection. In this case, $G$ is a finite group, so there are no non-trivial connections. Thus a gauge field is simply a $G$-bundle.

1) Here, by pure gauge theory, I think the authors mean that this is a theory whose fields are precisely $G$-bundles and nothing more (in particular, no "matter" fields).

2,3) The path-integral is one approach to quantizing a field theory. It says, in particular, we should compute the partition function of the theory by a suitably weighted integral over a space of fields of the theory. The weight is prescribed by the action, a certain function on the space of fields.

Fields in this theory are $G$-bundles. The collection of $G$ bundles on a manifold $M$ (spacetime) can naturally be organized in to a groupoid, whose objects are $G$-bundles and morphisms are isomorphisms of $G$-bundles. For example, the groupoid of $G$-bundles on the circle $S^1$ may naturally be identified with the quotient groupoid $G//G$ (where $G$ acts on itself by conjugation). This identification assigns to a $G$-bundle on $S^1$ its holonomy (which is well-defined up to conjugation).

The (exponentiated) action functional is given by the abelian character $\lambda:G \to T$. This assigns to a $G$-bundle on $S^1$ with holonomy $g$, the function $\lambda(g)$. Thus the partition function on $S^1$ is given by an ''integral'' over the groupoid $G//G$ weighted by the function $\lambda$. This really just means the sum of $\lambda(g)$ as $g$ ranges over $G$, and one must divide by $|G|$ to get the right notion of volume (we are integrating over $G//G$ not $G$). You can check that one gets the stated results.

There is a similar idea happening for the $0$-manifold $pt_+$. In general QFT, one expects to assign to a codimension 1 submanifold a vector space (the "Hilbert space" of the theory). Roughly speaking one might think of elements of this vector space as being functions on the space of fields - or better, sections of a line bundle (which again is associated to the action).

In this case, the function $\lambda$ is actually the same thing as a line bundle on the space of fields on a point (which is the groupoid $pt//G$). The space of sections of this line bundle is either 1-dimensional if the bundle is trivial ($\lambda =1$) or $0$-dimensional if the bundle is non-trivial ($\lambda \neq 1$).

I wish I had something reasonable to add about the Gauss Law. Perhaps someone more knowledgeable could weigh in?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ In principle the $G$-bundle is fixed while the path integral is over connections modulo gauge transformations. $\endgroup$ – Abdelmalek Abdesselam Oct 12 '19 at 19:40
  • 1
    $\begingroup$ @AbdelmalekAbdesselam I agree, that is what is usually meant by the path integral. But I think to do things carefully, one should still sum over the (discrete) set of topological types of bundles to define the partition function. In this case, there are no connections, and the sum is all that remains. $\endgroup$ – Sam Gunningham Oct 12 '19 at 20:06
  • $\begingroup$ Dear @SamGunningham, thank you for your answer! I spent some time on absorbing it, but still got some questions. In this comment I will focus on the "circle" part. --- If I understand correctly, a field here should mean a $G$-bundle with connection .. otherwise how would we define holonomy? If this is correct, is the space of $G$-bundles with connections still classified by $G//G$? I would really like to know some reference about this since ".. with connections" sounds differential for me, whereas $G//G$ sounds topological. --- Secondly, how can I think of $G//G$? I've always been treating.. $\endgroup$ – Student Oct 14 '19 at 3:36
  • $\begingroup$ ..double quotient $//$ like a black box. The only thing I am comfortable with is $*//G$, which I identify as the classifying bundle. In general, those are some infinite dimensional thing that can only be treated well topologically. But now we are doing integration over $G//G$.. how can I think about that? Is there any reference that treats this well? -- I hope my questions won't expand so fast that intimidate you. If that's the case, feel free to just throw some pointers to me. I'm willing to read more stuff before I ask more questions. Thank you! (I will ask for $pt_+$ later if it's ok). $\endgroup$ – Student Oct 14 '19 at 3:41
  • 1
    $\begingroup$ As for references, the best that I know of is that paper of Freed-Hopkins-Lurie-Teleman that you mentioned... You could also search google for lecture notes, notes from talks etc. E.g. I learned much of this stuff from these talks:web.ma.utexas.edu/users/benzvi/GRASP/lectures/NWTFT/nwtft.pdf $\endgroup$ – Sam Gunningham Oct 14 '19 at 9:24
6
$\begingroup$

Just to add something about Gauss' law to the excellent previous answer: The Hamiltonian of a pure gauge theory, as initially derived from the Lagrangean, typically has the structure $H=E^2 +B^2 - A_0 \nabla \cdot E$, where $E$ and $B$ are the electric and magnetic fields, and $A_0 $ is the temporal component of the gauge field. In addition, the Lagrangean contains no time derivative of $A_0 $, and therefore there is no conjugate momentum for $A_0 $ - it is not a dynamical degree of freedom. Instead, it acts as a Lagrange multiplier enforcing $\nabla \cdot E =0$. That is Gauss' law, a constraint on the remaining dynamical fields.

There are distinct ways of dealing with this. One avenue is to use the gauge freedom to choose $A_0 =0$. Then the non-dynamical degree of freedom is gone, but one has lost the knowledge about the Gauss law constraint. In $A_0 =0$ gauge, one must impose the constraint in addition to the dynamical equations generated by the Hamiltonian.

Another avenue is to choose a different gauge, but then the formalism still contains $A_0 $. The "equation of motion" for $A_0 $ is, again, not dynamical, but a constraint on $A_0 $, which one can then solve to eliminate $A_0 $ (this, e.g., generates the Coulomb interaction). In this manner of speaking, Gauss' law doesn't play quite such a central role - one thinks of $A_0 $ being constrained, as opposed to the other fields being constrained.

Either way, one eliminates two components of the gauge field: One by choice of gauge, one by constraint. Only the remaining components are dynamical degrees of freedom. Thus, light comes with two polarizations, even though one starts out with a 4-component field $A_{\mu } $.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.