though that's not surprising, given that the nlab entry on 'tensor category' specifically states that the literature isn't always consistent about the precise definition
@robjohn how about restricting the integral to the semi circular arc in the positive real half plane (instead of going full circle) and then collapse it to the imaginary axis $(-i,i)$ (the diameter of the circle)? would that work?
@robjohn I like my spaghetti code. I started reading Halmos online pdf about "How to write mathematics." Halmos level of mathematics is not what I am aiming for though. The code I put in my questions is for my self to remember what I was experimenting with. The questions I write are to get at least one upvote so that it stays at the top of my profile for a while, and so that I will find it more easiliy the next time I visit the site.
I posted an answer to a question that I misread. I came back hours later to a bunch of downvotes. I fixed my answer, and I think it is the simplest one, but some of the downvotes have persisted. Bleh!
Sometimes ... It's frustrating. I actually spent over an hour figuring out a differential geometry question (embarrassing — I should have seen the right approach in a second), but never got a response at all from the OP. The good news is that it motivated me to add a new exercise to my diff geo text :P
@Ted: It comes down to proving an a priori L^2 bound on flat connections in a Coulomb slice (that is, if $A$ is flat, a bound on the norm of 1-forms satisfying $d_A a + a\wedge a = 0$ and $d^*_A a = 0$)
I can get as many bootstrapping lemmas as I want but getting an actual bound to start with is proving difficult, though I expect it will end up being something obvious
It still ends up being the second equation of the two I wrote. This is one of those things where I'm sure there isn't just an analogy, it's just literally the same notion.
yeah. in relativistic electromagnetism, one can formulate everything in terms of a field strength tensor $F^{\mu \nu}$ determined by the 4-potential $A^\mu$ as $F^{\mu \nu}=\partial^\mu A^\nu -\partial^\nu A^\mu$
anyways, one has gauge freedom for that potential, which one usually fixes by a choice of gauge. for Coulomb gauge, it's that $\partial_i A^i = \nabla\cdot \mathbf{A} = 0$
so stuff like $a_\mu a^\mu = a_0^2-a_1^2-a_2^2-a_3^2$ is typical
anyways. what's not clear to me off the top of my head is what your $a$ would correspond to. as a 1-form, i guess i'd expect it to be the 4-potential $A^\mu$ expressed as a differential form
but then i'm also puzzled at how $a\wedge a$ wouldn't vanish trivially.
@TedShifrin Suppose if there is a map between two topological space each generated by different family of seminorms, how does one characterize the continuity of the map through the seminorms
@Semiclassical: It's a 1-form with values in a Lie algebra, essentially. The wedge product includes the Lie bracket. So for 1-forms it's trivial for any abelian Lie algebra - like that of U(1) - but not necessarily for interesting ones, like mine, the Lie algebra of SU(2)
@Semiclassical: Yes, I know. The curvature formula (where $A$ is a connection) is $F_{A+a} = F_A + d_A a + a \wedge a$. In your case, you're on the trivial bundle, and you've chosen $A$ to be the trivial connection so that $d_A = d$.
Do you believe there is value in trying to write/talk in a way that's comprehensible to both physicists & mathematicians in gauge theory, @MikeMiller? Or do you think it's a waste of time/effort?
I am probably going to stop thinking about translating equations between our languages. I'm not sure what any of the three of us talking about gauge theory gain from it...
The first equation is that the curvature (field strength tensor) vanishes. This isn't the normal object of study in Yang-Mills theory, it's the very simplest case.
@Danu: You asked earlier if there's value in increasing mathematician-physicist dialogue. I think there is. You could ask if there's value in this current back and forth. That's definitely false. So I'm going to go do some actual work.
@Danu: I take the previous thing back, because I realized you were thinking about $U(1)$. Flat $U(1)$ connections are in bijection with $H^1(M;i\Bbb R)$, where $M$ is the manifold you're doing this on; in $\Bbb R^3$ there are none.
Flat $U(1)$ connections on a given bundle, that is.
As SemiC said it's rather interesting for other gauge groups.
My initial response was annoyed about the tone. If you want to have a conversation, you shouldn't just tell me that something is uninteresting from your point of view. That's a good way to end a conversation. You should ask why it's interesting from mine.
And I should do the same to you, of course, which I'm not good at but should make more effort to be.
@Semiclassical: $(-1)^{k(n-k)}*d*$, where $*$ is determined by the Riemannian metric, $n$ is the dimension of your manifold, and you're acting on $k$-forms.
@Semiclassical: My mathematician's answer to why I need to gauge fix is that I want my equations to be elliptic PDEs. Normally I'm studying the ASD equation $d_A^+ a + (a \wedge a)^+ = 0$, where "$+$" denotes taking the self-dual part of the 2-form (this makes sense on an oriented Riemannian 4-fold). The linearization of this is just called $d_A^+$. (Here you may as well take $A$ to be the trivial connection for discussion.)
Now this is not an elliptic linear PDE. This is for more or less dumb reasons: Take a solution. Any gauge transformation of that solution is still a solution. This means (because there are a lot of gauge transformations) that I must have an infinite-dimensional space of solutions, and elliptic PDEs have a finite-dimensional space of solutions. At least to me, that's one of their most essential properties.
Well, how do I at least attempt to fix this? I should try and pick some "slice" of the space of connections that's more or less "orthogonal" to the action of the gauge group. I mean this more or less literally: at the base connection $A$, I can linearize the action of the gauge group to get a subset of $\Omega^1(\mathfrak g_P)$ (again, think of this as 1-forms with values in $\mathfrak{su}(2)$)
The subspace you get as the "linerization" of the gauge group action is the image of $d: \Omega^0(\mathfrak g_P) \to \Omega^1(\mathfrak g_P)$. And up to a finite-dimensional subspace, this is perpendicular to the kernel of $d^*$.
@Semiclassical: They're exactly the same notion. You're trying to restrict to a set of fields (?) that you can't gauge transform, at least not very much. That's exactly what I'm doing.
So I restrict to the kernel of $d_A^*$. Now the key point is that $d_A^* \oplus d_A^+: \Omega^1 \to \Omega^0 \oplus \Omega^2$ - this is a linear elliptic operator.
