I am trying to understand some calculations in a paper by Sidney Coleman. He is showing that certain distributions are positive. The paper can be found here.
What I am talking about is happening at the bottom of page 262. (Which is 4 pages into the document.)
We are looking at distributions in 2...
@AndyMiles Sorry, what I just said was nonsense. In any case, still no, not really. What I should have said was that you can choose this to be any section.
no. if there is some $n$ such that $z^n = 1$, then you immediately have $|z| = 1$, so at least you need $z$ to be in the unit circle around $0$ in $\Bbb C$.
any complex number outside the unit circle around $0$ in $\Bbb C$ cannot be in the set.
A more interesting question to ask if every complex number of norm $1$ is in the set. The answer to this is still no, but harder to see. :)
@MikeMiller I'm imagining a principal bundle to be more about the right action of the group than about the local trivialization. In some way, the local trivialization could be $M\times S$ with $S$ the underlying set of the group $G$. The fact that is important is that there's a right action of the group on $S$, and we don't really care about $S$ itself.
Yes, @AndyMiles, that's right. The local trivialization is completely unimportant, we just need to know it exists so that we haven't assembled the fibers in some wacky way.
Well, I mean, it's not completely unimportant, since we use it sometimes. But it's not the point of the idea.
@AndyMiles You should not actually think that there's a copy of $G$ above each point, but rather a $G$-torsor; like we're saying, we don't have a specified basepoint/identity element.
I guess I was wondering, then, if there was an $n \in \mathbb{R}$ for every $z \in \mathbb{C}$ that was not $n = 0$ that satisfies $z^n = 1$. If there was such an $n$, then all of $z \in \mathbb{C}$ would be in this set? @balarka
@MikeMiller I think of it more as a gauge group (because I don't know any algebraic geometry). The gauge is acting at each point, but there's no copy of the group at each point.
@AndyMiles When one says they're interested in QFT in math it can mean some very different things, I think :) The category theorists are really into it.
I ask because I do what some people call mathematical gauge theory, and it's of a very different flavor.
@MikeMiller Oh, cool. Well I left academia after the BSc, so don't expect a very specific interest like in a PhD. Besides my wish to have some understanding of string theory (who doesn't wish that..), I was totally blown when I learned that Witten and colleagues used TQFT to prove important mathematical stuff like topological invariants.
Instanton Floer homology. It's a way of assigning a graded abelian group (like singular homology) to a 3-manifold, that's functorial under cobordism; you do it by counting what we call instantons, but a physicist probably calls "Solutions to the ASD equation". I learned a while back that 'instanton' just means 'solution to the equations of motion'
so telling a physicist I do instantons, which I say to mathematicians at the right flavor of conference, would probably sound like saying "I do equations"
Well afaik an instanton is based on a solution to the classical equations of motion, but it's used for some tunneling amplitude between different vacuum states.
But that might be the QFT instanton. What's an instanton in Floer homology?
@AndyMiles It's probably a special case of the sort of instanton you mean. When I say 'instanton', I mean a solution to the ASD equation on a 4-manifold. The equation takes as input a connection on a principal bundle, and outputs $F_A^+$, the self-dual part of its curvature 2-form.
An instanton is a connection with $F_A^+ = 0$. We then consider them modulo gauge.
Just dropping by to let you know that your recent answer has stimulated me to get going again on Bredon's Topology and Geometry to learn homology theory :).
to be a symmetric matrix, does there only have to be 1 line of symmetry drawn across the matrix in any direction? for ex: is $\begin{bmatrix} a & b \\ a & b \end{bmatrix}$ symmetric?
whats the generic group table for this group: let $L = \{a,b,c\}$ and $G = \{L,\star\}$ is it $\begin{bmatrix} aa & ab & ac \\ ba & bb & bc \\ ca & cb & cc \end{bmatrix}$?
Because the $v$s considered before cannot express this, they cannot be a suitable semantics for intuitionism.
@idonutunderstand Indeed.
We therefore need to turn to other techniques of assigning meaning to our familiar formulas in order to grasp the intuitionistic principles of reasoning.
But if you haven't ever considered things like model theory for first-order logic, it will be quite hard to grasp these different semantics.
You might consider reading this (PDF); the semantics is in chapter 3 (Kripke semantics). I find it one of the best to get to grips with intuitionism. It might help you, or it may go way over your head.
In the latter case, I'd recommend you start with an undergraduate book on mathematical logic.
Let $X$ be a normed linear space and $K$ a bounded convex weak-$*$ closed subset of $X^{*}$. I need to show that $K$ possesses an extreme point; however, I am not entirely sure how to do this.
I suppose I could use Alaoglu's Theorem (Let $X$ be a normed linear spsace. Then the closed unit ball ...
Let $X$ be a normed linear space and $K$ a bounded convex weak-$*$ closed subset of $X^{*}$. I need to show that $K$ possesses an extreme point; however, I am not entirely sure how to do this.
I suppose I could use Alaoglu's Theorem (Let $X$ be a normed linear spsace. Then the closed unit ball ...
