So, I am having some difficulty here. Let me set the stage, but first let me apologize for the length and density of this post. (I offer no frills, however)..... Suppose that \(M\) is a symplectic manifold, i.e. there is a distinguished closed 2-form of maximal rank \(\Omega\) defined everywhere. Now we know that the set of all vector fields on \(M\), call it, say \(\mathfrak{X}(M)\), closes to a real Lie algebra \(\mathfrak{G}\). Set \(X \in \mathfrak{G}\) accordingly, noting that \(X\) is an arbitrary field on \(M\). We also know that the infinitesimal (left) action of some group on our manifold \(G \times M \to M\) induces the field \(X_M\) by the exponential map \(X_M = \frac{d}{dt}|_{t=0}\exp(X) \cdot m\) where \(m \in M\), and where \(t \in [0,1]\) is an arbitray parameter (call it time if you must), and the centre dot denotes the action. Notice that as \(t =0 \to t=1\), we recover the action \( g \cdot m = m' \in M\). In other words, provided we restrict ourselves to infinitesimal actions, we may recover a particular field on \(M\). So, one more piece of scenery: the field \(X_M\) is apparently said to be Hamiltonian iff, for our 2- form \(\Omega\), that it induces the interior product \(\imath_{X_M}\Omega = d \alpha\), that is this product is exact on 1-forms. Phew, did you bad guys need all that that? Anyway, to my question..... I am given the momentum map \(\mu: M \to \mathfrak{G}^{\ast}\) (where \(\mathfrak{G}^{\ast} = Map(\mathfrak{G}, R)\)) such that \(\imath_{X_M}\Omega = -d(\mu(X)) \,:= d(f_{X_M})\,\,\,\, \forall X \in \mathfrak{G}\). So, I do not get the middle expression: \(\mu\) acts on \(M\), whereas \(X \in \mathfrak{G}\). How can it be that \(-d(\mu(X)) \in \mathfrak{G}^{\ast}\)?
For every \(p\in M\), \(\mu(p)\in\mathfrak{G}^*\), let's denote it by \(\mu_p\in\mathfrak{G}^*\). Then since \(X\in\mathfrak{G}\), we have \(\mu_p(X)\in\mathbb{R}\). Now if you move \(p\) over \(M\), you will get a function \(p\mapsto\mu_p(X):M\to\mathbb{R}\). Call this function \(\mu(X)\), and take its exterior derivative, so \(d(\mu(X))\) is a one-form.
Temur's answered but I have a question about something you said. M is called parallelizable if it admits n=dim M linearly independent vector fields everywhere, ie at any given p in M there's n independent vectors definable and this varies smoothly over all of M. Is this synonymous with being symplectic? For instance, if I can define such n vector fields do I have a symplectic structure?
They are not the same. For instance the dimension must be even for a manifold to admit any symplectic structure.
Oh yeah, doh! Damn, just when people might be thinking I'm clever Please Register or Log in to view the hidden image!
temur, I love you! Your answer makes perfect sense, though I confess I am a little surprised. My (Spanish) authors seem to have a salt-shaker full of super- and sub-scripts which they sprinkle liberally over each page. They might have told me what you just did.... Fortunately, this is not a folly I was ever prone to ( *wink*). It seems that parallelizable manifolds are rather rare, the exceptions being those with the Lie group structure. Here it seems that the parallelizing vector fields are those generated by the (left) action of the group on itself, presumably at the identity. I don't have the details in front of me, lemme see if I can dig them up later. Also later, I want to ask a question about the coadjoint reps.
There's a string/gravity textbook by Orlin which is horrific in its notation. Unsurprisingly he talks about metrics and different coordinates and so initially you have \(g_{ab}\) and various transformations or redefinitions leads through to expressions like \(\hat{\hat{g}}_{\check{A}\check{B}}\). Indiies and tensors are hard enough normally, never mind the slew of hats, checks, tildes and bars used! Yeah, they are pretty rare, tori are such spaces through. It came up in my viva when the external examiner pointed out that you can't assume you can write elements of a cohomology as a linear combination of elements of \(\Lambda^{p}(T*M)\), ie \(\omega = \frac{1}{2!}\omega_{ab}dx^{a}\wedge dx^{b}\). Unfortunately that put a fork in about 10 pages of my thesis.... :bawl:
What, only 10 pages? You got off very lightly! (Mods may wish to move/delete the following, as it is totally off-topic, and slightly self-aggrandizing) When I was a young post-doc at Yale I was asked to external a PhD candidate, and I asked her to do a couple more experiments and re-write the entire thesis, paying due regard to English as it usually understood. The internal examiner (her advisor) never spoke to me again. But, I feel the following VERY strongly: PhD's are handed out like candy these days, at least in my speciality; I know many so-called "doctors of philosophy" who have absolutely no idea of the depth and breadth of their speciality. In fact they seem not even fully to understand the subject in which they are "doctors". So what does this say about the first degree? In the UK, at least, tertiary education is CRAP (I know - I have been at the coal-face) \end rant\ P.S. by edit: I just realized that Alpha may very understandably take the above as a denigration of his efforts and abilities to get his PhD. As might anyone else on this site, for that matter..... This was in no way intended, sorry if any offence was caused
The algebra was correct and since tori are parallelisable and used to construct orbifolds it was okay to leave in provided I made it clear that it was only true for parallelisable spaces. Ironically I was trying to prove a general result and thought "Hey, just write the cohomology elements in terms of the frame basis" and so while the general result was valid the way I proved it wasn't. Please Register or Log in to view the hidden image! You might be wondering how that even got into my thesis since surely my supervisor would point it out. I wouldn't be surprised in the slightest if my supervisor hasn't read my thesis or any of my published work. I think that says plenty.... I see what you mean. When writing up I looked through the theses from old students which are kept around and the variety is enormous. Some were 80 pages with the first 50 being book work while one was 500 pages crammed with vast quantities of work. We've had a few people who obviously couldn't cut it and dropped out in their 1st year. Generally they studied overseas and got the place on the word of their tutor, no interview or the like was asked for. I won't deny that there's tons of things which I probably should know but don't. I guess part of that is due to the fact you learn what you need as and when you realise you need it. Mind you one guy who was finishing up when I started couldn't remember even his 1st year (of the PhD) lecture courses. Didn't even seem familiar with basic terminology. Yet he's just finished his second PhD in another group in the same physics department?! I heard a good one when I went to my sisters masters graduation at LSE. Someone else got a PhD for, once you decoded the title, doing a statistical analysis on what non-British people ate at 3 London food court areas. While I can't deny they did 3 years of work I have to question the topic.... I can't argue much with that. Obviously there's plenty of good universities here in the UK but plenty more crap ones which teach nonsense or don't push students. I feel its my duty to write a scathing comment on a 2nd year's homework when they use \((a+b)^{2} = a^{2}+b^{2}\) or \(\frac{1}{a+b} = \frac{1}{a} + \frac{1}{b}\). AAARRRRGGGHHH!!!! Nah, its fine.
