I have 2 definitions here which I am struggling to bring into register, though my text assures me they are equivalent. Please help me. 1) Given the bundle \(P(G.M)\), where everything in sight is a manifold, and some tangent space \(T_p(P), \quad p \in P\), then a connection is a smooth assignment of a horizontal subspace \(H_p(P) \subset T_p(P)\) such that \(T_p(P) = H_p(P) \oplus V_p(P)\) where \(V_p(P)\) is the vertical subspace. 2) Given the bundle \(P(G,M)\), a connection is a 1-form \(\omega_p: T_p(P) \to \mathfrak{G}\) (the algebra for G). There are a few bells and whistles attached to do with preserving the right action \(R_g: P \times G \to P,\quad R_g(p) = pg \in P\), which am skating over (possibly at my peril?). The best I can do, on an extremely superficial and purely intuitional level is this: first consider \(V_p(P)\). Let us assume this is tangent to a typical fibre which we know is isomorphic to our structure group \(\pi^{-1}(m) \simeq G,\quad \pi: P \to M,\quad m \in M\). We know there is an isomorphism \(T_e(G) \simeq \mathfrak{G}\) so, since \(V_p(P)\) is tangent to \(G\), by transitivity I have that \(V_p(P) \simeq \mathfrak{G}\), more-or-less canonically. No such canonical mapping exists for the horizontal subspace, obviously, since it is tangent to \(P\) not \(G\). So in a manner of speaking this map needs to put in by hand, hence the 1-form \(\omega\). In a way this is nice: we know that that where,say, \(G = U(1),\quad \mathfrak{G} = \mathbb{R}^1\), so that in the case of a \(U(1)\) bundle we simply have our 1-form sending vectors to reals, as always. But my problem is this: in my first definition of a connection, I have it as a vector space \(H_p(P)\) whereas in my second I have it as an element in a covector space. How do I bring these into register?
Given a horizontal subspace \(H_p\), how do you construct the corresponding one form (at point p)? Well, it is just the linear projection \(\omega:T_p\to V_p\) whose kernel is \(H_p\). Given a one form \(\omega:T_p\to \mathfrak{G}\simeq V_p\), how do you find the corresponding horizontal space? Well, it is just the kernel of \(\omega\).
Unless I'm being silly (ie ignoring it's a matter of definition) is the algebra of U(1) not the set of purely imaginary numbers, as you have \(e^{i\theta} \in U(1)\), so \(i\theta \in u(1)\)?
Ah, I expected it to be something like that, not quite a matter of definition but close enough. In regards to the original post, I tend to think of the bundle first as a trivial bundle. The vertical part of the bundle is such that every point in the vertical space is associated with the same point in the base space and moving through the base space 'sweeps out' the total space itself. If the bundle isn't trivial then you'll get some kind of non-trivial variation of the vertical space as you move through the base space. Simplest examples would be having both the fibres and the base space as \(S^{1}\). Trivially any point in the base space has an associated circle fibre and as you move through the base circle the fibres sweep out a torus. This gives a simple distinction between the vertical space and the horizontal space of the total bundle. If you've got a non-trivial bundle then you get a twisted torus such as a Klein bottle.
I thank you both for your replies. Not for the first time on this forum, feel I have been a bit slow-witted. Specifically I was forgetting some elementary linear algebra. Suppose that \(V = U \oplus W\). Then define the projection \(\pi_w: V \to U\) by \(\pi_w(v) = u, \quad v = u+w\). This called the projection of \(V\) onto \(U\) along \(W\), and \(\text{Im}(\pi_w)= U,\quad \ker(\pi_w)= W\). But given \(V' = U \oplus X\), then, although \(\pi_x(v') = u\) we may not assume that \(\pi_w = \pi_x\). Making the obvious substitutions, we will have that \(\omega\) depends totally on the choice of horizontal subspace, which we know is not canonical, so I may, if I wish, identify \(\omega\) with this choice, which seems to answer my original question. This prompts the following, possibly equally dumb question. Linear algebra informs me there is always a second projection \(\pi_u:V (=U \oplus W) \to W\) in the general case, the projection of \(V\) onto \(W\) along \(U\), so may I have here a vertical projection \(p_v:T_p(P) \to H_p(P)\)? Well, one sees this is somehow related to the (right) action of \(G\) on \(P\), so that if, say, \(R_g(p) = pg:=p'\), one will have that \(T_p(P) \to H_{p'}(P)\). I don't like this, it looks like it should be \(H_p(P) \to H_{p'}(P)\), that is, shoving the horizontal space along the fibre. Then this is clearly not a projection. Help! Furthermore, since all fibres here are identical, how can I ensure that I "stay on" the same fibre? For example, the group of all automorphisms \(Aut(G):G \to G\) will, in general, tend to "mix up" fibres. May I specify a subgroup of "vertical" automorphisms, say \(Aut_v(G)\) of fibre-preserving maps? I have horrid feeling I am straying into gauge theory here.
Yes, but it is not very convenient to define connection using this since the range (of the projection) would be varying and so would not be a one form with values in some linear space. You mean \(Aut(P):P \to P\) ? In that case, the "vertical" (i.e., fibre-preserving) automorphisms are called gauge transformations, and the "horizontal" (i.e., G-equivariant) ones are called automorphisms of the principal bundle P.
Wow, thanks for that, it explains a lot to me. Your correction makes perfect sense - thanks again (will you marry me?) So since I seem to strayed into gauge theory territory, maybe someone can explain why (as I read) that the connection on a \(U(1)\) bundle is identically the vector potential \(A_{\mu}\) that is found in "ordinary" EM, which again by my reading, is de facto a gauge theory. Is this a question with only a physics answer, or is there some "pure" math to show why?
The main thing is that U(1) gauge theory with the Yang-Mills action gives Maxwell's equation, so the EM field can be described by a gauge theory. I think this is a result of many attempts to "geometrize" Maxwell's theory inspired by Einstein's GR. There can be many ways to geometrize Maxwell but this particular approach is justified by the fact that it leads to good theories like the Yang-Mills, Weinberg-Salam, and QCD. If I am not mistaken there is also relation between the phase of the quantum wave function and the internal U(1) variable, which is verified in the so-called Aharanov-Bohm experiment.
Ah well, maybe I shouldn't have asked. But anyway, let me just fly a kite. Assume for definiteness that my base manifold \(M\) is Minkowski spacetime, and that my structure group \(G\) is unitary with positive unit determinant and that the bundle \(P(G,M)\) is only locally trivial. OK, the math here is rather daunting - brain curdling, even - so let me work intuitionally for now. Define a curve \(C \in M\). Then at each point \(m \in M\) I will encounter the group \(G\) of continuous symmetries. Then at each point on my curve I have the choice of some symmetry given by \(g \in G\). Let's say I make the same choice at each point; I will say that in this circumstance the resulting cuve \(C' \in P(G,M)\) is in some vague sense "parallel" to the curve \(C \in M\). The fact that \(G\) is unitary ensures that whatever physics describes the curve \(C \in M\) applies equally to the curve \(C'\) An immediate problem arises: since \(P(G,M)\) is not globally trivial, all sorts of weird things seem to happen with this notion of parallelism (we knew they would, since parallelism is, in general, not a defined concept on manifolds globally). So let's work locally. The local bundle over an open subset \(U \sub M\) is defined as \(\pi^{-1}(U) \simeq U \times G\). Ah, that's nicer. So now our curve is deemed to be \(C \in U\) with \(C' \in \pi^{-1}(U)\). Intuition shows that the curve \(C' \in \pi^{-1}(U)\) is just the integral curve of elements of the horizontal subspaces \(H_p(P) \subset T_p(P) \quad \forall p \in P\). Let's now suppose there is another integral curve \(C''\) defined at the points \(R_g(p) = pg \equiv p' \in P\) similarly defined, but without the requirement for parallelism of the vague sort I talked about. Now since our derived curves are based on horizontal vectors, I can think of these as "crossing" the fibres at a certain "angle". Obviously the gradient 1-form is applicable here, so I arrive at the rather nice conclusion that, just the Levi-Civita connection preserves parallelism under arbitrary transport of a geometric object (tensor, generally) on a Riemann manifold, so the connection 1-form used in gauge theory preserves the gradient for any collection of curves in \(P(G,M)\), and thus preserves "parallelism" as I rather naughtily defined it Yes I know, I am a wind-bag. But am I getting the gist of gauge theory here?
I don't really understand the last (relatively) long paragraph but I have the feeling that you are on the correct path. It might help just to consider U(1) bundle or take G as some matrix group, and explicitly work out what you want to show.
Ok, I know I'm replying to an old post, but... The reason the two definitions are the same is the the kernel of the 1-form defines the complementary space to the vertical vectors: \(V_p \ = Ker \ T\pi: \ T_{p}P \rightarrow T_{\pi(p)} M\) \(H_p \ = Ker \ \omega: \ T_{p}P \rightarrow {\mathcal G}\) where \( {\mathcal G}\) is the Lie algebra of G. Then \(T_{p}P \ = \ V_p \oplus H_p\) The condition on \(\omega\) ensures the horizontal subspace is invariant under the right action of G.
Well informed posts and posters are very welcomed here, late or otherwise Please Register or Log in to view the hidden image! Welcome to sci, rwilsker.
Bizarrely, I just saw a post asking for a definition of holonomy in a loop quantum theory context. I did my best to summarize the Wikipedia page but expect that the question comes from someone who is decades away from understanding what he claims to be reading.
