# Connections on fibre bundles

1. Aug 13, 2006

### mma

I try to understand the notion of connections on fibre bundles from the lecture notes "http://arxiv.org/abs/math-ph/9902027" [Broken] by George Svetlichny.
On page 27 stands what the attached picture shows.

(I don't know whether the attached picture will be visible, so I copy the text here:
)

I don't see, why must have this projection function must have this form. Why must $$\Gamma$$ not depend on y, and why must it be linear function of ξ ?

Last edited by a moderator: Apr 22, 2017 at 12:10 PM
2. Aug 13, 2006

Staff Emeritus
The trivialization he is talking about is an open set U in the base cross the fibre F: U X F. The tangent space to U X F is just the cross product of tangent spaces
$$T_{(x,f)} = T_x(U) X T_f(F)$$

So let $(\xi,y)$ be a vector in the tangent space having components in the tangent spaces to U and F. This vector is vertical by definition if its component in the "U-direction" vanishes: $\xi = 0$. So it is only tangent to the fibre F. This is all unproblematical, the work he is doing here is on defining a selection of horizontal space, out of the inifnite choices of such available.

Now we have a projection map $\pi_{(x,f)}$ from our trivialization, inherited from the full bundle, and we can look to see what it does to the tangent spaces. It will induce a linear map the tangent space of the top $T_{(x,f)}(U X F)$ to the tangent space of the base $T_x(U)$. Linear because these are vector spaces and we can require this of the bundle. And we know that the vertical vector in the top tangent space has no U component so it will map into just the 0-vector in the U tangent space, but we want to define a "best possible" linear component in the U direction. By construction this component, a horizontal component, will be independent of our vertical component, y. So it will depend only on the horizontal vector $\xi$; of course this vanishes in the fibre as we have seen, but it will vary across T(U), and we will use this linear variation in a vector space to define our horizontal subspace. So we can express this linear variation by a linear factor multiplied by the vector $\xi$, and that factor will vary linearly as we vary the top space point (x,f), so write the factor as $\Gamma(x,f)$, and now our projection of tangent spaces reads
$$\pi_{(x,f)}(\xi,y) = (0+\Gamma(x,f)\xi)$$

The second term defines our new horizontal subspace.

Last edited: Aug 13, 2006
3. Aug 14, 2006

### mma

I hope that I understand you properly.

Does this figure show the situation correctly?

4. Aug 14, 2006

Staff Emeritus
Yes I believe so, although one of the components that it seems to me you have labelled v is one I would have labelled y. I could be misreading your diagram, though.

5. Aug 15, 2006

### mma

Thank you very much for your answers. Probably I will have other questions in the future about connections, but now I return to thinking for a little time. I will come again soon.

(sorry for my ugly letters. v appears only four times in the diagram: once in its $$v = (\xi, y)$$ definition expression, once as the label of the $$(\xi, y)$$ vector, and two times as the argunent of the projection maps. Other v-s are y-s.)

6. Aug 15, 2006

Staff Emeritus
That's cool, and I like your diagram. Press on!

7. Aug 16, 2006

### mma

Thank you!

Before I go further, a small supplement to your explanation.

I looked after this, and I found that in the theory of the differentiable manifolds, this induced map is called the tangent mapping and is proved that it is linear. So our requirement is fulfilled always automatically.

8. Aug 16, 2006

Staff Emeritus

You are right. I didn't have my references and didn't want to claim more than I could immediately show.

9. Aug 17, 2006

### mma

Could' somebody show an expressive geometrical or physical example for a trivial fiber bundle, where there is at least two kind of natural choices of the horizontal subsets?

10. Aug 17, 2006

### Doodle Bob

Keep in mind that, given a fibre bundle, there is no "natural" connection associated to it, even if it is a trivial fibre bundle. That's why the theory of connections (and gauge theory, for that matter) can be so challenging. However, in certain circumstances, one (or a family) of connections can be seen as the "best" one to use.

The examples that spring to mind involve seeing $R^3$ as an R-bundle over $R^2$.

Naturally, we have the "trivial" connection: At each point $P(x,y,z)$, the horizontal bundle is given by $span({\partial \over {\partial x}},{\partial \over {\partial y}})$, i.e. the plane passing through P and parallel to the xy-plane.

Another connection on this is given by: At each point $P(x,y,z)$, the horizontal bundle is given by
$span({\partial \over {\partial x}}+y{\partial \over {\partial z}},{\partial \over {\partial y}}+x{\partial \over {\partial z}})$. So, at every point, the horizontal bundle is the plane passing through P, spanned by (1) a line parallel to the xy-plane and passing through both P and the z-axis and (2) a line that passes through P, is perpendicular to the first line, and dips down on its leftside (viewing the line from the z-axis).

The geometry of this last connection is actually rather famous, known often as the Heisenberg group since it originated from Heisenberg's work in quantum physics and has a group structure to it that preserves the connection bundle.

11. Aug 18, 2006

### mma

Nice example, thanks!

This makes me remember to another (even more simple) example that I met before.

The Galilean (nonrelativistic) spacetime can be regarded as a fibre bundle with base space "time" (as a one-dimensional affine space) and with fiber "space" (as a 3-dimensional affine space). The total space of this bundle are the set of the (pointlike, instanteneous) events of the world (forming an 4-dimensional affine space). The vertical subsets are the 3-dimensional spaces of simultaneous events, but there is no absolute (observer-independent) way to define the horizontal subsets (i.e. the "space-points"). However, for every (pointlike) observer, it is natural to regard the world-lines parallel to his world-line (i.e. "space-points at rest" according to him) as horizontal subsets.

12. Aug 21, 2006

### mma

Later on this lecture note (referenced in #1) deals (among others) with the interpretation of Maxwell's equation as a gauge theory (on the top of page 67).

Could somebody summarize the basic objects of this model? Here stands only that the electromagnetic four-potential is regarded as a gauge potential of an invariant connection on a principal U(1) bundle.
But, what is exactly this bundle? What is the base-space, the fiber, the projection map, and what are the horizontal subspaces of the tangent spaces determined by the gauge potential?

Last edited: Aug 21, 2006
13. Aug 29, 2006

### mma

OK, I see that this was a stupid question.

But let' return back for a moment to the original question (that is to the first equation on page 27 of Svetlichny's lecture notes), I think that the point is that we have two kind of splitting of $$T_pE$$.
The first is defined by the $$h_U = (\pi, f)$$ trivialzation, more exactly, by the $$dh_{Up} = (d\pi_p,df_p)$$ tangent mapping of it. This splitting is $$T_pE = \Xi_{{h_U}p} \oplus V_pE$$ where $$\Xi_{{h_U}p}= dh_{Up}^{-1}(T_{\pi(p)}U \times \{0\})$$, $$V_pE = ker(d\pi) = dh_{Up}^{-1}(\{0\} \times T_{f(p)}F)$$.
The second splitting is $$T_pE = H_pE \oplus V_pE$$. We want to express the effect of the second splitting by means of the components in the first splitting. For the sake of readibility, the indices and other constant attributes will be omitted, so our formulae read $$T = \Xi \oplus V$$ and $$T = H \oplus V$$.

Let denote $$P(A,B)$$ the projection on $$A$$ along $$B$$. Let be $$\xi \in \Xi, y \in V$$. We want to calculate $$P(\Xi,V) \circ P(V,H)(\xi+y)$$ and $$P(V,\Xi) \circ P(V,H)(\xi+y)$$.

$$P(\Xi,V) \circ P(V,H) = 0$$ because $$P(V,H)(v) \in V (v \in T)$$. $$P(V,\Xi) \circ P(V,H) =P(V,H)$$ for the same reason. So, $$P(\Xi,V) \circ P(V,H)(\xi + y) = 0$$, $$P(V,\Xi) \circ P(V,H)(\xi + y) = P(V,H)(\xi + y) = P(V,H)(\xi) + P(V,H)(y) = P(V,H)(\xi) + y$$, because the linearity of $$P(V,H)$$ and because $$y \in V$$.

That is, in the first splitting (i.e, in our trivialization) $$P(V,H)(\xi,y) = (0, P(V,H)(\xi) + y) = (0, P(V,H)\vert_\Xi(\xi) + y)$$. Introducing the notations $$\pi^v = P(V,H)$$ and $$\Gamma=P(V,H)\vert_\Xi$$ and writing back the indices of them, we get literally the equation in question.

Last edited: Aug 29, 2006
14. Sep 5, 2006

### mma

Now I see clear what connections are. They are certain subspaces of the tangent space of the fiber bundle, and are declared to be 'horozontal'.
But, when I try to find out, what to do this with covariant derivative, I get in truoble. In the lecture notes of Svetlichny, there is a definition on p43 of covariant derivative, and this definition includes the projections on the vertical subspaces. Fine. But what are these horizontal subspaces for exaple when we speak of the covariant derivative of a vector field on a sphere? And at all, what is the fiber bundle in this case?

Last edited: Sep 5, 2006
15. Sep 6, 2006

### Doodle Bob

As Svetlichny most likely describes, if the fibre bundle is a vector bundle, then a connection (as a subbundle of the tangent bundle of the fibre space) can be used to create a covariant derivative on sections of the fibre bundle and vice versa.

In the case of the sphere S^n (and any other Riemannian manifold), the fibre space would simply be the tangent bundle, TS^n. I believe the process of construction would be:

2. Construct the Levi-Civita connection (i.e. covariant derivative) with respect to this metric
3. Use the covariant derivative to construct the horizontal subbundle of T(TS^n).

16. Sep 7, 2006

### mma

How can one imagine geometrically this horizontal subbundle of $$T(T(S^2)$$?

17. Sep 11, 2006

### mma

To pose this question alternatively, can we say anything about the Levi-Civita connection without using coordinates? I know that if our manifold is embedded into a higher dimensional Euclidean space, then yes: The parallelity of a vector field along a curve can be defined by projecting the derivative of the vector field on the manifold. If this projection is everywhere zero, then the vector field is parallel. And defining parallelity of the vector fields along curves means the definition of the Levi-Civita connection.
But can we give a coordinate-free definition of Levi-Civita connection in the case, when our manifold is a general Riemann-space and isn't embedded into an Euclidean space?

18. Sep 12, 2006

### mma

I expected an answer something like I've found now in this book: "http://www.amazon.com/Differential-Forms-Connections-W-Darling/dp/0521468000" [Broken]

This book seems to me very good!

Last edited by a moderator: Apr 22, 2017 at 12:45 PM