1
$\begingroup$

Let $(P,M,\pi)$ be a Principal $G$ bundle.

I am quoting following lines from the book Modern Differential geometry for Physicists by C J Isham.

The basic idea of parallel transport/covariant differentiation is to compare points in neighbouring fibers ina way that is not dependent on any particular local bundle trivialisation. This suggests looking for vector fields that point from one fiber to another.

I do not understand why does idea of comparing fibers of two nearby points should suggest looking for some vector fields. It is not clear what does it mean to say vector field that points from one fiber to another.

I thought I might understand if I go further but still it is not clear.

In search of some vector fields (which is not clear at this point what kind of vector fields they are) he defines vector field $X^A$ on $P$ given an element $A\in T_eG$. I am not defining it because definition is not relavent here. He then says with out proof that $$\pi_*\left( X^A_p\right)=0$$ I have checked this and it turns out to be true. Before saying this, he says

How ever, the particular vector fields are not suitable for our purposes since they do not point from one fiber to another. On the contrary, the vectors $X^A_p$ are tangent to the fiber at $p\in P$ and hence they point along the fiber, rather than away from it. Technically, $X^A_p$ is said to be a vertical vector i.e., it belongs to the vertical subspace $V_pP$ of $T_pP$ defined by

$$V_pP= \left\{ \tau\in T_pP : \pi_*\tau=0\right\}$$ where $\pi$ is the projection map of the bundle.

I do not understand what does it mean to say vectors are tangent to the fibers or point along the fibers.

$\endgroup$
  • $\begingroup$ Do you know what it means for a vector to be tangent to a submanifold? Well, being tangent to fibers means precisely that. $\endgroup$ – Mariano Suárez-Álvarez Dec 2 '17 at 15:06
  • $\begingroup$ I do not know what does it mean to say vector is tangent to a submanifold. Does it mean it is derivative of a curve at a point which lies entirely in the submanifold? @MarianoSuárez-Álvarez $\endgroup$ – user312648 Dec 2 '17 at 15:12
  • $\begingroup$ That is one possible definition, yes. If your confusion is about that, then essentally everything else in the question is irrelevant. I suggest you read up a bit more on manifolds, vector fields and so on before starting with connections on principal bundles! $\endgroup$ – Mariano Suárez-Álvarez Dec 2 '17 at 15:16
  • $\begingroup$ I have spend quite some time on manifolds, vector fields. Can you suggest what exactly would I need to understand connections on principal bundle. I now need to know what does it mean to say vector is tangent to submanifold. Anything else do you suggest? @MarianoSuárez-Álvarez some reference would also be useful. $\endgroup$ – user312648 Dec 2 '17 at 15:21
  • $\begingroup$ I would suggest before tackling connections on principal bundles, first consider Koszul connections on vector bundles, where the derivative is defined in terms of the familiar Newton quotient, and the need to relate adjacent fibers is very clear. Then consider the generalization to a general fiber bundle, the Ehresemann connection, and its associated connection form. Then consider how to make that definition $G$-equivariant for a principal bundle. A textbook which covers these which I liked is Michelson and Lawson's Spin Geometry, but different books will use different levels of abstraction... $\endgroup$ – ziggurism Dec 2 '17 at 15:26
1
$\begingroup$

Locally a fiber bundle looks like a product. So to understand local statements about fiber bundles, it is enough to study a product. For example, a line bundle over a 1-manifold looks like the trivial bundle $\mathbb{R}^2\overset{\pi}{\to}\mathbb{R}$, a bundle whose total space is the product $\mathbb{R}\times\mathbb{R}$, whose base is $\mathbb{R},$ and projection $\text{proj}_2\colon\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ which has value $\text{proj}_2(x,y)=x$.

In this trivialization, a point in the base looks like $(3,0)$, a fiber looks like $\pi^{-1}(3)=3\times\mathbb{R}=\{(3,y)\colon y\in\mathbb{R}\}.$ A vertical vector looks like $e_y=\partial_y$, and this vector is tangent to the fiber $\pi^{-1}(3).$ A vector pointing to an adjacent fiber may look like $e_x=\partial_x$ or $e_x+e_y=\partial_x+\partial_y$. Anything which is not vertical. The subbundle of vertical vectors is the set of all vectors $V(\mathbb{R}^2)=\{f(x,y)\partial_y\}.$

$\endgroup$
  • $\begingroup$ @cello I added some clarification about the bundle structure of $\mathbb{R}\times\mathbb{R}$. Hopefully it's clearer? Please feel free to ask if any aspect is not clear. $\endgroup$ – ziggurism Dec 2 '17 at 15:03
  • 1
    $\begingroup$ @cello, it is not a good idea to imply that an answer is not "very serious". Please do not do that. $\endgroup$ – Mariano Suárez-Álvarez Dec 2 '17 at 15:06
  • $\begingroup$ I tried to give a very concrete answer using low-dimensional easy-to-visualize spaces, to make it understandable. I thought this was a benefit, not a flaw. But if this answer doesn't meet your needs, please clarify what you would like to see. $\endgroup$ – ziggurism Dec 2 '17 at 15:09
  • 1
    $\begingroup$ I take back my words which sounds like I am saying that the answer is not serious. I really do not mean like that. Reading my comment for third time makes me feel like it means that and I ask for an apologise to user ziggurism. $\endgroup$ – user312648 Dec 2 '17 at 15:09
  • 1
    $\begingroup$ @cello Fixing a horizontal subbundle, as it's called, gives a unique way to transport between fibers. That's what is meant by a vector pointing to a neighboring fiber; a tangent vector to the total space which is complementary to the vertical (tangent to fiber) vectors. $\endgroup$ – ziggurism Dec 4 '17 at 5:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy