# Horizontal lift in a principal bundle vs. Parallel transport in Riemannian geometry

I am a physicist trying to understand the notion of holonomy in principal bundles.

I am reading about the horizontal lift of a curve in the base manifold of a principal bundle (or just fiber bundle) to the total space and would like to relate it to the "classic" parallel transport one comes across in Riemannian geometry. (I am guessing that the horizontal lift picture of the fiber bundle must be a generalization of the "classic" parallel transport of Riemannian geometry).

In particular, I would like to consider the example of parallel transporting a tangent vector (v) of S2 (embedded in ℝ3) in terms of horizontal lift. The picture I have for the "classic" parallel transport (in tangent bundle) is that one transports (infinitesimally) a given tangent vector of S2 as if it were in ℝ3 (Tℝ3) and then projects it back to the tangent space of S2.

To extend this picture in terms of the horizontal lift, I am considering the total space E as TS2, base manifold B as S2 and the fiber Ex at x as TxS2. The projection map then is clearly $\pi$: TS2 $\rightarrow$ S2: (x,v) $\rightarrow$ x. However, when trying to find the vertical space of the TxE, I am getting confused about d$\pi$:T(TS2)$\rightarrow$TS2. How do I make sense of T(TS2) in order to find ker(d$\pi$)?

I am not even sure how, with this approach, after completing a loop one will "move along the fiber"? And how does one eventually relate all this to the picture of moving the vector in ℝ3 and projecting back to tangent space of S2.

Any help would be greatly appreciated. Thank you!

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I may be able to provide a little help:

Trying to determine what $T(TS^2)$ looks like geometrically may be a little tough. Perhaps there is somebody out there with better geometric insight than myself, but I find it best to abstract away the internal $TS^2$ part. To be more precise, if we set $M = TS^2$ then $T(TS^2) = TM$ is just the tangent bundle of this new manifold $M$. This is in a sense "cheating," but maybe we can do a bit better.

I do not know about you, but one of the many ways I visualize $T_p M$ is as the collection of tangent vectors to M at the point p. In fact, if $\tilde V \in T_p M$ is a tangent vector, we may identify it with $\tilde V = (p,v)$ for $p \in M$ and $v \in T_p M$ is an "arrow." Hence an element of $\tilde V \in T_p (TS^2)$ looks like (p,v), where $p \in TS^2$ and v is an arrow at the point p. But we can break this down even further, and realize that p will look like $p = (x,w)$ for some $x \in S^2$ and $w \in T_x S^2$. In a sense, we may then think of elements in $T(TS^2)$ as triples (x,w,v).

You may already know this part, but let me re-iterate as it will be useful for the next part. For arbitrary manifolds M and N, and $f: M \to N$ a smooth map, we think of the pushforward/differential of f at p, $d_p f : T_p M \to T_{f(p)}N$ as the map which takes tangent vectors at p in M to tangent vectors at f(p) in N. At this point, I find the best way to think about how this map behaves is by defining tangent vectors as the differential of a smooth curve on M. If $\gamma:[0,1] \to M$ is a smooth curve with $\gamma(0) = p, \gamma'(0) = v \in T_p M$ then $f_* v$ is essentially $(f\circ \gamma)'(0)$; that is, we look at the image of the curve under f, and differentiate at zero.

Back to your question: how then do we compute $\ker d \pi$? We can intuitively guess what is happening using the previous paragraph. The projection map takes curves in $TS^2$ to curves in $S^2$ by just projecting onto the basepoint. In essence, it completely forgets about the tangent vectors themselves. Thus we expect that $d\pi$ will do something similar.

To see that this is true, it may be best to work with a coordinate system. Let $(U,x^i )$ be a coordinate chart on an arbitrary manifold M of dimension n. We know there is then an induced coordinate system on $TU$, which we shall write as $(TU, (x^i, X^i))$. Now every element of $v\in T_p(TU)$ may be written as
$$v = \sum_{i=1}^n \left[ a^i \left. \frac{\partial}{\partial x^i}\right|_{p} + b^i \left. \frac{\partial}{\partial X^i} \right|_p \right].$$
Since the differential is linear, we then have
$$(d_p \pi)v = \sum_{i=}^n \left[ a^i \left. d_p\pi\frac{\partial}{\partial x^i}\right|_{p} + b^i d_p\pi\left. \frac{\partial}{\partial X^i} \right|_p \right]$$
so it suffices to determine how $d_p \pi$ acts on the basis vectors $\frac\partial{\partial x^i}, \frac\partial{\partial X^i}$. If $\frac\partial{\partial y^i}, \frac\partial{\partial Y^i}$ is a basis for $T_{\pi(p)}(TU)$ then the standard change of basis formula tells us that
$$d_p\pi \left. \frac\partial{\partial x^i } \right|_p = \sum_{j=1}^n \left[\frac{\partial \pi^j}{\partial x^i}(p) \left. \frac{\partial}{\partial y^i} \right|_p + \frac{\partial \pi^j}{\partial X^i}(p) \left. \frac{\partial}{\partial Y^i} \right|_p \right]$$
and similarly for the $X^i$ and $Y^i$. This just says that the pushforward is precisely the Jacobian! Now computation of $d_p \pi$ is simple since we know exactly what $\pi$ does in coordinate. Namely, $\pi (p, w) = p$ so that
$$\frac{\partial\pi^j}{\partial x^i} = \delta^j_i, \qquad \frac{\partial \pi^j}{\partial X^i} = 0.$$
In block diagonal form, the coordinate representation is thus
$$d_p\pi = \begin{pmatrix} I_n & 0 \\ 0 & 0 \end{pmatrix}.$$
In the language of our triples, the differential of the projection thus takes vectors (x,w,v) to the vector (x,w): it forgets about the tangent vectors, exactly as we would expect!

This has become rather long-winded, so I will leave the my remaining comments as a reference. Namely, to see how horizontal lifts and parallel transport relate on general connects, take a look at Salamon's notes on differential topology. The information you are looking for is contained in Chapter 10.

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Thanks Kreizhn!
The triple (x,w,v) and the differential [ itex ] d \pi: (x,w,v) \rightarrow (x,w) [/ itex ] that you mention certainly makes sense (with v being 2 dimensional in order to keep the [ itex ] dim(T(TM)) = 2 dim(TM)[ / itex ]).

Thanks for pointing me to Salamon's notes. They seem very useful and I do hope I can find what I am looking for there.

For now, I have a feeling that to reconcile the horizontal lift picture of fiber bundle to the parallel transport of Riemannian geometry, we need to embed [ itex ] T(TS^2) [/ itex ] in ℝ3 and somehow the vertical space of [ itex ] T(TS^2) [/ itex ] must become normal to [ itex ] TS^2 [/ itex ] while the horizontal space can be chosen to be [ itex ] TS^2 [/ itex ] (which will be perpendicular to the vertical space due to the induced metric from ℝ3). I will have to think a bit more about it but thanks again for your help.

lavinia
Gold Member

The parallel translate of a frame around a closed curve is another - possibly different - frame. The linear map that changes the frames is an element of the holonomy group.

For orientable surfaces, the bundle of orthonormal frames is just the tangent unit circle bundle. A unit vector field maps a curve into this bundle and its derivative at a point lies in the tangent space to the unit circle bundle. The vector field is parallel at the point if the projection of the derivative onto the fiber - using the connection 1 form - is zero.

The pull back of the connection one form - using the vector field - gives you the covariant derivative (with a little algebra).

the orthogonal projection of a covariant derivative is a covariant derivative. This is a simple calculation. A vector field is parallel if its tendency to turn along the curve is orthogonal to the hypersurface. So for instance, a unit speed geodesic is just a curve whose acceleration is normal to the submanifold.

The tangent unit circle bundle of the 2 sphere is diffeomorphic to the projective space of lines in real 3 space.

A good exercise to see how this all works on the sphere is to construct its connection 1 form.

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