Jost explains (in Riemannian Geometry and Geometric Analysis) the term connection with the direct sum of the tangent space to a vector bundle $T_eE=V_e\oplus H_e$ as follows.

Let $E$ be a vector bundle with connection $\nabla$, $e\in\Gamma(E)$ and $x=\pi(e)\in M$, $M$ manifold. Considering the tangent space $T_e E$ there is a distinguished subspace, the vertical space $V_e$, namely the tangent space $T_eE_x$ to the fiber $E_x\subset E$.

However, there is no distinguished "horizontal space" $H_e$ complementary to $V_e$, i.e. satisfying $T_eE=V_e\oplus H_e$. If we have a connection $\nabla$, however, we can parallely transport $e$ for each $X\in T_xM$ along a curve $c(t)$ with $c(0)=x,~\dot{c}(0)=X$. Thus, for each $X$, we obtain a curve $e(t)$ in $E$. The subspace of $T_eE$ spanned by all tangent vectors to $E$ at $e$ of the form $$\left.\frac{\mathrm{d}}{\mathrm{d}t}e(t)\right|_{t=0}$$ then is the horizontal space $H_e$. In this manner, one obtains a rule for how the fibers in neighboring points are "connected" with each other.

First (and probably very naive), with this definition of $H_e$ it is not clear to me why the direct sum $T_eE=V_e\oplus H_e$ does hold.

  • Why is $\left.\frac{\mathrm{d}}{\mathrm{d}t}e(t)\right|_{t=0}\notin T_eE_x=V_e$ true?

Second, I cannot really make sense of the term connection with this direct sum in mind. Many other authors (e.g. J.M. Lee) refer the term connection to the isomorphism between fibers given by parallel transport along a curve $c$, $P_{c,t}:E_{c(0)}\to E_{c(t)}$ which seems to explain the connection better. So, I would like to know:

  • What exactly is meant by "a rule" for how the fibers are connected with each other?

I have read, that the concept of vertical and horizontal (sub-)bundle is closely related to an Ehresmann connection, but since I am unaware of this notion, I would prefer answers not using Ehresmann connection. Also, I hope the second question is not too soft, but I am highly interested in understanding this (alternative) notion of connection.


I don't have time to write a whole course here, but here are brief answers to your questions. First, the tangent vector $\dot e(0)$ projects to $\dot c(0)=X\in T_xM$, so $\dot e(0)$ cannot be a vertical vector. (In fact, you clearly get an isomorphism $H_e\to T_xM$ given by $\pi_{*e}$.) Second, starting with an element $e_0\in E_{c(0)}$, we get a unique horizontal lift of the curve $c(t)$, and so $e(t)\in E_{c(t)}$ gives us the parallel transport of $e_0\in E_{c(0)}$ to the appropriate element of $E_{c(t)}$.

  • $\begingroup$ By "$\pi_{*e}$" do you mean the push forward of the bundle's projection at $e\in E$, i.e. the projection $\pi_*:TE\to TM$? Also, since I don't understand the argument "$\pi_*(\dot{e}(0))=X\in T_xM\Rightarrow\dot{e}(0)\notin V_e$", what is the image of a vertical vector $v\in V_e$ under the projection $\pi_*$? $\endgroup$ – gofvonx Aug 20 '13 at 7:59
  • 1
    $\begingroup$ Yes, $\pi_*=d\pi$. And since the fiber, by definition, all projects to a point, $\pi_{*e}(\xi)=0$ whenever $\xi\in V_e$. $\endgroup$ – Ted Shifrin Aug 20 '13 at 10:42

Parallel transport along a curve $c$ will give you a lift of $c$ to a curve in the vector bundle $E$. Then the tangent vectors to the new curve (in $E$) will give you an element of the "horizontal" space $H$ at each point of the curve.

  • $\begingroup$ And why are the tangent vectors to the new curve not in $V$? Further, the "rule" how neighboring fibers are connected is again the isomorphism $P_{c,t}$ or is there a more obvious relation to the direct sum $T_eE=V_e\oplus H_e$? $\endgroup$ – gofvonx Aug 20 '13 at 10:16
  • $\begingroup$ @gofvonx: If the tangent vector were always in the vertical space, the curve itself would lie entirely in the tangent space by uniqueness of solution of ODE. Since the curve immediately leaves the fiber over $e$, the tangent vector must have a non-vertical component. $\endgroup$ – Mikhail Katz Aug 20 '13 at 12:45
  • $\begingroup$ Why the quotes around horizontal? Isn't this the standard terminology? $\endgroup$ – Teddy Baker Aug 29 '17 at 1:44

If $\xi\in T_eE_{\pi(e)}=V_e$ there exists $\gamma:I\to E$ such that $\gamma(t)\in E_{\pi(e)}\forall t\in I$ and $\dot{\gamma}(0)=\xi$. Therefore, $$\mathrm{d}\pi(\xi)=\left.\frac{\mathrm{d}}{\mathrm{d}t}\pi(\gamma(t))\right|_{t=0}=0$$ since $\pi(\gamma(t))\in M$ is constant. Further, $$\mathrm{d}\pi\left(\dot{e}(0)\right)=\dot{c}(0)=X\neq 0\Rightarrow \dot{e}(0)=\left.\frac{\mathrm{d}}{\mathrm{d}t}e(t)\right|_{t=0}\notin V_e.$$


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.