1
$\begingroup$

I'm trying to see the equivalence between the Ehresmann connection and the connection map, and having trouble getting the setup to be correct.

Suppose $M$ is a smooth manifold. Let $\pi:TM\to M$ denote tangent bundle, with differential $d\pi:TTM\to TM$. Then one has the canonical vertical subbundle $V=\ker{d\pi}$.

A horizontal subbundle $H$ is any subbundle which is complementary to $V$, that is, $TTM=H\oplus V$. Such an $H$ is equivalent to the existence of some $(0,2)$-tensor $\sigma$ on $TM$ ($\sigma:TTM\to TTM$) such that $\sigma^2=\sigma$ and $\sigma(TTM)=V$, and then letting $H=\ker{\sigma}$.

However, if $(M,g)$ is Riemannian with Levi-Civita connection $\nabla$, from what I've found in the literature, one typically defines a connection map $K:TTM\to TM$ by taking some $(\theta,\xi)\in TTM$, letting $\gamma(t)=(\alpha(t),\beta(t))\in TM$ with $\gamma(0)=\theta$, $\gamma'(0)=\xi$ and giving $$(\theta,\xi)\mapsto K_\theta(\xi)=\left(\nabla_{\alpha'(t)}\beta(t)\right)_{t=0}.$$ Then you define $H=\ker{K}$.

What's the correlation between $\sigma$ and $K$?

Also, if anyone has any references on the above that would be helpful. I seem to have only found a small section Sakai's Riemannian Geometry text, and a bit more in-depth section in Paternain's Geodesic Flows text. My interest in the topic is geared towards understanding the Sasakian metric on $TM$ better.

$\endgroup$
2
$\begingroup$

In general a connection on a fiber bundle $(B,\pi,M,F)$ is a smooth projection $\Phi : TB \rightarrow VB $ such that $\left. \Phi \right|_{VB}=id_{VB}$ and $\Phi \circ \Phi= \Phi$. After some manipulations you find that one can define $\chi = id_{TB}-\Phi: TB \rightarrow HB $ where $HB$ is called horizontal bundle and has the property that for every $b \in B$, $T_bB=H_bB \oplus V_bB$. In your case $B=TM$, however by contruction the map $\Phi$ which should be your "$\sigma$" is an element of $T^1_1(B)$ (i.e. a $(1,1)$ tensor) and not of $T_2(B)$.

In general on a manifold $M$ connections are introduced by a map $\nabla:\mathfrak{X}(M) \times \mathfrak{X}(M) \rightarrow \mathfrak{X}(M)$, linear in both argumet and satisfying the Libeniz rule $$\nabla_{fX}Y=f \nabla_XY, \space\ \nabla_{X}fY=X(f)+f \nabla_XY$$for al $X,Y \in \mathfrak{X}(M)$ and $f \in C^{\infty}(M)$. This "definition" comes up as a property once the concept of a principal connection on a principal bundle $(P,p,M,G)$ is introduced, then one can induce a connection on associate bundles to $(P,p,M,G)$. In particular given a representation of the Lie group $G$, on some vector space $V$, $\rho:G \rightarrow GL(V)$, one define $E=P \times_{\rho} V$ which is shown to be a vector bundle over $M$. From the induced connection on the vector bundle just described you contruct the connector $K : TE \rightarrow E $ such that $HE:= \textbf{ker}(K)$ and $T_eE=H_eE \oplus V_eE$ for all $e\in E$. Then you construct a covarian derivative $$\nabla_s X := K \circ Ts \circ X $$where $Ts$ is the tangent mapping and $s$ smooth section of $E$. Your comparison once again follows from taking $E=TM$, and using as a principal bundle the frame bundle $L(M)$. Hope to have been clear, for more information I recomend you to see Natural Operations in Differential Geometry Chaper III.

$\endgroup$
  • $\begingroup$ You're right on the $(1,1)$-tensor, I forgot I had just assumed smooth $M$ and not Riemannian $M$ yet. When I get I chance, I'll look through that book in more detail, it seems to have everything I want on the topic of connections, though it doesn't seem to mention anything about the Sasakian metric, which was my main focus of this topic. I don't have much detailed experience on Lie groups and principle bundles, so it'll probably take some effort to reconcile these ideas with my current knowledge. $\endgroup$ – Matt Oct 8 '18 at 9:25
  • $\begingroup$ Its a basic fact that smooth manifolds admits riemaniann metrics. You define locally a bunch of positive devinite bilinear forms and add tthem up with a partition of unity. $\endgroup$ – Baol Oct 8 '18 at 13:36
  • $\begingroup$ Indeed, which was why I stated the equivalent $(0,2)$-tensor in my post, but you corrected to a $(1,1)$-tensor, hence my comment. I'm not sure which of us is being pedantic here. $\endgroup$ – Matt Oct 8 '18 at 13:46
  • $\begingroup$ sorry i was off topic $\endgroup$ – Baol Oct 8 '18 at 13:47

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.