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.


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.

  • $\begingroup$ 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

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