Horizontal Subbundles and Connection Maps

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.

• 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.
– Matt
Oct 8, 2018 at 9:25
• 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.
– Baol
Oct 8, 2018 at 13:36
• 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.
– Matt
Oct 8, 2018 at 13:46
• sorry i was off topic
– Baol
Oct 8, 2018 at 13:47