Suppose $ \pi \colon E \to M$ is a smooth fiber bundle over some Riemannian manifold $M$. The vertical bundle at $\theta \in E$ is defined as the kernel of the differential of $\pi$, i.e., it is the tangent space of the fiber $E_{q}$.
Question: Being $M$ a Riemannian manifold, is there a canonical way to define a horizontal subbundle of $T_{\theta}TE$?
Naively, I would take the tangent space of $M$ at $\pi(\theta)$ as the horizontal subbundle, but I guess this would not be right. Why?
Suppose $M$ is a smooth manifold. Let $\pi:TM\to M$ denote tangent bundle, with differential $d\pi:TTM\to TM$. Then one clearly has the canonical vertical sub-bundle $V=\ker{d\pi}$.
A horizontal sub-bundle $H$ then, is any sub-bundle 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 $$(\theta,\xi)\mapsto K_\theta(\xi)=\left(\nabla_{\alpha'(t)}\beta(t)\right)_{t=0}.$$ One would then define $H=\ker{K}$.