[Math] How do we use an Ehresmann connection to define a semispray

Question

Let $M$ be a differentiable manifold, let $TM$ be its tangent bundle, and consider $TTM$, the double tangent bundle.

Let $V \subseteq TTM$ denote the vertical subbundle, which is determined in a canonical fashion. An Ehresmann connection is a choice of horizontal subbundle $H \subseteq TTM$ which is complimentary to $V$, in that the double tangent bundle admits the horizontal decomposition $TTM = V \oplus H$.

One may define an Ehresmann connection by way of a connection form $v$.¹ This is a bundle homomorphism $v : TTM \to TTM$ which satisfies $v^2 = v$ and $\operatorname{im}(v) = V$, and this generates the horizontal subbundle $H = \operatorname{ker}(v)$. One should think of $v$ as projecting onto the vertical subspace along $H$.

Suppose we are given an Ehresmann connection $H$ and connection form $v$. I would like to use these to generate a semispray. A semispray is a vector field on $TM$ (i.e., a section of $TTM$) which satisfies a certain compatibility condition with the tangent structure, and should somehow be compatible with the connection. I can see from Wikipedia how a semispray generates a torsion-free Ehresmann connection, but it is not clear to me how to use an Ehresmann connection (possibly with torsion) to generate a semispray.

_{1. The space $\mathcal C$ of connection forms is the subspace of $TTM$-valued $1$-forms $\Omega^1(TM, TTM)$ which satisfy $v^2 = v$ and $\operatorname{im}(v) = V$. Is there a concise, common name for the space $\mathcal C$? Does it have nice algebraic or topological structure?}

Answer 1

Isn't the spray associated to the connection just the geodesic differential equation? Let $\pi_M : TM \to M$ be bundle projection, $\pi_{TM} : TTM \to TM$ bundle projection.

A double tangent vector $w \in TTM$ represents parallel transport of $\pi_{TM}(w)$ if $v(w) = 0$. i.e. if $w$ is horizontal. You can identify the horizontal spaces with the tangent spaces of $M$ by taking the derivative of $\pi_M : TM \to M$. ($V$ is the kernel of these derivatives).

So take as your vector field on $TM$ the function $f : TM \to TTM$ where $\pi_{TM}(f(x))=x$ for all $x \in TM$ and $f(x)$ is the unique horizontal vector in $TTM$ such that $D\pi_M(f(x)) = x$.