Splitting of the double tangent bundle into vertical and horizontal parts, and defining partial derivatives

Question

Let $M$ be a manifold and $g$ a metric on $M$. Let $TM$ denote the tangent bundle of $M$, and denote points in $TM$ by $(x,v)$ where $v \in T_xM$.

The Levi-Civita connection of $(M,g)$ induces a splitting of the double tangent bundle $TTM = V \oplus H$, where $V$ is the vertical distribution, defined by $V_{(x,v)} = T_{(x,v)}T_xM$ (i.e. the tangent space to the fibre), and $H_{(x,v)}$ is the horizontal distribution, which is determined by the connection.

Suppose $A:TM \rightarrow TM$ is a map such that $A(x,v)\in T_xM$ (so the map $A(x,\cdot)$ is a map from $T_xM$ to itself for all $x \in M$).

How does one use the splitting described above to define "partial derivatives" $\nabla_xA$ and $\nabla_vA$, which should be maps:

$(\nabla_xA)(x,v):T_xM \rightarrow T_xM$,

$(\nabla_vA)(x,v):T_xM \rightarrow T_xM$.

These should have the property that if $\gamma(t)$ is a curve on $M$ and $u(t)$ is a vector field along $\gamma$ (so $u(t) \in T_{\gamma(t)}M$ for all $t$), and $\nabla_t$ denotes the covariant derivative along $\gamma$, then

$\nabla_t(A(\gamma,u)) = (\nabla_xA)(\gamma,u) \cdot \dot{\gamma} + (\nabla_vA)(\gamma,u) \cdot \nabla_t u$

(here on the LHS, $A(\gamma,u)$ is itself a vector field along $\gamma$, so the notation $\nabla_t(A(\gamma,u))$ is meaningful).

The expression above "makes sense" intuitively, but I can't get the formalism to work properly.

Answer 1

Let us write the isomorphism

$T_{(x,v)}TM = H_{(x,v)}TM \oplus V_{(x,v)})TM \cong T_xM \oplus T_xM$

by

$\xi \simeq (\xi^h,\xi^v)$,

so that $\xi^h \in T_xM$ and $\xi^v \in T_xM$.

Here the identification $H_{(x,v)}TM \cong T_xM$ is given by the restriction of $d_{(x,v)}\pi$ to $H_{(x,v)}TM$ (where $\pi:TM \rightarrow M$ is the projection), and the isomorphism $V_{(x,v)}TM \cong T_xM$ is canonical.

The key point is that under these identifications, if $z$ is a curve on $TM$, say $z(t)=(\gamma(t),u(t))$ then

$\dot{z}(0) \simeq (\dot{\gamma}(0),(\nabla_tu)(0))$.

So suppose $x \in M$ and $v,w,y \in T_xM$. Let $\gamma$ be a curve in $M$ such that $\gamma(0)=x$ and $\dot{\gamma}(0)=w$, and let $u$ be a vector field along $\gamma$ such that $u(0)=v$ and $(\nabla_tu)(0)=y$. Let $z(t)=(\gamma(t),u(t))$.

Think of $A$ as a map $TM \rightarrow TM$, so that the differential $dA$ is a map

$d_{(x,v)}A:T_{(x,v)}TM \rightarrow T_{(x,v)}TM$.

Then given $w\in T_xM$, if $\xi_w$ is the unique vector whose horizontal component is $w$ and whose vertical component is zero (i.e. $\xi^h = w$ and $\xi^v = 0$), then we define

$(\nabla_xA)(x,v)(w):=(d_{(x,v)}A(\xi_w))^v$,

and similarly if $\zeta_w$ is the unique vector whose horizontal component is zero and whose vertical component is $w$ (i.e. $\zeta^h = 0$ and $\zeta^v = w$), then we define

$(\nabla_vA)(x,v)(y):=(d_{(x,v)}A(\zeta_w))^v$.

Then it follows that

$d_{(x,v)}A(\xi) \simeq ((\nabla_xA)(x,v)(w),(\nabla_vA)(x,v)(y))$,

and these two maps have the properties you're looking for.

Answer 2

Denote $E=TM$. We have a vertical projection $P_{(x,v)}:T_{(x,v)}E\to E_x$ determined by the splitting. The covariant derivative of a vector field $(\gamma,u)$, regarded as a curve in $E$, is the vertical projection of its derivative. We also have linear maps $d\pi:T_{(x,v)}E\to T_xM$ (where $\pi:E\to M$ is the bundle projection), the horizontal lift $h:T_xM\to H_{(x,v)}$ such that $h\circ d\pi=id_H$, and the natural isomorphism $i:T_xM=E_x\to V_{(x,v)}$ (which is the only thing depending on the fact that $E$ is the tangent bundle and not an arbitrary vector bundle). In this notation, we have $\xi=h\circ d\pi(\xi)+i\circ P_{(x,v)}(\xi)$ for every $\xi\in T_{(x,v)}E$.

Define $\nabla_x A=P_{A(x,v)}\circ (dA)\circ h$ and $\nabla_v A=P_{A(x,v)}\circ (dA)\circ i$ where $dA:T_{(x,v)}E\to T_{A(x,v)}E$ is the ordinary derivative. Then, for a vector $\xi\in T_{(x,v)}E$ (the derivative of our vector field $(\gamma,u)$) we have $$ \begin{aligned} \nabla_t (A(\gamma,u)) &= P_{A(x,u)}\circ dA(\xi) = P_{A(x,u)}\circ dA \circ h\circ d\pi(\xi) +P_{A(x,u)}\circ dA \circ i\circ P_{(x,v)}(\xi) \cr &=(\nabla_xA)(d\pi(\xi)) + (\nabla_vA)(P_{(x,v)}(\xi)) =(\nabla_xA)\cdot\dot\gamma + (\nabla_vA)\cdot\nabla_t(\gamma,u) . \end{aligned} $$