Let $F(M)$ be a principal bundle over a smooth manifold $M$, with a connection $\nabla$. Let $\pi: F(M) \to M$ be the projection. I am working to prove a theorem related to the horizontal and vertical subspaces in this context, and I would like to know how to rigorously show that these subspaces are complementary.
Definitions:
Vertical subspace: The vertical subspace at a point $p$ in $F(M)$ is the kernel of the differential of the projection, i.e., $V_p F(M) = \ker d\pi_{F(M)}$.
Horizontal subspace (given connection $\nabla$): A curve $\{u_t\}$ in $F(M)$ is called horizontal if for each $e \in \mathbb{R}^d$ the vector field $\{u_te\}$ is parallel along $\{\pi u_t\}$ on $M$. The horizontal subspace at a point $p$ in $F(M)$ consists of the tangent vectors to such horizontal curves.
Theorem to prove:
The horizontal subspace is complementary to the vertical subspace in the tangent space of the bundle. Specifically:
- $V_p F(M) \cap H_p F(M) = \{0\}$.
- $\dim(V_p F(M)) + \dim(H_p F(M)) = \dim(T_p F(M))$.
Attempted proof:
I have tried to approach this by examining the properties of vectors that are both horizontal and vertical and by using parallel transport, but I found the proof to be elusive.
Could anyone provide guidance or a reference to a rigorous proof that the horizontal and vertical subspaces are complementary in this setting, given the above definitions? Thank you!
References
The following source states both the above definition of horizontal subspace and then states the theorem I have given as a fact:
Hsu, E. P. (2002). Stochastic Analysis on Manifolds. American Mathematical Society. p. 38.
Note
This question was originally asked on MathOverflow and was migrated to StackExchange. I am reposting it so that it gets a chance to be seen
$\newcommand\R{\mathbb{R}}$Here is a more detailed version:
Let $E(M)$ be a vector bundle with canonical projection $\pi: E(M) \rightarrow M$. Let $\nabla$ be a connection. In other words, for any $p \in M$, open neighborhood $O \subset M$ of $p$, and local section $s: O \rightarrow E(M)$ (so $\pi\circ s$ is the identity map on $O$), there is a linear map \begin{align*} T_pM &\rightarrow E_p(M)\\ X &\mapsto (\nabla_Xs)(p) \end{align*} that satisfies the following: For any smooth $\phi: O \rightarrow \R$, $$ (\nabla_X(s\phi))(p) = (\nabla_Xs)(p)\phi(p) + s(p)\langle d\phi(p), X\rangle. $$
Let $F(M)$ be the frame bundle of $E$ and $f = (f_1, \dots, f_m): O \rightarrow F(M)$ be a local section of $F(M)$ (so $\pi\circ f$ is the identity map on $O$). Then the connection defines a linear map \begin{align*} T_pM &\rightarrow gl(m)\\ X &\mapsto A, \end{align*} where $$ (\nabla_Xf_j)(p) = f_i(p)A^i_j. $$ This can be written in matrix form as $$ (\nabla_Xf)(p) = f(p)A. $$ This linear map defines a matrix of $1$-forms $\Gamma^i_j$ on $O$, where $$ \langle \Gamma^i_j,X\rangle = A^i_j $$ and therefore $$ \nabla_Xf_j = f_i\langle \Gamma^i_j,X\rangle $$ or in matrix form, $$ \nabla_Xf = f\langle\Gamma,X\rangle. $$ For each $X \in T_pM$, let $\gamma: (-\delta,\delta) \rightarrow M$ be a curve such that $\gamma(0)=p$ and $\gamma'(0)=X$. A map $f: (-\delta,\delta) \rightarrow F(M)$ is a frame along $\gamma$ if $\pi\circ f = \gamma$. The frame is parallel along $\gamma$ if $$ \nabla_{\gamma'}f = 0. $$
Given $p \in M$, $f_0 \in F_p(M)$, and $\gamma: (-\delta,\delta)$, there is a unique parallel frame $f: (-\delta,\delta)\rightarrow F(M)$ along $\gamma$ such that $f(0)=f_0$. This can be shown as follows.
Let $\bar f$ be a frame over $O$ such that $\bar f(p) = f_0$ and $\overline\Gamma$ be the matrix of connection $1$-forms asdefined above. Any frame $f$ along $\gamma$ can be written as $$ f(t) = \bar f(\gamma(t))L(t), $$ where $L: (-\delta,\delta) \rightarrow GL(m)$. If $f$ is parallel along $\gamma$, then \begin{align*} 0 &= \nabla_{\gamma'}f\\ &= \nabla_{\gamma'}(\bar f L)\\ &= (\nabla_{\gamma'}\bar f)L + \bar f L'\\ &= \bar f(\langle\overline\Gamma, \gamma'\rangle L + L'). \end{align*} In other words, $f$ is parallel along $\gamma$ if and only if $$ L' + \langle\overline\Gamma, \gamma'\rangle L = 0. $$ By the standard existence and uniqueness theorem of ODEs, there exists a unique solution $L: (-\delta,\delta) \rightarrow GL(m)$ such that $L(0)=I$. This implies the existence and uniqueness of the parallel frame $f$ along $\gamma$ such that $f(0)=f_0$.
The derivative of the parallel frame $f$ at $p$ is given by \begin{align*} f'(0) &= \left.\frac{d}{dt}\right|_{t=0}(\bar f(\gamma(t))L(t))\\ &= d\bar f_p\gamma'(0)L(0) + \bar f(p)L'(0)\\ &= d\bar f_p X - \bar f(p)\langle\overline\Gamma, X\rangle. \end{align*} This shows that the map \begin{align*} h_p: T_pM &\rightarrow T_{f_0}F(M)\\ X &\mapsto f'(0) \end{align*} is independent of the curve $\gamma$ (as long as $\gamma(0)=p$ and $\gamma'(0)=X$) and is linear. The horizontal subspace at $f_0$ is defined to be $$ H_{f_0} = h_p(T_pM). $$ Moreover, since $\pi\circ f(t) = \gamma(t)$, $$ d\pi_{f_0}h_p(X) = d\pi_{f_0}f'(0) = \left.\frac{d}{dt}\right|_{t=0}\pi(f(t)) = \gamma'(0) = X. $$ Therefore, $h_p$ is injective and the horizontal subspace is complementary to vertical subspace $V_{f_0}F(M) = T_{f_0}F_p(M) = \ker d\pi_{f_0}$.