I am self-studying https://sayanmuk.github.io/StochasticAnalysisManifolds.pdf and I am struggling with the definitions on page 37-38
Let us now see how the connection $\nabla$ manifests itself on the frame bundle $F(M)$ of $M$. A frame at $x$ is an $\mathbb{R}$-linear isomorphism $u : \mathbb{R}^d \rightarrow T_xM$. Let $e_1, \ldots, e_d$ be the coordinate unit vectors of $\mathbb{R}^d$. Then the tangent vectors $ue_1, \ldots, ued$ make up a basis (or equivalently, a frame) for the tangent space $T_x$. We use $F(M)_x$ to denote the space of all frames at $x$. The general linear group $\text{GL}(d, \mathbb{R})$ acts on $F(M)_x$ by $u \mapsto ug$, where $ug$ denotes the composition:
$$
\mathbb{R}^d \xrightarrow{g} \mathbb{R}^d \xrightarrow{u} T_xM.
$$
The frame bundle $F(M) = \{ x \in M \mid F(M)_x \}$ can be made into a differentiable manifold of dimension $d + \frac{d^2}{2}$, and the canonical projection $\pi : F(M) \rightarrow M$ is a smooth map. The group $\text{GL}(d, \mathbb{R})$ acts on $F(M)$ fiber-wise; each fiber $F(M)_x$ is diffeomorphic to $\text{GL}(d, \mathbb{R})$, and $M = F(M) / \text{GL}(d, \mathbb{R})$.
From here I can understand and relate to my previously acquired notions of associated bundles. However my problem starts in the following part:
A local chart $x = (x_i)$ on a neighborhood $O \subset M$ induces a local chart on $O_e = \pi^{-1}(O)$ in $F(M)$ as follows. Let $X_i = \frac{\partial}{\partial x_i}, 1 \leq i \leq d$, be the moving frame defined by the local chart. For a frame $u \in O_e$ we have $u e_i = e_j^i X_j$ for some matrix $e = (e_i^j) \in GL(d, \mathbb{R})$. Then $(x, e) = (x_i, e_i^j) \in \mathbb{R}^{d+d^2/2}$ is a local chart for $O_e$. In terms of this chart, the vertical subspace $V_u F(M)$ is spanned by $X_{kj} = \frac{\partial}{\partial e_k^j}, 1 \leq j, k \leq d$, and the vector fields ${ X_i, X_{ij}, 1 \leq i, j \leq d }$ span the tangent space $T_u F(M)$ for every $u \in O_e$. We will need the local expression for the fundamental horizontal vector field $H_i$.
Question: How is $u e_i = e_j^i X_j$? $e\in \mathbb{R}^d$ and u is a frame which I guess is $X_j$. Where does the group element $e_j^i\in GL(n,d)$ come from? I cannot make sense of this terminology.
Thanks in advance.
Best Answer
Edited after OP's comment
Let $M$ be a $d$-dimensional manifold. For each $p\in M$, denote by $\text{F}_p(M)$ the $\text{GL}(d,\mathbb{R})$-space of frames on $M$ at $p$, i.e. linear isomorphisms $\mathbb{R}^d\overset{\sim}{\to}T_pM$. This becomes a $\text{GL}(d,\mathbb{R})$-space with respect to the natural free right action $$\text{F}_p(M)\times\text{GL}(d,\mathbb{R})\to \text{F}_p(M),(u,A)\mapsto uA:=u\circ A.$$ Further, set $\text{F}(M):=\cup_{p\in M}\text{F}_p(M)$ and define $\pi\colon \text{F}(M)\to M$ by setting $\pi(\text{F}_p(M))=\{p\}$, for each $p\in M$. Clearly, this becomes a $\text{GL}(d,\mathbb{R})$-space wrt the fiberwise right action by $\text{GL}(d,\mathbb{R})$.
Pick a coordinate chart $(\mathcal{O},\varphi=(x_1,\ldots,x_d))$ on $M$, and consider the associated holonomic frame $X_i=\frac{\partial}{\partial x_i}$, $1\leq i\leq d$, so that, in particular, for each $p\in\mathcal{O}$, $$X_1|_p,\ldots,X_d|_p\ \text{is a basis of}\ T_pM.\tag{1}$$
Now, for each point $p\in\mathcal{O}$, and any frame $u$ on $M$ at $p$, i.e. $u\in\text{F}_p(M)$, since $u\colon\mathbb{R}^d\overset{\sim}{\longrightarrow}T_pM$ is a linear isomorphism, one gets that $$u(e_1),\ldots,u(e_d)\ \text{is a basis of}\ T_pM,\tag{2}$$ where $e_1,\ldots,e_d$ denotes the canonical basis of $\mathbb{R}^d$. Hence, concerning your question, from (1) and (2) it follows that there exists a unique $\Phi(u)=(\Phi_{ij}(u))\in\text{GL}(d,\mathbb{R})$ such that $$u(e_i)=\sum_{j=1}^d\Phi_{ji}(u)X_j|_p,\ \text{for}\ i=1,\ldots,d.\tag{3}$$ Notice also that, for any $A=(A_{ij})\in\text{GL}(d,\mathbb{R})$, since $uA:=u\circ A$ is still a frame on $M$ at $p$, i.e. $uA\in \text{F}_p(M)$, the latter allows to compute, for $i=1,\ldots,d$, $$\sum_{j=1}^d\Phi_{ji}(uA)X_j|_p\overset{(3)}{=}(uA)(e_i)=u(Ae_i)=u\left(\sum_{j=1}^dA_{ji}e_j\right)=\sum_{j=1}^dA_{ji}u(e_j)\overset{(3)}{=}\sum_{j,k=1}^dA_{ji}\Phi_{kj}(u)X_k|_p.$$ This proves that, for any $p\in\mathcal{O}$, $u\in\text{F}_p(M)$ and $A\in\text{GL}(d,\mathbb{R})$, the following holds $$\Phi(uA)=\Phi(u)\cdot A.$$ What above shows that each coordinate chart $(\mathcal{O},\varphi=(x_1,\ldots,x_d))$ on $M$ determines a bijection $$\Phi\colon\pi^{-1}(\mathcal{O})\longrightarrow\varphi(\mathcal{O})\times\text{GL}(d,\mathbb{R}),\ u\longmapsto((\varphi\circ\pi)(u),\Phi(u)),$$ which satisfies the following conditions:
In such a way, $\text{F}(M)$ gets equipped with an atlas of principal bundle charts formed by the charts $(\pi^{-1}(\mathcal{O}),\Phi)$ as $(\mathcal{O},\varphi)$ varies among the charts of $M$. Indeed, any two of these principal bundle charts are compatible as proved below.
Let $(\mathcal{O},\varphi=(x_1,\ldots,x_d))$ and $(\mathcal{O}',\varphi'=(x'_1,\ldots,x'_d))$ be two coordinate charts on $M$, with associated principal bundle charts $(\pi^{-1}(\mathcal{O}),\Phi)$ and $(\pi^{-1}(\mathcal{O}'),\Phi')$ respectively. For any $p\in\mathcal{O}\cap\mathcal{O}'$ and $u\in\text{F}_p(M)$, by definition one has $$u(e_i)=\sum_{j=1}^d\Phi_{ji}(u)\left.\frac{\partial}{\partial x_j}\right|_p\quad\text{and}\quad u(e_i)=\sum_{j=1}^d\Phi'_{ji}(u)\left.\frac{\partial}{\partial x'_j}\right|_p,\ \ \text{for}\ i=1,\ldots,d.$$ Since $\left.\frac{\partial}{\partial x_j}\right|_p=\sum_{k=1}^d\frac{\partial x'_k}{\partial x_j}(p)\left.\frac{\partial}{\partial x'_k}\right|_p$, from the latter it follows that, for any $p\in\mathcal{O}\cap\mathcal{O}'$ and $u\in\text{F}_p(M)$, $$\Phi'(u)=(D_{\varphi(p)}(\varphi'\circ\varphi^{-1}))\cdot \Phi(u),$$ where $D_{\varphi(p)}(\varphi'\circ\varphi^{-1})\in\text{GL}(d,\mathbb{R})$ denotes the Jacobian matrix of $\varphi'\circ\varphi^{-1}$ at the point $\varphi(x)$. Hence, the transition map $\Phi'\circ\Phi^{-1}\colon\varphi(\mathcal{O}\cap\mathcal{O}')\times\text{GL}(d,\mathbb{R})\to\varphi'(\mathcal{O}\cap\mathcal{O}')\times\text{GL}(d,\mathbb{R})$ is given by $$(\Phi'\circ\Phi^{-1})(x,A)=((\varphi'\circ\varphi^{-1})(x),(D_x(\varphi'\circ\varphi^{-1}))\cdot A),$$ and so it is smooth.