The joint exponential map: smoothness, invertible differential at the diagonal, and the small-velocity two-point inverse

(Openness and smoothness of $\mathrm{\Phi }$.) The set $\mathcal{D}=\{(q,v):(1,q,v)\in \mathcal{G}\}$ is the preimage of the open set $\mathcal{G}$ under the continuous map $(q,v)\mapsto (1,q,v)$; hence it is open in $TM$. The identity $\mathrm{\Phi }(q,v)=(q,{\gamma }_{q,v}(1))$ exhibits $\mathrm{\Phi }$ as the composition of the smooth map $(q,v)\mapsto (1,q,v)$ with the smooth evaluation of [L1] followed by the smooth map $(q,v)\mapsto (q,v)$; hence $\mathrm{\Phi }$ is smooth, proving (a).

(Coordinate form of $\mathrm{\Phi }$ and its differential.) Fix a chart $(U,x)$ around $p$, translate it so that $x(p)=0$, and write $TU\cong x(U)\times {\mathbb{R}}^4$, so that a tangent vector $v\in T_{q}M$ with $q\in U$ has components $\eta \in {\mathbb{R}}^4$. The coordinate representative of $\mathrm{\Phi }$ is $(\xi ,\eta )⟼(\xi ,e(\xi ,\eta )),e(\xi ,\eta ):=x({\exp }_{x^{-1}(\xi )}(\eta ))\mathrm{.}$ By [L1], $e$ is smooth on its domain, and $e(\xi ,0)=\xi$ for every $\xi$, so $\mathrm{\partial }e\mathrm{/}\mathrm{\partial }\xi (0,0)={\mathrm{i}\mathrm{d}}_{{\mathbb{R}}^4}$; and $\eta \mapsto e(0,\eta )=x({\exp }_p(\eta ))$ has derivative ${\mathrm{i}\mathrm{d}}_{{\mathbb{R}}^4}$ at $\eta =0$ by the identity $d({\exp }_p{)}_0={\mathrm{id}}_{T_{p}M}$ of [L1] read in the chart. Therefore the Jacobian matrix of the representative at $(0,0)$ is the block matrix $\left(\begin{array}{cc}I& 0\\ I& I\end{array}\right)$, i.e., in the identifications of (b) $d{\mathrm{\Phi }}_{(p,0)}(u,w)=(u,u+w)\mathrm{.}$

(Invertibility and the diffeomorphism.) The map $(u,w)\mapsto (u,u+w)$ is an isomorphism with inverse $(u,y)\mapsto (u,y-u)$; hence $d{\mathrm{\Phi }}_{(p,0)}$ is an isomorphism, and [L2] gives open neighborhoods $V_0$ of $(p,0)$ and $W_0$ of $(p,p)$ with $\mathrm{\Phi }{\mathrm{\mid }}_{V_0}:V_0\to W_0$ a diffeomorphism. In the bundle chart $(TU,Tx)$, the open neighborhood $V_0$ contains an exact product $x^{-1}(B_{r_0})\times B_{{\rho }_0}$ for some $r_0,{\rho }_0>0$, where the second factor records coordinate components of the tangent vector. Also, by [L5] in the product chart, $W_0$ contains $x^{-1}(B_{s_0})\times x^{-1}(B_{s_0})$ for some $s_0>0$.

(Product-like shape of $V$.) Take the exact product $V:=x^{-1}(B_r)\times B_{\rho }\subseteq V_0$ with $0<r<r_0$ and $0<\rho <{\rho }_0$, and set $W:=\mathrm{\Phi }(V)$. Since $\mathrm{\Phi }{\mathrm{\mid }}_{V_0}$ is a diffeomorphism and $V$ is open, $W$ is open and contains $(p,p)$. By [L5], choose $s>0$ small enough that $x^{-1}(\overline{B_s})\times x^{-1}(\overline{B_s})\subseteq W$ and $s<r$. Then the open product is also contained in $W$. The set $V$ has exactly the displayed product-like form, and $\mathrm{\Phi }{\mathrm{\mid }}_V:V\to W$ is a diffeomorphism. This proves the first two assertions of (c).

(The two-point inverse.) Define $v(q,r)$ for $(q,r)\in W$ as the second component of ${\mathrm{\Phi }}^{-1}(q,r)\in V$. Since ${\mathrm{\Phi }}^{-1}$ is smooth and the projection to the second factor is smooth, $v$ is smooth. The identity $\mathrm{\Phi }(q,v(q,r))=(q,r)$ unwinds to ${\exp }_q(v(q,r))=r$ and $(q,v(q,r))\in V$. For $q=r$ we get ${\exp }_q(v(q,q))=q={\exp }_q(0)$ with $(q,v(q,q)),(q,0)\in V$ and $\mathrm{\Phi }(q,v(q,q))=\mathrm{\Phi }(q,0)=(q,q)$, so $v(q,q)=0$ by injectivity of $\mathrm{\Phi }{\mathrm{\mid }}_V$. This proves the remaining assertions of (c).

(Lipschitz bound.) The closed product $x^{-1}(\overline{B_s})\times x^{-1}(\overline{B_s})$ is contained in $W$ by step 3.1. Let $M_0$ be the supremum of $\mathrm{\parallel }Dv\mathrm{\parallel }$ over its coordinate image $\overline{B_s}\times \overline{B_s}$, a compact convex set; it is finite because $v$ is smooth on the open set $W$. For fixed $q\in x^{-1}(B_s)$ and $r,r^{\mathrm{\prime }}\in x^{-1}(B_s)$, the straight coordinate segment from $x(r)$ to $x(r^{\mathrm{\prime }})$ lies in $B_s$, so [L4] applied to $t\mapsto v(q,x^{-1}(x(r)+t(x(r^{\mathrm{\prime }})-x(r))))$ gives $\mathrm{\mid }v(q,r^{\mathrm{\prime }})-v(q,r){\mathrm{\mid }}_x\le M_0\mathrm{\mid }x(r^{\mathrm{\prime }})-x(r){\mathrm{\mid }}_{\mathrm{e}}$. Taking $r=q$ and using $v(q,q)=0$ proves (d) with $\mathrm{\Lambda }=M_0$; also $\mathrm{\mid }x(q)-x(r){\mathrm{\mid }}_{\mathrm{e}}<2s$ gives $\mathrm{\mid }v(q,r){\mathrm{\mid }}_x\le 2\mathrm{\Lambda }s$. This proves (d) and completes the proof.