I have a rather basic question about the dimension of a graph mapped by the exponential map. The problem is I can't really visualize it. It starts like that:
Let $M$ be a smooth manifold of dimension $n$. $$T_{x}M = E_{1x} \oplus E_{2x}, \ \dim E_{1x}=k, \ \dim E_{2x}=n-k$$ $$B^{i}(\delta(x)) = \{ u \in E_{ix}: \lVert{u\rVert} \leq \delta(x) \}, \ i = 1,2, \ 0 < \delta(x) < \infty$$ $$B(\delta(x)) = B^{1}(\delta(x)) \times B^{2}(\delta(x)), \ U(x,\delta(x)) = \exp_{x} B(\delta(x))$$ For a.e. $x \in M$: $\exists$ $\phi(x) \in C^{1+\alpha}(B^{1}(\delta(x)))$ with $\phi(x): B^{1}(\delta(x)) \to E_{2x}$ and $\phi(x)(0) = 0$, $d\phi(x)(0) = 0$.
From this follows the existence of a submanifold $V(x) = \{\exp_{x}(u,\phi(x)u): u \in B^{1}(\delta(x))\}$, where, of course, $(u,\phi(x)u) = \operatorname{graph}(\phi(x))$.
My question is: Why is $\dim V(x) = k$ and not $n$? I am getting confused since $\exp_{x}$ is a diffeomorphism and $\operatorname{dom}(\exp_{x}) = B^{1}(\delta(x)) \times \phi(x)(B^{1}(\delta(x)))$ and not just $B^{1}(\delta(x))$. Shouldn't $\dim V(x)$ be the same like $\dim U(x,\delta(x)$?
Further implying question would be why $T_{x}V(x) = E_{1x}$?
Maybe someone even got an example in $\mathbb{R}^{3}$ or some good script/book/reference...
Thanks in advance.
The set $\operatorname{graph}(\phi(x)) = \{(y,\phi(x)(y)): y\in B^1(\delta(x))\}$ is a subset $T_xM$, and more specifically a smooth submanifold of dimension $k$. Its parametrization by an open subset of $\mathbb R^k$ is given to us.
The exponential map $\exp_x $ is a diffeomorphism in a neighborhood of $0\in T_x(M)$, and as any diffeomorphism, it maps $k$-dimensional submanifolds onto $k$-dimensional submanifolds. The set $V(x)$ is the image of $k$-dimensional submanifold $\operatorname{graph}(\phi(x))$ under the exponential map.
Concerning $T_xV(x)=E_{1x}$: for each nonzero vector $v\in T_xV(x)$ there is a curve $\gamma:(-\epsilon,\epsilon)\to V(x)$ such that $\gamma(0)=v$. The composition $\tilde\gamma = (\exp_x)^{-1}\circ \gamma$ takes values in $E_{1x}$, hence its derivative is in $E_{1x}$. The chain rule for $\gamma = \exp_x\circ \tilde \gamma$ does the rest.