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I will give two preceding theorems and the question, which uses both, follows afterwards:

Let $M$ be a smooth compact Riemannian manifold of dimension $n$ with a smooth measure $\mu$.

$T_{x}M = E_{1x} \oplus E_{2x}$, where $\dim E_{1x} = k$ and $\dim E_{2x} = n-k$.

$\textbf{Thm1}$ $\lim_{i \to \infty} d(E_{1x_{i}},E_{1x}) \to 0$, on a subset $N \subset M$, where $d(\cdot,\cdot)$ is the Grassmannian metric.

( One way to define the Grassmannian metric is $d(E_{1x_{i}},E_{1x}):=\sup_{v \in E_{1x_{i}},\|v\|=1}\inf\left\{\|v-w\|\colon w \in E_{1x}\right\}$, when $\dim E_{1x_{i}} = \dim E_{1x}$.)

Define $B^{i}(\delta(x)) = \{ u \in E_{ix}: \lVert{u\rVert}_{x} \leq \delta(x) \}, \ i = 1,2, \ 0 < \delta(x) < \infty$. We further know that $\delta_{N} = \inf_{x \in N} \delta(x) > 0$.

$\textbf{Thm2}$ For $\mu$-$a.e.$ $x \in M$ exists a function $\phi(x) \in C^{1}(B^{1}(\delta(x)),B^{2}(\delta(x)))$ and $\delta(x)$, which define a $k$ - dimensional submanifold of $M$ called $\textit{local stable manifold}$

$$V(x) = \left\{exp_{x}(u,\phi(x)u): u \in B^{1}(\delta(x)) \right\}$$.

The author further writes... (here $x \in N$) "By $\textbf{Thm1}$ we can deduce from $\textbf{Thm2}$ that for any $0 < q < \delta_{N}$, for sufficiently large $i$, $V(x_{i})$ has the form

$$V(x_{i}) = \left\{ exp_{x}(u,\chi_{i}(u)): u \in B^{1}(q) \right\} $$,

where $\chi_{i} \in C^{1}(E_{1x},E_{2x})$."

$\textbf{Question:}$ Because of the convergence result of $\textbf{Thm1}$, I can imagine intuitively, that the author wants to express $V(x_{i})$ by a function $\chi_{i}$ which has its domain on a subset of $E_{1x}$ rather than $E_{1x_{i}}$. But why is $\operatorname{dom}(\chi_{i}) = B^{1}(q)$ and not a bigger closed ball of $E_{1x}$ since by $\textbf{Thm2}$ $V(x_{i}) = \left\{exp_{x}(u,\phi(x_{i})u): u \in B^{1}(\delta(x_{i})) \right\}$ and $B^{1}(q) \subset B^{1}(\delta(x_{i}))$? How can you get all the points of $V(x_{i})$ by taking $q$ arbitrarily small for sufficiently large $i$?

Thanks in advance!

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Can you give a reference to the source? –  user53153 Feb 3 '13 at 0:08
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