Assume $M,N$ are two Riemannian manifolds and $f: M\rightarrow N$ is a smooth map. Suppose $dim M =m < dim N =n$. Let $\Sigma$ be the graph of $f$, that is, $\Sigma =(x, f(x))$ for $x\in M$. My question is that in the neighborhood of any point $(p, f(p))$ in $\Sigma$, can we choose the $n-m$ orthonormal basis $\{e_{i}\}$ of the normal bundle $N\Sigma$ in the product $M\times N$ such that the projection of $\{e_{i}\}$ into the tangent bundle of $M$ is 0.
When the rank of $f$ is constant in a neighborhood of a point on $\Sigma$, such orthonormal basis is easily to be found. What I am confused is that what's happening when the rank of $f$ is changing in a neighborhood.
Any references are very appreciated.
I think this can't be done, essentially for the rank problems you seem to be struggling with. Consider the map $f: \mathbb{R} \to \mathbb{R}^2$ given by $$ f(t) = \left\{ \begin{array}{lr} (e^{-1/t}, 0) & : t > 0\\ (0, e^{1/t}) & : t < 0 \\ (0, 0) & : t = 0 \end{array} \right. $$ This is obviously smooth for $t \neq 0$, and all of its derivatives are $0$ at $t = 0$, so one checks that it's smooth there as well.
The bundle $N\Sigma$ is parallelizable of rank $2$, and one can identify any of its fibers with the target space $\mathbb{R}^2$ by projection. For $t > 0$, the one-dimensional subspace projecting to $0$ in $T \mathbb{R}$ is the $y$-axis of $\mathbb{R}^2$, while for $t < 0$, it is the $x$-axis, and there's no hope then of getting even a continuous vector field spanning this subspace in a neighborhood of $t = 0$.