# John Lee Exercise 5.44

The following is an Exercise 5.44 in Lee's Introduction to Smooth manifolds

Suppose $$M$$ is a smooth manifold with boundary, $$f$$ is a boundary defining function, and $$p\in\partial M$$. Show that a vector $$v\in T_pM$$ is inward-pointing in and only if $$vf>0$$, outward-pointing if and only if $$vf<0$$, and tangent to $$\partial M$$ if and only if $$vf =0$$.

Let $$(x^i)$$ be a smooth coordinate on some neighborhood of $$p$$ and $$t$$ be a smooth coordinate on some neighborhood of $$f(p)$$. Write $$v = v^i{\partial\over\partial x^i}\bigg|_p$$ then $$vf = v^i{\partial f\over\partial x^i}\bigg|_p$$ and $$df_p(v) = df_p\left(v^i{\partial\over\partial x^i}\bigg|_p\right) = v^i{\partial f\over\partial x^i}(p){\partial\over\partial t}\bigg|_{f(p)}$$. I'm trying to use the fact that $$v\in T_pM$$ is inward-pointing if and only if $$v^n>0$$, outward-pointing if and only if $$v^n<0$$ and tangent to $$\partial M$$ if and only if $$v^n = 0$$ but I'm stuck. Please help.

Let $$p\in\partial M$$ and choose a smooth boundary chart $$(U,\varphi)$$ with local coordinate $$(x^i)$$. Since $$f$$ is a smooth function, $$f:U\to [0,\infty)$$ is a smooth function. For $$v\in T_pM$$, write $$v = v^i{\partial\over\partial x^i}\bigg|_p$$. The action of $$v$$ on $$f$$ is \begin{align*} vf & = v^i{\partial\over\partial x^i}\bigg|_p f = v^i{\partial f\over\partial x^i}(p) = v^n{\partial f\over\partial x^n}(p). \end{align*} Here, the last equality follows from the fact that the function $$f$$ along the axis $$x^i$$ for $$i =0,1,...,n-1$$ is contained in $$\partial M$$ so partial derivative is identically zero by definition. Since $$df_p\neq 0$$, this implies $${\partial f\over\partial x^n}(p)> 0$$ ($$\because$$ codomain is $$[0,\infty)$$). Hence the sign of $$vf$$ is completely determined by the sign of $$v^n$$ which is the desired result.
• Why is the function along the $n-1$ axis contained in the boundary and why is the nth dericative positive if the codomain is nonnegative? Commented Nov 30, 2023 at 9:46