Question about critical point of function on compact manifold 
How can I deduce that $f$ only has finitely many non-degenerate critical points by this function only has non-degenerate critical point? And how can I use the compact manifold’s properties to solve the question?
 A: $Hess(x)$ (the matrix in the question is the second differential). Let $U$ be a chart containing $x$ that we identify to an open subset of $\mathbb{R}^n$ the restriction of  $df$ to $U$ is a map $U\rightarrow\mathbb{R}^n$ and $Hess(x)$ is the differential of this map. The fact that $Hess(x)$ is invertible implies that $df$ is locally invertible, therefore there exists $U_x\subset U$ open such that the restriction $df_{\mid U_x}$ is injective.
Suppose that there exists an infinite numbers of critical points $x_1,...,x_n,...$ this sequence has an accumulation point $y$. $y$ is also critical since $df$ is continuous, but there does not exist a neighborhood $V$ of $y$ such that the restriction of $df$ to $V$ is invertible.  Contradiction.
A: In the $x_i$ coordinates of a patch $U$ containing $p$ at which 
$H[f](p) = \left ( \dfrac{\partial^2 f(p) }{\partial x_i \partial x_j} \right)\tag 1$
is invertible, we have
$df = \displaystyle \sum_1^n \dfrac{\partial f}{\partial x_i} dx^i = \sum_1^n f_{,i}dx^i, \tag 2$
where
$f_{,i} = \dfrac{\partial f}{\partial x_i}; \tag 3$
if also,
$df(p) = 0, \tag 4$
then in such a patch we may consider a map
$df:U \to \Bbb R^n \tag 5$
given by
$df(q) = (f_{, 1}(q), f_{, 2}(q), \ldots, f_{,n}(q)) \in \Bbb R^n; \tag 6$
it now follows from (4) and the invertability of $H[f]$ at $p$, via the inverse function theorem, that there exists a neighborhood $V$ of $0$ in $\Bbb R^n$ and a $C^\infty$ function
$\theta:V \to U_0, \; \theta(0) = p, \tag 7$
where $p \in U_0 \subset U$, $U_0$ open, such that
$\theta \circ df(q)  = q, \; \forall q \in U_0; \tag 8$ 
in particular, by (4) we see that if
$df(p') = 0, \; p' \in U_0, \tag 9$
then
$p' = \theta \circ df(p') = \theta(0) = p, \tag{10}$
that is, $p$ is the only point in $U_0$ at which $df = 0$; thus the zeroes of $df$, the critical points of $f$, are isolated:  each is contained in an open set containing no other.
Next, we remark that, since $f \in C^\infty(M, \Bbb R)$, $df \in C^\infty(TM, \Bbb R)$, hence is continuous; thus, choosing any Riemannian metric $\Bbb g$ on $T^\ast M$, the norm of $df$,
$\vert df \vert = \Bbb g(df, df) \tag{11}$
is also continuous; hence the inverse image of the closed set $\{0\} \subset \Bbb R$, $Z_{df} = \vert df \vert^{-1}(\{0\}) \subset M$ is also closed; but this is precisely the set of critical points of $f$; thus the set of regular points of $f$,
$R_f = \{q \in M \mid df(q) \ne 0 \} = M \setminus Z_{df} \tag{12}$
is open, being the complement of the closed set $Z_{df}$.
We may now construct an open cover of $M$ by taking, for each critical point $p_i$ of $f$, an open set $U_i$ in which $p_i \in U_i$ is the unique critical point of $f$ in $U$, and adding to this collection the open set $R_f$:
$M = R_f \cup \left (\bigcup U_i \right ); \tag{13}$
since $M$ is compact this open cover has a finite sub-cover, which contains only a finite number of the $U_i$; hence the number of critical points of $f$ must itself be finite.  
