Question

surface patch for S. regular surface and f: S Ra smooth EXERCISE 3.44. Let S be a function. Assume that the point p e S is a critical point of f, which means that dfp(v) 0 for all v e TpS. Define the Hessian of f atp in the direction v as Hess(f)p(v) (foy)"(0), where y is a regular curve in S with y(0) = p and y'(0) = v. Prove that the Hessian is well defined in the sense that it is independent of the choice of HINT: Work in local coordinates and use Exercise 3.5 on page 123. a diifeon EXERCISE 3.5. Let U C R2 be open andf : U Rsmooth. Denote the coordinates of R2 by fu, v}. Assume that the point q e U is a critical point of f, which means that dfg(w) = 0 for all wE R2, or equivalently, fu(a)= fu(a) =0. Let y: I U be a regular curve with (0) = q and (0)= = (a, b). Prove that (fo7'(0)=a2 fun(4) +2abfue(g) + 62for (9) a2 fun(g) +2abfuv (g) +b fr(g)

Answer 1

a regular curve and
R f:S-
a smooth function.
pES
is a critical point of . The Hessian is defined by

Hess(f(u) (f o"(0)
where
pES,UETS
and $\gamma$ is a regular curve with
(0) p,(0)= v
.

Suppose $\widetilde{\gamma}$ is another such curve, we have to show that .

o(0)fo (0)

Let $U \subset \mathbb{R}^2, q \in U$ and
O:U S
be a regular surface path such that $\sigma (q)=p$. Now $f \circ \sigma : U \rightarrow \mathbb{R}$ is a smooth function with q as a critical point (that is because p is a critical point and $\sigma$ is a diffeomorphism).

The two curves $\gamma$ and $\widetilde{\gamma}$ decend to a curve in U, call them $\alpha$ and $\widetilde{\alpha}$ , such that $\sigma \circ \alpha = \gamma, \sigma \circ \widetilde{\alpha} = \widetilde{\gamma}$ .

According to exercise 3.5, $(f \circ \sigma \circ \alpha)^{''}(0) = a^2 (f \circ \sigma)_{uu} (q) + 2ab (f \circ \sigma)_{uv} (q) + b^2 (f \circ \sigma)_{vv} (q)$ , where a and b depend only on v.

Since the right hand side of the equation does not depend on $\alpha$, we have that it is the same for $\widetilde{\alpha}$ . Hence $(f \circ \sigma \circ \alpha)^{''}(0) = (f \circ \sigma \circ \widetilde{\alpha})^{''}(0)$

$\implies (f \circ \gamma)^{''}(0) = (f \circ \widetilde{\gamma})^{''}(0)$

Therefore the Hessian is well defined.