2
$\begingroup$

On pages no. 52-56 of "Lectures on Morse Homology" by Augustin Banyaga and David Hurtubise presented a proof of Morse Lemma using homotopy method AKA Palais proof using Moser "path method". There are several statements I did not understand while trying to read the complete proof.

The proof starts like this: "By replacing f by f-f(p) and by choosing a suitable coordinate chart on M we may assume that the function f is defined on a convex neighborhood $U_0$ of $0\in R^m$ where $f(0)=0, df(0)=0$ and the matrix of the Hessian at $0\in R^m$, $M_0(f) = A = (\frac{\partial^2 f}{\partial x_i \partial x_j}(0))$, is a diagonal matrix with the first k diagonal entries equal to -1 and the rest equal to +1.

The matrix A induces a function $\widetilde{A}(x) = x^tAx=<Ax,x>=\sum_{j=1}^{m}\delta_j x_j^2$ where $\delta_j = \frac{\partial^2 f}{\partial^2 x_j}(0) = \pm1$ for all j=1,...,m. We want to prove that there are neighborhoods U and U' of $0$ with $U \subseteq U_0$ and a diffeomorphism $\varphi: U->U'$ such that:

$f\circ\varphi = \widetilde{A}$. (3.1)

The Idea of the path method is to interpolate f and $\widetilde {A}$ by a path such as, $f_t = \widetilde{A} + t(f-\widetilde{A})$ (3.2)

and to look for a smooth family $\varphi_t$ of diffeomorphisms such that $f_t\circ\varphi_t=f_0=\widetilde{A}$. (3.3)

Then $\varphi=\varphi_1$ will satisfy $f\circ\varphi=\widetilde{A}$.

We get $\varphi_t$ as a solution of the differential equations: $\frac{d\varphi_t}{dt}(x)=\xi_t(\varphi_t(x)); \varphi_0(x)=x$ where the smooth family $\xi_t$ is the tangent along the curves $t\mapsto\varphi_t(x)$. Taking the partial derivative with repect to t of both sides of (3.3) gives

$(\dot{f_t}\circ\varphi_t + (\xi_t\cdot f_t)\circ\varphi_t)(x)=0$ (3.4) for all $x \in U$ where $\dot {f_t}$ denotes $\frac{\partial f_t}{\partial t}$.

Thus, $(\dot{f_t} + \xi_t \cdot f_t)(y) = 0$ (3.5) for all $y \in U'$.

But $\dot{f_t} = f - \widetilde{A}$ by (3.2), and therefore (3.5) becomes $df_t(\xi_t)=g$ (3.6) where $g=\widetilde{A} - f$."

My questions regarding this part of the proof are these:

First, "Taking the partial derivative with respect to t of both sides of (3.3) gives (3.4)" - why is that so? I tried defining a function $h:R^{n+1}\rightarrow R^n$ such that $h(t,x)=f_t\circ\varphi_t$ and differentiating it using the chain rule, and what i got is:

$\frac{\partial h}{\partial t}(t,\varphi_t)+(\xi_t \circ \varphi_t) \cdot \frac{\partial h}{\partial x}(t,\varphi_t)$

and not:

$\frac{\partial h}{\partial t}(t,\varphi_t)+(\xi_t \circ \varphi_t) \cdot h(t,\varphi_t)$

as it seems I should get by (3.4).

Second, "and therefore (3.5) becomes $df_t(\xi_t)=g$" - to my understanding (3.5) should become: $\xi_t \cdot f_t=g$, meaning that $df_t(\xi_t)=\xi_t \cdot f_t$ - why is this the case here?

Moreover, the proof ends like this (there is a middle part to the proof which I omitted because I understood it...): "$B_0^t$ (which is defined by $\frac{\partial^2 f_t}{\partial x_i \partial x_j})(0)$) is non-degenerate for all $0\leq t\leq 1$, and there exists a neighborhood $\widetilde{U}$ of $0 \in R^m$ such that $B_x^t$ is also non-degenerate for all t. For $x \in U$, we have a unique solution $\xi_t$ of $<B_x^t \xi_t,x>=<G_x x,x>$ and this solution depends smoothly on both x and t. That is, we have a smooth solution to (3.6) defined on $\widetilde{U}$. Clearly, $\xi_t(0)=0$ since $B_0^t$ is non-degenerate. Hence, by shrinking $\widetilde{U}$ we can integrate $\xi_t$ and get a smooth family of diffeomorphisms $\varphi_t$ from a smaller neighborhood $U$ of $0$ to another neighborhood $U'$ of $0$ which satisfies $f_t \circ \varphi_t = f_0 = \widetilde{A}$."

I also have two questions regarding this part of the proof:

  1. It is said that "$B_0^t$ is non-degenerate for all $0\leq t\leq 1$, and there exists a neighborhood $\widetilde{U}$ of $0 \in R^m$ such that $B_x^t$ is also non-degenerate for all t". I don't understand why this statement is correct. What we know is that for every t between 0 and 1, there exists a neighborhood $\widetilde{U}_t$ of $0 \in R^m$ such that $B_x^t$ is also non-degenerate for all x in $\widetilde{U}_t$, but this statement says something stronger, which I do not really understand why this something is correct.

  2. At the very last sentence of the proof it is said that "by shrinking $\widetilde{U}$ we can integrate $\xi_t$ and get a smooth family of diffeomorphisms $\varphi_t$ from a smaller neighborhood $U$ of $0$ to another neighborhood $U'$ of 0 which satisfies $f_t \circ \phi_t=f_0=\widetilde{A}$." - what does it mean "by shrinking"? Does it have anything to do with the neighborhood I asked about in my third question?

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.