I am working on a problem for my differential geometry course. We are proving the following special case of the Morse lemma:

Let $U \subseteq \mathbb{R}^n$ be open and containing the origin $\mathbf{0} \in \mathbb{R}^n$ (denote coordinates on $U$ by $x = (x_1, \dots, x_n)$). Suppose we have smooth $f : U \to \mathbb{R}$, $f(\mathbf{0}) = 0$, with $\mathbf{0}$ a nondegenerate critical point of $f$ of index $1 \le \lambda \le n$. That is, we know that:

  1. $\frac{\partial f}{\partial x_i}(\mathbf{0}) = 0$, all $i$.

  2. The Hessian matrix $\left[ \frac{\partial^2f}{\partial x_i \partial x_j} \left(\mathbf{0}\right)\right]$ is nonsingular.

  3. $\left[ \frac{\partial^2f}{\partial x_i \partial x_j} \left(\mathbf{0}\right)\right]$ has $\lambda$ negative eigenvalues and $n - \lambda$ positive eigenvalues.

By transforming coordinates in a neighborhood of $\mathbf{0}$ to $y = (y_1, \dots ,y_n)$, prove that we can write:

$$f(y) = - \sum_1^\lambda (y_i)^2 + \sum_{\lambda + 1}^n (y_i)^2$$

Here is what I have so far. Because $f(\mathbf{0}) = 0$ and $\frac{\partial f}{\partial x_i}(\mathbf{0}) = 0$, all $i$, I know from a previous result in class that I am able to write

$$f(x) = \frac{1}{2} (x_1, \dots , x_n) \left[ \frac{\partial^2f}{\partial x_i \partial x_j} \left(\mathbf{0}\right)\right] (x_1, \dots, x_n)^T$$

in a sufficiently small spherical neighborhood $B$ of $\mathbf{0}$ ($B \subseteq U$). Then I can take the Hessian $\left[ \frac{\partial^2f}{\partial x_i \partial x_j} \left(\mathbf{0}\right)\right]$ and rewrite it in it's eigendecomposition

$$\left[ \frac{\partial^2f}{\partial x_i \partial x_j} \left(\mathbf{0}\right)\right] = Q^T \Lambda Q.$$

Here, $\Lambda$ is a diagonal matrix with the first $\lambda$ diagonal entries being the negative eigenvalues of the Hessian, and the last $n - \lambda$ diagonal values being the positive eigenvalues.

So it seems like I am getting close to finishing the proof, except $Q(x_1, \dots, x_n)^T$ is not quite the coordinate transformation I need. Roughly speaking, I need another transformation to ensure the diagonal elements of $\Lambda$ are $-1, -1, \dots ,1 ,1.$ But I'm not quite sure how to do this.

Hints or solutions are greatly appreciated.


Here's an example:

$$ \begin{pmatrix} -4 & 0 \\ 0 & 9 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} $$

  • $\begingroup$ Thanks! Now I see how to do it :-) $\endgroup$ – JZS Nov 15 '13 at 0:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.