Question on Morse Theory.

Question

I am studying Morse Theory, and I would like to know what a ‘non-degenerate smooth function’ means. Please help. Thanks!

Answer 1

Let $ M $ be a smooth $ n $-dimensional manifold and $ f: M \to \mathbb{R} $ a smooth function. We say that $ f $ is non-degenerate if and only if for each critical point $ p $ of $ f $, the Hessian matrix $ \left[ {\dfrac{\partial^{2} f}{\partial x_{i} \partial x_{j}}}(p) \right]_{i,j} $ computed at $ p $ is non-singular (i.e., has non-zero determinant), where $ (x_{1},\ldots,x_{n}) $ is a local-coordinate system around $ p $.

Important note: The definition of non-degeneracy is consistent because it does not depend on the local-coordinate system $ (x_{1},\ldots,x_{n}) $ used around a critical point $ p $. In other words, if $ (y_{1},\ldots,y_{n}) $ is another local-coordinate system around $ p $, then $$ \det \left( \left[ {\frac{\partial^{2} f}{\partial x_{i} \partial x_{j}}}(p) \right]_{i,j} \right) \neq 0 \iff \det \left( \left[ {\frac{\partial^{2} f}{\partial y_{i} \partial y_{j}}}(p) \right]_{i,j} \right) \neq 0. $$

Answer 2

$f$ is a non-degenerate smooth function if at each critical point $p$ of $f$, the Hessian matrix at $p$ is non-singular. For example, $f(x,y)=x^2-y^2$ is a non-degenerate smooth function, because the only critical point of $f$ is $(0,0)$ since $\nabla f=(2x,-2y)$. On the other hand, the Hessian of $f$ is given by $$ \left[ \begin{array}{cc} 2 & 0 \\ 0 & -2 \\ \end{array} \right]$$ is nonsingular at $(0,0)$.