Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the definition of the index of a Morse function in dimension one?

share|cite|improve this question

There's no difficulty in making a general definition, so I'll answer this for all dimensions, not just dimension $1$.

Let $M$ be a smooth $n$-dimensional manifold and suppose $f: M \longrightarrow \Bbb R$ is a Morse function on $M$. Let $p \in M$ be a critical point of $f$, i.e. $df_p$ is the zero map. Now choose local coordinates $(x^1, \dots, x^n)$ near $p$. Then we can define the Hessian of $f$ at $p$ with respect to these coordinates to be the $n \times n$ matrix whose $(i,j)$-entry is $$\frac{\partial^2 f}{\partial x^i \partial x^j}(p).$$ The Morse index of $f$ at the critical point $p$ is then defined to be the number of negative eigenvalues of this matrix.

One can show that since $f$ is Morse and $p$ is a critical point of $f$, the Morse index of $f$ at $p$ is independent of the choice of coordinates $(x^1, \dots, x^n)$.

share|cite|improve this answer
ok, but the Hessian matrix is : $\frac{\partial^2f}{\partial x^i \partial x^j}(p)$ right ? – Vrouvrou Apr 1 '13 at 21:33
@Vrouvrou: Yes, thanks for pointing that out. I typed the answer without thinking. – Henry T. Horton Apr 1 '13 at 21:35
and in dimension 1 the hessian matrix is $f''(p)$ , so the index is 0 when $f''(p)>0$ or 1 when $f''(p)<0$ ? – Vrouvrou Apr 1 '13 at 21:41
when n=1 , we have $ind_f(p)=1$ or =0 – Vrouvrou Apr 1 '13 at 21:58
@Vrouvrou: Yes. The case $f''(p)=0$ never happens because a Morse function is defined to have only nondegenerate critical points, i.e. the Hessian matrix at a critical point is always invertible. – Henry T. Horton Apr 1 '13 at 22:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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