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

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1 Answer 1

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 $(x^1, \dots, x^n)$.

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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
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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

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