Morse function definition and why index of critical point with respect to $-f$ is $n$ minus index with respect to $f$

Question

My book is From Calculus to Cohomology by Ib Madsen and Jørgen Tornehave.

In Chapter 12, we have that, for a smooth $n$-manifold $M$ that is a smooth submanifold of $\mathbb R^{n+k}$,

Corollary 12.17 If $M$ is compact and $n$ is odd, then its Euler characteristic $\chi(M)$ is zero.

In the proof of Corollary 12.17, it is claimed that for a Morse function $f$,

1. $-f$ is a Morse function

2. $-f$ has the same critical points as $f$

3. Since $-f$ is Morse, its critical points are nondegenerate such that $\text{Index}_{\{-f\}}(p)$ is defined for each critical point $p$ of $-f$ (and $f$)

4. $\text{Index}_{\{-f\}}(p) = n - \text{Index}_{\{f\}}(p)$ for each critical point $p$ of $f$ (and $-f$).

The definitions of $\text{Index}_{\{f\}}(p)$, Morse and nondegenerate are given in Definition 12.3, which refers to Proposition 12.2. Now, Definition 12.3 and Proposition 12.2 don't use the terms "Hessian" or "negative eigenvalue", but those terms seem to be relevant based on what I recall reading online.

Question 1. How do I prove (4)?

My thoughts:

5. Perhaps Theorem 12.6 is relevant. My understanding of Theorem 12.6's relevance is that the matrix for $f$ in Proposition 12.2 (could be a "Hessian") is a diagonal matrix each element to be $\pm 2$.

6. I probably don't understand the definition. With $T_pM \cong \mathbb R^n$, I think any subspace $V$ with dimension $\lambda$ is isomorphic to $\mathbb R^{\lambda}$, specifically by being isomorphic to $\mathbb R^{\lambda} \times \{0\}^{n-\lambda}$ or to $\prod_{i=1}^{n} A^i$ where $A^i=\mathbb R$ or $\{0\} = \mathbb R^0$, with $\lambda$ of the $A^i$'s equal to $\mathbb R$ and $n-\lambda$ of the $A^i$'s equal to $\{0\}$. Therefore, I think of $(d^2f)_p|_V = d^2_pf|_V$ as the "vanishing" (kinda like the "vanishing" in the submanifold sense. See here and here) of $n-\lambda$ of the $x^i$'s, where I think $x^i$'s represent the standard coordinates of $\mathbb R^n$.

   • This understanding seems to be consistent with Example 12.5: Vanishing of the last $n-\lambda$ $x^i$'s corresponds to making the $n-\lambda$ positive $2$'s become $0$. The matrix $\text{diag}(-2,-2,2)$ is not negative definite, and the matrix $\text{diag}(-2,-2,0)$ is negative definite.

Question 2. This is what I tried for $n=3$ and $\lambda=\text{Index}_{\{f\}}(p)=2$, for a critical point $p$ of a Morse function $f$ on a compact $n$-manifold $M$, $n=3$ and $\lambda=\text{Index}_{\{f\}}(p)=2$. Is this correct, and why?

7. We are given that $A$, the matrix for $f$ in Proposition 12.2 (could be a "Hessian"), is symmetric, invertible and not negative definite.

8. We are given that there exists a vector subspace $V$ of $T_pM$ with dimension $\lambda=2$ that represents vanishing of exactly $n-\lambda=3-2=1$ of the $x^i$'s such that at least of the following vector subspaces $\{x^1 = 0\}, \{x^2 = 0\}, \{x^3 = 0\}$ is $V$ and such that $d^2_pf|_V$ is negative definite.

9. We deduce that $-A$ is the matrix for $-f$ in Proposition 12.2 (could be a "Hessian").

10. We have to show that $-A$ not negative definite.

11. We have to show that there exists a vector subspace $W$ of $T_pM$ with dimension $n-\lambda=3-2=1$ that represents vanishing of exactly $\lambda=2$ of the $x^i$'s such that at least of the following vector subspaces $\{x^1 = 0, x^2 = 0\}, \{x^2 = 0, x^3 = 0\}, \{x^3 = 0, x^1 = 0\}$ is $W$ and such that $d^2_p(-f)|_W = -d^2_pf|_W$ is negative definite.

12. We have to show that for any vector subspace $S$ of $T_pM$ with dimension $2$, $d^2_p(-f)|_S = -d^2_pf|_S$ is not negative definite.

13. We have that the vanishing of a coordinate $x^i$ corresponds not to all the elements of row $i$ and column $i$ of the matrices $A$ and $-A$ becoming zero but to those elements deleted. (For example, $\text{diag}(-2, -2, 0)$ is not negative definite, but $\text{diag}(-2, -2)$ is negative definite.)

14. We have that $d^2_pf$ is negative definite if and only if $A$ is negative definite (Just checking that we mean negative definite matrix and not negative definite function or something).

15. By (5), $A$ is $\text{diag}(-2, -2, 2)$, $\text{diag}(-2, 2, -2)$ or $\text{diag}(2, -2, -2)$. Let's say $A = \text{diag}(-2, -2, 2)$

16. By (9) and (15), $-A = \text{diag}(2,2,-2)$.

17. By (15), $V = \{x^3 = 0\}$ because $\text{diag}(-2, -2)$ is negative definite.

18. Therefore: We show (11) by choosing $W = \{x^1 = 0, x^2 = 0\}$ because $\text{diag}(-2)$ is negative definite. We show (12) by showing that $\text{diag}(2,-2)$, $\text{diag}(2,-2)$ and $\text{diag}(2,2)$ are not negative definite. We show (10) by showing $-A$ in (16) is not negative definite.

Answer 1

The index is defined as number of negative eigenvalues of the Hessian (the matrix of second partial derivatives of $f$) evaluated at a nondegenerate critical point. It's easy to check that if $c$ is nondegenerate critical point of $f$ then $c$ is also a nondegenerate critical point of $-f$, and the number of negative eigenvalues of $-f$ is the number of positive eigenvalues of $f$---this is simply because of linearity of the partial derivatives.

Edit: The notation $d_p^2$ in your definition refers to the Hessian as a bilinear form. The maximal dimension of the eigenspace on which it is negative definite is the same thing as the number of negative eigenvalues of the corresponding matrix, this is linear algebra.