# Thursday January **Recall:** For $M^n$ a closed smooth manifold, consider a smooth map $f: M^n \to \RR$. Definition (Non-degenerate Critical Points) : A critical point $p$ of $f$ is *non-degenerate* iff $\det( H\definedas \frac{\del^i f}{\del x_i \del x_j}(p)) \neq 0$ in some coordinate system $U$. Proposition (The Morse Lemma) : For any non-degenerate critical point $p$ there exists a coordinate system around $p$ such that \begin{align*} f(x_1, \cdots, x_n) = f(p) - x_1^2 - x_2^2 - \cdots - x_\lambda^2 + x_{\lambda+1}^2 + \cdots + x_n^2 .\end{align*} $\lambda$ is called the *index of $f$ at $p$*. Lemma (Relating Index to Eigenvalues) : $\lambda$ is equal to the number of *negative* eigenvalues of $H(p)$. Proof : A change of coordinates sends $H(p) \to A^t H(p) A$, which (exercise) has the same number of positive and negative values. > Exercise: show this assuming that $A$ is invertible and not necessarily orthogonal. > Use the fact that $A^t H A$ is diagonalizable. This means that $f$ can be written as the quadratic form \begin{align*} \left[\begin{array}{ccccc} -2 & 0 & 0 & 0 & 0 \\ 0 & -2 & 0 & 0 & 0 \\ 0 & 0 & \ddots & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 2 \end{array}\right] .\end{align*} ## Proof of Morse Lemma Suppose that we have a coordinate chart $U$ around $p$ such that $p\mapsto 0\in U$ and $f(p) = 0$. ### Step 1 Claim : There exists a coordinate system around $p$ such that \begin{align*} f(x) = \sum_{i,j=1}^n x_i x_j h_{ij}(x) ,\end{align*} where $h_{ij}(x) = h_{ji}(x)$. Proof : Pick a convex neighborhood $V$ of $0\in \RR^n$. ![Image](figures/2020-01-09-11:38.png)\ Restrict $f$ to a path between $x$ and $0$, and by the FTC compute \begin{align*} I = \int_0^1 \frac{df(tx_1, tx_2, \cdots, tx_n) }{dt} ~dt = f(x_1, \cdots, x_n) - f(0) = f(x_1, \cdots, x_n) .\end{align*} since $f(0) = 0$. We can compute this in a second way, \begin{align*} I = \int_0^1 \dd{f}{x_1} x_1 + \dd{f}{x_2} x_2 + \cdots \dd{f}{x_n} x_n ~dt \implies \sum_{i=1}^n x_i \int_0^1 \dd{f}{x_i} ~dt = f(x) .\end{align*} We thus have $f(x) = \sum_{i=1}^n x_i g_i(x)$ where $\dd{f}{x_i}(0) = 0$, and $\dd{f}{x_i} = x_1 \dd{g_1}{x_i} + \cdots + g_i + x_i \dd{g_i}{x_i} + \cdots + x_n \dd{g_n}{x_i}$. When we plug $x = 0$ into this expression, the only term that doesn't vanish is $g_i$, and thus $\dd{f}{x_i}(0) = g_i(0)$ and $g_i(0) = 0$. Applying the same result to $g_i$, we obtain $g_i(x) = \sum_{j=1}^n x_j h_{ij}(x)$, and thus $f(x) = \sum_{i, j =1}^n x_i x_j h_{ij}(x)$. We still need to show $h$ is symmetric. For every pair $i, j$, there is a term of the form $x_i x_j h_{ij} + x_j x_i h_{ji}$. So let $H_{ij}(x) = \frac{h_{ij}(x) + h_{ji}(x)}{2}$ (i.e. symmetrize/average $h$), then $f(x) = \sum_{i, j = 1}^n x_i x_j H_{ij}(x)$ and this shows claim 1. ### Step 2: Induction Assume that in some coordinate system $U_0$, \begin{align*} f(y_1, \cdots, y_n) = \pm y_1^2 \pm y_2^2 \pm \cdots \pm y_{r-1}^2 + \sum_{i, j \geq r} y_i y_j H_{ij}(y_1, \cdots, y_n) .\end{align*} Note that $H_{rr}(0)$ is given by the top-left block of $H_{ij}(0)$, which is thus looks like ![Image](figures/2020-01-09-11:41.png)\ Note that this block is symmetric. Claim (1) : There exists a linear change of coordinates such that $H_{rr}(0) \neq 0$. We can use the fact that $\frac{\del^2 f}{\del x_i \del x_j} (0) = H_{ij}(0) + H_{ji}(0) = 2 H_{ij}(0)$, and thus $H_{ij}(0) = \frac{1}{2} \left( \frac{\del f}{\del x_i \del x_j} \right)$. Since $H(0)$ is non-singular, we can find $A$ such that $A^t H(0) A$ has nonzero $rr$ entry, namely by letting the first column of $A$ be an eigenvector of $H(0)$, then $A = [\vector v, \cdots]$ and thus $H(0)A = [\lambda \vector v, \cdots]$ and $A^t[\lambda \vector v] = [\lambda \norm{\vector v}^2, \cdots ]$. So \begin{align*} \sum_{i,j\geq r} y_i y_j H_{ij}(y_1, \cdots, y_n) &= y_r^2 H_{rr}(y_1, \cdots, y_n) + \sum_{i > r} 2y_i y_r H_{ir}(y_1, \cdots, y_n) \\ &= H_{rr}(y_1, \cdots, y_n) \left( y_r^2 + \sum_{i > r} 2y_i y_r H_{ir}(y_1, \cdots, y_n)/H_{rr}(y_1, \cdots, y_n) \right) \\ &= H_{rr}(y_1, \cdots, y_n) ( \left( y_r + \sum_{i > r}^n y_i H_{ir}(y_1, \cdots, y_n) / H_{rr}(y_1, \cdots, y_n) \right)^2 \\ &\cdot \sum_{i > r}^n y_i^2 \left( H_{ir}Y/H_{rr}(Y) \right)^2 \\ &\cdot \sum_{i, j > r}^n H_{ir}(Y)H_{jr}(Y)/H_{rr}(Y))^2 \\ &\quad\text{by completing the square} .\end{align*} > Note that $H_{rr}(0) \neq 0$ implies that $H_{rr} \neq 0$ in a neighborhood of zero as well. Now define a change of coordinates $\phi: U \to \RR^n$ by \begin{align*} z_i = \begin{cases} y_i & i\neq r \\ \sqrt{ H_{rr}(y_1, \cdots, y_n) } \left( y_r + \sum_{i> r} y_i H_{ir}(Y)/H_{rr}(Y) \right) & i=r \end{cases} .\end{align*} This means that $$ f(z) = \pm z_1^2 \pm \cdots \pm z_{r-1}^2 \pm z_r^2 + \sum_{i, j \geq r+1}^n z_i z_j \tilde{H}(z_1, \cdots, z_n) .$$ > Exercise: show that $d_0\phi$ is invertible, and by the inverse function theorem, conclude that there is a neighborhood $U_2 \subset U_1$ of 0 on which $\phi$ is still invertible. $\qed$ Corollary : The nondegenerate critical points of a Morse function $f$ are isolated. Proof : In some neighborhood around $p$, we have $$ f(x) = f(p) - x_1^2 - \cdots - x_\lambda^2 + x_{\lambda + 1}^2 + \cdots + x_n^2 ,$$ Thus $\dd{f}{x_i} = 2x_i$, and so $\dd{f}{x_i} = 0$ iff $x_1 = x_2 = \cdots = x_n = 0$. Corollary : On a closed (compact) manifold $M$, a Morse function has only finitely many critical points. We will need these facts to discuss the $h\dash$cobordism theorem. For a closed smooth manifold, $\del M = \emptyset$, so $M$ will define a cobordism $\emptyset \to \emptyset$. Definition (Morse Function) : Let $W$ be a cobordism from $M_0 \to M_1$. A *Morse function* is a smooth map $f: W\to [a, b]$ such that 1. $f\inv(a) = M_0$ and $f\inv(b) = M_1$, 2. All critical points of $f$ are non-degenerate and contained in $\mathrm{int}(W) \definedas W\setminus \del W$. > So $f$ is equal to the endpoints only on the boundary. > Next time: existence of Morse functions. This is a fairly restrictive notion, but they are dense in the $C^2$ topology on (?).