Morse lemma

If $n=0$, the manifold is locally a point, so $f$ is locally constant at $p$. The Hessian acts on the zero vector space, hence $\lambda =0$ by [F1], and the displayed formula is exactly $f=f(p)$ with both sums empty.

Assume the theorem proved in dimensions $<n$, where $n>0$. Choose local coordinates $u=(u^1,\dots ,u^n)$ centered at $p$ and write $g:=f\circ u^{-1}-f(p)$. By [L1], after a linear change of the $u$-coordinates the Hessian matrix of $g$ at $0$ is diagonal with entries in $\{1,-1,0\}$. Since $p$ is nondegenerate and has index $\lambda$, [F1] and [L2] force exactly $\lambda$ negative diagonal entries, exactly $n-\lambda$ positive diagonal entries, and no zero entry. Reorder the coordinates so the first diagonal entry is negative when $\lambda >0$ and positive when $\lambda =0$; in particular ${\mathrm{\partial }}^{2}g\mathrm{/}\mathrm{\partial }(u^1{)}^2(0)\mathrm{\ne }0$. [F1, L1, L2, given, assume-case[ positive-dimension], construct]

Apply [L3] to the first coordinate $u^1$, taking the remaining variables as parameters. After shrinking the chart there are new coordinates $(v^1,y)$ with $g(v^1,y)=\epsilon (v^1{)}^2+H(y)$, where $\epsilon \in \{\pm 1\}$, $y\in {\mathbb{R}}^{n-1}$, and $0$ is a critical point of $H$.

By [L2], the Hessian of $g$ in the $(v^1,y)$ chart still has index $\lambda$. By [L4], the Hessian of $H$ at $0$ is the restriction to the $y$-coordinates, and the split $v^1$-direction contributes one negative square exactly when $\epsilon =-1$. Therefore ${\mathrm{Hess}}_0(H)$ is nondegenerate, with index $\lambda -1$ when $\epsilon =-1$ and index $\lambda$ when $\epsilon =1$.

Apply the induction hypothesis to $H$ on ${\mathbb{R}}^{n-1}$. It yields local coordinates $(v^2,\dots ,v^n)$ putting $H$ into its Morse normal form, and adjoining $v^1$ contributes one additional negative square exactly when $\epsilon =-1$. Therefore the full expression for $g$ has exactly $\lambda$ negative squares and $n-\lambda$ positive squares.

Combining steps 2.1 and 4.1 proves the displayed normal form for dimension $n$, and step 1.1 covers the base case.