Morse lemma

Let $f:M\to \mathbb{R}$ be smooth, and let $p$ be a critical point whose Hessian is nondegenerate. There are smooth local coordinates $y_1,\dots ,y_n$, centered at $p$, in whichThe integer $\lambda$ is the Morse index. This is an exact local identity, rather than a quadratic approximation with a remainder.^[ref]

To see the mechanism, apply the integral form of Taylor's theorem twice at the critical point, writing $f-f(p)={\sum }_{i,j}{x}_i{x}_j{h}_{ij}(x)$ with smooth symmetric coefficients. Nondegeneracy allows successive smooth completion of squares; the resulting change of coordinates has derivative given by an invertible linear map, so the inverse function theorem makes it a valid coordinate system. Sylvester's law of inertia makes the number of negative squares independent of the coordinates.

For $f(x,y)=x^2-y^2$, the origin has index one and is a saddle point. For $f(x)=x^3$, the origin is critical but its Hessian is zero; no smooth coordinate change makes it a nondegenerate quadratic critical point. The normal form implies that every nondegenerate critical point is isolated, and supplies the local model for the Morse handle-attachment theorem.

At a nondegenerate critical point placed at the origin, integral Taylor theorem gives $f(x)-f(0)=x^{T}A(x)x$, where $A(x)$ is a smooth symmetric matrix and $A(0)=\frac{1}{2}\mathrm{Hess}f(0)$. A constant linear change makes $A(0)$ diagonal with entries $\pm 1$. Successive completion of squares, with the Schur complement at each step, produces a smooth invertible triangular matrix $B(x)$ such that $x^{T}A(x)x={\sum }_i{ϵ}_i(B(x)x{)}_i^2$. The pivots keep their nonzero signs near zero. The map $x\mapsto B(x)x$ has invertible derivative at zero, so the inverse function theorem makes it a coordinate system. The signs are the Hessian matrix signature by Sylvester's law of inertia.