Question about index of critical points. I don't really understand what index of a critical point is and I am trying to do a very simple example.  I was wondering if someone could help me figure out what the index of the critical point $(0,0)$ is of the function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ given by $f(x, y) = x^2 + y^2$.  Thanks in advance.
 A: The index of a critical point of a real-valued function $f$ is the number of negative eigenvalues of the Hessian of $f$ at that critical point.
For your function, we have the following:
$$DF_{(x,y)} = \begin{pmatrix} 2x & 2y \end{pmatrix}.$$
Hence the only critical point of $f$ is $(0,0)$.
The Hessian of $f$ is
$$\mathrm{Hess}(f)(x,y) = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}.$$
So at the critical point $(0,0)$, we have
$$\mathrm{Hess}(f)(0,0) = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix},$$
which clearly has eigenvalues $2$ and $2$, which are positive. Hence
$$\mathrm{ind}(f)(0,0) = 0.$$
In this example, the Hessian matrix turned out to be constant; in general you need to plug in the critical point in order to get a matrix with real number entries. Additionally, the Hessian here was already diagonal so we could read the eigenvalues immediately; in general we need to do the usual linear algebra to find the eigenvalues of the Hessian. Finally, in this example the function was defined on Euclidean space; in general one has a real-valued function on a smooth manifold and we need to use local coordinates to write out the Hessian.
A: At a non-degenerate critical point, the tangent plane to $z = f(x, y)$ is horizontal.  It's a $2$-dimensional vector space, so there are $2$ independent directions one could travel an infinitesimal distance away from the critical point.  How many of those bend downward?
This is measured by the number of negative eigenvalues of the Hessian
$$
H(f) = 
\begin{bmatrix}
\frac{\partial^2 f}{\partial x \partial x} & \frac{\partial^2 f}{\partial x \partial y} \\
\frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y \partial y}
\end{bmatrix}.
$$
In your example,
$$
H(f) = 
\begin{bmatrix}
2 & 0 \\
0 & 2
\end{bmatrix}.
$$
The eigenvalue $\lambda = 2$ occurs (with multiplicity $2$), so there are $0$ negative eigenvalues.  This means that the critical point has index $0$.
Geometrically, when you view $z = f(x, y) = x^2 + y^2$ as the paraboloid surface in $\mathbb{R}^3$, the Morse function is just projection onto the $z$-coordinate.  The critical point of degree zero corresponds to the introduction of the $0$-dimensional disk (a point) in the iterative construction of (a homotopy equivalent representative of) the surface.  Since the paraboloid is contractible, this is the end of the story.
