Critical values/points: Morse Theory and Calculus

Simon found a promising series of lectures on Morse theory on YouTube.  I tend to watch such things a couple times since my life is rife with interruption lately (of the cute, cuddly infant sort).  I was watching it again this morning and he was defining what a critical point/value was.  It’s where the derivative is not of maximal rank, a definition I had heard before.  I wondered how this definition meshes with the usual one we teach our calculus students.  I have a tendency to attempt to reduce beautiful ideas to trite calculations and this is no exception.

Suppose we consider an embedding of the open interval (1,2) into \mathbb{R}^2 by f(x)=3-(x-3/2)^2.   The image of f is the set of ordered pairs (x, f(x)) and we can treat this as a 1-manifold, M, by using the projection map on the first coordinate as our chart.  Now we can consider the projection map on the second coordinate as a continuous function from our manifold M to \mathbb{R}.  This will represent the height of the curve above the x-axis.


We can calculate the differential of this map which will be a map between tangent spaces.  So for each point p\in M we have a differential of f, D_p f: T_pM\to \mathbb{R} (really this should map to T_{f(p)}\mathbb{R} but this is again just \mathbb{R}).  This will be the 1\times 1 matrix given by the derivative of f at p.   I’ve drawn a part of the tangent bundle of \mathbb{R} and the image of the differential of f in various fibers.  Notice that the maximal rank of D_p f is one since we’re mapping from 1-dimensional tangent spaces (of M) to other 1-dimensional tangent spaces (of \mathbb{R}).


There’s only one place where the rank of D_p f is zero.  That’s at p=(3/2, 3) where the image of D_p f is the zero vector in the tangent space at y=3 in \mathbb{R}.

In calculus 1, we’re only ever worried about maps between 1-manifolds so we can rephrase the definition of a critical point as a point where the rank of the differential is zero, or equivalently, where the differential is the zero map.  Since the differential is just the 1\times 1 matrix with the derivative of f in it, this amounts to business as usual.

Hold your applause, this isn’t intended to blow your mind.  It’s just meant to connect one definition to another. If I felt like drawing more pictures we could go to 3D with a function f:\mathbb{R}^2\to\mathbb{R}.  Suppose it’s something like a gaussian centered at something like (2,2) and we take the image of the gaussian on an open disc in the plane.  The resulting surface can be given one chart, the projection map on the first two coordinates, making it a 2-manifold.  Then we can take the projection map on the third coordinate and have that represent the height of the surface above the xy plane.  Then the differential is just the 2\times1 matrix the components of which are the components of the gradient.  That matrix takes tangent vectors to the surface and maps them to tangent vectors of \mathbb{R}.  Maximal rank in this case is again one.  For the gaussian, there’s a peak and at that point the differential of the height function will have rank zero.  So this point will be a critical point and the resulting value (under f) will be a critical value.

Again, not very deep but I wanted to task myself with sketching out the details and this blog seems to hold me more accountable than I hold myself.


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