Focal points in abstract sense

Question

Guillemin and Pollack define a focal point of a hypersurface $X \subset \mathbb{R}^n$ to be a critical value of the normal bundle map $h: N(X) \to \mathbb{R}^n$ defined by $h(x,v) =x+v$.

What does this really mean intuitively?

I'm asked to find the focal points of the parabola $y=x^2$ in $\mathbb{R}^2$. At any point $(x,x^2) \in X$ (the parabola), the normal to the tangent space should be the line $y= -x/2$. But I'm not sure where to go from here.

Answer 1

The idea given by @Ted Shifrin is correct. I just give some illustration here.
The key point is to comprehend the derivative of the normal bundle map. Although it has a linear form, $dh_{x,v}$ is not equal to $h$ since the parametrization of $N(X)$ is not simply the identity( even though $dimN(X)=n$. See p.71 of Pollack). In order to clarify the structure of $dh_{x,v}$, let $x=(t,t^2)$. Then for $t\neq 1$, $v$ can be expressed as $s\cdot(1,-\frac{1}{2t})$, which implies $$ h(t,s)=(t,t^2)+(s,-\frac{s}{2t})=(t+s,t^2-\frac{s}{2t}) $$ In this form, the domain $h$ is set to be open in $R^2$, which leads to the derivative $$ dh_{t,s}=(1,1;2t+\frac{s}{2t^2},-\frac{1}{2t}) $$ Hence the critical point occurs iff $2t+\frac{s}{2t^2}+\frac{1}{2t}=0$. Incorporate it into $h(t,s)$, we know the critical value can be expressed as $$ h(t,s(t))=(t-(t+4t^3),t^2+\frac{t+4t^3}{2t})=(-4t^3,\frac{1}{2}+3t^2) $$ that is the curve we requires, with some appropriate modification on $t=0$.

Answer 2

From Milnor's Morse Theory Chapter 6 (for an embeddeed submanifold $M \subseteq \mathbb{R}^n$), "a focal point of $M$ is a point in $\mathbb{R}^n$ where nearby normals intersect".

This intuition can be observed from the example worked out in Jiayu's answer for the parabola $y = x^2$. Compare the resulting curve with the original parabola to see this picture. Here are 2 more examples

1. For the $x$-axis embedded in $\mathbb{R}^2$, the normal bundle in $\mathbb{R}^2 \times \mathbb{R}^2$ looks like $$\{(x,0,0,\lambda) : x,\lambda \in \mathbb{R}\}$$ So the resulting "endpoint map" is the identity map $(x,\lambda) \mapsto (x,\lambda)$. Its Jacobian is the identity everywhere so this embedding has no focal points. This makes sense, as the normal spaces of points on the $x$-axis are parallel, so they never intersect.

2. For the unit circle in $\mathbb{R}^2$, we can parametrize it as $\theta \mapsto (\cos\theta, \sin\theta)$. In this case the normal bundle looks like $$\{(\cos\theta,\sin\theta,\lambda\cos\theta,\lambda\sin\theta)\}$$ In this case the determinant of the Jacobian works out to be $1+\lambda$, so the focal points are of the form $(\theta,\lambda=-1)$, but this is just the origin. This makes sense : at two nearby points (in fact for any pair of distinct points) on the circle, the orthogonal lines do meet at the origin.