## Sunday, 8 May 2016

### Persistence of transverse intersections

In Example 1.4.9 we discuss the persistence of transverse intersections as an application of the Impicit Function Theorem. Someone asked me about this in the second revision class. I will try to elucidate the final conclusion in this example from the notes.

"Persistence" of the isolated intersection of two curves in $$\mathbb{R}^2$$ in this example, means that if we "perturb" the curves slightly, then there remains to be a unique isolated intersection of these two curves near the original isolated intersection.

We represent the two curves by differentiable functions that parametrize these curves: $$f,g:\mathbb{R}\to\mathbb{R}^2$$. We assume the intersection to be at $$f(0)=g(0)$$. We now consider a parametrized family of functions $$f_\lambda,g_\lambda:\mathbb{R}\to\mathbb{R}^2$$, representing "perturbations of the original curves", where $$\lambda$$ serves as the "small parameter" so that $$f_0=f$$ and $$g_0=g$$. We furthermore assume that the perturbations are such that  the derivatives $$Df_\lambda(0)$$ and $$Dg_\lambda(0)$$ are continuous in $$\lambda$$ near $$\lambda=0$$.

In the example it is proposed to consider the function $$h_\lambda(s,t):\mathbb{R}^2\to\mathbb{R}^2$$ defined as $$h_\lambda(s,t):=f_\lambda(s)-g_\lambda(t).$$ By construction $$h_0(0,0)=(0,0)$$ and indeed the intersection points of the curves represented by $$f_\lambda$$ and$$g_\lambda$$ are given by $$h_\lambda^{-1}(0,0)$$.

It follows that $$Dh_\lambda(s,t)=(Df_\lambda(s),-Dg_\lambda(s))$$, as in the notes. This two-by-two matrix is non-singular (ie has no zero eigenvalue, or - equivalently - is invertible) if and only if the two-dimensional vectors $$Df_\lambda(s)$$ and $$Dg_\lambda(t)$$ are not parallel (ie not real multiples of each other).

We now use this to analyze the intersection at $$\lambda=0$$: when $$Df_0(0)$$ and $$Dg_0(0)$$ (which are the tangent vectors to the respective curves at the intersection point) are not parallel, then $$Dh_0(0,0)$$ is invertible and the intersection of the two curves at $$f_0(0)=g_0(0)$$ is isolated (there is a neighbourhood of this point, where there is no other intersection).

Considering a small variation of $$\lambda$$, we note that by application of the Implicit Function Theorem to $$h_0$$,  for sufficiently small $$\lambda$$ there exist continuous functions $$s(\lambda)$$ and $$t(\lambda)$$ so that $$(s(\lambda),t(\lambda))$$ is the element of $$h_\lambda ^{-1}(0,0)$$ near $$(0,0)=(s(0),t(0))$$. This unique "continuation" of the original solution $$0,0$$ is of course also isolated since if $$Dh_0(0,0)$$ is invertible then so is $$Dh_\lambda(s(\lambda),t(\lambda))$$ by continuity of all the dependences; so the vectors $$Df_\lambda(s(\lambda))$$ and $$Dg_\lambda(t(\lambda))$$ will not be parallel for sufficiently small $$\lambda$$.

#### 1 comment:

1. Thanks for sharing such an informative post. It will be quite helpful for the students to know and understand fractions of this type.