I am working on a practice question for differential geometry and reviewing the provided solutions - but I am having a bit of trouble connecting the ideas of transverse planes and regular curves,

The question is this:

Prove that if all normal lines of a connected surface $\sum$ meet a fixed straight line, then every point of $\sum$ has a neighbourhood that is contained in a surface of revolution.

The solution proceeds as follows:

Let $r$ be a fixed line which is met by the normals of the surface $\sum$. Consider the plane $Q$ which is perpendicular to the line $r$ and contains a point $p \in \sum$.

Here is where my confusion is, the solution then says:

Since the normal at $p$ meets the line $r$, it must be that $Q$ is transverse to the tangent plane at $p$. Thus, in a neighbourhood of $p$, $Q \cap \sum$ is regular curve.

Looking up the definition of transverse, it says that smooth manifolds intersect transversally if at every point of intersection, their separate tangent spaces at that point together generate the tangent space of the ambient manifold at that point.

I'm having a bit of trouble with this definition - if $Q$ and $T_p \sum$ are transverse, does this mean that the tangent vectors of $p$ in $Q$ and $T_p \sum$ generate the tangent space at $p$? (Would this be the whole space?) - what is the 'ambient manifold' in this case? Is it the curve where $Q$ and $T_p \sum$ intersect?

Why does this then mean that $Q \cap \sum$ is a regular curve?

Thanks for any help you can give me.


In your case the ambient space is $\mathbb{R}^3$. Hence $T_p\mathbb{R}^3$ is just $\mathbb{R}^3$ with center shifted to $p$ (it is not formal description, just how to feel it).

If we are given with two surfaces $\Sigma_1,\Sigma_2$ containing $p$ then we call them trasverse at $p$ if $T_p\Sigma_1 \cup T_p\Sigma_2$ generates $T_p\mathbb{R}^3.$ Intuitively, two surfaces in $\mathbb{R}^3$ are transvere at $p,$ if their tangent spaces at $p$ are different.

In your case $\Sigma_2$ is just a plane $Q,$ so it's tangent space at $p$ is just $Q$ in $T_p\mathbb{R}^3.$ Hence now transversality at $p$ means that $Q\cup T_p\Sigma$ generates $T_p\mathbb{R}^3.$ It holds since $Q$ contains normal vector to $\Sigma$ (Crutial thing is that $n\not\in T_p\Sigma$).

Everything is smooth here, hence transversality at $p$ implies that we have similar situation in neighberhoud $V$ around that point (simply we cannot have the situation where at $p$ plane $Q$ contain normal vector and eveywhere around it is tangent to $\Sigma$).

Now we jump to the regularity issue. As we deal with the surface, in order to define one, we require charts and hence regular maps from open subset of $\mathbb{R}^2$ to $\mathbb{R}^3.$ Let $$\phi:\mathbb{R}^2\supset U\rightarrow\Sigma\subset R^3$$ be a homeomorphism such that $\phi(x)=p.$ We say that $\phi$ is regular at $x$ if $d_x\phi:T_x\mathbb{R}^2\rightarrow T_p\mathbb{R}^3$ has maximal rank (that is 2). Equivalently Jacoby matrix of $\phi$ at $x$ has rank 2.

In similar manner we would say that $\gamma=\Sigma\cap Q\cap V$ is a regular curve if there exists homeomorphism $$\psi:(a,b)\rightarrow\gamma$$ such that $\psi$ is regular everywhere (its Jacoby matrix at every point has rank 1, equvalently $\psi'(x)\neq 0$ for every $x\in(a,b)$).

The question is how can we find such chart $\psi.$ It can be constructed from $\phi,$ but details require implicit function theorem.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.