1
$\begingroup$

Let $M$ be a set in $\mathbb{R}^n$ such that $M$ is locally a graph of some differentiable function (not necessarily $C^1$ ). Let $p\in \bar M$. We can define a tangent cone $C(M,p)$ as a set of all tangent vectors at point $p$, where tangent vector $v\in\mathbb{R}^n$ is defined by the following property:

(1) $v$ is tangent iff $\exists (p_n)_{n\in\mathbb{N}}\subset M $ : $p_n\rightarrow p, \ \lim_{n \to \infty} \frac{p_n-p}{|p_n-p|}=\frac{v}{|v|}$

How to prove that in this case $C(M,p)$ is a tangent space $T_pM$ in the standard sense (equivalence class of curves) ?

$\endgroup$
2
$\begingroup$

Without some extra conditions on the closure $\overline M$ (stratifications come to mind), there may be trouble. For instance, suppose $M$ is the famous graph of $y=\sin\frac{1}{x}$ for $x>0$, and let $p$ be the origin. As limits $\frac{p_n-p}{|p_n-p|}$ one gets all unit vectors $(a,b)$ with $0<a\le1$. However, there is no curve (even continuous) in $M$ that reaches the origin. If you want that such a curve exists, then you can change the curve for $x\ge1$ into a loop that reaches the origin from the semiplane $x>0$ with some slope you choose, and still the cone has much more vectors.

$\endgroup$

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.