I have proved that it is for $a>0$ by constructing six parametrizations that are diffeomorphisms. When $a=0$, these parametrizations lose their diffeomorphic character as the derivative fails to exist at the origin.

However, the definition of a manifold is that ${\bf there \ exists}$ a local diffeomorphism from an open neighborhood of the set to $\mathbb{R}^k$ everywhere.

So just because the specific parametrizations that I constructed fails doesn't disprove that the set isn't a manifold. Obviously at the vertex of a cone, the object cannot map smoothly to $\mathbb{R}^2$; but how would I formally prove this using the definitions of manifolds, diffeomorphisms, parametrizations, smoothness, etc?


If it were a manifold, the point $P$ (the origin) would have a basis neighbourhood on the induced topology (which is an intersection of a ball with the cone) homeomorphic to an open connected set of $\mathbb{R}^2$. However, taking away the point $P$ from such a basis neighbourhood would leave a not-connected set, whereas taking away the image of the point $P$ from the chart codomain will leave a connected set.

If you are only asking about the upper part of the cone, then it indeed admits a structure as a smooth manifold, since there is a global homeomorphism to $\mathbb{R}^2$ (any topological manifold with one single chart is obviously a smooth manifold). Just not one which makes it a submanifold of $\mathbb{R}^3$.


May be this will help: From my understanding, this is a question as to do with the dimension of a topology. In other words as Aloizio Macedo said it if $a=0$ at the origin there is a problem.

Using the definition reported by Tasty Romeo on the answer to the dimension of a topology

A collection $\mathcal{A}$ of subsets of the space $X$ is said to have order $m+1$ if some point of $X$ lies in $m+1$ elements of $\mathcal{A}$, and no point of $X$ lies in more than $m+1$ elements of $\mathcal{A}$.

Given a collection $\mathcal{A}$ of subsets of $X$, a collection $\mathcal{B}$ is said to refine $\mathcal{A}$, or to be a refinement of $\mathcal{A}$, if for each element $B$ of $\mathcal{B}$, there is an element $A$ of $\mathcal{A}$ such that $B \subseteq A$.

With these notions we can define the dimension of topological space [Munkres]:

A space $X$ is said to be finite dimensional if there is some integer $m$ such that for every open covering $\mathcal{A}$ of $X$, there is an open covering $\mathcal{B}$ of $X$ that refines $\mathcal{A}$ and has order at most $m+1$. The topological dimension of $X$ is defined to be the smallest value of $m$ for which this statement holds; we denote it by $\dim X$.


Now it is evident that at the exception of the origin the dimension of this topological space is 2. What is it at the origin ?

  • $\begingroup$ Would you be wiling to share how you did formally build your proof ? $\endgroup$ – Nicolas Lussier-Clément Jul 10 '18 at 20:25

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