Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I think the cone (or what is also called the "half cone") is a differential manifold but not a smooth manifold.

Can anyone help me understand this the nuts and bolts way?

How explicitly can I write down the differential structure on the cone?

Also what is the natural metric on it?

(Is that what is called the ``Sasakian Metric"? If yes then can one kindly explain that or give an expository reference on this?)

share|improve this question

2 Answers 2

up vote 1 down vote accepted

There seems to be fair bit of confusion.

First of, there are different notions of "cone". The standard one refers to a cone on a topological space (see wiki article), and is defined to be $CX = (X \times [0,1])/(X \times \{0\})$. This can, in particular, be applied to manifolds. What one gets is a topological space which, in general, is not a manifold, but can, in some special cases, be given a structure of a smooth manifold (see for example the discussion here).

Now, if one removes the "cone point" $X\times \{0\}$, what remains is just the product $X \times (0,1]$, which, as any product of a manifold and a manifold with boundary, is a manifold with boundary. The boundary is $X\times \{1\}$, which can be identified with $X$. This is the kind of cone one talks about in Sasakian geometry.

This is a completely separate issue, but the distinction between smooth and C^1 manifolds is largely nonexistent, due to a theorem of Whitney.

As for Sasaki manifolds, those are very special types of Riemannian manifolds. As the wiki referenced above explains, one way to define when a given Riemannian manifold $X$ is Sasaki, is to use the Riemannian metric on $X$ to put a metric on the cone, and see if the resulting metric makes the cone Kähler. (The article uses $X \times (0, \infty)$, which is diffeomorphic to the interior of the manifold we have, so this amounts to the same thing.) Usually it will not (Kähler metrics are quite special), but sometimes it will.

It is done this way because Kähler manifolds are more well known - they appear as projective or affine complex varieties. The meaning of this construction is that Sasaki manifolds are boundaries of (non-compact) Kähler manifolds.

Now, with these things sorted out, one can actually ask how does one put the Riemannian metric on the cone $X \times (0, \infty)$. The formula is in the wiki article - it calls it the "cone metric". If it is unclear what it means, please ask - here as a comment, or elsewhere.

share|improve this answer
    
@Max Thanks a lot for this detailed reply. Let me give a reference to the issue that I am puzzled about. Look at this link books.google.co.in/… –  Anirbit Nov 27 '10 at 22:26
    
@Max Here a statement is being made that the half cone is not a smooth manifold. This is the thing I want to understand. If the half cone has a $C^1$ structure then how does it escape the Whitney's theorem? –  Anirbit Nov 27 '10 at 22:28
    
Here you have yet another problem. What the book calls half-cone is what I would call a cone on $S^1$, as this is what this space is homeomorphic to (for the definition of a cone given above in the answer). It is also homeomorphic to the unit disc, and hence - as abstract topological space - has a smooth ($C^{\infty}$) structure. However, here it actually comes as a subset in $\mathbb{R}^3$, and this subset is not an embedded submanifold (one way to see this is to note it has no tangent space at the vertex). In fact, the "obvious" map from the unit disc to this subset has no derivative at 0. –  Max Nov 28 '10 at 0:40
    
@Max Then what does that book mean by saying that the cone over $S^1$ is not a smooth manifold when you are saying that it has a $C^\infty$ structure? Does that book only mean that it is not a smooth submanifold of $\mathbb{R}^3$ though abstractly it is a smooth manifold? Is the differential structure given by Ronaldo in his answer the smooth structure? –  Anirbit Nov 28 '10 at 9:15
    
@Anirbit 1)It is not very clear what the book means. Books are not always perfectly written (or even right). 2) Yes, that would be a reasonable way to interpret it. 3) What he did is take the standard R^2 and map it to the (infinite) cone (that's what I was calling "the "obvious" map". The map is a smooth embedding away from the origin. –  Max Nov 28 '10 at 11:36

The differentiable structure of the cone can be viewed as the same as the one of $S^1\times\mathbb R$, or $\mathbb{R}^2 - \{0\}$ . That is, you construct a homeomorphism between $\mathbb{R}^2 - \{0\}$ and the cone (as subsets of $\mathbb{R}^3$) and say that this is a diffeomorphism, in the most intuitive way: $(x,y)\mapsto (x,y,\sqrt{x^2 + y^2})$. With this imposition, the differentiable structure of $\mathbb{R}^2 - \{0\}$ is carried to the cone.

I think the "natural" metric is the one induced from $\mathbb{R}^3$: $\tilde g = i^*g$, where $i$ is the inclusion map from the cone to $\mathbb{R}^3$ and $g$ is the Euclidean metric of $\mathbb{R}^3$.

share|improve this answer
    
Thanks for the answer. I am aware of this differential structure. The point I am puzzled about is how to show that there exists no differential structure which is $C^n$ and $n \geq 2$. Also is the metric you referred to the Sasakian metric? –  Anirbit Nov 27 '10 at 15:09
    
I don't know about the Sasakian metric... –  Ronaldo Dec 3 '10 at 18:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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