# Unique smooth structure on Euclidean space

1. Jan 4, 2016

### JonnyG

I was doing more reading in John Lee's "Introduction to smooth manifolds" and he mentioned that for every $n \in \mathbb{N}$ such that $n \neq 4$, the smooth structure that can be imposed on $\mathbb{R}^n$ is unique up to diffeomorphism, but for $\mathbb{R}^4$, there are uncountably many smooth structures, none of which are diffeomorphic to another. Why is this so? What makes 4-space so special? I know that the full answer to this question is beyond my scope (for now), but I was hoping for a laymens answer.

2. Jan 4, 2016

### WWGD

In a way, dimension 4 is large enough to allow for strange/interesting things but too small so that you cannot "undo" the strangeness. Maybe you can read up on intersection forms for 4-manifolds, which largely help determine the characteristics of a 4-manifold.

https://en.wikipedia.org/wiki/Intersection_form_(4-manifold)

Last edited: Jan 4, 2016
3. Jan 4, 2016

### andrewkirk

Is it possible to describe in a post a smooth structure that is not equivalent to the structure imposed by the metric that is, in Cartesian coordinates, everywhere the 4x4 identity matrix, which I would think of as the 'natural' smooth structure of $\mathbb{R}^4$?
Is the smooth structure imposed by the Minkowski metric a different smooth structure? If so, why is it impossible to construct an analogous structure in $\mathbb{R}^5$ using a metric with signature -++++ or, if it is, why is that structure not equivalent to the natural structure when the equivalence does not hold in four dimensions?

4. Jan 4, 2016

### lavinia

A metric does not determine a smooth structure. It determines a geometry on a manifold that already has a smooth structure.

5. Jan 4, 2016

### andrewkirk

Are you saying that there is more than one different smooth structure possible for a given manifold with specified metric?

6. Jan 5, 2016

### lavinia

What you call a "given manifold" is a topological manifold, not a smooth manifold. Without a smooth structure it has no tensor fields of any type.
There are even topological manifolds that can not be given any smooth structure so for them, the idea of tensor field makes no sense.

A metric does not exist outside of a smooth structure. It exists on a given smooth structure,

So to talk about possible smooth structures for a given manifold with specified metric makes no sense. One can only talk about possible metrics on a manifold with a given smooth structure.

However, one can specify properties of a metric and ask whether a given smooth structure admits a metric with these properties. For instance, it is possible to ask questions such as,"What smooth structures on a given topological manifold admit a metric of positive scalar curvature or positive sectional curvature?" But this means that one starts with the smooth structure and then searches for a metric. For instance, some exotic 7-spheres do not admit metrics of positive scalar curvature.

* An exotic 7-sphere is a smooth manifold that is homeomorphic to the standard 7-sphere but not diffeomorphic to it.

Last edited: Jan 5, 2016
7. Jan 5, 2016

### andrewkirk

Exactly. So specifying a metric specifies the smooth structure as well as additional structure. Thus specifying a metric for a manifold specifies a smooth structure for it, just as specifying the city in which someone lives also specifies the country in which they live.
Hence I'm afraid I'm still mystified by your comment that 'A metric does not determine a smooth structure.' Would you say that my saying I live in Sydney does not indicate that I live in Australia?

8. Jan 5, 2016

### Staff: Mentor

I don't want to get between the lines. I just want to remark something as an interested reader.
Firstly, no. Sydney could be in Nova Scotia.
Secondly. My understanding was, that in the cases you talk about, i.e. manifolds, they are per definition topological objects and the rest comes after it. To start with a metric can get you some very weird geometries that will have nothing to do with the intended examination of the manifold. One can do that, it simply is for other purposes. Open sets were first.
(That's my reading here. Correct me, if I'm wrong.)

9. Jan 5, 2016

### andrewkirk

The reference to the metric is simply the quickest and most accessible way I know to identify a smooth structure. It is also easier for me as I am more familiar with differential geometry than I am with the taxonomy of smooth manifolds. I suspect that is also true for most readers of this forum. I am well aware that the metric provides more information than the smooth structure and that there is a many-to-one map between metrics and smooth structures.

Say we were having a debate over whether tall people walked with a slower stride cadence (strides per second). You say they do and I say they don't. We are in a busy square where there are lots of people walking past, some tall, some not. I say to you 'Look at that man with the Purple Fedora hat. See how he has a fast stride.'

Would you reply 'But the facts that he is male and is wearing a hat and that it's a Fedora and its colour are all irrelevant to his height! Don't you know that tall people can wear all sorts of different colour hats, and some of them don't even wear hats at all?'?

10. Jan 5, 2016

### Staff: Mentor

I was thinking similar, too, as on many occasions it starts: "Be M a manifold with a norm induced or metric induced topology" - and all is said.
I was just checking Wiki on Riemann manifold. " .... with a positive definite bilinear form called metric tensor .... which isn't a metric in the definition of metric spaces." Maybe it's about this.

11. Jan 16, 2016

### lavinia

Another question is why is there a unique smooth structure on $R^n$ when $n≠4$? For dimensions $n≥5$ there is a single proof due to Stallings. Here is a reference.

http://www.maths.ed.ac.uk/~aar/papers/stallings2.pdf

A key ingredient is the "Engulfing Lemma" which fails below dimension 5.

Dimensions 1 2 and 3 are proved differently. Don't know the proof for dimension,3, but for dimension 1 one can prove it by showing that a smooth parameterization of a coordinate neighborhood of the manifold can always be extended to a larger neighborhood. This is done by showing that one can smoothly glue together overlapping parameterizations.

For dimension 2 there is a proof based on the existence of Isothermal coordinates on any smooth surface. One shows that the coordinate transformations between two isothermal coordinate patches satisfy the Cauchy-Riemann equations so the surface is a complex manifold. One then appeals to the Uniformization Theorem that says that any simply connected Riemann surface is conformally equivalent to either the complex plane or to the upper half plane.

https://en.wikipedia.org/wiki/Uniformization_theorem

12. Jan 16, 2016

### WWGD

Good point, it is actually constructed by pulling back the standard Euclidean metric by the chart maps. Since these chart maps are diffeomorphisms, they preserve positive definiteness . These metrics are pulled back locally and then glued together using a partition of unity subordinate to the cover (the P.O. Unity exists by paracompactness/2nd countability of a manifold). This process uniquely defines the metric in the manifold. A different chart set would give rise to a different metric. If there were inequivalent smooth structures, I don't know what the relation between the associated metrics would be.

13. Jan 16, 2016

### WWGD

You are right, this is not your standard metric distance function, this is a metric tensor, a smooth assignment of a bilinear form at each tangent space.

Understanding 4-maniolds can be done to a good degree by studying their intersection form, which describes properties of the intersection of 2-submanifolds-surfaces (actually, instead of the submanifolds, one uses their representatives in 2nd homology). You then have a partition of the intersection form, a bilinear form, into even and odd forms. This gives rise to conditions that allow , e.g., the existence of a quadratic form.

Last edited: Jan 16, 2016
14. Jan 17, 2016

### lavinia

Here is a paraphrasing of a sketched construction of an exotic $R^4$ given in this paper

http://projecteuclid.org/download/pdf_1/euclid.jdg/1214437666

Donaldson apparently proved that for closed smooth simply connected 4 manifolds with negative definite intersection form, there is a basis of homology classes that diagonalizes the form. Interestingly, there is an example of a symmetric bilinear form over the integers, called $E_8$, which is negative definite but not diagonalizable(over the integers). This form, by Donaldson's theorem, can not be the intersection form of a smooth simply connected 4 manifold.

One uses Donaldson's theorem on a manifold whose intersection form is $E_8⊕E_8⊕H⊕H⊕H$ where $H$ is the two dimensional form over $Z$ represented by the matrix that interchanges the two basis vectors. (Note that $H$ is not negative definite since over the rational numbers one can choose a basis so that it has the diagonal matrix, $e_1⋅e_1 = -1,e_2⋅e_2 = 1$). This 4 manifold is an algebraic variety called the Kummer surface. It is the zeros of the homogeneous complex polynomial, $z_1^4 + z_2^4 + z_3^4 + z_4^4$ in the projective space, $CP^3$. One shows that the three $H$'s are represented by the homeomorphic image of a manifold,$N$, that can be cut out and then replaced with a 4 ball. The resulting manifold is simply connected and has intersection form, $E_8⊕E_8$ which is also not diagonalizable. By Donaldson's theorem this manifold cannot be smooth. This, by itself, is interesting, an example of a manifold with no smooth structure.

This manifold , $N$, is homeomorphic to the the triple connected sum of $S^2$ x $S^2$ with itself minus an open 4 ball. ( Notice that the intersection form of $S^2$ x $S^2$ is $H$) One shows that its image,$X$, in the Kummer surface has an open neighborhood, $U$, such that $U-X$, is homeomorphic to $S^3$ x $R$. However it can not be diffeomorphic to the standard $S^3$ x $R$ and so is an exotic $S^3$ x $R$. Again interesting. Freedman has a separate paper where he publishes this result.

Now that one has this exotic $S^3$ x $R$, the author finds a diffeomorphic copy of it inside another manifold. Again this copy is an open neighborhood of the image of the triple connected sum of $S^2$ x $S^2$ with itself minus an open 4 ball inside this new manifold. In this case though after removing the triple connected sum minus a 4 ball, one is left with an open simply connected smooth 4 manifold with trivial second Z homology group. This is an exotic $R^4$.

Last edited: Jan 17, 2016
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