# Diffeomorphic manifolds

1. Jan 24, 2008

### WWGD

what property do they share that non-diffeomorphic manifolds do not
share?. I have thought that if A,B were non-diffeomorphic (with dimA=dimB=n),
certain functions (i.e, with their respective coord. representations) from A
into IR^n that are differentiable , would not be differentiable, in the same sense
as with A, as functions from B into IR^n.

Any other one?.

Thanks.

2. Jan 25, 2008

### mathboy

I diffeomorphism is a differentiable homeomophism. So all the topological invariants like connectedness, compactness, normality, etc... are preserved by diffeormorphisms.

3. Jan 25, 2008

### WWGD

Right. I should have specified: say M,N are homeomorphic, but not
diffeomorphic, and M',N' are diffeomorphic. What properties do M',N'
share that M,N do not share?.

4. Jan 25, 2008

### mathboy

Properties involving derivatives: e.g. boundaries, orientability, critical points, handles.

5. Jan 25, 2008

### WWGD

Thanks, Mathboy. Any chance you have some examples?. I can think maybe of
two incompatible atlases for a single manifold M, one atlas orientable, another one
not orientable, yet M has the same topology, so the two copies of M with different
atlases, are homeomorphic.
Can you think of any other?

6. Jan 26, 2008

### gel

Orientability is purely a topological property and does not require any atlas, so this can't be done.

7. Jan 26, 2008

### gel

Even finding manifolds that are homeomorphic but not diffeomorphic is not easy. Wikipedia has a page on exotic spheres.

8. Jan 26, 2008

### mathwonk

I understood the question to be, how do you tell if two manifolds are niot diffeomorphic, when all their topologiucal invariants, such as homology, etc... are the same.

Milnor's method is the following: if the manifold of interest occurs as the boundary of another manifold, it may be that one differentiable structure extends to a compatible differentiable structure on the larger manifold but the other does not.

thus he found a class of triangulated manifolds with boundary homeomorphic to the 7 sphere, and which would have a compatible differentiaBLE STRUCTURE provided the boundary manifold were actually diffeomorphic to the 7 sphere.

however he showed the combinatorial pontrjagin classes of the larger manifold were not always integers, which would be the case if these manifolds did have differentiable structures compatible with the given triangulation.

i.e. when there is a differentiable structure compatible with the triangulation, then the pontragin classes coincide with the hirzebruch classes of the tangent bundle, which is defined for differentiable manifolds, and those are integers.

if you think about it, extending maps to be differentiable is not so easy. e.g. think of two discs, and a homeomorphism betwen their boundary circles. it is easy to extend this to a hmeomorphism between the discs by extending long radial lines, i.e. map the center to the center, and map each radius along a boundary point of one circle to the radius along the corresponding boundary point of the other.

but suppose the map of circles is a diffeomorphism, then this process usually does not give a map of discs which is differentiable at the ceNTER, unless the original diffeo of boundary circles was a rotation.

this illustration occurs in a film of a talk by milnor some 40 or 50 years ago.

9. Jan 27, 2008

### WWGD

Well, thanks to all, but I was thinking of a specific case: of IR^4 and (IR^4)',
with (IR^4)' being an exotic IR^4, which is homeo. but not diffeo. to std. IR^4
(with id. chart), and, in general, if we somehow knew from the start that we had
M=(X,S),M'=(X,S') with:

X as the topology of the space (meaning that both spaces
are, as topological spaces, equivalent, i.e, M,M' are homeomorphic), and

S,S' , the respective differentiable--say C^oo--structures on M,M', such
that M,M' are not diffeomorphic;

the example of exotic and original IR^4 is just used to show--only case I know of--
that we are not starting with a false premise --so we won't end up with an
"If pigs fly" issue .

Then, the question would be (and I think this is somewhat different from where
wonk was getting at what would different properties would we find to be
different in IR^4 and in (IR^4)',and/or, more generally, what properties would
be different in M,M'.

.

And, while we are at it: does anyone know of manifolds that are C^0
but not C^1?. ( I guess the general case of a C^k-but-not-C^(k+1)
would be too difficult), meaning that M admits an atlas in which the
overlaps are continuous, but, for any atlas, the overlaps are not
differentiable; there is always a pair of overlaping charts where the
exchange is not differentiable.
This last issue came up re a recent result I saw that
proved to me that the cone (right-circular cone: circular base, vertex
perp. to the base, etc.) was not an example of a C^0-but-not-C^1
mfld.
Result is not that big of a deal: an n- manifold (n=/4)with a 1-chart
atlas, is a smooth mfld., and diffeomorphic (globally) to IR^n. The cone has
a 1-chart atlas.

It is 4 a.m, and as cold as hell, so I hope this makes sense.

10. Jan 27, 2008

### WWGD

11. Jan 27, 2008

### gel

I assume that the polynomial invariants mentioned here Simon Donaldson would answer this question, although I don't know much about them. Interesting question though

12. Jan 27, 2008

### gel

13. Jan 27, 2008

### mathwonk

i think i gave a very precise answer to exactly what you were asking, but in the compact case.

i.e. in that case the difference was that the two manifolds behave differently wrt extensions of smooth structure from themselves to manifolds they bound.

14. Jan 27, 2008

### mathwonk

if you are interested in the newer approaches sucha s by seiberg and witten, they show that two smooth 4-manifolds which are homeomorphic but not diffeomorphic, can sometimes be given different algebraic structures so that they have different "plurigenera".

i.e. an algebraic manifold is in particular a complex manifold with a cotangent bundle. the powers of the determinant of this cotangent bundle have finite dimensional vector spaces of sections. these dimensions are called plurigenera, by analogy with the fact that on a riemann surface, the dimension of the space of one forms i the genus of the 2-manifold.

thus the result is that two algebraic complex surfaces with different plurigenera cannot be diffeomorphic, but they can be homeomorphic.

again this is for compact manifolds, and i do not know the theory for non differentiable models of R^4.

indeed if you actually read the links provided above on donaldson and seiberg witten invariants they seem to apply to the compact case.

further search suggests the cases are related, through theory of h cobordism. presumably one uses trivial constructions such as the fact that a compact 4-sphere becomes a non compact R^4 upon removal of one point.

the word cobordism suggests also that the original idea of milnor i described above, is again central, namely of regarding one manifold as a boundary of another. ("bord" = border = boundary.)

see:

http://en.wikipedia.org/wiki/H-cobordism_theorem

Last edited: Jan 27, 2008
15. Jan 27, 2008

### mathwonk

applications of these ideas in complex algebraic geometry yield results such as the following: if a compact algebraic surface S (hence S is a smooth 4 manifold) is diffeomorphic to an algebraic surface T which has a zariski open set which is holomorphically isomorphic to a zariski open set of P^2, then S also has such a zariski open set which is holomorphically isomorphic to a zariski open set of P^2.

briefly, an algebraic surface is rational if (and only if) it is diffeomorphic to a rational surface.

Last edited: Jan 27, 2008
16. Jan 27, 2008

### mathwonk

let me try again. a diff'ble 4 - manifold M is a topological 4 - manifold, with a sheaf of functions on it which is locally diffeomorphic to the sheaf of smooth functions on a ball in R^4.

examples of this are for example P^2, the complex projective 2 plane. P^2 also has a sheaf of holomorphic functions which is a subsheaf of its sheaf of smooth functions.

more generally, any complex projective algebraic "surface" has a sheaf of holomorphic functions, locally isomorphic to the sheaf of holomorphic functions on a ball in C^2, and which is contained in its sheaf of smooth functions.

the tangent bundle and cotangent bundle of the smooth 4 manifold inherit a complex structure when the manifold has a complex structure, and this enables us to assign some numbers.

i.e. the determinant bundle of the cotangent bundle has only finite number of independent holomorphic sections, whereas it has an infinite number of smooth ones.

so to each holomorphic subsheaf of the sheaf of smooth functions on M, we get a number, the maximum number of independent holomorphic sections of the determinant of the cotangent bundle,

i.e. the number of independent holomorphic 2 forms is finite, whereas it is infinite for smooth 2 forms.

then we can say that if two 4 manifolds are diffeomorphic, then any way of choosing a holomorphic subsheaf of their sheaves of smooth functons, will yield the same number of independent holomorphic 2 forms.

but if they are only homeomorphic and not diffeomorphic, there can be holomorphic subsheaves of the sheaves of smooth functions which yield different numbers of independent holomorphic 2 forms.

i.e. smooth 4 manifolds which are only homeomorphic can have holomorphic structures with different "holomorphic genus", but all holomorphic structures on diffeomorphic 4 manifolds will have the same holomorphic genus.

in a nutshell, different smooth structures on real 4 manifolds are detected by their different relationship with compatible complex structures on complex 2 manifolds.

Last edited: Jan 27, 2008
17. Jan 27, 2008

### gel

Thanks mathwonk, that explanation of plurigenera seems clear enough to me. I'll have a read through the link I posted for more details.

It's not obvious why the plurigenera will not depend on the choice of complex structure though, but hopefully it will be clearer when I have some time to think about it in more detail.

18. Jan 27, 2008

### mathwonk

it is indeed far from obvious. that was a big result that followed from the use of the seiberg witten invariants, which in turn were an easier tool than the earlier very important donaldson polynomials.

19. Apr 10, 2008

### WWGD

Hi, Wonk, I was reviewing some of the entries, and I had a follow up; I am not
sure I understood clearly this issue of extending smooth structures of bounding
submflds N,N'. Please tell me if this is the correct set up :

We have M,M' homeo, but not diffeo. Then N,N' are submflds of both M,M'.

Then we have that we can extend smoothly the structures of N,N' into M,

but we cannot extend the structures of N,N' into M'. Seems like an

h-cobordism thing.

Thanks.

20. Apr 10, 2008

### wofsy

Two atlasses can be incompatible but the two differentiable structures on the manifold diffeomorphic.

Incomtibility of atlasses does not by itself mean that the manifolds are not diffeomorphic.

Last edited: Apr 10, 2008