Is there a 4D compact smooth manifold with specific properties?

[lavinia]

I am looking for an example of a 4 dimensional compact smooth manifold that has the following properties

- it is orientable

- it can be smoothly embedded in R^8

- its Euler characteristic is odd

- its second Stiefel-Whitney class is zero

 

[simeonsen_bg]

Cp^2

 

[quasar987]

afaik, the second stiefel-whitney class of CP^2 is nonvanishing (corollary 11.15 of Milnor-Stasheff).

 

[simeonsen_bg]

[Sorry, I confused in fact 'spin' with 'symplectic'.]

from MathWorld:

'A spin structure exists if and only if the second Stiefel-Whitney class
w2 of the tangent bundle of the manifold vanishes.'

So, it is enough to find a spin 4-manifold with b_2 odd to complete the
example. I am not an expert, but seems that there are lots of such
spaces.

 

[lavinia]

I am beginning to think that there is no example.

I am not sure if this argument is right but here goes.

Embed the manifold in R^8. The Stiefel-Whitney classes of the normal bundle must cancel the Stiefel-Whitney classes of the tangent bundle.

Since the tangent bundle is orientable and has zero second Stiefel-Whitney class the 4'th Stiefel Whitney class of the normal bundle must cancel the 4'th Stiefel-Whitney class of the tangent bundle and so can not be zero because the Euler characteristic of the manifold is odd.

But the Thom class of the normal bundle is zero because R^8 has no cohomology (with compact supports) except in dimension 8.

 

[quasar987]

Could you explain the last part of your argument please? (But the Thom class of the normal bundle is zero because R^8 has no cohomology (with compact supports) except in dimension 8.)

And also, why does w_1(\tau_M)w_3(\tau_M^{\perp})+w_3(\tau_M)w_1(\tau_M^{\perp})=0??

 

[lavinia]

Could you explain the last part of your argument please? (But the Thom class of the normal bundle is zero because R^8 has no cohomology (with compact supports) except in dimension 8.)

And also, why does w_1(\tau_M)w_3(\tau_M^{\perp})+w_3(\tau_M)w_1(\tau_M^{\perp})=0??

the first Whitney class of the tangent bundle is zero because the manifold is orientable. Since the sum of the normal and tangent bundles is trivial - because the tangent bundle of euclidean space is trivial - the normal bundle must also be orientable. So each term in the sum of the products of the odd whitney classes in the fourth mod 2 cohomology group is zero.

Poincare duality says that the Thom class of the normal bundle is dual to the homology class of the embedded manifold.But the embedded manifold is homologous to zero. Thus the Euler class of the normal bundle is zero and reducing mod 2, the 4'th Whitney class of the normal bundle is also zero.

If one compactifies R^8 into an 8 dimensional sphere, then one sees that the Thom class is null homologous since the 8 sphere has zero cohomology except in dimension 8.

 

[simeonsen_bg]

You may be right. Let me notice that by the Whitney embedding theorem every smooth n-manifold embeds smoothly in 2n-Euclidean space. So your second condition is superfluous.
Now, as your manifold is orientable with w2=0, it is a spin manifold. And that’s what i found in a book by Stephen Hawkins ‘himself’, a proof that every closed spin 4-manifold has even Euler characteristic - see attachment.

I am beginning to think that there is no example.

I am not sure if this argument is right but here goes.

Embed the manifold in R^8. The Stiefel-Whitney classes of the normal bundle must cancel the Stiefel-Whitney classes of the tangent bundle.

Since the tangent bundle is orientable and has zero second Stiefel-Whitney class the 4'th Stiefel Whitney class of the normal bundle must cancel the 4'th Stiefel-Whitney class of the tangent bundle and so can not be zero because the Euler characteristic of the manifold is odd.

But the Thom class of the normal bundle is zero because R^8 has no cohomology (with compact supports) except in dimension 8.

 

[lavinia]

You may be right. Let me notice that by the Whitney embedding theorem every smooth n-manifold embeds smoothly in 2n-Euclidean space. So your second condition is superfluous.
Now, as your manifold is orientable with w2=0, it is a spin manifold. And that’s what i found in a book by Stephen Hawkins ‘himself’, a proof that every closed spin 4-manifold has even Euler characteristic - see attachment.

Thanks. I think you can get this result without the Atiyah-Singer Index theorem. The arguments I gave show that the manifold can not be embedded in R^8. But the Whitney Embedding Theorem says that any smooth closed 4 manifold can be embedded in R^8. Therefore the Euler characteristic must be even.