# Obstruction Theory and Embeddings

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2019 Award

## Main Question or Discussion Point

Does the use of Chern classes and other characteristic classes extend to the existence of embeddings? For example, it is known that $S^n$ , the n-sphere does not embed in $\mathbb R^n$ or $\mathbb R^m$ for $m<n$ (I think this is a corollary of Borsuk Ulam). Can this be determined by Chern classes (or some other general theory, e.g., (co)homology, homotopy theory, etc.)? I think there are some results using the normal bundle of the embedded object, which must be trivial, so that the bundle sum of the embedded figure and its normal complement must be trivial in the ambient space. Are there anything other results?

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lavinia
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My knowledge is that obstructions to embeddings and more strongly to immersions depend upon Stiefel- Whitney classes .
However, a manifold may be immersible without being embeddable. The simplest examples are the Klein bottle and the Projective Plane both of which can be immersed in
$R^3$ but not embedded.

As with an embedding, an immersion has a well defined normal bundle. If the immersion is into Euclidean space then the sum of its normal bundle and the tangent bundle
is trivial so by the Whitney product formula their total Stiefel-Whitney classes are inverse to each other. This is the key fact.

Here is a classic example: (See Milnor, Characteristic Classes)

The total Stiefel-Whitney class of the real projective space, $P^n$ is $( 1+x)^{n+1}$ where $x$ is the generator of the first $Z_{2}$ cohomology.
Suppose n is a power of 2. Then the total Whitney class is $1 + x + x^n$ and the total Whitney class of its inverse is $1 + x + ... + x^{n-1}$.

So if $P^n$ is immersible in $R^ {n+m}$, $m$ must be greater than or equal to $n-1$

Another Example:

Suppose a manifold is immersible in a Euclidean space and has a trivial normal bundle.
Then all of its Stiefel-Whitney classes are zero. In particular it must have even Euler characteristic.

All orientable surfaces are all embeddable in 3 space which means that their Euler characteristics are all even because their normal bundle is a trivial line bundle.

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Gold Member
2019 Award
Thanks, lavinia, one of these days I will follow up on my pledge to seriously learn characteristic classes; they seem to be helpful in a lot of areas.

lavinia
Gold Member
Thanks, lavinia, one of these days I will follow up on my pledge to seriously learn characteristic classes; they seem to be helpful in a lot of areas.
I find them fascinating. In some sense, they all come from a theorem of Gauss and Bonnet proved in the 19'h century. It says that the sum of the intrior angles of a geodesic triangle minus π on a surface is equal to the integral of the Gauss curvature over the interior of the triangle. This theorem can be used to show that the integral of the Gauss curvature over the whole surface is its Euler characteristic.

In modern terms the Gauss Bonnet theorem says that the Gauss curvature times the volume element is the Euler class of the tangent bundle. Today we know that any oriented sphere bundle has an Euler class. For Riemannian manifolds there is an n- form built from the curvature forms that integrates to the Euler characteristic of the manifold. This again is the Euler class of the tangent bundle. So the Gauss Bonnet theorem generalizes to higher dimensions.

More generally one has such a differential form in the curvature for any oriented Riemannian vector bundle.

All Chern classes and Pontryagin classes Euler classes of some bundle.

There are also "secondary characteristic classes" built from the Chern and Pontryagin forms and the Euler form that are not integer cohomology classes but take values in R/Z. These classes were discovered by Chern and Simons and some are obstructions to conformal immersions of Riemannian manifolds in Euclidean space.

Finally, the top Stiefel-Whitney class may be thought of as a kind of Euler class. It is the orientation class for a vector bundle using $Z_{2}$ coefficients. It always exists because any vector bundle can be oriented over $Z_{2}$.

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