3
$\begingroup$

The two following theorems appear to be contradictory. Both are proved in Milnor's "Topology from the differentiable viewpoint." I'm sure I'm overlooking something incredibly trivial:

  1. Let $f$ be a smooth map from a smooth compact oriented manifold $K$ to a smooth connected oriented manifold $N$, both of dimension $m-1$. If $K$ is the oriented boundary of some smooth compact oriented $m$-manifold (with boundary) $Y$ and if $f$ extends to a smooth map from $Y$ to $N$ then the degree of $f$ (i.e. the sum of the signs of the determinants of the Jacobians at the elements of the preimage of any regular value) is zero.

  2. (Poincaré-Hopf) If $M$ is a smooth compact manifold and $V$ is a smooth vector field on $M$ with isolated singularities then the degree of the vector field is equal to the Euler characteristic of $M$.

Now consider the setting of the second statement and assume $M$ is oriented. delete a small open ball around each zero of $V$. This gives a manifold with boundary $Y$ and by normalizing $V$ we get a map $g$ from $Y$ to the $(m-1)$-sphere. Now if $K$ is the boundary of $M$ then the first statement says the degree of $g$ restricted to $K$ is zero while the second statement says it is the Euler characteristic of $M$.

  • 1
    $K$ is the domain of $g$, there is no extension. The natural way to extend the definition of $g$ to all of $Y$ would be to trivialize the tangent bundle, and if you could do that, certainly the Euler characteristic would be zero.2011-08-29
  • 0
    thanks for your help ryan. maybe this doesn't fix the problem but let's assume M was imbedded in some Euclidean space. Now g makes sense on all of Y regardless of the whether the tangent bundle is trivial right?2011-08-29
  • 0
    $g$ does make sense, but not of something that you can take the degree of, since the domain and range have the wrong dimensions.2011-08-29
  • 0
    but I only speak of the degree of g restricted to K. K has dimension m-1 and the target space is the m-1 sphere. im clearly missing something basic and thank you for your patience.2011-08-29
  • 0
    maybe the cofusion is a result of my bad notation: the M from the first proposition is not the M from the counter example2011-08-29
  • 0
    Could you be precise on what you're using the embedding in Euclidean space for? You're starting to lose me.2011-08-29
  • 0
    yeah I guess I wanted the imbedding so that the vector values of V are honest-to-god vectors in euclidean space. then V can be viewed as a map from Y to punctured Euclidean space. then we normalize to get g. I was trying to remedy the flaw you pointed out with your first comment2011-08-29
  • 0
    It doesn't seem like the remedy is doing much good -- once you make such extensions, there's nothing to take the degree of, unless you go all the way to co-dimension zero submanifolds of Euclidean space.2011-08-29
  • 0
    if I could make edits to my question from my phone I would change all instances of M in the first proposition to K and all instances of X to M and m to m-1. then all the notation would agree.2011-08-29
  • 0
    er sorry I meant change X to Y2011-08-29
  • 0
    Okay, so your edits are done. But still, you don't have a well-defined map from $Y$ to $S^{m-1}$. To get such a map you need a definition, and you haven't supplied one (because there isn't one to define, basically). Your map is only defined $\partial Y \to S^{m-1}$2011-08-29
  • 0
    V is a vector field on M and and im restricting V to Y and then sending a point in Y to the normalized vector value of V at that point.2011-08-29
  • 0
    But that normalized vector is just a vector in a tangent space to the manifold. It's not in $S^{m-1}$. You've completely forgotten that step. And you can't fix that step, since "that step" is finding a trivialization of the tangent bundle. If one existed, the Euler characteristic would be zero.2011-08-29
  • 0
    I see! thank you. I was assuming that M was imbedded in R^m! how silly2011-08-29
  • 0
    $V$ is a manifold, so unless you're using some strange convention I'm unaware of, there's no way to naturally consider a manifold to be a function.2011-08-29
  • 0
    thanks for your time I knew my mistake must be very stupid bit this has been very helpful2011-08-29
  • 0
    No problem. It's a subtle theorem, especially when you try to see why it fails if you "perturb" the hypothesis a little.2011-08-29
  • 0
    so on the bright side have I at least proved that a compact m manifold without boundary does not embed in R^m? (-:2011-08-29
  • 0
    That's a very fundamental theorem. :)2011-08-29

0 Answers 0