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How to determine whether the preimage of a point is a imbedding submanifold?

  1. Apr 7, 2012 #1
    How to determine whether the preimage of a point is a imbedding submanifold??

    Dear Folks:
    It is well known that the preimage of a regular point is a imbedding submanifold, but is it possible that the preimage of a critical point is also a imbedding submanifold?? More generally, is there a prosedure to determine when the preimage of a point is a imbedding ??
    Many thanks!!
     
  2. jcsd
  3. Apr 8, 2012 #2

    quasar987

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    Re: How to determine whether the preimage of a point is a imbedding submanifold??

    It is possible that the preimage of a critical point is an embedded submanifold. For instance, consider the map f:R²-->R : (x,y) --> y² and let M:=graph(f)={(x,y,z) in R³ | z=f(x,y)=y²} This is the manifold that you get by taking a parabola parabolla in the yz plane and "sliding" it along the x axis so as to have one such parabola standing on each point (x,0,0). Now let h:M-->R be the height function h(x,y,z)=z. Then h-1(0)=R x {0} x {0} is an embedded submanifold, but each point of h-1(0) is critical because the derivative of h vanishes there.
     
  4. Apr 9, 2012 #3

    mathwonk

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    Re: How to determine whether the preimage of a point is a imbedding submanifold??

    This example only works with a special definition of the inverse image of a point, and to some extent an artificial one. In algebraic geometry, and even in analytic geometry, when one studies sets more complicated than manifolds, i.e. when one studies both manifolds and non manifolds, one does not assume the structure of an inverse image is defined only by the set of points but by the function defining the map. Thus in this example the correct structure on the inverse image is defined by setting z= y^2 = 0. this defines an analytic variety which is not the same as the manifold defined by y=0, but the "double" of that variety. The situation is clarified by looking at y^2 = t and letting t-->0. These are all manifolds for t≠0, but as t-->0 they come together to form a variety which is actually not a manifold anywhere, but everywhere singular. It is only when one takes the "reduced" structure on this variety that one obtains a manifold. So the answer is that yes the inverse image set of a critical point can look like a manifold, but only if one ignores the finer structure contained in the mapping.
     
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