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Example of transversal submanifolds?

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  1. Feb 25, 2016 #1
    Hi every body
    I study transversal submanifolds,I have an example but I didn't find some quatities.
    In Rn+p we consider the submanifold N=Rn+1, we know that P is tottaly geodesic and another manifolds M wich is the sphere of dimension n+1 of radius r and centrer p0. then the transversal intersection of M and P depend on the value of r. If the distance between N and p0 is less then r then M and N are transverse and the intersection is a sphere L of dimension n. in this situation I want to define the principale curvature of the inclusion L on M and the vector normal to L in M.
    Another question for a normal vector to M (define below) in Rn+p how we can define the shape operator of the second fondamental form.
    Thanks friends
     
  2. jcsd
  3. Feb 26, 2016 #2

    Ssnow

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    more simple, you can imagine two copies of a sphere that intersect in a circle, this circle is a transversal submanifold ...
     
  4. Feb 26, 2016 #3
    But my Idea is to intersection between a totally geodesic submanifold and another one, I think my example is the simplest one.
     
  5. Feb 26, 2016 #4
    "more simple, you can imagine two copies of a sphere that intersect in a circle, this circle is a transversal submanifold ..."

    I would not say the circle is a "transversal" submanifold. The only manifolds in this picture are the circle, the two spheres, and presumably 3-space where I presume all this is taking place. The circle of intersection between two spheres (if any) is not transversal to any of these four manifolds.

    We get a circle of intersection when the distance between two spheres' centers is greater than 0 but less than the sum of their radii. Assuming this to be the case, then the two spheres are transverse to each other.

    In general: Let Mm, Nn be two compact submanifolds of another manifold Wp. *See below. W is called the "ambient" manifold.

    Then M, N are defined to be transversal xactly when, at every point Q where they intersect, their tangent spaces TQ(M) ≈ Rm and TQ(N) ≈ Rn at the point Q span the tangent space TQ(W) ≈ Rp, or notationally

    TQ(M) + TQ(N) = TQ(W).​

    Of course, this is possible (though of course not guaranteed) only if the dimensions m, n, p satisfy

    m + n ≥ p.​

    So: Transversal submanifolds whose dimensions don't add up to at least the dimension of the ambient manifold cannot have an intersection point!

    The intersection of transversal compact submanifolds is also a compact manifold, and its dimension is p - (m+n) as long as m + n ≥ p, and otherwise the intersection is empty.

    It is also a colossally important fact that for any two submanifolds M, N of W that are as above but not necessarily transverse, there exist submanifolds M', N' that are arbitrarily close (in the C1 metric) to M, N respectively, such that M' and N' are transverse.

    And, once two submanifolds M', N' are transverse, then any sufficiently small C1 perturbation of them preserves transversality.
    When we have

    m + n = p

    exactly, then the intersection is a discrete set of points, and since we have assumed M and N to be compact, this must be a finite number of points. If as often happens M, N and the ambient manifold are oriented, then each point can be assigned +1 or -1, and these signed numbers added up to get an integer that doesn't depend on how the perturbation was chosen. It's called the intersection number of M and N. If you don't have oriented manifolds or don't care about signs, you can still do the arithmetic modulo 2, to get the mod 2 intersection number, which will be either 0 or 1.

    This can be used, e.g., to define the linking number of two oriented closed curves in R3: If C1 and C2 are two simple closed curves in R3, then we can always choose a surface D whose boundary is C1; then their mod 2 linking number

    Lk2(C1, C2)​

    can be defined as the mod 2 intersection number of D and C2. It will be, of course, just 0 or 1.

    If the curves are oriented then with more care one can similarly define the (just plain) linking number

    Lk(C1, C2),​

    which will be an integer and give more information than the mod 2 linking number. This will also be equal to the famous linking integral of Gauss.
    __________________________
    * A superscript here denote the dimensions of the manifold, and will not be repeated in future references.
     
  6. Feb 26, 2016 #5

    lavinia

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    Zinq ,a question about linking number in 3 space.

    In the case of 2 closed loops that link each other can one always deform one of the loops so that is it a great circle in the 3 sphere, say the z-axis plus the point at infinity then project the z-axis and the second loop onto the xy-plane and compute the linking number as the winding number of the projected second loop around the origin?

    This certainly works in some cases.
     
    Last edited: Feb 26, 2016
  7. Feb 26, 2016 #6
    Lavinia, I hadn't known that, but saw something like it only very recently in Wikipedia: Each of the two simple closed curves can be continuously deformed (i.e., through a homotopy) in the complement of the other so that it itself is unknotted. A proof sketch is given there: https://en.wikipedia.org/wiki/Linking_number. But it's also very intuitive: First make any knotting in one curve occur in a very small region, and then pull it tight to make "local" knots disappear.

    At that point, S3 can be continuously deformed diffeomorphically (e.g., through a smooth isotopy) so that one simple closed curve (but usually not both) becomes a great circle. Quickly switching viewpoints back to R3 can be done so it turns that curve (minus a point) into the z-axis.

    At this point none of the actions taken can have changed the linking number, which just as you say can now be computed by projecting the other curve (which now misses the z-axis) orthogonally to R2 and finding its winding number about the origin.

    That's a very nifty way of thinking of linking — thanks for mentioning it!
     
    Last edited: Feb 26, 2016
  8. Feb 26, 2016 #7

    lavinia

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    I will try to one up with a case where it doesn't work.
     
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