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A Transversality unlocks the secrets of the manifold?

  1. Feb 20, 2016 #1
    Hi everybody
    I seen in a book a quote of H. E. Winkelnkemper "Transversality unlocks the secrets of the manifold.".
    Can someone explain to me this citation and why transversality has all this importance in geometry?
    thank you in advance.
  2. jcsd
  3. Feb 21, 2016 #2
    I can't but I sent an email to professor H. E. Winkelnkemper about your question, maybe he will respond.

    Googling the quote I found it in a book, maybe the same book?


    https://books.google.com/books?id=e...y unlocks the secrets of the manifold&f=false

    Very dense stuff, good luck!
  4. Feb 21, 2016 #3
    Thank you for your help, if you have any response from professor H. E. Winkelnkemper, please inform me.
    Keep in touch.
  5. Feb 21, 2016 #4


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    it seems to be explained just by reading the rest of the page, or maybe the chapter, of the book by Hirsch in which it appears. But if you hear from Elmar, tell him hi for me. Ask if he still remembers the barbecue in Harvard, MA in 1970.
  6. Feb 22, 2016 #5


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    Tranversality is one of the important ideas in differential topology. It arises in many contexts but the two that I know best are when two submanifolds intersect transversally and when a smooth function is traverse to a submanifold of another. In the first case, the intersection is another smooth submanifold. In the second, the inverse image of the submanifold is a submanifold. Transversality thus allows one to study mappings and intersection theory(often called cohomoloy theory) in the category of smooth manifolds.

    Underlying the idea of transversality is the all important Implicit Function Theorem. This theorem guarantees for instance that the inverse image of a submanifold is a submanifold. I would rather say that the Implicit Function Theorem unlocks the secrets of manifolds. In some sense, a great deal of differential topology is ingenious applications of the Implicit Function Theorem.

    I would be happy to illustrate these ideas with some examples.
  7. Feb 22, 2016 #6


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    I will read them and it'll make me happy, too. Could be a nice idea for an insight.
  8. Feb 23, 2016 #7
    thanks for these explanations, I sdee that Transversality is one of the key in differenial topology, this why we have this quote.
  9. Feb 23, 2016 #8
    Here's one simple example. Suppose you have a directed (oriented) continuous closed curve* in the plane that misses the origin, but is allowed to intersect itself.

    In simple cases you can easily count how many times such a curve goes around the origin. But what about in the general case? Such a curve can be arbitrarily complicated, with any number of self-intersections — even infinitely many.

    But by using transversality, you can always assign to such a curve an integer that tells you how many times the curve goes around the origin, and in what sense (clockwise or counterclockwise).

    It works roughly like this: Such a curve C can always be approximated arbitrarily closely by a smooth curve C1; pick such an approximation. Now pick a ray from the origin, say the positive x-axis. Finally, by a second arbitrarily small approximation we can pick another smooth curve C2 that intersects the ray transversely. (In other words, at any point p where the ray and C2 intersect, the tangent line of the ray and the tangent line of C2 span the entire 2-dimensional tangent space of the plane at point p.)

    This can be shown to imply that the curve C2 and the ray intersect at only finitely many points; and to each intersection point we can assign the integer +1 or -1 according as whether the two tangent lines are jointly oriented as the plane is, or with the opposite orientation.

    Finally finally, we add up all these +1's and -1's to get a single integer: that is the number of times the original continuous curve C goes around the origin, sometimes called the winding number of C about the origin. It turns out, miraculously, that this winding number is independent of all the choices and approximations that were made.
    * A closed curve is just a curve that ends at the same point where it begins.
    Last edited: Feb 23, 2016
  10. Feb 23, 2016 #9
    Thakno yu for your example, it is very helpful fr me cause it gives for me an idea how transversality is used.
  11. Feb 24, 2016 #10


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    This post illustrates the simplest case of transversality of a map to a submanifold, the case of a map between manifolds of the same dimension and a zero dimensional submanifold. This idea is a little different than transverse intersection of two submanifolds.

    If ##f:M \rightarrow N## is a smooth map of two manifolds of the same dimension, then a point ##p##
    in ##N## is called a "regular value" of ##f## if for every point,##x##, in ##f^{-1}(p)## the differential ##df_{x}:TM_{x} \rightarrow TN_{p}## is an isomorphism of vector spaces. At a regular value the map is transversal to the zero dimensional manifold, ##p##. (In the general case of a higher dimensional submanifold and two manifolds of possibly different dimensions the differential need not be an isomorphism but its image plus the tangent space of the submanifold must span the tangent space of ##N##. This generalizes the idea of a regular value.)

    - If ##f^{-1}(p)## is empty, ##p## is still called a regular value.

    The Number of Points in the Inverse Image of a Regular Value:

    The Inverse Function Theorem says that for a regular value, ##p##, around any point ##x## in ##f^{-1}(p)## there is an open neighborhood that is mapped diffeomorphically into an open neighborhood of ##p##. This means that each ##x## is isolated from all of the others in ##f^{-1}(p)##. That is: ##f^{-1}(p)## is a discrete set.

    If ##M## is compact this means that ##f^{-1}(p)## is actually finite and one might ask if there is any significance to the number of points. It is not hards to prove - try it as an exercise - that there is an open neighborhood of ##p## for which the number of points in the inverse image is constant. If two such neighborhoods overlap then since they share regular values, the number of points is the same for both. One can imagine piecing overlapping neighborhoods together to get a maximal domain in which the count is constant. The regular values get partitioned into disjoint connected open sets and on each,the number of points in ##f^{-1}(p)## is constant.


    If all of the regular values form a connected set then the number of points in ##f^{-1}## is the same on all of them. For example the regular values of a complex polynomial (viewed as a mapping of the Riemann sphere) form a connected set, in fact the entire sphere minus finitely many points, because the polynomial has only finitely many critical points. Since clearly many of these regular values have non-empty preimages, it follows that every regular value has a non-empty preimage and therefore that the polynomial is surjective. This proves the Fundamental Theorem of Algebra. (This proof can be found in Milnor's Topology from the Differentiable Viewpoint.)

    Brouwer Degree

    If the manifolds ##M## and ##N## are oriented then ##df## is either orientation preserving or reversing at each ##x## in the preimage of a regular value. If instead of counting the number of points in ##f^{-1}(p)## ,one subtracts the number of points where ##df## is orientation reversing, one gets a different number called the Brouwer degree of ##f##. (Again see Milnor's Topology from the Differentiable Viewpoint). This number remarkably is the same for all regular values whether or not they are in the same connected set. Further, the Brouwer degree is a homotopy invariant. That is: two homotopic functions have the same Brouwer degree.

    Examples and Applications.

    - The Brouwer degree of the identity map of an oriented manifold into itself is one since the the identity is orientation preserving and there is only one point in any inverse image.

    - The Brouwer degree of a complex polynomial of degree,n, on the Riemann sphere is n since the polynomial is n to 1 away from critical points and it is orientation preserving. In this case, the number of points and the degree is the same.

    - The antipodal map on an even dimensional sphere is orientation reversing so its Brouwer degree is -1. (For a sphere centered at the origin of Euclidean space, it is the map that sends each point to its negative) This means that the antipodal map on an even dimensional sphere can not be homotopic to the identity map. This provides a classical proof that an even dimensional sphere cannot have a vector field with no zeros. (See Milnor again but the proof is not hard. One shows that a non-zero vector field allows one to construct a homotopy from the identity to the antipodal map. This is a famous proof.)

    - For any surface in 3 space, the Gauss map of the surface into the unit sphere is defined by sliding the unit normal at each point to the origin. For instance, the Gauss map on the unit sphere is the identity map. If the surface is compact without boundary, the the Brouwer degree of the Gauss map is one half the Euler characteristic of the surface. For instance, the Brouwer degree for the sphere is 1 and its Euler characteristic is 2. For a torus shaped like an inner tube lying on the xy-plane, the Brouwer degree is zero. Each regular value has two inverse points one in the outer part of the torus where the Gauss curvature is positive and the Gauss map is orientation preserving and one on the inner part of the torus, the part that is saddle shaped, where the Gauss curvature is negative and the Gauss map is orientation reversing. So the Euler characteristic of the torus is zero. Note that the torus is separated into two regions, the regions of positive and negative Gauss curvature by two parallel circles of critical points, one which is mapped to the north pole, the other to the south pole. The two poles are the only non-regular values. This means that the regular values form a connected set so one could have concluded without keeping track of orientations that the number of points in any inverse image is always 2.

    The Euler characteristic of any manifold is a combinatorial invariant so the Brouwer degree of the Gauss map of any embedding whatsoever of a closed surface in 3 space will always be the same. Warp a sphere in any way you like, the Brouwer degree is still 1 even if it is warped the so that inverse images have more that one point. (This would happen for instance if the sphere was distorted into a shape that had regions of negative Gauss curvature. Why?) Warp the torus to your heart's content. The Brouwer degree will always be zero.

    - If every point of ##f## is a critical point then its Brouwer degree is zero because the inverse image of any regular value is always empty. For instance, the constant map, ##f(x) = p## for all ##x## in ##M## has Brouwer degree zero. This also means that the Brouwer degree is zero for any map that is homotopic to a constant map. So in particular, the identity map of a closed compact manifold into itself is never homotopic to a constant.

    This observation leads to a proof of Brouwer's fixed point theorem. This theorem says that any map of a closed ball into itself must have a fixed point. This famous proof starts by assuming by contradiction that there are no fixed points so ##x## and ##f(x)## are always different. Then define a map,##H##, from the closed ball onto its boundary sphere by sending x to the intersection point of the sphere with the directed line segment beginning at ##f(x)## and passing through ##x##. This map is the identity on the boundary sphere. It also defines a homotopy of the identity map with the constant map and thus, a contradiction. To see this, map ##S##x##I## ,the boundary sphere Cartesian product the unit interval, into the ball by sending ##S##x##t## to the sphere of radius, ##t##. At time one it is the boundary sphere. At time zero, it is the center of the ball. Follow this map by ##H##. The composed map is the identity at time 1 and maps the sphere to a single point at time zero.

    - Back to complex polynomials. Let ##f## be a complex polynomial of degree, ##n##. On a very large circle,##C##, centered at the origin, ##f## is dominated by the term of degree ##n## and so is close to a rotation ##n## times around zero and an expansion of by the power,##n##. The image of the circle is a curve that winds around the origin n times so projecting it back onto the circle along radial lines through the origin, produces a map of the circle into itself of degree, ##n##. (Note that if the radius of the circle is chosen to be sufficiently large, its image under ##f## will not intersect zero so the projection is well defined.) Now suppose that ##f## has no roots. This means that the radial projection of any point ##f(x)## onto the circle is well defined. In particular, the image under ##f## of the disk bounded by ##C## can be projected onto the circle. But this, just as in the proof of Brouwer's Fixed Point Theorem, is a homotopy between a map of non-zero degree, in this case degree ##n## ,and the constant map which has degree ##0##. This is impossible so ##f## must have a root and the Fundamental Theorem of Algebra is proved again.

    Last edited: Feb 25, 2016
  12. Feb 24, 2016 #11


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    First I want to say thank you for the comprehensive and detailed work you have done. I really appreciate it and as promised I not only read but work through it. I have tried to figure it out on a quite common example. But I have some difficulty with a special and crucial point (your exercise) that makes me think I fundamentally misunderstood something.

    I considered the projection ##π: O(3,ℝ) → SO(3,ℝ)## defined by ##π(A) = (\det A)^{-1} A##. Then
    $$ p=\begin{pmatrix}-1 & 0 & 0\\0 & 0 & -1\\0 & -1 & 0\end{pmatrix}$$ is regular, since ##p## and ##-p## are the only preimages and multiplication by ##-1## doesn't affect the tangent spaces at these points. Both, ##O(3,ℝ)## and ##SO(3,ℝ)## are compact.

    If I now consider open neighborhoods
    $$U_ε(p) = \left\{ \begin{pmatrix}-1 & 0 & 0\\0 & -ε & ε-1\\0 & -ε-1 & ε\end{pmatrix} \right\} $$
    then there are still two preimages for all points in it but infinitely many for ##U_ε(p)## as a whole.
    I understand that there might be another neighborhood fulfilling the existence quantor but I cannot see how.
    What am I missing here?
  13. Feb 24, 2016 #12


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    No you have the right idea. All that is said is that the preimage of any point is finite. The preimage of the entire neighborhood will be infinite - in fact an open set. Did I understand your question?
  14. Feb 24, 2016 #13


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    Yes. Thanks for the fast answer. Still stuff to read ...
  15. Feb 24, 2016 #14


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    BTW: These examples are fundamental in differential topology and the proofs are known to all mathematicians who use it and probably most who do not.
    I once sat in on a lecture where the guest speaker contradicted Brouwer's Fixed Point Theorem. Everyone in the audience gasped.

    These theorems can be proved with non-differentable techniques using algebraic topology. The interplay between the two is interesting in itself and goes way beyond these theorems.

    Milnor's book is really good and basic. I can't recommend it enough.
    Last edited: Feb 24, 2016
  16. Feb 24, 2016 #15


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    That's what I just thought: Well, less than 60 pages, I'll have to read it.
    I must admit that my education has been rather algebra (and informatic) sided plus it's been quite a while now. I was always happy to turn to the tangent spaces. Like: SU(2), SO(3) - who cares all SL(2). Not to mention how hard it is for me to follow all these index fetishes here. But I want to understand topology better and the interplay you mentioned. That's another reason I liked your post. It get's me hooked.
  17. Feb 25, 2016 #16


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    You mean the Einstein notation for tensors?
    I know the feeling.
  18. Mar 16, 2016 #17


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    I am curious, what happened next?
  19. Mar 16, 2016 #18


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    A few people yelled out. The speaker stopped, went silent, put down his chalk. The lecture was over.
  20. Mar 16, 2016 #19
    fresh_42, I just noticed your post #11. Where you write: "If I now consider open neighborhoods", the matrices you wrote in terms of ε do not comprise an open neighborhood in SO(3) (or in O(3)). This can be seen right away because an open neighborhood in either of those groups would have to be 3-dimensional, but your family of matrices is clearly only 1-dimensional.
    Last edited: Mar 16, 2016
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