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Is the Möbius band transverse orientable in $\Bbb R^4$?

The definition of transverse orientability as per T. Frankel in "Geometry of Physics": Let $V^p$ be a p-dimensional submanifold of a manifold $M^n$, the tangent space at each point x of $V^p$ is of the form $M_x^n=V_x^p\oplus N^{n-p}$, where the vectors in N are transverse to $V^p$. If each transversal $N^{n-p}$ can be oriented continuously as a function of each point in $V^p$, then $V^p$ is said to be transversally oriented.

The Möbius band, Mö embedded in $\Bbb R^3$ is not transverse orientable as one can not choose a section of the frame bundle of the space orthogonal to the surface of Mö with a consistent global orientation like one can trivially if it were embedded in $Mö\times\Bbb R$.

Suppose we embed Mö in $\Bbb R^4$. If I were to visualize myself riding along the the closed path running longitudinally along the center of the band, I would see both transverse vectors of the frame rotate by 180 degrees. The resulting sign of the determinant of the transformation of the orthogonal frame would not change. Could one then integrate an odd two-form over Mö after pulling it back from $\Bbb R^4$ using the inclusion map?

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    $\begingroup$ Could you please include the definition of 'transverse orientable'? $\endgroup$ – Michael Albanese Sep 11 '16 at 19:47
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    $\begingroup$ @MichaelAlbanese I think it means coorientable in which case the answer is clearly negative. $\endgroup$ – Moishe Kohan Sep 11 '16 at 19:51
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Consider the Möbius band as the non orientable line bundle $L$ over the circle $S^1$. For every $n$, the Witney sum of the trivial bundle $E_n$ over $S^1$ with $L$, $E_n+L$ is non trivial, as its determinant bundle is $L$. In the case $n=2$ this implies that the normal bundle of $M$ restricted to the circle is not trivial, as $N+L$ is trivial, because $L$ is the tangent bundle of the Möbius band restricted to the cercle.

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  • $\begingroup$ But wouldn't the tangent bundle of the Möbius band restricted to the circle be two dimensional? In the case where the normal bundle of the Möbius band is 2 dimensional as in when considering it in $\Bbb R^4$, then the determinant of $N+L$ is $L$. I really like the elegance of your answer, but I am still confused mainly because of my unfamiliarity of Witney Sums. $\endgroup$ – dualredlaugh Sep 13 '16 at 1:51
  • $\begingroup$ Yes. It Is 2 dimension al, but non trivial by the very definition of the Mobius band $\endgroup$ – Thomas Sep 14 '16 at 9:03

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