The Unapologetic Mathematician

Mathematics for the interested outsider

Compatible Orientations

Any coordinate patch (U,x) in a manifold M is orientable. Indeed, the image x(U)\subseteq\mathbb{R}^n is orientable — we can just use du^1\wedge\dots\wedge du^n to orient \mathbb{R}^n — and given a choice of top form on x(U) we can pull it back along x to give an orientation of U itself.

But what happens when we bring two patches U and V together? They may each have orientations given by top forms \omega_U and \omega_V. We must ask whether they are “compatible” on their overlap. And compatibility means each one picks out the same end of \Lambda*_k(U\cap V) at each point. But this just means that — when restricted to the intersection U\cap V\omega_U=f\omega_V for some everywhere-positive smooth function f.

Another way to look at the same thing is to let \omega_U be the pullback x^*(du^1\wedge\dots\wedge du^n)=dx^1\wedge\dots\wedge dx^n, and \omega_V=dy^1\wedge\dots\wedge dy^n. Then we must ask what this function f is. It must exist even if the orientations are incompatible, since \omega_V is never zero, but what is it?

A little thought gives us our answer: f is the Jacobian determinant of the coordinate transformation from one patch to the other. Indeed, we use the Jacobian to change bases on the cotangent bundle, and transforming between these top forms amounts to taking the determinant of the transformation between the 1-forms du^i and dv^j.

So what does this mean? It tells us that if the Jacobian of the coordinate transformation relating two coordinate patches is everywhere positive, then the coordinates have compatible orientations. On the other hand, if the coordinate transformation’s Jacobian is everywhere negative, then the coordinates also have compatible orientations. Why? because even though the sample orientations differ, we can just use \omega_U and -\omega_V, which do give the same orientation everywhere.

The problem comes up when the Jacobian is sometimes positive and sometimes negative. Now, it can never be zero, but if the intersection has more than one component it may be positive on one and negative on the other. Then if you pick orientations which coincide on one part of the overlap, they must necessarily disagree on the other part, and no coherent orientation can be chosen for the whole manifold.

I won’t go into this example in full detail yet, but this is essentially what happens with the famous Möbius strip: glue two strips of paper together at one end and we can coherently orient their union. But if we give a half-twist to the other ends before gluing them, we cannot coherently orient the result. The Jacobian is positive on the one overlap and negative on the other.

August 29, 2011 - Posted by | Differential Topology, Topology


  1. […] patch . Since each one also comes with its own orientation form, we can ask whether they’re compatible or […]

    Pingback by Orientable Atlases « The Unapologetic Mathematician | August 31, 2011 | Reply

  2. […] we can set up the same definitions and come up with an orientation on each point of . If and are compatibly oriented, then and must be compatible as […]

    Pingback by Oriented Manifolds with Boundary « The Unapologetic Mathematician | September 16, 2011 | Reply

  3. […] patch, we can assume that by restricting each patch to their common intersection. We’ve already determined that the forms differ by a factor of the Jacobian […]

    Pingback by The Hodge Star on Differential Forms « The Unapologetic Mathematician | October 6, 2011 | Reply

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