A real vector can be thought of as an oriented line segment. Linear algebra and multivariable calculus can be taken pretty far just by considering these types of objects (obviously there are generalizations to other vector spaces, but let's not worry about those).

Using these oriented line segments, we can do several important things. We can

  1. describe oriented subspaces of $\Bbb R^n$
  2. build a calculus on them such that derivatives of paths tell you not only the tangent space but also "which way a particle would move on this path"
  3. in integration, it allows us to find oriented areas/ volumes/ etc
  4. Lots and lots more examples exist

To me, though, the idea of orientation is just something extra and possibly unnecessary. $\Bbb R^n$ isn't naturally endowed with an orientation -- we have to add one. So could we devise a similar set of objects without the orientation? So you'd just have a line segment, the derivative of a path would just give you the tangent space at a point, and integrals would just give you areas/ volumes/ etc.

As I understand it, Descartes described geometry with just line segments. And later Hermann Grassmann's father thought that a natural product of two line segments is just an area segment (like a bivector but without the orientation).

This question partly comes from me thinking it would be nice if we could take integrals and not have to worry about the part of of the function above or beneath the $x$-axis -- we'd just get an area. And it'd be nice if we didn't have to add an orientation onto a space to use differential area/ volume/ etc elements. And it'd be nice if we could use elementary calculus methods to integrate on non-orientable spaces.

So my questions are: Could we develop a consistent system with these properties? Would it be at all useful (maybe it'd be more complicated and completely subsumed by vector math)? Has it already been done and I just haven't heard about it (if so, could you provide a source)?

  • $\begingroup$ There are a few things you are confusing in your motivation. You are confusing the orientation of a vector with an orientation of $\mathbb{R}^n$. The set $\mathbb{R}^n$ does come with an special point $0=(0,0,0,...,0)$ and once we define translations (addition for its elements) that immediately gives each point different from $0$ its "orientation", which I think a more appropriate name is direction. A different thing is an orientation of $\mathbb{R}^n$, which is a choice of an equivalence class of bases that transform with positive determinant among them. $\endgroup$ – Alamos May 2 '15 at 19:34
  • $\begingroup$ Instead of $\mathbb{R}^n$ you may want to consider affine space, in which no point is special like $(0,0,0,...,0)$. $\endgroup$ – Alamos May 2 '15 at 19:35
  • $\begingroup$ No, orientation doesn't come from the choice of directions of vectors, it is choice of the vectors themselves. Non-zero vectors inevitably define a direction. Now, your last question has an answer. what I am saying is that your motivation is based in a misunderstanding. $\endgroup$ – Alamos May 2 '15 at 21:40
  • $\begingroup$ I realize that the orientation of a space is chosen independently of the choice of directions of vectors -- that's why I said "$\Bbb R^n$ isn't naturally endowed with an orientation". But the orientation of any $k$-blade is defined the same way. In particular, the orientation of space is given -- in Clifford calculus -- as the orientation of positive unit pseudoscalar of the space. Obviously which unit pseudoscalar is the positive one is a choice made by the mathematician. But a choice must be made to build a volume form which is necessary to integrate volumes. $\endgroup$ – user236266 May 2 '15 at 21:48
  • $\begingroup$ But whether part of my motivation is based on a misunderstanding or not doesn't answer the main question. $\endgroup$ – user236266 May 2 '15 at 21:54

It sounds like you're looking for the real projective space. This can be imagined as taking a sphere and identifying ("gluing") antipodal points together. In this sense, a vector and its negative (opposite orientation) are one and the same.

That said, I'm less clear on whether this enables the kind of "orientationless" calculus that you're looking for. For instance, the real projective space is an orientable manifold in its own right.

Perhaps instead of a tangent space associated with a point at a manifold, one could consider the projective space instead and build up a projective bundle.

  • $\begingroup$ I'll look into that. Thanks. $\endgroup$ – user236266 May 2 '15 at 21:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.