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I know about the notion of orientation in an abstract vector space, yet I am stuck on this seemingly trivial issue.

Let $V$ be a subspace of $\mathbb{R}^{d >3}$. Suppose that $\dim V = 3$. If I have two orthonormal vectors $e_{1}, e_{2}$ in $V$, then there are two canonical ways to define an orthonormal basis of $V$: pick $e_{3}$ so that $(e_{1}, e_{2}, e_{3})$ follows – or does not follow – the right hand rule.

Note that I can define $e_{3}$ without any reference to $\mathbb{R}^{d}$. In other words, if I have to repeat the process of completing a pair of vectors into a basis for different three-dimensional subspaces, then I can perform such task consistently: I can choose to follow (or not follow) the right hand rule once and for all.

On the other hand, when $\dim V = 2$ the situation is different. If I have multiple two-dimensional vector subspaces, then I have no means to compare the choices of orientation that I make on each of them.

Could anybody explain why this happens? Is it simply because the cross product in dimension two does not exist?

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    Not completely sure what you're asking, but if it matters, "the right-hand rule" isn't intrinsic to a three-dimensional inner product space, it depends on a choice of orientation. Separately, an orientation on a three-dimensional space does not induce an orientation on two-dimensional subspaces. – Andrew D. Hwang Aug 06 '22 at 19:28
  • @AndrewD.Hwang I agree, I misused the term “intrinsic”. What I am trying to say is that if I have two orthonormal vectors in a 3D subspace and I know which one is the first and which the second, then I can use my right hand to define a third. Clearly there is nothing canonical about my choice of using the right hand. I could use the left one instead. My point is that, if I have to repeat this process for many different subspaces, I can decide to use the same hand for all of them. I cannot do the same in dimension two. –  Aug 06 '22 at 20:12
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    Yes, though speaking of one's "right hand" amounts to choosing an orientation of $V$. (!) Martin Gardner has a good discussion of the issue in The New Ambidextrous Universe. Apparently even Immanuel Kant believed the right hand was distinguished by fitting into a right-handed glove, but later realized all we can say is that a hand and a glove are either consistently or inconsistently oriented, not that either is "right" or "left". IIRC, in $\mathbf{R}^d$ a right hand can be rotated to a left hand. (That said, I may still be misunderstanding the substance of the question.) – Andrew D. Hwang Aug 06 '22 at 20:34
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    @AndrewD.Hwang Thanks for the interesting comment! The problem I have in mind is the following. Suppose you have a curve on a Riemannian hypersurface. Let $T$ be the curve’s tangent and let $N$ be the surface normal along the curve. Suppose you are given a one-dimensional subspace $\mathcal{H}{t}$ in the orthogonal complement of $T(t)$ and $N(t)$ for all $t$. Can we then define, for each $t$, a unit vector $H(t)$ in $\mathcal{H}{t}$ in such a way that $(T, H, N)$ is a right-handed frame for the distribution along $\gamma$ spanned by $T$, $N$, and $H$? –  Aug 06 '22 at 20:56
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    That seems not to be true. For a prospective counterexample, consider the product of a Moebius strip and a line. In coordinates $(t, x, y)$ with $0 \leq t \leq 1$, attach the $(x,y)$ plane at $t = 0$ to the plane at $t = 1$ by $(0, x, y) \sim (1, x, -y)$. The $(t, y)$ plane descends to a Moebius strip. We have $T = \partial_t$, $N =\partial_x$, and $H$ is spanned by $\partial_y$, so the standard frame gets reflected under parallel transport around the circle. Does that satisfy your hypotheses and not satisfy the desired conclusion? – Andrew D. Hwang Aug 06 '22 at 21:34
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    If the ambient space and hypersurface are orientable, however, the answer does appear to work in your favor. – Andrew D. Hwang Aug 06 '22 at 21:37
  • If $V$ is a 3D inner product space and $e_1,e_2$ are orthonormal, there are two choices of $e_3$ that make $e_1,e_2,e_3$ orthonormal. If $V$ doesn't come equipped with an orientation to begin with, there is no sense in which either choice is "left" or "right." – anon Aug 06 '22 at 22:03
  • @AndrewD.Hwang Thanks, that’s helpful! –  Aug 06 '22 at 22:43
  • @runway44 Alright, but why can’t I define an orientation using the “right hand rule”? I mean, there are two choices of $e_{3}$ that make the basis orthonormal, but only one of the two makes it “look like” the standard basis of $\mathbb{R}^{3}$. –  Aug 06 '22 at 22:58
  • That's meaningless. The right-hand rule assumes there already is an orientation (the one described by your right hand). The axioms of a vector space do not equip vector spaces with human hands or human eyes to say what things "look like," so your spatial intuitions do not apply. – anon Aug 06 '22 at 23:08
  • @runway44 Thanks a lot for the help! –  Aug 07 '22 at 08:08
  • @AndrewD.Hwang In view of runway44 comments and yours, I wonder if, under the assumption of the ambient manifold and the hypersurface being oriented, you see any way to induce an orientation on the subspace spanned by $T(t)$, $N(t)$, and $H(t)$. –  Aug 07 '22 at 08:16
  • If $M'$ is a hypersurface in $M$, and if $N$ denotes a continuous unit normal field on $M'$, then an orientation (represented by a local orthonormal frame) induces an orientation on $M$ by appending $N$ at the end. The converse is a little more work, but amounts to rotating a local orthonormal frame for $M$ so that it's last element is $N$, then showing the resulting orientation on $M'$ is independent of choices. (To emphasize, a "frame" here is an ordered set of tangent vectors.) <> If $\dim M > 3$, however, we need more data, amounting to a normal frame for $M'$. – Andrew D. Hwang Aug 07 '22 at 12:47
  • @AndrewD.Hwang Yes, makes sense! I wonder about the higher codimensional case. Is there a way to induce an orientation on a subspace of codimension higher than one, given an orientation of the ambient vector space? –  Aug 07 '22 at 16:46
  • In a word, "no." However, an orientation on any two of the (finite-dimensional, real vector spaces) $V$, $V'$, and $V \oplus V'$ uniquely induces an orientation on the third. (That's the origin of the final sentence about needing a normal frame for the submanifold $M'$.) – Andrew D. Hwang Aug 07 '22 at 17:19
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    @AndrewD.Hwang OK great, thanks a lot for the helpful discussion. –  Aug 07 '22 at 17:25

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Based on clarifications in the comments: If $V$ and $V'$ are complementary finite-dimensional subspaces of a real vector space, then orientations on any two of $V$, $V'$, and the direct sum $V \oplus V'$ determine an orientation on the third.

An orientation on any one space, by contrast, does not determine an orientation on either of the other two.

Perhaps surprisingly, "the right-hand rule" does not "promote" an orientation on a plane $P$ to an orientation of a containing three-dimensional space $V$: The meaning of "right hand" amounts to an orientation on the three-space $V$. Martin Gardner's The New Ambidextrous Universe (pp. 151 ff., especially 153-154 of the 1990 edition), discusses the issue in detail, including Immanuel Kant's (temporary) belief that a (chiral) right hand could only fit one arm of a (bilaterally symmetric) torso.

Andrew D. Hwang
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