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Consider a $3$-dimensional real inner product space, then each bivector can be written in the form $v\wedge w$ with $v,w\in V$ and is interpreted as an oriented area. The inner product of two bivectors $A=v\wedge w$ and $B=x\wedge y$ is defined by$$A\cdot B=\det\begin{pmatrix}v\cdot x&v\cdot y\\ v\cdot y&w\cdot y\end{pmatrix}$$ On the one hand, the inner product defines an angle $$\theta_1=\arccos\frac{A\cdot B}{\sqrt{A\cdot A}\sqrt{B\cdot B}}$$ between $A$ and $B$ and on the other hand, we can consider the angle $\theta_2$ between the planes spannend by $v,w$ and $x,y$.

Are they equal?

My issue is that I don't know how to express the angle $\theta_2$ in a way that is suitable for solving the problem...

Filippo
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  • Yes, they are equal. To be clear, you must interpret the two planes as oriented, and then interpret the angle between them as acute or obtuse accordingly. I wrote a generalization here recently. – anon Jul 11 '22 at 06:45
  • @runway44 Thank you for your comment. In your answer, you assume that $\Pi_1$ and $\Pi_2$ intersect in a codimension 1 subspace. But in my case, the codimension is 2, isn't it? – Filippo Jul 11 '22 at 06:59
  • @runway44 Because the dimension of the intersection is 1 and the codimension of the intersection is the dimension of the inner product space minus the dimension of the intersection, isn't it? – Filippo Jul 11 '22 at 07:09
  • In 3D, two 2D planes intersect in a 1D line. The 1D line has codimension 1 in the planes, which is what I was talking about (or at least meant), which should be clear from the next sentences about extending bases. – anon Jul 11 '22 at 16:39
  • @runway44 Thank you for the clarification. I like the first paragraph, because I can easily visualize the definition for two intersecting planes and it is clear why the orientations matter. But isn't that angle different from the angle between the normals? – Filippo Jul 11 '22 at 17:02
  • Imagine an oriented plane is rotated ever so slightly around an axis through it. We expect the angle to be small. In 3D, the normal vectors (chosen according to orientation) point in approximately the same direction; the angle between them is likewise small. If you were to reflect one of the normals to its opposite, the angle would be approximately a reflex angle (180 degrees). If you project the whole picture orthogonally to a 2D plane perpendicular to the axis of rotation, it is easy to see the normal lines are just the planes rotated by a right angle, so they are the same. – anon Jul 11 '22 at 18:33
  • @runway44 I was about to send you a simple counterexample, but I realized that I made a mistake - one really needs to careful with the orientations. Thank you for your patience. – Filippo Jul 11 '22 at 19:14
  • @runway44 Okay, I have read your answer. As far as I understand, there are at least three different but equivalent ways to define the angle $\phi$ between the planes: 1) As the angle between the normals, 2) by the procedure described in the first paragraph of your answer (i.e. by considering orthonormal and positively oriented bases of the planes with one common vector which lies in the intersection of the planes) and 3) by $\phi=\arccos\lambda$, where $\lambda$ is the distortion factor defined in your answer. Do you agree? – Filippo Jul 11 '22 at 19:30
  • I agree, with the caveat that when it comes to the generalizations, the normal vectors will only be available when both subspaces are codimension 1 in the overarching space. – anon Jul 12 '22 at 03:45

1 Answers1

1

I think the answer is yes:

Consider the hodge star with respect to one of the two orientations (the final result is independent of our choice). Then $*A=v\times w$ and $*B=x\times y$, so the fact that the hodge star is isometric implies the desired result: $$\theta_1=\arccos\frac{A\cdot B}{\sqrt{A\cdot A}\sqrt{B\cdot B}}=\arccos\frac{*A\cdot *B}{\sqrt{*A\cdot *A}\sqrt{*B\cdot *B}}=\angle(v\times w,x\times y)=\theta_2$$

Filippo
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