Consideration of vector bases easily establishes that in projective space of dimension n-1, subspaces of dimensions k-1 and m-1 must meet in a common subspace of dimension at least k+m-n-1, when this is non-negative. Here n = 5, k = m = 3, so there must be a point in common somewhere. But it depends a little on what you mean by "parallel": distinct planes in 4-space might meet in a line at infinity, only a point at infinity, or in a finite line or finite point. In the second case, they're only "semi-parallel" --- but that would hardly qualify as "skew" to your student, I don't suppose. WFL On 3/22/11, Dan Asimov <dasimov@earthlink.net> wrote:
Fred's query reminds me of the time in 1994 that I gave a lecture -- about 4-dimensional space -- to the Hampshire College summer math program for mathematically talented high school students.
Some brilliant youngster asked me whether there could be skew (2D) planes in 4-space. (Skew meaning neither parallel nor intersecting.) I thought about it for a few seconds . . . and then a few more seconds . . . and had to admit I didn't know.
But I went home and soon figured it out. Can you?
--Dan
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