0

Is a subspace of GA(n) closed under the geometric product? Say we let a k-blade represent a subspace of GA(n), where k < n. Does that also represent a subalgebra of GA(n)?

I can see that we won't get any higher dimensional elements. But I'm not sure that the product of 2 blades in this subspace will always stay in the subspace. Can you give me an example of how subspaces are NOT closed under the geometric product? Or if it is, can you point in the direction of how to prove it?

  • Blades represent subspaces of the base vector space that the geometric algebra is built upon. In this case (if I take your notation correctly), a $k$-blade here would represent a $k$-dimensional subspace in $\mathbb R^n$. – Muphrid Jun 03 '14 at 15:48
  • Right. Let me clarify: If you had a blade A which represented a subspace A', blade B which represented a subspace B', where both A' and B' are subspaces of a subspace C' represented by C. Then would AB represent a subspace of C'? –  Jun 03 '14 at 15:53

1 Answers1

1

In general, blades representing subspaces do not multiply under the geometric product to produce other blades.

A simple counterexample would be two vectors. All vectors are blades, and any two vectors lie in a plane spanned by those vectors: that plane is formed by the wedge product. But the geometric product of those two vectors will (unless the vectors are orthogonal) produce a scalar term also. The result is not a blade.

Muphrid
  • 19,902
  • Yeah, sorry. I should have taken the time to think about it before posting the question. I'm just going through MacDonald's book on GA and trying to prove one of the theorems, I noticed that if this were true it would be really simple. –  Jun 03 '14 at 18:25
  • Can't hurt to ask the real question instead, if you're still having trouble with it. – Muphrid Jun 03 '14 at 19:14
  • I figured it out. Thanks, though. –  Jun 03 '14 at 19:19
  • Actually, I do have a small question (though I don't know if you're going to read this, again): Alan MacDonald keeps saying things like "X is in subspace Y (or X is zero)". But isn't zero in EVERY subspace? And if so, is there a reason why he should consider the multivector zero different from the vector zero -- even though he considers vectors (1-blades) in GA(n) the same as vectors (n-tuples) in R(n)? –  Jun 03 '14 at 20:12
  • Perhaps you can point to a specific passage? I have Linear and Geometric Algebra on hand and Vector and Geometric Calculus close by. – Muphrid Jun 03 '14 at 21:09
  • Nevermind, going back through it, I realize it's only when he's discussing blades that he says this. So clearly Macdonald doesn't consider 0 to be a blade -- I guess because it's hard to tell WHICH blade it actually is. –  Jul 09 '14 at 22:41