Are the wedge rules derivable
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A dated working note (PhL, 11.8.15, with a resolution added 1.15.16) from the Wedge World tensor wedge documentation. It concludes the wedge rules are not derivable from the tensor product rules but are imposed by fiat, with the antisymmetrized (1/k!) sum over permutations as a candidate definition that satisfies them. It also records that associativity of the wedge remains an open issue.
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Are the wedge rules derivable? PhL 11.8.15
This issue has been resolved. The wedge product rules are not derivable from the tensor product rules. They are by fiat, and then our candidate wedge product of vectors form satisfies these by-fiat rules and is therefore a viable candidate. 1.15.16
I claimed they are, but now I am not so sure. Here are the two sets of rules:
v1(v2 + v2')v3.....vk = v1v2v3 .....vk + v1v2'v3 .....vk
v1(sv2)v3 ..... vk = s(v1v2v3 .....vk) s = scalar (5.3.1)
v1^(v2 + v2')^v3^.....^vk = v1^v2^v3^ .....^vk + v1^v2'^v3^ .....^vk
v1^(sv2)^v3^ .....^ vk = s(v1^v2^v3^ .....^vk) s,s1,s2 = scalar (6.2.3)
Prior to my fancy (6.1.2) and (6.1.3) forms, all I know about the action of ^ is this:
a ^ b ≡ (ab - ba)/2 . dim(V) = n (4.3.1)
This instruction DOES NOT SAY how to take the wedge product of more than two vectors, so I have no way to "check" the wedge rules stated above!
Fact: You cannot "verify" the rules (5.3.1) if you have no method of wedge three vectors!
OK, I will go correct my claim.
I have rewritten item 2 starting at (6.2.3) and state clearly that these rules cannot be derived from the rules, I had that dead wrong. I made the change without altering equation numbers. But on 1.15.16 this statement is below (7.2.3) so I guess things were renumbered.
Status: I have dealt with the "rules" problem and am done with that.
The associativity of ^ is still a live issue, however.
Question on 1.15.16. Suppose you accept the candidate definition of a wedge product of k vectors given in (7.1.3). Why is it that you can't "derive" the wedge rules assuming that definition? I will try to do that right here.
v1^ v2^ .....^ vk = (1/k!) ΣP (-1)S(P) ( vP(1) vP(2) ..... vP(k)) // candidate
v1^(v2 + v'2)^v3^.....^vk = v1^v2^v3^ .....^vk + v1^v'2^v3^ .....^vk // rule want to show
Well, I think you CAN derive the wedge rules if you assume the candidate form. I think my point is that you assume the wedge rules by fiat, and THEN the candidate is viable and meets those rules. Yes, I say just that near (7.2.4).