Proof that wedge product is associative
DOCX · 24.9 KB
Open DOCX file
Working notes by Phil (dated 11.8.15, with a 1.15.16 update) from the Wedge World project. He tries to derive associativity of the wedge product from antisymmetrized tensor-product expansions, testing (v1^v2)^v3 against v1^v2^v3, and hits a wrong-answer result when assuming distributivity. He notes he could not succeed and that associativity became an axiom. The text breaks off mid-argument.
AI-written summary; may contain errors. This description is approximate.
Extracted text (machine-read; may contain errors)
Proof that wedge product is associative PhL 11.8.15
1.15.16 I flailed away trying to derive ^ associativity from other things, but did not succeed. In the end this became an axiom.
Not sure which form is best to use.
vj^ vj^ .....^ vj = (1/k!) ΣP (-1)S(P) ( vP(j) vP(j) ..... vP(j))
v1^ v2^ .....^ vk = (1/k!) ΣP (-1)S(P) ( vP(1) vP(2) ..... vP(k))
Then consider:
v1^ v2^ .....^ vk = (1/k!) ΣP (-1)S(P) ( vP(1) vP(2) ..... vP(k))
vk+1^ vk+2^ .....^ vk+k' = (1/k'!) ΣQ (-1)S(Q) ( vQ(k+1) vQ(2+2) ..... vQ(k+k'))
Then
(v1^ v2^ .....^ vk) ^ (vk+1^ vk+2^ .....^ vk+k' )
= []^[]
= (1/k!) (1/k'!)ΣP,Q (-1)S(P)+S(Q)
( vP(1) vP(2) ..... vP(k)) ^ ( vQ(k+1) vQ(2+2) ..... vQ(k+k'))
This is the wedge product of two tensors. But I don't want to use that work because it already assumes that ^ is associative.
Consider a simpler case:
(ab) ^ (cd)
(ab) ^ (cd)
This is the wedge of two rank-2 tensors. Can I use the and ^ "rules" here? Try an expansion?
Well what does my tensor product of tensors say in this case?
Wedge Product of 2 Tensors
T1^T2 = ΣII T1IT2I (eI^eI) = ΣI (T1T2)IeI
T1^T2 = Σ'I [alt(T1T2)]I eI eI ≡ ei^ ei .....^ ei
where
[alt(T1T2)]I = ΣP (-1)S(P) T1P(I)T2P(I) (6.11.9)
T1 = (ab) T2 = (cd) k1 = 2 k2 = 2
T1P(I) = (ab)ii = aibi but add P
T2P(I) = cidi but add P
[alt(T1T2)]I = ΣP (-1)S(P) T1P(I)T2P(I)
= ΣP (-1)S(P) aP(i)bP(i)cP(i)dP(i)
= aibicidi + signed permutations
(ab) ^ (cd) = Σiiii aibicidi (ei^ ei^ ei^ ei)
Back up and look at a simple example,
v1 ^ v2 ^ v3 =
(v1v2v3 - v1v3v2 + v3v1v2 - v3v2v1 + v2v3v1 - v2v1v3)/6
v1^ v2 = (v1 v2 - v2 v1)/2
Then consider
(v1 ^ v2) ^ v3 = (1/2)( v1 v2 - v2 v1)^ v3
= (1/2) ( A - B)^v3
We have a rule here if A and B are vectors, but I don't know if we have a rule where A and B are tensors! Since ^ is suppose to be linear, it seems possible that we then have
= (1/2) [ A ^ v3 - B ^ v3] = (1/2) [ (v1 v2) ^ v3 - (v2 v1) ^ v3 ]
But where do I go from here?
Suppose it were true that ^ distributes over . Then write as
= (1/2)[ (v1 ^ v3) v2 + v1 (v2 ^ v3) - (v2 ^ v3) v1 - v2 (v1 ^ v3) ]
= (1/2) [ (1/2)(v1v3 - v3v1) v2 + v1(1/2)(v2v3 - v3v2)
- (1/2)(v2v3 - v3v2) v1 - v2 (1/2)(v1v3 - v3v1)
= (1/4) [ v1v3 v2 - v3v1 v2 + v1v2v3 - v1v3v2
- v2v3 v1+ v3v2 v1 - v2 v1v3 + v2 v3v1 ]
= (1/4) [ - v3v1 v2 + v1v2v3
v3v2 v1 - v2 v1v3 ] = the wrong answer!
Back up. Is associative?
(abc....)ijk.... = aibjck....
[(ab)(c...)]ijk.... [AB]ijk.... = Aij Bk... = Aij Bk... = (ab)ij(c...)k...
= ajbjck.....
and yes seems to be associative. I added this to Chapter 2.
Start over. Consider:
a ^ b ^ c versus (a ^ b) ^ c
Just write it all out
v1^ v2^ v3 = (1/k!) ΣP (-1)S(P) ( vP(1) vP(2). vP(3))
= (v1v2v3 - v1v3v2 + v3v1v2 - v3v2v1 + v2v3v1 - v2v1v3)/6
(v1 ^ v2) ^ v3
= (1/2)[v1v2 - v2v1] ^ v3
Then
2 (v1 ^ v2) ^ v3 = (v1v2)^ v3 - (v2v1)^ v3
This is the critical step. What happens next? If I assume the tensor product formulas, then I know
(v1v2) ^ v3 =
= Σa<b<c {TabSc - TacSb + TcaSb - TcbSa + TbcSa - TbaSc} (ea^ eb^ ec) (6.10.16)
= Σa<b<c { (v1)a(v2)b (v3)c + permutations } (ea^ eb^ ec)
= Σa<b<c [alt(v1v2v3)]abc (ea^ eb^ ec)
But this looks the same as v1^v2^v3. How can that possibly be? I know that
(v1^v2) ^ v3 = (1/2)(v1v2 - v2v1)^ v3
= (1/2) [ (v1v2)^ v3 - (v2v1)^ v3 ] // how do I know this step?
Then I would have to have
(v2v1)^ v3 = - (v1v2)^ v3
Are there any "mixed" rules?
I now question my claim of (6.2.3) about the ^ rules. I suspect I have goofed here, ouch!
v1^(v2 + v2')^v3
[v1^(v2 + v2')] ^v3
= [v1^v2 + v1^v2'] ^v3 // using (4.3.4)
***************************
Start over this time with basis vectors only, based on this rule
(ej ^ ej ^ .... ^ ej)ii...i = (1/k!) ΣP (-1)S(P) δP(j)i δP(j)i... δP(j)i
Then an example here would be
(e1 ^ e2 ^ e3)iii = (1/3!)ΣP (-1)S(P δP(1)i δP(2)i... δP(3)i
(e1 ^ e2)ii = (1/2!)ΣP (-1)S(P δP(1)i δP(2)
[ (e1 ^ e2) ^ e3]iii
8. Associativity. Our candidate expansion (6.1.3) states that
v1^ v2^ v3 = (1/k!) ΣP (-1)S(P) ( vP(1) vP(2) vP(3))
= (v1v2v3 - v1v3v2 + v3v1v2 - v3v2v1 + v2v3v1 - v2v1v3)/6
Equation (6.1.3) says nothing about this object
(v1^ v2) ^ v3
which is the wedge product of n