Prove a Theorem Three
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Phil's dated working notes (11.12.15, annotated 1.15.16) trying to prove what he calls Theorem Three of Appendix C: antisymmetrizing T^ and S^ gives the same result as antisymmetrizing TS. He tries direct substitution, hits notation problems with permutation composition, checks the k=2, k'=2 case by hand, then uses a rearrangement trick on the permutation sum. He also restates it as Alt[Alt(T)Alt(S)] = Alt(TS).
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Prove a Theorem PhL 11.12.15
1.15.16 I think this is Theorem Three of current Appendix C. So all resolved, this is now obs.
I think the following is true:
(T^S)(v1,v2....vk, vk+1....vk+k')
= (1/(k+k')!) ΣP (-1)S(P) T(vP(1), vP(2) ....vP(k))S(vP(k+1), vP(k+2) ....vP(k+k')) E1
= (1/(k+k')!) ΣP (-1)S(P) T^(vP(1), vP(2) ....vP(k))S^(vP(k+1), vP(k+2) ....vP(k+k')) E2
If you antisymmetrize each T and S object first, then antisymmetrize again, seems the result is the same as if you just antisymmetrize once. So I want to show that the last two lines above are equal, that is the theorem in question.
Theorem is then this, where I add equal factors on both sides: (think as LHS = RHS)
ΣP (-1)S(P) T^(vP(1), vP(2) ....vP(k))S^(vP(k+1), vP(k+2) ....vP(k+k')) LHS
= ΣP (-1)S(P) T(vP(1), vP(2) ....vP(k))S(vP(k+1), vP(k+2) ....vP(k+k')) (*) RHS
The obvious starting point in a proof is to make use of these equations,
T^(v1,v2....vk) = (1/k!) ΣQ (-1)S(Q)T(vQ(1),vQ(2)....vQ(k)) (8.5.10)
S^(vk+1,vk+2....vk+k') = (1/k'!) ΣR (-1)S(R)T(vR(k+1),vR(k+2)....vR(k+k')) (8.5.10)
Note that P acts on {1,2,...(k+k')} whereas Q is on {1,2...k} and R is on {k+1,k+2...k+k'} .
Then try replacing the arguments such as v1 → vP(1) we get
T^(vP(1), vP(2) ....vP(k)) = (1/k!) ΣQ (-1)S(Q)T(vQP(1),vQP(2)....vQP(k)) (**)
S^(vP(k+1), vP(k+2) ....vP(k+k')) = (1/k'!) ΣR (-1)S(R)S(vRP(k+1),vRP(k+2)....vRP(k+k'))
STOP. Look at vQP(1). Suppose P(1) = k+k'. We get then Q(k+k'), but Q cannot act on such a number, so the above does not make sense!! Let's try going back to write
T^(vi,vi....vi) = (1/k!) ΣQ (-1)S(Q)T(vQ(i),vQ(i)....vQ(i)) (8.5.10)
Then can we set vi = vP(i) ? If so, then this is
T^(vP(i),vP(i)....vP(i)) = (1/k!) ΣQ (-1)S(Q)T(vQ(P(i)),vQ(P(i))....vQ(P(i)))
This has the same problem. We could have P(i1) = ik+k' . Then what is Q( ik+k') ?
Now I know that (**) itself is reasonable, it is just my notation that has a problem. It is nothing more than antisym in whatever arguments you specify.
T^(vP(1), vP(2) ....vP(k)) = (1/k!) ΣQ (-1)S(Q)T(vQP(1),vQP(2)....vQP(k)) (**)
= (1/k!) [ T(vP(1), vP(2) ....vP(k)) + signed permutations ]
= (1/k!) [ T(vP(1), vP(2) ....vP(k)) - T(vP(2), vP(1) ....vP(k)) ] + ... ]
= (1/k!) ΣQ (-1)S(Q)T(vP(Q(1)),vP(Q(2)).....vP(Q(k)))
Now I think this makes more sense. So maybe we have
T^(v1,v2....vk) = (1/k!) ΣQ (-1)S(Q)T(vQ(1),vQ(2)....vQ(k))
T^(vP(1),vP(2)....vP(k)) = (1/k!) ΣQ (-1)S(Q)T(vP(Q(1)),vP(Q(2))....vP(Q(k)))
= (1/k!) [ T(vP(1),vP(2)....vP(k)) - T(vP(2),vP(1)....vP(k)) + ... ]
I think it is OK to write
vP(Q(1)) = v(PQ)(1) = vPQ(1)
In this order, there is nothing illegal happening! So let's go with this
T^(vP(1),vP(2)....vP(k)) = (1/k!) ΣQ (-1)S(Q)T(vPQ(1),vPQ(2)....vPQ(k))
S^(vP(k+1), vP(k+2) ....vP(k+k')) = (1/k'!) ΣR (-1)S(R) S(vPR(k+1),vPR(k+2)....vPR (k+k'))
Again everything is "legal" here. So insert these into (*) to get
LHS = ΣP (-1)S(P) T^(vP(1), vP(2) ....vP(k))S^(vP(k+1), vP(k+2) ....vP(k+k'))
= ΣP (-1)S(P) (1/k!) ΣQ (-1)S(Q)T(vPQ(1),vPQ(2)....vPQ(k))
(1/k'!) ΣR (-1)S(R) S(vPR(k+1),vPR(k+2)....vPR (k+k'))
From a group viewpoint, we have in some sense P Q and P R . Any permutation within R makes sense as a permutation within the P group, but not vice versa. We know that PQ ϵ P for example. That is to say, you can extend the meaning of R and Q to the full set {1,2,...(k+k')}.
OK, now try this particular ordering of LHS and must suppress the factorials for now
LHS'1 = ΣP (-1)S(P)ΣQ (-1)S(Q)T(vPQ(1),vPQ(2)....vPQ(k))
{ΣR (-1)S(R)S(vPR(k+1),vPR(k+2)....vPR (k+k'))}
Since P is not an element of the group R, I cannot use the r.t. with R→P'R
{ΣR (-1)S(R)S(vPR(k+1),vPR(k+2)....vPR (k+k'))}
{ΣR (-1)S(P'R)S(vPP'R(k+1),vPP'R(k+2)....vPP'R (k+k'))}
This is wrong because P' is not an element of the group R. So this makes it perhaps less likely that the r.t. is going to solve this little problem!
Let's try a different ordering,
LHS'1 = ΣQ (-1)S(Q)ΣR (-1)S(R)
{ ΣP (-1)S(P)T(vPQ(1),vPQ(2)....vPQ(k))S(vPR(k+1),vPR(k+2)....vPR (k+k'))} }
Now inside {} we hvae R and Q both fixed, and we have in effect R and Q but elements of the P group. So we could do a r.t. at this point. For example, we could take P→PQ-1 = PX to get
LHS'1 = ΣQ (-1)S(Q)ΣR (-1)S(R)
{ ΣP (-1)S(PX)T(vPXQ(1),vPXQ(2)....vPXQ(k))S(vPXR(k+1),vPXR(k+2)....vPXR (k+k'))} }
or
LHS'1 = ΣQ (-1)S(Q)ΣR (-1)S(R)
{ ΣP (-1)S(PQ)T(vP(1),vP(2)....vP(k))S(vPXR(k+1),vPXR(k+2)....vPXR (k+k')) }
= ΣQ ΣR (-1)S(R) { ΣP (-1)S(P)T(vP(1),vP(2)....vP(k)) S(vPXR(k+1),vPXR(k+2)....vPXR (k+k')) }
= ΣP (-1)S(P)T(vP(1),vP(2)....vP(k) ΣR (-1)S(R) X = Q-1
{ΣQ S(vPXR(k+1),vPXR(k+2)....vPXR (k+k')) }
= ΣP (-1)S(P)T(vP(1),vP(2)....vP(k) ΣR (-1)S(R) X = Q-1
{ΣX S(vPXR(k+1),vPXR(k+2)....vPXR (k+k')) }
I see no thread opening in this direction. Cannot use r.t. in {...}.
Plan B. Try a sample case and see if theorem is really true. Maybe sample case will suggest general method to prove the theorem.
T^(v1,v2) = (1/2) ΣQ (-1)S(Q)T(vQ(1),vQ(2)) = (1/2) [ T(v1,v2) - T(v2,v1) ]
S^(v3,v4) = (1/2) ΣR (-1)S(R)S(vR(3),vR(4)) = (1/2) [ S(v3,v4) - S(v4,v3) ]
Then
(T^S)(v1,v2,v3,v4) = (1/4!) ΣP (-1)S(P) T^(vP(1),vP(2)) S^(vP(3),vP(4))
= (1/4!)(1/2)2 ΣP (-1)S(P) [T(vP(1),vP(2)) - T(vP(2),vP(1)) ] [S(vP(3),vP(4)) - T(vP(4),vP(3)) ]
What do I know about T(v1,v2) ? Nothing really. So list off the four terms in the above
T1 = (1/4!)(1/2)2 ΣP (-1)S(P)T(vP(1),vP(2))S(vP(3),vP(4)) = 4! = 24 terms
T2 = - (1/4!)(1/2)2 ΣP (-1)S(P)T(vP(1),vP(2))S(vP(4),vP(3))
T3 = - (1/4!)(1/2)2 ΣP (-1)S(P)T(vP(2),vP(1))S(vP(3),vP(4))
T4 = (1/4!)(1/2)2 ΣP (-1)S(P)T(vP(2),vP(1))S(vP(4),vP(3))
Are these all the same?? Consider the last term
T4 = (1/4!)(1/2)2 ΣP (-1)S(P)T(vP(2),vP(1))S(vP(4),vP(3))
Let P'{1,2,3,4} = {2,1,4,3}. Then rewrite T4 as note that (-1)P' = 1
T4 = (1/4!)(1/2)2 ΣP (-1)S(PP')T(vPP'(2),vPP'(1))S(vPP'(4),vPP'(3))
= (1/4!)(1/2)2 ΣP (-1)S(P)T(vP(1),vP(2))S(vP(3),vP(4))
= T1
So yes, all four terms are the same and we end up with
(T^S)(v1,v2,v3,v4) = 4*T1
= (1/4!) ΣP (-1)S(P)T(vP(1),vP(2))S(vP(3),vP(4))
and in this case my theorem is indeed valid!
Plan B2. Rewrite the above in general notation and try to locate the gimmick that makes it work.
T^(v1,v2) = (1/2) ΣQ (-1)S(Q)T(vQ(1),vQ(2))
S^(v3,v4) = (1/2) ΣR (-1)S(R)S(vR(3),vR(4))
Then
P[T^(v1,v2)] = T^(vP(1),vP(2)) P{1,2,3,4} (*)
P[ (1/2) ΣQ (-1)S(Q)T(vQ(1),vQ(2))]
= (1/2) ΣQ (-1)S(Q) P[T(vQ(1),vQ(2))]
But
P[T(vQ(1),vQ(2))] = T(vPQ(1),vPQ(2)) just like (*)
OK, so this justifies what I want to be true, namely
T^(vP(1),vP(2)) = (1/2) ΣQ (-1)S(Q)T(vPQ(1),vPQ(2))
So here is our same sample case in formal notation,
(T^S)(v1,v2,v3,v4) = (1/4!) ΣP (-1)S(P) T^(vP(1),vP(2)) S^(vP(3),vP(4))
= (1/4!) ΣP (-1)S(P) (1/2) ΣQ (-1)S(Q)T(vPQ(1),vPQ(2))(1/2) ΣR (-1)S(R)S(vPR(3),vPR(4))
= ΣQR { (1/4!)(1/2)2ΣP (-1)S(P) [ (-1)S(Q)(-1)S(R)T(vPQ(1),vPQ(2))S(vPR(3),vPR(4)) ] }
= Σn=14 { Tn }
It is as if
{ (1/4!)(1/2)2ΣP (-1)S(P) [ (-1)S(Q)(-1)S(R)T(vPQ(1),vPQ(2))S(vPR(3),vPR(4)) ] }
= the same for all QR choices.
Decluttering, it is as if
(-1)S(QR) ΣP (-1)S(P) T(vPQ(1),vPQ(2))S(vPR(3),vPR(4)) ]
is the same for all Q,R choices. Let's make sure this is really true for the example:
choice 1: (-1)0 ΣP (-1)S(P) T(vP(1),vP(2))S(vP(3),vP(4)) ] (*)
Q=I R=I
choice 2: (-1)1 ΣP (-1)S(P) T(vP(1),vP(2))S(vP(4),vP(3)) ]
Q=I R=3↔4 = (-1)1 ΣP (-1)S(PR) T(vPR(1),vPR(2))S(vPR(4),vPR(3)) ] r.t.
= (-1)1(-1) ΣP (-1)S(P) T(vP(1),vP(2))S(vP(3),vP(4)) ] = (*)
choice 3: (-1)1 ΣP (-1)S(P) T(vP(2),vP(1))S(vP(3),vP(4)) ]
Q=1↔2 R=I = (-1)1 ΣP (-1)S(PQ) T(vPQ(2),vPQ(1))S(vPQ(3),vPQ(4)) ] r.t
= (-1)1 ΣP (-1)S(PQ) T(vP(1),vP(2))S(vP(3),vP(4)) ] = (*)
choice 4: (-1)S(QR) ΣP (-1)S(P) T(vPQ(1),vPQ(2))S(vPR(3),vPR(4)) ]
Q=1↔2 R=3↔4 (-1)2 ΣP (-1)S(P) T(vPQ(1),vPQ(2))S(vPR(3),vPR(4)) ]
(+1) ΣP (-1)S(PQ) T(vP(1),vP(2))S(vPQR(3),vPQR(4)) ] r.t. #1
(+1) ΣP (-1)S(PQ) T(vP(1),vP(2))S(vP(4),vP(3)) ]
(-1) ΣP (-1)S(P) T(vP(1),vP(2))S(vP(4),vP(3)) ]
(-1) ΣP (-1)S(PR) T(vPR(1),vPR(2))S(vPR(4),vPR(3)) ] r.t #2
(+1) ΣP (-1)S(P) T(vP(1),vP(2))S(vP(3),vP(4)) ] = (*)
Now maybe I have the trick. Try the general case again
= ΣP (-1)S(P) (1/k!) ΣQ (-1)S(Q)T(vPQ(1),vPQ(2)....vPQ(k))
(1/k'!) ΣR (-1)S(R) S(vPR(k+1),vPR(k+2)....vPR (k+k'))
= (1/k!)(1/k!) ΣQ,R *
[(-1)S(QR) ΣP (-1)S(P)T(vPQ(1),vPQ(2)....vPQ(k))S(vPR(k+1),vPR(k+2)....vPR (k+k')) ]
Want to show that [...] is the same for all Q,R choices.
Do a r.t. application like this where X = Q-1 and P → PX. Note that S(X) = S(Q).
[ ] = [(-1)S(QR) ΣP (-1)S(PX)T(vPXQ(1),vPXQ(2)....vPXQ(k))S(vPXR(k+1),vPXR(k+2)....vPXR (k+k')) ]
= [(-1)S(R) ΣP (-1)S(P)T(vP(1),vP(2)....vP(k))S(vPXR(k+1),vPXR(k+2)....vPXR (k+k')) ]
Now claim that vPXR(k+1) = vPR(k+1) because X = Q-1 has no effect on the k+1 index. Note that we are dealing here with extended meanings for Q and R! So we get
= [(-1)S(R) ΣP (-1)S(P)T(vP(1),vP(2)....vP(k))S(vPR(k+1),vPR(k+2)....vPR (k+k')) ]
Now do a second r.t application taking P → PY where Y = R-1. Note that S(Y) = S(R).
= [(-1)S(R) ΣP (-1)S(PY)T(vPY(1),vPY(2)....vPY(k))S(vPYR(k+1),vPYR(k+2)....vPYR (k+k')) ]
= [ ΣP (-1)S(P)T(vPY(1),vPY(2)....vPY(k))S(vP(k+1),vP(k+2)....vP (k+k')) ]
Now claim that vPY(1) = vP(1) because Y has not effect on index 1,2...k So
= [ ΣP (-1)S(P)T(vP(1),vP(2)....vP(k))S(vP(k+1),vP(k+2)....vP (k+k')) ]
Thus we have shown that
[(-1)S(QR) ΣP (-1)S(P)T(vPQ(1),vPQ(2)....vPQ(k))S(vPR(k+1),vPR(k+2)....vPR (k+k')) ]
= [ ΣP (-1)S(P)T(vP(1),vP(2)....vP(k))S(vP(k+1),vP(k+2)....vP (k+k')) ]
which is independent of Q and R. Therefore
E2 = ΣP (-1)S(P) T^(vP(1), vP(2) ....vP(k))S^(vP(k+1), vP(k+2) ....vP(k+k'))
= ΣP (-1)S(P) [ (1/k!) ΣQ (-1)S(Q)T(vPQ(1),vPQ(2)....vPQ(k)) ]
[ (1/k'!) ΣR (-1)S(R)T(vPR(k+1),vPR(k+2)....vPR(k+k')) ]
= (1/k!) (1/k'!) ΣQ,R
{ ΣP (-1)S(P)(-1)S(Q)(-1)S(R)T(vPQ(1),vPQ(2)....vPQ(k))T(vPR(k+1),vPR(k+2)....vPR(k+k')) }
= (1/k!) (1/k'!) {ΣQ,R 1}
{ ΣP (-1)S(P)T(vP(1),vP(2)....vP(k))T(vP(k+1),vP(k+2)....vP(k+k')) }
= ΣP (-1)S(P)T(vP(1),vP(2)....vP(k))T(vP(k+1),vP(k+2)....vP(k+k'))
= E1
Plan B3. How would you express the above stuff using the function Alt and alt functions?
[Alt(X)](v1,v2....vk) ≡ (1/k!) ΣP (-1)S(P)X(vP(1),vP(2)....vP(k))
Consider then.
f(v1,v2....vk+k') ≡ T(v1,v2....vk)S(vk+1,vk+2....vk+k')
Then
[Alt(f)](v1,v2....vk+k') = (1/k!) ΣP (-1)S(P)X(vP(1),vP(2)....vP(k))
= (1/k!) ΣP (-1)S(P)f (vP(1),vP(2)....vP(k))
= (1/k!) ΣP (-1)S(P)T(v1,v2....vk)S(vk+1,vk+2....vk+k')
Meanwhile we have
T^(v1,v2....vk) = (1/k!) ΣQ (-1)S(Q)T(vQ(1),vQ(2)....vQ(k)) (8.5.10)
= [Alt(T)](v1,v2....vk)
S^(vk+1,vk+2....vk+k') = (1/k'!) ΣR (-1)S(R)T(vR(k+1),vR(k+2)....vR(k+k')) (8.5.10)
= [Alt(S)](vk+1,vk+2....vk+k')
To summarize so far,
[Alt(f)](v1,v2....vk+k') = [Alt(TS)](v1,v2....vk+k')
= (1/k!) ΣP (-1)S(P)T(v1,v2....vk)S(vk+1,vk+2....vk+k')
= (T^ ^ S^)(v1,v2....vk+k') (T^ ^ S^) = [Alt(TS)]
T^(v1,v2....vk) = [Alt(T)](v1,v2....vk') T^ = [Alt(T)]
S^(vk+1,vk+2....vk+k') = [Alt(S)](vk+1,vk+2....vk+k') S^ = [Alt(S)]
Now here is what I think my theorem says:
[Alt(f)](v1,v2....vk+k') = (1/k!) ΣP (-1)S(P)T^(v1,v2....vk)S^(vk+1,vk+2....vk+k')
= [Alt(T^S^)](v1,v2....vk+k')
so that my theorem is basically
[Alt(T^S^)] = [Alt(TS)]
which says
Alt[Alt(T)Alt(S)] = Alt(TS)
Is this a Spivak theorem?? Hold and note that
T(v1,v2....vk)S(vk+1,vk+2....vk+k') = [T S](v1,v2..vk+k')
because I showed this in 6.7.1
(TS)(v1,v2....vk, vk+1....vk+k') = T(v1,v2....vk) S(vk+1,vk+2....vk+k') (6.7.1)
This is true for ANY functions in the right size Lk and Lk'. Thus, it is true for both T and T^ functions since these are all in Lk whether or not in Λk. So
(TS)(v1,v2....vk, vk+1....vk+k') = T(v1,v2....vk) S(vk+1,vk+2....vk+k')
(T^S^)(v1,v2....vk, vk+1....vk+k') = T^(v1,v2....vk) S^(vk+1,vk+2....vk+k')
Then my theorem claims
Alt(T^S^) = Alt[Alt(T)Alt(S)] = Alt(TS)
I'll bet this is also true
Alt(T^S) = Alt[Alt(T)S)] = Alt(TS)
Spivak does have this theorem,
I will come back to that in a moment. Here are some of my other results,
T^S = Σi<i<....<i [alt(T S)]ii...i (λi^ λi ......^ λi) .
with (8.10.10)
[alt(T S)]ii...i = ΣP (-1)S(P) TP(i)P(i)...P(i)S P(i)P(i)..P(i)
But these are chapter 8 tensors, so I would write
(T^^S^) = Σi<i<....<i [alt(T^ S^)]ii...i (λi^ λi ......^ λi) .
with (8.10.10)
[alt(T^ S^)]ii...i = ΣP (-1)S(P) T^P(i)P(i)...P(i)S^ P(i)P(i)..P(i)
Or,
(T^^S^) = Σii....i [T^ S^ ]ii....i(λi^ λi ......^ λi)
(T^^S^) = Σi<i<....<i [alt(T^ S^)]ii...i (λi^ λi ......^ λi)
I think my translation of Spivak's theorem is this
Alt(Alt(T^ S^) R^) = Alt(T^ S^ R^)
Do I have a proof for this baby? Here is what I do know
Alt(T^R^) = Alt[Alt(T)Alt(R)] = Alt(TR)
*************************** local scrap heap ************************************
Then inserting this into LHS we get
LHS = ΣP (-1)S(P) (1/k!) ΣQ (-1)S(Q)T(vQP(1),vQP(2)....vQP(k))
* (1/k'!) ΣR (-1)S(R)T(vRP(k+1),vRP(k+2)....vRP(k+k'))
In order to show LHS = RHS, we have to cause the capital subscript letters on each of the v arguments to be the SAME. The tools I have are to use the rearrangement theorem in one or more perm sums.
The puzzle is this: one subscript involves Q, the other R, how to you ever make them the same? For the moment suppress the factorials to declutter things
LHS' = ΣP (-1)S(P)ΣQ (-1)S(Q)T(vQP(1),vQP(2)....vQP(k))
* ΣR (-1)S(R)T(vRP(k+1),vRP(k+2)....vRP(k+k'))
Lots of ways to order the sums. Here is one way
ΣP (-1)S(P)ΣQ (-1)S(Q) T(vQP(1),vQP(2)....vQP(k)) { ΣR (-1)S(R)T(vRP(k+1),vRP(k+2)....vRP(k+k')) }
Inside the R sum, P and Q are constants. One could for example take R→RP-1 :
{ ΣR (-1)S(R)T(vRP(k+1),vRP(k+2)....vRP(k+k')) }
= { ΣR (-1)S(RP)T(vR(k+1),vR(k+2)....vR(k+k')) }
Then LHS' becomes
ΣP ΣQ (-1)S(Q) T(vQP(1),vQP(2)....vQP(k)) {ΣR (-1)S(R)T(vR(k+1),vR(k+2)....vR(k+k'))}
At this point we can slide the R sum to the left to get
ΣR (-1)S(R)T(vR(k+1),vR(k+2)....vR(k+k')) ΣP {ΣQ (-1)S(Q) T(vQP(1),vQP(2)....vQP(k)) }
Now within the Q sum, P and R are constant. So lets take Q→QP-1 so that
{ΣQ (-1)S(Q) T(vQP(1),vQP(2)....vQP(k)) }
= {ΣQ (-1)S(QP) T(vQ(1),vQ(2)....vQ(k)) }
We then have LHS' being this:
ΣR (-1)S(R)T(vR(k+1),vR(k+2)....vR(k+k')) ΣP (-1)S(P){ΣQ (-1)S(Q) T(vQ(1),vQ(2)....vQ(k)) }
But now ΣP (-1)S(P) = 0 and we then have LHS' = 0, oops!