Bad proof of tiny theorem
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A brief working note by Phil dated 11.12.15, from the Wedge World tensor wedge material. It defines notation for permuting the arguments of a function (Qf) and states a 'tiny theorem' about antisymmetrized sums over permutations with sign (-1)^S(Q). The note then gives a proof it marks as bad and asks why it fails, suspecting that the permutation P acts on argument positions in an order-dependent way, so the index composition lands on the wrong side.
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Bad proof of tiny theorem PhL 11.12.15
C.1 Notation to describe the permutation of function arguments.
Let f(v1,v2,...vk) be a function of k arguments and let Q be a permutation of {1,2..k}.
Then define the permuted function Qf as,
Q[f(v1,v2,...vk)] = f(vQ(1),vQ(2),...vQ(k)) . (C.1.1)
Example: If Q{1,2,3} = {2,1,3} then we think of Q(1) = 2, Q(2) = 1 and Q(3) = 3 so that
Q[f(v1,v2,v3)] = f(vQ(1),vQ(2),vQ(3)) = f(v2,v1,v3) .
Suppose P is some other permutation of {1,2..k}. Then
P{Q[f(v1,v2,...vk)]} = P[f(vQ(1),vQ(2),...vQ(k))] = f(vQ(P(1)),vQ(P(2)),...vQ(P(3)))
Example: If P{1,2,3} = {2,1,3} then
P{Q[f(v1,v2,v3)]} = P[f(vQ(1),vQ(2),vQ(3))] = f(vQ(P(1)),vQ(P(2)),...vQ(P(3)))
= f(vQ(2),vQ(1),...vQ(3)))
Notice therefore that P[f(vQ(1),vQ(2),vQ(3))] = f(vQ(2),vQ(1),...vQ(3))) merely swaps the first two arguments of the function f. In abbreviated notation,
PQ[f(v1,v2,...vk)]} = P[f(vQ(1),vQ(2),...vQ(k))] = f(vQP(1)),vQP(2)),...vQP(3))) (C.1.2)
Tiny Theorem. If
f(v1,v2,...vk) = ΣQ(-1)S(Q) F(vQ(1),vQ(2),...vQ(k)) Q = permutations of {1,2..k}
then if P is some particular permutation,
f (vP(1),vP(2),...vP(k)) = ΣQ(-1)S(Q) F( vP(Q(1)),vP(Q(2)),...vP(Q(k))) . (C.1.3)
In those subscripts, the summation index goes on the right. The right side of (C.1.3) is then
f (vP(1),vP(2),...vP(k)) – f (vP(2),vP(1),...vP(k)) + etc.
THIS IS THE WAY IT SHOULD BE. so why does the following proof fail?
I think below when you apply P to the sum, it wants to do 1↔2, but it does not know where these indices are because that is a function of each Q in the sum. So throw this proof out please!
Proof :
f(v1,v2,...vk) = ΣQ(-1)S(Q) F(vQ(1),vQ(2),...vQ(k))
P[f(v1,v2,...vk)] = f(vP(1),vP(2),...vP(k)) // (C.1.1)
P [ΣQ(-1)S(Q) f( vQ(1),vQ(2),...vQ(k)) ] = ΣQ(-1)S(Q) P[f( vQ(1),vQ(2),...vQ(k))]
= ΣQ(-1)S(Q) f(vQP(1)),vQP(2)),...vQP(3))) //(C.1.2) bad line
Therefore f(vP(1),vP(2),...vP(k)) = ΣQ(-1)S(Q) f(vQP(1)),vQP(2)),...vQP(3))).
But this is wrong!!!!!!! The summation index needs to go on the right. So why is my proof no good?