proof of a lemma regarding Hodge star v 2 (CH2)
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Short note dated 7.15.15 by Phil proving the lemma *(*dxI) = (-1)^(kn+k) dxI, so that **α = (-1)^(kn+k) α for any k-form. It defines signs S and S' by calibrating against the volume form dV, proves a sublemma on commuting the complementary product past dxI by moving each factor through n-k terms, and shows SS' = (-1)^(kn+k). A closing comment derives an explicit expression for S'.
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Proof of a Lemma PhL 7.15.15
Show that for a k-form: *(*dxI) = (-1)kn+k dxI Sjamaar p 28 Ex 2.12 (CH2) (*)
[ If α = any k-form, this shows that **α = (-1)kn+k α .]
1. Definitions:
dxI ≡ dxa dxb .... dxq a < b < ... <q a k-form (1.1)
Πh dxi ≡ dx1 ....dxa ....dxb........dxq ....dxn red = holes (1.2)
*dxI ≡ S Πh dxi (1.3)
2. Calibration for S:
dxI(*dxI) = dV or dxI S Πh dxi = dV
dxI(Πh dxi) = S dV (2.1)
3. Compute *(Πh dxi):
(Πh dxi) = dx1 ....dxa ....dxb........dxq ....dxn
*(Πh dxi) = S' dxa dxb .... dxq = S'dxI (3.1)
4. Calibration for S' :
(Πh dxi)*(Πh dxi) = dV
(Πh dxi)S'dxI = dV
(Πh dxi)dxI = S' dV (4.1)
5. Sublemma: (this will be proved in a later item below)
(Πh dxi)dxI = (-1)kn+k dxI(Πh dxi) . (5.1)
6. Therefore using (4.1) on the LHS of sublemma, and (2.1) for RHS of sublemma, we have
S' dV = (-1)kn+kS dV
or
SS' = (-1)kn+k (6.1)
7. Now consider
*(*dxI) = *( S Πh dxi) // using (1.3) for *dxI
= S *(Πh dxi) // factor out constant
= SS'dxI // using (3.1) for *(Πh dxi)
= (-1)kn+k dx // using (6.1) (7.1)
This concludes the proof since we have shown (*).
8. Proof of sublemma: (Πh dxi)dxI = (-1)kn+k dxI(Πh dxi)
Start with the LHS
(Πh dxi)dxI = dx1 ....dxa ....dxb........dxq ....dxn dxa dxb .... dxq
Now let's start by moving dxa all the way to the left.. It has then to pass through n-k dxi factors, so we pick up (-1)n-k. We then have
(Πh dxi)dxI = (-1)n-kdxa dx1 ....dxa ....dxb........dxq ....dxn dxb .... dxq
Now move dxb in this same manner and pick up another factor (-1)n-k , so
(Πh dxi)dxI = (-1)2n-2k dxa dxb dx1 ....dxa ....dxb........dxq ....dxn .... dxq
Do this for all k factors up through dxq and end up with
(Πh dxi)dxI = (-1)kn-kk dxa dxb .... dxq dx1 ....dxa ....dxb........dxq ....dxn
= (-1)kn-kk dxI (Πh dxi)
Finally, note that (-1)kk = (-1)k so
(-1)kn-kk = (-1)kn-k = (-1)kn+k since (-1)2k = 1
Therefore
(Πh dxi)dxI = (-1)kn+kdxI (Πh dxi) (8.1)
and the sublemma is proved.
Comments: Although the derivation involved signs S and S', we never had to obtain an expression for S or S' separately, we only needed an expression for the product SS'. In my original document I showed that
S = (-1)a+b+..+q (-1)k(k+1)/2 note that S2 = 1 (1.13)
for
dxI = dxa dxb .... dxq a < b < ... < q a k-form (2)
Since SS' = (-1)kn+k, we could then deduce an expression for S'
S' = (-1)a+b+..+q (-1)k(k+1)/2 (-1)kn+k
(-1)a+b+..+q (-1)-k(k+1)/2 (-1)kn+k
(-1)a+b+..+q (-1)kn-k(k-1)/2