proof of a lemma regarding Hodge star (CH2)

Proof of a Lemma regarding *(*dxI) Lemma: Show that *(*dxI) = (-1)kn+k dxI (1) Take an example dxI = dxa dxb .... dxq a < b < ... <q a k-form, k dxi factors (2) Note that this is in Rn, so we have k and n to worry about. Then *dxI = S Πh dxi where we have a hole for each of dxa dxb ....dxq and S = a certain sign. (3) Then sign S is selected so that dxI(*dxI) = +dx1....dxn = dV. (4) Inserting (2) and (3) into (4), dxa dxb .... dxq S Πh dxi = dV (5) and therefore dxa dxb .... dxq Πh dxi = SdV (6) 1. Now how do we find S? I did this just for interest. The resulting expression for S is not needed in Section 2 below. Combine (2) and (3) above to get dxI (*dxI) = S dxa dxb .... dxq Πh dxi (1.1) The sign is found by right-sliding each of the dxi listed into the correct position in Π dxi. Write dxa dxb .... dxq Πh dxi = dxa dxb .... dxq [ dx1 ....dxa ....dxb........dxq ....dxn ] (1.2) Here red shows the holes. Let's start by sliding dxa into its red hole. We first have to slide dxa through k-1 other dx's to get it to the right, so this gives LHS(1.2) = (-1)k-1dxb .... dxq dxa [ dx1 ....dxa ....dxb........dxq ....dxn ] (1.3) Now we have to slide dxa another (a-1) positions to get it into its proper hole. So, = (-1)k-1 (-1)a-1dxb .... dxq [ dx1 .....dxb........dxq ....dxn ] . (1.4) Now we slide dxb first to the right, picking up (-1)k-2 and then we pick up (-1)b-1 so = (-1)k-1 (-1)a-1(-1)k-2 (-1)b-1 .... dxq [ dx1 ......dxq ....dxn ] . (1.5) We repeat this little process for each of the k elements of the dxI group. When we get to the last item dxq, since it is all by itself, we don't have a "first sign" since no sliding is necessary. So I think the k-signs are these (** means exponentiation) (-1)k-1(-1)k-2 .....(-1)k-k = Πi=0k-1 (-1)i = (-1)**(Σi=0k-1 i) = (-1)** (k-1)k/2 = (-1)k(k-1)/2 (1.6) where I use this classic result (1.7) Can this be simplified? Write out some cases: k (-1)k(k-1)/2 0 + 1 + 2 - 3 - 4 + 5 + 6 - (1.8) I see no simplification. The other sign contributions are this set of k factors (-1)a-1(-1)b-1.... (-1)q-1 = (-1)a+b+..+q (-1)k (1.9) The total k related factor is then (-1)k(k-1)/2 (-1)k = (-1)k+k(k-1)/2 = (-1)k(k+1)/2 (1.10) My conclusion is that dxa dxb .... dxq Πh dxi = (-1)a+b+..+q (-1)k(k+1)/2[ dx1 ......dxn ] = (-1)a+b+..+q (-1)k(k+1)/2dV (1.11) Now recall from (4) and (1.1) and (1.11) that dV = dxI (*dxI) = S dxa dxb .... dxq Πh dxi = S (-1)a+b+..+q (-1)k(k+1)/2dV (1.12) The conclusion then is that S = (-1)a+b+..+q (-1)k(k+1)/2 (1.13) Example 1: n = 6, k = 2, dxI = dx2dx5 , *dxI = S dx1dx3dx4dx6 (1.14) Claim of the above is that S = (-1)a+b+..+q (-1)k(k+1)/2 = (-1)2+5 (-1)3 = (-1)10 = +1 and then we are supposed to have dxI = dx2dx5 *dxI = dx1dx3dx4dx6 Let's manually check to verify that dxI (*dxI) = dV dxI (*dxI) = dx2dx5 dx1dx3dx4dx6 = + dx5 dx1dx2dx3dx4dx6 = + dx1dx2dx3dx4dx5 dx6 This example succeeds. Example 2: n = 4 and k = 2, dxI = dx1dx2, *dxI = S dx3dx4. Claim of the above is that S = (-1)a+b+..+q (-1)k(k+1)/2 = (-1)1+2(-1)3 = (-1)6 = +1 Let's manually check to verify that dxI (*dxI) = dV dxI (*dxI) = dx1dx2dx3dx4 = dV This example succeeds and agrees with Sja page 24 A. Example 3: n = 4 and k = 2, dxI = dx1dx3, *dxI = S dx2dx4. Claim of the above is that S = (-1)a+b+..+q (-1)k(k+1)/2 = (-1)1+3(-1)3 = (-1)7 = -1 Let's manually check to verify that dxI (*dxI) = dV dxI (*dxI) = dx1dx3(-dx2dx4) = dV This example succeeds and agrees with Sja p 24B So this is the best result I can get S = (-1)a+b+..+q (-1)k(k+1)/2 note that S2 = 1 (1.13) dxI = dxa dxb .... dxq a < b < ... < q a k-form (2) Example 4 n = 4 and k = 1, dxI = dx1, *dxI = S dx2dx3dx4 Claim of the above is that S = (-1)a+b+..+q (-1)k(k+1)/2 = (-1)1 (-1)1(1+1)/2 = (-1)1+1 = +1 Let's manually check to verify that dxI (*dxI) = dV dxI (*dxI) = dx1dx2dx3dx4 = dV This example succeeds and agrees with Sja p 24C 2. What is the next step? Recall that dxI = dxa dxb .... dxn (2) (*dxI) = S [ dx1 ....dxa ....dxb........dxq ....dxn ] = S Πh dxi (3) where S is the sign which makes this be true dxI(*dxI) = dV (4) Now consider *(*dxI) = *( S [ dx1 ....dxa ....dxb........dxq ....dxn ] ) (2.1) = S *{ dx1 ....dxa ....dxb........dxq ....dxn } (2.2) = S *{ S' dxa dxb .... dxq } = SS' dxa dxb .... dxq = SS' dxI (2.3) where S' is the sign which causes this to be true (*dxI)*(*dxI) = dV (2.4) Our goal is to show that SS' = (-1)kn+k since then (2.3) gives the desired result (1). (2.4a) Now write out (2.4) making use of (3) for (*dxI) and (2.3) for *(*dxI) to get S [ dx1 ....dxa ....dxb........dxq ....dxn ] S [ S' dxa dxb .... dxq ] = dV (2.5) Since S appears twice, we get S' [ dx1 ....dxa ....dxb........dxq ....dxn ] [ dxa dxb .... dxq ] = [dx1......dxn] . (2.6) Now let's start by moving dxa all the way to the left and sits then after S'. It has then to pass through n-k dxi factors, so we pick up (-1)n-k. We then have S' (-1)n-k dxa [ dx1 ....dxa ....dxb........dxq ....dxn ] [dxb .... dxq ] = dV (2.7) Now do this next with dxb and pick up another (-1)n-k. Do this k times and we then have S' (-1)kn-kk [dxa dxb .... dxq ] [ dx1 ....dxa ....dxb........dxq ....dxn ] = dV (2.8) or S' (-1)kn-kk dxa dxb .... dxq Πh dxi = dV (2.9) or dxa dxb .... dxq Πh dxi = S' (-1)kn-kk dV (2.10) Finally, we use our previous result (6), dxa dxb .... dxq Πh dxi = SdV (6) Comparing the last two results gives S' (-1)kn-kk dV = SdV (2.11) so that S'S = (-1)kn-kk (2.12) Now what can you say about (-1)kk ? If k is even, then k2 is certainly even 22 = 4 42 = 16 If k is odd, then k2 is odd. 12= 1 32= 9 52 = 25 Therefore (-1)kk = (-1)k ! (2.15) and then S'S = (-1)kn-k Now multiply by 1 = (-1)2k to get S'S = (-1)kn+k and we have then achieved our goal as stated in (2.4a) and thus (1) is proved. Comments: Try a summary of the above derivation. S is defined by (4) and S' by (2.4). From (2.3) we find that we have solved the problem if we can show that SS' = (-1)kn+k. We have this result S' (Πhdxi )dxI = dV (2.6) After sliding dxI to the left, this becomes dxI (Πhdxi ) = S' (-1)kn-kk dV (2.10) But our earlier result (6) said dxI (Πhdxi ) = SdV and this then gives the solution S'S = (-1)kn-kk. We never have to use the explicit formula for S which was developed above in Section 1.