Problem 1. (max. 20 = 10+10 points) (a) Prove the versions of the graded product rule which concern the exterior derivative of the wedge product of three or of four differential forms. More specifically, assume that n is a positive integer, that U is some open subset of R", and finally, that, for each j = 1, 2, 3, 4, wj is a kj-form on U. Show that: (i) in the case of three differential forms, we have that d( w1 A W2 A W3) = ( dw1 ) A w2 Aws + ( - 1) k21 . W1 1 ( dw2) Aw3 + ( - 1) kitk z . W1 / W2 / ( dw3) . (ii) In the case of four differential forms, we have that d ( w1 A w2 A w3 A WA ) = ( dw1 ) A w2 A w3 AwA + ( - 1) 151 . W1 A (dw2) A w3 A w4 + + ( - 1 ) kitk z . W1 A w2 A ( dw3) A wA + ( - 1) (1+k2+k3 . W1 / W2 Aw3 / (dw4). (b) Formulate yourselves now the analogous version of the graded product rule in the case where we have five differential forms. That is, assume that we also consider a fifth differential form, w5, on U, which is a k5-form. The left-hand side of the identity we expect to have should be d ( W 1 A W 2 A W 3 A WA A W5). The right-hand side should be sign1 . (dw1) Aw2/w3/w4 Aws + sign2.w1/(dw2)Aw3/w4 Aws + sign3.W1/w2/(dw3)AW4 AW5 + sign4 . W1 A W2 A w3 A (dw4) Aws + signs . W1 A W2 A w3 A WA A (dw5) where, for each j = 1, 2, 3, 4, 5, sign; = (-1)mj for some non-negative integer m; (so essentially sign; E {-1, +1} for each j, or in other words it is a choice of sign). So what you have to do here is to determine sign; for each j = 1, 2, 3, 4, 5