Spivak's Wedge Product Definition v2 REVIEWED 9_9_15

Examples of Spivak type wedge products. PhL 9.3.15 These are the rules I shall use: Example 1. Two elements of Λ1. ω ^ η = [ ωη - η ω] = ωη - η ω Note that ωη and η ω are elements of T2 but not of Λ2. Note that ωη - η ω is an element of both T2 and Λ2. How to verify now that this result is really in Λ2 ? Do I have a projection operator? Alt (ωη - η ω) = Alt(ωη) - Alt(ηω) = (1/2) [ ωη - ηω ] - (1/2) [ ηω - ωη ] = (ωη - η ω) This is like Spivak's little page 78 rule which I am running backwards. Example 2: Element of Λ1 with an element of Λ2. As our element of Λ2 use α = ω2ω3 - ω3ω2 where these ωi are in Λ1. As our element of Λ1 use ω1. Then: ω1 ^ α = ω1 ^ [ ω2ω3 - ω3ω2] = Alt { ω1 [ ω2ω3 - ω3ω2] } The two terms inside the Alt are really equal, so we will just double the first term, so then = 2 Alt { ω1 ω2 ω3 } ___________________________________________________________________ Note added 9.9.15. Why are they equal? I ignore the 1/3! here: A = Alt { ω1 ω2 ω3 } = εijk ωi ωj ωk factor omitted? B = - Alt { ω1 ω3 ω1 } = - ???? Write A = Alt { ω1 ω2 ω3 } = ω1 ω2 ω3 + other terms B = - Alt { ω1 ω3 ω1 } = - ω1 ω3 ω2 + other terms Maybe write in terms of a sum over permutations A = ΣP (-1)S ωP(1) ωP(2) ωP(3) where { P(1),P(2).....} is P {1,2...}. Then the above two lines for A,B are reasonable. (they were anyway). So we then have A = ΣP (-1)S ωP(1) ωP(2) ωP(3) B = - ΣP (-1)S ωP(1) ωP(3) ωP(2) = .. OK stop. I know it is right, later I will write down a better tool for doing this. __________________________________________________________________________ Now we use Alt { ω1 ω2 ω3 } = [ ω1 ω2 ω3 + 5 more terms ] = εijk ωi ωjωk We then find that ω1 ^ α = 2 [ ω1 ω2 ω3 + 5 more terms ] = [ ω1 ω2 ω3 + 5 more terms ] = εijk ωiωjωk This result can also be written as ω1 ^ ω2 ^ ω3 = εijk ωiωjωk implied sum on i,j and k. But we won't always have such nicely named vectors like ωi so better just leave these results as ω ^ η = ωη - η ω ω ^ (a ^ b) = ωab - ωba + 4 more terms Comment: We have to represent α as ab - ba in order to talk about α being "alternating" and therefore a viable element of Λ2. If you don't do this and just use α, you cannot write out the result. That is to say, if you are interested in ω ^ α, you cannot express this as a sum of terms as I have done above just using the symbols ω and α. So you really need vectors to get explicit expressions! Example 3: The generalization of Example 2 is going to be this: ω1 ^ ω2 ^ ω3 ^ .... ωk = εabc...q ωaωbωc.....ωq. k factors Example 4. Let's now try to wedge an element of Λs with an element of Λt . α = ω1 ^ ω2 ^ ω3 ^ .... ωs β = ωs+1 ^ ωs+2 ^ ωs+3 ^ .... ωs+t where as usual we just represent these elements this way so we can have something to chew on. I guess we need all these vectors different so we don't zero out things. Think of a very high n for dim(V). We know that α = εijk...q ωiωjωk ...ωq s factors β = εIJK...Q ωs+Iωs+Jωs+K .... ωs+Q t factors And we also know that α ^ β = εijk...z ωiωjωk ...ωz s+t factors That is about all we can say. Now would one relate β ^ α to α ^ β ? I think we just have to slide the two groups through each other to get from one to the other! ************* Maybe rename the vectors so that β = εIJK...Q ωs+Iωs+Jωs+K .... ωs+Q = εIJK...Q ηIηJηK .... ηs s factors α = εIJK...Q ηt+Iηt+Jηt+K .... ωt+Q t factors Then certainly we get β ^ α = εijk...z ηiηjηk ...ηz t+s factors So how would w rearrange this thing to get the α ^ β expression shown above? Comments 9.9.15. I have just reread the above. Here I am scrupulously avoiding "arguments" of the Spivak type. There are no arguments at all. I find that if I want to evaluate wedge products say of an element of Λ1 with an element of Λ2 (Example 2 above), I can't write out the resulting wedge product unless I assume that the Λ2 element is given by something like α = ω2ω3 - ω3ω2 where ω2 and ω3 are elements of Λ1. I claim that ω2ω3 - ω3ω2 is an element of Λ2 because I interpreted Spivak's Alt formula shown at the start in a completely different way, one with no arguments. This whole approach now seems very wobbly and I have trouble really supporting it. Plan A . Why note represent an arbitrary element of Λ2 as an expansion on basis vectors. Example 2A: Element of Λ1 with an element of Λ2 . I am NOT going to use i < j here because it makes problems down the road. So for use here I will write. α = Σi,j αij ei ^ ej Then it really is in terms of those required vectors, but it is at the same time a general element of Λ2. Then let ω1 be the element of Λ1 same as in example 2. Now we already know that ω ^ η = ωη - η ω for any two vectors ω and η each of which is in Λ1 Then we can say ω = ei and = ej and we have ei ^ ej = eiej - ejei And then α = Σi,j αij [eiej - ejei] = Σi,j αij eiej - Σi,j αij ejei = Σi,j αij eiej - Σj,i αji eiej = Σi,j (αij - αji) eiej This shows that only the antisymmetric component of αij makes a contribution to α. We can maintain a symmetric component of α if we wish. How do I verify this to be an element of Λ2? I just repeat what I did in Example 1. Apply the Alt operator and find that Alt(q) = q. Now we are ready to continue with this Example 2a: ω1 ^ α = ω1 ^ { Σi,j αij [eiej - ejei]} = Σi,jαij { ω1 ^ [eiej - ejei] } = Σi,j αij Alt { ω1 [eiej - ejei] } = Σi,jαij 2 Alt { ω1 ei ej } = Σi,jαij [ ω1 ei ej - ω1 ej ei + 4 more signed terms ] So at least I have a result! Can this be simplified? I don't think so. Example 2B: Let's repeat the above, but write ω1 = Σkakek Then we start over again: ω1 ^ α = (Σkakek) ^ { Σi,j αij [eiej - ejei]} = Σkak Σi,jαij ek ^ (eiej - ejei) = Σkak Σi,jαij Alt { ek [eiej - ejei] } = Σkak Σi,jαij 2 Alt { ek eiej } = Σkak Σi,jαij [ ekeiej + 5 other signed terms ] = Σkak Σi,jαij [ eiejek + 5 other signed terms ] Now write this as ω1 ^ α = Σijk Tijk ΣP (-1)S eP(i)eP(j)eP(k) Tijk ≡ αijak Now use the result of "theorem on sums" to rewrite this as ω1 ^ α = Σijk [ ΣP (-1)STP(i)P(j)P(k) ] eiejek where now the coefficient is ΣP (-1)STP(i)P(j)P(k) = Tijk - Tikj + Tkij - Tkji + Tjki - Tjik = αijak - αikaj + αkiaj - αkjai + αjkai - αjiak If we consider αij to be antisymmetric as noted above, then we can combine terms to get ΣP (-1)STP(i)P(j)P(k) = 2 [ αijak + αkiaj + αjkai ] and then our result is ω1 ^ α = 2 Σijk [ αijak + αkiaj + αjkai ] eiejek Is this coefficient really antisymmetric under i↔j? The first term is. The second term does this αkiaj → αkjai = - αjkai = minus the third term So yes, the coefficient is totally antisym! Comment: This is the first time ever that I have computed the wedge product of a general vector with a general bivector using a non-function-based definition of the wedge product: I first considered this for the case ω and η were vectors so k = l = 1. I then considered this for the case that ω was a vector and η was a general bivector! Here I will do the first case again showing general vectors: ω = Σiaiei η = Σjbjej ω ^ η = Σijaibj [ eiej - ej ei] = Σijaibj [ eiej - ej ei] = Σij [aibj- ajbi] eiej I think I can extend this method to compute the wedge product of any two multivectors. Maybe the function-based approach is the dual space? Consider the simple space V with elements like ei or x. The space V* is the dual space of functionals on V, such as f(x). Now consider the direct product space V x V which has elements like (x,y). The space V* x V* would be the space of linear functionals on V x V such as f(x,y) In general we can define Vk with elements like (v1, v2....vk) . The dual space would be V*k with functionals like f(v1, v2....vk). Do any of my authors talk about things in this way? Spivak: Well, he is in fact quite clear on this. Right on his opening page: So for Spivak, a k-tensor is a functional in this dual space, though he does not mention dual spaces. Benn-Tucker. They thing the same way except they have their perverse thing about V ↔ V*.