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************************** 1. If V and W are two sets (which could be vector spaces), then VxW is called the Cartesian product of the two sets. The elements of this Cartesian product set have the form (v,w) where v ϵ V and w ϵ W. No structure of any kind is implied. For example, (v,w)+(v',w') and α(v,w) are undefined. The object VxW is not a tensor product of V and W. 2. Holding that idea for a moment, one can define F(S) to be the set of all functions f which map a set S into the reals. The elements of F(S) are functions f where f : S → R. One can declare that within F(S), one can add two functions so that f(s) + g(s) = h(s) = (f+g)(s) and one can scale a function by a real so that (αf)(s) = α f(s), where here (αf) is a function in F(S) different from f. The set of functions F(S) is called a free vector space. Each function f in F(S) just assigns a real number to each point in the set S. One can define a basis for F(S) as the set of basis functions δs(x) = 1 if x = s, otherwise 0. Then the most general f in F(S) has the form f(x) = Σs fs δs(x) for all x in S where fs is the real number assigned to point s in S. If the set S is infinite (like Z, the integers) this sum must only have a finite number of terms for which fs ≠ 0. The above construction works for any field K, not just the reals R. Note that the values of δs(x), 0 and 1, belong to any field K. Some authors restrict the form of the functions f(s) to be finite linear combinations of the elements of S which are weighted by real numbers. In this case, the most general element of F(S) has the form f(s') = Σsfs(s')s where the sum runs over the set S and where the fs are real numbers. A more practical way to write this is f(s') = Σifi(s')si where the elements of S are indexed and the set is then S = {si}. Suppressing the argument, one has just f = Σifisi. It is straightforward to add and scale functions of this free vector space F(S): f + g = (Σifisi)+(Σifisi) = Σi(fi+gi)si = h, and (αf) = α(Σifisi) = Σi(αfi)si where α is a real number (scalar). Technical notes on the free vector space One must have f(s) ≠ 0 for all s ϵ S in order that f(s) be included in F(S). S can be infinite (like the integers) but the sums f(S) must be finite, using some subset of S. Since f: S→ R and since f = Σifisi with fi real, one really needs si ϵ R, a fact often not stated. An alternative is to use f : S → X where X is some unspecified space which includes the si. This is akin to having a polynomial like p(x) = 2 + x3 - 3x5 where the powers of x are inert carriers of the coefficients and can be regarded as lying in some unspecified space X. We said f: S→ R, the reals, but more generally f: S → K, a field. In this case, in the above discussion the word "real" should be replaced by "element of field K". Often one sees discussion of a "free module" rather than a "free vector space". A module is a vector space defined over a ring instead of over a field. Since a ring is a more general object than a field, the discussion in terms of "free modules" is more general and presentations try to be as general as possible. As a reminder, here are the definitions of a field and a ring: Closed under Associative Identity exists Inverse exists such that: Commutative + (a+b)+c=a+(b+c) "0": 0+g = g+0 = g "-g": g+(-g) = (-g)+g = 1 a+b = b+a (ab) c=a (bc) "1": 1g = g1 = g "g-1": gg-1 = g-1g = 1 ab = ba and: a (b+c) = ab + ac (distributive property) FIELD Closed under Associative Identity exists Inverse exists such that: Commutative + (a+b)+c=a+(b+c) "0": 0+g = g+0 = g "-g": g+(-g) = (-g)+g = 1 a+b = b+a (ab) c=a (bc) [ "1" may exist] [g-1 may exist] ab = ba ? and: a (b+c) = ab + ac (distributive property) RING 3. We shall now combine steps 1 and 2 together. For a free vector space, instead of using a generic set S, we use S = V x W which is the Cartesian product of Step 1. Furthermore, we shall assume that V and W are not just sets but are vector spaces with vectors vi and wi. We now combine items 1 and 2 above to discuss F(VxW) which is a free vector space over the set of elements (v,w) ϵ VxW, as opposed to s ϵ S. So F(VxW) is the set of functions F: VxW → R. A typical element of F(VxW) might be 2(v1,w1) - 3(v2,w2+w3) where the elements (v,w) of VxW take real values. Here it is The Quotient Consider two vector spaces V and W of dimension n and n'. The Cartesian product of V with W is written VxW and the elements of this spaces are all pairs of the form (v,w) where v ϵ V and w ϵ W. Nothing else is implied. Sometimes writers talk about F(VxW) being the set of all functions f: VxW → K where K is a field like the reals R, so functions are of the form k = f((v,w)). F(VxW) is called a "free vector space" which implies that one can add functions to get a new function in the space, f(v,w) + g(v,w) = h(v,w). **************************** The components coefficients R Fij = Σij (qm)i Tij(q'n)j Tmn = Σab(ea)m(e'b)nFab (qiq'j)ca Apply (eie'j) If the basis vectors ei and e'i are more complicated, then whatever they are, we know we can find alternate bases Ei and E'i such that Σk (Ei)k(ej)k = δi,j for V, and Σk (E'i)k(e'j)k = δi,j for W. The basis {Ei} is the "dual basis" of {ei} and always exists and can be computed from the {ei} (see **). In more compact language, we have Ei ej = δi,j and E'i e'j = δi,j. We can then start over with T = Σij Fij (eie'j) . We then "dot" both sides with (EaEb) to get (EaE'b)T = (EaE'b) [Σij Fij (eie'j)] = Σij Fij (EaE'b) (eie'j) = Σij Fij ( Ea ei) (E'b e'j) = Σij Fij δa,iδb,j = Fab . In this case, the T expansion coefficients Fij onto the (eie'j) basis are the projections of the tensor T onto the dual basis element (EiE'j) . Notice that we quietly defined (vw)(v'w') ≡ (vv')(ww'). We have assumed a Cartesian metric tensor in this little discussion, but things can be generalized for curvilinear spaces and covariant notation. In this case one usually writes T = Σij Fij (eie'j) (EaE'b) = (eae'b) v v' = Σrs g(V)rsvrvs g(V)rs = covariant metric tensor for V w w' = Σrs g(W)rswrws g(W)rs = covariant metric tensor for W Then we get (eae'b)T = Fab and index positions match on both sides, indicating a covariant equation. As discussed in tensor doc, if the ei are the tangent base vectors of a transformation, then the ei are the reciprocal base vectors. **************