# MATH 101A: ALGEBRA I PART C: TENSOR PRODUCT AND MULTILINEAR ALGEBRA. This is the title page for the notes on tensor products and multilinear algebra.

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1 MATH 101A: ALGEBRA I PART C: TENSOR PRODUCT AND MULTILINEAR ALGEBRA This is the title page for the notes on tensor products and multilinear algebra. Contents 1. Bilinear forms and quadratic forms definition of bilinear form quadratic form nondegenerate bilinear forms nonsingular bilinear forms 3 2. Matrix of a bilinear form matrix as homomorphism dual bases basis for homomorphisms 7 3. Naturality and contravariant functors natural transformations contravariant functors unnaturality of the dual 10 0

2 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA 1 1. Bilinear forms and quadratic forms To save time, I am talking about bilinear forms and quadratic forms at the same time. We assume throughout that R is a commutative ring. (This will make E R F into an R-module for R-modules E, F.) 1.1. definition of bilinear form. Definition 1.1. If E, F are R-modules, a bilinear form on E F is an R-bilinear map f : E F R. I.e., (1) f(x, ) : F R is linear for each x E and (2) f(, y) : E R is linear for each y F. In the case E = F, f is called a bilinear form on E. Example 1.2. E = F = R n and f : E F R is given by n f(x, y) = x i y i. In this example you can see why R needs to be commutative: i=1 f(x, ry) = x i ry i = r x i y i = rf(x, y) holds because x i r = rx i. This is an example of a symmetric bilinear form on E where symmetric means f(x, y) = f(y, x). Example 1.3. E = R n = F and f is given by f(x, y) = i<j x i y j x j y i. This is an example of an alternating form which means f(x, y) = f(y, x). Note that symmetric and alternating only apply to the case E = F. Example 1.4. Take R = R and E = F = C 0 (I), the ring of continuous functions I = [0, 1] R considered as an R-module. Then f is given by f(φ, ψ) = 1 This is a symmetric bilinear form. 0 φ(x)ψ(x) dx.

3 2 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA The notation f(x, y) = x, y is sometimes used, especially in this last example. Here is a general example where E F. Example 1.5. Let E be any R-module and let F = Hom R (E, R). Let be given by f : E Hom R (E, R) R f(x, g) = g(x). This is R-linear in x by definition and it is R-linear in g because of the way the R-module structure of Hom R (E, R) is defined, namely by pointwise addition and scalar multiplication given by rg(x) = g(rx) This definition only works if R is commutative: s(rg)(x) = rg(sx) = g(rsx) = (rs)g(x) quadratic form. I gave two definitions of a quadratic form and showed that one implies the other. Definition 1.6. (This definition requires 1 2 on E is a function f : E R so that R.) A quadratic form f(x) = 1 g(x, x) 2 for some symmetric bilinear form g on E. The second definition works for any ring R. Definition 1.7. A quadratic form on E is a function f : E R so that (1) f(rx) = r 2 f(x) for every r R, x E. (2) The function g(x, y) := f(x + y) f(x) f(y) is a symmetric bilinear form on E. To see that the second definition implies the first, suppose that x = y. Then g(x, x) = f(2x) 2f(x) = 4f(x) 2f(x) = 2f(x). This implies that f(x) = 1g(x, x) if 1 R. Conversely, f(x) := 2 2 g(x, x) is easily seen to satisfy the second definition. 1 2

4 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA nondegenerate bilinear forms. Definition 1.8. Given a bilinear form f : E F R and S E, I defined S := {y F f(x, y) = 0 x S} Similarly, if T F then T := {x E f(x, y) = 0 y T } Proposition 1.9. Suppose E = F and f : E E R is symmetric or alternating. Then S = S for any S E. It is easy to see that S is always a submodule of F and T is always a submodule of E. However, in class I only pointed this out in the case of F E and E F. These are important because they are the kernels of the maps and φ f : E Hom R (F, R) ψ f : F Hom R (E, R). Definition We say that f : E F is nondegenerate on the left if F = 0, i.e., φ f is a monomorphism. We can that f is nondegenerate on the right if E = 0, i.e., ψ f is a monomorphism. Lang uses the notation E 0 = F and F 0 = E. Proposition Every bilinear form f : E F R induces a nondegenerate (on both sides) bilinear form by the equation f : E/E 0 F/F 0 R f(x + E 0, y + F 0 ) = f(x, y) nonsingular bilinear forms. Definition A bilinear form f : E F R is nonsingular on the left if φ f : E Hom R (F, R) is an isomorphism. It is nonsingular on the right if ψ f : F Hom R (E, R) is an isomorphism. f is called nonsingular if it is nonsingular on both sides.

5 4 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA We discussed why nonsingularity on the left and right were different. It is because a module is not always equal to its double dual. Definition The dual E of any R module E is defined by E = Hom R (E, R). If f : E F R is nonsingular on the left then φ f : E = F. Nonsingularity on the right would mean that ψ f : F = Hom R (F, R) = F But there are well-known examples where this is not true. There are also easy examples: If F is a torsion module over a PID R then F = 0. The well-known example is V a countably infinite dimensional vector space over Q. Then V is uncountable dimensional and V is even bigger dimensional. There is a related isomorphism: Proposition The mapping which sends a bilinear form f to φ f : E F and to ψ f : F E gives isomorphisms of R-modules: L 2 (E F, R) = Hom R (E, Hom R (F, R)) = HomR (F, Hom R (E, R)) Proof. The inverse of the first mapping is given by sending φ : E F to f(x, y) = φ(x)(y). This proposition is about all possible f. For a fixed nonsingular f, we get a different formula: Theorem If f : E F R is nonsingular then we get an induced isomorphism of R-modules End R (E) = L 2 (E F, R) where A End R (E) is mapped to the function fa given by fa(x, y) = f(ax, y).

6 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA 5 Remark On the second day, I pointed out that this theorem requires only that f : E F R be nonsingular on the left. Proof. We know that L 2 (E F, R) = Hom R (E, F ) and that a bilinear form g maps to the homomorphism φ g : E F given by φ g (x) = g(x, ). The definition of nonsingular on the left means that φ f : E F is an isomorphism. But, F = E implies that We conclude that Hom R (E, F ) = Hom R (E, E) = End R (E). L 2 (E F, R) = Hom R (E, F ) = Hom R (E, E) = End R (E). The only question is: What is the formula for this isomorphism? The isomorphism is given by composition with the inverse φ 1 f : F E. So, g corresponds to A = φ 1 f φ g which, when applied to x E gives Ax = φ 1 f (φ g(x)). Applying φ f to both sides we got: f(ax, ) = φ g (x) = g(x, ) which is the formula in the theorem. The significance of this theorem is that it shows the need for a second bilinear form, namely f, in order to express g in matrix form.

7 6 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA 2. Matrix of a bilinear form Suppose that E = R n and F = R m are free R-modules with bases {a 1,, a n } and {b 1,, b m }. If f : E F R is any bilinear form, we get a bunch of scalars: f(a i, b j ) = g ij R. These scalars determine the bilinear form f uniquely since arbitrary elements x E, y F are given by x = x i a i, y = y j b j. Then ( f(x, y) = f xi a i, ) y j b j = i This can be written in matrix form as f(x, Y ) = t XGY x i y j f(a i, b j ) = j i x i y j g ij where X, Y are column vectors with coordinates x i, y j, t X is the transpose and G = (g ij ) is the n m matrix with ij-entry equal to g ij matrix as homomorphism. At this point I tried to explain that there are hidden nonsingular forms in the matrix equation. This is because the product of a row matrix and a column matrix is the their dot product: f(x, Y ) = X, GY E where X, Z E = x i z i is the dot product in E. This is a nonsingular symmetric bilinear form. The bilinear form can also be written in terms of the dot product in F : f(x, Y ) = t GX, Y F The point is that, given a fixed nonsingular form on E or F we can express bilinear forms as homomorphisms: Proposition 2.1. If E, F are free and finitely generated over R then a bilinear form f on E F is given by f(x, Y ) = X, GY E where G = (g ij ) Hom R (F, E) is uniquely determined by f. This is just a rewording of what I explained earlier (that the matrix determines f). I just needed to explain how matrices give homomorphisms. This is just standard linear algebra but over arbitrary commutative rings. j

8 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA dual bases. First, we need the dual basis on F. If b 1,, b m is the basis for F then we have the dual basis b 1,, b m F given by ( ) b j (x) = b j xi b i = x j I.e., b j picks out the jth coordinate of x. Theorem 2.2. The b i form a basis for F = R m. Proof. First, the b i generate F. To show this let f : F R be any homomorphism. Then f is determined by the numbers f(b i ) = c i R and f = c i b i by the following: ( ) f(x) = f xi b i = x i f(b i ) = x i c i = c i b i (x). This shows that f = c i b i. Therefore, these dual elements b i generate F and give an epimorphism φ : R m F. Next we have to show that φ is 1-1, i.e., its kernel is zero. So, suppose that φ(x 1,, x m ) = x i b i = 0 F. If we apply this element of F to the basis element b j we get xi b i (b j ) = x i δ ij = x j = 0. Since this is true for all j, x = (x 1,, x m ) = (0,, 0). So, φ is an isomorphism and the b i form a basis for F basis for homomorphisms. Corollary 2.3. If E = R n and F = R m are free with bases {a i }, {b j } then Hom R (F, E) = R nm is free with basis the functions a i b j given by if y = y j b j. a i b j(y) = y j a i This corollary says that a homomorphism is given by a linear combination g = g ij a i b j i,j and the action of g is given by ( ) g(x) = g xj b j = i,j x j g ij a i = (a 1,, a n )GX (matrix multiplication) where G = (g ij ) and X = (x j ).

9 8 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA I gave a (much longer than necessary) categorical proof of this corollary using the following lemma. Lemma 2.4. Hom R (M, A B) = Hom R (M, A) Hom R (M, B) Proof. This follows from the universal property of the product A B = A B namely, a homomorphism into the product f : M A B is given uniquely by the projections p 1 f : M A and p 2 f : M B. So, the isomorphism in the lemma is given by and the inverse is given by f (p 1 f, p 2 f) (f, g) s 1 f + s 2 g where s 1 : A A B and s 2 : B A B are the inclusion maps. Proof of Corollary. Since E = R n we have by the lemma that Hom R (F, E) = Hom R (F, n R) = Hom R (F, R) = (F ) n. n Since the ith inclusion map s i : R E = R n is given by s i (r) = ra i the basis elements of Hom R (F, E) are s i b j = a i b j.

10 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA 9 3. Naturality and contravariant functors Somehow, I got into a long explanation about contravariant functors and naturality natural transformations. The concept of naturality is made precise with the definition of a natural transformation. One example I used was the commutator subgroup G = [G, G] of a group G vs. its center Z(G). I claimed that G was natural but Z(G) was not natural. This comes from the fact that any homomorphism G H sends G into H but does not necessarily send Z(G) to Z(H). Definition 3.1. Suppose that F, G : A B are functors between categories A, B. Then a natural transformation η : F G is defined to be an operation which assigns to each object A A a morphism in B of the form η A : F A GA so that, for all morphisms f : A B in A, we have the following commuting square of morphisms in B. F A F f F B η A η B GA Gf GB For example, if A = B = Gps, the inclusion map is a natural transformation from G to G. This is a natural transformation from the commutator subgroup functor to the identity functor. Another example is the torsion submodule tm of a module M over a PID R. The following commuting diagram shows that tm is a natural submodule of M and M/tM is a natural quotient module. tn N N/tN tm M M/tM The two squares in this diagram mean we have two natural transformations t id id/t Natural is not the same as functorial. It means more. In order to have naturality, we need two functors and a natural transformation. The dual E of E is thus functorial but not natural. But it is contravariantly functorial.

11 10 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA 3.2. contravariant functors. Definition 3.2. If A and B are categories, a contravariant functor F : A B is a pair of mappings (1) a mapping F : Ob(A) Ob(B), i.e., for each A A we have F A B. (2) for every morphism f : A B in A we get a morphism satisfying the two conditions: (a) F id A = (id A ) = id F A (b) (f g) = g f. F f = f : F B F A Example 3.3. Suppose that X is any fixed R-module. Then we have a contravariant functor sending M to Hom R (M, X). Hom R (, X) : R-Mod Z-Mod Example 3.4. If X is a topological space, let C 0 (X) be the ring of all continuous functions X R. This is a contravariant functor C 0 : T op Rings from the category of topological spaces and continuous maps to the category of rings and ring homomorphisms. Duality E E = Hom R (E, R) is a contravariant functor. So, E is functorial and its elements behave in a functorial way. However, the dual basis is not natural unnaturality of the dual. Theorem 3.5. There is no natural isomorphism E = E in the category of finitely generated free R-modules and isomorphism. First, what does this theorem say? We take the category C whose objects are f.g. free R-modules and whose morphisms are isomorphisms. By a natural isomorphism φ E : E E we mean an isomorphism which makes the following diagram commute for any isomorphism f : E = F of f.g. free R modules: E f F φ E E f F φ F

12 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA 11 Proof. I proved this by contradiction. But the proof assumes the 1+1 0, i.e., the characteristic of the ring is not 2. Suppose there is such a natural isomorphism. Let E = F = R 2 be free on two generators v, w. Then E is free on the dual basis elements v, w. Let f : E E be the isomorphism given by f(v) = v and f(w) = v ± w (doing two cases at once). Then f (w ) = ±w since f w (xv + yw) = w f(xv + yw) = w (xv + yv ± yw) = ±y and f (v ) = v + w since f v (xv + yw) = v f(xv + yw) = v (xv + yv ± yw) = x + y. Now suppose that there is some isomorphism φ : E E which makes the diagram commute. Then φ = f φf. Suppose that φ(v) = xv + yw. Then φ(v) = f φf(v) = f φ(v) = f (xv + yw ) = xv + xw ± yw. In order for this to be equal to f(v) = xv + yw we must have y = x + y = x y since the same isomorphism φ must work for both functions f. This implies that x = 0 and 2y = 0. But, for φ to be an isomorphism, we must have y invertible in the ring. So, 2 = 0. When you looked at this counterexample, you can see one of the reasons that the dual basis is not natural: When we changed the second basis element from w to v + w, it was the dual basis element v which changed. This dual element v depends not on v but on your choice of the other basis element w (or, in higher dimensions, vi depends on all v j where j i). The reason is that the other basis elements must span the kernel of vi by definition. So, they are the ones that determine what vi are up to a scalar multiple.

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