Tensor, Tor, UCF, and Kunneth

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1 Tensor, Tor, UCF, and Kunneth Mark Blumstein 1 Introduction I d like to collect the basic definitions of tensor product of modules, the Tor functor, and present some examples from homological algebra and topology. The two main theorems I will cover are the universal coefficient theorem and the Kunneth theorem. The universal coefficient theorem for homology is used to translate back and forth between singular homology with Z-coefficients and singular homology with A-coefficients, where A is any abelian group. Changing coefficients is pervasive throughout topology, making the theorem a handy tool to use. Perhaps in the future I ll make the dual to this document and describe the UCF and Kunneth formulas for cohomology, but I won t use L A TEX I ll use xetal. That horrible joke aside, here goes. 2 Tensor Product Throughout the document any ring R is assumed to be a unital ring. Suppose that M and N are R-modules. Definition The tensor product M R N is an abelian group, constructed as a quotient group in the following way: Take the Z-span of the set M N, i.e. form the set of all finite Z-linear combinations of elements (m, n) M N. Observe that Z-span{M N} is an abelian group. Let H be the subgroup of Z-span{M N} generated by elements of the form: (m, rn) (rm, n) (m 1, n) + (m 2, n) (m 1 + m 2, n) (m, n 1 ) + (m, n 2 ) (m, n 1 + n 2 ) (Verify that H is actually a subgroup) We define M R N. = Z-span{M N}/H. If (m, n) is a coset representative in M R N, then standard notation is that (m, n) + H is written m n, and is called a simple tensor. A more general element x M R N is a finite linear combination of simple tensors, e.g. x = d i=1 k i(m i n i ), where k i Z. The relations defined by the subgroup H in the definition give the basic properties to work with tensors. The first relationship just says that scalars pull past a tensor, e.g. m (rn) = (rm) n. The second and third give a distributive property for tensors m 1 n + m 2 n = (m 1 + m 2 ) n. As a group, the zero element is represented by any m n with (m, n) H. For example, m 0 = 0 = 0 n for any m and n. Proof: (m, 0 N ) = (m, 0 R 0 N ) (0 R m, 0 N ) = (0, 0). Be aware that 0 n and m 0 are not necessarily the only choices of representative which equal zero in M R N. A simple example arises by picking r Ann R (m). Then, pick an n N, where rn 0. Superficially, the tensor m rn doesn t look like 0, since neither m nor rn equals 0, but by passing r across the tensor, we see that m rn = rm n = 0. 1

2 In addition to being an abelian group, it is not so hard to assign an R-module structure to M R N. Some care must be taken defining the module structure when R is not commutative. When R is commutative, it is particularly easy: On a simple tensor we define the module structure by r (m n) = (rm) n = m (rn). Then extend the action linearly over linear combinations of simple tensors. If you re new to module theory, pay close attention to the ring R. For R-modules M and N, you can always form M Z N, just using the underlying abelian group structure of M and N. However, the theory is made richer by considering the tensor product as also an R-module and not just an abelian group. As we will see, the cases when R = Z and when R is a field tend to be a bit more straightforward. By definition a Z-module is just an abelian group, so whenever you see that the ring is Z, the objects are just abelian groups and the module homomorphisms are just group homomorphisms. e.g. A Z B is just the tensor product of two abelian groups. Whenever a module has coefficients in a field, you get a vector space. e.g. V R W is the tensor product of two real vector spaces V and W, and any module homomorphism is just a linear map of vector spaces. Example Let V be a finite dimensional vector space over the field k with basis {e 1,..., e n }. Suppose v, w V, with v = x i e i and w = y i e i. Using the properties of the tensor product we compute v w = ( x i e i ) ( y i e i ) = (x i y j )e i e j V k V. i j By sending the coefficient of e i e j to the ith row and jth column of a matrix, we get what is called the outer product of two vectors. For example, let k = R, = 2 ( ) = For real (column) vectors, the outer product u v may be written uv T. product, u T v.. Compare to the inner Example If V and W are vector spaces over k, with bases {e 1,..., e n } and {f 1,..., f n } respectively, then V k W is an (nm)-dimensional vector space over k with basis formed by the tensors {e i f j }. For example, if V = R 2 and W = R 3, then we can get a basis for R 2 R R 3 by tensoring together one basis vector from each. Say, ( 1 0 ) = ( From the previous example, it is clear that the outer product of real vectors, u v is the transpose of v u. So for a general choice of a, b M R M, a b b a. The exceptions are called symmetric tensors (think symmetric matrices.) 2 )

3 Example x x is a symmetric tensor, as is x y + y x. This second example is familiar from linear algebra. Take x and y to be vectors in a real vector space. As before, may be interpreted as the outer product of the two vectors. Let A = x y, so that y x = A T. The matrix A + A T is of course a symmetric matrix. Note: If we take a complex vector space, and define outer product using the complex conjugate, we get a Hermitian matrix. Example Sometimes it s easy to compute a tensor product just by inspecting the elements. For example, Z/2 Z Z/3 = 0. Note that 1 1 is a generator, since for any simple tensor a b Z/2 Z/3, a b = ab (1 1) since we can pull integers past the tensor. Now, Thus, Z/2 Z Z/3 is trivial. 1 1 = (3 2) 1 (Distribute the tensor over ±) = = = = 0 0 = 0. Example For any R-module M, R R M = M since R pulls past the tensor. Example The tensor product of two R-algebras A and B has an R-algebra structure, it is called the tensor algebra. Theorem (The Universal Property of Tensor Products) The mantra: Every bi-linear map on the product induces a unique linear map on the tensor. Given R-modules M, N, and P, and an R-bi-linear map f : M N P, then there exists a map ˆf such that the following diagram commutes: M N M R N f The map simply takes an element (m, n) M N and sends it to m n, and extend linearly for sums. Lemma Let M and N be R-modules M R N = N R M ( α M α ) R N = α (M α R N) Example P Z/m Z Z/n = Z/d, where d = gcd(m, n). Proof: Consider the multiplication map µ : Z/m Z/n Z/d, defined by µ(a, b). = (ab) mod d. This map is bilinear, and so by the universal property of tensor products, it extends to a well-defined linear map µ : Z/m Z/n Z/d. We need to show that µ is an isomorphism of abelian groups. It s certainly surjective since d is less than or equal to either m or n, and 1 a = a 1 a mod d. Let a b ker µ, so that ab mod d = 0 and there exists an integer k such that ab = kd. Recall that running the Euclidean algorithm backwards allows you to write the greatest common divisor of ˆf 3

4 m and n as d = mx + ny for some integers x and y. Now use the properties of the tensor product to compute a b = ab(1 1) = (kmx + kny)(1 1) = kmx kny Therefore, µ is injective, completing the proof that it is an isomorphism. = 0 Compare the last example to the direct product of abelian groups, Z/m Z/n = Z/(mn) if and only if gcd(m, n) = 1. So when m and n are relatively prime, we see that Z/m Z/n = Z/(mn), but Z/m Z Z/n is trivial. 3 Tor Functor Invariably, a topologist finds herself in a situation requiring the use of some homological algebra. This short section introduces a tiny bit of homological algebra we will use later. For the time being, I m going to omit some basic definitions like projective module, exact sequence, functor. However, skimming through some of the examples and properties of the Tor functor should be enough to do the homology computations using the universal coefficients theorem in the following section. Let M be a module over a ring R. M is said to be a free R-module if there exists a basis {m 1,..., m n } for M, meaning that the m i s are linearly independent and every m M can be written as a unique linear combination of the m i s with coefficients from R, i.e. for all m M, m =! n i=1 r im i. Say that {m 1,..., m n } is basis for the free R-module M, then define the map M R n by sending m i to the element with a 1 in the ith spot and zeros everywhere else. After extending linearly, one can show that this map is an isomorphism, and so any free module is isomorphic to a direct sum of copies of the underlying ring. Every free R-module is also a projective R-module, but the converse isn t necessarily true. It depends on the ring R. Example For a Z-module (an abelian group) being projective is equivalent to being a free abelian group. For example, Z is a free abelian group, and every subgroup of Z is also a free abelian group. Any subgroup is free since a subgroup has the form mz for an integer m. Of course mz is generated over the integers by m, so it is free (in fact it is isomorphic to Z.) Similarly, for all n > 0, any subgroup of Z n is a free abelian group. There is a more general result in the same flavor when M is an R-module, and R is a principal ideal domain. As an example, both Z and any field are principal ideal domains, so the following result covers the case of abelian groups and vector spaces. Lemma For a principal ideal domain R, every projective R-module is free. Also, every submodule of a free R-module is free. The tensor product is a right-exact functor... 4

5 Example For A an abelian group, Z/n Z A = A/nA. Proof: 0 Z n Z Z/n 0 is a short exact sequence. Now apply Z A to get the right exact sequence: Z A n 1 Z A Z/n A 0. Since Z Z A = A, we get the exact sequence: A n A Z/n A 0. Now apply the first isormophism theorem to get the result. A complex is a sequence of modules and module homomorphisms M i i Mi 1 i 1 M 1 1 M0 0, where i i 1 = 0 for all i. Although the tensor product is not an exact functor, it does take an exact sequence of modules to a complex of modules as follows: Suppose that M i 2 1 M1 M0 is an exact sequence of right R-modules. Let N be a left R-module, and apply R N. This gives a sequence M i R N i id Mi 2 id M2 M1 R N 1 id M1 M0 R N, which in general is no longer an exact sequence, but it does satisfy the property that applying two consective maps always yields 0. That is, ( i id Mi ) ( ) i 1 id Mi 1 = 0 for each i, and therefore this sequence is a complex. In homological algebra, once you ve got your hands on a complex, the next thing to do is compute homology. Why? Note that the condition that successive maps of a complex equal zero ( = 0) is equivalent to the condition that image is a sub-module of kernel, allowing for the computation of homology in the first place. As a side note, if you ve had a first course in topology, you might be led to believe that the word homology implies that there is some topology going on in the background, but this isn t necessarily the case. In a topology class when you learn about simplicial, singular, or cellular homology theories, you are seeing specific examples of this abstract algebraic process (the homology theories from topology tell you how to construct a chain complex from a topological space, and from there it s all algebra to compute homology. So a priori, all you need to compute homology is a chain complex, no topology required.) In any case, since tensor takes an exact sequence of modules to a complex of modules, we can compute the homology groups, which in this case are called Tor groups (or modules.) Definition Let M and N be R-modules. T ori R (M, N) is the derived functor of the tensor product. This means the following: Begin by computing a projective resolution of M by R-modules, i.e. an exact sequence of the f i f 2 f 1 ɛ form P i P1 P0 M 0, where each Pi is a projective module. Next, take each P i and tensor it with N, and take each map f i : P i P i 1 and tensor it with. the identity map on N. We may use the notation (f i ) = fi id N. This yields the complex P i R N (f i) P i 1 R N (f i 1) (f 1) P 0 R N 0. i Define T or R 0 (M, N). = M R N. T or R i (M, N). = ker(f i id N )/Im(f i+1 id N ). By lemma 3.0.2, for R a principal ideal domain, we can replace the term projective resolution by free resolution in the definition of Tor. 5

6 Example The sequence 0 Z m Z Z/m 0 is a free resolution of Z/m. In fact, one can construct a two term free resolution of any abelian group as in the following example. Let A be an abelian group generated by elements {a 1,..., a n }. Let s construct a free resolution of A. Let e i be the ith standard basis vector of Z n, i.e. the element with a 1 in the ith spot and a 0 everywhere else. Consider the sequence, 0 ker ɛ j Z n ɛ A 0, where j is inclusion, and ɛ is defined by ɛ(e i ). = a i for each i. Since ker ɛ is a subgroup of Z n, ker ɛ is a free abelian group (as explained above). Therefore, we have obtained a two term free resolution of A. (Note: A slight modification is required if A is not finitely generated, but the procedure is basically the same.) Let B be an abelian group, and tensor it with the free resolution to compute Tor. 0 ker ɛ B j id B Z n B ɛ id B 0, This seqeuence is a complex, so we can compute Tor by taking homology (kernel mod image). For abelian groups A and B, T or Z 0 (A, B) = A Z B T or Z 1 (A, B) = ker(j id B ) T or Z i (A, B) = 0, i 2. Let A be an abelian group, m Z. A free resolution of Z/m is given by 0 Z m Z Z/m 0. Tensor the resolution by A to get 0 Z A m Z A 0. By definition, m = m id A. Then, m (1 a) =. m a = 1 ma. There is an isomorphism, Z Z A = A, by sending n a na. Using this isomorphism, we see that, Z A m Z A is isomorphic to A m A. Therefore, Using the last example, we compute: T or Z 0 (A, Z/m) = A Z Z/m = A/mA T or Z 1 (A, Z/m) = ker(a m A) T or Z i (A, Z/m) = 0, i 2. T or Z 0 (Z/n, Z/m) = Z/n Z Z/m = Z/gcd(m, n) T or Z 1 (Z/n, Z/m) = ker(z/n m Z/n) = Z/gcd(m, n) T or Z i (Z/n, Z/m) = 0, i 2. Example Let R be a pid. Let M be a finitely generated R-module generated by a minimal generating set {m 1,..., m n }. Define the map ɛ : R n M by sending the ith basis element of R n to m i. Since the kernel of ɛ is a submodule of R n, and every submodule of a free module over a pid, is itself a free module, we get the following free resolution 0 ker ɛ R n ɛ M 0. Just as was the case for abelian groups, we see that all Tor modules vanish past the first one. (If M is not finitely generated, you can adapt this argument to get the same result.) 6

7 Example The Tor functor doesn t pick up any information from vector spaces. Why? To compute Tor with coefficients in a field, you start with a free resolution. Well, a free module is one with a basis, and of course every vector space is defined by such a basis. So for any vector space V, you can always find a length 0 free resolution, namely V itself. Therefore, if V and W are vector spaces over a field k, then T ori k (V, W ) = 0 for all i 1. By the same logic, any projective module (and hence any free module) has a length 0 free resolution: Lemma If M or N is a projective R-module, then T or R i (M, N) = 0 for all i 1. These examples illustrate that for vector spaces T ori k vanishes for all i 1, and for a pid R, T ori R vanishes for all i 2. Then, the utility of higher Tor groups only really start to come into play with modules over a ring which is not a principal ideal domain. Lemma For any ring R, and any R-module M, a free resolution of M exists. Proof. We suppose M is finitely generated, but the same proof up to small modifications works for the infinite case too. Let M be generated as an R-module by the set {m 1,..., m n0 }. Just as in our construction of free modules for abelian groups, we begin by taking the map ɛ : R n 0 M which sends the ith basis element of R n 0 to the ith generator m i. When R is a pid, the sequence 0 ker ɛ R n 0 M 0 is a free resolution since it is an exact sequence, and ker ɛ is a free module. But when R is not a pid, we lose the free-ness of ker ɛ. That s okay though, we can just repeat the same process for ker ɛ. Take a map 1 : R n 1 ker ɛ which sends the basis elements of R n 1 to the generators of ker ɛ. Then, 0 ker 1 R n 1 ker ɛ 0 is a short exact sequence. We now have the exact sequence, 0 ker 1 R n 1 1 R n 0 ɛ M 0 If ker 1 happens to be a free module, then we have computed a free resolution of M. If not, repeat as many times as necessary ad infinitum. 4 Homology with Coefficients Let s recall the topological definition of singular homology with coefficients. Let i be the convex hull of the standard basis vectors e 0,..., e i R i. For example 0 is a point, 1 is a line segment, 2 a filled in triangle, 3 a solid tetrahedron, and so on into higher dimensions. Let X be a topological space, a singular i-simplex is a continuous function σ : i X define the ith chain group with coefficients in G by C i (X; G) =. { finite gσ : g G, σ a singular i-simplex} Note that the chain group is a purely algebraic object, it s the free group formed by the singular simplices. So while singular simplices have some topological meaning, general elements of the chain group can no longer be interpreted as maps from some i to X. It s neat that adding a layer of pure algebra (by constructing chain groups) on top of the topological info from the simplices, is so useful for studying topology. 7

8 The differentials i : C i (X; G) C i 1 (X; G) are defined as for ordinary homology, and so we have for i 0, H i (X; G). = ker( i )/im( i ). There is an alternate way to define the chain groups by using the tensor product. This method is a bit more general, since it applies to any algebraic complex. Suppose (C, ) is a complex of R-modules. To compute homology with coefficients in the group G, apply Z G to C, and then take homology. 4.1 Universal Coefficient Theorem In this section we apply the algebra results of the first two sections to topology. The universal coefficients theorem gives a computational tool to compute the homology of a topological space X with coefficients in some group G, from the homology with Z-coefficients. Z is the universal coefficient group, so to speak, it determines the homology for any other coefficient group. To state the theorem, the choice of homology theory isn t so important e.g. cellular or simplicial theory could be used instead of singular. Theorem (The Universal Coefficient Theorem) For X a topological space, and A an abelian group, the following sequence is exact for all i 0: 0 H i (X; Z) Z A H i (X; A) T or Z 1 (H i 1 (X; Z), A) 0. Further, this sequence splits, but not naturally. To say that the sequence splits means that H i (X; A) = (H i (X; Z) Z A) T or Z 1 (H i 1 (X; Z), A). To see how integral homology is computed in the first place, check out document intro alg top. Example Let K be the Klein bottle whose integral homology is Z, i = 0 H i (K; Z) = Z Z/2, 1 0, else Let s use the UCF to compute homology with coefficients in the field Z/p for p prime. H 0 (K; Z/p) = (H 0 (K; Z) Z Z/p) T or Z 1 (H 1 (K; Z), Z/p) = (Z Z Z/p) T or Z 1 (0, Z/p) = Z/p H 1 (K; Z/p) = (H 1 (K; Z) Z Z/p) T or Z 1 (H 0 (K; Z), Z/p) = (Z Z/2) Z Z/p T or Z 1 (Z, Z/p) We can distribute the tensor across the sum to simplify. Also, since Z is a free abelian group, the Tor term vanishes (lemma ) 8

9 H 1 (K; Z/p) = (Z Z Z/p) (Z/2 Z Z/p) = Z/p Z/gcd(2, p). At this stage it depends on if p = 2 or if p is an odd prime. For p = 2, we get that H 1 (K; Z/2) = Z/2 Z/2. If p is odd then H 1 (K; Z/p) = Z/p. Finally, H 2 (K; Z/p) = (H 2 (K; Z) Z Z/p) T or Z 1 (H 1 (K; Z), Z/p) = (0 Z Z/p) T or Z 1 (Z Z/2, Z/p) = T or Z 1 (Z Z/2, Z/p) Just like the tensor product, Tor is additive. H 2 (K; Z/p) = T or Z 1 (Z Z/2, Z/p) = T or Z 1 (Z, Z/p) T or Z 1 (Z/2, Z/p) = 0 T or Z 1 (Z/2, Z/p) (Tor vanishes on free abelian groups) = Z/gcd(2, p) 0. Again, we break the computation up: For p = 2, H 2 (K; Z/2) = Z/2, and for p odd, H 2 (K; Z/p) = It might seem strange that the dimension 2 homology of the Klein bottle vanishes when the coefficients are Z or Z/p (with p odd,) but does not vanish when the coefficients are Z/2. This is a consequence of more general topological machinery which says that the top dimensional homology of a closed connected manifold is related to the orientability of the manifold. The canoncial example of a non-orientable manifold is the mobius strip - if a 2-dimensional person lived on the mobius strip and took a lap around it, they would end up mirror reversed after going through the twist. The Klein bottle may be realized by taking two mobius strips and gluing them along the boundary, and so the Klein bottle is non-orientable. It turns out that for non-orientable closed connected manifolds of dimension n, the n-dimensional homology is zero. Since the Klein bottle is a non-orientable 2-manifold (surface), its integral homology is zero in dimension 2. But why didn t it vanish when coefficients were Z/2? We can reason intuitively that since +1 equals 1 in Z/2, the homology has no chance of detecting the orientation - it can t tell the difference between positive and negative! Theorem Kunneth Theorem From the perspective of topology, the Kunneth theorem allows one to compute the homology of a product of topological spaces X Y, from the homology of X and the homology of Y. On the algebraic side, the machinery for the Kunneth formula is similar to that for the universal coefficient theorem. We saw properties of the Tor functor for R-modules: for example, Tor vanishes when R is a field (vector spaces), there is a first Tor term for R = Z (abelian groups) and for R a principal ideal domain, and when R is not a pid the higher Tor terms play a role. Because of this, we will see that the Kunneth formula is easiest for vector spaces, slightly more complicated when the ring is a pid, and quite tricky for other rings R (we need spectral sequences! yuck.) 9