tensor product of vector spaces

THE TENSOR PRODUCT OF VECTOR SPACES TOM COATES Abstract. My presentation of tensor products in class today was extremely confusing. Here is a (hopefully) clearer version. 1.Motivation One of the most powerful ideas in 20th-century mathematics — an idea which will come up in a lot of classes as you study more math — is that one can study the geometry of a space X(which could be a metric space, or a topological space, or a manifold, or . . . ) by studying the functions on X. For example, one could study the space [0,1] by studying the vector space1of bounded continuous functions C([0,1]). This suggests a question: we know how to take the product X×Yof two spaces, but how is the vector space of functions on X×Yrelated to the vector spaces of functions on Xand functions on Y. The tensor product is the answer to this question: roughly speaking, we will define the tensor product of two vector spaces so that Functions (X×Y) =Functions (X)⊗Functions (Y). The “roughly speaking” in the last sentence is because this statement will be true only for XandYfinite sets. Exercise. Once you have read this note, read through it again and work out why the statement isn’t true for infinite sets. What happens if we replace Functions byFunctions- which-are-non-zero-at-only-finitely-many-points ? Or by Continuous-functions ? One can define the tensor product of vector spaces in a number of different ways — Halmos uses a different definition, for example. His definition is significantly simpler than the one we are about to develop. The reason that I want to use this definition is that it works in a very general setting: the same construction gives the tensor product of infinite-dimensional Date : February 17, 2005. 1This is slightly misleading: one should study C([0,1]) not as a vector space but as an algebra . This means that we should think of C([0,1]) as a vector space equipped with a multiplication map C([0,1])× C([0,1])− → C ([0,1]) (f(t), g(t))/mapsto− → f(t)g(t). But the basic point remains: one can study [0 ,1] by looking at C([0,1]). 1 vector spaces, the tensor product of modules over a ring (once one knows what modules and rings are), etc. 2.Construction From now on, think about two finite dimensional vector spaces VandW. We will regard Vas the vector space of functions on some finite set S, and Was the vector space of functions on some finite set T. Example. In all our examples, we will take S={1,2,3,4} T={1,2,3} so we can think of S×Tas a 4 ×3 grid T 1• • • • 2• • • • 3• • • • S 1 2 3 4 2.1.Functions on the product S×T.Given a function f(s) onSand a function g(t) on the set T, we can form a function h(s, t) =f(s)g(t) on the set S×T. Example. If f(s) =/braceleftBigg 1s= 2 0 otherwiseand g(t) =/braceleftBigg 1t= 1 0 otherwise then we get h(s, t) =/braceleftBigg 1 (s, t) = (2 ,1) 0 otherwise Every function on S×Tcan be written as a linear combination of functions which are 1 in exactly one place and 0 everywhere else, so we see that every function on S×Tcan be written as a linear combination of products of functions on Sand functions on T: (1) H(s, t) =aαβfα(s)gβ(t) for any function H:S×T→kand some choice of scalars aαβand functions fα:S→k, gβ:T→k. Exercise. Show that H(s, t) =  1 (s, t) = (2 ,1) 5 (s, t) = (1 ,3) 0 otherwise is not the product of a function on Sand a function on T. Express Has a linear combination of products, as in (1). Since we are thinking about Vas functions on SandWas functions on W, and we want V⊗Wto be functions on S×T, this suggests that elements of V⊗Wshould be built as linear combinations of pairs of elements ( f, g), where f∈Vandg∈W. Here we think of the pair ( f, g) as representing the function (s, t)/mapsto− →f(s)g(t) onS×T. 2.2.The free vector space generated by V×W.The free vector space generated by V×Wis a precise version of “all linear combinations of pairs of elements ( f, g), where f∈V andg∈W”. It is defined to be the vector space over kwith basis {δ(f,g): (f, g)∈V×W} So in other words, elements of the free vector space Fgenerated by V×Whave the form α1δ(v1,w1)+. . .+αnδ(vn,wn) for some n, some choice of scalars α1, . . . , α n, and some choice of ndistinct elements (v1, w1), . . . , (vn, wn)∈V×W. Example. Letv,v/primebe distinct elements of Vandw,w/primebe distinct elements of W. Then/parenleftbig 3δ(v,w)−δ(v,w/prime)/parenrightbig + 6/parenleftbig δ(v,w/prime)+δ(v/prime,w/prime)/parenrightbig = 3δ(v,w)+ 5δ(v,w/prime)+ 6δ(v/prime,w/prime) To find out why I am using the notation δ(v,w), read the footnote2. Note that we could use this construction to make “the free vector space generated by the set P” for any set P — we never needed to use the fact that VandWare vector spaces. 2As Toly pointed out at the end of class, another way to think about the free vector space generated by V×Wis as a vector space of functions {F:V×W→k|Fis non-zero at only finitely many points of V×W}. A basis for this vector space is given by the “delta functions” δ(v,w ):V×W→ k (x, y)/mapsto→/braceleftBigg 1 if ( x, y) = ( v, w) 0 otherwise. These match up with the basis elements δ(v,w )used above. 2.3.The subspace of relations Z.To reiterate, we are thinking of elements f∈Vas functions on the set Sand elements g∈Was functions on the set T. So the free vector space Fgenerated by V×Wlooks rather like the functions on S×T, where we regard the element α1δ(f1,g1)+. . .+αnδ(fn,gn)∈F as representing the function on S×Twhich is (s, t)/mapsto→α1f1(s)g1(t) +. . .+αnfn(s)gn(t). ButFis not quite the functions on S×T, because (af1(s) +bf2(s))g(t) =af1(s)g(t) +bf2(s)g(t) as functions on S×T, but δ(af1+bf2,g)/negationslash=aδ(f1,g)+bδ(f2,g) inF. In other words, if we want to turn Finto the vector space of functions on S×Tthen we need to impose the relations δ(af1+bf2,g)=aδ(f1,g)+bδ(f2,g) δ(f,ag 1+bg2)=aδ(f,g1)+bδ(f,g2) But we know how to do this from class: we make a subspace ZofFwhich contains all the things that we want to be zero ( i.e.all the relations that we want to hold) Z=span/braceleftbig δ(af1+bf2,g)−aδ(f1,g)−bδ(f2,g), δ(f,ag 1+bg2)−aδ(f,g1)−bδ(f,g2):a, b∈k, f, f 1, f2∈V, g, g 1, g2∈W/bracerightbig and then define V⊗W=F/Z 2.4.Why this does exactly what we want. Write v⊗w=δ(v,w)+Z Then the fact that δ(af1+bf2,g)−aδ(f1,g)−bδ(f2,g)∈Z means that δ(af1+bf2,g)+Z=aδ(f1,g)+bδ(f2,g)+Z or in other words that (2) ( af1+bf2)⊗g=a(f1⊗g) +b(f2⊗g). Similarly, (3) f⊗(ag1+bg2) =a(f⊗g1) +b(f⊗g2). 2.5.A basis for the tensor product. At the end of class, I claimed that if {x1, . . . , x n} is a basis for Vand{y1, . . . , y m}is a basis for Wthen B={xi⊗yj: 1≤i≤n,1≤j≤m} is a basis for V⊗W. To get some practice working with tensor products, let us first see why the set Bspans V⊗W. We need to take an arbitrary element of V⊗Wand write it as a linear combination of elements of B. But, since V⊗W=F/Z, we know that anything in V⊗Wis of the form α1δ(v1,w1)+. . .+αnδ(vn,wn)+Z. Put another way, anything in V⊗Wis of the form α1v1⊗w1+. . .+αnvn⊗wn But we can write3 vi=bγ ixγ wi=c/epsilon1 iy/epsilon1 and so our element of V⊗Wis α1(bγ 1xγ)⊗(c/epsilon1 1y/epsilon1) +. . .+αn(bγ nxγ)⊗(c/epsilon1 ny/epsilon1) Now we can apply equations (2) and (3) repeatedly to get α1bγ 1c/epsilon1 1(xγ⊗y/epsilon1) +. . .+αnbγ nc/epsilon1 n(xγ⊗y/epsilon1) This is a linear combination of the xi⊗yj’s, so we’re done. In class on Friday, we will prove that the set Bis LI. 3Summation convention!