Tensor product of vector spaces

Tensor product of vector spaces Construction Let V,W be vector spaces over K = R or C. Let F denote the vector space freely generated by the set V W and let N F denote the subspace spanned by the elements of the form (αv 1 +v 2,w) α(v 1,w) (v 2,w), (v,αw 1 +w 2 ) α(v,w 1 ) (v,w 2 ), (1) where v,v 1,v 2 V, w,w 1,w 2 W and α K. By definition, the tensor product of V and W is the quotient vector space V W := F/N. Elements For v V and w W, denote v w := [(v,w)] theequivalence class of (v,w). These are special elementsof V W, called puretensor products. Since the pairs (v,w) span F, the pure tensor products v w span V W. 1. Pure tensor products fulfil (αv 1 +v 2 ) w = α(v 1 w)+v 2 w, v (αw 1 +w 2 ) = α(v w 1 )+v w 2. (2) Proof. (αv 1 +v 2 ) w = [(αv 1 +v 2,w)] def of = [α(v 1,w)+(v 2,w)] choose another representative of the same class = α[(v 1,w)]+[(v 2,w)] def of linear structure on F/N = α(v 1 w)+v 2 w. 2. Every element of V W is a finite sum of pure tensor products. Proof. The elements of F are finite sums of the form i α i(v i,w i ) for some α i K, v i V and w i W. Passing to classes, we find that the elements of V W are of the form [ ] α i(v i,w i ) = α i[(v i,w i )] def of linear structure on F/N i i = i α i(v i w i ) notation v w = i (α iv i ) w i. property 2 1

Universal property Let X be a further vector space. For every bilinear mapping f : V W X, there exists a unique linear mapping f : V W X such that f(v w) = f(v,w) for all v V and w W. (3) Proof. (Existence) Define a linear mapping ˆf : F X by ( ) ˆf α i(v i,w i ) := α if(v i,w i ). 1 i i Bilinearity of f implies ˆf(N) = 0: ˆf ( (αv 1 +v 2,w) α(v 1,w) (v 2,w) ) = f(αv 1 +v 2,w) αf(v 1,w) f(v 2,w) = 0, and similarly for the other type of spanning vectors. Hence, ˆf descends to a linear mapping f : F/N V W X, f([x]) := ˆf(x). By construction, f(v w) = f ( [(v,w)] ) notation v w = ˆf ( (v,w) ) def of f = f(v,w). def of ˆf (Uniqueness) Since V W is spanned by pure tensor products, every linear mapping is determined by its values on such elements. Tensor product of operators For every pair of linear operators A on V and B on W, there exists a unique linear operator A B on V W such that Proof. Define a mapping (A B)(v w) = (Av) (Bw). f A,B : V W V W, f A,B (v,w) := (Av) (Bw). f A,B is bilinear: f A,B (αv 1 +v 2,w) = ( A(αv 1 +v 2 ) ) (Bw) def of f A,B = (αav 1 +Av 2 ) (Bw) A is linear = α(av 1 ) (Bw)+(Av 2 ) (Bw) property (2) 1 ˆf is the linear extension of the mapping defined on the basis V W by f. 2

= αf A,B (v 1,w)+f A,B (v 2,w), def of f A,B and similarly for a linear combination in the second argument. Hence, by the universal property of the tensor product, there exists a unique linear mapping f A,B : V W V W such that Put A B := f A,B. f A,B (v w) = f A,B (v,w) (Av) (Bw). Remark: There are many operators on V W which are not of the form A B with A and B being operators on V and W, respectively. For example, if W = V, the mapping V V V V defined by (v,w) (w,v) induces an operator on V V (check this) which is not of that form. Tensor product of linear functionals For every pair of linear functionals α V and β W, there exists a unique linear functional α β (V W) such that (α β)(v w) = α(v)β(w) Proof. Check that the mapping f α,β : V W K, f α,β (v,w) := α(v)β(w) is bilinear and put α β := f α,β. Remark: Consider the mapping f : V W (V W), f(α,β) := α β. One can check that it is bilinear. Hence, it induces a linear mapping f : V W (V W) and this mapping is defined by the condition f(α β) = α β α V, β W. This equation has to be read with care: on the left hand side, the tensor sign has its genuine meaning as the class of the pair (α,β), whereas on the right hand side, it stands for the tensor product of linear functionals defined above. In particular, f is not the identical mapping (it cannot be, because it runs between distinct spaces). However, one can check that f is an isomorphism. Therefore, one may use it to identify V W with (V W) in a natural way. This justifies (a posteriori) the use of the same notation for the elements of these two spaces. Product bases Let {v i : i I} and {w i : j J} be bases in V and W, respectively. Then, {v i w j : i I,j J} is a basis in V W. In particular, dim(v W) = dim(v)dim(w). 3

Proof. Denote B := {v i w j : i I,j J}. It is clear that the linear span of B contains all pure tensor products v w. Hence, B spans V W. To see that B is linearly independent, consider the dual bases {α i : i I} in V and {β j : j J} in W, respectively. They are defined by the conditions α k (v i ) = δ ki, β l (w j ) = δ lj for all i,k I and j,l J. Given numbers c ij K, i I, j J, with only finitely many c ij 0, such that i,j c ijv i w j = 0, then ) 0 = (α k β l ) ( i,j c ijv i w j = c ij(α k β l )(v i w j ) = c ijα k (v i )β l (w j ) = c kl i,j i,j for all k I, l J. Thus, B is linearly independent, indeed. Tensor product of Hilbert spaces Construction Let H and K be Hilbert spaces over C. 2 Let H alg K denote the tensor product as vector spaces, i.e., H alg K = F/N, where F is the vector space freely generated by H K and N is the subspace spanned by the elements of the form (1). The scalar products, H on H and, K on K define a mapping ŝ : F F K by ŝ ( (ϕ,ψ),(ϕ,ψ ) ) := ϕ,ϕ H ψ,ψ K and extension by antilinearity in the first argument and linearity in the second one. I.e., ( ŝ α i(ϕ i,ψ i ), ) i j α j(ϕ j,ψ j) = i,j α iα j ϕ i,ϕ j H ψ i,ψ j K. Antilinearity of, H and, K in the first argument implies ŝ(n F) = 0: ŝ ( (αϕ 1 +ϕ 2,ψ) α(ϕ 1,ψ) (ϕ 2,ψ),(ϕ,ψ ) ) = αϕ 1 +ϕ 2,ϕ H ψ,ψ K α ϕ 1,ϕ H ψ,ψ K ϕ 2,ϕ H ψ,ψ K = 0, and similarly for the other type of spanning vectors. By an analogous argument, linearity of, H and, K in the second argument implies ŝ(f N) = 0. It follows that ŝ descends to a mapping, : (F/N) (F/N) (H K) (H K) C, [x],[y] := ŝ(x,y), 2 To treat the case of real Hilbert spaces, omit complex conjugation in what follows. 4

which is antilinear in the first argument and linear in the second one. We compute ϕ ψ,ϕ ψ = [(ϕ,ψ)],[(ϕ,ψ )] notation ϕ ψ = ŝ ( ϕ,ψ),(ϕ,ψ ) ) def of, = ϕ,ϕ H ψ,ψ K. def of ŝ (4) We claim that, is a scalar product. It remains to show that 1. [x],[y] = [y],[x] : by (anti)linearity, it suffices to check this for pure tensor products, ϕ ψ,ϕ ψ = ϕ,ϕ H ψ,ψ K = ϕ,ϕ H ψ,ψ K = ϕ ψ,ϕ ψ ; 2. [x],[x] = 0 implies [x] = 0: We can write [x] = i α iϕ i ψ i for some α i C and nonzero ϕ i H, ψ i K. By means of the orthonormalization procedure we may turn the ϕ i into an orthonormal system in H. Then, i α iϕ i ψ i, j α jϕ j ψ j = i,j α iα j ϕ i ψ i,ϕ j ψ j (anti)linearity = i,j α iα j ϕ i,ϕ j H ψ i,ψ j K Fml (4) = i α i ψ i 2 K. ϕ i,ϕ j H = δ ij It follows that α i = 0 and hence [x] = 0. Now, having a scalar product on H alg K, we can define the tensor product of Hilbert spaces H K to be the completion of H alg K in the corresponding norm. Remarks. 1. For pure tensor products, the norm square fulfils ϕ ψ 2 H K = ϕ ψ,ϕ ψ = ϕ,ϕ H ψ,ψ K = ϕ 2 H ψ 2 K and hence the norm fulfils ϕ ψ H K = ϕ H ψ K (5) 2. If both H and K have finite dimension, H alg K is already complete. Hence, in this situation, H K = H alg K. 5

Tensor product of operators For every pair of bounded linear operators A on H and B on K, there exists a unique bounded linear operator A B on H K such that (A B)(ϕ ψ) = (Aϕ) (Bψ) for all ϕ H, ψ K. (6) Proof. By the universal property of the tensor product of vector spaces, there exists a unique linear operator A B on H alg K satisfying (6). One can show that it is bounded and hence continuous, see Exercise 1 of the course. As a consequence, it can be extended uniquely to a bounded linear operator on H K, denoted by the same symbol. Since pure tensor products belong to H alg K, the extension fulfils (6). Since it is uniquely determined by its restriction to H alg K and since the latter is uniquely determined by (6), A B is uniquely determined by (6). Orthonormal product bases Let {φ i : i I} and {ξ j : j J} be orthonormal bases in H and K, respectively. Then, {φ i ξ j : i I,j J} is an orthonormal basis in H K. Proof. Denote B := {φ i ξ j : i I,j J}. B is orthonormal and hence linearly independent: φ i ξ j,φ k ξ l,= φ i,φ k ξ j,ξ l = δ ik δ jl = δ (i,j),(k,l). The span of B is dense in H K: let B denote the closure of the span of B. We have to show B = H K. Since pure tensor products span the dense subspace H alg K, it suffices to show that B contains ϕ ψ for all ϕ H and ψ K. Given ϕ and ψ, denote Then, ϕ n := i n φ i,ϕ φ i, ψ n := j n ξ j,ψ ξ j. ϕ n ψ n = i,j n φ i,ϕ ξ j,ψ φ i ξ j so that ϕ n ψ n B for all n. By the triangle inequality, we have ϕ n ψ n ϕ ψ ϕ n ψ n ϕ n ψ + ϕ n ψ ϕ ψ. Since ϕ n 2 converges, it is bounded for large n. Hence, by (5), Analogously, Hence, ϕ n ψ n ϕ n ψ 2 = ϕ n (ψ n ψ) 2 = ϕ n 2 ψ n ψ 2 n 0. ϕ n ψ ϕ ψ 2 = (ϕ n ϕ) ψ 2 = ϕ n ϕ 2 ψ 2 n 0. This shows that ϕ ψ B, as asserted. ϕ n ψ n n ϕ ψ. 6

Tensor product of L 2 -spaces Let C 0 (R 3 ) denote the continuous functions on R 3 with compact support. The mapping F : C 0 (R 3 ) C 0 (R 3 ) C 0 (R 3 R 3 ), ( F(ϕ,ψ) ) ( x, y) := ϕ( x)ψ( y), induces a Hilbert space isomorphism from L 2 (R 3 ) L 2 (R 3 ) onto L 2 (R 3 R 3 ). Proof. F is bilinear: ( F(αϕ1 +ϕ 2,ψ) ) ( x, y) = (αϕ 1 +ϕ 2 )( x)ψ( y) = αϕ 1 ( x)ψ( y)+ϕ 2 ( x)ψ( y) = α ( F(ϕ 1,ψ) ) ( x, y)+ ( F(ϕ 2,ψ) ) ( x, y). = ( αf(ϕ 1,ψ)+F(ϕ 2,ψ) ) ( x, y), and similarly for a linear combination in the second argument. Moreover, for all ϕ,ϕ,ψ,ψ C 0 (R 3 ), we have ( ) F(ϕ,ψ),F(ϕ,ψ ) = ( F(ϕ,ψ) ( x, y) F(ϕ,ψ ) ) ( x, y)d 3 xd 3 y R 3 R 3 = ϕ( x) ψ( y) ϕ ( x)ψ ( y)d 3 xd 3 y R 3 R 3 = ϕ( x) ϕ ( x)d 3 x ψ( y) ψ ( y)d 3 y R 3 R 3 = ϕ,ϕ ψ,ψ. (7) As a consequence of (7), F is bounded and hence continuous. It follows that it extends to a continuous bilinear mapping F : L 2 (R 3 ) L 2 (R 3 ) L 2 (R 3 R 3 ). By the universal property of the tensor product of Hilbert spaces, there exists a unique continuous linear mapping F : L 2 (R 3 ) L 2 (R 3 ) L 2 (R 3 R 3 ) such that F(ϕ ψ) = F(ϕ,ψ) for all ϕ,ψ L 2 (R 3 ). By continuity and by (7), F is isometric. To see that F is an isomorphism of Hilbert spaces, it remains to show that it is surjective. By isometry, the image im( F) of F is closed. Hence, for proving surjectivity, it suffices to show that im( F) is dense in L 2 (R 3 R 3 ). To see this, let ξ C 0 (R 3 R 3 ) be orthogonal to im( F). Then, for all ϕ,ψ C 0 (R 3 ), 0 = F(ϕ ψ),ξ = F(ϕ,ψ),ξ = ϕ( x) ψ( y) ξ( x, y)d 3 xd 3 y R 3 R 3 7

We read off that the mapping ( ) = ϕ( x) R 3 ψ( y) ξ( x, y)d 3 y d 3 x. R 3 x ψ( y) ξ( x, y)d 3 y R 3 is orthogonal to all elements of C 0 (R 3 ). Being continuous, it must vanish then. Hence, for all x, the mapping y ξ( x, y) is orthogonal to all elements of C 0 (R 3 ). As before, being continuous, it must vanish then. Thus, ξ = 0. This proves that im( F) is dense in L 2 (R 3 R 3 ). 8