My friend and gifted mathematical physicist David McAnally died suddenly last week. I attended David’s funeral today. It’s hard to believe that he’s gone. It hadn’t ever crossed my mind that this shy, gentle man, who loved mathematics and physics would pass on so soon. David was 44.
For the last couple of months David had been teaching me a crash course on general relativity. He’d gone through the material but decided to give me one more lecture - scheduled for last Tuesday afternoon. Unfortunately, he passed away unexpectedly that morning.
I sat in that empty lecture room last week, reminiscing and trying to get my head around this awful thing.
The first time I saw the axioms for tensor products of vector spaces I was confused to say the least. I kept thinking about why the particular axioms were chosen. Recently I have developed a better understanding of where the axioms come from.
We begin our considerations with 2 by 2 matrices [tex]A[/tex] and [tex]B[/tex] over a field [tex]\mathbb{F}[/tex], where
[tex] A= \left ( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right ) \quad \textrm{and} \quad B= \left ( \begin{array}{cc} b_{11} & b_{12} \\\ b_{21} & b_{22} \end{array} \right )[/tex]
The tensor (or Kronecker) product of [tex]A[/tex] and [tex]B[/tex] is defined as
[tex] A \otimes B = \left ( \begin{array}{cc} a_{11} B & a _{12} B \\\ a_{21} B & a_{22} B \end{array} \right ) [/tex]
[tex] = \left ( \begin{array}{cccc}a_{11} b_{11} & a_{11} b _{12} & a _{12} b_{11} & a _{12}b _{12} \\ a_{11} b_{21} & a_{11} b_{22} & a _{12} b_{21} & a _{12}b_{22} \\ a_{21} b_{11} & a_{21} b _{12} & a_{22} b_{11} & a_{22}b_{12} \\ a_{21} b_{21} & a_{21} b_{22} & a_{22} b_{21} & a_{22} b_{22} \end{array} \right )[/tex].
The tensor product of [tex]B[/tex] with [tex]A [/tex] is
[tex] B \otimes A = \left ( \begin{array}{cc} b_{11} A & b _{12}A \\ b_{21} A & b_{22}A \end{array} \right )[/tex]
[tex] = \left ( \begin{array}{cccc} b_{11} a_{11} & b_{11} a _{12} & b _{12} a_{11} & b _{12} a _{12} \\ b_{11} a_{21} & b_{11} a_{22} & b _{12} a_{21} & b _{12} a_{22} \\ b_{21} a_{11} & b_{21}a _{12} & b_{22} a_{11} & b_{22}a _{12} \\ b_{21}a_{21} & b_{21}a_{22} & b_{22} a_{21} & b_{22}a_{22} \end{array} \right )[/tex] .
Clearly, [tex]A \otimes B \ne B \otimes A[/tex] , so that the tensor product is non-commutative [1] (it is however associative).
We now consider the effect of multiplication by a scalar [tex] \nu \in \mathbb F[/tex] . We set
[tex] \nu A = \nu \left ( \begin{array}{cc} a_{11} & a _{12} \\\ a_{21} & a_{22} \end{array}\right ) = \left ( \begin{array}{cc} \nu a_{11} & \nu a _{12} \\\ \nu a_{21} & \nu a_{22} \end{array} \right) [/tex]
now
[tex] \nu A \otimes B = \left ( \begin{array}{cc} \nu a_{11} & \nu a _{12} \\\ \nu a_{21} & \nu a_{22} \end{array}\right) \otimes \left ( \begin{array}{cc} b_{11} & b _{12} \\\ b_{21} & b_{22} \end{array}\right) [/tex]
[tex] = \left ( \begin{array}{cc} \nu a_{11} B & \nu a _{12} B \\\ \nu a_{21} B & \nu a_{22} B \end{array} \right ) [/tex]
[tex] = \left ( \begin{array}{cccc} \nu a_{11} b_{11} & \nu a_{11} b _{12} & \nu a _{12} b_{11} & \nu a _{12}b _{12} \\ \nu a_{11} b_{21} & \nu a_{11} b_{22} & \nu a _{12} b_{21} & \nu a _{12}b_{22} \\ \nu a_{21} b_{11} & \nu a_{21} b _{12} & \nu a_{22} b_{11} & \nu a_{22}b_{12} \\ \nu a_{21} b_{21} & \nu a_{21} b_{22} & \nu a_{22} b_{21} & \nu a_{22} b_{22} \end{array} \right )[/tex]
[tex] = \nu \left ( \begin{array}{cccc} a_{11} b_{11} & a_{11} b _{12} & a _{12} b_{11} & a _{12}b _{12} \\ a_{11} b_{21} & a_{11} b_{22} & a _{12} b_{21} & a _{12}b_{22} \\ a_{21} b_{11} & a_{21} b _{12} & a_{22} b_{11} & a_{22}b_{12} \\ a_{21} b_{21} & a_{21} b_{22} & a_{22} b_{21} & a_{22} b_{22} \end{array} \right )[/tex]
[tex] = \nu (A \otimes B) = A \otimes \nu B [/tex] .
For arbitrary matrices [tex]A = a_{ij}, B = b_{kl}[/tex] the tensor product takes the form
[tex]A \otimes B = \left( \begin{array}{ccc} a_{11} B & \cdots & a_{1j} B \\ \vdots & \ddots & \vdots \\ a_{i1}B & \cdots & a_{ij}B\end{array} \right)[/tex]
For column vectors, [tex]u=\left( \begin{array}{c}u_1 \\\ \vdots \\\ u_j\end{array} \right), v= \left( \begin{array}{c} v_1 \\\ \vdots \\\ v_k \end{array} \right)[/tex] the tensor product is of the same general form.
A column vector is simply a [tex]k \times 1[/tex] matrix and the tensor product of [tex]u[/tex] with [tex]v[/tex] is
[tex]u \otimes v =\left( \begin{array}{c}u_1 v\\\ \vdots \\\ u_j v\end{array} \right)[/tex].
We can begin abstracting a bit now. Let [tex]V [/tex] and [tex]W[/tex] be two finite dimensional vector spaces. We define the tensor product [tex]V \otimes W[/tex] as the vector space spanned by all formal products [tex]v \otimes w[/tex] , [tex]v \in V[/tex] , [tex]w \in W [/tex] where [tex]\otimes [/tex] satisfies
(i) [tex] ( v_1 + v_2 ) \otimes w = v_1 \otimes w + v_2 \otimes w[/tex]
(ii) [tex] v \otimes ( w_1 + w_2 ) = v \otimes w_1 + v \otimes w_2 [/tex]
(iii) [tex] \nu ( v \otimes w ) = ( \nu v ) \otimes w = v \otimes ( \nu w ) [/tex],
for all [tex]v, v_1, v_2 \in V, w,w_1,w_2 \in W, \nu \in \mathbb{F}[/tex].
The first two parts of the defintion are straightforward, we want [tex]\otimes[/tex] to be linear in both the first and the second spaces (i.e. bilinear). The third axiom is the most confusing one. As you can see from the results derived from the Kronecker product of matrices, the third axiom makes sense if we are going to model tensor products algebraically.
Note that a general element of [tex]V \otimes W[/tex] is of the form [tex]\sum_{i,j} \alpha_{ij} v_i \otimes w_j[/tex] and cannot in general be expressed in the simpler form [tex]v \otimes w [/tex], for some [tex]v,w \in V[/tex].
Example, let [tex]V[/tex] be a 2-dimensionl vector space with basis [tex]\{e_1, e_2\}[/tex], then consider the space [tex]V \otimes V[/tex]. A typical element of [tex]V \otimes V[/tex] is
[tex]e_1 \otimes e_2 + e_2 \otimes e_1[/tex].
It cannot be expressed as a tensor product of two vectors, for suppose that there are vectors [tex]v, v’ \in V[/tex] such that
[tex] v \otimes v’ = e_1 \otimes e_2 + e_2 \otimes e_1 [/tex].
Now since [tex]\{e_1,e_2\}[/tex] is a basis of [tex]V[/tex] we can express [tex]v,v’[/tex] in terms of the basis vectors as
[tex]v = \alpha_1 e_1 + \alpha_2 e_2,\quad v’ =\beta_1 e_1 + \beta_2 e_2[/tex]
where [tex]\alpha_1,\alpha_2, \beta_1, \beta_2 \in \mathhbb{F}[/tex], so that
[tex]v \otimes v’ = (\alpha_1 e_1 + \alpha_2 e_2) \otimes (\beta_1 e_1 + \beta_2 e_2)[/tex]
[tex] =\alpha_1 \beta_1 \ e_1 \otimes e_1 + \alpha_1 \beta_2 \ e_1 \otimes e_2 + \alpha_2 \beta_1 \ e_2 \otimes e_1 + \alpha_2 \beta_2 \ e_2 \otimes e_2 [/tex]
[tex]= e_1 \otimes e_2 + e_2 \otimes e_1[/tex].
Now by equating coeffecients we must have [tex]\alpha_1 \beta_1=0[/tex], which implies that either [tex]\alpha_1 \beta_2 =0[/tex] or [tex]\alpha_2 \beta_1 =0[/tex], so that there is no solution, hence there are no vectors [tex]v,v’[/tex] such that [tex] v \otimes v’ = e_1 \otimes e_2 + e_2 \otimes e_1 [/tex].
The tensor product comes up a lot in quantum mechanics, for example in the addition of angular momenta.
It is also the basis of entanglement in quantum mechanics, a state such as
[tex] | \uparrow \ \rangle \otimes |\downarrow \ \rangle + |\downarrow \ \rangle\otimes |\uparrow\ \rangle [/tex]
is called an entangled state because it has no decomposition, just replace [tex]e_1[/tex] with [tex]| \uparrow \ \rangle[/tex] (spin up) and [tex]e_2[/tex] with [tex]| \downarrow \ \rangle[/tex] (spin down) in the previous argument.
A state such as
[tex] |\uparrow\ \rangle \otimes |\uparrow \ \rangle + |\downarrow\ \rangle\otimes |\uparrow \ \rangle [/tex] is not entangled because it can be expressed as [tex] (1|\uparrow \ \rangle + 1|\downarrow \ \rangle) \otimes ( 1|\uparrow \ \rangle + 0|\downarrow \ \rangle )[/tex].
[1]. To see how [tex]A \otimes B[/tex] and [tex]B \otimes A[/tex] are related, we introduce the permutation operator [tex]T[/tex] where,
[tex] T = \left ( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right )[/tex].
Note that [tex]T^2 =I[/tex], so that [tex]T = T^{-1}[/tex]. A quick calculation shows that
[tex] A \otimes B = T \cdot (B \otimes A ) \cdot T^{-1} [/tex].
