In this section the universal mapping property is used to establish a number basic isomorphisms involving tensor products.

Theorem Given vector spaces \(V_1\) and \(V_2\) over \(K\), there is a unique isomorphism, \(V_1\otimes V_2\cong V_2\otimes V_1\), such that for any \(v_1\in V_1\) and \(v_2\in V_2\), \(v_1\otimes v_2\mapsto v_2\otimes v_1\).

Proof Consider the bilinear function \(f:V_1\times V_2\mapto V_2\otimes V_1\) defined by \(f(v_1,v_2)=v_2\otimes v_1\) on elements \(v_1\in V_1\) and \(v_2\in V_2\). That \(f\) is indeed bilinear is a consequence of the bilinearity of the tensor product \(v_2\otimes v_1\). From the universal mapping property it follows that there is a linear map \(L:V_1\otimes V_2\mapto V_2\otimes V_1\) such that \(L(v_1\otimes v_2)=v_2\otimes v_1\). But likewise we could have started with a bilinear map from \(V_2\times V_1\mapto V_1\otimes V_2\) to end up with a linear map, \(L’:V_2\otimes V_1\mapto V_1\otimes V_2\), inverse, at least on pure tensors, to \(L\). That \(L’L=\id_{V_1\otimes V_2}\) on the whole of \(V_1\otimes V_2\) and indeed that \(LL’=\id_{V_2\otimes V_1}\) on the whole of \(V_2\otimes V_1\) follows since both \(L\) and \(L’\) are linear and the tensor product space is spanned by the linear sum of pure products.\(\blacksquare\)

Note that while \(V_1\otimes V_2\cong V_2\otimes V_1\) it is certainly not the case that \(v_1\otimes v_2=v_2\otimes v_1\) for arbitrary \(v_1\in V_1\) and \(v_2\in V_2\). The generalisation of the this to a tensor product of \(r\) vector spaces says that for any permutation \(\sigma\) of the numbers \(1,\dots,r\) there is a unique isomorphism,
\begin{equation}
V_1\otimes\dots\otimes V_r\cong V_{\sigma(1)}\otimes\dots\otimes V_{\sigma(r)},
\end{equation}
such that, \(v_1\otimes\dots\otimes v_r\mapsto v_{\sigma(1)}\otimes\dots\otimes v_{\sigma(r)}\), for any \(v_i\in V_i\).

Now let us consider associativity of the tensor product.

Theorem For vector spaces \(V_1\), \(V_2\) and \(V_3\) over \(K\) there is a unique isomorphism,
\begin{equation}
(V_1\otimes V_2)\otimes V_3\cong V_1\otimes(V_2\otimes V_3),
\end{equation}
such that for any \(v_1\in V_1\), \(v_2\in V_2\) and \(v_3\in V_3\), \((v_1\otimes v_2)\otimes v_3\mapsto v_1\otimes(v_2\otimes v_3)\).

Proof The function \(f:V_1\times V_2\times V_3\mapto(V_1\otimes V_2)\otimes V_3\), given by \(f(v_1,v_2,v_3)=(v_1\otimes v_2)\otimes v_3\) is clearly trilinear so by the universal mapping property we have linear map \(V_1\otimes V_2\otimes V_3\mapto(V_1\otimes V_2)\otimes V_3\) such that \(v_1\otimes v_2\otimes v_3\mapsto(v_1\otimes v_2)\otimes v_3\). Choosing bases for \(V_1\), \(V_2\) and \(V_3\) it’s clear that this maps one basis to another and so is an isomorphism. Similarly, we find \(V_1\otimes V_2\otimes V_3\cong V_1\otimes(V_2\otimes V_3)\) and the result follows.\(\blacksquare\)

Theorem For vector spaces \(V_1\), \(V_2\) and \(V_3\) over \(K\) there is a unique isomorphism,
\begin{equation}
V_1\otimes(V_2\oplus V_3)\cong(V_1\otimes V_2)\oplus(V_1\otimes V_3),
\end{equation}
such that for any \(v_1\in V_1\), \(v_2\in V_2\) and \(v_3\in V_3\), \(v_1\otimes (v_2,v_3)\mapsto (v_1\otimes v_2,v_1\otimes v_3)\).

Proof Here we need a bilinear function \(V_1\times(V_2\oplus V_3)\mapto(V_1\otimes V_2)\oplus(V_1\otimes V_3)\) so let us define a function \(f\) according to \(f(v_1,(v_2,v_3))=(v_1\otimes v_2,v_1\otimes v_3)\). That this is bilinear is demonstrated as follows,
\begin{align*}
f(av_1+bv_1′,(v_2,v_3))&=((av_1+bv_1′)\otimes v_2,(av_1+bv_1′)\otimes v_3)\\
&=(av_1\otimes v_2+bv_1’\otimes v_2,av_1\otimes v_3+bv_1’\otimes v_3)\\
&=(av_1\otimes v_2,av_1\otimes v_3)+(bv_1’\otimes v_2,bv_1’\otimes v_3)\\
&=a(v_1\otimes v_2,v_1\otimes v_3)+b(v_1’\otimes v_2,v_1’\otimes v_3)\\
&=af(v_1,(v_2,v_3))+b(v_1′,(v_2,v_3)),
\end{align*}
and
\begin{align*}
f(v_1,a(v_2,v_3)+b(v_2′,v_3′))&=f(v_1,(av_2+bv_2′,av_3,bv_3′))\\
&=(v_1\otimes(av_2+bv_2′),v_1\otimes(av_3+bv_3′))\\
&=(v_1\otimes av_2+v_1\otimes bv_2′,v_1\otimes av_3+v_1\otimes bv_3′)\\
&=(av_1\otimes v_2+bv_1\otimes v_2′,av_1\otimes v_3+bv_1\otimes v_3′)\\
&=(av_1\otimes v_2,av_1\otimes v_3)+(bv_1\otimes v_2′,bv_1\otimes v_3′)\\
&=a(v_1\otimes v_2,v_1\otimes v_3)+b(v_1\otimes v_2′,v_1\otimes v_3′)\\
&=af(v_1,(v_2,v_3))+bf(v_1,(v_2′,v_3′)).
\end{align*}
Then by the universal mapping property there is a linear map \(V_1\otimes(V_2\oplus V_3)\mapto(V_1\otimes V_2)\oplus(V_1\otimes V_3)\) such that \(v_1\otimes (v_2,v_3)\mapsto (v_1\otimes v_2,v_1\otimes v_3)\). Choosing bases for \(V_1\), \(V_2\) and \(V_3\) we see that this maps one basis to another so is an isomorphism.\(\blacksquare\)

More generally, we have that for a vector spaces \(U\), together with a (possibly infinite) set of spaces \(V_i\),
\begin{equation}
U\otimes(\bigoplus_iV_i) \cong\bigoplus_i(U\otimes V_i),
\end{equation}
with the isomorphism being the obvious extension of the one from the previous theorem.

Theorem For vector spaces \(V_1\) and \(V_2\) over \(K\) with respective dual spaces \(V_1^*\) and \(V_2^*\) there is a unique isomorphism,
\begin{equation}
V_1^*\otimes V_2^*\cong(V_1\otimes V_2)^*,
\end{equation}
such that, \(f_1\otimes f_2\mapsto(v_1\otimes v_2\mapsto f_1(v_1)f_2(v_2))\), for any \(v_1\in V_1\), \(v_2\in V_2\), \(f_1\in V_1^*\) and \(f_2\in V_2^*\).

Proof Define a function \(V_1^*\times V_2^*\mapto\mathcal{L}(V_1,V_2;K)\) by \((f_1,f_2)\mapsto((v_1,v_2)\mapsto f_1(v_1)f_2(v_2))\). That this is bilinear is clear so by the universal mapping property there is unique linear map \(V_1^*\otimes V_2^*\mapto\mathcal{L}(V_1,V_2;K)\) such that \(f_1\otimes f_2\mapsto((v_1,v_2)\mapsto f_1(v_1)f_2(v_2))\). Now this is a linear mapping between vector spaces of the same dimension, \(\dim V_1\dim V_2\), and moreover its image contains the bilinear functions we’ve already observed form a basis for \(\mathcal{L}(V_1,V_2;K)\), namely, \((v_1,v_2)\mapsto \alpha_{i_1}^{(1)}(v_1)\alpha_{i_2}^{(2)}(v_2)\), where the \(\alpha_{i_1}^{(1)}\) and \(\alpha_{i_2}^{(2)}\) are dual bases of \(V_1\) and \(V_2\) respectively. Thus this linear map is an isomorphism, which when combined with the isomorphism, \(\mathcal{L}(V_1,V_2;K)\cong(V_1\otimes V_2)^*\), gives us the isomorphism we sought.\(\blacksquare\)

We have of course also the obvious generalisation of this isomorphism,
\begin{equation}
V_1^*\otimes\dots\otimes V_r^*\cong(V_1\otimes\dots\otimes V_r)^*.
\end{equation}

Theorem For vector spaces \(V_1\) and \(V_2\) over \(K\), with \(V_1^*\) dual space of \(V_1\), there is a unique isomorphism,
\begin{equation}
V_1^*\otimes V_2\cong\mathcal{L}(V_1,V_2),
\end{equation}
such that, \(f_1\otimes v_2\mapsto(v_1\mapsto f_1(v_1)v_2)\), for any \(v_1\in V_1\), \(v_2\in V_2\) and \(f_1\in V_1^*\).

Proof The function defined by, \((f_1,v_2)\mapsto(v_1\mapsto f_1(v_1)v_2)\), is clearly a bilinear function from \(V_1^*\times V_2\) to \(\mathcal{L}(V_1,V_2)\). Therefore, by the universal mapping property, there is a unique linear map from \(V_1^*\otimes V_2\) to \(\mathcal{L}(V_1,V_2)\) given by, \(f_1\otimes v_2\mapsto(v_1\mapsto f_1(v_1)v_2)\). Now both \(V_1^*\otimes V_2\) and \(\mathcal{L}(V_1,V_2)\) have dimension \(\dim V_1\dim V_2\) and choosing bases \(e_i^{(1)}\) and \(e_i^{(2)}\) for \(V_1\) and \(V_2\) respectively, with \(\alpha_i^{(1)}\) the dual basis of \(V_1^*\), then this map takes the basis elements, \(\alpha_i^{(1)}\otimes e_j^{(2)}\), of \(V_1^*\otimes V_2\), to the linear maps \(v_1\mapsto\alpha_i^{(1)}(v_1)e_j^{(2)}\). Considering these maps applied to the basis elements of \(V_1\) we see that their matrix representations are the matrices with \(1\) in the \(j\)th row and \(i\)th column with zeros everywhere else. These matrices form a basis of \(\mathcal{L}(V_1,V_2)\) so we see that our linear map takes a basis to a basis and is therefore an isomorphism.\(\blacksquare\)

In the case of \(V_1=V_2=V\), it is of interest to establish the element of \(V^*\otimes V\) which corresponds to \(\id_V\). Denoting by \(e^i\) the dual basis of the basis \(e_i\) of \(V\) then this is the element \(\sum_ie^i\otimes e_i\) of \(V^*\otimes V\).

Consider the function \(V^*\times V\mapto K\) given by \((f,v)\mapsto f(v)\). This is clearly bilinear so induces a unique linear map \(V^*\otimes V\mapto K\), given by \(f\otimes v\mapsto f(v)\). This, understood as a linear map \(\mathcal{L}(V)\mapto K\), is just the trace, now given a basis-free (`canonical’) definition. To see that this really does coincide with the trace as previously encountered, consider an arbitrary element of \(V^*\otimes V\). It has the form, \(\sum_{ij}A_i^je^i\otimes e_j\), for some scalars, \(A_i^j\), and corresponds to the linear operator on \(V\) such that, \(e_k\mapsto\sum_{ij}A_i^je^i(e_k)e_j=\sum_jA_k^je_j\), that is, the linear operator represented by the matrix \(\mathbf{A}\) with elements \(A_i^j\). The trace of this linear operator is then \(\sum_{ij}A_i^je^i(e_j)=\sum_iA_i^i\) in accordance with our previous definition.

More generally, we have the notion of contraction. If in some tensor product space, \(V_1\otimes\dots\otimes V_r\), we have, \(V_j=V_i^*\), for some \(i\) and \(j\), then the contraction with respect to \(i\) and \(j\) is a linear mapping,
\begin{equation}
V_1\otimes\dots\otimes V_r\mapto\bigotimes_{k\neq i,j}^rV_r,\label{eq:abstract contraction}
\end{equation}
formed as a composition of a permutation of the tensor factors such that the \(i\) and \(j\) spaces are in the first two positions with the remaining order unchanged followed by the map formed as the tensor product of the map \(V^*\otimes V\mapto K\) discussed above tensored with the identity for the remaining factors followed by the trivial isomorphism corresponding to \(K\otimes V\cong V\).