Tensor Products of Z-Modules III (Tensor Product of R-Vector Spaces over Z)

Structure as an Abelian Group.

Let $V$ and $W$ be vector spaces over the field of real numbers, $\mathbb{R}$. The tensor product $V{\otimes }_{\mathbb{Z}}W$ is formed by treating $V$ and $W$ only as abelian groups (i.e., $\mathbb{Z}$-modules). The resulting object is, by definition, an abelian group whose elements are finite sums of simple tensors $v\otimes w$ for $v\in V,w\in W$. The relations are those of a $\mathbb{Z}$-module tensor product:

$$(v_1+v_2)\otimes w=v_1\otimes w+v_2\otimes w$$ $$v\otimes (w_1+w_2)=v\otimes w_1+v\otimes w_2$$ $$(nv)\otimes w=v\otimes (nw)\text{ for any integer }n\in \mathbb{Z}$$

Absence of a Natural R-Vector Space Structure.

There is no natural way to define scalar multiplication by an element $r\in \mathbb{R}$ on a simple tensor $v\otimes w$. A proposed definition such as $r\cdot (v\otimes w)=(rv)\otimes w$ is not well-defined because it must be consistent with $r\cdot (v\otimes w)=v\otimes (rw)$. However, the axioms of the tensor product over $\mathbb{Z}$ do not guarantee that $(rv)\otimes w=v\otimes (rw)$ for $r\notin \mathbb{Z}$. For example, in $\mathbb{R}{\otimes }_{\mathbb{Z}}\mathbb{R}$, the elements $(\sqrt{2}\cdot 1)\otimes 1$ and $1\otimes (\sqrt{2}\cdot 1)$ are distinct. Since the proposed scalar multiplication is ambiguous, $V{\otimes }_{\mathbb{Z}}W$ does not carry a natural $\mathbb{R}$-vector space structure.

Structure as a Q-Vector Space.

Definition (Torsion-Free and Divisible Modules). A $\mathbb{Z}$-module $A$ is torsion-free if for any non-zero integer $n$, the map $a\mapsto na$ is injective. A $\mathbb{Z}$-module $A$ is divisible if for any non-zero integer $n$, the map $a\mapsto na$ is surjective.

A $\mathbb{Z}$-module is a vector space over the rational numbers $\mathbb{Q}$ if and only if it is both torsion-free and divisible. Any $\mathbb{R}$-vector space $V$, when viewed as a $\mathbb{Z}$-module, has these properties. It is torsion-free because if $nv=0$ for $n\ne 0$, then multiplication by $1/n\in \mathbb{R}$ implies $v=0$. It is divisible because for any $v\in V$ and non-zero $n\in \mathbb{Z}$, the equation $nx=v$ has the solution $x=(1/n)v\in V$.

Theorem (Structure of the Tensor Product). Let $V$ and $W$ be vector spaces over $\mathbb{R}$. The abelian group $V{\otimes }_{\mathbb{Z}}W$ is a vector space over $\mathbb{Q}$.

Proof. Since $V$ and $W$ are $\mathbb{Q}$-vector spaces, the tensor product $V{\otimes }_{\mathbb{Z}}W$ is also a $\mathbb{Q}$-vector space. Scalar multiplication by $q=m/n\in \mathbb{Q}$ is well-defined on a simple tensor by $q(v\otimes w)=(qv)\otimes w=v\otimes (qw)$. This is well-defined because for any $v^{\prime }\in V$ such that $v=n{v}^{\prime }$, $q(v\otimes w)=m(v^{\prime }\otimes w)=(m{v}^{\prime })\otimes w=((m/n)n{v}^{\prime })\otimes w=(qv)\otimes w$. $\square$

Isomorphism and Dimension.

Theorem (Isomorphism over Q). If $V$ and $W$ are vector spaces over $\mathbb{Q}$, there is a canonical isomorphism of $\mathbb{Q}$-vector spaces:

$$V{\otimes }_{\mathbb{Z}}W\cong V{\otimes }_{\mathbb{Q}}W$$

Proof. The $\mathbb{Z}$-bilinear map $\varphi :V\times W\to V{\otimes }_{\mathbb{Q}}W$ sending $(v,w)$ to $v\otimes w$ induces a $\mathbb{Z}$-module homomorphism $\Phi :V{\otimes }_{\mathbb{Z}}W\to V{\otimes }_{\mathbb{Q}}W$. This map is also $\mathbb{Q}$-linear. The map $\psi :V\times W\to V{\otimes }_{\mathbb{Z}}W$ is not only $\mathbb{Z}$-bilinear but also $\mathbb{Q}$-bilinear, since for $q\in \mathbb{Q}$, $(qv)\otimes w=v\otimes (qw)$. This induces a homomorphism $\Psi :V{\otimes }_{\mathbb{Q}}W\to V{\otimes }_{\mathbb{Z}}W$. The maps $\Phi$ and $\Psi$ are mutually inverse. $\square$

Theorem (Dimension of the Tensor Product). Let $V$ and $W$ be $\mathbb{R}$-vector spaces with dimensions $d_V={\dim }_{\mathbb{R}}(V)$ and $d_W={\dim }_{\mathbb{R}}(W)$. The dimension of $V{\otimes }_{\mathbb{Z}}W$ as a $\mathbb{Q}$-vector space is:

$${\dim }_{\mathbb{Q}}(V{\otimes }_{\mathbb{Z}}W)=(d_V\cdot d_W)\cdot \mathfrak{c}$$

where $\mathfrak{c}=|\mathbb{R}|$ is the cardinality of the continuum.

Proof. The dimension of $\mathbb{R}$ as a vector space over $\mathbb{Q}$ is ${\dim }_{\mathbb{Q}}(\mathbb{R})=\mathfrak{c}$. For any $\mathbb{R}$-vector space $V$, its dimension over $\mathbb{Q}$ is ${\dim }_{\mathbb{Q}}(V)={\dim }_{\mathbb{R}}(V)\cdot {\dim }_{\mathbb{Q}}(\mathbb{R})=d_V\cdot \mathfrak{c}$. Using the isomorphism $V{\otimes }_{\mathbb{Z}}W\cong V{\otimes }_{\mathbb{Q}}W$ and the rule for the dimension of a tensor product of vector spaces:

$$\begin{array}{rl}{\dim }_{\mathbb{Q}}(V{\otimes }_{\mathbb{Z}}W)& ={\dim }_{\mathbb{Q}}(V{\otimes }_{\mathbb{Q}}W)\\ & ={\dim }_{\mathbb{Q}}(V)\cdot {\dim }_{\mathbb{Q}}(W)\\ & =(d_V\cdot \mathfrak{c})\cdot (d_W\cdot \mathfrak{c})\\ & =(d_V\cdot d_W)\cdot {\mathfrak{c}}^2\\ & =(d_V\cdot d_W)\cdot \mathfrak{c}\end{array}$$

The final step uses the fact that ${\mathfrak{c}}^2=\mathfrak{c}$ in cardinal arithmetic. $\square$

Summary and Examples.

Example ($\mathbb{R}{\otimes }_{\mathbb{Z}}\mathbb{R}$). Let $V=\mathbb{R}$ and $W=\mathbb{R}$, so $d_V=1$ and $d_W=1$. The tensor product $\mathbb{R}{\otimes }_{\mathbb{Z}}\mathbb{R}$ is a $\mathbb{Q}$-vector space with dimension:

$${\dim }_{\mathbb{Q}}(\mathbb{R}{\otimes }_{\mathbb{Z}}\mathbb{R})=(1\cdot 1)\cdot \mathfrak{c}=\mathfrak{c}$$

Example (${\mathbb{R}}^2{\otimes }_{\mathbb{Z}}{\mathbb{R}}^3$). Let $V={\mathbb{R}}^2$ and $W={\mathbb{R}}^3$, so $d_V=2$ and $d_W=3$. In contrast to the $\mathbb{R}$-vector space ${\mathbb{R}}^2{\otimes }_{\mathbb{R}}{\mathbb{R}}^3$, which has dimension 6, the tensor product over $\mathbb{Z}$ is a $\mathbb{Q}$-vector space with dimension:

$${\dim }_{\mathbb{Q}}({\mathbb{R}}^2{\otimes }_{\mathbb{Z}}{\mathbb{R}}^3)=(2\cdot 3)\cdot \mathfrak{c}=6\mathfrak{c}=\mathfrak{c}$$