tensor product and dual spaces

Let $k$ be a field and $V$ be a vector space over $k$. Recall that

$$V^{*}=\{f:V\to k|f\text{ is linear}\}$$

denotes the dual space of $V$ (which is also a vector space over $k$).

Proposition. Let $V$ and $W$ be vector spaces. Consider the map $\varphi :V^{*}\otimes W^{*}\to {(V\otimes W)}^{*}$ such that

$$\varphi (f\otimes g)(v\otimes w)=f(v)g(w).$$

Then $\varphi$ is a monomorphism. Moreover if one of the spaces $V$, $W$ is finite dimensional, then $\varphi$ is an isomorphism.

Proof. One can easily check that $\varphi$ is a well defined linear map, thus it is sufficient to show that $\mathrm{Ker}(\varphi )=0$. So assume that $F\in V^{*}\otimes W^{*}$ is such that $\varphi (F)=0$. It is clear that $F$ can be (uniquely) expressed in the form

$$F=\sum _{i,j}{\alpha }_{i,j}{f}_i\otimes g_j,$$

where $(f_i)$ is a basis of $V^{*}$, $(g_j)$ is a basis of $W^{*}$ and ${\alpha }_{i,j}\in k$. Then for any $v\in V$ and $w\in W$ we have:

$$0=\varphi (F)(v\otimes w)=\varphi (\sum _{i,j}{\alpha }_{i,j}{f}_i\otimes g_j)(v\otimes w)=$$

$$=\sum _{i,j}{\alpha }_{i,j}\varphi (f_i\otimes g_j)(v\otimes w)=\sum _{i,j}{\alpha }_{i,j}{f}_i(v)g_j(w).$$

Since $w\in W$ is arbitrary then we can write this equality in the form:

$$0=\sum _{i,j}{\alpha }_{i,j}{f}_i(v)g_j=\sum _j(\sum _i{\alpha }_{i,j}{f}_i(v))g_j$$

and since $(g_j)$ are linearly independent we obtain that ${\sum }_i{\alpha }_{i,j}{f}_i(v)=0$ for all $j$. Again since $v\in V$ was arbitrary we obtain that ${\sum }_i{\alpha }_{i,j}{f}_i=0$ for all $j$. Now since $(f_i)$ are linearly independent we obtain that ${\alpha }_{i,j}=0$ for all $i,j$. Thus $F=0$.

Now assume that . Let ${(v_i)}_{i=1}^q$ be a basis of $V$ and let ${(v_i^{*})}_{i=1}^q$ be an induced basis of $V^{*}$. Moreover let ${(w_p)}_{p\in P}$ be a basis of $W$. We wish to show that $\varphi$ is onto, so let $f:V\otimes W\to k$ be an element of ${(V\otimes W)}^{*}$. Define $F\in V^{*}\otimes W^{*}$ by the formula:

$$F=\sum _{i=1}^q{v}_i^{*}\otimes g_i,$$

where $g_i:W\to k$ is such that $g_i(w_p)=f(v_i\otimes w_p)$. Then for any $v_j$ from ${(v_i)}_{i=1}^q$ and for any $w_p$ from ${(w_p)}_{p\in P}$ we have:

$$\varphi (F)(v_j\otimes w_p)=\varphi (\sum _{i=1}^q{v}_i^{*}\otimes g_i)(v_j\otimes w_p)=\sum _{i=1}^q\varphi (v_i^{*}\otimes g_i)(v_j\otimes w_p)=$$

$$=\sum _{i=1}^q{v}_i^{*}(v_j)g_i(w_p)=g_j(w_p)=f(v_j\otimes w_p)$$

and thus $\varphi (F)=f$. $\mathrm{\square }$

Remark. The map $\varphi$ from the previous proposition is very important in studying algebras and coalgebras (more precisly it is an essence in defining dual (co)algebras). Unfortunetly $\varphi$ does not have to be an isomorphism in general. Nevertheless, the spaces ${(V\otimes W)}^{*}$ and $V^{*}\otimes W^{*}$ are always isomorphic (see this entry (http://planetmath.org/TensorProductOfDualSpacesIsADualSpaceOfTensorProduct) for more details).

Title	tensor product and dual spaces
Canonical name	TensorProductAndDualSpaces
Date of creation	2013-03-22 18:31:51
Last modified on	2013-03-22 18:31:51
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	6
Author	joking (16130)
Entry type	Theorem
Classification	msc 15A69