# tensor product and dual spaces

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

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

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

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

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

Proof. One can easily check that $\phi$ is a well defined linear map, thus it is sufficient to show that $\mathrm{Ker}(\phi)=0$. So assume that $F\in V^{*}\otimes W^{*}$ is such that $\phi(F)=0$. 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:

 $0=\phi(F)(v\otimes w)=\phi(\sum_{i,j}\alpha_{i,j}f_{i}\otimes g_{j})(v\otimes w)=$
 $=\sum_{i,j}\alpha_{i,j}\phi(f_{i}\otimes g_{j})(v\otimes w)=\sum_{i,j}\alpha_{% i,j}f_{i}(v)g_{j}(w).$

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 $\mathrm{dim}_{k}V=q<+\infty$. 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 $\phi$ 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:

 $\phi(F)(v_{j}\otimes w_{p})=\phi(\sum_{i=1}^{q}v_{i}^{*}\otimes g_{i})(v_{j}% \otimes w_{p})=\sum_{i=1}^{q}\phi(v_{i}^{*}\otimes g_{i})(v_{j}\otimes w_{p})=$
 $=\sum_{i=1}^{q}v_{i}^{*}(v_{j})g_{i}(w_{p})=g_{j}(w_{p})=f(v_{j}\otimes w_{p})$

and thus $\phi(F)=f$. $\square$

Remark. The map $\phi$ from the previous proposition is very important in studying algebras   and coalgebras (more precisly it is an essence in defining dual (co)algebras). Unfortunetly $\phi$ 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 TensorProductAndDualSpaces 2013-03-22 18:31:51 2013-03-22 18:31:51 joking (16130) joking (16130) 6 joking (16130) Theorem msc 15A69