# 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 |