Different definitions of a tensor

Question

Some Linear Algebra textbooks defines tensors as the k-linear map: \begin{equation}T:V^{\times p}\to\mathbb{F}\end{equation} While others define it using the dual: \begin{equation}T:V^{\times p}\times (V^*)^{\times q}\to\mathbb{F}\end{equation}

How does one differ from another, and why does one use the dual? I know it works, but fundamentally, why?

Answer 1

Here's a reasonably general way to do it. We'll start with $p$ modules all over the same commutative ring $R$ with identity. [The ring could possibly be a field $F$ and we could, if we wished, fix a particular vector space $V$ and take any of the modules to be $V$ or its dual.]I'll write out the details for the case $p=2$ and let you write out what happens for general $p$. Let $U$ and $W$ be $R-$modules. Let $M$ be the set of functions from $U$ X $W$ to $R$ that are 0 except for finely many $(u,w).$ Then, with pointwise addition and scalar multiplication,$M$ is an $R-$module.Let $G$ be the subset of $M$ consisting of all members of the form $r(u,w)-(ru,w)$ or $r(u,w)-(u,rw)$ or $r(u_1+u_2,w)-r(u_1,w)-r(u_2,w)$ or $r(u,w_1+w_2)-r(u,w_1)-r(u,w_2)$. Let $H$ be the set of all finite $R-$linear combinations of members $G.$ Then $H$ is an $R-$ submodule of$M.$ Define $$U \otimes_R W= \text {the quotient $R-$ module } M/H.$$

Answer 2

The space of $(q,p)$-tensors is isomorphic to the space of multilinear maps \begin{equation} V^{\times p}\times (V^*)^{\times q}\to\mathbb{F} \end{equation} For example, $V$ is naturally isomorphic to the space of linear maps $$ V^*\to\mathbb F$$ but you cannot do this with any of the $(q,p)$ for $q=0$. \begin{equation} V^{\times p}\to\mathbb{F} \end{equation} because these are canonically isomorphic to $(V^*)^p$.