general expression for isomorphism of tensor product (I am still waiting for an answer to the following question. Thank you.)
While I was reading postings relating to tensors, I came across the following explanation from Tensors as matrices vs. Tensors as multi-linear maps 
"Let $V$ be a (finite-dimensional) real vector space.  A $(1, 1)$-tensor over $V$ is any of the following equivalent objects:


*

*A linear transformation $V \to V$,

*An element of $V^{\ast} \otimes V$,

*A linear map $V \otimes V^{\ast} \to \mathbb{R}$."


Can anybody kindly show the general expression (all three above) for general types of tensor $(n,m)$ other than $(1,1)$-tensor?  The order of $(n,m)$ for $V$ and $V^*$ is always confusing to me.  Thank you in advance.  
 A: If V is finite dimensional and U, W are any vector spaces, then there is an equivalence:


*

*Linear maps $W \otimes V \rightarrow U$

*Linear maps $W \rightarrow U \otimes V^{\star}$
Going from 1. to 2. uses the finite dimension: If $\{v_i\}$ is a basis for $V$ and $\{v_i^*\}$ is the dual basis, then a map $\phi:W \otimes V \rightarrow U$ becomes $\tilde{\phi}: W \rightarrow U \otimes V^{\star}$ via $w \mapsto \sum_i \phi(w \otimes v_i) \otimes v_i^*$.
Going from 2. to 1. works more generally: a map $W \rightarrow U \otimes V^*$ gives $W \otimes V \rightarrow U \otimes V^* \otimes V \rightarrow U$, where the last arrow is contraction on the last two spaces.
Anyway, what this means is you can shuffle things around when you are talking about maps between tensor products, these are all the same: (for finite dimensional $V$)
-A linear map $V^{* \otimes m} \otimes V^{\otimes n} \rightarrow \mathbb{R}$
-A linear map $V^{* \otimes m} \rightarrow V^{*\otimes n}$
-A linear map $V^{\otimes n} \rightarrow V^{\otimes m}$
-A linear map $\mathbb{R} \rightarrow V^{* \otimes n} \otimes V^{\otimes m}$
-A tensor of type $(n,m)$
Note that a linear map $\mathbb{R} \rightarrow V$ can be thought of as the same thing as a vector in $V$ (so your second bullet point above is a case of this equivalence). Also, as for the $(m,n)$ convention, you're just gonna have to memorize it.
