Tensor product.vector space,equivalent definitions Let $V$ be a real vector space. There are 2 definitions of $V \otimes V$. One is the set of all multilinear maps $L(V^*,V^*,R)$,and the other is the qutionet group $G/H$,where $G$ is free abelian group generated by $V \times V$,and $H$ is its subgroup generated by elements of type $(xr,y)-(x,ry)$,$(x+y,z)-(x,z)-(y,z)$ and $(x,y+z)-(x,y)-(x,z)$. 
The first definition is used in geometry while the second in module theory.
Are these two definitions somehow related, or the $\bigotimes$ sign is a mere coincidence?
 A: They're the same. The fact you want to use to show this is that if $f$ is a bilinear map from $V \times V$ to $R$, then it extends to a map on your group $G$, and vanishes on all of the elements of $H$, so it induces a linear map on the tensor product $V \otimes V$.
Knowing this, we want the linear maps $L(V \otimes V, R)$ to correspond to the bilinear maps $L(V, V, R)$. 
This is true since $(V \otimes V)^* = L(V, V, R)$, so $V \otimes V = (V \otimes V)^{**} = L(L(V, V, R), R) = L(V^*, V^*, R)$.
What's going on in the last line is basically proving that $V^* \otimes V^* = (V \otimes V)^*$. It's not hard to see. Basically, you use that linear maps on the tensor product are bilinear maps on the product, and a bilinear map with one element fixed is linear. You want a linear functional on the space of bilinear maps, which, since fixing an element in one slot makes a bilinear map linear, is equivalent to a bilinear map on linear functionals.
A: These definitions coincide if $V$ is a finite dimensional vector space. If we denote by $V\otimes V$ the tensor product via the quotient construction, then you can construct an isomorphism $\Phi\colon V\otimes V\rightarrow L(V^*,V^*,\mathbb R)$ as follows:
The natural map $V\times V\rightarrow L(V^*,V^*,\mathbb R),\; (v,w) \mapsto [(\varphi,\psi)\mapsto \varphi(v)\cdot \psi(w)]$ is bilinear and hence induces a linear map
$$\Phi\colon V\otimes V\longrightarrow L(V^*,V^*,\mathbb R),\quad v\otimes w\longmapsto [\Phi(v\otimes w)\colon (\varphi,\psi)\mapsto \varphi(v)\cdot \psi(w)].$$
If $V$ is $n$-dimensional, you can construct an inverse: Let $v_1,\dotsc,v_n$ be a basis in $V$ and $\varphi_1,\dotsc,\varphi_n$ its dual basis in $V^*$. For each $v\in V$ we have $v = \sum_{i=1}^n\varphi_i(v)\cdot v_i$ and for each $\varphi\in V^*$ we have $\varphi = \sum_{i=1}^n\varphi(v_i)\cdot \varphi_i$. Now consider the linear map
$$\Psi\colon L(V^*,V^*,\mathbb R)\longmapsto V\otimes V,\quad b\longmapsto \sum_{i,j=1}^nb(\varphi_i,\varphi_j)\cdot v_i\otimes v_j.$$
A short calculation shows that $\Phi$ and $\Psi$ are inverse to each other: For $v,w\in V$ we have
$$\begin{align*}
\Psi\bigl(\Phi(v\otimes w)\bigr) &= \sum_{i,j=1}^n \varphi_i(v)\cdot \varphi_j(w)\cdot v_i\otimes v_j\\
&= \left(\sum_{i=1}^n\varphi_i(v)\cdot v_i\right)\otimes \left(\sum_{j=1}^n \varphi_j(w)\cdot v_j\right)\\
&= v\otimes w\\ 
&= \operatorname{id}_{V\otimes V}(v\otimes w)
\end{align*}$$
Conversely, for each $b\in L(V^*,V^*,\mathbb R)$ and $\varphi,\psi\in V^*$ we have
$$\begin{align*}
\Phi\bigl(\Psi(b)\bigr)(\varphi,\psi)&= \sum_{i,j=1}^n b(\varphi_i,\varphi_j) \cdot\Phi(v_i\otimes v_j)(\varphi,\psi)\\
&= \sum_{i,j=1}^n b(\varphi_i,\varphi_j)\cdot \varphi(v_i)\cdot \psi(v_j)\\
&= b\left(\sum_{i=1}^n\varphi(v_i)\cdot \varphi_i, \sum_{j=1}^n \psi(v_j)\cdot \varphi_j\right)\\
&= b(\varphi,\psi)\\
&= \operatorname{id}_{L(V^*,V^*,\mathbb R)}(b)(\varphi,\psi).
\end{align*}$$
