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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?

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  • $\begingroup$ $G$ should be the vector space with basis the elements of $V\times V$. $H$ should be the subspace generated by elements of the form $(rx,y) - r(x,y)$, $(rx,y) - (x,ry)$, $(x+y,z) - (x,z) - (y,z)$, $(x,y+z) - (x,y) - (x,z)$ for $x,y,z\in V$ and $r\in \mathbb R$ (you were missing one relation). $\endgroup$ – Claudius Sep 11 '15 at 9:29
  • $\begingroup$ Yes, thank you. I took this definition from Louis Rowen's book,so he is is also missing it.Please,did you consider my last two questions about $\Phi$ below? $\endgroup$ – user122424 Sep 11 '15 at 14:11
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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.

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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 $$\begin{align*} \Phi\bigl(\Psi(b)\bigr)(\varphi,\psi)&= \sum_{i,j=1}^n b(\varphi_i,\varphi_j) \cdot\Phi(v_i\otimes v_j)(\varphi,\psi)\\ &= \sum_{i,j=1}^n b(\varphi_i,\varphi_j)\cdot \varphi(v_i)\cdot \psi(v_j)\\ &= b\left(\sum_{i=1}^n\varphi(v_i)\cdot \varphi_i, \sum_{j=1}^n \psi(v_j)\cdot \varphi_j\right)\\ &= b(\varphi,\psi)\\ &= \operatorname{id}_{L(V^*,V^*,\mathbb R)}(b)(\varphi,\psi). \end{align*}$$

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  • $\begingroup$ You should also show that $\Phi$ is well-defined, i.e. that it does not depend on the choice of representatives and that it vanishes on $H$. $\endgroup$ – user122424 Sep 11 '15 at 9:07
  • $\begingroup$ $\Phi$ is induced by a bilinear map, which is immediate from the construction of $V\otimes V$. It’s not that $\Phi$ is implicitly assumed to be well-defined. $\endgroup$ – Claudius Sep 11 '15 at 9:34
  • $\begingroup$ Sorry,my English is not good enough to understand. We do not need that $\Phi$ is well-defined or is it a consequence of the construction of $V\otimes V$?How precisely this follows?Is it immediate that $\Phi$ is $0$ on $H$? $\endgroup$ – user122424 Sep 11 '15 at 9:46
  • $\begingroup$ I mean that you give formula for $\Phi$, so one should show that it is well-defined, or I'm false? $\endgroup$ – user122424 Sep 11 '15 at 11:38
  • $\begingroup$ It is a consequence of $V\otimes V$ that $\Phi$ (being induced from a bilinear map) is well-defined. In general, you have the following universal property of $V\otimes V$: If $f\colon V\times V\rightarrow W$ is a bilinear map into a vector space $W$, there exists a unique linear map $\hat f\colon V\otimes V\rightarrow W$ such that $f(v,w) = \hat f(v\otimes w)$ for all $v,w\in V$. To construct $\hat f$, you define $h\colon G\rightarrow W$ by $h(v,w) = f(v,w)$ for all $(v,w)\in V\times V$ and extend $\mathbb R$-linearly. Since $f$ is bilinear, $h$ vanishes on $H$ and hence induces $\hat f$. $\endgroup$ – Claudius Sep 11 '15 at 18:17

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