I just started to read about tensor products and tensors and I understand that a tensor product $V \otimes W$ is a space used to replace bilinear maps $V \times W \to U$ with linear maps $V \otimes W \to U$. Now I have tried to solve a few simple exercises and I realize that I have no idea how to actually "work" with those tensor products/tensors.

Let's look at $$u_1 = \begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}, u_2 = \begin{pmatrix} 0\\ -1\\ 1 \end{pmatrix}, u_3 = \begin{pmatrix} 1\\ 0\\ -1 \end{pmatrix}$$ as a basis of $V = \mathbb{R}^3$. Let $G$ be a tensor in $V^* \otimes V^*$, given by $$G(x,y) = x_1 y_3 + x_2 y_2,$$ where $x = (x_1, x_2, x_3), y = (y_1, y_2, y_3)$ in $\mathbb{R}^3$. Express $G$ in terms of the basis $(u_i^* \otimes u_j^*)$ of $V^* \otimes V^*$.

Looking at the solution, this exercise should be really easy to solve, as we can write $$G = e_1^* \otimes e_3^* + e_2^* \otimes e_2^*.$$ Why can we write $G$ in this way? Maybe I am lacking some basic knowledge of tensors, because without the solution, I would have no idea how to even approach this exercise and even with the solution, I have no idea why it is true.

  • $\begingroup$ You notice the subscripts, right? Can you use that observation to extrapolate some simple calculational rules and then justify why they work? $\endgroup$ – anon Jun 19 '12 at 9:28
  • $\begingroup$ I have no idea where to start and what I'm trying to obtain really. $\endgroup$ – Huy Jun 19 '12 at 10:08
  • $\begingroup$ If a bilinear map $f$ separates as $f(x,y)=\alpha(x)\beta(y)$, with $\alpha,\beta\in V^\vee$ linear functionals, then the element of $V^\vee\otimes V^\vee$ representing it is in fact the pure tensor $\alpha\otimes\beta$. Using $x_i=e_i^*x$ and $y_i=e_i^*y$, since the bilinear map $G$ can be written $(x,y)\mapsto e_1^*(x)e_3^*(y)+e_2^*(x)e_2^*(x)$, it follows that the tensor representing it is $e_1^*\otimes e_3^*+e_2^*\otimes e_2^*$. $\endgroup$ – anon Jun 19 '12 at 10:17
  • $\begingroup$ Surely you have encountered situations where one set of coordinates is better suited than another to solve a problem. So you need to know how vector and tensor coordinates change. This is what the exercise is about. All you need to do is express the dual of the canonical base $(e_1^*,e_2^*,e_3^*)$ in terms of the dual base $(u_1^*,u_2^*,u_3^*)$. This means finding $a_{ij}$'s such that $e_i^* = \sum_j a_{ij} u_j^*$. Then use the bilinearity of $\otimes$: for example $e_1^*\otimes e_3^* = (\sum_j a_{1j} u_j^*) \otimes (\sum_k a_{3k} u_k^*) = \sum_{j,k} a_{1j}a_{3k} (u_j^* \otimes u_k^*)$. $\endgroup$ – AFK Jun 19 '12 at 14:22
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    $\begingroup$ @YBL: I know exactly how to change coordinates, my problem is that I don't understand why $G = e_1^* \otimes e_3^* + e_2^* \otimes e_2^*$. $\endgroup$ – Huy Jun 19 '12 at 16:11

It seems to me if you already know its representation in terms of $e_i^\ast\otimes e_j^\ast$, and you want to know it in terms of $u_i^\ast\otimes u_j^\ast$, isn't it straightforward to apply change of bases?

All you should need for that is the change of basis matrix between the bases $E=\{e_1\dots e_n\}$ and $U=\{u_1\dots u_n\}$, the matrix $A$ that computes $Av_E=v_U$ where $v_E$ means "$v$ in terms of $E$" and $v_U$ means "$v$ in terms of $U$".

$$ A=\begin{pmatrix}\frac{1}{2}&\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}\\\frac{1}{2}&\frac{1}{2}&-\frac{1}{2} \end{pmatrix} $$


$$ A^{-1}=\begin{pmatrix}1&1&0\\0&-1&1\\1&0&-1 \end{pmatrix} $$ The rest should flow from the transformation rules for tensors, right?

  • $\begingroup$ No. I don't understand how I can represent $G$ in terms of $e_i^* \otimes e_j^*$. Once I understand how to obtain this representation and why that is the correct representation, I know how to proceed. $\endgroup$ – Huy Jun 19 '12 at 16:08
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    $\begingroup$ @Huy Oh, it sounded like you were taking that for granted, and couldn't get the written question. $e_i^\ast$ is presumably the projection from $V$ to $F$ of the $i$'th coordinate. Isn't it obvious that $(x,y)\mapsto x_1 y_3 + x_2 y_2=e_1^\ast(x)e_3^\ast(y)+e_2^\ast(x)e_2^\ast(y)$ defines a multilinear map from $V\times V\rightarrow F$ which gives rise to a linear map $e_1^\ast\otimes e_3^\ast+e_2^\ast\otimes e_2^\ast:V \otimes V\rightarrow F$ matching $G$ on $V\times V$? $\endgroup$ – rschwieb Jun 19 '12 at 16:37

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