4
$\begingroup$

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.

  • 0
    @anon: I don't really understand your comment yet, I'll try again later.2012-06-19

1 Answers 1

1

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} $

and

$ 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?

  • 1
    @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 $V\times V$?2012-06-19