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edited Jan 25, 2018 at 9:47

Ok, I think I have something that works. It seems to me that the Haagerup tensor norm coincides in this case with the projective norm. If I am not mistaken, the Haagerup tensor norm of a tensor $v$ can be written as infimum of $\sqrt{\sum_{i} \|x_i\|^2}\cdot \sqrt{\sum_{i} \|y_{i}\|^2}$ over all possible decompositions $v=\sum_{i} x_i \otimes y_i$. By working with $\frac{x_i}{c_i}$ and $c_i x_i$ for appropriate numbers $c_i$, we see that it is equal toIt seems to me that the Haagerup tensor norm coincides in this case with the projective norm. If I am not mistaken, the Haagerup tensor norm of a tensor $v$ can be written as infimum of $\sqrt{\sum_{i} \|x_i\|^2}\cdot \sqrt{\sum_{i} \|y_{i}\|^2}$ over all possible decompositions $v=\sum_{i} x_i \otimes y_i$. By working with $\frac{x_i}{c_i}$ and $c_i x_i$ for appropriate numbers $c_i$, we see that it is equal to the projective norm.

EDIT: I will only compute the projective norm.

Since everything is real, we can work with $\mathbb{R}^2 \widehat{\otimes} \mathbb{R}^2$, where $\mathbb{R}^2$ is endowed with the $\ell_{\infty}$-norm. This is not very good, as the projective tensor product worksinteracts nicely forwith the $\ell_1$ norm. However, $\ell_1^{2}$ and $\ell_{\infty}^2$ are isometric, so we can work with $\ell_1^2$. The resultThis isometry is given by $\ell_{\infty}^2 \ni (x_1,x_2) \mapsto (\frac{x_1+x_2}{2}, \frac{x_1-x_2}{2}) \in \ell_{1}^2$. Our tensor is given by $(1,0) \otimes (1,t) + (0,1)\otimes (1,-t)$, so the image under this isometry is given by $(\frac{1}{2}, \frac{1}{2})\otimes (\frac{1+t}{2}, \frac{1-t}{2}) + (\frac{1}{2}, - \frac{1}{2})\otimes (\frac{1-t}{2}, \frac{1+t}{2})$. We want to use the fact that $\ell_1^2\widehat{\otimes} \ell_1^2$ is isometric to a vector valued $\ell_1$ space. We rewrite our tensor as $(1,0)\otimes (\frac{1}{2}, \frac{1}{2}) + (0,1)\otimes (\frac{t}{2}, \frac{t}{2})$ and the $\ell_1$-norm is equal to $1+t$.

So far, I don't know how to handle the case of the Haagerup tensor norm.

Ok, I think I have something that works. It seems to me that the Haagerup tensor norm coincides in this case with the projective norm. If I am not mistaken, the Haagerup tensor norm of a tensor $v$ can be written as infimum of $\sqrt{\sum_{i} \|x_i\|^2}\cdot \sqrt{\sum_{i} \|y_{i}\|^2}$ over all possible decompositions $v=\sum_{i} x_i \otimes y_i$. By working with $\frac{x_i}{c_i}$ and $c_i x_i$ for appropriate numbers $c_i$, we see that it is equal to the projective norm. Since everything is real, we can work with $\mathbb{R}^2 \widehat{\otimes} \mathbb{R}^2$, where $\mathbb{R}^2$ is endowed with the $\ell_{\infty}$-norm. This is not very good, as the projective tensor product works nicely for $\ell_1$ norm. However, $\ell_1^{2}$ and $\ell_{\infty}^2$ are isometric, so we can work with $\ell_1^2$. The result is $1+t$.

EDIT: I will only compute the projective norm.

Since everything is real, we can work with $\mathbb{R}^2 \widehat{\otimes} \mathbb{R}^2$, where $\mathbb{R}^2$ is endowed with the $\ell_{\infty}$-norm. This is not very good, as the projective tensor product interacts nicely with the $\ell_1$ norm. However, $\ell_1^{2}$ and $\ell_{\infty}^2$ are isometric, so we can work with $\ell_1^2$. This isometry is given by $\ell_{\infty}^2 \ni (x_1,x_2) \mapsto (\frac{x_1+x_2}{2}, \frac{x_1-x_2}{2}) \in \ell_{1}^2$. Our tensor is given by $(1,0) \otimes (1,t) + (0,1)\otimes (1,-t)$, so the image under this isometry is given by $(\frac{1}{2}, \frac{1}{2})\otimes (\frac{1+t}{2}, \frac{1-t}{2}) + (\frac{1}{2}, - \frac{1}{2})\otimes (\frac{1-t}{2}, \frac{1+t}{2})$. We want to use the fact that $\ell_1^2\widehat{\otimes} \ell_1^2$ is isometric to a vector valued $\ell_1$ space. We rewrite our tensor as $(1,0)\otimes (\frac{1}{2}, \frac{1}{2}) + (0,1)\otimes (\frac{t}{2}, \frac{t}{2})$ and the $\ell_1$-norm is equal to $1+t$.

So far, I don't know how to handle the case of the Haagerup tensor norm.

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created Jan 24, 2018 at 15:13

Mateusz Wasilewski

Ok, I think I have something that works. It seems to me that the Haagerup tensor norm coincides in this case with the projective norm. If I am not mistaken, the Haagerup tensor norm of a tensor $v$ can be written as infimum of $\sqrt{\sum_{i} \|x_i\|^2}\cdot \sqrt{\sum_{i} \|y_{i}\|^2}$ over all possible decompositions $v=\sum_{i} x_i \otimes y_i$. By working with $\frac{x_i}{c_i}$ and $c_i x_i$ for appropriate numbers $c_i$, we see that it is equal to the projective norm. Since everything is real, we can work with $\mathbb{R}^2 \widehat{\otimes} \mathbb{R}^2$, where $\mathbb{R}^2$ is endowed with the $\ell_{\infty}$-norm. This is not very good, as the projective tensor product works nicely for $\ell_1$ norm. However, $\ell_1^{2}$ and $\ell_{\infty}^2$ are isometric, so we can work with $\ell_1^2$. The result is $1+t$.