The question below has been posted on Stackexchange few days ago but I decided to share it on MO also. Hope this is not a misuse.


Fix $t\geqslant1$ and define $u\_t=\pmatrix{1 & 0\\0 & t}\otimes\pmatrix{1 & 0\\0 & 0}+\pmatrix{1 & 0\\0 & -t}\otimes\pmatrix{0 & 0\\0 & 1}\in\mathcal{M}\_2\otimes\mathcal{M}\_2$. Calculate $\|u\_t\|\_{\pi}$ and $\|u\_t\|\_h$, where these are the (Banach space) projective and the Haagerup tensor norms, respectively.



projective and Haagerup tensor norms
Here is a computation of the Haagerup norm. It is not particularly instructive, since it relies very heavily on a special feature of the case at hand. I for one would be very glad to learn some more robust techniques for computing in Haagerup tensor products.


The first step, as pointed out by Mateusz Wasilewski, is to use the isometric embedding $\mathbb{C}^2\otimes\_h \mathbb{C}^2 \hookrightarrow M\_2(\mathbb C)\otimes\_h M\_2(\mathbb C)$. 


The next step is to use the isometric embedding $\mathbb{C}^2\otimes\_h \mathbb{C}^2 \hookrightarrow \mathbb{C}^2 \ast \mathbb{C}^2$, where $\ast$ denotes the unital free product of unital $C^\*$-algebras. (This embedding extends the obvious multiplication map $\mathbb C^2\otimes\_{\mathrm{alg}} \mathbb C^2\to \mathbb C^2 \* \mathbb C^2$; see, for instance, Theorem 5.13 of [Pisier's book](https://mathscinet.ams.org/mathscinet-getitem?mr=2006539) on operator spaces.)


The next step is to identify $\mathbb C^2 \*\mathbb C^2 \cong C^\*(S\_2 \*S\_2)$, where $S\_2$ is the two-element group. This is done by identifying $\mathbb C^2\cong C^\*(S\_2)$, and comparing the universal properties. If, for the sake of definiteness, we send the element $(1,-1)\in \mathbb C^2$ to the generator of $S\_2$, then the isomorphism $\mathbb C^2 \* \mathbb C^2 \cong C^\*(S\_2 \* S\_2)$ sends $u\_t^\*u\_t$ to $\frac{1}{2}(1+t^2 + a - t^2bab)$, where $a$ and $b$ are the respective generators of the two canonical copies of $S\_2$ in $S\_2\ast S\_2$. 


Now comes the special feature: $S\_2\ast S\_2$ is virtually unique among free products in having a very tractable representation theory. Identifying this group with the infinite dihedral group $\mathbb Z \rtimes S\_2$ and applying Mackey's theory for semidirect products, one finds that every irreducible unitary representation of $S\_2\ast S\_2$ occurs as a subrepresentation of one of the two-dimensional representations 
$$
\pi\_z(a) = \begin{bmatrix} 0 & z \\ \overline{z} & 0\end{bmatrix},\quad \pi\_z(b) = \begin{bmatrix} 0 & 1\\ 1& 0\end{bmatrix}
$$ 
where $z$ ranges over the unit circle in $\mathbb C$. (For Mackey's theory, see e.g. Chapter 4, and particularly Theorem 4.40, in [Kaniuth and Taylor's book](https://mathscinet.ams.org/mathscinet-getitem?mr=3012851) on induced representations of locally compact groups.) 


Using the formula for the image of $u\_t^\*u\_t$ in $C^\*(S\_2\*S\_2)$ found above, we end up with
$$
\|u\_t\|\_h = \sup\_{|z|=1} \|\pi\_z(u\_t^\* u\_t)\|^{1/2}=\sup\_{|z|=1} \left\| \frac{1}{2}\begin{bmatrix} 1+t^2 & z-t^2 \overline{z} \\ \overline{z}-t^2 z & 1+t^2 \end{bmatrix} \right\|\_{M\_2(\mathbb C)}^{1/2}.
$$
Computing the spectrum of this $2\times 2$ matrix, one finds that the supremum occurs at $z=i$, giving $\|u\_t\|\_h = (1+t^2)^{1/2}$.


There must be a better way to do this.


