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This questions arises from the answer to this previous question, which leaded me to 'relax' the statement that I wanted to prove.

Suppose that a matrix $M\in\mathrm{SL}_2(\mathbb{C})\setminus\{\pm I\}$ has diagonal elements complex conjugates of each other and that their product is less than $1$. That is, $$M=\begin{pmatrix} a & b\\ c & \bar{a}\end{pmatrix},$$ with $|a|^2<1$ and $\det M=1$.

Also, suppose that there exists a matrix $T\in\mathrm{SL}_2(\mathbb{C})$ such that $T^{-1}MT\in\mathrm{SU}_2$.

Then, I think it is possible to find $T$ diagonal, i.e.,

When can find such $T$ with $T=\mathrm{diag}(p, 1/p)$.

I would like to prove this statement and have tried by contradiction with no success.

A couple of preliminary ideas:

  • Since there exists a matrix $T\in\mathrm{SL}_2(\mathbb{C})$ such that $T^{-1}MT\in\mathrm{SU}_2$, then $\mathrm{tr}M\in (-2,2)$. This is easy to prove using the properties of the trace and those of unitary matrices.

  • Therefore, we can deduce that $\mathrm{Re}(a)\in (-1,1)$. Also, since $|a|^2<1$ and the determinants are equal to $1$, we obtain that $bc<0$.

Since I want to prove that such diagonal matrix can be found, I can argue by contradiction and suppose that $T$ cannot be diagonal, that is, $$T=\begin{pmatrix} t_{11} & t_{12}\\ t_{21} & t_{22}\end{pmatrix}\in\mathrm{SL}_2(\mathbb{C}),$$ with at most one $t_{ij}=0$, for $i,j\in\{1,2\}$.

However, this doesn't seem to be the right approach. I don't really see how to use those conditions stated above for $M$, in order to deduce something about $T$: the only idea that I've had is to actually compute $T^{-1}MT$, which I know is unitary, but I don't know how to use the other conditions (and their implications).

Ideas or clues are very welcome.

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What you want is true, even without the hypothesis that there exists $T$ with $T^{-1}MT\in \mathrm{SU}_2$ (which of course ends up being true afterwards). We can also require $p$ real.

So we assume that $|a|<1$, $\det M=1$, and nothing else.

For any $p\ne0$, $$ \begin{bmatrix} p&0\\0&1/p\end{bmatrix}\begin{bmatrix} a&b\\ c&\bar a\end{bmatrix}\begin{bmatrix} 1/p&0\\0&p\end{bmatrix} =\begin{bmatrix} a&b\,p^2\\ c/ p^2&\bar a\end{bmatrix}. $$ We want this to be a unitary. We have $$ \begin{bmatrix} a&b\,p^2\\ c/ p^2&\bar a\end{bmatrix}^*\begin{bmatrix} a&b\,p^2\\ c/ p^2&\bar a\end{bmatrix} =\begin{bmatrix} |a|^2+|c|^2/|p|^4 &\bar a(bp^2+\bar c/\bar p^2)\\ a(\bar b \bar p^2+c/p^2)& |a|^2+|b|^2|p|^4. \end{bmatrix} $$ For this latter matrix to be the identity, we need $bp^2+\bar c/\bar p^2=0$. Multiplying by $\bar p^2$, $$ 0=b|p|^4+\bar c,\ \ \ \text{ so } |p|^4=-\bar c/b=-\frac{(|a|^2-1)/\bar b}b=\frac{1-|a|^2}{|b|^2}. $$

Also, $$ |a|^2+|c|^2/|p|^4=|a|^2+|c|^2/(-\bar c/b)=|a|^2-bc=1 $$ and $$ |a|^2+|b|^2|p|^4=|a|^2+|b|^2(-\bar c/b)=|a|^2-\overline{bc}=\overline{|a|^2-bc}=1. $$ So any $p\in\mathbb C$ with $$ |p|=\left(\frac{1-|a|^2}{|b|^2}\right)^{1/4} $$ will work.

As an example, $$ \begin{bmatrix} 1/12^{1/4}&0\\0&12^{1/4}\end{bmatrix} \begin{bmatrix} 1/2&3\\-1/4&1/2\end{bmatrix} \begin{bmatrix} 12^{1/4}&0\\0&1/12^{1/4}\end{bmatrix} =\begin{bmatrix} 1/2&\sqrt3/2\\ -\sqrt3/2&1/2\end{bmatrix}. $$

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  • $\begingroup$ I don't follow your choice of $M$. Your $M$ is not an element of $\mathrm{SL}_2(\mathbb{C})$. Also, one of the extra conditions on $M$ is that $|a|^2<1$, from which it follows that $bc<0$. Still, the element $M\in\mathrm{SL}_2(\mathbb{C})$ that I'm considering has $\frac{1}{2}\mathrm{tr} M\in(-1,1)$, so it is always diagonalizable. $\endgroup$ – Edu Feb 6 at 15:53
  • $\begingroup$ Take a look now. $\endgroup$ – Martin Argerami Feb 6 at 19:03
  • $\begingroup$ Very nice construction. Just for the sake of clarity, your are conjugating the other way round than what I was doing (which is irrelevant) and in your first matrix equation I think there shouldn't be the $\bar{p}^2$ but just $p^2$ instead. Thank you again. $\endgroup$ – Edu Feb 6 at 20:15

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