The matrix representation $U$ for the group $SU(2)$ is given by

$$U = \begin{bmatrix} \alpha & -\beta^{*} \\ \beta & \alpha^{*} \\ \end{bmatrix}$$

where $\alpha$ and $\beta$ are complex numbers and $|\alpha|^{2}+|\beta|^{2}=1$.

This can be derived using the unitary of $U$ and the fact that $\text{det}\ U=1$.

Is any complex $2\times 2$ matrix with unit determinant necessarily unitary?

Consider the following argument:

$$\text{det}\ (U) = 1$$ $$(\text{det}\ U)(\text{det}\ U) = 1$$ $$(\text{det}\ U^{\dagger})(\text{det}\ U) = 1$$ $$\text{det}\ (U^{\dagger}U) = 1$$ $$\text{det}\ (U^{\dagger}U) = \text{det}\ (U)$$ $$U^{\dagger}U = U$$ $$U^{\dagger}= 1$$

Where's my mistake in this argument?

• Two matrices with the same determinant are not necessarily equal, which you've assumed 2 steps from the bottom. – John Hughes Nov 17 '16 at 13:27
• Any help with my first question about the unitarity (or not) of a matrix with unit determinant? – nightmarish Nov 17 '16 at 13:35
• Consider $\begin{bmatrix}3 & 1 \\ 2 & 1\end{bmatrix}$. – John Hughes Nov 17 '16 at 14:26
• Got it, thanks! – nightmarish Nov 17 '16 at 14:29
• This is becoming a chameleon question (see meta.stackexchange.com/questions/43478/…). – John Hughes Nov 17 '16 at 16:20

The mistake is in passing from $\text{det}\ (U^{\dagger}U) = \text{det}\ (U)$ to $U^{\dagger}U = U$. There is no reason for that: $$\text{det} A=\text{det} B\not\Rightarrow A=B\quad (!)$$
Your mistake is that the determinant is not inversible and $$det(U{}^TU) = det(U)$$ definitely doesn't imply $$U{}^TU = U$$