The power equation in $\mathrm{PSL}_2(\mathbb{C})$

Question

Problem. Show that the equation $$x^2=\begin{pmatrix}-1&1\\0&-1\end{pmatrix}$$ has no solutions in $\mathrm{SL}_2(\mathbb{C})$. From this, show that any power equation $x^m=a$ is solvable in $\mathrm{PSL}_2(\mathbb{C}).$

For the first part, I showed by elementary i.e. put $x=(a_{ij})$ but I don't know any other way. Please help me with the following part in $\mathrm{PSL}_2(\mathbb{C}).$ Furthermore, I wonder if it's still correct to replace $\mathrm{PSL}_2(\mathbb{C})$ with $\mathrm{PSL}_2(\mathbb{H})$ (resp. $\mathrm{PSL}_2(D)$) where $\mathbb{H}$ is a real quaternion division ring (resp. $D$ is a division ring).

Thank you for all your support.

Answer 1

Let $a \in \operatorname{SL}_2(\mathbb{C})$. Up to conjugation by an element of $\operatorname{SL}_2(\mathbb{C})$, $a$ has one of the following forms:

1. $a = \begin{pmatrix} \lambda & 0 \\ 0 & \frac{1}{\lambda} \end{pmatrix}$ where $\lambda \neq 0, 1, -1$. This case corresponds to the situation $a$ has two distinct eigenvalues. In this case, let $\lambda^{\frac{1}{m}}$ be an arbitrary $m$-th root of $\lambda$. Then $$ x = \begin{pmatrix} \lambda^{\frac{1}{m}} & 0 \\ 0 & \frac{1}{\lambda^{\frac{1}{m}}} \end{pmatrix} $$ belongs to $\operatorname{SL}_2(\mathbb{C})$ and satisfies $x^m = a$.
2. $a = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. In this case, let $$ x = \begin{pmatrix} 1 & \frac{1}{m} \\ 0 & 1 \end{pmatrix}. $$ Then $x$ belongs to $\operatorname{SL}_2(\mathbb{C})$ and $x^m = a$.
3. $a = \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix}$. In this case, you can't neccesary find $x \in \operatorname{SL}_2(\mathbb{C})$ with $x^m = a$ but in $\operatorname{PSL}_2(\mathbb{C})$, the matrix $a$ is equivalent to $\begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$ which has $$ x = \begin{pmatrix} 1 & -\frac{1}{m} \\ 0 & 1 \end{pmatrix} $$ as an $m$-th root.

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BTW, to see quickly that $$ x^2 = \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix} $$ has no solution in $\operatorname{SL}_2(\mathbb{C})$, assume by contradiction that we can find such an $x$. Then by Cayley-Hamilton, we have $$ x^2 -\operatorname{tr}(x) x + \det(x)I = 0 \iff \operatorname{tr}(x) x = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. $$ Taking the trace of both sides, we get that $\operatorname{tr}^{2}(x) = 0$ which immediately leads to a contradiction.