# Decompose elements in $SL_2$ as a pair of elements in $SL_2^*$.

I have a question about decomposing elements in $SL_2$ as a pair of elements in $SL_2^*$. Here $SL_2^*$ is the dual Poisson Lie group of $SL_2$ which is defined as follows.

Let $G$ be a Poisson-Lie group and $\mathfrak{g}$ its Lie algebra. Let $\langle \cdot, \cdot \rangle$ be the Killing form of on $\mathfrak{g}$. It is defined by \begin{align*} \langle x, y \rangle = \text{tr}(ad(x)ad(y)), \end{align*} where $x, y \in \mathfrak{g}$.

Let $\pi_{>0}$, $\pi_{<0}$ be projections of $\mathfrak{g}$ onto subalgebras spanned by positive and negative roots, $\pi_0$ be the projection onto the Cartan subalgebra $\mathfrak{h}$, and let $R = \pi_{>0} - \pi_{<0}$.

The double of $\mathfrak{g}$ is defined as $D(\mathfrak{g}) = \mathfrak{g} \oplus \mathfrak{g}$ which is equipped with an invariant non-degenerate bilinear form \begin{align} \langle \langle (\xi, \eta), (\xi', \eta') \rangle \rangle = \langle \xi, \xi' \rangle - \langle \eta, \eta' \rangle. \label{eq: bilinear form on Drinfeld double} \end{align} Define subalgebras $\mathfrak{d}_{\pm}$ of $D(\mathfrak{g})$ by $\mathfrak{d}_+ = \{ (\xi, \xi): \xi \in \mathfrak{g} \}$ and $\mathfrak{d}_- = \{ (R_+(\xi), R_-(\xi)): \xi \in \mathfrak{g} \}$, where $R_{\pm} \in \text{End}(\mathfrak{g})$ is given by $R_{\pm} = \frac{1}{2}(R \pm id)$. The Drinfeld double of $G$ is $D(G)=G \times G$.

The diagonal subgroup $\{(X,X): X \in G\}$ is a Poisson-Lie subgroup of $D(G)$ naturally isomorphic to $G$. The Lie algebra of $\{(X,X): X \in G\}$ is $\mathfrak{d}_+$.

The group $G^*$ whose Lie algebra is $\mathfrak{d}_-$ is a Poisson-Lie subgroup of $D(G)$ called the dual Poisson-Lie group of $G$.

In the case of $SL_2$, we have \begin{align*} G=\{ \left( \left( \begin{matrix} a & b \\ c & d \end{matrix} \right), \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \right): a,b,c,d \in \mathbb{C}, ad-bc=1 \} \end{align*} \begin{align} SL_2^*=G^* = \{ \left( \left( \begin{matrix} x & y \\ 0 & x^{-1} \end{matrix} \right), \left( \begin{matrix} x^{-1} & 0 \\ z & x \end{matrix} \right) \right): x \in \mathbb{C}^*, y, z \in \mathbb{C} \}. \end{align}

Generic elements $x\in SL_2$ can be decomposed as $x=x_+\cdot x_-^{-1}$, $(x_+,x_-) \in SL_2^*$.

Let \begin{align*} x = \left(\begin{array}{cc} a & b\\ c & d \end{array}\right). \end{align*}

The decomposition which satisfies $diag(x_+) \cdot diag(x_-) = 1$ and $x=x_+ \cdot x_-^{-1}$, $(x_+, x_-) \in SL_2^*$, is \begin{align} & x_+ = \left(\begin{array}{cc} \frac{1}{\sqrt{d}} & \frac{b}{\sqrt{d}}\\ 0 & \sqrt{d} \end{array}\right),\\ & x_- = \left(\begin{array}{cc} \sqrt{d} & 0\\ - \frac{c}{\sqrt{d}} & \frac{1}{\sqrt{d}} \end{array}\right). \end{align} But there are elements in $x_+, x_-$ which involve the square root of $d$. Are there some mistakes in my computations? Thank you very much.