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I know that every coset representative $g\in SL(2,\mathbb{C})$ for $SL(2,\mathbb{C})/SU(2)$ can be chosen of the form

$$ g = \left( \begin{array}{cc} \sqrt{t} & \frac{z}{\sqrt{t}}\\ 0 & \frac{1}{\sqrt{t}} \end{array} \right), z\in \mathbb{C}, t>0, z=x+iy $$ and then $g$ is mapped onto $z+kt$ in the quaternionic upper half plane

$$\mathcal{H}^c = \{z+kt = x+iy +kt | x,y\in \mathbb{R}, t>0\}$$

the elements are equal to quaternions from

$$ \mathbb{H} = \mathbb{R} \bigoplus \mathbb{R} i \bigoplus \mathbb{R} j \bigoplus \mathbb{R} k$$

with the $j$-coordinate equal to zero. An exercise says: Show that the invariant arc length on $\mathcal{H}^c$ is given by

$$ds ^2 = (dx^2 +dy^2 +dt^2 )t^{-2}$$

This would proof that $SL(2,\mathbb{C})/SU(2)$ can be identified with the hyperbolic $3$-space, wouldn't it? Do I have to start with a Riemannian metric on $SL(2,\mathbb{C})/SU(2)$? What is the metric on $SL(2,\mathbb{C})$?

I would appreciate any explanation of basic knowledge that is behind a possible solution here, since I am not very familiar with quaternions or Lie groups. Thank you in advance!

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You have shown a map $SL(2, \mathbb C)/SU(2) \simeq \mathbb R^3 \to SL(2, \mathbb C)$. Presumably, the metric you are trying to get is the pullback of the (real part of) the Killing form $B$ on $SL(2, \mathbb C)$. At the identity, where $T_e SL(2, \mathbb C)$ is identifiable with traceless matrices, this is just given by (possibly a multiple of) $B(X,Y) = Re~tr(XY)$. Then you make this globally invariant under the action of $SL(2, \mathbb C)$ on itself via left translation: $B_g = L_{g^{-1}}^* B_e$ where $L_g^{-1}$ is left translation by $g^{-1}$.

The general context for this is as follows: given $G$ semisimple and $K \subset G$ maximal compact, you can get a Cartan decomposition of the Lie algebra $\mathfrak g$ of $G$ as $\mathfrak k \oplus \mathfrak k^\perp$ where $\mathfrak k$ is the Lie algebra of $K$ and the orthogonal complement is taken using the Killing form. This satisfies $$ [\mathfrak k, \mathfrak k^\perp] \subset \mathfrak k^\perp, ~~ [\mathfrak k^\perp, \mathfrak k^\perp] \subset \mathfrak k $$ and $K \times \mathfrak k^\perp \to G, ~ (k, X) \mapsto k\exp X$ is a diffeomorphism of manifolds. Then $K$ acts on $\mathfrak k^\perp$ via the adjoint action and there is an isomorphism $$ T (G/K) \simeq G \times_{Ad} \mathfrak k^\perp $$ where on the right hand side we view $G$ as a principal $K$-bundle over $G/K$. Then the metric on $G/K$ is given by $\langle [g, X], [g, Y]\rangle = B_e(X,Y)$. I think if you follow out the identifications you will also arrive at the same metric (I tried the computations and they seemed pretty laborious though-- maybe I missed a shortcut).

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Thank you, I will need time to understand this, I really appreciate it. – Max Aug 12 '12 at 18:30

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