Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

1 Answer 1

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).

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.