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I am trying to write a map $s$, such that $s: SL(2,\mathbb{C})/SU(2) \to SL(2,\mathbb C)$.

Using Iwasawa decomposition, I see that any element $g$ in $SL(2,\mathbb{C})$ can be written as $g = su$, where $u$ is an element of $SU(2)$ and $s$ is a matrix that is a product of two matrices $a$ and $k$, where $a$ is a member of the subgroup of diagonal matrices with positive diagonal components, and $k$ is a member of the subgroup of upper triangular nilpotent matrices.

However, I don't exactly see how to formulate this map; I see that it should come from this decomposition, but I am a bit stuck after that.


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Try the polar decomposition and identify $SL(2,\mathbb{C})/SU(2)$ with the positive definite, skew-Hermitian matrices $P$ of determinant one. This gives you an obvious map $s:P \to SL(2,\mathbb{C})$. – t.b. Oct 11 '11 at 17:16
Well, I think you need to specify which properties your map needs to have. Otherwise one can probably construct many different maps, for example a trivial one: $s(g) = 1$ for $g\in SL(2,\mathbb C)/SU(2)$ and $1$ is the identity element of $SL(2,\mathbb C)$. – Heidar Oct 11 '11 at 17:17
tb: I guess my problem is that I don't see the 'obvious map'. It is just s: P -> SL(2,C) such that p -> sp for s in SU(2)? – Mary Oct 11 '11 at 19:11
You have too many $s$'es. The set $P$ of positive definite skew-Hermitian matrices is a subset of the group $G = SL(2,\mathbb{C})$. By uniqueness of the polar decomposition every $g$ is uniquely $g= pu$ with $p = \sqrt{a^\ast a}\in P$ and $u = p^{-1}a \in SU(2)$. This gives you a unique coset representative $p$ for each $SU(2)$-coset and hence the projection $\pi:SL(2,\mathbb{C}) \to SL(2)/SU(2)$ restricts to a bijection of the set $P$ with the quotient $SL(2,\mathbb{C})/SU(2)$. Then the map is just the inclusion $s: P \to SL(2,\mathbb{C})$ which maps $p$ to $p$, so $s(p) = p$. – t.b. Oct 11 '11 at 19:32
Which you can then compose with $(\pi|_P)^{-1}: SL(2,\mathbb{C})/SU(2) \to P$ if you wish. – t.b. Oct 11 '11 at 19:35

I don't think that the Iwasawa decomposition can help you, because $B \cap SU(2) \neq \{1\}$, so you have to work with the subset of $B$, where the diagonal entries are strictly positive.

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