understanding the homogeneous space $SL(2,\mathbb C)\times \mathbb C\cong SU(2)\times S^1$?

Question

How to prove that $SL(2,\mathbb C)\times \mathbb C$ is diffeomorphic to $SU(2)\times S^1$? Can this also be interpreted geometrically?

Can we actually consider $\times $ above as scalar multiplication?

I in fat found this example here I want to understand it.

Answer 1

I think that there is quite a bit of misunderstandig, partly caused by rather nasty typos in the text you have copied. I think the text is about homogeneous line bundles on $\mathbb CP^1$ (which are a special case of vector bundles associated to a prinicipal fiber bundle as mentions in the comments of @ziggurism). Here $\mathbb CP^1$ is viewed at the same time as $G/B$, where $G=SL(2,\mathbb C)$ and $B$ is the Borel subgroup of upper triangular matrices and as $K/L$ where $K=SU(2)$ and $L=B\cap K$. Indeed, since the subgroup $K\subset G$ acts transitively on $G/B=\mathbb CP^1$ it readily follows that the inclusion of $K$ into $G$ induces a diffeomorphism $K/(K\cap B)\to G/B$.

Now in both cases, you can construct homogeneous line bundles over $\mathbb CP^1$. In the first interpretation you need a 1-dimensional representation (i.e. a character) $\chi:B\to\mathbb C^*$ and you define $G\times_{\chi}\mathbb C$ (missing of the subscript $\chi$ in the third line of the text you have copied is one of the nasty typos) as $(G\times\mathbb C)/B$ where $B$ acts via $(g,z)\cdot b:=(gb,\chi(b^{-1})z)$. The projection of the homogeneous bundle maps the orbit of $(g,z)$ to $gB\in G/B$. In the second interpretation, you take $\tilde\chi:=\chi|_L:L\to \mathbb C^*$, which is a character of $L$, and then in the same way consider $(K\times\mathbb C)/L$ with the analogous projection to $K/L$. As above, it easily follows that the inclusion $K\times\mathbb C\to G\times\mathbb C$ induces an isomorphism $K\times_{\tilde\chi}\mathbb C\to G\times_\chi\mathbb C$, and this is what the text you have copies claims.

Answer 2

This is false. $SU(2)$ is $S^3$ topologically, so its product with $S^1$ is compact. $SL(2, \mathbb{C})$ and $\mathbb{C}$ are both not compact, so neither is their product. Nor do the dimensions match.

Answer 3

The page you posted does not claim that $SL(2,\mathbb C)\times \mathbb C\cong SU(2)\times S^1$, so it's not clear what exactly you would like explained.

But here is a comment that could perhaps be useful: the Lie group $GL_n(\mathbb C)$ is homeomorphic to $SU_n\times \mathbb C^k$ via QR decomposition (so of course it cannot be homeomorphic to $SU_n$ alone), but it is homotopy equivalent to its maximal compact subgroup $SU_n$ via Gram-Schmidt orthogonalization. Similar remarks relate $\mathbb C^\times$ and $U(1)$, or indeed, relate any Lie group to its maximal compact subgroup. I believe that may be helpful to understand what the author is doing.