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Assuming that a map of the form:$a\in SL(2,\mathbb{C}) \rightarrow m[a]$ with$m[a](z)=\frac{a_1z + a_2}{a_3z+a_4}$ is a group homomorphism, it is easy to show that this mapping is not bijective, hence why the groups are homomorphic to each other. We can find$a$ and$a'$ so that$m[a](z)=m[a'](z)$ and the elements of$SL(2,\mathbb{C})$ that satisfy this are$\{I,-I\}$. And the Kernel of it is isomorphic to$\mathbb{Z}_2$.

But I am curious to know if the Moebius group is simply connected.How can one show it? Would it be possible to consider curves that connect$\text{diag}(1,1)\in SL(2,\mathbb{C})$ and$\text{diag}(-1,-1)\in SL(2,\mathbb{C})$

askedNov 15 at 0:12
imbAF's user avatar
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    $\begingroup$Your Möbius group is exactly the group $PSL(2, \mathbb{C}) := SL(2, \mathbb{C})/\{I, -I\}$. $SL(2, \mathbb{C})$ is connected and the projection $SL(2, \mathbb{C}) \to PSL(2, \mathbb{C})$ is a covering map of degree 2, hence $PSL(2, \mathbb{C})$ cannot be simply connected. (In fact as $SL(2, \mathbb{C})$ is simply connected, this implies that the fundamental group of $PSL(2, \mathbb{C})$ is the kernel of the above covering map, which is $\mathbb{Z}_2$.)$\endgroup$CommentedNov 15 at 1:00
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    $\begingroup$At least the first part of the comment is provedhere andhere, i.e., that $SO(3)$ is not simply connected. Hence $PSL(2,\Bbb C)$ is not simply connected. Probably quicker is to find details about the proof given in the first comment, namely that the projection $SL(2,\Bbb C)\rightarrow PSL(2,\Bbb C)$ is a covering map of degree $2$. Then we are done.$\endgroup$CommentedNov 15 at 12:31
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    $\begingroup$The Möbius group is the group ofMöbius transformations. For a diagonal matrix, this is $f(z)=\frac{az}{d}$ with $ad=1$ in $\Bbb C$, which means that $I$ and $-I$ have the same Möbius transformation, namely $f(z)=z$. So the group is $SL(2,\Bbb C)/\{I,-I\}$ rather than $SL(2,\Bbb C)$ - if this was your question.$\endgroup$CommentedNov 15 at 14:37
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    $\begingroup$And $SL(2,\Bbb C)/\{I,-I\}$ is the quotient group, andnot "without the $\{I,-I\}$". They are still there, but belong to the same equivalence class in the quotient.$\endgroup$CommentedNov 15 at 14:52
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    $\begingroup$Of course one should justify this. But this is done anyway always before starting with further properties, see for examplethis duplicate.$\endgroup$CommentedNov 15 at 19:11

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