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(Just so we're clear: that the Lie group of planar translations $R^2$ is isomorphic to a quotient of the 2D Euclidean Lie group $E(2)$ and the circle group $U(1)$.)

I am trying to prove that $R^2 = E(2)/U(1)$ directly but not getting very far. It seems as if it should be true, and trivially so, but I am new to Lie groups and am not quite sure what implies what yet. For instance, it is known that $E(2)$ is the semidirect product of $R^2$ and $U(1)$ (or $O(2)$ if you like); and I reckon that implies that E(2) is a principal fibre bundle with base $R^2$ and fibre $U(1)$, is any of that true? And does it imply there is a quotient relationship?

If so - can you give explicit homomorphisms $f$ and $g$ such that the following sequence is exact:

$1\rightarrow U(1)\xrightarrow f E(2)\xrightarrow g R^2\rightarrow1$?

(i.e. Im $f$ = Ker $g$)

I thought that $f$ is the map that takes each element of $U(1)$ to the element of $E(2)$ with the same rotation but $\underline{0}$ translation - i.e. if a general element of E(2) is expressed as $(M,\underline{v})$ for a rotation matrix $M$ and translation vector $\underline{v}$ [with $(M,\underline{a}).(N,\underline{b})=(MN,\underline{a}+M\underline{b})]$ then $f$ embeds $U(1)$ in E(2) as

$f:R\mapsto (R, \underline{0})$

for $R\in U(1)$, $R$ a rotation matrix. But presumably that would mean that $g$ would map the whole of $E(2)$ to $R^2$ by taking $(R,\underline{v})\mapsto (0,\underline{v})$ (Thaking Im $F$, the pure rotations, to the identity element $e_{R^2}=\underline{0}$ - unfortunately, this doesn't appear to be a homomorhpism.

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    The map $g$ is rarely a homomorphism. Yes, in this case $\mathbb{R}^2$ has the structure of a Lie group, but in more general settings, $\mathbb{R}^2$ is replaced by a smooth manifold. In this particular case, the most natural choice of $g$ (mapping $(M, v)$ to $v$) is *not* a homomorphism, and I suspect there is no such homomorphism $g$ which fits there, but I haven't proven it.2012-08-18
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    Ah, I thought that homomorphisms were required in the exact sequence defining the quotient! Thanks so much! If they aren't, is it true that they simply have to embed the group isomorphically? i.e. in this case, if $g:E(2)\rightarrow R^2$ mapped $(R,\underline{v})\mapsto \underline{v}$, does that suffice to define the quotient?2012-08-18
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    Well, I'm not sure what you mean by "isomorphic" in this context, but they are *not* usually embeddings - they are almost **never** injective. In your particular example, it's not injective (the inverse image of, say, $\{0\}$ is all of $O(2)\subseteq E(2)$).2012-08-18
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    Thanks for your help! One final question then - what are the actual constraints on $f$ and $g$ in the exact sequence, if they don't have to be homomorphisms? Is it still true (e.g.) that $f$ is injective and $g$ is surjective?2012-08-18
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    Yes, those are both true. I think of it only as an exact sequence "in spirit" - a more proper name is a Fiber bundle. Further, $f$ is always an injective homomorphism.2012-08-18
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    Thank you so much, you've been so helpful!2012-08-18

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