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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. – Jason DeVito Aug 18 '12 at 1:55
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? – Erik Pan Aug 18 '12 at 2:01
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 $O(2)\subseteq E(2)$). – Jason DeVito Aug 18 '12 at 2:11
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? – Erik Pan Aug 18 '12 at 2:32
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. – Jason DeVito Aug 18 '12 at 2:50

The group of distance preserving applications from the Euclidean plane to itself is exactly the group of applications of the form $x \mapsto Ax + b$ with $A \in O(2)$ and $b \in \mathbb{R}^2$. So you should have $R^2 \simeq E(2)/O(2)$ not $E(2)/U(1)$ where $R^2$ is the translation group. The problem with only quotienting by $U(1) \simeq SO(2)$ is that you still have the reflections in your group.

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Ah, that's a subtle distinction I hadn't made, thanks. Can you possibly tell me please what forms the homomorphisms of the exact sequence would take in that case? Have I made the right guess about $f$? – Erik Pan Aug 18 '12 at 1:57

Your main problem is that in the semidirect product $E(2)=\mathbf R^2\rtimes O(2)$, you are trying to quotient by the non-normal subgroup $O(2)$ rather than by the normal subgroup $\mathbf R^2$. The subgroup $O(2)$ is the stabiliser of the origin $0$; its conjugate by $g\in E(2)$ is the stabiliser of $g(0)$, which is a different subgroup whenever $g(0)\neq0$, so it really is a non-normal subgroup. Any attempt to understand the nonexistent quotient $E(2)/O(2)$ is doomed to fail.

What does hold, and is easy to understand, is $O(2)\cong E(2)/\mathbf R^2$, through the usual map that associates to an affine transformation a linear transformation (its "differential") of the (tangent) vector space associated to the affine space.

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I think that this answer can be improved by pointing out that the quotient is still a smooth manifold, just not a Lie group (unless I've missed something?). – Joel Bosveld May 20 at 3:51

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