Sorry to inundate the feed with a question quite similar to my last, but again I've been drawing pictures for quite a while with little success. Does anyone have any idea how to represent the product group $\textrm{SO}(2)\times \mathbb{Z}_2$ by a solid (corresponding to rotations only)?
Solid whose rotational symmetry group corresponds to $\textrm{SO}(2)\times \mathbb{Z}_2$
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0A cone? $\phantom{}$ – 2011-12-02
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0Doesn't that only work for SO(2), and not account for the x Z_2 part? However, everything I'm saying could be entirely wrong :) – 2011-12-02
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0Two identical cones glued base to base (for the $\mathbb{Z}_2$ factor). – 2011-12-02
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0But that's not abelian is it? Isn't rotating about the axis from cone point to cone point then flipping the glued cones is in general different from flipping then rotating? – 2011-12-02
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0What do you mean by a solid and how does the group act on solids? – 2011-12-02
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1Solid is just some 3 dimensional shape, e.g. cylinder, cone, cube, torus...the book I'm working with doesn't bother to define it, so I'd imagine the standards are pretty loose. I guess to be more precise, they're looking for a solid whose rotational symmetry group (set of isometric, orientation-preserving mappings sending the solid to itself) is isomorphic to SO(2) x Z_2. Anyone can feel free to phrase that better... – 2011-12-02
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1A circular rubber band that you can flip inside out? Not sure how it's possible to get an abelian subgroup of SO(3) that has SO(2) as a strict subgroup. – 2011-12-02
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0@p.s. Problem is that you can still rotate that about an orthogonal plane, thus screwing up commutativity again... – 2011-12-02
1 Answers
As the comment by p.s. mentioned, your first concern is to find the subgroup isomorphic to $\mathrm{SO}_2\times\mathbb{Z}/2$, and there aren't any inside $\mathrm{SO}_3$: any subgroup of $\mathrm{SO}_3$ isomorphic to $\mathrm{SO}_2$ is the set of rotations around a fixed axis, and there are no other rotations that commute with all of them. You can find an appropriate subgroup in $\mathrm{O}_3$ (add the reflection in the plane perpendicular to the axis), but I don't think it is the subgroup of $\mathrm{O}_3$ defined by any solid (because a solid fixed by all rotations will automatically also be symmetric with respect to reflection in a plane containing the axis). This argument suggests that you won't even have much luck either if your solid is situated in a higher-dimensional space (in which case the $\mathrm{O}_3$ of the $3$-space containing you solid becomes the image by restriction of a subgroup of $\mathrm{SO}_n$).