Suppose that an $n$-dimensional abelian Lie group has no nonzero element $A$ such that $A+A=0$ (using + for the group operation). Does it follow that the group is isomorphic to the vector space $\mathbb{R}^n$? This seems likely to me, since a simply connected abelian Lie group is a vector space, and the examples that are not simply connected have the topology of a direct product of a torus with a vector space. Is it necessary to assume that the group is connected?
Abelian Lie group with no nonzero element $A$ such that $A+A=0$
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linear-algebra
lie-groups
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0You need to assume that the group is connected (otherwise you would be asserting an isomorphism between $\Bbb{Z}_2$ and $\Bbb{R}^n$ for some $n$). Apart from that, you are right. – 2017-01-08
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0@RobArthan: I'm not quite following you. $\mathbb{Z}_2$ has the element 1, for which 1+1=0, so doesn't it fail to be a counterexample? But maybe $\mathbb{Z}\times\mathbb{R}$ works, since it's a one-dimensional manifold, but not isomorphic to $\mathbb{R}^1$? – 2017-01-08
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1Sorry, read $\Bbb{Z}_3$ (or any discrete abelian group $G$ such that $2G \neq 0$) where I wrote $\Bbb{Z}_2$. – 2017-01-08
1 Answers
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Yes, let $X$ such an Abelian Lie group, the universal covering map $p:\hat X=R^n\rightarrow X$ is a surjective morphism of Lie group. Suppose that $p$ is not injective, you have $p(B)=0$, $B\neq 0$. Since $p$ is a covering, there exists a neighborhood $0\in U\subset \hat X$ such that the restriction of $p$ to $U$ is injective. There exists an integer $n$ such that ${B\over 2^n}\in U$, this implies that $p({B\over 2^n})\neq 0$. You can take $m We conclude that if $p$ is not injective, there exists $A\in X, A\neq 0, 2A=0$. So if there does not exist an element $A$ such that $2A=0$, $p$ is injective and is henceforth an isomorphism.
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0Nice, thanks. Is there an assumption hidden in your argument that $X$ is connected? – 2017-01-08
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0Yes, assume that $X$ is connected. – 2017-01-08
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0At what step in your argument was that assumption required? – 2017-01-08
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1When I use the fact that the universal cover is $R^n$. – 2017-01-08