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What is an example of a Lie group which does not have a fixed point- free homeomorphism of order 2?

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The group $\mathbb{R}$ works. To see that, note that any homeomorphism $\mathbb{R} \to \mathbb{R}$ of order two must be decreasing, so its graph intersects the line $y = x$, so $f$ has a fixed point.

As pointed out by John Ma in the comments, we cannot take the Lie group to be compact, since any compact Lie group contains a non-trivial torus, and therefore an element of order two. Multiplication by that element then gives a homeomorphism of order two.

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    $\begingroup$ Thanks for saving me that last minute of typing (+1). Discarding 3 lines is fine :-) $\endgroup$ – Jyrki Lahtonen Oct 27 '15 at 7:47
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    $\begingroup$ Just curious, is there any compact example? $\endgroup$ – user99914 Oct 27 '15 at 7:54
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    $\begingroup$ $SO(3)$ is $\mathbb S^3 /\{\pm 1\}$, so it's a double cover of a Lens space $\mathbb S^3/\mathbb Z_4$ so it seems that it has such a homeomorphism (I do not know the answer too). $\endgroup$ – user99914 Oct 27 '15 at 8:29
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    $\begingroup$ @JohnMa $SO(3)$ has an element of order 2 so there is a free action by $Z/2Z$ $\endgroup$ – Ali Taghavi Oct 27 '15 at 9:02
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    $\begingroup$ Every compact Lie groups must have a toric subgroup $\mathbb S^1$, that $-1$ in $\mathbb S^1$ will be an order two element. @AliTaghavi $\endgroup$ – user99914 Oct 27 '15 at 9:13

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