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Jul 20, 2017 at 11:54 comment added YCor But anyway the compact-open topology for $Homeo(\mathbf{R}^2)$ is, if I'm correct, a group topology (coinciding with compact convergence on compact subsets), and hence in restriction to your subgroup is also a group topology.
Jul 20, 2017 at 10:37 comment added YCor I'm sorry, you're right.
Jul 20, 2017 at 9:55 comment added Ali Taghavi @YCor What about if we replace the compact open topology by another one: For example We identify $G$ with certain subspace of $\mathbb{R}^{\infty}$. with the product topology?(The identification of the space of polynomials with the space of their coefficient).
Jul 20, 2017 at 9:04 comment added Ali Taghavi @YCor I think In the first version of the question, I had also pointed out to compact open topology. May be I did not understand your comment, correctly?
Jul 20, 2017 at 8:25 history edited Ali Taghavi CC BY-SA 3.0
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Jul 20, 2017 at 8:10 comment added YCor When a group $G$ is given "is $G$ a topological group" makes no sense. One should rather ask whether $G$ admits interesting/natural group topologies.
Jul 20, 2017 at 6:20 history edited Ali Taghavi CC BY-SA 3.0
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Jul 20, 2017 at 6:13 comment added Ali Taghavi @abx Very great comment. Thank you!
Jul 20, 2017 at 6:09 comment added abx This group has been heavily studied in algebraic geometry. For a nice survey, you can have a look at Algebraic Automorphisms of Affine Space by H. Kraft, Progress in Math. 80 (Birkhäuser), p. 81-105.
Jul 20, 2017 at 6:02 history asked Ali Taghavi CC BY-SA 3.0