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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeOct 4th 2010

    I guess I should know this, but when asked today I couldn’t give a good answer:

    What can we say about putting two inequivalent structures of a Lie group on a given bare group?

    A maybe related basic question: are the exotic 4\mathbb{R}^4s still in any way canonically Lie groups?

    • CommentRowNumber2.
    • CommentAuthorTodd_Trimble
    • CommentTimeOct 4th 2010

    Of course: as bare groups, \mathbb{R} and ×\mathbb{R} \times \mathbb{R} are isomorphic.

    As for exotic 4\mathbb{R}^4: no, they are not.

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeOct 4th 2010

    Of course: as bare groups, \mathbb{R} and ×\mathbb{R} \times \mathbb{R} are isomorphic.

    Ah, of course. Thanks. We identify the underlying sets using the Peano curve and then of course the group structure of \mathbb{R} induces a wild (smoothly speaking) group structure on ×\mathbb{R} \times \mathbb{R}.

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeOct 4th 2010

    Wait, let me see. I am looking for an example (or not) of two different Lie group structures whose underlying bare group structure coincides.

    • CommentRowNumber5.
    • CommentAuthorMike Shulman
    • CommentTimeOct 4th 2010

    I think Todd meant to consider both \mathbb{R} and ×\mathbb{R}\times\mathbb{R} with their usual group structures. They are isomorphic because both are vector spaces of dimension 2 02^{\aleph_0} over \mathbb{Q}.

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeOct 4th 2010

    Ah, right. Thanks.

    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTimeOct 5th 2010

    aded a remark along these lines to Lie group