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I have been trying to find for days a non-abelian nilpotent Lie Group that is not simply connected nor product of Lie Groups, but haven't been able to succeed. Is there an example of this, or hints to this group, or is it fundamentally impossible?

Cheers and thanks.

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The typical answer is a sort of Heisenberg group, presented as a quotient (by a normal subgroup) $$ H \;=\; \{\pmatrix{1 & a & b \cr 0 & 1 & c\cr 0 & 0& 1}:a,b,c\in \mathbb R\} \;\bigg/\; \{\pmatrix{1 & 0 & b \cr 0 & 1 & 0\cr 0 & 0& 1}:b\in \mathbb Z\} $$

Edit: To certify the non-simple-connectedness, note that the group of upper-triangular unipotent matrices is simply connected, and that the indicated subgroup is discrete, so this Heisenberg group has universal covering group isomorphic to that discrete subgroup, and $\pi_1$ of the quotient (the Heisenberg group) is isomorphic to that covering group.

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    $\begingroup$ Thanks, I see that it is non abelian and nilpotent, but it is not clear to me that it is not simply connected, nor a product, could you please elaborate? $\endgroup$
    – rabipelais
    Jan 13, 2013 at 20:56
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$SL(2;\mathbb{R})$ is nonabelian, has infinite cyclic fundamental group, and is not a product since (I believe) $\mathfrak{sl}(2;\mathbb{R})$ does not decompose as a direct sum of Lie algebras.

Edit: Wait, do you want just a nonabelian, non-product Lie group with nontrivial $\pi_1$, or do you want it nilpotent as well?

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  • $\begingroup$ Thanks for the answer, I forgot to write nilpotent as well (as in the title). If I am not mistaken, SL(2) is not nilpotent, is it? Because the center of SL are the diagonal matrices with constant entries. $\endgroup$
    – rabipelais
    Jan 13, 2013 at 20:26
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    $\begingroup$ You are not mistaken. $[\mathcal{sl}(2),\mathcal{sl}(2)] = \mathcal{sl}(2)$. $\endgroup$
    – Neal
    Jan 13, 2013 at 20:32

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