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Does there exist a complex analytic Lie group which doesn't have faithful representations in $GL(N,\mathbb R)$, viewed as a real Lie group?

There are examples of complex Lie groups which do not allow faithful complex representations, like tori $\mathbb C^n/\mathbb Z^{2n}$, but such tori have many faithful real representations.

Also there are examples of real Lie groups without faithful linear representations, like the universal cover of $SL(2,\mathbb R)$ (but they are not complex analytic Lie groups).

How about complex Lie groups without faithful real representations?

flag	asked Apr 22 2011 at 15:01 mathreader 420●2●11

Richard Borcherds · Accepted Answer · 2011-04-22 16:17:08Z UTC

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Take the complex Heisenberg group of 3 by 3 upper triangular unipotent complex matrices, and mod out by a subgroup $Z\times Z$ in the center.

answered Apr 22 2011 at 16:17

Richard Borcherds
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Thank you for your answer. Could you please explain why this group will not admit faithful representations? – mathreader Apr 22 2011 at 19:41

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In an irreducible representation the center must act by scalars, but in the case of the Heisenberg it must also act by trace zero operators because the center of this Lie algebra is generated by brackets. Therefore the center acts like zero in an irreducible representation, and thus it acts with all eigenvalues zero in any representation. But when a compact $1$-dimensional Lie group acts faithfully its Lie algebra always has a nonzero eigenvalue. – Tom Goodwillie Apr 22 2011 at 21:28

@Tom: Thanks! This is a really nice explanation! – mathreader Apr 22 2011 at 23:52

Complex Lie group without faithful real representations?

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