MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 13 down vote accepted

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.

share|cite|improve this answer
Thank you for your answer. Could you please explain why this group will not admit faithful representations? – mathreader Apr 22 '11 at 19:41
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 '11 at 21:28
@Tom: Thanks! This is a really nice explanation! – mathreader Apr 22 '11 at 23:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.