2
$\begingroup$

Consider all the finite-dimensional irreducible representations of a group.

  1. For each finite-dimensional irreducible representation of a group, is there one and only one corresponding representation of the Lie algebra?

  2. For a Lie group, are there always finite-dimensional irreducible representations of $n \times n$ matrices for all values of $n$?

  3. For a Lie group, can there only be one finite-dimensional irreducible representation of $n \times n$ matrices for each value of n?

$\endgroup$

closed as off-topic by Daniel Robert-Nicoud, G-man, Michael Albanese, Cameron Williams, Mark Viola Nov 20 '15 at 0:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Daniel Robert-Nicoud, G-man, Michael Albanese, Cameron Williams, Mark Viola
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ For 1, for non simply connected Lie groups, representations of the Lie algebra don't generally lift to the group. So I would say no. Look for instance at $G=S^1$, $\frak{ g}=\mathbb{R}$ acting on $\mathbb{R}$. $\endgroup$ – Tim kinsella Nov 18 '15 at 20:24
3
$\begingroup$
  1. Every representation of a Lie group $G$ naturally induces a corresponding representation of the corresponding Lie algebra $\mathfrak{g}$, by differentiation. I'm not sure how to interpret "only one" in this question.

  2. No. For example, $SO(3)$ has irreducible representations only in odd dimensions $1, 3, 5, \dots$.

  3. No. For example, $SU(2) \times SU(2)$ has two nonisomorphic irreducible $2$-dimensional representations, one for each of its factors.

$\endgroup$

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