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I've been asked this immediately after been asked to show that the structure constants $c_{ijk}$ are totally antisymmetric, so I suppose there must be a connection, although I can't figure out where.

I'm really confused about this. A quick search took me to this: Example of two-dimensional non-abelian Lie algebra?

and to this: Geometry and Quantum Field Theory, from which I quote:

An example of a Lie group of dimension 2 with a non-abelian Lie algebra is the matrix Lie group $$G=\left\{\begin{pmatrix}a&b\\0&1\end{pmatrix}\bigg|\,a\in\mathbb{R}^+,\,b\in\mathbb{R}\right\}$$ In fact, it is not hard to show that, up to isomorphism, this is the only connected non-abelian Lie group of dimension 2 (...).

The author refers first to the algebra and then to the group, so that makes my confusion worse.

So are there, in fact, non-abelian Lie groups of dimension 2? Or is it the algebra that can be non-abelian? If so, how should I argue that, however, the group must be abelian?

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Cal Gibson
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  • Lie groups are topological groups where the group action is analytic (or at least smooth); Lie algebras are vector spaces with an extra operation $[\cdot, \cdot]$ satisfying the Jacobi formula. The link between the two is the tangent space $T_1 G$ at the origin in a Lie group $G$ is a Lie algebra, the operation being the Lie bracket for vector fields $X, Y\in T_1 G$. It's not a one-to-one correspondence, though, and the situation with infinite dimension, Lie algebras over finite fields, etc. is more complicated. – anomaly Oct 19 '15 at 15:32
  • Where did you get the idea that there is no 2-d non-abelian Lie group? – Ben Grossmann Oct 19 '15 at 15:38
  • Actually, my guess is that you were asked to show that a particular Lie Group has no $2$-$D$ Lie subgroup. – Ben Grossmann Oct 19 '15 at 15:40
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    Oh no, I was asked this verbatim in a workshop sheet, so then yes, I guess they wanted to ask something else. – Cal Gibson Oct 19 '15 at 16:21
  • @CalGibson I'm assuming you're actually on the same course as I am since we have the same issue 9 hours apart. I am pretty sure the lecturer meant to ask: "Why are there nonabelian Lie groups with dimension one?" If you look at the notes there is a comment under One Dimensional Groups that states "all 1 dim Lie Groups are abelian" with Proof: (See exercise). So I'm going to prove that and just state he must have made a typo by referring to an the abelian dim 2 matrix. – Alexander McFarlane Oct 20 '15 at 01:11

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The Lie Group $$ G = \left\{\pmatrix{a&b\\0&1} : a \in \Bbb R^+,b \in \Bbb R\right\} $$ is indeed non-abelian. In particular, note that

$$ \pmatrix{1&1\\0&1}\pmatrix{2&0\\0&1} \neq \pmatrix{2&0\\0&1}\pmatrix{1&1\\0&1} $$ The Lie algebra of this group is also non-abelian (i.e. non-trivial).

I'm not sure where you're getting the idea that there's no such Lie group.

Ben Grossmann
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