Lie Algebra of the reals with addition

[holy_toaster]

The following question may be trivial, but I just can't get it figuered out:

Consider the real numbers R with the addition operation + as a Lie Group (R,+). What is the Lie Algebra of this Lie Group? Is it again (R,+), this time considered as a vector space? If so, what is the exponential map and the adjoint representation of the Lie group on the Lie Algebra?

 

[micromass]

Now, the set (R,+) is isomorphic to SUT₂(R) (= the set of all 2x2 upper triangular matrices with diagonal entries =1). An isomorphism is given by

\mathbb{R}\rightarrow SUT_2(\mathbb{R}):c\rightarrow \left(\begin{array}{cc} 1 & c\\ 0 & 1\end{array}\right).

So the Lie Algebra of (R,+) will be the Lie-algebra of SUT₂(R). But this Lie algebra is well known to be the upper triangular 2x2-upper triangular matrices with diagonal elements zero. The exponential map being the normal exponential map. Of course, this Lie-algebra can also be written as (R,+) with Lie bracket [a,b]=0. The exponential is then

exp(a)\cong exp\left(\begin{array}{cc}0 & a\\ 0 & 0\end{array}\right)=I+\left(\begin{array}{cc} 0 & a\\ 0 & 0\end{array}\right)\cong a

So the exponential map is simply the identity map.

 

[holy_toaster]

Aha. Thanks that helps. But if I calculate now the adjoint representation of the group over the Lie Algebra, I get that it is the trivial representation, i.e. every group element acts as the identity. Can that be true?

 

[micromass]

Aha. Thanks that helps. But if I calculate now the adjoint representation of the group over the Lie Algebra, I get that it is the trivial representation, i.e. every group element acts as the identity. Can that be true?

Yes, that is to be expected. The adjoint representation of any abelian Lie group is trivial (according to wiki). The thing is that the conjugation action G->Aut(G) is trivial since the group is abelian. Thus it is not surprising that adjoint representation is also trivial.

 

[holy_toaster]

Yes, I see. Thank you.