# Lie algebra of the bounded continuous functions

I can think of the set of bounded, continuous functions from $\mathbb R \to \mathbb R$ as a group, with composition as addition of functions. In other words, this group has the rule that the composition of two elements in the group, $f(x)$ and $g(x)$, is their point-wise sum $f(x) + g(x)$. Since this set is path connected, the group is continuous, i.e. it is a lie group.

What is the lie algebra of this group?

I am trying to use this approach to find a relationship between Lie theory and Fourier series, but I realize this might not work. So I have decided to just ask the question in this form. If anyone knows of any connections between Lie theory and Fourier series however, I am interested to hear about that.

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Matt, the usual way in which one says what you mean by «this group has the rule that the composition of two elements in the group is their point-wise sum» is «This is a group with operation given by pointwise sum». – Mariano Suárez-Alvarez Aug 30 '10 at 22:03
@Qiaochu Yuan: The neutral element is the zero function. The group is additive. (At least, that is how I understand the question.) – Pierre-Yves Gaillard Aug 31 '10 at 4:29
I think the answer is contained in one of Mariano Suárez-Alvarez's comments below: "Well sure: if «with composition as addition of functions» does mean that the operation of the group is addition, then any sensible definition should result in a vector space with the zero bracket :P". This is so for any Banach space. The Lie group is the space itself. – Pierre-Yves Gaillard Aug 31 '10 at 4:40
Sorry, I misread the definition. It is very confusing to say "composition as addition of functions" since composition and addition are different operations of functions. One should say "the group operation is pointwise addition." – Qiaochu Yuan Aug 31 '10 at 5:20
@Qiaochu Yuan: You were not the only one to be confused: see the answers and the other comments. – Pierre-Yves Gaillard Aug 31 '10 at 5:42

In any case, your «Lie group of bounded continuous functions $\mathbb R\to\mathbb R$ under composition» is not really a group... You'd have to restrict your attention to the group of homeomorphisms (or something along that line) to actually get a group.
@Mariano: I think you have misread part of Matt's question: he is taking the set of all continuous functions from $\mathbb{R}$ to $\mathbb{R}$ under pointwise addition. This is indeed a commutative group, even an $\mathbb{R}$-vector space. (The rest of what you say I agree with; I'm not sure you can extract a Lie algebra in this situation, and since the group is commutative, I suspect that even if this works, one will end up with an identically zero Lie algebra.) – Pete L. Clark Aug 30 '10 at 21:11