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If $G$ is an abstract Lie group, then one defines the Lie algebra as the tangent space $T_eG$ at the identity. I understand $T_eG$, to be exactly the derivations of functions defined on neighborhoods of $e \in G$.

Concretely though I don't really see what happens. For example, if $G = SL(n)$ then all matrices with trace zero form the Lie algebra, but in what sense are these derivations? And why don't the matrices with nonzero trace act as derivations?

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Since $SL(n)$ is a submanifold of $\mathbb{R^{n \times n}}$, it's easier to just think about vectors.

The Lie algebra $\mathfrak{sl}(n)$ is just the set of vectors based at the identity that are tangent to $SL(n)$; that is, the tangents to curves in $SL(n)$. If $M: (-\epsilon, \epsilon) \to SL(n)$ is a curve with $M(0) = e$, then we have $\det M(t) = 1$ for every $t \in (-\epsilon, \epsilon)$. Differentiating this at $t=0$ (using e.g. Jacobi's formula) gives $\operatorname{tr}(M'(0))=0$; so any tangent vector $\xi \in \mathfrak{sl}(n)$ must be a trace-free matrix.

Since the trace-free matrices form a vector space of dimension $n^2-1=\dim SL(n) = \dim \mathfrak{sl}(n)$, we must conversely have that every trace-free matrix based at the identity is in to $\mathfrak{sl}(n)$.

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  • $\begingroup$ @user95274: if you don't have an embedding of the group into $GL(n,\mathbb R)$ then it doesn't even make sense to talk about Lie algebra elements as matrices. You only get such concrete realizations of the lie algebra by starting with a concrete realization of the group. $\endgroup$ Sep 17, 2013 at 4:59

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