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Assume $A$ is a finite dimensional associative $\mathbb{R}-$algebra, with identity $1( \neq 0)$, let $A^{\times}$ be the set of all invertible elements of $A$, then it's easy to see that $A^{\times}$ is a Lie group and open in $A$. My question is: what is its Lie algebra?

The answer seems to be $A$, with Lie bracket $[a, b]=ab-ba$, but I don't know why is it so.

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  • $\begingroup$ If $A$ is associative then the Lie bracket is indeed $[a,b]=a.b-b.a$ (the Jacobi identity follows from associativity of the product $a.b$ in $A$. However, there may be no invertible elements in $A$. Consider the $2$-dimensional real algebra with basis $(e_1,e_2)$ and $e_1.e_1=e_2$ and the other products zero. $\endgroup$ – Dietrich Burde Dec 7 '13 at 20:44
  • $\begingroup$ Of course $A$ is associative. $\endgroup$ – Lao-tzu Dec 8 '13 at 1:01
  • $\begingroup$ You also need your algebra to have a unit, otherwise take any vector space with identically zero multiplication. (In many contexts, that "algebras" are associative and have units would be the default assumption, but when Lie algebras are in play...clearly not.) $\endgroup$ – Pete L. Clark Dec 8 '13 at 1:16
  • $\begingroup$ OK, I will make this clear. $\endgroup$ – Lao-tzu Dec 8 '13 at 1:19
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I suggest thinking of the canonical embedding of $A$ as a subalgebra of $\operatorname{End} A$, $x \mapsto (y \mapsto xy)$.

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