Given a finite dimmensional associative algebra $A$, it can be proven that the unit group $$A^{\ast}:=\{a\in A \mid \exists\, b \in A \mbox{ such that } ab=ba=1_A\} $$ is always a Lie group (The unit group of a finite dimensional associative algebra is a Lie group?), but I have problems trying to identify its Lie algebra. I think that its Lie algebra $(\mathfrak{a},[.,.])$ should be isomorphic (as Lie algebras) to $(A,[.,.]_c)$, where $[a,b]_c=ab-ba$ (as the case $A=M_n(\mathbb{K})$). It is true?? I hope somebody can help me

  • $\begingroup$ You should add the assumption of the linked question and answer that $\Bbb K$ is a complete valued field (as otherwise the concept of "Lie group" makes little sense to begin with). $\endgroup$ – Torsten Schoeneberg Feb 11 at 1:31
  • $\begingroup$ Well, of course $\mathbb{K}=\mathbb{R} $ or $\mathbb{C}$. $\endgroup$ – GaSa Feb 11 at 1:36
  • $\begingroup$ Of course ... (Many people are investigating $p$-adic Lie groups, but of course that need not be your business.) Your conjecture sounds plausible, but what have you tried? And what construction of the Lie algebra of a Lie group are you using in this generality -- invariant vector fields, tangent space, invariant derivations ...? $\endgroup$ – Torsten Schoeneberg Feb 11 at 1:39
  • $\begingroup$ I have tried considering $\mathfrak{a}$ as tangent space but the problem is choose a basis on $A$ (based in the case $A=M_n(\mathbb{K}$)). $\endgroup$ – GaSa Feb 11 at 1:43

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