Closed subgroup of $GL({cal A})$
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I'm trying to prove that for any algebra ${cal A}$ , $Aut({cal A}) = {a in End({cal A}) | a([x, y]) = [a(x), a(y)], forall x, y in {cal A}}$ is a closed subgroup of $GL({cal A})$ . I can prove that it is a subgroup just showing that $acirc a^{'-1}$ is in $Aut({cal A})$ which is direct computation. Nevertherless, I don't know how to show that thi group is closed. I've been thinking about proving that it is the isotropy group of $GL({cal A})$ , but I don't get it. I tried to impose that $[x, y]$ is the fixed point for the isotropy subgroup, but I just get: $$a([x, y]) = a(x)a(y) - a(y)a(x) = xy - yx$$ But this means nothing because $a in Aut({cal A})$ is not imposing that $[x, y]$ is fixed. How do I solve it?
group-theory lie-groups lie-al...
