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I'm currently working on a problem for homework in my Banach algebras course and I've run into a bit of an issue with terminology. Let $\mathcal{A}$ be a Banach algebra, then $\mathcal{A}^{-1}_0$ denotes the subgroup of invertible elements that can be written as a finite product of exponentials. That is to say that if $a\in\mathcal{A}^{-1}_0$, then $a=e^{b_1}\cdots e^{b_n}$, where $b_1,\ldots,b_n\in\mathcal{A}$.

My question is for clarification as to what the following means (I do not want help with the actual proof, just clarification): Show that $\mathcal{A}^{-1}_0$ consists of all invertible elements which can be connected to the identity by a continuous path of invertible elements.

I'm aware that such an idea exists in Lie theory but I've never studied that in any meaningful way so paths of elements isn't a well-defined concept for me.

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Well, as a Banach algebra, $\mathcal A$ has a norm, so naturally a topology, hence 'continuous path' does make sense here.

One direction of the statement is easy: if $a=e^{b_1}e^{b_2}\dots e^{b_n}$, then the continuous function $[0,1]\to\mathcal A,\ \,t\mapsto e^{tb_1}e^{tb_2}\dots e^{tb_n}\ \,$ connects $1$ to $a$.

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  • $\begingroup$ Oh I see what continuous path means now. I wasn't really sure what that term meant. Thanks. $\endgroup$ – Cameron Williams Jan 26 '14 at 23:58
  • $\begingroup$ Though, I cannot see yet the other direction. Can you share with me if you find it? $\endgroup$ – Berci Jan 27 '14 at 0:01
  • $\begingroup$ I will if/when I figure it out! $\endgroup$ – Cameron Williams Jan 27 '14 at 0:02
  • $\begingroup$ I guess it would require some differentiation somehow.. Probably the topology induced by the norm is nice enough that every continuous path can be transformed arbitrarily close to a differentiable path. $\endgroup$ – Berci Jan 27 '14 at 0:06

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