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I'm trying to reconcile this proof that I've read that a group of invertible elements in a commutative (complex) Banach algebra have 1 or infinite connected components with this example I'm looking at. The example is the Banach algebra, $\mathcal B$ with basis elements $\{a,e\}$, with $a^2=e$. The singular values of this Banach algebra are, I think $\{\lambda(a+e):\lambda\in\mathbb C\}$ and $\{\lambda(a-e):\lambda\in\mathbb C\}$, which should create $4$ quadrants.

The gist of the proof is that $\sigma(a)=\{1,-1\}$, so define a logarithm branch on $\Omega\supseteq\sigma(a)$. Extend this to $\mathcal B$, basically by its power series. Now $a'=\log(a)\in\mathcal B$, so that $\exp(\lambda a'),\lambda\in[0,1]$ is a path connecting $a$ to $e$.

I believe this proof, but it's hard to see where it breaks down and crosses the singular boundary on my example, because it's hard to get my hands on the extension of $\log$ and $\exp$. Any help? What am I missing?

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If your example $\mathcal{B}$ were a real Banach algebra instead of a complex Banach algebra, then you would be right that there are four connected components, since $\mathcal{B}$ can be identified with $\mathbb{R}^2$ and the invertible elements split into four quadrants. But over $\mathbb{C}$, you have the complement in $\mathbb{C}^2$ of two (complex) one-dimensional subspaces, and this complement is actually connected! To see why, note that, for instance, the complement of $\mathbb{C}\times\{0\}$ in $\mathbb{C}^2$ is connected, since this complement is just $\mathbb{C}\times(\mathbb{C}\setminus\{0\})$ and $\mathbb{C}\setminus\{0\}$ (unlike $\mathbb{R}\setminus\{0\}$) is connected. Similarly, the complement of $\mathbb{C}\times\{0\}\cup\{0\}\times\mathbb{C}$ is also connected since it is $(\mathbb{C}\setminus\{0\})\times(\mathbb{C}\setminus\{0\})$. By more complicated arguments, you can show that the complement of any finite union of proper subspaces of a complex vector space is connected.

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  • $\begingroup$ Thanks! I had realized that two sets where planes, but for some reason was picturing $\mathbb C^2$ like 3 real dimensions. I should have figured that out, but I was looking for an issue in my understanding of these theorems. $\endgroup$ – T.J. Gaffney Jun 26 '16 at 3:41

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