# Set of invertible operators in B(H) is connected. Is it true? Is there a reference?

Suppose $H$ is a Hilbert space, $B(H)$ is the algebra of bounded linear operators on it, $K(H)$ is ideal of compact operators in $B(H)$, $Inv(B(H)/K(H))$ is the topological group of invertible operators in $B(H)/K(H)$, $Inv(B(H)/K(H))_0$ --- connected component of $id$ in $Inv(B(H)/K(H))$. $ind\colon Inv(B(H)/K(H))\to \mathbb{Z}$ --- Fredholm index.

I want to find a reference for the following fact:

Fact 1. If $H$ is infinite-dimensional and separable then $ind$ is locally-constant and provides an isomorphism between $Inv(B(H)/K(H))/Inv(B(H)/K(H))_0$ and $\mathbb{Z}$.

In Murphy's textbook, where I've read almost all I know about Fredholm index, there is Atkinson's theorem, the fact, that Fredholm index is locally constant and $ind(ab)=ind(a)+ind(b)$. But in order to check the fact 1, I need also the fact, stated in the head of the question, or, equivalently, the following:

Fact 2. If $H$ is infinite-dimensional and separable then the set $\{a\in B(H)/K(H)\mid ind(a)=0\}$ is connected.

Where can I find the references? Where can I read that?

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The proof of 1 as e.g. in the quoted paper by Kuiper is really elementary fact; it's a linear algebra version of Hilbert's Grand Hotel argument (aka Hotel California). As to 2, recall that any Fredholm operator $T:X\to Y$ of can be written as a direct sum $0\oplus T_1: X_0\oplus X_1\to Y_0\oplus Y_1$, with $T_1:X_1\to Y_1$ invertible, $X_0$ and $Y_0$ finite dimensional, $\mathrm{ind}(T)=\dim(X_0)-\dim(Y_0)$; if $\mathrm{ind}(T)=0$ there is an isomorphism $A:X_0\to Y_0$ that produces a path $[0,1]\ni s \mapsto (sA)\oplus T_1$ from $T$ to an invertible operator. – Pietro Majer Sep 27 '10 at 15:27
@ Pietro Majer: You have only prooved, that any connected component of $Inv(B(H)/K(H))$ with 0 Fredholm index contains an invertible operator. Fact 2 doesn't follow from this. – Fiktor Sep 27 '10 at 19:07

## 3 Answers

You can use some spectral theory to show that the set of unitary operators in $B(H)$ is path-connected. Then the (path-)connectedness of the invertibles follows easily from the polar decomposition. See Theorem 5.29 and Corollary 5.30 in Douglas's Banach Algebra Techniques (1st edition).

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Thank you, this is very good reference, because 1. it is a good-written book, 2. I'm already referencing it. – Fiktor Sep 27 '10 at 16:32

Kuiper's theorem gives the contractibility of the invertible operators in $B(H)$. Start with

http://en.wikipedia.org/wiki/Kuiper's_theorem

Check out papers of Boris Mitjagin [Mitiagin] in the 1970s for more information.

EDIT 9/27: In "A course in functional analysis" (Springer TTM 96), Conway proves in Theorem XI.4.1 that the components of the semi-Fredhlom operators on $\ell_2$ are determined by the index.

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I recommend the paper by P. de la Harpe, {\it Initiation a l'algebre de Calkin}, pp. 180-219 in Alg`ebres d'Op\'erateurs, Springer Lect. Notes in Math. 725, 1979; all these facts (and much more) are available there.

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