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Let $f$ be an invertible element of $C({\mathbb{T}}; C_b(r,1))$, that is, there exists a $f^{-1}\in C({\mathbb{T}}; C_b(r,1))$ such that for all $z\in {\mathbb{T}}$, $f(z)f^{-1}(z)=1$ in $C_b(r,1)$. Here $C_b(r,1)$ denotes the $C^*$-algebra of complex-valued bounded continuous functions on the open interval $(r,1)$, where $r$, fixed, belongs to $[0,1)$, and $\mathbb{T}$ denotes the unit circle with center $0$. Then $f$ induces a bounded operator $M_f$ on $L^2({\mathbb{T}}; L^2(r,1))$ in a natural manner: for each $z$ in $\mathbb{T}$, we have the multiplication operator corresponding to $f(z)$ in $C_b(r,1)$ going from $L^2(r,1)$ to $L^2(r,1)$. Using the projection $P:L^2({\mathbb{T}}; L^2(r,1)) \rightarrow H^2({\mathbb{T}}; L^2(r,1))$, we can consider the Toeplitz operator $T_f $ on $H^2({\mathbb{T}}; L^2(r,1))$ given by $T_f g = P (M_f g)$ for $g$ in $H^2({\mathbb{T}}; L^2(r,1))$.

My question is this: Is $T_f$ is Fredholm?

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I'm a bit confused by $C(r,1)$...do you mean all bounded continuous functions? –  Mike Jury Jun 5 '13 at 11:37
Yes, sorry. You are right, it should be bounded continuous functions. I have rephrased it now. –  Amol Sasane Jun 5 '13 at 11:41
Maybe I'm missing something, but it seems like the answer should be no...if we take $f(z)=z\otimes 1$, doesn't this produce a shift of infinite multiplicity? –  Mike Jury Jun 5 '13 at 11:51
What does $H^2$ mean in this context? Usually it means $L^2$ functions whose Fourier coefficients in negative degree vanish; does that have some interpretation here? –  Paul Siegel Jun 5 '13 at 12:54
@Mike Jury: Many thanks! (The question was prompted by the following consideration. Any $f$ invertible in $C({\mathbb{T}};C_b(r,1))$ has a well-defined winding number associated with it---we can fix any $R\in (r,1)$ and look at the winding number $w(f)$ of $z\mapsto (f(z))(R):\mathbb{T}\rightarrow \mathbb{C}\setminus \{0\}$. This integer $w(f)$ is independent of $R$. I was hoping that the associated Toeplitz $T_f$ would be Fredholm with a Fredholm index equal to $-w(f)$.) –  Amol Sasane Jun 5 '13 at 14:09

1 Answer 1

up vote 3 down vote accepted

This is too long for a comment; but maybe it can help. (I am not an expert in $C^*$-algebras, though, so it is very possible that I am all wet. If anyone can clean this up I would be grateful). Per the comments, we have that $T_f$ is not Fredholm but $f$ does have a winding number. This winding number may have the following $K$-homology interpretation: let us write simply $C(X)$ for $C_b(r,1)$. Then $C(\mathbb{T}; C(X))$ may be identified with the algebra $C(\mathbb{T})\otimes C(X)$ (these algebras are commutative, hence nuclear, so there is only one $C^*$ norm on the tensor product). Likewise, it seems (though this should be checked carefully) that the $C^*$ algebra generated by the $T_f$'s should be isomorphic to $\mathcal T\otimes C(X)$, where $\mathcal T$ denotes the usual Toeplitz algebra (continuous symbols). On the other hand, since $C(X)$ is nuclear, it is exact, and hence preservers exact sequences under tensoring. Thus by tensoring the usual exact sequence for the Toeplitz algebra with $C(X)$, we have an exact sequence $$ 0\to \mathcal{K}\otimes C(X) \to \mathcal T\otimes C(X)\to C(\mathbb{T})\otimes C(X)\to 0. $$ This sequence represents an element of the $K$-homology group $Ext(C(\mathbb{T})\otimes C(X), \mathcal{K}\otimes C(X))$. If there is any justice in the world, the quotient map here should be your "symbol map" $T_f\to f$, and it seems likely that this extension generates a copy of $\mathbb{Z}$ in the $Ext$ group; the pairing of this $Ext$ element with $K^1(C(\mathbb T)\otimes C(X))$ should recover the winding number you describe in the comments. All of this needs to be checked of course, and I am not sure if the non-seperability of $C(X)$ breaks anything.

EDIT: On further reflection, it seems like getting the pairings to work out is really a job for $KK$-theory, which is beyond me; additionaly the non-seperability may be an issue there.

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