Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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
1  
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
2  
@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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.