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At the first page of the following article http://arxiv.org/pdf/1004.1018v1.pdf [edit: the formula on the arXiv differs from the formula in the published paper, and the formula displayed below is the published version], I read "The degree of a Hölder continuous function $f:\mathbb{S}^1 \to \mathbb{S}^1$ of exponent $\alpha$ is expressed by an analytic formula..." $$ \text{deg} (f)=\frac{1}{(2i \pi)^{2k}} \int \overline{f(z_0)} \frac{f(z_1)-f(z_0)}{z_1-z_0} \ldots \frac{f(z_0)-f(z_{2k})}{z_0-z_{2k}} dz_0 \ldots dz_{2k} $$ whenever $\alpha(2k+1)>1$.

Problem is, I don't know how to understand this formula to make it work. For example, taking $f=id$ and $k=1$ and integrating over $\mathbb{S}^1$,

$$ \text{deg} (f)= \frac{1}{(2i \pi)^2} \int_{\mathbb{S}^1} \overline{z_0} dz_0 \int_{\mathbb{S}^1} dz_1 \int_{\mathbb{S}^1} dz_2=0 $$ which is obviously false. The formula comes from "Non-commutative geometry" http://www.alainconnes.org/docs/book94bigpdf.pdf by Connes, p 215 which seems really interesting but doesn't help me much.

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  • $\begingroup$ I don't think it helps, but the formula in that paper has $(2\pi i)^{2k}$ on the bottom of the fraction. $\endgroup$
    – Dan Ramras
    Jul 24, 2015 at 4:06
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    $\begingroup$ the published formula differs from the arXiv version that you cite by complex conjugation of $f(z_0)$. $\endgroup$ Jul 24, 2015 at 14:11
  • $\begingroup$ More specifically, the published version has $\overline{f(z_0)}$ as the first term of the integrand; the other appearances of $f(z_0)$ are as written (no conjugation). I will edit the question accordingly. $\endgroup$
    – Dan Ramras
    Jul 24, 2015 at 19:24
  • $\begingroup$ so this is an example of how much ambiguity ... can bring. $\endgroup$
    – Fan Zheng
    Dec 29, 2015 at 2:37

2 Answers 2

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First, let me apologize for the confusion regarding the pre-factor. It should read $(-1)^{k}(2\pi i)^{-2k-1}$. This can be seen in several ways. One way is to compare the constant $c_k$ appearing in Proposition 3.b on page 215 of Connes' book (http://www.alainconnes.org/docs/book94bigpdf.pdf) with those coming from the pairing with $K$-theory in Proposition 3.a on page 230. Beware that there is a missing $2\pi i$ coming from that the cyclic $1$-cocycle $\tau$ on page 215 needs to be normalized by $(2\pi i)^{-1}$ to reproduce the winding number.

Another way to see the pre-factor is in the origin of the formula. It comes from the following index computation. Really, it is what the odd index pairing in cyclic cohomology boils down to once all normalization constants are cancelled. We consider a projection $P$ and a unitary operator $u$ on a Hilbert space $\mathcal{H}$ such that the commutator $[P,u]$ belongs to the Schatten class $\mathcal{L}^{2p}(\mathcal{H})$. The operator $T:=PuP:P\mathcal{H}\to P\mathcal{H}$ is Fredholm, its inverse in the Calkin algebra is represented by $Pu^{-1}P$. To compute the index of $PuP$, one can use the computations in Proposition 4 on page 303 of Connes' bok. Following the proof of this proposition, $$\mathrm{ind}(PuP)=(-1)^p\mathrm{Tr}(\gamma e[F,e]^{2p}),$$ where $\gamma=\begin{pmatrix} 1&0\\0&-1\end{pmatrix}$, $e=\begin{pmatrix}P&0\\0&P\end{pmatrix}$ and $F=\begin{pmatrix}0&U^*\\U&0\end{pmatrix}$ are operators on $\mathcal{H}\oplus \mathcal{H}$. After some computations, we arrive at the expression $$(-1)^p\mathrm{Tr}(\gamma e[F,e]^{2p})=(-1)^p\mathrm{Tr}((2P-1)([U^*,P][U,P])^p).$$ If $[U^*,P]\in \mathcal{L}^{2p-1}(\mathcal{H})$ one can proceed and write $$(-1)^p\mathrm{Tr}((2P-1)([U^*,P][U,P])^p)=\frac{(-1)^{p}}{2}\mathrm{Tr}((2P-1)[(2P-1),U^*][P,U]([P,U^*][P,U])^{p-1})=(-1)^{k-1}\mathrm{Tr}(U^*[P,U]([P,U^*][P,U])^{k}),$$ where $k=p-1$.

The winding number of a mapping $f:S^1\to S^1$ can be computed from an index as above by taking $u$ to be point wise multiplication by $f$ and $P$ to be the Szegö projection. To be precise, the index formula for Toeplitz operators on the circle says that $\mathrm{deg}(f)=-\mathrm{ind}(PuP)$. The Szegö projection on the circle is defined by $$Pg(z):=\frac{1}{2\pi i}\int_{S^1} \frac{g(w)}{w-z}\mathrm{d}w=\frac{1}{2\pi}\int_{0}^{2\pi} \frac{g(\mathrm{e}^{i\theta})}{1-\mathrm{e}^{-i\theta} z}\mathrm{d}\theta.$$ The integral should be interpreted as a non tangential interior limit, i.e. $Pg(z):=\lim_{r\to 1^-} Pg(rz)$. For $a\in C^\alpha(S^1)$, the operator $[P,a]$ has an absolutely integrable integral kernel given by $\frac{a(w)-a(z)}{w-z}$. Note that if $f:C^\infty(S^1)$ is smooth, $[P,f]\in\mathcal{L}^{2p}(L^2(S^1))$ for any $p>0$ and all traces considered above are well defined. More generally, for $f\in C^\alpha(S^1)$, $[P,f]\in\mathcal{L}^{\frac{1}{\alpha},\infty}(L^2(S^1))\subseteq \cap_{p>\frac{1}{\alpha}} \mathcal{L}^p(L^2(S^1))$. Using the integral formula in Theorem 2.4 from http://arxiv.org/pdf/1004.1018v1.pdf, we arrive at the expression $$\mathrm{deg}(f)=(-1)^{k}\int_{(S^1)^{2k+1}} \overline{f(z_0)}\frac{f(z_1)-f(z_0)}{2\pi i(z_1-z_0)}\prod_{j=1}^{k}\frac{\overline{f(z_{2j})}-\overline{f(z_{2j-1})}}{2\pi i(z_{2j}-z_{2j-1})}\frac{f(z_{2j+1})-f(z_{2j})}{2\pi i(z_{2j+1}-z_{2j})}\mathrm{d}z_0\mathrm{d}z_1\cdots \mathrm{d}z_{2k},$$ where we identify $z_{2k+1}=z_0$. This gives the pre-factor $(-1)^k(2\pi i)^{-(2k+1)}$, where the number of $2\pi i$ correspond to the number of commutators with the Szegö projection.

There is a tricky point to this formula. The Szegö projection is defined from an interior limit. However, for $a\in C^\alpha(S^1)$ the commutator $[P,a]$ is defined from an absolutely integral kernel. The latter fact makes it unnecessary to include the interior limit in the expression for the mapping degree. However, when computing it makes life easier. To consider an example of this, take $f(z)=z$. The case $k=0$ holds formally as the integrand $\overline{f(z_0)}\frac{f(z_1)-f(z_0)}{2\pi i(z_1-z_0)}$ has the limit $\frac{1}{2\pi i}\overline{f(z_0)}f'(z_0)$ as $z_1\to z_0$ (this can be made rigorous using for instance Theorem 3.9 in Barry Simon's book on trace ideals). For $k=1$, we have \begin{align*} \mathrm{deg}(f)&= \frac{-1}{(2\pi i)^3}\int_{(S^1)^3} \overline{z_0}\frac{z_1-z_0}{z_1-z_0}\frac{\overline{z_2}-\overline{z_1}}{z_{2}-z_{1}}\frac{z_0-z_{2}}{z_{0}-z_{2}}\mathrm{d}z_0\mathrm{d}z_1\mathrm{d}z_{2}\\ &=\frac{-1}{(2\pi i)^3}\int_{(S^1)^3} \overline{z_0}\frac{\overline{z_2}-\overline{z_1}}{z_{2}-z_{1}}\mathrm{d}z_0\mathrm{d}z_1\mathrm{d}z_{2}. \end{align*} If we perform the integral over $z_2$, keeping in mind that it is defined from an interior limit, we arrive at $$\frac{-1}{(2\pi i)^3}\int_{(S^1)^3} \overline{z_0}\frac{\overline{z_2}-\overline{z_1}}{z_{2}-z_{1}}\mathrm{d}z_0\mathrm{d}z_1\mathrm{d}z_{2}=\frac{1}{(2\pi i)^2}\int_{(S^1)^2} \overline{z_0}\overline{z_1}\mathrm{d}z_0\mathrm{d}z_{1}.$$ Here we can apply the residue theorem arriving at the identity $\mathrm{deg}(f)=1$.

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a bit of searching produced this formula on page 37 of Magnus Goffeng's thesis:



this should be the same formula as in the cited publication (it refers to the same proposition from Connes), but you'll notice that the function and its reciprocal alternate from term to term.

So let's try this example of the identity, $u(z)=z$, with $k=1$,

$${\rm deg}(u)=\frac{1}{(2\pi i)^2} \oint \frac{1}{z_0 z_1 z_2} dz_0 dz_1 dz_2=2\pi i$$

I would have expected a degree of 1, presumably another factor of $2\pi i$ is missing from Goffeng's formula. I would also have expected a factor $(-1)^k$ instead of just a minus sign (to get a $k$-independent degree); apart from these details, the formula seems to make sense.


update, December 2015: Magnus Goffeng has now confirmed that, indeed, the factor $-(2\pi i)^{-2k}$ should read $(-1)^k(2\pi i)^{-2k-1}$.

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