4
$\begingroup$

I have a question regarding the classical trace $\text{Tr} \colon \Psi^{-\infty}(S^1)\to \mathbb C$ on pseudodifferential operators of infinite negative order (i.e. smoothing operators), defined over the unit circle $S^1$ and acting on smooth functions. The definition of $\text{Tr}$ is given by $$ \text{Tr}(A) = \int_{S^1} k_A(x,x) $$ where $k_A$ denotes the Schwartz kernel of $A$.

I know that $\text{Tr}$ vanishes on commutators, and I would like to use this fact to study the trace of the an operator $C$ of the form $$ C(\psi) := f(\theta)\frac{\partial}{\partial \theta}\big(g(\theta)e^{\triangle} \,\psi\big) $$ where $f(\theta)$ and $g(\theta)$ are fixed smooth functions on $S^1$ (more details are known about these functions though they are rather technical in nature and I hope that these will not be relevant). Furthermore, $e^\triangle$ denotes the heat operator associated with a Laplace type operator $\triangle$. The question that I would like to understand is whether or not $C$ can generically be written as a (sum of) commutator(s). Hints and suggestions for survey reading would be helpful.

$\endgroup$

1 Answer 1

4
$\begingroup$

Take $f=g=1$, $\Delta$ the usual Laplace operator on $\mathbb S^1$, then $$ C_0=\partial_\theta e^{\partial_\theta^2}, $$ can be identified to the diagonal infinite matrix $ (ik e^{-k^2})_{k\in \mathbb Z} $ acting on $\ell^2(\mathbb Z)$ and has indeed a null trace. On the other hand, your $C$ is such that $$ C=f\partial_\theta g e^{\partial_\theta^2}= f[\partial_\theta, g ]e^{\partial_\theta^2} +fg \partial_\theta e^{\partial_\theta^2}=(fg'+fg) C_0. $$ Take now $ f=\sum_{k\ge 1}k^{-2} e^{ik\theta},\quad g=\sum_{k\ge 1}k^{-2} e^{-ik\theta}, $ so that we have $$ C(e^{im\theta})=\sum_{k,l\ge 1}k^{-2}l^{-2} e^{ik\theta}\partial_\theta\bigl(e^{-il\theta} e^{-m^2} e^{im\theta}\bigr)= \sum_{k,l\ge 1}k^{-2}l^{-2} e^{ik\theta}i(m-l)e^{-m^2}e^{i(m-l)\theta}, $$ and \begin{multline} \text{trace } C=\sum_{m\in \mathbb Z}e^{-m^2}\sum_{k=l\ge 1}k^{-2}l^{-2}i(m-l) =i\sum_{m\in \mathbb Z}e^{-m^2}\sum_{k\ge 1}k^{-4}(m-k) \\=i\sum_{m\in \mathbb Z}e^{-m^2}m\zeta(4) - i\sum_{m\in \mathbb Z}e^{-m^2}\zeta(3) =- i\zeta(3)\sum_{m\in \mathbb Z}e^{-m^2}\not=0. \end{multline} An iff condition for a null trace of $C$ can be found writing $$ f=\sum_k f_ke^{ik\theta},\quad g=\sum_l g_le^{il\theta}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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