For the n Heisenberg($\Bbb C^n\times\Bbb R$) it is known that the heat kernel $q_s(z,t)=c_n\int_{\Bbb R} e^{-i\lambda t}\Big( \frac{\lambda}{\sinh(\lambda s)}\Big)^n e^{-\frac{\lambda|z|^2\coth(\lambda s)}{4}}d\lambda$ satisfied the followin estimate: $$q_s(z,t)\leq C s^{-n-1} e^{\frac{A}{s}|(z,t)|^{1/2}} $$ with some positive constants $C$ and $A$ and $(z,t)\in \Bbb C^n\times\Bbb R$ with $|(z,t)|=(|z|^4+|t|^2)^{\frac{1}{4}}$.

Now for the one quaternion Heisenberg it is known that the heat kernel $p_s(x,t)=c\int_{\Bbb R^3} e^{-i\lambda .t}\Big( \frac{|\lambda|}{\sinh(|\lambda| s)}\Big)^2 e^{-\frac{ |\lambda||x|^2\coth(|\lambda| s)}{4}}d\lambda$ where $(x,t)\in Q\times \Bbb R^3$ and $Q$ the set of quaternions.

My question can we have the same estimate for $p_s(x,t)$. Thank you in adavance



Your Answer

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

Browse other questions tagged or ask your own question.