I have some basic questions on the relation of looping and delooping in the stable and unstable homotopy categories. I state them it in the motivic setting, but if somebody has an answer for an analogue question in the classic setting it would be appreciated.

Let $X$ be a scheme, in Voevodsky motivic homotopy theory we have an unstable homotopy category of $\mathbf{H}(X)$, a stable homotoy category $\mathbf{SH}(X)$ and an infinite suspension $\Sigma^\infty \colon \mathbf{H}(X) \to \mathbf{SH}(X)$. We have a looping given by the internal Hom object $\underline{\mathrm{Hom}}_{\mathbf{H}(X)}(\mathrm{Th}(\mathcal{O}), \underline{\phantom{a}}\ )$ in $\mathbf{H}(X)$ and also a looping $\underline{\mathrm{Hom}}_{\mathbf{SH}(X)}(\mathrm{Th}(\mathcal{O}), \underline{\phantom{a}}\ )=\mathrm{Th}(-\mathcal{O})\wedge \underline{\phantom{a}}\ $ in $\mathbf{SH}(X)$. I would like to understand the relation between $\Sigma^\infty \underline{\mathrm{Hom}}_{\mathbf{H}(X)}(\mathrm{Th}(\mathcal{O}), \underline{\phantom{a}}\ )$ and $\mathrm{Th}(-\mathcal{O})\wedge \underline{\phantom{a}}\ $ and if for some simple objects they are the same.

To be more precise, let $L$ be a line bundle in $X$.

  1. What is the relation between $\Sigma^\infty \underline{\mathrm{Hom}}_{\mathbf{H}(X)}(\mathrm{Th}(\mathcal{O}), \mathrm{Th}(L))$ and $\mathrm{Th}(-\mathcal{O})\wedge \mathrm{Th}(L) $? And between $\mathrm{Th}(\mathcal{O})\wedge \Sigma^\infty \underline{\mathrm{Hom}}_{\mathbf{H}(X)}(\mathrm{Th}(\mathcal{O}), \mathrm{Th}(L))$ and $ \mathrm{Th}(L)$? Are they isomorphic?
  2. If they are not isomorphic, it seems to me that they do have the same cohomology. To fix ideas: is their $K$-theory $K_0(\Sigma^\infty \underline{\mathrm{Hom}}_{\mathbf{H}(X)}(\mathrm{Th}(\mathcal{O}), \mathrm{Th}(L)))$ and $K_0(\mathrm{Th}(L-\mathcal{O}))$ isomorphic? What about $K_0(\mathrm{Th}(\mathcal{O})\wedge \Sigma^\infty \underline{\mathrm{Hom}}_{\mathbf{H}(X)}(\mathrm{Th}(\mathcal{O}), \mathrm{Th}(L)))$ and $K_0(\mathrm{Th}(L))$, are they isomorphic?

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.