6
$\begingroup$

Voevodsky defined the slice filtration on the motivic stable homotopy category $SH(S)$ over a Noetherian scheme $S$. In the article Open Problems in the Motivic Stable Homotopy Theory, I, Section 2, he defined $SH^{eff}(S)$ to be the smallest triangulated subcategory in $SH(S)$ which is closed under direct sums and contains suspension spectra of spaces. It follows from a result of Neeman that the inclusion functor $i_n:\Sigma^n_TSH^{eff}(S)\to SH^{eff}(S)$ has a right adjoint $r_n$. Since $i_n$ is full, we have a natural isomorphism $\mathrm{id}\simeq r_n\circ i_n$. Denote $f_n=i_n\circ r_n$. Applying the counit $f_{n+1}\to\mathrm{id}$ to $f_n$, we get a natural transformation $f_{n+1}\circ f_n\to f_n$. Voevodsky claims that $f_{n+1}=f_{n+1}\circ f_n$ so that we get a natural transformation $f_{n+1}\to f_n$. The slice functor $s_n$ is defined to be the cofiber of this map. Denoting the inclusion functor $\Sigma^n_TSH^{eff}(S)\to \Sigma^{n-1}_TSH^{eff}(S)$ by $j_n$, (so that $i_{n+1}=i_n\circ j_{n+1}$), I calculated that $$f_{n+1}=i_{n+1}\circ r_{n+1}=i_n\circ j_{n+1}\circ r_{n+1}=i_n\circ\mathrm{id}\circ j_{n+1}\circ r_{n+1}$$ $$\simeq i_n\circ r_n\circ i_n\circ j_{n+1}\circ r_{n+1}=f_n\circ f_{n+1}$$ but don't see why $f_{n+1}=f_{n+1}\circ f_n$. Note that $f_{n+1}=f_n\circ f_{n+1}$ also induces a natural transformation $f_{n+1}\to f_n$.

Am I missing something here? Or is there a typo in that paper?

$\endgroup$

1 Answer 1

6
$\begingroup$

The key here is that $SH^{eff}(S)$ is closed under suspensions, so there's an inclusion $j_{n+1}:\Sigma^{n+1}_T SH^{eff}(S)\subseteq \Sigma^n_T SH^{eff}(S)$. Hence you can write $i_{n+1}=i_n \circ j_{n+1}$ and $r_{n+1} = l_{n+1}\circ r_n$ where $l_{n+1}$ is the right adjoint of $j_{n+1}$ (which exists because $j_{n+1}$ commutes with colimits). So

$f_{n+1}f_n = i_{n+1}r_{n+1}i_nr_n = i_n j_{n+1} l_{n+1} r_n i_n r_n = i_n j_{n+1}l_{n+1}r_n = f_{n+1}\,.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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