# Triangulated structure on $\mathbf{SH}(S)$: $\mathbb{P}^1$-suspension versus classical suspension

I am studying the construction of the motivic stable homotopy category of schemes $\mathbf{SH}(S)$ following Riou's paper Categorie homotopiquement stable d'un site suspendu avec intervalle (click to see the paper) and I have a problem due to my weakness in homotopy theory.

Riou proves that $\mathbf{SH}(S)$ is a triangulated category out of a result of Quillen in his book Homotopical Algebra . The result Riou uses is the following (cf. Corollaire 3.4 in Riou's paper)

Corollary 3.4: Let $C$ be a pointed model category and let $H$ be its homotopy category. Assume that the suspension functor $\Sigma \colon H\rightarrow H$ is an autoquivalence of categories. Then $H$ is a triangulated category.

Riou then proves that $\mathbf{SH}(S)$ is a triangulated category in the following result (I state it for the case of the stable homotopy category of schemes):

Theorem 3.10: Recall there is an isomorphism $\mathbb{P}^1\simeq S^1\wedge \mathbb{G}_m$ in $H_\bullet (S)$. Then the functor $\underline{\phantom{a}}\wedge S^1\colon \mathbf{Spt}^{\mathbb{P}^1}\to \mathbf{Spt}^{\mathbb{P}^1}$ induces an equivalence of categories $\mathbf{SH}(S)\rightarrow \mathbf{SH}(S)$. The category $\mathbf{SH}(S)$ is therefore canonically triangulated.

Lemma 3.11: If the suspension functor $\mathbf{SH}^{S^1}(S)$ is an equivalence of categories then so is the suspension in $\mathbf{SH}^{S^1\wedge\mathbb{G}_m}(S)=\mathbf{SH}(S)$.

He assumes that the suspension functor in $\mathbf{SH}$ is $\underline{\phantom{a}}\wedge \mathbb{P}^1\colon \mathbf{Spt}^{\mathbb{P}^1}\to \mathbf{Spt}^{\mathbb{P}^1}$. My question is the following:

Question: Quillen's suspension on a model category $C$ is a very concrete one he defines in terms of the model structure (cf. I.2 Theorem 2 of Homotopical algebra) It is natural to expect that the suspension on $\mathbf{SH}$ is $\underline{\phantom{a}}\wedge \mathbb{P}^1$. But: why is Quillen suspension functor $\underline{\phantom{a}}\wedge \mathbb{P}^1$ in $\mathbf{SH}(S)$? How does one prove such a thing?

• This is not true: suspension is always $-\wedge S^1$. I think that's also what Riou means (he would say $T$-suspension'' otherwise). Aug 17, 2015 at 13:31
• You are right. I was assuming that suspension was $\underline{\phantom{a}}\wedge T$, not Riou, because he was talking about $T$-spectra. You are right, thank you very much Aug 17, 2015 at 14:18
• "Hovey: Spectra and symmetric spectra in general model categories" is a general strategy and results about how to invert a good endofunctor (left Quillen) $F: C\rightarrow C$ of a good model category $C$ (simplicial, closed monoidal model category, left proper, combinatorial or cellular,...). If I'm not wrong Hoveys's motivation was a generalization of Morel-Voevodsky stable category. Aug 17, 2015 at 17:42
This question is answered by Hoyois' first comment: suspension is always $\underline{\phantom{a}}\wedge S^1$. One should write $\mathbb{P}^1$-suspension otherwise.