First off, this is a vague question about a survey which is, I guess, meant to be vague. So bear with me

In Morel's "Motivic Homotopy Theory" survey he mentioned the following fact in motivating the Tate circles. One notes that $\mathbb{P}^1$ is equivalent to $S^1 \wedge \mathbb{G}_m$ while, if we follow our topological intuition, $\mathbb{P}^1$ ought to be $S^2$ which is equivalent $S^1 \wedge S^1$. So we need to keep track of this difference between the "usual" topological situation and count the number of $\mathbb{G}_m$'s.

Now Morel justified the first equivalence by saying that $\mathbb{A}^1$ is invertible in the the unstable homotopy category. With just this information, is there an "intuitive", or maybe "geometric" (i.e. without going to the details of the motivic unstable category) way to see how this fact leads to the equivalence above?

Any other comments about these two-circles phenomena are much appreciated!


1 Answer 1


So I learned this today (credits to Marc Hoyois). The point is that we can think of $\mathbb{P^1}$ as gluing of two copies of the affine line along $\mathbb{G_m}$. This is a homotopy pushout diagram so, because $\mathbb{A}^1$ is contractible you are taking the homotopy pushout of $\star \leftarrow \mathbb{G}_m \rightarrow \star$ which will give you a suspension of $\mathbb{G}_m$, i.e. $S^1 \wedge \mathbb{G}_m$.


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