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13 votes
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Realization Functor From $SH$ to Derived Category of $Gal$-Modules

The most general functor of this form was constructed by Ayoub in La réalisation étale et les opérations de Grothendieck. Ayoub considers the ∞-category $DA^{et}(S,\Lambda)$ which is defined exactly …
Marc Hoyois's user avatar
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7 votes
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Blow-ups in Motivic Homotopy Theory

I assume that $Z$ is also smooth over $k$. Then the base change of the map $$Bl_Z(X)-\sigma(Z)\to X$$ to $Z$ is the map $$\mathbb{P}(N_{X,Z})-\sigma(Z) \to Z$$ where $\mathbb{P}(N_{X,Z})$ is the …
Marc Hoyois's user avatar
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4 votes
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Algebraic cobordism (of a point) outside the geometric diagonal

The answer to both question is yes, provided you invert $p$ in characteristic $p$ (though conjecturally this is not necessary). In fact, as far as I can see, all of Spitzweck's computations apply to f …
Marc Hoyois's user avatar
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18 votes
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Category of motivic spectra

Here's a direct link to the book by Hovey–Palmieri–Strickland. The category of motivic spectra is known to satisfy the axioms of Definition 1.1.4 in the book when the base is a countable field of cha …
Marc Hoyois's user avatar
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8 votes
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Ring structure for the motivic spectrum/complex that represents singular cohomology?

Singular cohomology is represented by an $E_\infty$ motivic ring spectrum. That spectrum is $\mathbf{R}f(H\mathbb{Z})$ where $f$ is right adjoint to the stable topological realization functor and $H\m …
Marc Hoyois's user avatar
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7 votes
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When is the Thom spectrum of a virtual vector bundle effective?

Yes. A bit more generally, if $\xi$ is a perfect complex of rank $\geq 0$, then $Th(\xi)$ is effective (even very effective): the question is Nisnevich-local on $X$ and $\xi$ is locally a complex of …
Marc Hoyois's user avatar
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5 votes

A question about the vanishing of motivic cohomology in negative Tate twist

Here's an easy way to see this (which is more or less an elaboration of Mikhail's answer): Suppose $i > 0$ and $n\in \mathbb Z$. We have the Thom isomorphism $$ H^n(X, \mathbb Z(-i)) \cong H^{n+2i}_X …
Marc Hoyois's user avatar
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25 votes

Why is the motivic category defined over the site of smooth schemes only?

It's worth noting first that smooth schemes are essentially the smallest possible category from which one can define the motivic homotopy category: to make sense of $\mathbb A^1$-homotopy and of the N …
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14 votes
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The vanishing of $MGL^{2n+i,n}(X)$; do spectra of smooth projective varieties generate $SH_{...

(1) is true if $char(k)=0$. This follows from a combination of results. First of all, it is true over any field that the spectrum $MGL$ is connective, which means that $$MGL^{p,q}(X)=0$$ if $p>q+dim …
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10 votes
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Is $MGL$ an $H\mathbb{Z}$-algebra?

$MGL$ does not admit a structure of $H\mathbb Z$-module. There are many ways to prove this. As Sean said in the comments, if it were true over $\mathbb C$, topological realization would imply that $MU …
Marc Hoyois's user avatar
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15 votes
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Who proved the motivic 6-functor formalism?

My understanding is that Scholze was citing Gallauer for the universal property of the 6FF of motivic spectra (an unpublished result as far as I know), not for the existence of the 6FF. This result wa …
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