4
$\begingroup$

Let $X$ be a smooth scheme over a field $k$, and let $Z\subset X$ be a closed sub-scheme of codimension $2$. Let $Bl_Z(X)$ denote the blow-up of $X$ at $Z$, and let $\pi\colon Bl_Z(X)\to X$ denote the projection. Suppose I have a section $\sigma\colon Z\to Bl_Z(X)$ of $\pi$ over $Z$ (i.e. $\pi\sigma=1_Z$).

Question: (when) is the map $Bl_Z(X)\setminus\sigma(Z)\to X$ a weak equivalence?

The references to blow-up theorems which I have found (e.g. Voevodsky's Seattle lectures) suggest that it becomes an equivalence after suspension, but I'd like to avoid suspending if possible.

I'm also happy to restrict the choice of $Z$ and $X$. The case I'm most interested has $X$ being an iterated blow-up of affine space at (proper transforms) of linear sub-spaces, and $Z$ being a (proper transform of a) linear sub-space.

Improve this question
$\endgroup$
2
  • $\begingroup$ I'm confused about your question: if $Z$ is already a Cartier divisor, then you seem to be asking whether $X-Z\to X$ is a weak equivalence. Clearly this is not always the case... $\endgroup$
    – Marc Hoyois
    Commented Apr 23, 2014 at 10:18
  • $\begingroup$ Marc, good point. The examples I'm thinking about are all for $Z$ of codimension $>1$. $\endgroup$
    – Jesse Wolfson
    Commented Apr 23, 2014 at 10:35

1 Answer 1

Reset to default
7
$\begingroup$

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 bundle of lines in the normal bundle $N_{X,Z}$. The second map is an equivalence iff $Z$ has codimension $2$. If not, the two sides have different $\ell$-adic cohomology after pulling back to a geometric point, for example. So the first map can only be a weak equivalence when the codimension is exactly $2$.

But if the codimension is $2$, I don't know if your map an equivalence. I haven't heard of any progress towards removing the suspension in Voevodksy's blow-up theorem...

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thanks for the answer, and you're right that the question I wrote requires codimension 2. My intuition is that $Bl_Z(X)\setminus\sigma(Z)$ is obtained from X by gluing in an affine bundle over $Z$. Since affine spaces are $\mathbb{A}^1$-contractible, my intuition was that we could contract away the fibers over $Z$ and recover $X$. In higher codimension, I need to delete a hyperplane section over $Z$ to get the right picture, but this is the situation I had in mind. Do you know if there is reason to be suspicious of this intuition? $\endgroup$
    – Jesse Wolfson
    Commented Apr 23, 2014 at 23:38
  • $\begingroup$ Actually, this intuition is wrong, and this is not the question I should have asked. Thanks for helping clear this up. $\endgroup$
    – Jesse Wolfson
    Commented Apr 24, 2014 at 3:21
  • 1
    $\begingroup$ If you remove a hyperplane section from the exceptional divisor, then I agree that it looks like the map could be an $\mathbb{A}^1$-equivalence, but you need a finer topology than the Nisnevich one. Maybe this will be useful: Unstable motivic homotopy categories in Nisnevich and cdh-topologies. $\endgroup$
    – Marc Hoyois
    Commented Apr 24, 2014 at 10:41
  • 1
    $\begingroup$ I think it's unlikely that it will ever be an $\mathbb{A}^1$-equivalence. Consider the blow up of $\mathbb{A}^2$ at a point. This is the total space of the tautological line bundle over $\mathbb{P}^1$. If we remove a point from the exceptional divisor, we get something whose topological realization looks like $S^3\vee S^2$. FWIW, the situation I was considering turns out to involve blow-ups of affine space where we remove the fiber over a hyperplane section of the exceptional divisor (I only saw the hyperplane at first). After removing the whole fiber, we get an $\mathbb{A}^1$-equivalence. $\endgroup$
    – Jesse Wolfson
    Commented Apr 25, 2014 at 21:42
  • $\begingroup$ Ah, that's a great example. In this case the blow-up square is trivially a pushout (without suspending), but there's no reason for it to remain a pushout when you remove a point... I missed this point before. $\endgroup$
    – Marc Hoyois
    Commented Apr 28, 2014 at 10:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .