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This question is motivated by this other question and this answer, which show that irreducible algebraic varieties and more generally integral schemes are contractible as topological spaces.

What are examples of connected non-contractible schemes? I expect some gluing involved in the answer. But could such a scheme also be affine?

Pedro
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  • $Spec(\mathbb{C}[z]/(z(z-1)))$. – SphericalTriangle Mar 21 '18 at 15:16
  • I was thinking of connected spaces and completely disregarded that trivial example, yes. My bad. Thanks for pointing it out! – Pedro Mar 21 '18 at 15:22
  • The obstruction in that example is not really being disconnected. But to where the generic points of the components are allowed to travel during the contraction. With that in mind, you can produce examples for your new problem. – SphericalTriangle Mar 21 '18 at 15:33
  • @SphericalTriangle You mean that one could still come up with examples like $\operatorname{Spec}(\mathbb{C}[x,y]/(x-1,x+1,y-1,y+1))$? I don't completely follow you when you say "the generic points of the components are allowed to travel during the contraction". Perhaps you could elaborate it a bit as an answer (if you want). Thanks again – Pedro Mar 21 '18 at 15:33
  • Because that sentence is precisely the part of the problem that I am not seeing I guess. From a naive topological perspective, the impression I have, is that $\operatorname{Spec}(\mathbb{C}[x,y]/(x-1,x+1,y-1,y+1))$ should be as much non-contractible as $\mathbb{C}P^{1}$. But this last space is contractible. So that "moving the generic points" part is precisely what I am missing here I guess – Pedro Mar 21 '18 at 15:40
  • Assume that $X$ is irreducible, all its points are closed except for one generic point $g$. During $f:X\times [0,1]\to X$ continuous $g$ must be fixed. $f(g,t)=g$ for all $t\in[0,1]$. Now look at $Spec(\mathbb{C}[x,y]/(xy))$ – SphericalTriangle Mar 21 '18 at 15:50
  • @SphericalTriangle I think I see what you mean, thanks! – Pedro Mar 21 '18 at 16:05

1 Answers1

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Take a union $X$ of a line $L$ and a conic $C$ in $\mathbb{A}^2$ (over some field) which intersect at two points $p$ and $q$. Then I claim $X$ is not contractible, and in fact that $H_1(X)$ is nontrivial. To prove this, cover $X$ by the open sets $U=X\setminus\{p\}$ and $V=X\setminus\{q\}$. Note that $U$ and $V$ are path-connected, but $U\cap V$ is disconnected (its connected components are $L\setminus\{p,q\}$ and $C\setminus\{p,q\}$). The Mayer-Vietoris sequence $$H_1(X)\to H_0(U\cap V)\to H_0(U)\oplus H_0(V)$$ then tells us that $H_1(X)$ must be nontrivial, since the map $H_0(U\cap V)\to H_0(U)\oplus H_0(V)$ is the same on every path-component of $U\cap V$ and thus is not injective.

The intuition here is that the space $X$ is like a circle, obtained by gluing together two contractible sets $L$ and $C$ which intersect at two points, just like a circle is obtained by gluing together two intervals that intersect at two points. In fact, with a bit more work you can show that $X$ is weakly homotopy equivalent to a circle.

More generally, I suspect the following statement is true but have not carefully worked out a proof. Let $X$ be a sober Noetherian topological space. Let $P$ be the smallest collection of irreducible closed subsets of $X$ such that every irreducible component of $X$ is in $P$ and if $A,B\in P$, then every irreducible component of $A\cap B$ is in $P$. We can consider $P$ as a poset with respect to inclusion. Then $X$ is weakly homotopy equivalent to the nerve of $P$.

(Alternatively, you could consider the poset $Q$ which consists simply of all irreducible closed subsets of $X$, or equivalently the poset of points of $X$ with respect to the specialization order. It should also be true that $X$ is weak equivalent to the nerve of $Q$. This construction is somewhat more natural, but the poset $P$ above has the advantage of being finite and thus nice to compute with.)

Eric Wofsey
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