7
$\begingroup$

Let $k$ be an algebraically closed field and $\mathbb A_k^n$ be the affine $n$-space. We can give the Zariski topology on it. Then consider $X=\mathbb A_k^m \times \mathbb A_k^n$ under product topology, where each of $\mathbb A_k^m$ and $\mathbb A_k^n$ are given the Zariski topology. And also consider $X=\mathbb A_k^{m+n}$ under the Zariski topology. My question is : For which $m,n\ge 1$, are these two topological spaces homeomorphic ?

Note that I am not asking when two topologies are same ... I'm asking when they are not homeomorphic ...

$\endgroup$

migrated from mathoverflow.net Sep 1 '18 at 3:19

This question came from our site for professional mathematicians.

  • $\begingroup$ Essentially never. More precisely, $X$ won't be homeomorphic to $\mathbb{A}^m\times \mathbb{A}^n$ unless $m$ or $n$ is zero. $\endgroup$ – Donu Arapura Aug 28 '18 at 1:06
  • $\begingroup$ @DonuArapura: can you please explain why is that ? $\endgroup$ – user521337 Aug 28 '18 at 1:31
  • $\begingroup$ HInt: What is the intersection of open sets? $\endgroup$ – Jan-Christoph Schlage-Puchta Aug 28 '18 at 21:01
12
$\begingroup$

Clearly the natural map $|\mathbb A^{n+m}| \to |\mathbb A^n| \times |\mathbb A^m|$ is not a homeomorphism unless $n = 0$ or $m = 0$ (it's not even a bijection, because of points like $(x-y) \in \operatorname{Spec} k[x] \otimes k[y]$). We show that in fact there is no homeomorphism, using the following ad hoc notion of dimension:

Definition. Let $X$ be a topological space. Then the intersection dimension $\operatorname{int. dim}(X)$ of $X$ is the supremum of all $n$ for which there exist irreducible closed sets $V_0,\ldots,V_n \subseteq X$ such that

  1. any intersection $V_I = \bigcap_{i\in I} V_i$ for $I \subseteq \{0,\ldots,n\}$ nonempty is either irreducible or empty;
  2. the intersection $V_I$ for any $I \subseteq \{0,\ldots,n\}$ with $|I| = n$ is exactly one point;
  3. the intersection $V_{\{0,\ldots,n\}}$ is empty.

I feel that I have seen some variant of this somewhere before (probably under a different name), but I do not recall where. References are welcome!

Remark. If there exist sets $V_0,\ldots,V_n$ with properties 1, 2, and 3 above, then the same is true for any $m \leq n$. Indeed, we may take the sets $V_0, \ldots, V_{m-1}, V_{\{m,\ldots,n\}}$.

Lemma 1. Let $k$ be a field. Then $\operatorname{int. dim}(\mathbb A^n_k) = n$.

Proof. The hyperplanes $V(x_1+\ldots+x_n), V(x_1), \ldots, V(x_n)$ show that $\operatorname{int. dim}(\mathbb A^n_k) \geq n$. Conversely, let $V_0, \ldots, V_m$ be irreducible closed subsets satisfying 1, 2, and 3 above. Conditions 2 and 3 imply that no $V_i$ is contained in any intersection of the others, so by induction on $|I|$ the dimension of $V_I$ is at most $n-|I|$ (because all intersections are irreducible). Taking $|I| = m$ gives $m \leq n$, showing that $\operatorname{int. dim}(\mathbb A^n_k) \leq n$. $\square$

More generally, this argument shows that if $X$ is a topological space of dimension $n$, then $\operatorname{int. dim}(X) \leq \dim X$. It's not clear even for smooth proper varieties that the two are equal (in the projective case we might be able to use some Bertini argument, using tangency to reduce the intersections to one point).

Lemma 2. Let $X$ and $Y$ be topological spaces with a closed point. Then $$\operatorname{int. dim}(X \times Y) = \max(\operatorname{int. dim}(X), \operatorname{int. dim}(Y)).$$

Proof. By the Lemma of this post, a subset $Z \subseteq X \times Y$ is closed and irreducible if and only if $Z = V \times W$, with $V \subseteq X$ and $W \subseteq Y$ closed and irreducible.

Let $m$ be the maximum of the intersection dimensions of $X$ and $Y$; without loss of generality let's say $m = \operatorname{int. dim}(X)$. Then there exist sets $V_0,\ldots,V_m \subseteq X$ satisfying properties 1, 2, and 3 of the definition above. Choose a closed point $y \in Y$, and set $W_i = \{y\}$ for all $i$. Then the sets $Z_i = V_i \times W_i$ are closed and irreducible. Moreover, the finite intersections $Z_I = \bigcap_{i \in I} Z_i$ equal $V_I \times W_I$. Since properties 1, 2, and 3 hold for the $V_i$ and properties 1 and 2 for the $W_i$, this shows properties 1, 2, and 3 for the $Z_i$. Therefore, $\operatorname{int. dim}(X \times Y) \geq m$.

Conversely, if $Z_0, \ldots, Z_n$ satisfy properties 1, 2, and 3, then we must have $Z_i = V_i \times W_i$ for some irreducible closed subsets $V_i \subseteq X$, $W_i \subseteq Y$. Again we have $Z_I = V_I \times W_I$, so both $V_i$ and $W_i$ satisfy 1 and 2. Since $Z_{\{0,\ldots,n\}} = \varnothing$, we conclude that either $V_{\{0,\ldots,n\}} = \varnothing$ or $W_{\{0,\ldots,n\}} = \varnothing$. Hence, either the $V_i$ or the $W_i$ satisfy the conditions above, so $\operatorname{int. dim}(X \times Y) \leq m$. $\square$

Corollary. There exists a homeomorphism $|\mathbb A^{n+m}_k| \stackrel\sim\to |\mathbb A^n_k| \times |\mathbb A^m_k|$ if and only if $n = 0$ or $m = 0$.

Proof. By Lemma 1 and Lemma 2, we have $$\operatorname{int. dim}(|\mathbb A^n_k| \times |\mathbb A^m_k|) = \max(n,m),$$ whereas $$\operatorname{int. dim}(|\mathbb A^{n+m}_k|) = n + m.$$ These numbers are equal if and only if $n = 0$ or $m = 0$. Clearly the stated homeomorphisms exist in this case. $\square$

We can actually mix and match fields, i.e. consider things like $|\mathbb A^{100}_{\mathbb C}| \times |\mathbb A^{17}_{\bar{\mathbb F_3}}|$. The same proof shows that this cannot be homeomorphic to any space of the form $|\mathbb A^{117}_k|$ for a field $k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.