# how do I show that Segre embedding is a homeomorphism?

I have to check that the diagonal of space projective $\mathbb P^n$ is closed in the product $\mathbb P^n$x$\mathbb P^n$. So I need to show that Segre embedding is a homeomorphism. How do I do this?

the projective spaces over k, with k a algebraically closed field. And I using the Zariski topology.

Specifically my doubt is Proposition 7.11(b) and Corollary 7.13 of Gathmann. Link below: pdf

• I'm not sure what the Segre embedding being a homeomorphism has to do with the diagonal being closed in $\mathbb{P}^n\times\mathbb{P}^n$... Commented Jun 25, 2017 at 19:51
• The image of the diagonal will be a closed, if the Segre embedding is a homeomorphism, so the diagonal will be a closed. Commented Jun 25, 2017 at 20:08
• The Segre embedding is a map $\Bbb P^n\times \Bbb P^m \to \Bbb P^{(n+1)(m+1)-1}$. This is, at a first glance, rather far away from checking that the diagonal is closed, which you can prove much more easily in other ways (take an affine patch, compute, etc). If your question is something more along the lines of "I want to prove that the diagonal embedding of $\Bbb P^n\hookrightarrow \Bbb P^n\times\Bbb P^n$ via the Segre embedding", you should show your thought process for how the Segre embedding would be involved in your proof. Commented Jun 25, 2017 at 20:30
• You should also clarify what definitions and framework you're working with. Are your projective spaces over $\mathbb{C}$? Are you using the Euclidean topology or the Zariski topology? Commented Jun 25, 2017 at 20:53
• I want to prove that the projective space is a variety following the steps of Andreas Gathmann's algebraic geometry. I have already shown that the projective space is a pre-variety, now it is enough to show that the diagonal is closed. The Gathmann says that the diagonal image of the Segre embedding is closed, and concludes that the diagonal is closed. Commented Jun 25, 2017 at 22:12

Gathmann says that the Segre embedding is a morphism because it is given locally in affine charts by polynomial maps. So lets make the technique of going to affine coordinates very precise.

We have the following lemma, which follows from Remark 4.6, lemma 4.7 and the fact that for a ringed space $(Z,O_Z)$ and $U\subseteq Z$ open the inclusion $(U,O_{Z|U})\to (Z,O_Z)$ is a morphism:

Let $(X,O_X), (Y,O_Y)$ be ringed spaces, $f:X\to Y$ a set map $(U_i)_{i\in I}$ an open cover of $X$ and $V_i\subseteq Y$ open such that $f(U_i)\subseteq V_i$ for all $i\in I$. Then $f:(X,O_X)\to (Y,O_Y)$ is a morphism if an only if $f_{|U_i}:(U_i,O_{X|U_i})\to (V_i,O_{Y|V_i})$ is a morphism for all $i\in I.$

Now lets apply this lemma in the case where $X,Y$ are prevarieties:

Assume $f:X\to Y$ is a set map and you want to show that $f$ is a morphism. Choose an affine open cover $(V_i)_{i\in I}$ of $Y$ and form the sets $f^{-1}(V_i)$. Now assume you can "see" that the $f^{-1}(U_i)$ are open. Then you can regard them as prevarieties and find open affine covers $(U_{ij})_{j\in I_i}$ of the $f^{-1}(U_i)$. Forming now the restrictions $f_{ij} :U_{ij}\to V_i$, by the above lemma $f$ is a morphism iff the $f_{ij}$ are.

How to check this? Lets be very precise:

As the $U_{ij}$ and $V_i$ are affine varieties there exists embedded affine varieties $A_{ij}$, $B_i$ (which now really "live" in affine spaces) and isomorphisms $\phi _{ij}:U_{ij}\to A_{ij}$, $\psi _i:V_i\to B_i$. If one now forms the maps $\tilde{f}_{ij}=\psi_i\circ f_{ij}\circ \phi_{ij}^{-1}$ $\require{AMScd}$ $\require{AMScd}$ \begin{CD} U_{ij} @>f_{ij}>> V_i\\ @V \phi _{ij} V V @VV \psi _{i} V\\ A_{ij} @>>\hat{f_{ij}}> B_i \end{CD} then the $f_{ij}$ are morphisms iff the $\tilde{f}_{ij}$ are morphisms. As the $A_{ij}$, $B_i$ are embedded affine varieties, this happens iff the $\tilde{f}_{ij}$ are polynomial maps, so summarizing $f$ is a morphism iff all $\tilde{f}_{ij}$ are polynomial maps.

For your particular case, the the Segre embedding $f:\mathbb P^n \times \mathbb P^m\to X\subseteq\mathbb P^N$ restricts bijectively to the affine open subvartieties $f_{ij}:U_{ij}\to V_{ij}$, where

$$U_{ij}=\{((x_0:\ldots :x_n),(y_0:\ldots :x_m))\in \mathbb P^n \times \mathbb P^m \,: x_i,y_j\neq 0\}\,\, \text{and}\\ V_{ij}=\{(z_{00}:\ldots :z_{nm})\in X : z_{ij}\neq 0\}$$

and the isomorphisms are given by

$$\phi_{ij}:U_{ij}\to \mathbb A^n\times \mathbb A^m=\mathbb A^{n+m},\,(x,y)\mapsto (\frac{x_0}{x_i},\ldots,\hat{\frac{x_i}{x_i}},\ldots ,\frac{x_n}{x_i},\frac{y_0}{y_j},\ldots,\hat{\frac{y_j}{y_j}},\ldots ,\frac{y_m}{y_j})\\ \\ \psi _{ij}:V _{ij}\to\text{im}\, \,\psi _{ij} \subseteq \mathbb A ^N\, , (z_{00}:\ldots:z_{nm})\mapsto (\frac {z_{00}}{z_{ij}},\ldots,\hat{\frac {z_{ij}}{z_{ij}}},\ldots,\frac {z_{nm}}{z_{ij}})\, ,$$ where $\hat{\,}$ are ommitted. Now calculating the $\hat{f_{ij}}$, $\hat{f_{ij}}^{-1}$ yields that they are polynomial maps, so $f$ is an isomorphism.

A counterexample comes from $$\Psi(\mathbb{P}^1 \times \mathbb{P}^1) \cong Q= V(xw-yz) \subset \mathbb{P}^3$$, as $$Q$$ contains curves (such as the 3-uple embedding of $$\mathbb{P}^1$$ in $$\mathbb{P}^3$$, $$\nu_3(\mathbb{P}^1) \cong V(xz-y^2, xw-yz, yw-z^2)$$ which do not correspond to a curve in $$\mathbb{P}^1 \times \mathbb{P}^1$$ in the product topology.

That is, the Segre embedding $$\Psi(\mathbb{P}^n \times \mathbb{P}^m) \hookrightarrow \mathbb{P}^{nm + n + m}$$ is not a homeomorphism onto its image in general if we put the product topology on $$\mathbb{P}^n\times \mathbb{P}^m$$. This raises the question of what topology you’re putting on the domain $$\mathbb{P}^n \times \mathbb{P}^m$$.

• This misses the point - it is well known that the topology on the product of varieties is not the product topology. Commented May 4, 2022 at 18:16