# Segre embedding of $\mathbb{P}^n\times \mathbb{P}^m\times\mathbb{P}^q$

How could I define the segre embedding of three projective spaces? Usually in references only the Segre embedding of two projective spaces are defined:

$\mathbb{P}^n \times \mathbb{P}^m \rightarrow \mathbb{P}^N$, $N=(n+1) (m+1)-1$ where the coordinates of $\mathbb{P}^N$ are $[u_{00},u_{01},...,u_{n+1,m+1}]=[x_0y_0,x_0y_1,...,x_{n+1}y_{m+1}]$ and the $u_{ij}$ satisfies $u_{ij}u_{kl}-u_{il}u_{kj}=0$.

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The general Segre embedding discussed in EGA I$_{\text{new}}$, 9.8 (which I quite like because it is coordinate-free and transparent) can be generalized without effort as follows: Let $S$ be a scheme, and consider quasi-coherent sheaves $\mathcal{E}_1,\dotsc,\mathcal{E}_n$ on $S$. Then there is a closed immersion of $S$-schemes

$$\mathbb{P}(\mathcal{E}_1) \times_S \dotsc \times_S \mathbb{P}(\mathcal{E}_n) \hookrightarrow \mathbb{P}(\mathcal{E}_1 \otimes_{\mathcal{O}_S} \dotsc \otimes_{\mathcal{O}_S} \mathcal{E}_n),$$

given on $X$-valued points by $((\mathcal{L}_1,s_1),\dotsc,(\mathcal{L}_n,s_n)) \mapsto (\mathcal{L}_1 \otimes \dotsc \otimes \mathcal{L}_n,s_1 \otimes \dotsc \otimes s_n)$ (where $\mathcal{L}_i$ is invertible on $X$, $p : X \to S$ is the structural morphism and $s_i : p^* \mathcal{E}_i \to \mathcal{L}_i$ is an epimorphism). In particular, for all $k_1,\dotsc,k_n \in \mathbb{N}$ there is a closed immersion

$$\mathbb{P}^{k_1}_S \times_S \dotsc \times_S \mathbb{P}^{k_n}_S \hookrightarrow \mathbb{P}^{(k_1+1) \cdot \dotsc \cdot (k_n+1)-1}_S.$$

It corresponds to the graded quasi-coherent ideal $\subseteq \mathcal{O}_S[(x_{i_1,\dotsc,i_n})_{0 \leq i_p \leq k_p}]$ generated by the relations $x_{i_1,\dotsc,i_{p-1},a,i_{p+1},\dotsc,i_n} \cdot x_{i_1,\dotsc,i_{p-1},b,i_{p+1},\dotsc,i_n} = x_{i_1,\dotsc,i_{p-1},b,i_{p+1},\dotsc,i_n} \cdot x_{i_1,\dotsc,i_{p-1},a,i_{p+1},\dotsc,i_n}$.

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Of course, this coincides with the suggestion by Fredrik Meyer. – Martin Brandenburg Jan 7 '13 at 22:44
Thanks Martin and Fredrik. I'm gonna try to translate to a physicist point of view what you said, because I don't know anything about schemes.My knowledge of AG is at best restricted to the first capter of Hartshorne. Let us consider for example $\mathbb{P}^2\times \mathbb{P}^2 \times \mathbb{P}^2$ with coordinates $[w_0:w_1,w_2;x_0:x_1:x_2;y_0:y_1:y_2]$. The Segre embedding of $\mathbb{P}^2 \times \mathbb{P}^2$ is a map defined by $[x_0:x_1:x_2;y_0:y_1:y_2]\rightarrow [u_{00}:...:u_{22}]=[x_0y_0:..:x_2y_2]$ – user55451 Jan 8 '13 at 18:54
Now I define the Segre embedding of $\mathbb{P}^2 \times \mathbb{P}^8$ by, $[w_0:w_1,w_2;u_{00}:...:u_{22}]\rightarrow [z_{000}:...:z_{222}]=[w_0u_{00}:...:w_2z_{22}]$ where the $z_{ijk}$ are coordinates of a $\mathbb{P}^{26}$. Is that correct? What would be now the image? Thank you so much for help. – user55451 Jan 8 '13 at 18:55
I have written down the defining relations of the image. And as I said, you don't have to apply the usual Segre embedding twice. You can (and should) do it on one step. – Martin Brandenburg Jan 9 '13 at 4:16

You could do the usual Segre embedding twice. That is, first embed $\mathbb{P}^n \times \mathbb{P}^m \times \mathbb{P}^q \to \mathbb{P}^{N} \times \mathbb{P}^q$, with $N=(n+1)(m+1)-1$ and then embed $\mathbb{P}^n \times \mathbb{P}^q \to \mathbb{P}^M$, with $M=(N+1)(q+1)-1=(n+1)(m+1)(q+1)-1$.

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thanks, Fredrik. I tought about this, but the problem is that the variables of $\mathbb{P}^N$ are constrained by $u_{ij}u_{kl}−u_{il}u_{kj}=0$, and I don't know how to deal with this. – user55451 Jan 7 '13 at 15:15
The variables don't satisfy this (it is just $\mathbb{P}^N$), they only satisfy it on the image of the embedding. – Martin Brandenburg Jan 7 '13 at 22:45