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Let $V_1$, $V_2$ be vector spaces of dimension n+1 and m+1 respectively and $\mathbb{P}(V_1)$ and $\mathbb{P}(V_2)$ their projectivization.

Rigorously, the Segre map is the function

$\sigma_{n,m} : \mathbb{P}(V_1) \times \mathbb{P}(V_2) \rightarrow \mathbb{P}^{(n+1)(m+1)-1}: ([x_0,...,x_n],[y_0,...,y_m]) \rightarrow [x_0y_0, x_0y_1,...,x_0y_m,x_1y_0,...,x_iy_j,...,x_ny_m]$.

But there exists another definition used in Quantum Physics viewed in the light of algebraic geometry, which is (from https://en.wikipedia.org/wiki/Segre_embedding)

$\sigma_{n,m} : \mathbb{P}(V_1) \times \mathbb{P}(V_2) \rightarrow \mathbb{P}(V_1 \otimes V_2) : ([v],[w]) \rightarrow [v \otimes w]$

where $\otimes$ refers to the tensor product of vector spaces.

I don't understand how to show that these two definitions are equivalent, if they are so.

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  • $\begingroup$ $\otimes$ is used as the symbol for the outer product $\endgroup$
    – JMP
    Feb 15, 2022 at 9:48
  • $\begingroup$ According to en.wikipedia.org/wiki/Segre_embedding, I think that $\otimes$ refers to the tensor product in this second definition of the Segre map. Am I wrong ? $\endgroup$
    – Baloo
    Feb 15, 2022 at 9:52
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    $\begingroup$ Can you be more precise about what your definition of $\sigma_{n,m}$ in the second setting is? The first thing that shows up on the Wikipedia page is just the coordinate map you've written down. $\endgroup$
    – KReiser
    Feb 15, 2022 at 10:08
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    $\begingroup$ Did you try choosing a basis of $V_1$ and $V_2$ and decomposing $v\otimes w$ in the corresponding basis of $V_1\otimes V_2$? $\endgroup$ Feb 15, 2022 at 10:22
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    $\begingroup$ Your first formula for $\sigma_{n,m}$ is a little awkward, because it actually depends on a choice of basis for $V_1$ and $V_2$, so it makes more sense to define it as a map $\mathbb{P}^{n}\times \mathbb{P}^{m}\to \mathbb{P}^{(n+1)(m+1)-1}$. $\endgroup$ Feb 15, 2022 at 10:25

1 Answer 1

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If $a_0,\cdots,a_n$ are a basis for $V_1$ and $b_0,\cdots,b_m$ are a basis for $V_2$, then $a_i\otimes b_j$ are a basis for $V_1\otimes V_2$. In particular, if you have a pair of vectors $\sum_i x_ia_i$ and $\sum_j y_jb_j$ where the $x_i$ and $y_j$ are scalars, then you get a vector $\sum_{i,j} x_iy_j (a_i\otimes b_j)$. So the map in coordinates from sending $([v],[w])\to [v\otimes w]$ has as its coordinate representation exactly the description you've started with.

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  • $\begingroup$ Thanks for your answer. I think that I understand better the link between these two definitions :) $\endgroup$
    – Baloo
    Feb 15, 2022 at 11:51

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