# How to show that these two versions of the Segre map are equal?

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.

• $\otimes$ is used as the symbol for the outer product
– JMP
Feb 15, 2022 at 9:48
• 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 ? Feb 15, 2022 at 9:52
• 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. Feb 15, 2022 at 10:08
• Did you try choosing a basis of $V_1$ and $V_2$ and decomposing $v\otimes w$ in the corresponding basis of $V_1\otimes V_2$? Feb 15, 2022 at 10:22
• 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 $\mathbb{P}^{n}\times \mathbb{P}^{m}\to \mathbb{P}^{(n+1)(m+1)-1}$. Feb 15, 2022 at 10:25

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.