The Segre map

On first hand, the Segre map is the function

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

and the Segre variety $\Sigma_{n,m}$ is the image of the Segre map $$\Sigma_{n,m}=\sigma_{n,m}(\mathbb{P}^n \times \mathbb{P}^m).$$

Let $Z_{i,j}=x_iy_j$. The Segre variety is also defined as the common zero locus of the quadratic polynomials $$Z_{{i,j}}Z_{{k,l}}-Z_{{i,l}}Z_{{k,j}},$$ where $Z_{i,j}$ is understood to be the natural coordinate on the image of the Segre map.

The k-secant of a projective variety

The k-secant of $X\subset \mathbb{P}^n$ is $$ \sigma_k(X)=\overline{\underset{x1,...,x_k\in X}{\cup} \mathbb{P}^{k-1}_{x_1,...,x_k}} $$ where $\mathbb{P}^{k-1}_{x_1,...,x_s}$ is a projective space of dimension $s-1$ passing through $x_1,...,x_s$. If needed to better understand this definition, another equivalent approach to define k-secant is the following.

Let $Y\subset \mathbb{P}^n$. The joint of X and Y is $$J(X,Y)=\overline{\underset{x\in X, y\in Y, x\neq y}{\cup} \mathbb{P}^1_{x,y}}$$ where $\mathbb{P}^1_{x,y}$ is the projective line containing x and y.

The 2-secant of X, also simply named the secant of X, is $ \sigma(X)=J(X,X) $ and the k-secant is $$\sigma_k(X)=J(X,\sigma_{k-1}(X)).$$


What is the set of polynomials whose common zero locus is the k-secant of the Segre variety ?


1 Answer 1


Consider the matrix of size $(m+1)$ by $(n+1)$ with entries $x_iy_j$. Then the equations of the $k$-secant variety are all minors of this matrix of size $k + 1$.

  • $\begingroup$ Thanks. Could you please tell me where I can find a proof, or how to prove it myself ? $\endgroup$
    – Baloo
    Feb 17, 2022 at 12:35
  • 1
    $\begingroup$ The main idea is to think of the ambient space as of the space of matrices; then the $k$-secant variety is the variety of matrices of rank $k$. $\endgroup$
    – Sasha
    Feb 17, 2022 at 19:54
  • $\begingroup$ Ok, thanks a lot ! $\endgroup$
    – Baloo
    Feb 18, 2022 at 8:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .