Consider the Segre embedding $(\mathbb P^1)^n \rightarrow \mathbb P^{2^n-1}$. What is the ideal corresponding to the image of this embedding? It is known that it is generated by quadratic relations. Is there a proper reference where the relations are written explicitly for this particular map in terms of $x_i$'s and $y_i$'s ?


Let $(x_i:y_i)$ be homogeneous coordinates for the $i$-th copy of $\mathbb{P}^1$. Let $(z_I)$ be homogeneous coordinates for $\mathbb{P}^{2^n - 1}$, where $I$ ranges over all subsets of $[n] := \{1,2,\dots,n\}$. (The first psychological hurdle to overcome is possibly a bias toward wanting a linear ordering for coordinates. This gave me trouble when I was first learning algebraic geometry. But the point is that the most useful combinatorial gadget to index the coordinates on $\mathbb{P}^{2^n-1}$ is subsets of $[n]$, not the elements of $[2^n]$.)

The Segre map can be described as $$z_I = \prod_{i \in I} x_i \prod_{j \notin I} y_j.$$ Then we must determine the condition on subsets $I,J,K,L$ of $[n]$ such that $z_I z_L = z_J z_K$ for all points in the image of the Segre map. Then if we think of $I \cup L$, resp. $J \cup K$, as a multiset (counting an element $a \in I \cap L$, resp. $b \in J \cap K$, as having multiplicity 2 in $I \cup L$, resp. $J \cup K$), the condition for $z_I z_L - z_J z_K$ to belong to the homogeneous ideal defining the image of the Segre map is that $I \cup L = J \cup K$ as multisets.

As a concrete example, consider $n = 5$, $I = \{1,2,3\}$, $L = \{1,5\}$, $J = \{1,2,5\}$, $K = \{1,3\}$. Then $z_I z_L = z_J z_K$ on the image of the Segre map because $I \cup L = J \cup K = \{1,1,2,3,5\}$ as multisets.

It should also be pointed out that there is a natural bijection between subsets of $[n]$ and sequences of $0$'s and $1$'s of length $n$. Given $I \subseteq \ [n]$, associate to it the sequence $(a_1,a_2,\dots,a_n)$ where $a_i = 1$ if $i \in I$ and $a_i = 0$ if $i \notin I$. Thus, some people index homogeneous coordinates on $\mathbb{P}^{2^n - 1}$ by using sequences of $0$'s and $1$'s of length $n$. In that case, the multiset formalism is equivalent to adding the sequences. So in the concrete example given above, you would have $I \leftrightarrow (1,1,1,0,0)$, $L \leftrightarrow (1,0,0,0,1)$, $J \leftrightarrow (1,1,0,0,1)$, $K \leftrightarrow (1,0,1,0,0)$ and $I \cup L = J \cup K \leftrightarrow (2,1,1,0,1)$.

  • $\begingroup$ Does this prove that the ideal defining the Segre is generated by these quadratic binomials? (I seems to me that it only proves that the ideal contains these binomials) $\endgroup$ – emeu Oct 21 '15 at 11:51
  • $\begingroup$ Just to be sure that I understand correctly: does these binomials correspond to all 2x2 minors of all possible flattenings of the tensors of format 2x2x...x2 ? $\endgroup$ – emeu Oct 21 '15 at 11:53
  • $\begingroup$ No, I did not attempt to prove that these binomials generate the ideal. I'm only answering the original question asking for an explicit description of the generators (without giving a proof that they do, in fact, generate). $\endgroup$ – Michael Joyce Oct 21 '15 at 11:53
  • $\begingroup$ Is there a reference to cite ? $\endgroup$ – Jack Oct 21 '15 at 16:45
  • $\begingroup$ I believe it is in several of the standard algebraic geometry texts. It's certainly in Hartshorne, though almost certainly as an exercise there. $\endgroup$ – Michael Joyce Oct 21 '15 at 20:20

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