Consider $H_1,H_2,H_3\subset\mathbb{P}^{2m+1}$ three general linear subspaces of projective dimension $m$.

Then there exists a quadric hypersurface $Q^{2m}\subset\mathbb{P}^{2m+1}$ containing $H_1,H_2,H_3$. This is just a dimension count.

Does there exist a smooth quadric hypersurface $Q^{2m}\subset\mathbb{P}^{2m+1}$ containing $H_1,H_2,H_3$?

The answer is positive when $m = 1$, since if $Q^2$ is a cone then the three lines must intersect, and if $Q^2$ is the union of two planes or a double plane then the three lines must intersect as well.


1 Answer 1


This is true when $m$ is odd and false otherwise.

Indeed, let $\mathbb{P}^{2m+1} = \mathbb{P}(V)$. Assuming $H_1 \cap H_2 = \emptyset$ (by genericity), we have $H_1 = \mathbb{P}(V_1)$, $H_2 = \mathbb{P}(V_2)$ and $$ V = V_1 \oplus V_2. $$ Furthermore, assuming $H_3 \cap H_1 = H_3 \cap H_2 = \emptyset$, we see that $H_3 = \mathbb{P}(V_3)$, where $V_3 \subset V_1 \oplus V_2$ is the graph of an isomorphism $V_1 \to V_2$. Thus, choosing bases appropriately we may assume $$ V_1 = \langle e_1,\dots,e_{m+1} \rangle,\quad V_2 = \langle e_{m+2},\dots,e_{2m+2} \rangle,\quad V_3 = \langle e_1 + e_{m+2}, \dots, e_{m+1} + e_{2m+2} \rangle. $$ Let $A = (a_{ij})$ be the matrix of the quadric $Q$. It follows that $a_{ij} = 0$ when either $1 \le i,j \le m+1$ or $m+2 \le i,j \le 2m+2$, and $$ b_{ij} = a_{i,m+1+j} $$ is a skew-symmetric matrix. It remains to note that $$ \det(A) = \pm \det(B)^2, $$ and that a skew-symmetric matrix of odd size is always degenerate.

  • $\begingroup$ So, if I got correctly you argument, $B$ is an $(m+1)\times (m+1)$ matrix. If $m$ is odd then $m+1$ is even, $\det(B)\neq 0$ and so $\det(A)\neq 0$ and the quadric is smooth. If $m$ is even then $m+1$ is odd, $\det(B)= 0$ and so $\det(A)= 0$ and the quadric is singular. Is this right? $\endgroup$
    – SMB
    Nov 17, 2018 at 21:35
  • $\begingroup$ Yes, precisely. $\endgroup$
    – Sasha
    Nov 17, 2018 at 21:36

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