Linear subspaces in quadric hypersurfaces












3












$begingroup$


Consider $H_1,H_2,H_3subsetmathbb{P}^{2m+1}$ three general linear subspaces of projective dimension $m$.



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



Does there exist a smooth quadric hypersurface $Q^{2m}subsetmathbb{P}^{2m+1}$ 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.






ag.algebraic-geometry linear-algebra projective-geometry grassmannians quadrics







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$endgroup$

















    3












    $begingroup$


    Consider $H_1,H_2,H_3subsetmathbb{P}^{2m+1}$ three general linear subspaces of projective dimension $m$.



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



    Does there exist a smooth quadric hypersurface $Q^{2m}subsetmathbb{P}^{2m+1}$ 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.






    ag.algebraic-geometry linear-algebra projective-geometry grassmannians quadrics







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Consider $H_1,H_2,H_3subsetmathbb{P}^{2m+1}$ three general linear subspaces of projective dimension $m$.



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



      Does there exist a smooth quadric hypersurface $Q^{2m}subsetmathbb{P}^{2m+1}$ 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.






      ag.algebraic-geometry linear-algebra projective-geometry grassmannians quadrics







      share|cite|improve this question











      $endgroup$




      Consider $H_1,H_2,H_3subsetmathbb{P}^{2m+1}$ three general linear subspaces of projective dimension $m$.



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



      Does there exist a smooth quadric hypersurface $Q^{2m}subsetmathbb{P}^{2m+1}$ 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.






      ag.algebraic-geometry linear-algebra projective-geometry grassmannians quadrics




      ag.algebraic-geometry linear-algebra projective-geometry grassmannians quadrics






      share|cite|improve this question















      share|cite|improve this question













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      edited Nov 18 '18 at 7:04









      Kugelblitz

      1185




      1185










      asked Nov 17 '18 at 21:01









      SMBSMB

      1135




      1135






















          1 Answer
          1






          active

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          5












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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 '18 at 21:35










          • $begingroup$
            Yes, precisely.
            $endgroup$
            – Sasha
            Nov 17 '18 at 21:36











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          1 Answer
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          1 Answer
          1






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          active

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          active

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          5












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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 '18 at 21:35










          • $begingroup$
            Yes, precisely.
            $endgroup$
            – Sasha
            Nov 17 '18 at 21:36
















          5












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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 '18 at 21:35










          • $begingroup$
            Yes, precisely.
            $endgroup$
            – Sasha
            Nov 17 '18 at 21:36














          5












          5








          5





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 17 '18 at 21:37

























          answered Nov 17 '18 at 21:20









          SashaSasha

          20.4k22655




          20.4k22655












          • $begingroup$
            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 '18 at 21:35










          • $begingroup$
            Yes, precisely.
            $endgroup$
            – Sasha
            Nov 17 '18 at 21:36


















          • $begingroup$
            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 '18 at 21:35










          • $begingroup$
            Yes, precisely.
            $endgroup$
            – Sasha
            Nov 17 '18 at 21:36
















          $begingroup$
          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 '18 at 21:35




          $begingroup$
          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 '18 at 21:35












          $begingroup$
          Yes, precisely.
          $endgroup$
          – Sasha
          Nov 17 '18 at 21:36




          $begingroup$
          Yes, precisely.
          $endgroup$
          – Sasha
          Nov 17 '18 at 21:36


















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