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canonical representation of three skew lines in $mathbbP^3$

Disjoint Lines in Projective SpaceAn exercise involving the twisted cubicProjections of a rational normal curve of $mathbbP^4$ (Exercise 3.9 in Harris' _Algebraic Geometry_)Euler number of a surface in the complex projective spaceExercise 2.12 in Harris - Algebraic Geoemetry: a first courseimage of Segre-Veronese as a tuple of polynomialsDeriving the Quadratic Polynomials Defining the Twisted Cubic$mathcalExt^i(mathcalO_L_1, mathcalO_L_2)$ and $textExt^i(mathcalO_L_1, mathcalO_L_2)$ for two lines in $mathbbP^3$Projection of the twisted cubicProjective equivalence: Linear subspaces under the action of $PGL_n$Determining the image of a map

2

$begingroup$

Consider three skew (non-intersecting) lines $L,M,N$ in $mathbbP^3$.
Each line is given by two equations of the form $alpha_i,j^top z = 0, , i=1,2,3, j = 1,2$, where $z=(z_0,z_1,z_2,z_3)$ are the homogeneous coordinates of $mathbbP^3$ and $alpha_i,j in mathbbP^3$.

What i want is to show that up to projective equivalence the equations for the three lines can be written as (this is possible according to the hint of
exercise 2.12 in Harris, Algebraic Geometry - a first course)

$L: , z_0=z_1=0 \
M: , z_2=z_3 = 0 \
N: , z_0=z_3, , z_1=z_2.$

Since say $L, M$ are non-intersecting, this implies that $alpha_i,j$ for every $i=1,2, j=1,2$ are in general position. So they are projectively equivalent to $(1,0,0,0), (0,1,0,0),(0,0,1,0),(0,0,0,1)$. This gives us
the description that we want for $L$ and $M$. But how about $N$?

Note here that the most general result that we have is projective equivalence between two sets of 5 points in general position. In the context of the present question, i am concerned about two things:

1) The fact that the lines $L,M,N$ are skew does not imply in general that
every group of 5 points among the $alpha_i,j$ will be in general position.

2) The $alpha_i,j$ are six!

Question: What am i missing here? How can the above two issues be addressed?

algebraic-geometry

share|cite|improve this question

asked Apr 7 '15 at 18:33

ManosManos

14k33288

$endgroup$

add a comment | 

2

$begingroup$

Consider three skew (non-intersecting) lines $L,M,N$ in $mathbbP^3$.
Each line is given by two equations of the form $alpha_i,j^top z = 0, , i=1,2,3, j = 1,2$, where $z=(z_0,z_1,z_2,z_3)$ are the homogeneous coordinates of $mathbbP^3$ and $alpha_i,j in mathbbP^3$.

What i want is to show that up to projective equivalence the equations for the three lines can be written as (this is possible according to the hint of
exercise 2.12 in Harris, Algebraic Geometry - a first course)

$L: , z_0=z_1=0 \
M: , z_2=z_3 = 0 \
N: , z_0=z_3, , z_1=z_2.$

Since say $L, M$ are non-intersecting, this implies that $alpha_i,j$ for every $i=1,2, j=1,2$ are in general position. So they are projectively equivalent to $(1,0,0,0), (0,1,0,0),(0,0,1,0),(0,0,0,1)$. This gives us
the description that we want for $L$ and $M$. But how about $N$?

Note here that the most general result that we have is projective equivalence between two sets of 5 points in general position. In the context of the present question, i am concerned about two things:

1) The fact that the lines $L,M,N$ are skew does not imply in general that
every group of 5 points among the $alpha_i,j$ will be in general position.

2) The $alpha_i,j$ are six!

Question: What am i missing here? How can the above two issues be addressed?

algebraic-geometry

share|cite|improve this question

asked Apr 7 '15 at 18:33

ManosManos

14k33288

$endgroup$

add a comment | 

$begingroup$

Consider three skew (non-intersecting) lines $L,M,N$ in $mathbbP^3$.
Each line is given by two equations of the form $alpha_i,j^top z = 0, , i=1,2,3, j = 1,2$, where $z=(z_0,z_1,z_2,z_3)$ are the homogeneous coordinates of $mathbbP^3$ and $alpha_i,j in mathbbP^3$.

What i want is to show that up to projective equivalence the equations for the three lines can be written as (this is possible according to the hint of
exercise 2.12 in Harris, Algebraic Geometry - a first course)

$L: , z_0=z_1=0 \
M: , z_2=z_3 = 0 \
N: , z_0=z_3, , z_1=z_2.$

Since say $L, M$ are non-intersecting, this implies that $alpha_i,j$ for every $i=1,2, j=1,2$ are in general position. So they are projectively equivalent to $(1,0,0,0), (0,1,0,0),(0,0,1,0),(0,0,0,1)$. This gives us
the description that we want for $L$ and $M$. But how about $N$?

Note here that the most general result that we have is projective equivalence between two sets of 5 points in general position. In the context of the present question, i am concerned about two things:

1) The fact that the lines $L,M,N$ are skew does not imply in general that
every group of 5 points among the $alpha_i,j$ will be in general position.

2) The $alpha_i,j$ are six!

Question: What am i missing here? How can the above two issues be addressed?

algebraic-geometry

share|cite|improve this question

asked Apr 7 '15 at 18:33

ManosManos

14k33288

$endgroup$

share|cite|improve this question

asked Apr 7 '15 at 18:33

ManosManos

14k33288

share|cite|improve this question

asked Apr 7 '15 at 18:33

ManosManos

14k33288

asked Apr 7 '15 at 18:33

ManosManos

14k33288

asked Apr 7 '15 at 18:33

ManosManos

14k33288

1 Answer
1

active

oldest

votes

5

$begingroup$

Let usthink in terms of linear algebra. You have $3$ $2$-dimensional subspaces $S$, $T$ and $U$ in $V=k^4$ which intersect pairwise trivially.

Since $Scap T=0$, we have $V=Soplus T$. Let $u_1,u_2$ be a basis of $U$, and suppose that $u_1=s_1+t_1$ and $u_2=s_2+t_2$ with the $s_i$ in $S$ and the $t_i$ in $T$. Since $Ucap S=Ucap T=0$, the four of $s_1$, $s_2$, $t_1$, $t_2$ are non-zero.

• If $s_1$ and $s_2$ are linearly dependent, so that for example $u_2=alpha u_1$, then $alpha u_1-u_2=alpha t_1-t_2in Tcap U=0$ so that in fact also $t_2=alpha t_1$, which is absurd because it tells us that $u_1=u_2$.




• If $t_1$ and $t_2$ are linearly dependent, we can do the same.




• If neither of these two cases occur, then $s_1,s_2,t_1,t_2$ is a basis of $V$ and if $x_1,x_2,x_3,x_4$ are coordinates with respect to it, the equations of our subspaces are
  beginalign
  &x_3=x_4=0 && textfor $S$, \
  &x_1=x_2=0 && textfor $T$ and \
  &x_1-x_3=x_2-x_4=0 && textfor $U$.
  endalign

share|cite|improve this answer

edited Apr 7 '15 at 18:56

answered Apr 7 '15 at 18:50

Mariano Suárez-ÁlvarezMariano Suárez-Álvarez

112k7157289

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just found it :-) How did I come up with it? I just explored the possibilities.
  $endgroup$
  – Mariano Suárez-Álvarez
  Apr 7 '15 at 19:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dear Mariano, there seems to be some mix-up in your first bullet. It is perfectly possible for $s_1$ and $s_2$ to be dependent without having $u_2=alpha u_1$. For example take $u_1=s_1+t_1, u_2=2s_1+t_1$. So your conclusion " $s_1,s_2,t_1,t_2$ is a basis of $V$ " seems to me unwarranted.
  $endgroup$
  – Georges Elencwajg
  Apr 7 '15 at 21:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GeorgesElencwajg, but in that example $Uni 2u_1-u_2=2(s_1+t_1)-(2s_1+t_1)=t_1in T$, so that $t_1=0$, from which we see that $u_1=s_1in Scap T=0$, which is impossible. SOmething more is indeed needed, though :-) I'll try to fix it later.
  $endgroup$
  – Mariano Suárez-Álvarez
  Apr 8 '15 at 0:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right Mariano: my objection was not valid:sorry. I have upvoted your answer!
  $endgroup$
  – Georges Elencwajg
  Apr 8 '15 at 7:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A geometer should post a geometric answer!
  $endgroup$
  – Mariano Suárez-Álvarez
  Apr 8 '15 at 7:29
  

  
  
  
  

 | 
show 3 more comments

Your Answer

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5

$begingroup$

Let usthink in terms of linear algebra. You have $3$ $2$-dimensional subspaces $S$, $T$ and $U$ in $V=k^4$ which intersect pairwise trivially.

Since $Scap T=0$, we have $V=Soplus T$. Let $u_1,u_2$ be a basis of $U$, and suppose that $u_1=s_1+t_1$ and $u_2=s_2+t_2$ with the $s_i$ in $S$ and the $t_i$ in $T$. Since $Ucap S=Ucap T=0$, the four of $s_1$, $s_2$, $t_1$, $t_2$ are non-zero.

• If $s_1$ and $s_2$ are linearly dependent, so that for example $u_2=alpha u_1$, then $alpha u_1-u_2=alpha t_1-t_2in Tcap U=0$ so that in fact also $t_2=alpha t_1$, which is absurd because it tells us that $u_1=u_2$.




• If $t_1$ and $t_2$ are linearly dependent, we can do the same.




• If neither of these two cases occur, then $s_1,s_2,t_1,t_2$ is a basis of $V$ and if $x_1,x_2,x_3,x_4$ are coordinates with respect to it, the equations of our subspaces are
  beginalign
  &x_3=x_4=0 && textfor $S$, \
  &x_1=x_2=0 && textfor $T$ and \
  &x_1-x_3=x_2-x_4=0 && textfor $U$.
  endalign

share|cite|improve this answer

edited Apr 7 '15 at 18:56

answered Apr 7 '15 at 18:50

Mariano Suárez-ÁlvarezMariano Suárez-Álvarez

112k7157289

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just found it :-) How did I come up with it? I just explored the possibilities.
  $endgroup$
  – Mariano Suárez-Álvarez
  Apr 7 '15 at 19:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dear Mariano, there seems to be some mix-up in your first bullet. It is perfectly possible for $s_1$ and $s_2$ to be dependent without having $u_2=alpha u_1$. For example take $u_1=s_1+t_1, u_2=2s_1+t_1$. So your conclusion " $s_1,s_2,t_1,t_2$ is a basis of $V$ " seems to me unwarranted.
  $endgroup$
  – Georges Elencwajg
  Apr 7 '15 at 21:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GeorgesElencwajg, but in that example $Uni 2u_1-u_2=2(s_1+t_1)-(2s_1+t_1)=t_1in T$, so that $t_1=0$, from which we see that $u_1=s_1in Scap T=0$, which is impossible. SOmething more is indeed needed, though :-) I'll try to fix it later.
  $endgroup$
  – Mariano Suárez-Álvarez
  Apr 8 '15 at 0:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right Mariano: my objection was not valid:sorry. I have upvoted your answer!
  $endgroup$
  – Georges Elencwajg
  Apr 8 '15 at 7:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A geometer should post a geometric answer!
  $endgroup$
  – Mariano Suárez-Álvarez
  Apr 8 '15 at 7:29
  

  
  
  
  

 | 
show 3 more comments

5

$begingroup$

Let usthink in terms of linear algebra. You have $3$ $2$-dimensional subspaces $S$, $T$ and $U$ in $V=k^4$ which intersect pairwise trivially.

Since $Scap T=0$, we have $V=Soplus T$. Let $u_1,u_2$ be a basis of $U$, and suppose that $u_1=s_1+t_1$ and $u_2=s_2+t_2$ with the $s_i$ in $S$ and the $t_i$ in $T$. Since $Ucap S=Ucap T=0$, the four of $s_1$, $s_2$, $t_1$, $t_2$ are non-zero.

• If $s_1$ and $s_2$ are linearly dependent, so that for example $u_2=alpha u_1$, then $alpha u_1-u_2=alpha t_1-t_2in Tcap U=0$ so that in fact also $t_2=alpha t_1$, which is absurd because it tells us that $u_1=u_2$.




• If $t_1$ and $t_2$ are linearly dependent, we can do the same.




• If neither of these two cases occur, then $s_1,s_2,t_1,t_2$ is a basis of $V$ and if $x_1,x_2,x_3,x_4$ are coordinates with respect to it, the equations of our subspaces are
  beginalign
  &x_3=x_4=0 && textfor $S$, \
  &x_1=x_2=0 && textfor $T$ and \
  &x_1-x_3=x_2-x_4=0 && textfor $U$.
  endalign

share|cite|improve this answer

edited Apr 7 '15 at 18:56

answered Apr 7 '15 at 18:50

Mariano Suárez-ÁlvarezMariano Suárez-Álvarez

112k7157289

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just found it :-) How did I come up with it? I just explored the possibilities.
  $endgroup$
  – Mariano Suárez-Álvarez
  Apr 7 '15 at 19:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dear Mariano, there seems to be some mix-up in your first bullet. It is perfectly possible for $s_1$ and $s_2$ to be dependent without having $u_2=alpha u_1$. For example take $u_1=s_1+t_1, u_2=2s_1+t_1$. So your conclusion " $s_1,s_2,t_1,t_2$ is a basis of $V$ " seems to me unwarranted.
  $endgroup$
  – Georges Elencwajg
  Apr 7 '15 at 21:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GeorgesElencwajg, but in that example $Uni 2u_1-u_2=2(s_1+t_1)-(2s_1+t_1)=t_1in T$, so that $t_1=0$, from which we see that $u_1=s_1in Scap T=0$, which is impossible. SOmething more is indeed needed, though :-) I'll try to fix it later.
  $endgroup$
  – Mariano Suárez-Álvarez
  Apr 8 '15 at 0:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right Mariano: my objection was not valid:sorry. I have upvoted your answer!
  $endgroup$
  – Georges Elencwajg
  Apr 8 '15 at 7:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A geometer should post a geometric answer!
  $endgroup$
  – Mariano Suárez-Álvarez
  Apr 8 '15 at 7:29
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

Let usthink in terms of linear algebra. You have $3$ $2$-dimensional subspaces $S$, $T$ and $U$ in $V=k^4$ which intersect pairwise trivially.

Since $Scap T=0$, we have $V=Soplus T$. Let $u_1,u_2$ be a basis of $U$, and suppose that $u_1=s_1+t_1$ and $u_2=s_2+t_2$ with the $s_i$ in $S$ and the $t_i$ in $T$. Since $Ucap S=Ucap T=0$, the four of $s_1$, $s_2$, $t_1$, $t_2$ are non-zero.

• If $s_1$ and $s_2$ are linearly dependent, so that for example $u_2=alpha u_1$, then $alpha u_1-u_2=alpha t_1-t_2in Tcap U=0$ so that in fact also $t_2=alpha t_1$, which is absurd because it tells us that $u_1=u_2$.




• If $t_1$ and $t_2$ are linearly dependent, we can do the same.




• If neither of these two cases occur, then $s_1,s_2,t_1,t_2$ is a basis of $V$ and if $x_1,x_2,x_3,x_4$ are coordinates with respect to it, the equations of our subspaces are
  beginalign
  &x_3=x_4=0 && textfor $S$, \
  &x_1=x_2=0 && textfor $T$ and \
  &x_1-x_3=x_2-x_4=0 && textfor $U$.
  endalign

share|cite|improve this answer

edited Apr 7 '15 at 18:56

answered Apr 7 '15 at 18:50

Mariano Suárez-ÁlvarezMariano Suárez-Álvarez

112k7157289

$endgroup$

share|cite|improve this answer

edited Apr 7 '15 at 18:56

answered Apr 7 '15 at 18:50

Mariano Suárez-ÁlvarezMariano Suárez-Álvarez

112k7157289

answered Apr 7 '15 at 18:50

Mariano Suárez-ÁlvarezMariano Suárez-Álvarez

112k7157289

answered Apr 7 '15 at 18:50

Mariano Suárez-ÁlvarezMariano Suárez-Álvarez

112k7157289