Interesection of two “homogenous lines”












0












$begingroup$


Suppose we have two lines given by $a_{1}x + b_{1}y + c_{1}z = 0$ and $a_{2}x + b_{2}y + c_{3}z = 0$, where $(x,y,z) in mathbf{C^3}$.



What I am confused about, is that clearly $(0,0,0)$ is a point on both lines.



However if these two lines are distinct, then I should think that this is the only point of intersection, since I keep hearing people say "two distinct lines intersect at most at a point".



However, these two lines must intersect in the complex projective space $mathbf{P^2}$; either they interect when $z = 1$ in $mathbf{C^2}$, or at the line at infinity when $z = 0$. And if they intersect at a point $(a,b,c)$ then they also intersect at all nonzero complex scalar multiples of this point.



How do I reconcile the fact that I've been hearing for years that two distinct lines intersect at most at one point, with the fact that we know there are (a whole scalar multiples) of points with coordinates not all zero, that lie on both lines.






algebraic-geometry algebraic-curves







share|cite|improve this question









$endgroup$












  • $begingroup$
    $ax+by+cz=0$ in $Bbb C^3$ determines a plane, not a line. $ax+by+cz=0$ in $Bbb P^2_{Bbb C}$ determines a line. Remember, lines through the origin in $Bbb C^n$ correspond to points in $Bbb P^{n-1}_{Bbb C}$.
    $endgroup$
    – KReiser
    Jan 21 at 6:52










  • $begingroup$
    When you say a plane, you mean something isomorphic to $mathbf{R^2}$? When you say line in $mathbf{P^2}$, each point of the line is an equivalence class, so the points of the lines are themselves lines but in $mathbf{C^3}$?
    $endgroup$
    – trynalearn
    Jan 21 at 6:56












  • $begingroup$
    When I say a plane in $Bbb C^3$, I mean two complex dimensions. When I talk of a line in $Bbb P^2$, one can say that the points of this line represent lines in $Bbb C^3$, but you should be clear about where stuff lives every time you say the word line - I think not keeping those things straight is contributing to your confusion.
    $endgroup$
    – KReiser
    Jan 21 at 7:03












  • $begingroup$
    $(0,0,0)inmathbb C^3$ does not correspond to any point in $mathbb P^2$.
    $endgroup$
    – amd
    Jan 21 at 22:10
















0












$begingroup$


Suppose we have two lines given by $a_{1}x + b_{1}y + c_{1}z = 0$ and $a_{2}x + b_{2}y + c_{3}z = 0$, where $(x,y,z) in mathbf{C^3}$.



What I am confused about, is that clearly $(0,0,0)$ is a point on both lines.



However if these two lines are distinct, then I should think that this is the only point of intersection, since I keep hearing people say "two distinct lines intersect at most at a point".



However, these two lines must intersect in the complex projective space $mathbf{P^2}$; either they interect when $z = 1$ in $mathbf{C^2}$, or at the line at infinity when $z = 0$. And if they intersect at a point $(a,b,c)$ then they also intersect at all nonzero complex scalar multiples of this point.



How do I reconcile the fact that I've been hearing for years that two distinct lines intersect at most at one point, with the fact that we know there are (a whole scalar multiples) of points with coordinates not all zero, that lie on both lines.






algebraic-geometry algebraic-curves







share|cite|improve this question









$endgroup$












  • $begingroup$
    $ax+by+cz=0$ in $Bbb C^3$ determines a plane, not a line. $ax+by+cz=0$ in $Bbb P^2_{Bbb C}$ determines a line. Remember, lines through the origin in $Bbb C^n$ correspond to points in $Bbb P^{n-1}_{Bbb C}$.
    $endgroup$
    – KReiser
    Jan 21 at 6:52










  • $begingroup$
    When you say a plane, you mean something isomorphic to $mathbf{R^2}$? When you say line in $mathbf{P^2}$, each point of the line is an equivalence class, so the points of the lines are themselves lines but in $mathbf{C^3}$?
    $endgroup$
    – trynalearn
    Jan 21 at 6:56












  • $begingroup$
    When I say a plane in $Bbb C^3$, I mean two complex dimensions. When I talk of a line in $Bbb P^2$, one can say that the points of this line represent lines in $Bbb C^3$, but you should be clear about where stuff lives every time you say the word line - I think not keeping those things straight is contributing to your confusion.
    $endgroup$
    – KReiser
    Jan 21 at 7:03












  • $begingroup$
    $(0,0,0)inmathbb C^3$ does not correspond to any point in $mathbb P^2$.
    $endgroup$
    – amd
    Jan 21 at 22:10














0












0








0





$begingroup$


Suppose we have two lines given by $a_{1}x + b_{1}y + c_{1}z = 0$ and $a_{2}x + b_{2}y + c_{3}z = 0$, where $(x,y,z) in mathbf{C^3}$.



What I am confused about, is that clearly $(0,0,0)$ is a point on both lines.



However if these two lines are distinct, then I should think that this is the only point of intersection, since I keep hearing people say "two distinct lines intersect at most at a point".



However, these two lines must intersect in the complex projective space $mathbf{P^2}$; either they interect when $z = 1$ in $mathbf{C^2}$, or at the line at infinity when $z = 0$. And if they intersect at a point $(a,b,c)$ then they also intersect at all nonzero complex scalar multiples of this point.



How do I reconcile the fact that I've been hearing for years that two distinct lines intersect at most at one point, with the fact that we know there are (a whole scalar multiples) of points with coordinates not all zero, that lie on both lines.






algebraic-geometry algebraic-curves







share|cite|improve this question









$endgroup$




Suppose we have two lines given by $a_{1}x + b_{1}y + c_{1}z = 0$ and $a_{2}x + b_{2}y + c_{3}z = 0$, where $(x,y,z) in mathbf{C^3}$.



What I am confused about, is that clearly $(0,0,0)$ is a point on both lines.



However if these two lines are distinct, then I should think that this is the only point of intersection, since I keep hearing people say "two distinct lines intersect at most at a point".



However, these two lines must intersect in the complex projective space $mathbf{P^2}$; either they interect when $z = 1$ in $mathbf{C^2}$, or at the line at infinity when $z = 0$. And if they intersect at a point $(a,b,c)$ then they also intersect at all nonzero complex scalar multiples of this point.



How do I reconcile the fact that I've been hearing for years that two distinct lines intersect at most at one point, with the fact that we know there are (a whole scalar multiples) of points with coordinates not all zero, that lie on both lines.






algebraic-geometry algebraic-curves




algebraic-geometry algebraic-curves






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 21 at 6:40









trynalearntrynalearn

662314




662314












  • $begingroup$
    $ax+by+cz=0$ in $Bbb C^3$ determines a plane, not a line. $ax+by+cz=0$ in $Bbb P^2_{Bbb C}$ determines a line. Remember, lines through the origin in $Bbb C^n$ correspond to points in $Bbb P^{n-1}_{Bbb C}$.
    $endgroup$
    – KReiser
    Jan 21 at 6:52










  • $begingroup$
    When you say a plane, you mean something isomorphic to $mathbf{R^2}$? When you say line in $mathbf{P^2}$, each point of the line is an equivalence class, so the points of the lines are themselves lines but in $mathbf{C^3}$?
    $endgroup$
    – trynalearn
    Jan 21 at 6:56












  • $begingroup$
    When I say a plane in $Bbb C^3$, I mean two complex dimensions. When I talk of a line in $Bbb P^2$, one can say that the points of this line represent lines in $Bbb C^3$, but you should be clear about where stuff lives every time you say the word line - I think not keeping those things straight is contributing to your confusion.
    $endgroup$
    – KReiser
    Jan 21 at 7:03












  • $begingroup$
    $(0,0,0)inmathbb C^3$ does not correspond to any point in $mathbb P^2$.
    $endgroup$
    – amd
    Jan 21 at 22:10


















  • $begingroup$
    $ax+by+cz=0$ in $Bbb C^3$ determines a plane, not a line. $ax+by+cz=0$ in $Bbb P^2_{Bbb C}$ determines a line. Remember, lines through the origin in $Bbb C^n$ correspond to points in $Bbb P^{n-1}_{Bbb C}$.
    $endgroup$
    – KReiser
    Jan 21 at 6:52










  • $begingroup$
    When you say a plane, you mean something isomorphic to $mathbf{R^2}$? When you say line in $mathbf{P^2}$, each point of the line is an equivalence class, so the points of the lines are themselves lines but in $mathbf{C^3}$?
    $endgroup$
    – trynalearn
    Jan 21 at 6:56












  • $begingroup$
    When I say a plane in $Bbb C^3$, I mean two complex dimensions. When I talk of a line in $Bbb P^2$, one can say that the points of this line represent lines in $Bbb C^3$, but you should be clear about where stuff lives every time you say the word line - I think not keeping those things straight is contributing to your confusion.
    $endgroup$
    – KReiser
    Jan 21 at 7:03












  • $begingroup$
    $(0,0,0)inmathbb C^3$ does not correspond to any point in $mathbb P^2$.
    $endgroup$
    – amd
    Jan 21 at 22:10
















$begingroup$
$ax+by+cz=0$ in $Bbb C^3$ determines a plane, not a line. $ax+by+cz=0$ in $Bbb P^2_{Bbb C}$ determines a line. Remember, lines through the origin in $Bbb C^n$ correspond to points in $Bbb P^{n-1}_{Bbb C}$.
$endgroup$
– KReiser
Jan 21 at 6:52




$begingroup$
$ax+by+cz=0$ in $Bbb C^3$ determines a plane, not a line. $ax+by+cz=0$ in $Bbb P^2_{Bbb C}$ determines a line. Remember, lines through the origin in $Bbb C^n$ correspond to points in $Bbb P^{n-1}_{Bbb C}$.
$endgroup$
– KReiser
Jan 21 at 6:52












$begingroup$
When you say a plane, you mean something isomorphic to $mathbf{R^2}$? When you say line in $mathbf{P^2}$, each point of the line is an equivalence class, so the points of the lines are themselves lines but in $mathbf{C^3}$?
$endgroup$
– trynalearn
Jan 21 at 6:56






$begingroup$
When you say a plane, you mean something isomorphic to $mathbf{R^2}$? When you say line in $mathbf{P^2}$, each point of the line is an equivalence class, so the points of the lines are themselves lines but in $mathbf{C^3}$?
$endgroup$
– trynalearn
Jan 21 at 6:56














$begingroup$
When I say a plane in $Bbb C^3$, I mean two complex dimensions. When I talk of a line in $Bbb P^2$, one can say that the points of this line represent lines in $Bbb C^3$, but you should be clear about where stuff lives every time you say the word line - I think not keeping those things straight is contributing to your confusion.
$endgroup$
– KReiser
Jan 21 at 7:03






$begingroup$
When I say a plane in $Bbb C^3$, I mean two complex dimensions. When I talk of a line in $Bbb P^2$, one can say that the points of this line represent lines in $Bbb C^3$, but you should be clear about where stuff lives every time you say the word line - I think not keeping those things straight is contributing to your confusion.
$endgroup$
– KReiser
Jan 21 at 7:03














$begingroup$
$(0,0,0)inmathbb C^3$ does not correspond to any point in $mathbb P^2$.
$endgroup$
– amd
Jan 21 at 22:10




$begingroup$
$(0,0,0)inmathbb C^3$ does not correspond to any point in $mathbb P^2$.
$endgroup$
– amd
Jan 21 at 22:10










1 Answer
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$begingroup$

The projective plane it defined as
$$
mathbb{P}^2 = frac{mathbb{C}^3 backslash {(0,0,0)} }{sim}
$$
where $(a,b,c) sim (d,e,f)$ if there exists $lambda in mathbb{C}^ast$ such that $(a,b,c) = (lambda d,lambda e,lambda f)$. Hence a point
$(a:b:c)$ is an equivalence class that encodes the vector $(a,b,c)in mathbb{C}^3$ and all it's nonzero multiples i.e. a line through the origin (except for the origin itself).



As you know, a line in $mathbb{P}^2$ has a homogeneous equation and the intersection of two (distinct) such lines can be described by the solutions of a linear system
$$
left{ begin{matrix}
a_1x+b_1y+c_1z =0 \
a_2x+b_2y+c_2z =0
end{matrix}right.,
$$
which in turn gives you a line $Lsubset mathbb{C}^3$ through the origin, since the system above is homogeneous of rank $2$. Therefore the quotient $(Lbackslash {(0,0,0)})/sim$ gives you a point in $mathbb{P}^2$.



Let me know if it's clear...






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    $begingroup$

    The projective plane it defined as
    $$
    mathbb{P}^2 = frac{mathbb{C}^3 backslash {(0,0,0)} }{sim}
    $$
    where $(a,b,c) sim (d,e,f)$ if there exists $lambda in mathbb{C}^ast$ such that $(a,b,c) = (lambda d,lambda e,lambda f)$. Hence a point
    $(a:b:c)$ is an equivalence class that encodes the vector $(a,b,c)in mathbb{C}^3$ and all it's nonzero multiples i.e. a line through the origin (except for the origin itself).



    As you know, a line in $mathbb{P}^2$ has a homogeneous equation and the intersection of two (distinct) such lines can be described by the solutions of a linear system
    $$
    left{ begin{matrix}
    a_1x+b_1y+c_1z =0 \
    a_2x+b_2y+c_2z =0
    end{matrix}right.,
    $$
    which in turn gives you a line $Lsubset mathbb{C}^3$ through the origin, since the system above is homogeneous of rank $2$. Therefore the quotient $(Lbackslash {(0,0,0)})/sim$ gives you a point in $mathbb{P}^2$.



    Let me know if it's clear...






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The projective plane it defined as
      $$
      mathbb{P}^2 = frac{mathbb{C}^3 backslash {(0,0,0)} }{sim}
      $$
      where $(a,b,c) sim (d,e,f)$ if there exists $lambda in mathbb{C}^ast$ such that $(a,b,c) = (lambda d,lambda e,lambda f)$. Hence a point
      $(a:b:c)$ is an equivalence class that encodes the vector $(a,b,c)in mathbb{C}^3$ and all it's nonzero multiples i.e. a line through the origin (except for the origin itself).



      As you know, a line in $mathbb{P}^2$ has a homogeneous equation and the intersection of two (distinct) such lines can be described by the solutions of a linear system
      $$
      left{ begin{matrix}
      a_1x+b_1y+c_1z =0 \
      a_2x+b_2y+c_2z =0
      end{matrix}right.,
      $$
      which in turn gives you a line $Lsubset mathbb{C}^3$ through the origin, since the system above is homogeneous of rank $2$. Therefore the quotient $(Lbackslash {(0,0,0)})/sim$ gives you a point in $mathbb{P}^2$.



      Let me know if it's clear...






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The projective plane it defined as
        $$
        mathbb{P}^2 = frac{mathbb{C}^3 backslash {(0,0,0)} }{sim}
        $$
        where $(a,b,c) sim (d,e,f)$ if there exists $lambda in mathbb{C}^ast$ such that $(a,b,c) = (lambda d,lambda e,lambda f)$. Hence a point
        $(a:b:c)$ is an equivalence class that encodes the vector $(a,b,c)in mathbb{C}^3$ and all it's nonzero multiples i.e. a line through the origin (except for the origin itself).



        As you know, a line in $mathbb{P}^2$ has a homogeneous equation and the intersection of two (distinct) such lines can be described by the solutions of a linear system
        $$
        left{ begin{matrix}
        a_1x+b_1y+c_1z =0 \
        a_2x+b_2y+c_2z =0
        end{matrix}right.,
        $$
        which in turn gives you a line $Lsubset mathbb{C}^3$ through the origin, since the system above is homogeneous of rank $2$. Therefore the quotient $(Lbackslash {(0,0,0)})/sim$ gives you a point in $mathbb{P}^2$.



        Let me know if it's clear...






        share|cite|improve this answer









        $endgroup$



        The projective plane it defined as
        $$
        mathbb{P}^2 = frac{mathbb{C}^3 backslash {(0,0,0)} }{sim}
        $$
        where $(a,b,c) sim (d,e,f)$ if there exists $lambda in mathbb{C}^ast$ such that $(a,b,c) = (lambda d,lambda e,lambda f)$. Hence a point
        $(a:b:c)$ is an equivalence class that encodes the vector $(a,b,c)in mathbb{C}^3$ and all it's nonzero multiples i.e. a line through the origin (except for the origin itself).



        As you know, a line in $mathbb{P}^2$ has a homogeneous equation and the intersection of two (distinct) such lines can be described by the solutions of a linear system
        $$
        left{ begin{matrix}
        a_1x+b_1y+c_1z =0 \
        a_2x+b_2y+c_2z =0
        end{matrix}right.,
        $$
        which in turn gives you a line $Lsubset mathbb{C}^3$ through the origin, since the system above is homogeneous of rank $2$. Therefore the quotient $(Lbackslash {(0,0,0)})/sim$ gives you a point in $mathbb{P}^2$.



        Let me know if it's clear...







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 11 at 14:33









        Alan MunizAlan Muniz

        2,2061828




        2,2061828






























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