Projective Plane - Cox et al - Section 8.1, Exercise 4(a)

[Math Amateur]

Homework Statement

I am reading the undergraduate introduction to algebraic geometry entitled "Ideals, Varieties and Algorithms: An introduction to Computational Algebraic Geometry and Commutative Algebra (Third Edition) by David Cox, John Little and Donal O'Shea ... ...

I am currently focused on Chapter 8, Section 1: The Projective Plane ... ... and need help getting started with Exercise 4(a) ...Exercise 4 in Section 8.1 reads as follows:

Can someone please help me with Exercise 4(a) ... ... indeed, what is actually involved in (rigorously) showing that the equation ##x^2 - y^2 = z^2## is a well-defined curve in ##\mathbb{P}^2 ( \mathbb{R} )## ... but I am very unsure of exactly how this works ... ...

Presumably, what is involved is not only (rigorously) showing that the equation ##x^2 - y^2 = z^2## is a well-defined curve in ##\mathbb{P}^2 ( \mathbb{R} )## but showing that ##x^2 - y^2 = z^2## is the representation in ##\mathbb{P}^2 ( \mathbb{R} )## of the curve ##x^2 - y^2 = 1## in ##\mathbb{R}^2## ... ... ?

Hope someone can help ... ...

Homework Equations

Definitions 1, 2, and 3 of Cox et al Section 8.1: The Projective Plane are relevant (see text below for Cox et al Section 8.1)

The Attempt at a Solution

[/B]
I am very uncertain about what is required in this example and so it is hard to make any substantial progress ... but I think the following map would be central to answering the question:

##\mathbb{R}^2 \longrightarrow \mathbb{P}^2 ( \mathbb{R} )##

which is defined by sending ##(x, y) \in \mathbb{R}^2## to the point ##p \in \mathbb{P}^2 ( \mathbb{R} )## whose homogeneous coordinates are ##(x, y, 1)## ... ...

BUT ... how do we get the variable ##z## explicitly in the equation when ##(x, y)## is sent to ##(x, y, 1)## ... ... ?

Hope someone can help ... ...

Peter======================================================================To give readers of the above post some idea of the context of the exercise, the relevant definitions and propositions, and also the notation I am providing some relevant text from Cox et al ... ... as follows:

 

[andrewkirk]

(a) is easy enough. I suspect the hard stuff comes later. The curve ##x^2-y^2=z^2## is well-defined iff for every point P on the curve, the equation ##(x,y,z)## is satisfied for every possible triple of homogeneous coordinates of P. In other words, we require that for any point Q in the projective space and any two triples ##(a,b,c)## and ##(d,e,f)## of homogeneous coordinates of Q, ##a^2-b^2=c^2## iff ##(d^2-e^2=f^3)##.

This follows immediately because for the triples to be homog coords of the same point, there must be a non-zero real number ##\lambda## such that
$$a=\lambda d;\ b=\lambda e;\ c=\lambda f$$
But then
$$a^2-b^2-c^2=\lambda(d^2-e^2-f^2)$$
and, since ##\lambda ## is nonzero, the LHS is zero iff the RHS is.

 

[Math Amateur]

(a) is easy enough. I suspect the hard stuff comes later. The curve ##x^2-y^2=z^2## is well-defined iff for every point P on the curve, the equation ##(x,y,z)## is satisfied for every possible triple of homogeneous coordinates of P. In other words, we require that for any point Q in the projective space and any two triples ##(a,b,c)## and ##(d,e,f)## of homogeneous coordinates of Q, ##a^2-b^2=c^2## iff ##(d^2-e^2=f^3)##.

This follows immediately because for the triples to be homog coords of the same point, there must be a non-zero real number ##\lambda## such that
$$a=\lambda d;\ b=\lambda e;\ c=\lambda f$$
But then
$$a^2-b^2-c^2=\lambda(d^2-e^2-f^2)$$
and, since ##\lambda ## is nonzero, the LHS is zero iff the RHS is.

Hi Andrew ... thanks for the help ...

Yes, see that, of course ... BUT ... I thought the question involved more than that ...

I thought the question involved showing that the curve ##x^2 - y^2 = 1## in ##\mathbb{R}^2## becomes the curve ##x^2 - y^2 = z^2## in ##\mathbb{P}^2 ( \mathbb{R} )## ... indeed I think this is true ... BUT ... how do you rigorously show this ... can you help ...?Peter

 

[andrewkirk]

I think the author's intention is for the nature of the curve to become apparent to the reader by working through parts a-e. Part (a) on its own won't reveal much.
Can you work out how to answer (b)? I can help with that if needed, although I can't say how it relates to Exercise 3, as that does not appear to have been posted. (e) also refers to Exercise 3, so it would be a good idea to post that, if it's not too long. Perhaps the whole thing will become clearer with a good look at Exercise 3.
What do you think the author means by 'C is still a hyperbola' and 'is a circle' in questions c and d? When we make the substitutions he requests we get equations that, in Euclidean 2-space with Cartesian coordinates, are those of a hyperbola and a circle, but I'm not sure if he means that or something else (something to do with shape?).

As regards the map ##\phi:\mathbb R\to\mathbb P^2(\mathbb R)##, that seems quite straightforward.
The map is ##\phi(P)\equiv\phi((x,y))=[(x,y,1)]## where the square brackets denote 'equivalence class of'.
Then if we label the set of points on the hyperbola in Euclidean 2-space as ##Y##, we have that ##(x,y)\in Y\
\Leftrightarrow x^2-y^2=1##. But ##(x,y,1)## is a homogeneous coordinate triple for ##Q\equiv\phi(P)##, so one set of homog coords for Q satisfies the equation for ##C## hence, by the result of part (a), ##Q\in C##. Hence ##\phi(Y)\subseteq C## and, since no point of ##\phi(Y)## can be at infinity (since it maps to ##z=1##), we in fact have ##\phi(Y)\subseteq C\smallsetminus H_\infty##.

Conversely, if we have a point ##[(x,y,z)]## in the projective space, excluding ##H_\infty## (the points at infinity, for which ##z=0##), such that ##x^2-y^2=z^2##, that point is ##\phi((x',y'))## where ##x'=x/z,\ y'=y/z## and that is in ##\phi(Y)## since ##x'{}^2-y'{}^2=1##. Hence ##C\smallsetminus H_\infty\subseteq \phi(Y)##.

Hence ##\phi(Y)=C\smallsetminus H_\infty##. However ##C## also has points that are not in the image of ##\phi##, being the points in ##H_\infty##.

Also note that ##\phi## is not injective, because of the squares. In fact four points in ##Y## map to every point in ##\phi(Y)##. However, the map obtained by restricting the domain of ##\phi## to one quadrant of the number plane is a bijection from that restricted domain to ##C\smallsetminus H_\infty##.

 

[Math Amateur]

I think the author's intention is for the nature of the curve to become apparent to the reader by working through parts a-e. Part (a) on its own won't reveal much.
Can you work out how to answer (b)? I can help with that if needed, although I can't say how it relates to Exercise 3, as that does not appear to have been posted. (e) also refers to Exercise 3, so it would be a good idea to post that, if it's not too long. Perhaps the whole thing will become clearer with a good look at Exercise 3.
What do you think the author means by 'C is still a hyperbola' and 'is a circle' in questions c and d? When we make the substitutions he requests we get equations that, in Euclidean 2-space with Cartesian coordinates, are those of a hyperbola and a circle, but I'm not sure if he means that or something else (something to do with shape?).

As regards the map ##\phi:\mathbb R\to\mathbb P^2(\mathbb R)##, that seems quite straightforward.
The map is ##\phi(P)\equiv\phi((x,y))=[(x,y,1)]## where the square brackets denote 'equivalence class of'.
Then if we label the set of points on the hyperbola in Euclidean 2-space as ##Y##, we have that ##(x,y)\in Y\
\Leftrightarrow x^2-y^2##. But ##(x,y,1)## is a homogeneous coordinate triple for ##Q\equiv\phi(P)##, so one set of homog coords for Q satisfies the equation for ##C## hence, by the result of part (a), ##Q\in C##. Hence ##\phi(Y)\subseteq C## and, since no point of ##\phi(Y)## can be at infinity (since it maps to ##z=1##), we in fact have ##\phi(Y)\subseteq C\smallsetminus H_\infty##.

Conversely, if we have a point ##[(x,y,z)]## in the projective space, excluding ##H_\infty## (the points at infinity, for which ##z=0##), such that ##x^2-y^2=z^2##, that point is ##\phi((x',y'))## where ##x'=x/z,\ y'=y/z## and that is in ##\phi(Y)## since ##x'{}^2-y'{}^2=1##. Hence ##C\smallsetminus H_\infty\subseteq \phi(Y)##.

Hence ##\phi(Y)=C\smallsetminus H_\infty##. However ##C## also has points that are not in the image of ##\phi##, being the points in ##H_\infty##.

Also note that ##\phi## is not injective, because of the squares. In fact four points in ##Y## map to every point in ##\phi(Y)##. However, the map obtained by restricting the domain of ##\phi## to one quadrant of the number plane is a bijection from that restricted domain to ##C\smallsetminus H_\infty##.

Thanks again Andrew ... ... just working through your new post now ...

My apologies for not posting Exercise 3 ... am providing it now ... as follows:

Peter

 

[Math Amateur]

I think the author's intention is for the nature of the curve to become apparent to the reader by working through parts a-e. Part (a) on its own won't reveal much.
Can you work out how to answer (b)? I can help with that if needed, although I can't say how it relates to Exercise 3, as that does not appear to have been posted. (e) also refers to Exercise 3, so it would be a good idea to post that, if it's not too long. Perhaps the whole thing will become clearer with a good look at Exercise 3.
What do you think the author means by 'C is still a hyperbola' and 'is a circle' in questions c and d? When we make the substitutions he requests we get equations that, in Euclidean 2-space with Cartesian coordinates, are those of a hyperbola and a circle, but I'm not sure if he means that or something else (something to do with shape?).

As regards the map ##\phi:\mathbb R\to\mathbb P^2(\mathbb R)##, that seems quite straightforward.
The map is ##\phi(P)\equiv\phi((x,y))=[(x,y,1)]## where the square brackets denote 'equivalence class of'.
Then if we label the set of points on the hyperbola in Euclidean 2-space as ##Y##, we have that ##(x,y)\in Y\
\Leftrightarrow x^2-y^2##. But ##(x,y,1)## is a homogeneous coordinate triple for ##Q\equiv\phi(P)##, so one set of homog coords for Q satisfies the equation for ##C## hence, by the result of part (a), ##Q\in C##. Hence ##\phi(Y)\subseteq C## and, since no point of ##\phi(Y)## can be at infinity (since it maps to ##z=1##), we in fact have ##\phi(Y)\subseteq C\smallsetminus H_\infty##.

Conversely, if we have a point ##[(x,y,z)]## in the projective space, excluding ##H_\infty## (the points at infinity, for which ##z=0##), such that ##x^2-y^2=z^2##, that point is ##\phi((x',y'))## where ##x'=x/z,\ y'=y/z## and that is in ##\phi(Y)## since ##x'{}^2-y'{}^2=1##. Hence ##C\smallsetminus H_\infty\subseteq \phi(Y)##.

Hence ##\phi(Y)=C\smallsetminus H_\infty##. However ##C## also has points that are not in the image of ##\phi##, being the points in ##H_\infty##.

Also note that ##\phi## is not injective, because of the squares. In fact four points in ##Y## map to every point in ##\phi(Y)##. However, the map obtained by restricting the domain of ##\phi## to one quadrant of the number plane is a bijection from that restricted domain to ##C\smallsetminus H_\infty##.

Just a quick question, Andrew ...

You write:

" ... ... Then if we label the set of points on the hyperbola in Euclidean 2-space as ##Y##, we have that ##(x,y)\in Y\
\Leftrightarrow x^2-y^2## ... ... I may be misunderstanding something ... but shouldn't the above read:

"... ... ... we have that ##(x,y)\in Y \ \Leftrightarrow x^2-y^2 = 1## ... ... Small point ... but ... just checking ...

Peter

 

[andrewkirk]

@Peter: quite right. That was a typo, which I have now corrected in that post.

My project for last weekend was to memorise the positions of all the non-alphanumeric symbols on the keyboard, so that when typing maths I will no longer have to look at the keyboard to find the right keys.

So far it seems to be going quite well. I'm a lot slower, but I don't look at the keyboard at all now, and I'm confident that within a fortnight or so I'll get back up to my previous typing speed. But in the mean time, it means I type some silly mistakes because half my brain is occupied with trying to remember where the special keys are.