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.