Picturing a line in $\mathbb RP^2$

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I'm currently reading that line in $\mathbb RP^2$ is the set of all points $[a:b:c] \in \mathbb RP^2$ satisfying $\alpha X + \beta Y + \gamma Z =0$ for some constants $\alpha, \beta, \gamma$, not all zero.

However, I'm having difficulty understanding what this is.

In Euclidean space $\mathbb R^3$, the equation $\alpha X + \beta Y + \gamma Z =0$ cuts out a plane through the origin.


For a concrete instance, let's take $\alpha = \beta =0$ and $\gamma =1$. I would picture the solutions to $Z=0$ as the $(x,y)$-plane.

But in $\mathbb RP^2$, any line through the origin that lies on the plane is actually a point.

So, what exactly is a line in $\mathbb RP^2$?

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One can picture $\mathbb{RP}^2$ by the map $$ \Phi : \mathbb R^2 \to \mathbb{RP}^2, \ \ \ \Phi (x, y) = [x, y, 1].$$ That is, for each $(x, y)$, $\Phi(x, y)$ is the line in $\mathbb R^3$ through $(x, y, 1)$. $\Phi$ is injective but not surjective. Using $\Phi$ we think of $\mathbb {RP}^2$ as $\mathbb R^2$ together with the infinities: those points in $\mathbb{RP}^2$ of the form $[x, y, 0]$.

Now consider a line $\ell: \alpha x+\beta y +\gamma=0$ in $\mathbb R^2$. What is $\Phi (\ell)$? Think of $\alpha x+\beta y+\gamma = \alpha x+ \beta y+\gamma(1)$, $\Phi(\ell)$ is contained in $$\tag{1} \{ [a, b, c] \in \mathbb {RP}^2: \alpha a + \beta b + \gamma c = 0\}.$$

Thus we define a line in $\mathbb{RP}^2$ to be something of the form in (1). If $L$ is a line in $\mathbb{RP}^2$, then $\Phi^{-1}(L)$ is either an ordinary line in $\mathbb R^2$, or is empty (when $L$ is defined by $\alpha = \beta =0$ and $\gamma =1$. In this case $L$ is called the line at infinity).