What's going on with the points where this birational map isn't regular?

Question

Let $X\subset\mathbb{P}^3$ be the quadric surface cut out by $F(x,y,z,w)=x^2+y^2-z^2-w^2$. Let $p$ be the point $(1:0:1:0)\in X$, and let $Y$ be the plane $\{x=0\}$. Stereographic projection from $p$ gives a birational map $f$ from $X$ to $Y$. If I have calculated correctly, the map $X\mapsto Y$ is given by

$$(x:y:z:w)\mapsto (0:y:z-x:w)$$

and the inverse $f^{-1}$ is given by

$$(0:r:s:t)\mapsto (r^2-s^2-t^2:2rs:r^2+s^2-t^2:2st)$$

The original map is regular away from $p$. It seems to me that the inverse is regular away from $(0:1:0:\pm 1)$. What I want to know is why, from a geometric perspective, these are the only two points where the map is not regular.

Geometrically, what is stopping $f^{-1}$ from being regular at $(0:1:0:\pm 1)$ and only these points?

In more detail, here is my thinking and what I still want to understand:

$f$ induces a bijection from points $q$ of $X$ to points $q'$ of $Y$ such that $q'$ is the intersection of $Y$ with the line through $p,q$ and $q$ is the intersection of the line through $p,q'$ with $X$ other than $p$. Any point $q$ of $X$ other than $p$ will give us a unique line through $p,q$ that hits $Y$ in a unique point $q'$, and this same line will determine $q$ from $q'$. Furthermore, every line through $p$ will hit $Y$ somewhere, and unless it is a tangent to $X$ at $p$ it will also hit $X$ at a second point $q$. It seems to me that this means that $f$ gives an isomorphism from $X\setminus\{p\}$ to $Y$ minus the $\mathbb{P}^1$'s worth of points where the tangents to $X$ at $p$ hit it.

If I have calculated correctly, the points of $Y$ where the tangents to $X$ at $p$ hit it are given by $\{s=0\}$. So $f$ is giving a bijection from $X\setminus\{p\}$ to $Y\setminus\{s=0\}$, which is an affine plane.

This explains why $f$ is regular away from $p$ and why $f^{-1}$ is regular away from $\{s=0\}$. What it does not explain is why $f^{-1}$ is also regular on $\{s=0\}$ except at $(0:1:0:\pm 1)$.

Any point $q'$ of $Y$ determines a line through $p,q'$. This line either hits $X$ twice or else is tangent to $X$ at $p$. I would have thought that whenever it is tangent to $X$ at $p$, then $f^{-1}(q')$ would be $p$; so that $f^{-1}$ should really be regular on all of $Y$. So,

What is going wrong when I take $q'=(0:1:0:\pm 1)$?

Thanks in advance.

Answer 1

Let me give my point of view; sorry if I repeat anything you said.

The map $f^{-1}$ sends a point $q$ to the second intersection point (i.e. other than $p$) of the line $\overline{pq}$ with the quadric $X$. If $\overline{pq}$ is tangent to $X$ at $p$, then as you say that second point is just $p$ again. In other words, $f^{-1}$ contracts $\{s=0\}$ to the point $p$.

This only breaks down if the line $\overline{pq}$ is actually contained in $X$. Now through any point on a quadric there are exactly two lines; the points $(0:1:0:\pm 1)$ are the intersections of the two lines $$L_1 = \{ x=z, \, y=w \}\\ L_2 = \{x=z, y=-w \}$$ on $X$ through $p$ with the plane $Y$.

From an intrinsic point of view, we resolve $f^{-1}$ by blowing up those two points; the morphism we get then contracts one $(-1)$ curve on the resulting surface to give $\mathbf P^1 \times \mathbf P^1$, i.e. the quadric.

Answer 2

First let me confirm that all your calculations are correct: bravo!
An important point is that the tangent plane at $p$ to $X$ is given by $T_p(X)=\{x=z\}$ and that it cuts the quadric in the lines $L=\{x=z,y=w\}$ and $M=\{x=z,y=-w\}$.
These lines cut the plane $Y=\{x=0\}$ in the points $l=(0:1:0:1)$ and $m=(0:1:0:-1)$.
It is then clear that $f^{-1}$ is not defined at $l$ since that point originates from any point of $L\setminus \{p\}$.
The same goes for $m$.

If you take another point $q'\neq l,m$ of the line $N'$ given by $\{s=0\}$ and join it to $p$, you obtain a line which is tangent to $X$ at $p$ but not included in $X$ and so quite logically $f^{-1}$ sends $q'$ to $p$.
However $f^{-1}$ is thus seen not to be injective on $N'\setminus \{l,m\}$.

And herein lies the paradox/difficulty: to say that $f$ is defined on some open subset of $X$ and $f^{-1}$ on some open subset of $Y$ does not mean that $f$ is an isomorphism between the open subsets: you still have to shrink these open subsets in order to get an isomorphism.

In the present case the restriction of $f$ is an isomorphism between $X\setminus (L\cup M)$ and $Y\setminus N'$: both these open subsets are isomorphic to $\mathbb A^2$ and the geometric picture should now be clearer.