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Is $\mathbb{C}P^n$ birationally equivalent to $\mathbb{C}^n$? This may seem like a trivial question.

Based on the standard definition of a birational map or its equivalent characterization (GTM52-Hartshorne-page 26), we can conclude that $\mathbb{C}P^n$ must indeed be birationally equivalent to $\mathbb{C}^n$. However, when attempting to express a birational map between them, I encounter a problem.

Question: Can someone help provide an explicit expression of a birational map between $\mathbb{C}P^n$ and $\mathbb{C}^n$? (Related post: How to show that $\Bbb{P}^n$ is birational to $\Bbb{A}^n$ but they are not isomorphic?)

For a better understanding of the notion, we list here the definition of meromorphic map and bimeromorphic map in complex analytic setting.

In the book- K. Ueno, Classification theory of algebraic varieties and compact complex spaces, definition 2.2 and definition 2.7 read as follows:

Let $X$ and $Y$ be two irreducible and reduced complex analytic spaces. A map $\varphi$ of $X$ into the power set of $Y$ is a meromorphic map of $X$ into $Y$, denoted by $\varphi: X \dashrightarrow Y$, if $X$ satisfies the following conditions:

The graph $\mathcal{G}(\varphi):=\{(x, y) \in X \times Y: y \in \varphi(x)\}$ of $\varphi$ is an irreducible analytic subset in $X \times Y$; The projection map $p_X: \mathcal{G}(\varphi) \to X$ is a proper modification, where $\mathcal{G}(\varphi)$ is equipped with the reduced structure.

Additionally, a meromorphic map $\varphi: X \dashrightarrow Y$ of complex analytic spaces is called a bimeromorphic map if $p_Y : \mathcal{G}(\varphi) \rightarrow Y$ is also a proper modification.

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    $\begingroup$ Did you understand the construction in the case $n=1$? What difficulty did you encounter generalizing this construction to $n>1$? $\endgroup$
    – Moishe Kohan
    Commented Apr 19 at 15:06
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    $\begingroup$ The map is $\mathbb C^n\to\mathbb{CP}^n$ which sends $(x_1,\dots,x_n)$ to $(x_1:\cdots:x_n:1)$. $\endgroup$
    – Kenta S
    Commented Apr 19 at 15:16
  • $\begingroup$ @MoisheKohan I cannot write the right expression of a birational map from $\mathbb{C}^1$ to $\mathbb{C}^1$ and its inverse (as a rational map). Can you give an answer? Thank you! $\endgroup$
    – Lelong Wang
    Commented Apr 20 at 6:14
  • $\begingroup$ @KentaS What you write is a morphism and we can check that the closure (in $\mathbb{C}^n\times \mathbb{P}^n$) of the graph of your morphism cannot projection onto the $\mathbb{P}^n$. In my knowledge on a bimeromorphic map, the corresponding projection must be surjective. Contradiction. $\endgroup$
    – Lelong Wang
    Commented Apr 20 at 6:20
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    $\begingroup$ Ok, it's an awful definition. Assuming that for properness of analytical morphisms Ueno uses the standard notion of properness (he never discuses this issue in the book), with his definition, $C^n$ is not bimeromorphic to $CP^n$. I do not know who should be blamed for this mess, maybe Remmert... $\endgroup$
    – Moishe Kohan
    Commented Apr 21 at 8:53

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The maps you're looking for are $\Bbb A^n \to \Bbb P^n$ by $(x_1,\cdots,x_n)\mapsto [1:x_1:\cdots:x_n]$ and $\Bbb P^n \dashrightarrow \Bbb A^n$ by $[x_0:\cdots:x_n]\mapsto (x_1/x_0,\cdots,x_n/x_0)$ (and this last one is defined on $D(x_0)$). It's easy to see that $\Bbb A^n\to\Bbb P^n\dashrightarrow \Bbb A^n$ is the identity, while on the open set $D(x_0)$ where $\Bbb P^n\dashrightarrow \Bbb A^n$ is defined, we have $\Bbb P^n\dashrightarrow \Bbb A^n\to\Bbb P^n$ is also the identity. This shows these two maps are rational inverses, and each is a birational map.

Ueno's terminology that you bring up in your edit is pretty far away from the sorts of formulas you're looking for here, and you've been lead astray. Typically in algebraic geometry, we don't worry about a map from $X$ to the power-set of $Y$ - if we wanted to express things in the form that Ueno is using here, given a birational map $f:X\dashrightarrow Y$, we'd take the closure of the graph $(x,f(x))\subset X\times Y$ as $x$ runs over the points where $f$ is defined. This doesn't prove that the projection from the closure of the graph is surjective - this case right here is a counterexample - but you can recover Ueno's "map to the power set" via the closure of the graph if you want. (The surjectivity requirement is nonsense.) Essentially nobody uses "map to the power set" in algebraic geometry - what sort of variety or scheme structure are you putting on that?

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  • $\begingroup$ Great, thank you. I am confused by '' recover Ueno's … via the closure of the graph" Denoted by $f$ the $\Bbb A^n \to \Bbb P^n$ by $(x_1,\cdots,x_n)\mapsto [1:x_1:\cdots:x_n]$. Notet that "the graph $(x, f(x)) \subset \Bbb A^n \times \Bbb P^n$ as $x$ runs over…" is isomorphic to $\Bbb A^n$ here, thus $\Bbb A^n$ is a closed subset of $ \Bbb A^n \times \Bbb P^n$. So the projection from "the closure of the graph…as $x$ runs over the points where $f$ is defined" to $\Bbb P^n$ is not surjective. Is it? $\endgroup$
    – Lelong Wang
    Commented Apr 21 at 11:25
  • $\begingroup$ @LelongWang Right, that portion of Ueno's definition is very broken and you shouldn't use it. Sorry for the mis-type in the original version of the answer. $\endgroup$
    – KReiser
    Commented Apr 21 at 15:00

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