I am reading the book "Algebraic Geometry and Statistical Learning Theory" by Sumio Watanabe and have a question regarding Remark 3.16 (1) on page 95.

He defines the blow-up of $ \mathbb{R}^2$ with center $V=\lbrace 0\rbrace$ as $B_V(\mathbb{R}^2)=\overline{\lbrace (x,y,(x:y))\in\mathbb{R}^2\times\mathbb{P}^1\vert (x,y)\in\mathbb{R}^2\setminus V\rbrace }$ where $\mathbb{P}^1$ denotes the real projective line. Then after using the typical identification of $(x:y)$ with $(1:z)$ where \begin{align} z=\begin{cases}\frac{y}{x}\,,\quad x\neq 0\\ \infty\,,\quad x=0\end{cases}\end{align} some easy computation shows that $B_V(\mathbb{R}^2) = \lbrace (x,y,z)\vert y=zx\rbrace\subset\mathbb{R}^2\times\mathbb{P}^1$. This approach is understandable for me. However in the remark the author writes that the above blow-up is equivalent to the substitution \begin{align} x=u=st\,,\quad y=uv=s \end{align} and then gluing the two coordinates $(u,v)$ and $(s,t)$. Unfortunately there are no more comments regarding this procedure. As far as I understand this, he defines two sets

\begin{align} U_1=\lbrace (u,v)\vert u,v\in\mathbb{R}\rbrace\,\quad U_2=\lbrace(s,t)\vert s,t\in\mathbb{R}\rbrace \end{align}

and then sets $U=U_1\sqcup U_2/\sim$ where $\sim$ is defined by the above two equations. But I have no idea how $U$ is the same as $B_V(\mathbb{R}^2)$. Could someone please explain me this approach?

Best regards

  • $\begingroup$ Something is wrong with your definition of $B_V(\mathbb R^2)$. Perhaps you mean this? $$B_V(\mathbb{R}^2)=\overline{\lbrace (x,y,(x:y))\in\mathbb{R}^2\times\mathbb{P}^1\mid (x,y) \in\mathbb{R}^2\setminus V\rbrace }$$ Or perhaps instead $x \in \mathbb R \setminus V$? $\endgroup$ – Lee Mosher Aug 30 '18 at 12:49
  • $\begingroup$ @LeeMosher Yes of course, thank you for your correction. $\endgroup$ – Squeezelemma Aug 30 '18 at 12:51

You can define the blow-up by $$ Bl_0(\mathbb{R}^2) = \{(x,y;u:t) \in \mathbb{R}^2\times \mathbb{P}^1 \mid xt=yu \} $$ Note that this implies that $(x:y) = (u:t)$.

You'll get two charts: $\{ u=1 \}$ and $\{t=1\}$. For the first you get $$Bl_0(\mathbb{R}^2) \cap \{ u=1 \}= \{(x,y;1:t) \in \mathbb{R}^2\times \mathbb{P}^1 \mid xt=y \} = \{(x,y,t) \in \mathbb{R}^3 \mid xt=y \}$$ with coordinates $(x,t)$ and the blow-up map is given by the projection to $\mathbb{R}^2$: $$ \pi(x,y;u:t) = (x,y). $$ In coordinates this means the composition $$ (x,t) \mapsto (x,xt;1:t) \mapsto (x,xt) $$ The other chart is similar. The gluing data comes from the gluing data of $\mathbb{P}^1$, $u=1/t$.

Let me know if you need to discuss it further.

  • $\begingroup$ Thanks for your answer. Now I see what my mistake was. But could you please give a quick explanation of the blow-up map. It wasn't introduced in the book yet. $\endgroup$ – Squeezelemma Aug 30 '18 at 13:22
  • $\begingroup$ I've expanded that a little more. $\endgroup$ – Alan Muniz Aug 30 '18 at 13:26
  • $\begingroup$ There is a nice description of the blow-up in this paper: ams.org/journals/bull/2003-40-03/S0273-0979-03-00982-0/… See the begining of chap. 0 $\endgroup$ – Alan Muniz Aug 30 '18 at 13:30
  • $\begingroup$ Great, this is really helpful. I think it is clear now. $\endgroup$ – Squeezelemma Aug 30 '18 at 13:32
  • $\begingroup$ Unfortunately I have to ask another question. Since I have little to no experience with differential geometry, I still couldn't figure how to glue the charts. Could you please give me an explanation about how to do that? $\endgroup$ – Squeezelemma Sep 12 '18 at 8:08

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