4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.