# Blow-ups of $\mathbb{P}^{n-3}$ and $(\mathbb{P}^1)^{n-3}$

Let us consider the points $$p_1=[1:0:...:0],p_2 = [0:1:...:0],...,p_{n-2} =[0:...:0:1],\\ p_{n-1}=[1:1:...:1]\in\mathbb{P}^{n-3}$$ and the blow-up $X = Bl_{p_1,...,p_{n-1}}\mathbb{P}^{n-3}$.

Furthermore, consider $$p_1 = ([0:1],...,[0:1]), p_2 = ([1:0],...,[1:0]), p_3=([1:1],...,[1:1])\in (\mathbb{P}^1)^{n-3}$$ and the blow-up $Y = Bl_{p_1,p_2,p_3}(\mathbb{P}^1)^{n-3}$. Note that for the Picard numbers we have $$\rho(X) = n = \rho(Y).$$ Does there exist a small $\mathbb{Q}$-factorial transformation $f:X\dashrightarrow Y$?

• What does it mean for a rational map to be $\mathbb Q$-factorial? – user47305 Dec 21 '14 at 16:55
• A small $\mathbb{Q}$-factorial modification of X is a birational map $g:X\dashrightarrow Y$, where $Y$ is a normal, projective, and $\mathbb{Q}$-factorial, and $g$ is an isomorphism in codimension $1$. – user58018 Dec 21 '14 at 17:49
• I assume you are asking whether $X$ and $Y$ are isomorphic away from a codim $\geq 2$ subset? – Piotr Achinger Dec 21 '14 at 17:49
• Exactly. That's it! – user58018 Dec 21 '14 at 17:50
• Maybe you know this, but maybe it can be helpful. (It's not an answer.) The space $\overline M_{0,n}$ can be obtained from $\mathbf P^{n-3}$ by a sequence of blow ups, where the first step is blowing up at $(n-1)$ points in general position. This is in Kapranov, "Chow quotients of Grassmannian I". It can also be obtained by iterated blowing up of $(\mathbf P^1)^{n-3}$ where the first step is blowing up at the three points you picked. This is in Tavakol, "The Chow ring of the moduli space of curves of genus zero". – Dan Petersen Dec 21 '14 at 21:30

Here is an explicit way to construct a small modification.

Consider the points $p_1,...,p_{n-3}$. We have $n-3$ codimension two linear subspaces $H_{i_1,...,i_{n-4}}^{n-5} = \left\langle p_{i_1},...,p_{i_{n-4}}\right\rangle$. For any choice of $i_1,...,i_{n-4}$ we define $\{j_1,j_2\} = \{0,...,n-3\}\setminus\{i_1-1,...,i_{n-4}-1\}$. Then, the projection from $H_{i_1,...,i_{n-4}}^{n-5}$ is the rational map $$\begin{array}{lccc} \pi_{i_1,...,i_{n-4}}: & \mathbb{P}^{n-3} & \dashrightarrow & \mathbb{P}^1 \\ & \left[x_0:...:x_{n-3}\right] & \mapsto & [x_{j_1}:x_{j_2}] \end{array}$$ We get a rational map $$\begin{array}{lccc} g: & \mathbb{P}^{n-3} & \dashrightarrow & (\mathbb{P}^1)^{n-3} \\ & x=\left[x_0:...:x_{n-3}\right] & \mapsto & (\pi_{1,...,n-4}(x),...,\pi_{2,...,n-3}(x)) \end{array}$$ Note that the hyperplane $W = \left\langle p_1,...,p_{n-3}\right\rangle = \{x_{n-3}=0\}$ is mapped by $g$ to the point $q_1=([1:0],...,[1:0])\in (\mathbb{P}^1)^{n-3}$. Furthermore, this is the only divisor contracted by $g$. Therefore, blowing-up $q_1\in (\mathbb{P}^1)^{n-3}$ we get a small transformation $g_1:X_{n-3}^{n-3} = Bl_{p_1,...,p_{n-3}}\mathbb{P}^{n-3}\dashrightarrow Y_1^{n-3} = Bl_{q_1}(\mathbb{P}^1)^{n-3}$ mapping $\widetilde{W}$ (the strict transform of $W$) to the exceptional divisor $E_{q_1}$, while the exceptional divisors $E_{p_1},...,E_{p_{n-3}}$ are mapped to the strict transforms of the $n-3$ divisors in $(\mathbb{P}^1)^{n-3}$ obtained by fixing one the factors.

Furthermore, $g([0:...:0:1]) = ([0:1],...,[0:1])$ and $g([1:...:1]) = ([1:1],...,[1:1])$. Let $\mathcal{U}\subset X_{n-3}^{n-3}$ and $\mathcal{V}\subset Y_1^{n-3}$ be the two open subsets on which $g_1$ is an isomorphism. Now, by applying the universal property of the blow-up we get that $g_{1|\mathcal{U}}$ lifts to an isomorphism $f:Bl_{p_{n-2},p_{n-1}}\mathcal{U}\rightarrow Bl_{q_2,q_3}\mathcal{V}$. Since $g_1$ is an isomorphism in codimension one we conclude that $f$ induces a small transformation $f:X = Bl_{p_1,...,p_{n-1}}\mathbb{P}^n\dashrightarrow Y=Bl_{q_1,q_2,q_3}(\mathbb{P}^1)^{n-3}$ mapping $E_{p_{n-2}}$ to $E_{q_2}$, and $E_{p_{n-3}}$ to $E_{q_3}$.

I think the answer is yes. Let us forget about the points $p_{n-1} = [1:\ldots:1] \in \mathbb{P}^{n-3}$ and $p_3 = ([1:1], \ldots, [1:1])\in (\mathbb{P}^1)^{n-3}$ for the moment. Now the two blow-ups are toric varieties, with an action of the torus $T=\mathbb{G}_m^{n-3}$. To show that they are isomorphic in codimension $1$, we need to show that the corresponding fans have the same rays. Now $\mathbb{P}^{n-3}$ has rays $e_1, \ldots, e_{n-3}$ and $-e_1-\ldots-e_{n-3}$ where the $e_i$ are standard basis vectors of the co-character lattice $N$ of $T$. Blowing up the $T$-invariant points adds the rays $-e_1, \ldots, -e_{n-3}$ and $e_1 + \ldots + e_{n-3}$. For $(\mathbb{P}^1)^{n-3}$, we start with the rays $\pm e_1, \ldots, \pm e_{n-3}$ and the blow-up at $p_1$ and $p_2$ introduces the rays $\pm (e_1 + \ldots e_{n-3})$. We conclude that if we forget to blow upthe two non-$T$-invariant points, the two varieties we get are isomorphic in codimension $1$.

To get what we want, we observe that both varieties are toric varieties, and on both sides we blow up a point in the dense orbit. We can use $T$ to move that point on both sides to the locus where the birational isomorphism is defined.

• Thank you for the answer. However, to make your argument work don't we need the birational isomorphism $g$ between the two toric varieties to be toric? If so, then we can move $g([1:...:1])$ to $([1:1],...,[1:1])$ via $T$ and then conclude. How do we know that $g$ is toric? Thanks. – user58018 Dec 22 '14 at 16:21
• It is indeed toric - it's the identity on the torus, and the analysis of fans is used to show that it extends to the open orbits of the boundary divisors. – Piotr Achinger Dec 24 '14 at 12:47
• I like the trick with moving "extra" points, but it would still be nice to have a purely geometric construction. – Anton Fonarev Dec 24 '14 at 14:56