Let $f:\mathbb{P}^n_1\dashrightarrow\mathbb{P}^n_2$ be the standard Cremona transformation based on $p_1,...,p_{n+1}\in\mathbb{P}^n_1$ and $q_1,...,q_{n+1}\in\mathbb{P}^n_2$. That is, $f$ is the rational map associated to the linear system of hypersurfaces of degree $n$ having multiplicity $n-1$ at $p_1,...,p_{n+1}$.

Let $D\subset\mathbb{P}^n$ be an hypersurface of degree $d$ having multiplicities $m_1,...,m_{n+1}$ at $p_1,...,p_{n+1}$. Let us assume that $D$ is not contracted by $f$. Is there a formula for the degree of $f(D)$, and for the multiplicities of $f(D)$ at $q_1,...,q_{n+1}$?

For instance, by an explicit computation, I got that when $d = 1$ and $D$ is an hyperplane spanned by $n-2$ of the $p_i$'s and another two general points then $f(D)$ is an irreducible quadric whit a codimension three linear space as vertex.


For simplicity of notation we may assume $\mathbb{P}^n_1=\mathbb{P}^n_2=\mathbb{P}^n$, and $q_i=p_i$. Let $P$ be the variety obtained by blowing up $\mathbb{P}^n$ at $p_1,\ldots ,p_{n+1}$, $E_i$ the exceptional divisor above $p_i$, $H$ the pull back of a hyperplane section. Then $f$ induces an involution of $P$ (which I still denote $f$), and one has in $\mathrm{Pic}(P)$ : $$f_*h=nH-\sum (n-1)E_i\qquad f_*E_i=H-\sum_{j\neq i}E_j$$ By computing $f_*(dH-\sum m_iE_i)$ you get $\deg f(D)=dn-\sum m_i$ and $\mathrm{mult}_{p_i}f(D)=d(n-1)-\sum_{j\neq i}m_j$.

Note that your final computation is incorrect, the hyperplane through $n$ of the $p_i$ is contracted to a point (think of the standard Cremona transformation for $n=2$).

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  • $\begingroup$ Thanks for the answer. Is it possible to do a similar computation in order to compute the multiplicities of $f(D)$ along the linear subspaces of type $\left\langle p_{i_1},...,p_{i_j}\right\rangle$? $\endgroup$ – user47036 Jul 18 '14 at 14:42

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