Let $X:=\mathbb P(a_0,a_1, \ldots, a_n)$ be a well formed weighted projective variety. Let $-K_X$ be its anticanonical divisor, then how to express its volume ${\rm vol}(-K_X)=(-K_X)^n$ in terms of $a_0,\ldots, a_n$?

In principle, this could be computed by toric geometry, but the data seems too complicated to compute (especially to write the dual lattice). Besides, I was wondering if there is a generally formula for the self intersection $D^n$ of the torus invariant divisor $D = \sum_{i=0}^n c_i D_i$?

  • $\begingroup$ For $n=1. $. if you write the linear Hamiltonian $S^{1}$-action with weights $(a_{0},a_{1})$ on $\mathbb{C}^{2}$ then the Euler class of the associated bundle over the symplectic quotient (which is topologically a $2$-sphere) will be degree $-1/(a_{0}a_{1})$. Then in principal one can compute using the Duistermaat-Heckman theorem. $\endgroup$
    – Nick L
    Commented Nov 3, 2017 at 10:58
  • $\begingroup$ Let $X$ be $\text{Proj}\ k[x_0,\dots,x_n]$ with $\text{deg}(x_i)=a_i$. Let $a$ be $\text{lcm}(a_i)_i$, let $b_i$ equal $a/a_i,$ and let $b$ be $b_0\cdots b_n.$ Let $Y$ be $\text{Proj}\ k[y_0,\dots,y_n]$ with $\text{deg}(y_i) = a.$ Denote $f:X\to \mathbb{P}^n$ by $f^*y_i = x_i^{b_i}.$ Denote by $\mathcal{O}_X(1),$ resp. $\mathcal{O}_Y(1),$ the ample generator of the Picard group of the smooth locus $X^o$, resp. $Y^o=Y$. Then $f^*\mathcal{O}_Y(1)$ equals $\mathcal{O}_X(a).$ Thus, $(c_1(\mathcal{O}_X(1)))^n_X = b/a^n = 1/(a_0\cdots a_n),$ $\text{vol}(X) = (a_0+\dots +a_n)^n/(a_0\cdots a_n).$ $\endgroup$ Commented Nov 3, 2017 at 11:41
  • $\begingroup$ @Li Yultong, Jason Starr : I am sorry, this comment comes too late. However I hope it will be of some interest (for the community). Many computations in Jason Starr's answer below have been done in : Classes d'idéaux et groupe de Picard des fibrés projectifs tordus ... researchgate.net/.../226119679 Article in K-Theory 2(5):559-578 · January 1989 (=weighted) projective bundles . Notice that it concerns weighted projective BUNDLES. In particular the smooth locus is well determined there. Unfortunately the work is done in French. $\endgroup$
    – Al-Amrani
    Commented Mar 16, 2018 at 19:57

1 Answer 1


I am just rewriting my comment above as an answer.

Let $S=k[x_0,\dots,x_n]$ be the $\mathbb{Z}_{\geq 0}$-graded $k$-algebra with every $x_i$ homogeneous of degree $a_i.$ Denote by $X$ the associated projective $k$-scheme, $X =\text{Proj}\ S.$ Denote by $a$ the least common multiple of $(a_0,\dots,a_n).$ For every $i,$ denote by $b_i$ the integer such that $a_i\cdot b_i$ equals $a.$ Denote by $b$ the product $b_0\dots b_n.$

Let $R=k[y_0,\dots,y_n]$ be the $\mathbb{Z}_{\geq 0}$-graded $k$-algebra with every $y_i$ homogeneous of degree $a.$ Denote by $Y$ the associated projective $k$-scheme, $Y=\text{Proj}\ R.$ This is $k$-isomorphic to $\mathbb{P}^n_k.$ There is a unique homomorphism of $\mathbb{Z}_{\geq 0}$-graded $k$-algebras, $$f^*:R \to S, \ \ y_i\mapsto x_i^{b_i}.$$ The inverse image of the irrelevant (prime) ideal is primary for the irrelevant (prime) ideal. Thus, there is an induced morphism of $k$-schemes, $$f:X\to Y.$$ This morphism is finite, and it is flat over an open subscheme that includes the open $Y_*=D_+(y_0\dots y_n).$ In fact, because of the well-formedness hypothesis, the restriction of the morphism over $Y_*$ is naturally a torsor for the finite, flat, commutative group scheme $\Gamma=\mu_{b_0}\times \dots \times \mu_{b_n}$ acting by $$(\zeta_0,\dots,\zeta_n)\cdot [x_0,\dots,x_n] = [\zeta_0\cdot x_0,\dots, \zeta_n\cdot x_n].$$

Assume that $(a_0,\dots,a_n)$ is well-formed, i.e., the greatest common divisor of any $n$ of the $n+1$ weights equals $1$. In particular, this implies that the smooth locus $X^o$ of $X$ is a dense open subscheme whose complement has codimension $\geq 2$. Moreover, the Picard group of $X^o$ is generated by the ample invertible sheaf $\mathcal{O}_X(1)|_{X^o}$ that is the restriction of the rank $1$, reflexive, coherent sheaf $\mathcal{O}_X(1) = \widetilde{S[1]}.$ In particular, for every integer $d\geq 0,$ $$H^0(X^o,\mathcal{O}_X(d)|_{X^o}) = H^0(X,\mathcal{O}_X(d)) = S_d.$$

The Picard group of $Y$ is generated by an ample invertible sheaf $\mathcal{O}_Y(1)$ whose vector space of global sections is the free $k$-vector space with basis $y_0,\dots,y_n.$ Since $f^*(y_i)$ has degree $a,$ the pullback $f^*\mathcal{O}_Y(1)$ is an ample invertible sheaf on $X$ whose restriction to $X^o$ equals $\mathcal{O}_X(a)|_{X^o}.$ In particular, $f^*\mathcal{O}_Y(-(a_0+\dots+a_n))|_{X^o}$ is isomorphic to $\omega_{X^o/k}^{\otimes a}.$ Thus, the $n$-fold self-intersection on $X$ of $c_1(f^*\mathcal{O}_Y(a_0+\dots +a_n))$ equals $(a_0+\dots+a_n)^n$ times the $n$-fold self-intersection on $X$ of $f^*c_1(\mathcal{O}_Y(1)).$

The $n$-fold self-intersection on $Y$ of $c_1(\mathcal{O}_Y(1))$ is the unique class whose cap product with $[Y]$ equals the class of every $k$-point of $Y$. Thus, the $n$-fold self-intersection on $X$ of $f^*c_1(\mathcal{O}_Y(1))$ equals the class of every fiber of $f$ over any element of $Y_*.$ Since the morphism is a torsor for the finite, flat, commutative group scheme $\Gamma$ of length $b = b_0\cdots b_n,$ it follows that the $n$-fold self-intersection $X$ of $f^*c_1(\mathcal{O}_Y(1))$ equals $b.$

Putting the pieces together, there is a unique invertible $\mathcal{O}_X$-module, $f^*\mathcal{O}_Y(a_0+\dots + a_n)$, whose restriction to $X^o$ equals $(\omega_{X^o/k}^\vee)^{\otimes a}.$ The $n$-fold self-intersection of $c_1(f^*\mathcal{O}_Y(a_0+\dots+a_n))$ equals $(a_0+\dots+a_n)^n b.$ Thus, considered as a rational number, the $n$-fold self-intersection of $c_1(\omega_{X/k}^\vee)$ equals $(a_0+\dots+a_n)^nb/a^n.$ Finally, using the fact that $b_i/a$ equals $1/a_i,$ this gives, $$\left( c_1(\omega_{X/k}^\vee) \right)^n_X = \frac{(a_0+\dots+a_n)^n}{a_0\cdots a_n}.$$

  • $\begingroup$ Really great answer! $\endgroup$
    – diverietti
    Commented Nov 3, 2017 at 20:45
  • $\begingroup$ @Jason Starr Thank you so much for your detailed answer! I just want to double check if I understand correctly: suppose $D_i$ is the torus invariant divisor associated to $x_i$, then it is $\mathbb Q$-linear equivalent to $\mathcal{O}_X(a_i)$ (view later as a $\mathbb Q$-divisor), is that right? So combine this fact and your calculation above, one can get a formula for $(\sum c_iD_i)^n = \frac{(\sum c_ia_i)^n}{a_0\dots a_n}$. $\endgroup$
    – Li Yutong
    Commented Nov 4, 2017 at 4:46
  • $\begingroup$ Besides, I want to understand where the property well-formed used. The example in my mind is $\mathbb P(1,a,a)$ which is just $\mathbb{P}^2$. So which part of your argument breaks if I use $X:=\mathbb P(1,a,a)$? The complement of $X_0$ in $X$ is still of codimension $\geq 2$ because $X$ is normal. It seems that $f$ is still a finite morphism. Then where goes wrong? $\endgroup$
    – Li Yutong
    Commented Nov 4, 2017 at 4:51
  • $\begingroup$ @LiYutong. For $X=\mathbb{P}(1,a,a),$ the generator of the Picard group of $X^o$ is not the restriction of $\mathcal{O}_X(1).$ In fact, $\mathcal{O}_X(1)$ is isomorphic to $\mathcal{O}_X.$ Also, $\omega_{X/k}$ is not isomorphic to $\mathcal{O}_X(-a_0-a_1-a_2).$ That holds only on the locus where the variables are well-formed, i.e., the open complement of the vanishing locus of the degree-$1$ variable. This is an open affine space whose complement is (the support of) an ample divisor. So computations on that open affine say little about the Picard group of $X.$ $\endgroup$ Commented Nov 4, 2017 at 11:59
  • $\begingroup$ @JasonStarr Sorry for keep asking... Suppose $X^0 \subseteq X$ is the smooth locus of a not necessarily well-formed WPS $X=\mathbb P(a_0,\cdots,a_n)$, its complement has codim $\geq 2$ because $X$ is normal. Let $D_i$ be the toric invariant divisor corresponding to the weight $a_i$. Let $f: X \to Y$ be the morphism you defined above. Then is it ture $aD_i|_{X^0} \sim_{\mathbb Q}a_i(f^*\mathcal O_{Y}(1))|_{X^0}$? If this holds, it seems that the argument can just go through without difficulty... $\endgroup$
    – Li Yutong
    Commented Nov 5, 2017 at 2:38

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