Wonderful compactifications of the moduli space of points in affine and projective space

Abstract

We introduce and study smooth compactifications of the moduli space of n labeled points with weights in projective space, which have normal crossings boundary and are defined as GIT quotients of the weighted Fulton–MacPherson compactification. We show that the GIT quotient of a wonderful compactification is also a wonderful compactification under certain hypotheses. We also study a weighted version of the configuration spaces parametrizing n points in affine space up to translation and homothety. In dimension one, the above compactifications are isomorphic to Hassett’s moduli space of rational weighted stable curves.

Appendix

Lemma 6.1

Let $Z\hookrightarrow X$ be an embedding of smooth varieties, both flat over a variety Y. For any geometric point $y\in Y$ let $Z_y$ and $X_y$ be the fibers of $Z\rightarrow Y$ and $X\rightarrow Y$ over y. Then the blowup $\mathrm{Bl}_Z X$ of X at Z is flat over Y and we have an isomorphism

Proof

Let $\mathscr {I}$ be the ideal sheaf corresponding to the embedding $Z\hookrightarrow X$. Since Z and X are smooth, that embedding is regular. Consequenlty, is a locally free, hence flat, sheaf of -modules for all $n\geqslant 0$. By the hypothesis is flat over $\mathscr {O}_Y$, consequently is also flat over $\mathscr {O}_Y$. Then, by the exact sequence

we deduce by induction that is also flat over $\mathscr {O}_Y$ for all $n\geqslant 0$. Now, the lemma follows by [13, Lemma 1]. $\square$

Lemma 6.2

Let Z be a smooth subvariety of a smooth variety Y and let $\pi :\mathrm{Bl}_Z Y\rightarrow Y$ be the blowup, with exceptional divisor $E=\pi ^{-1}(Z)$.

• (i) Let V be a smooth subvariety of Y, not contained in Z, and let $\widetilde{V}\subset \mathrm{Bl}_Z Y$ be its strict transform. Then:

(i$_1$):

If V meets Z transversally (or is disjoint from Z), then $\widetilde{V}=\pi ^{-1}(V)$ and $\mathscr {I}_{\pi ^{-1}(V)}=\mathscr {I}_{\widetilde{V}}$. Moreover,

$$\begin{aligned} N_{\widetilde{V}/\mathrm{Bl}_Z Y}\cong \pi ^*\!N_{V/Y}. \end{aligned}$$

(i$_2$):

If $V\supset Z$, then $\mathscr {I}_{\pi ^{-1}(V)}=\mathscr {I}_{\widetilde{V}}\cdot \mathscr {I}_E$. Moreover,

$$\begin{aligned} N_{\widetilde{V}/\mathrm{Bl}_Z Y}\cong \pi ^*\!N_{V/Y}\otimes \mathscr {O}(E). \end{aligned}$$

Also, if Z has codimension 1 in V, the projection from $\widetilde{V}$ to V is an isomorphism.

• (ii) Let $Z_1, Z_2$ be smooth subvarieties of Y intersecting transversally.

(ii$_1$):

Assume $Z_1\cap Z_2 \supseteq Z$. Then their strict transforms $\widetilde{Z_1}$ and $\widetilde{Z_2}$ intersect transversally and $\widetilde{Z_1}\cap \widetilde{Z_2}= \widetilde{Z_1\cap Z_2}$; in particular, if $Z_1\cap Z_2 = Z$, then $\widetilde{Z_1}\cap \widetilde{Z_2}= \varnothing$.

(ii$_2$):

Assume Z intersects transversally with $Z_1$ and $Z_2$, as well as with their intersection $Z_1\cap Z_2$. Then their strict transforms $\widetilde{Z_1}$ and $\widetilde{Z_2}$ intersect transversally and $\widetilde{Z_1}\cap \widetilde{Z_2}= \widetilde{Z_1\cap Z_2}$.

(ii$_3$):

If $Z_1\supseteq Z$ and $Z_2$ intersects transversally with Z, then the intersections

$$\begin{aligned} \widetilde{Z_1}\cap \widetilde{Z_2}\quad \text {and}\quad (E\cap \widetilde{Z_1})\cap \widetilde{Z_2} \end{aligned}$$

are transversal. Moreover, $\widetilde{Z_1}\cap \widetilde{Z_2}=\widetilde{Z_1\cap Z_2}$.

Proof

(i) is standard; the proof of (ii) follows from [18, Lemma 2.9]. $\square$

1.1 Proofs of Lemmas 4.24 and 4.25

In order to prove Lemma 4.24, we consider the intersections of the iterated strict transforms of $\delta _I$ with the centers of each of the blowups in the sequence . In addition, we distinguish two types of subvarieties $\delta _J\cap \delta _I$ of $\delta _I$ under the identification (5) of $\delta _I$ with

as follows:

• (a) Let $\delta _J \in \mathscr {H}^I_1$ (i.e. J contains I). In this case .

• (b) Let $\delta _J \in \mathscr {H}^I_2$ (i.e. J is disjoint from I). In this case .

Clearly, we have an equality of sets

where $\mathscr {H}_{\mathscr {A}_+(I^\mathrm{c})}$ is defined in Definition 4.15 for input data $(d,n-|I|+1, \mathscr {A}_+(I^\mathrm{c}))$. We give $\mathscr {H}_{\mathscr {A}_+(I^\mathrm{c})}$ an order $(\lessdot )$ compatible with the order $(\prec )$ of Definition 4.20, that is:

• for any $\delta _J \in \mathscr {H}^I_1\cup \mathscr {H}^I_2$, $\delta _J\cap \delta _I \lessdot \delta _{J'}\cap \delta _I$ if and only if .

Lemma 6.3

The ordered set $(\mathscr {H}_{\mathscr {A}_+(I^\mathrm{c})},\lessdot )$ satisfies the second condition of Theorem 3.9.

Proof

The proof is very similar to the Proof of Lemma 4.16, so we omit it. $\square$

For any let $M_i$ be the cardinality of the set $\mathscr {H}^I_1\!\cup \cdots \cup \mathscr {H}^I_i$. For any $k\in \mathbb {Z}$ such that $0<k\leqslant M_4$, we define $P_{I,k}$ to be k-th step in the sequence of blowups with respect to the order $(\prec )$. Moreover, for any , we denote by $V^{(k)}$ the iterated dominant transform of in $P_{I,k}$. In particular $\delta _J^{(M_i)}\!=\delta _J^{[i]}$ for any $\delta _J\in \mathscr {H}_{\mathscr {A}}$.

Lemma 6.4

• (i) Let $\delta _J \in \mathscr {H}^I_1\cup \mathscr {H}^I_2$. For each k such that $0\leqslant k\leqslant M_2$ the subvariety ${\delta _J}^{(k)}$ of $P_{I,k}$ either contains or intersects transversally with the center of the blowup $P_{I,k+1}\rightarrow P_{I,k}$.

• (ii) $({\delta _I\cap \delta _J})^{(k)}= {\delta _I}^{(k)}\cap {\delta _J}^{(k)}$ for any $\delta _J\in \mathscr {H}_2^I$ and $0\leqslant k\leqslant M_1$. Moreover, each intersection ${\delta _I}^{(k)}\cap {\delta _J}^{(k)}$ is transversal.

• (iii) ${\delta _I}^{(M_1)}\cap {\delta _J}^{(M_1)}=\varnothing$ for any $\delta _J\in \mathscr {H}_3^I$. Therefore ${\delta _I}^{(k)}\cap {\delta _J}^{(k)}=\varnothing$ for all $k>M_1$ as well.

• (iv) $({\delta _I\cap \delta _J})^{(k)}= {\delta _I}^{(k)}\cap {\delta _J}^{(k)}$ for any $\delta _J\in \mathscr {H}_2^I$ and $M_1< k\leqslant M_2$. Moreover, each intersection ${\delta _I}^{(k)}\cap {\delta _J}^{(k)}$ is transversal.

Proof

(i) By Lemma 4.21, $\mathscr {H}^I_1\cup \mathscr {H}^I_2$ is a building set. Also, observe that the order $(\prec )$ on $\mathscr {H}^I_1\cup \mathscr {H}^I_2$ is inclusion preserving. Therefore, the claim follows from Proposition 3.8 (ii).

(ii) Let k be such that $0\leqslant k\leqslant M_1$. By part (i), the iterated strict transform of each in $P_{I,k}$ must either intersect the blowup center inside $P_{I,k}$ transversally or it must contain that center. Also, observe that for we have that the intersection $\delta _I\cap \delta _J$ is transversal. Moreover, the iterated strict transform of $\delta _I$ in $P_{I,k}$ contains the corresponding center for all the above k. Therefore, by a repeated use of Lemma 6.2 (ii$_1$) and (ii$_3$) we deduce (ii).

(iii) For any $\delta _J\in \mathscr {H}_3^I$, by definition, the set J overlaps with I. Consider an element $j\in J\cap I$ and set . Since ${\delta _{J'}}^{(M_1)}\supseteq {\delta _J}^{(M_1)}$, it is enough to show that ${\delta _I}^{(M_1)}\cap {\delta _{J'}}^{(M_1)}=\varnothing$. While $\delta _{J'}$ is not necessarily an element of $\mathscr {H}_{\mathscr {A}}$, the same argument as in Lemma 4.16 shows that the set , with $\mathscr {H}_1^I$ given an ascending dimension order and $\delta _J$ listed last, is a building set. Now the intersection $\delta _I\cap \delta _J'=\delta _{I\cup J'}$ belongs to $\mathscr {H}_1^I$; assume it is the m-th element of that set with respect to $(\prec )$ for some m with $0<m\leqslant M_1$. Consider the iterated strict transforms $\delta _I^{(m-1)}$ and $\delta _{J'}^{(m-1)}$ of $\delta _I$ and $\delta _{J'}$ respectively under . Since the intersection $\delta _I\cap \delta _{J'}$ is transversal, by using the argument of the previous paragraph we see that $\delta _I^{(m-1)}\cap \delta _{J'}^{(m-1)}=\delta _{I\cup J'}^{(m-1)}$. Now consider the m-th blowup $P_{I,m}\rightarrow P_{I,m-1}$. By Lemma 6.2 (ii), we deduce that $\delta _I^{(m)}\cap \delta _{J'}^{(m)}=\varnothing$, so ${\delta _I}^{(M_1)}\cap {\delta _{J'}^{(M_1)}}=\varnothing$.

(iv) By Lemma 4.21, $\mathscr {H}^I_1\cup \mathscr {H}^I_2$ is a building set and it is straightforward to see that the set $\mathscr {H}^I_1\cup \mathscr {H}^I_2\cup \{\delta _I\}$ is a building set as well. Since the latter is a subset of $\mathscr {H}_{\mathscr {A}}$ we can consider it ordered with order ($\prec$). Now, observe that the order $(\prec )$ on $\mathscr {H}^I_1\cup \mathscr {H}^I_2 \cup \{\delta _I\}$ is inclusion preserving. Therefore, by Proposition 3.8 (ii), $\delta _I^{(k)}$ intersects transversally with the center of the k-th blowup for all k as in the statement and so does $\delta _J^{(k)}$ where $\delta _J\in \mathscr {H}_2^I$. The proof is then very similar to the proof of (ii). $\square$

Proof of Lemma 4.24

By Lemma 6.4 (iii), we see that the iterated dominant transform $\delta _I^{[3]}\subset \mathbf P _I^{[3]}$ is isomorphic to the iterated dominant transform $\delta _I^{[2]}\subset \mathbf P _I^{[2]}$. By Lemma 6.4 (ii) and (iv), the latter is, in turn, equal to the iterated blowup of $\delta _I$ at the ordered building set $(\mathscr {H}_{\mathscr {A}_+(I^\mathrm{c})},\lessdot )$. The lemma now follows by Lemma 6.3 and Theorem 3.9 (ii). $\square$

Proof of Lemma 4.25

(i) First, note that the normal bundle of in is isomorphic to

Clearly $\mathscr {O}_{\mathbb {P}^{d(n-1)-1}}(-1)$ pulls back to $\mathscr {O}_\mathbf{P ^{[1]}}(-1)$ on $\mathbf P ^{[1]}$, so $\mathscr {O}_{\mathbb {P}^{d(n-|I|)-1}}(-1)$ pulls back to $\mathscr {O}_{{\delta _I}^{[1]}}(-1)$ on ${\delta _I}^{[1]}$. Therefore, by a repeated application of Lemma 6.2 (i$_2$) we see that the normal bundle of the iterated strict transform of $\delta _I$ in $\mathbf P ^{[1]}$ is isomorphic to

Now, by Lemma 6.4 (iii) and (iv), the intersection of each iterated strict transform of $\delta _I$ in the sequence of blowups $\mathbf P ^{[3]}\rightarrow \mathbf P ^{[1]}$ with every blowup center (corresponding to $\mathscr {H}_2^I \cup \mathscr {H}_3^I$) is transversal (even empty). Therefore, by applying Lemma 6.2 (i$_1$) we complete the proof.

The proof of part (ii) is identical to the proof of (i) and is therefore omitted. $\square$