Smoothing of multiple structures on embedded Enriques manifolds

Abstract

We show that given an embedding of an Enriques manifold of index d in a large enough projective space, there will exist embedded multiple structures with conormal bundle isomorphic to the trace zero module of the universal covering map, the universal cover being either a hyperkähler or a Calabi–Yau manifold. We then show that these multiple structures (also known as d-ropes) can be smoothed to smooth hyperkähler or Calabi–Yau manifolds respectively. Hence we obtain a flat family of hyperkähler (or Calabi–Yau) manifolds embedded in the same projective space which degenerates to an embedded d-rope structure on the given Enriques manifold of index d. The above shows that these d-rope structures on the embedded Enriques manifold are points of the Hilbert scheme containing the fibres of the above family. We show that they are smooth points of the Hilbert scheme when $d=2$.

Appendix

Here we discuss the condition that an embedding $Y\hookrightarrow \mathbb {P}^N$ of the Enriques manifold satisfies $N\ge 2\dim (Y)+1$. It is known that any variety Y can be embedded inside $\mathbb {P}^{2\dim (Y)+1}$ (see [36]). We show that $N\ge 2\dim (Y)+1$ is always the case if Y is an Enriques manifold of index two and X is one of the known classes of hyperkähler manifolds of dimension six .

Proposition 5.1

Let Y be an Enriques manifold of dimension six and index two whose universal double cover X is either $K3^{[3]}$ or $K^3(A)$. Let $Y\hookrightarrow \mathbb {P}^N$ be an embedding. Then $N\ge 13$.

Proof

It is enough to show that Y can not be embedded inside $\mathbb {P}^{12}$. For the sake of contradiction, assume $Y\hookrightarrow \mathbb {P}^{12}$ is a degree d embedding. We have that $\pi :X\rightarrow Y$ is the universal cover. Let $\mathscr {N}$ be the normal bundle of the embedding $Y\hookrightarrow \mathbb {P}^{12}$ and K be the canonical bundle of Y.

From the Euler sequence on $\mathbb {P}^{12}$ we get that

$$\begin{aligned} c_t(T_{\mathbb {P}^{12}})=(1+ht)^{13} \end{aligned}$$

(5.1)

where $h\in A^1(\mathbb {P}^{12})$ is the class of a hyperplane.

Suppose $\mathscr {N}$ is the normal bundle for the embedding. From the following exact sequence

$$\begin{aligned} 0\rightarrow T_Y\rightarrow T_{\mathbb {P}^{12}}\vert _Y\rightarrow \mathscr {N}\rightarrow 0 \end{aligned}$$

(5.2)

we see that $c_t(T_Y)\cdot c_t(\mathscr {N})=c_t(T_{\mathbb {P}^{12}}\vert _Y)$. That gives us

$$\begin{aligned} \left( 1-Kt+\displaystyle \sum _{i=2}^{i=6}c_i(Y)t^i\right) \left( 1+\displaystyle \sum _{i=1}^{i=6}c_i(\mathscr {N})t^i\right) =\displaystyle \sum _{i=0}^{i=6}\left( {\begin{array}{c}13\\ i\end{array}}\right) H^it^{6-i} \end{aligned}$$

(5.3)

where $H\in A^1(Y)$ is the class of a hyperplane section.

Expanding the equality above we get the following equations

$$\begin{aligned}&c_1(\mathscr {N})=13H+K \end{aligned}$$

(5.4)

$$\begin{aligned}&c_2(\mathscr {N})=\left( {\begin{array}{c}13\\ 2\end{array}}\right) H^2+Kc_1(\mathscr {N})-c_2(Y) \end{aligned}$$

(5.5)

$$\begin{aligned}&c_3(\mathscr {N})=\left( {\begin{array}{c}13\\ 3\end{array}}\right) H^3+Kc_2(\mathscr {N})-c_2(Y)c_1(\mathscr {N})-c_3(Y) \end{aligned}$$

(5.6)

$$\begin{aligned}&c_4(\mathscr {N})=\left( {\begin{array}{c}13\\ 4\end{array}}\right) H^4+Kc_3(\mathscr {N})-c_2(Y)c_2(\mathscr {N})-c_3(Y)c_1(\mathscr {N})-c_4(Y) \end{aligned}$$

(5.7)

$$\begin{aligned}&c_5(\mathscr {N})=\left( {\begin{array}{c}13\\ 5\end{array}}\right) H^5+Kc_4(\mathscr {N})-c_2(Y)c_3(\mathscr {N})-c_3(Y)c_2(\mathscr {N})-c_4(Y)c_1(\mathscr {N})-c_5(Y) \end{aligned}$$

(5.8)

$$\begin{aligned}&\left( {\begin{array}{c}13\\ 6\end{array}}\right) H^6=c_6(\mathscr {N})-Kc_5(\mathscr {N})+c_2(Y)c_4(\mathscr {N})+c_3(Y)c_3(\mathscr {N})\nonumber \\&\quad +c_4(Y)c_2(\mathscr {N})+c_5(Y)c_1(\mathscr {N})+c_6(Y) \end{aligned}$$

(5.9)

Next we put the values of $c_i(\mathscr {N})$ for $i=1,\dots ,5$ in  (5.9) and pull them back to X to get the following (recall that the odd Chern classes of X are trivial and $c_6(\mathscr {N})=Y.Y=d^2$)

$$\begin{aligned} \begin{aligned}&\left( {\begin{array}{c}13\\ 6\end{array}}\right) \dfrac{(\pi ^*H)^6}{2}=\left( \dfrac{(\pi ^*H)^6}{2}\right) ^2+\left( {\begin{array}{c}13\\ 4\end{array}}\right) \dfrac{(\pi ^*H)^4}{2}c_2(X)+\left( {\begin{array}{c}13\\ 2\end{array}}\right) \dfrac{(\pi ^*H)^2}{2}c_4(X)\\&\quad -\left( {\begin{array}{c}13\\ 2\end{array}}\right) \dfrac{(\pi ^*H)^2}{2}\left( c_2(X)\right) ^2\\&\quad +\dfrac{1}{2}\left( \left( c_2(X)\right) ^3-2c_2(X)c_4(X)+c_6(X)\right) . \end{aligned} \end{aligned}$$

(5.10)

Let $q(-,-)$ and c be the Beauville–Bogomolov–Fujiki form and Fujiki constant respectively on X. We know the value of the constants $a_i$ (see  (2.1)) if X is of type $K3^{[3]}$ or $K^3(A)$.

Suppose X is of the form $K3^{[3]}$. The following Riemann–Roch formula has been proven by Ellingsrud, Göttsche and Lehn (see [11]) when X is of $K3^{[3]}$ type. Fujiki constant takes the value 15.

$$\begin{aligned} \chi (L) = \left( {\begin{array}{c}\frac{1}{2}q(L)+4\\ 3\end{array}}\right) . \end{aligned}$$

(5.11)

Comparing degree-two terms of the above expression and the usual Riemann–Roch we find that

$$\begin{aligned} c_2(X)L^4=108q(L). \end{aligned}$$

(5.12)

Note that this gives us the value of the constant $C(c_2)$ as well (see Theorem  2.4 for the definition).

$$\begin{aligned} C(c_2)=108 \end{aligned}$$

(5.13)

Comparing degree-one terms of  (5.11) and the usual Riemann–Roch expression we get

$$\begin{aligned} 3(c_2(X))^2L^2-c_4(X)L^2=3120q(L). \end{aligned}$$

(5.14)

Recall that we have the following equality on hyperkähler manifolds of dimension 2n for a line bundle L (see [28] and [9], proof of Theorem 3.2)

$$\begin{aligned} \int _X\sqrt{\text {td}(X)}L^{2n-4}=(2n-4)!\left( {\begin{array}{c}n\\ n-2\end{array}}\right) \lambda (L)^{n-2}\int _X\sqrt{\text {td}(X)} \end{aligned}$$

(5.15)

where $\lambda (L)$ is the characteristic value given by the following expression

$$\begin{aligned} \lambda (L)=\dfrac{12c}{(2n-1)C(c_2(X))}q(L). \end{aligned}$$

(5.16)

Comparing degree-one terms of the left hand side and right hand side of  (5.15) and using  (5.13) and the fact that $\int _X\sqrt{\text {td}(X)}=\dfrac{9}{16}$ (see [34], Proposition 19) we get the following equality

$$\begin{aligned} 7(c_2(X))^2L^2-4c_4(X)L^2=6480q(L). \end{aligned}$$

(5.17)

We solve  (5.14) and  (5.17) to get that $(c_2(X))^2L^2=1200q(L)\text { and }c_4(X)L^2=480q(L)$.

Suppose X is of the form $K^3(A)$. The Riemann–Roch formula for this case is given below, it was calculated by Britze–Nieper (see [8]). The value of the Fujiki constant in this case is 60.

$$\begin{aligned} \chi (L) = 4 \left( {\begin{array}{c}\frac{1}{2}q(L)+3\\ 3\end{array}}\right) . \end{aligned}$$

(5.18)

We carry out the same computations ($\int _X\sqrt{\text {td}(X)}=\dfrac{2}{3}$ by [34], Proposition 21) to get that

$$\begin{aligned} (c_2(X))^2L^2=1920q(L)\text { and }c_4(X)L^2=480q(L). \end{aligned}$$

(5.19)

Now, $c_2(X)^3-2c_2(X)c_4(X)+c_6(X)$ can be calculated form the Hodge diamond (see [34], Appendix B). The Hodge diamonds of $K3^{[3]}$ and $K^3(A)$ are given below (see [5]).

Let $\chi ^p(X)=\sum (-1)^qh^{p,q}(X)$. Then [34], Appendix B gives the following equations.

$$\begin{aligned}&c_2(X)^3=7272\chi ^0(X)-184\chi ^1(X)-8\chi ^2(X), \end{aligned}$$

(5.20)

$$\begin{aligned}&c_2(X)c_4(X)=1368\chi ^0(X)-208\chi ^1(X)-8\chi ^2(X), \end{aligned}$$

(5.21)

$$\begin{aligned}&c_6(X)=36\chi ^0(X)-16\chi ^1(X)+4\chi ^2(X). \end{aligned}$$

(5.22)

We set $L=\pi ^*H$ and $x=q(\pi ^*H)$. We put the values of $(c_2(X))^2$, $c_4(X)L^2$ and $c_2(X)^3-2c_2(X)c_4(X)+c_6(X)$ calculated above in  (5.9) to get an equation in x.

If X is of the form $K3^{[3]}$ the equation we obtain is the following

$$\begin{aligned} 51480x^3=225x^6+154440x^2-112320x+21120. \end{aligned}$$

(5.23)

When X is of the form $K^3(A)$ the equation we obtain is given below

$$\begin{aligned} 51480x^3=900x^6+102960x^2-56160x+8664. \end{aligned}$$

(5.24)

That concludes the proof since neither  (5.23) nor  (5.24) has positive even integer solution.$\square$

Now we proceed to prove the same when the universal cover X is O’Grady’s six dimensional hyperkähler manifold $\mathscr {M}_6$. In order to do that we follow the same procedure but first we need $\int _{\mathscr {M}_6}\sqrt{\text {td}(\mathscr {M}_6)}$ which we calculate below.

Lemma 5.2

$\int _{\mathscr {M}_6}\sqrt{\text {td}(\mathscr {M}_6)}=\dfrac{2}{3}$.

Proof

We know the Hodge diamond of $\mathscr {M}_6$ by Mongardi-Rapagnetta-Saccà’s computation (see [27]).

Hodge diamond of $\mathscr {M}_6$

$$ \begin{array}{ccccccccccccc} &{} &{} &{} &{} &{} &{} 1 &{} &{} &{} &{} &{} &{} \\ &{} &{} &{} &{} &{} 0 &{} &{} 0 &{} &{} &{} &{} &{} \\ &{} &{} &{} &{} 1 &{} &{} 6 &{} &{} 1 &{} &{} &{} &{} \\ &{} &{} &{} 0 &{} &{} 0 &{} &{} 0 &{} &{} 0 &{} &{} &{} \\ &{} &{} 1 &{} &{} 12 &{} &{} 173 &{} &{} 12 &{} &{} 1 &{} &{} \\ &{} 0 &{} &{} 0 &{} &{} 0 &{} &{} 0 &{} &{} 0 &{} &{} 0 &{} \\ 1 &{} &{} 6 &{} &{} 173 &{} &{} 1144 &{} &{} 173 &{} &{} 6 &{} &{} 1 \\ &{} 0 &{} &{} 0 &{} &{} 0 &{} &{} 0 &{} &{} 0 &{} &{} 0 &{} \\ &{} &{} 1 &{} &{} 12 &{} &{} 173 &{} &{} 12 &{} &{} 1 &{} &{} \\ &{} &{} &{} 0 &{} &{} 0 &{} &{} 0 &{} &{} 0 &{} &{} &{} \\ &{} &{} &{} &{} 1 &{} &{} 6 &{} &{} 1 &{} &{} &{} &{} \\ &{} &{} &{} &{} &{} 0 &{} &{} 0 &{} &{} &{} &{} &{} \\ &{} &{} &{} &{} &{} &{} 1 &{} &{} &{} &{} &{} &{} \\ \end{array}$$

As before $\chi ^p(X)=\sum (-1)^qh^{p,q}(X)$. Then we have $\chi ^0=4$, $\chi ^1=-24$, $\chi ^2=348$.

It has been proven by Sawon (see [34]) that $\int _{\mathscr {M}_6}\sqrt{\text {td}(\mathscr {M}_6)}\!=\!-\dfrac{1}{48^3\cdot 3!}\left( s_2^3\!+\!\dfrac{12}{5}s_2s_4\!+\!\dfrac{64}{35}s_6\right)$ where

$$\begin{aligned} s_2^3=&-58176\chi ^0+1472\chi ^1+64\chi ^2,\quad s_2s_4=-18144\chi ^0-928\chi ^1-32\chi ^2,\\ s_6=&-6552\chi ^0-784\chi ^1-56\chi ^2. \end{aligned}$$

A direct computation yields the result.

Once we know the value of $\int _{\mathscr {M}_6}\sqrt{\text {td}(\mathscr {M}_6)}$ we can carry out the procedure explained in the proof of Proposition  5.1.

Proposition 5.3

Let Y be an Enriques manifold and $Y\hookrightarrow \mathbb {P}^N$ be a closed embedding. Assume that the universal cover X of Y is O’Grady’s six dimensional hyperkähler manifold $\mathscr {M}_6$. Then $N\ge 13$.

Proof

Recall that on $\mathscr {M}_6$ the Riemann–Roch expression is the following (see [9])

$$\begin{aligned} \chi (L)=4\left( {\begin{array}{c}\lambda (L)+3\\ 3\end{array}}\right) \end{aligned}$$

(5.25)

where $\lambda (L)$ is the characteristic value given by  (5.16).

Comparing degree-one terms of this expression and the usual Riemann–Roch expression, we get

$$\begin{aligned} \dfrac{1}{12}c_2(\mathscr {M}_6)\cdot \dfrac{1}{24}L^4=4(\lambda (L))^2 \end{aligned}$$

(5.26)

Using $c_2(\mathscr {M}_6)L^4=C(c_2(\mathscr {M}_6))q(L)$ and the value of the Fujiki constant (which is 60), we get that $C(c_2(\mathscr {M}_6))=288$. We use them to get the following two equations; the first one is obtained by comparing degree-two terms of  (5.25) and the usual Riemann–Roch expression, the second one is a consequence of  (5.15).

$$\begin{aligned}&3(c_2(\mathscr {M}_6))^2L^2-c_4(\mathscr {M}_6)L^2=5280q(L)\end{aligned}$$

(5.27)

$$\begin{aligned}&7(c_2(\mathscr {M}_6))^2L^2-4c_4(\mathscr {M}_6)L^2=11520q(L) \end{aligned}$$

(5.28)

We solve the above two equations to get $(c_2(\mathscr {M}_6))^2L^2=1920q(L)$ and $c_4(\mathscr {M}_6)L^2=480q(L)$. Moreover the values of $(c_2(\mathscr {M}_6))^3$, $c_2(\mathscr {M}_6)c_4(\mathscr {M}_6)$ and $c_6(\mathscr {M}_6)$ have been calculated in [27], Corollary 6.8. We put these values in  (5.10) by setting $L=\pi ^*(H)$ where H is the class of a hyperplane section of Y and $\pi$ is the map from the universal cover. As before, setting $x=q(\pi ^*(H))$, we obtain

$$\begin{aligned} 51480x^3=900x^6+102960x^2-56160x+8640 \end{aligned}$$

(5.29)

which has no positive integer solution. That concludes the proof.$\square$