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\title{Corrigendum: Finiteness of polarized K3 surfaces and hyperkähler manifolds}
\alttitle{Corrigendum : Finitude des surfaces K3 et des variétés hyperkähleriennes polarisées}

\author[\initial{D.} \lastname{Huybrechts}]{\firstname{Daniel} \lastname{Huybrechts}}
\address{Mathematisches Institut, Universität Bonn,\\ Endenicher Allee 60, 53115 Bonn, Germany}
\email{huybrech@math.uni-bonn.de}
\editor{Chenyang Xu}

\datereceived{2019-08-28}
\dateaccepted{2019-10-05}



\begin{document}

\maketitle

In the proof of Proposition~2.8 in~\cite{HAHL} we consider isometric embeddings
\[
\varphi\colon T(S_0)\,\hookrightarrow T(S_0)\oplus\ZZ\cdot e
\]
with $\varphi(\sigma)\in \CC\cdot\sigma\oplus\CC\cdot\bar\sigma\oplus\CC\cdot e$. At this point Lemma~2.5 is evoked, which, however, assumes the more restrictive and unrealistic condition (2.1) $\varphi_\CC(\sigma)=\lambda\cdot \sigma+\mu\cdot e$, i.e. $\varphi_\CC(\sigma)\in\CC\cdot\sigma\oplus\CC\cdot e$.

This is fixed by instead using the following lemma which is proved with essentially the same techniques as the original~\cite[Lemma~2.5]{HAHL}.

\begin{lem}\label{lem:CMlambda}
Assume that $\sigma\in T\otimes\CC$ defines a general Hodge structure of K3 type on $T$ such that $K_\sigma$ is a subfield of a CM field $K$. Then there exist at most finitely many isometric embeddings $\varphi\colon T\,\hookrightarrow T\oplus\ZZ\cdot e$ satisfying
\begin{equation}\label{eqn:varphibis}
\varphi_\CC(\sigma)=\lambda\cdot \sigma+\lambda'\cdot\bar\sigma+\nu\cdot e.
\end{equation}
\end{lem}

\begin{proof}
Consider an isometric embedding $\varphi$ satisfying~\eqref{eqn:varphibis}. We think of $\varphi$ in terms of the integral matrix $(a_{ij}| b_j)$, where $\varphi(\gamma_i)=\sum a_{ij}\cdot \gamma_j+b_i\cdot e$ with $\gamma_i$ a basis of $T(S)$ as above. Then writing $\sigma=\sum\mu_i\cdot\gamma_i$ the image $\varphi_\CC(\sigma)$ corresponds to $(a_{ij}|b_j)\cdot (\mu_i)$ and~\eqref{eqn:varphibis} becomes the system of equations $\sum a_{ij}\mu_j=\lambda\mu_i+\lambda'\bar\mu_i$, $i=1,\ldots,n$, and $\sum b_j\mu_j=\nu$. This shows $\lambda,\lambda',\nu\in K$. Indeed, there is at most one solution $\lambda,\lambda'$, which then is contained in $K$, unless $\det\left(
\begin{smallmatrix}\mu_i&\bar\mu_i\\
\mu_j&\bar\mu_j
\end{smallmatrix}\right)=0$, i.e.\ $\mu_i\bar\mu_j\in\RR$, for all $i\ne j$. But then $1=\sigma_1=\sum(\mu_i\gamma_i.\gamma_1)$ implies $\bar\mu_j=\sum\mu_i\bar\mu_j(\gamma_i.\gamma_1)\in\RR$ for all $j$, which would yield the contradiction $\sigma=\bar\sigma$. Eventually use that $K_\sigma=\QQ(\mu_i)$ and that the CM field $K$ is closed under complex conjugation.

Next observe that, as $a_{ij},b_j\in\ZZ$, there exists an $N\in \ZZ$ independent of $\varphi$ such that $N\lambda,N\lambda',N\nu\in\ko_{K}$. As $\varphi$ is an isometry, one also has $(\sigma.\bar\sigma)=(\varphi(\sigma).\varphi(\bar\sigma))$, which translates into
\begin{equation}\label{eqn:varphi2}
|\lambda|^2+|\lambda'|^2+|\nu|^2(d/(\sigma.\bar\sigma))=1.
\end{equation}



As any embedding $g\colon K\,\hookrightarrow \CC$ commutes with complex conjugation, $g$ applied to~\eqref{eqn:varphi2} also shows $|g(\lambda)|^2+|g(\lambda')|^2+|g(\nu)|^2(d/g(\sigma.\bar\sigma))=1$. Observe that $g(\sigma.\bar\sigma)>0$ and that, therefore, the last summand is non-negative. Indeed, choose $z\in \CC$ such that $(z\sigma,\bar z\bar\sigma)=1$. Then $|g(z)|^2\,g(\sigma.\bar\sigma)=(g(z\sigma),g(\bar z\bar \sigma))=1$, as $g$ commutes with complex conjugation. Hence, for $N\lambda,N\lambda'\in\ko_K$ one has $|g(N\lambda)|\leq N$ and $|g(N\lambda')|\leq N$ for all $g\colon K\,\hookrightarrow\CC$. By Minkowski theory there are only finitely many such $N\lambda,N\lambda'\in{\mathcal O}_{K}$.

As there are only finitely many possibilities for $\lambda$ and $\lambda'$, it suffices to show that they determine $\varphi$ essentially uniquely. Indeed, if $\varphi$ and $\varphi'$ are both isometric embeddings satisfying~\eqref{eqn:varphibis} with the same $\lambda,\lambda'$, then $\sigma\in{\rm Ker}(\psi-\psi')\otimes\CC$, where $\psi,\psi'\colon T\to T$ are the compositions of $\varphi$, $\varphi'$ with the projection to $T$. Hence, using the assumption that the Hodge structure is general, one finds $\psi=\psi'$. Using that $\varphi$ and $\varphi'$ are both isometric embeddings allows one to conclude.
\end{proof}



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