---
abstract: 'We show that if $\alpha$ is a positive $(2,2)$-form then so is $\alpha^2$. We also prove that this is no longer true for forms of higher degree.' 
address:
- 'Instytut Matematyki, Uniwersytet Jagielloński, Łojasiewicza 6, 30-348 Kraków, Poland'
- 'Instytut Matematyki, Politechnika Krakowska, Warszawska 24, 31-155 Kraków, Poland'
author:
- Zbigniew Błocki
- Szymon Pliś
title: 'Squares of positive $(p,p)$-forms'
---

[^1]

namedef[subjclassname@2010]{}[ Mathematics Subject Classification]{}

Introduction {#introduction .unnumbered}
============

Recall that a $(p,p)$-form $\alpha$ in $\mathbb C^n$ is called [*positive*]{} (we write $\alpha\geq 0$) if for $(1,0)$-forms $\gamma_1,\dots,\gamma_{n-p}$ one has $$\alpha\wedge i\gamma_1\wedge\bar\gamma_1\wedge\dots\wedge
       i\gamma_{n-p}\wedge\bar\gamma_{n-p}\geq 0.$$ This is a natural geometric condition, positive $(p,p)$-forms are for example characterized by the following property: for every $p$-dimensional subspace $V$ and a test function $\varphi\geq 0$ one has $$\int_V\varphi\,\alpha\geq 0.$$

It is well known that positive forms are real (that is $\bar\alpha=\alpha$) and if $\beta$ is a (1,1)-form then $$\label{cond}
\alpha\geq 0,\ \beta\geq 0\ \Rightarrow\ \alpha\wedge\beta\geq 0.$$ It was shown by Harvey and Knapp [@HK] (and independently by Bedford and Taylor [@BT]) that does not hold for all $(p,p)$ and $(q,q)$-forms $\alpha$ and $\beta$, respectively.  We refer to Demailly’s book [@De], pp.  129-132, for a nice and simple introduction to positive forms. 

Dinew [@D] gave an explicit example of $(2,2)$-forms $\alpha,\beta$ in $\mathbb C^4$ such that $\alpha\geq 0$, $\beta\geq 0$ but $\alpha\wedge\beta<0$. We will recall it in the next section.  The aim of this note is to show the following, somewhat surprising result:

\[thm1\] Assume that $\alpha$ is a positive $(2,2)$-form.  Then $\alpha^2$ is also positive. 

It turns out that this phenomenon holds only for $(2,2)$-forms:

\[thmp\] For every $p\geq 3$ there exists a $(p,p)$-form $\alpha$ in $\mathbb C^{2p}$ such that $\alpha\geq 0$ but $\alpha^2<0$.

We do not know if similar results hold for higher powers of positive forms. 

The paper is organized as follows: in Section \[s1\] we present Dinew’s criterion for positivity of (2,2)-forms in $\mathbb C^4$ which reduces the problem to a certain property of $6\times 6$ matrices.  Further simplification reduces the problem to $4\times 4$ matrices.  We then solve it in Section \[s2\].  This is the most technical part of the paper.  Higher degree forms are analyzed in Section \[s3\] where a counterpart of Dinew’s criterion is showed and Theorem \[thmp\] is proved. 

Dinew’s Criterion {#s1}
=================

Without loss of generality we may assume that $n=4$. Let $\omega_1,\dots,
\omega_4$ be a basis of $(\mathbb C^4)^\ast$ such that $$dV:=i\omega_1\wedge\bar\omega_1\wedge\dots\wedge i\omega_4\wedge
     \bar\omega_4=\omega_1\wedge\dots\wedge\omega_4\wedge
     \bar\omega_1\wedge\dots\wedge\bar\omega_4>0.$$ Set $$\Omega_1:=\omega_1\wedge\omega_2,\ \ \ \Omega_2:=\omega_1\wedge\omega_3,\ \ \ 
    \Omega_3:=\omega_1\wedge\omega_4,$$ $$\Omega_4:=\omega_2\wedge\omega_3,\ \ \ \Omega_5:=-\omega_2\wedge\omega_4,\ \ \
   \Omega_6:=\omega_3\wedge\omega_4.$$ Then $$\Omega_j\wedge\Omega_k=
    \begin{cases}\omega_1\wedge\dots\wedge\omega_4,&\text{if }k=7-j\\
     0,&\text{otherwise.}\end{cases}$$ Every $(2,2)$-form $\alpha$ we can associate with a $6\times 6$-matrix $A=(a_{jk})$ by $$\alpha=\sum_{j,k}a_{jk}\,\Omega_j\wedge\bar\Omega_k.$$ For $$\beta=\sum_{j,k}b_{jk}\,\Omega_j\wedge\bar\Omega_k$$ we have $$\label{prod}
\alpha\wedge\beta=\sum_{j,k}a_{jk}\,b_{7-j,7-k}\,dV.$$

The key will be the following criterion from [@D]:

\[din\] $\alpha\geq 0$ iff $\bar zAz^T\geq 0$ for all $z\in\mathbb C^6$ with $z_1z_6+z_2z_5+z_3z_4=0$.

For $\gamma_1=b_1\omega_1+\dots+b_4\omega_4$, $\gamma_2=c_1\omega_1+\dots+c_4\omega_4$ we have $$\begin{aligned}
i\gamma_1\wedge\bar\gamma_1\wedge i\gamma_2\wedge\bar\gamma_2
  &=\sum_{j,k,l,m=1}^4b_j\bar b_kc_l\bar c_m\,\omega_j\wedge\omega_l
     \wedge\bar\omega_k\wedge\bar\omega_m\\
  &=\sum_{\substack{j<l\\ k<m}}(b_jc_l-b_lc_j)\overline{(b_kc_m-b_mc_k)}
     \,\omega_j\wedge\omega_l\wedge\bar\omega_k\wedge\bar\omega_m.\end{aligned}$$ It is now enough to show that the image of the mapping $$\begin{aligned}
\mathbb C^8\ni(b_1,\dots,b_4,c_1,\dots,c_4)\longmapsto&\\
  (b_1c_2-b_2c_1,b_1c_3-b_3c_1,b_1&c_4-b_4c_1,b_2c_3-b_3c_2,
      -b_2c_4+b_4c_2,b_3c_4-b_4c_3)\in\mathbb C^6\end{aligned}$$ is precisely $\{z\in\mathbb C^6:z_1z_6+z_2z_5+z_3z_4=0\}$.

Using Theorem \[din\] and an idea from [@D] we can show:

\[prop\] The form $$\alpha_a=\sum_{j=1}^6\Omega_j\wedge\bar\Omega_j
     +a\,\Omega_1\wedge\bar\Omega_6+\bar a\,\Omega_6\wedge\bar\Omega_1$$ is positive if and only if $|a|\leq 2$.

We have $$\bar zAz^T=|z|^2+2\Re(a\bar z_1z_6)
     \geq 2|z_1z_6|+2|z_2z_5+z_3z_4|+2\Re(a\bar z_1z_6).$$ If $z_1z_6+z_2z_5+z_3z_4=0$ and $|a|\leq 2$ we clearly get $\bar zAz^T\geq 0$. If we take $z_1,z_6$ with $\bar z_1z_6=-\bar a$, $|z_1|=|z_6|$ and $z_2,\dots,z_5$ with $z_2z_5+z_3z_4=-z_1z_6$ then $\bar zAz^T=2|a|(2-|a|)$.

By $$\alpha_a\wedge\alpha_b=2(3+\Re(a\bar b))dV.$$ Therefore $\alpha_2,\alpha_{-2}$ are real and positive but $\alpha_2\wedge\alpha_{-2}<0$.

In view of Theorem \[din\] we see that Theorem \[thm1\] is equivalent to the following:

\[thm2\] Let $A=(a_{jk})\in\mathbb C^{6\times 6}$ be hermitian and such that $\bar zAz^T\geq 0$ for $z\in\mathbb C^{6}$ with $z_1z_6+z_2z_5+z_3z_4=0$. Then $$\sum_{j,k=1}^{6}a_{jk}\,a_{7-j,7-k}\geq 0.$$

We will need the following technical reduction:

\[lem\] For every $(2,2)$-form $\alpha$ in $\mathbb C^4$ we can find a basis $\omega_1,\dots,\omega_4$ of $(\mathbb C^4)^\ast$ such that $$\label{cc1}
\alpha\wedge\omega_1\wedge\omega_2\wedge
     \bar\omega_1\wedge\bar\omega_j
    =\alpha\wedge\omega_1\wedge\omega_2\wedge
         \bar\omega_2\wedge\bar\omega_j=0$$ for $j=3,4$.

We may assume that $\alpha\neq 0$, then we can find $\omega_1,\omega_2\in(\mathbb C^4)^\ast$ such that $$\label{cc}
\alpha\wedge\omega_1\wedge\omega_2\wedge\bar\omega_1\wedge
\bar\omega_2=\alpha\wedge i\omega_1\wedge\bar\omega_1\wedge
i\omega_2\wedge\bar\omega_2\neq 0.$$ By $V_1$ denote the subspace spanned by $\omega_1$, $\omega_2$ and by $V_2$ the subspace of all $\omega\in(\mathbb C^4)^\ast$ satisfying with $\omega_j$ replaced by $\omega$. Then $\dim V_1=2$, $\dim V_2\geq 2$ and by we infer $V_1\cap V_2=\{0\}$, hence $(\mathbb C^4)^\ast=V_1\oplus V_2$.

Proof of Theorem \[thm2\] {#s2}
=========================

By Lemma \[lem\] we may assume that the matrix from Theorem \[thm2\] satisfies $$a_{26}=a_{36}=a_{46}=a_{56}=0$$ and $$a_{62}=a_{63}=a_{64}=a_{65}=0.$$ Then $$\sum_{j,k=1}^{6}a_{jk}\,a_{7-j,7-k}=
     \sum_{j,k=2}^{5}a_{jk}\,a_{7-j,7-k}
     +2(a_{11}a_{66}+|a_{16}|^2).$$ Therefore Theorem \[thm2\] is in fact equivalent to the following result:

\[thm4\] Let $A=(a_{jk})\in\mathbb C^{4\times 4}$ be hermitian and such that $\bar zAz^T\geq 0$ for $z\in\mathbb C^{4}$ with $z_1z_4+z_2z_3=0$. Then $$\label{aa}
\sum_{j,k=1}^{4}a_{jk}\,a_{5-j,5-k}\geq 0.$$

Write $$A=\left(\begin{matrix}
      \lambda_1&a&b&\alpha\\ 
      \bar a&\lambda_2&\beta&-c\\ 
      \bar b&\bar\beta&\lambda_3&-d\\ 
      \bar\alpha&-\bar c&-\bar d&\lambda_4
      \end{matrix}\right).$$ It satisfies the assumption of the theorem if and only if for every $z\in\mathbb C^4$ of the form $z=(1,\zeta,w,-\zeta w)$ one has $\bar zAz^T\geq 0$. We can then compute $$\begin{aligned}
\bar zAz^T=&\lambda_1+2\Re(a\zeta)+\lambda_2|\zeta|^2\\
   &+2\Re[\big(b-\alpha\zeta+\beta\bar\zeta+c|\zeta|^2\big)w]\\
   &+\big(\lambda_3+2\Re(d\zeta)+\lambda_4|\zeta|^2\big)|w|^2.\end{aligned}$$ Therefore $A$ satisfies the assumption iff $\lambda_j\geq 0$, $$\label{war1}
|a|\leq\sqrt{\lambda_1\lambda_2},\ \ \ 
|b|\leq\sqrt{\lambda_1\lambda_3},\ \ \
|c|\leq\sqrt{\lambda_2\lambda_4},\ \ \
|d|\leq\sqrt{\lambda_3\lambda_4},$$ and for every $\zeta\in\mathbb C$ $$\label{war2}
\left|b-\alpha\zeta+\beta\bar\zeta+c|\zeta|^2\right|^2
\leq\left(\lambda_1+2\Re(a\zeta)+\lambda_2|\zeta|^2\right)
\left(\lambda_3+2\Re(d\zeta)+\lambda_4|\zeta|^2\right).$$ In our case is equivalent to $$4\Re(a\bar d+b\bar c)\leq 
     2(\lambda_1\lambda_4+\lambda_2\lambda_3)
     +2(|\alpha|^2+|\beta|^2).$$ We will in fact prove something more: $$\label{aa2}
4\Re(a\bar d+b\bar c)\leq(\sqrt{\lambda_1\lambda_4}
   +\sqrt{\lambda_2\lambda_3})^2+(|\alpha|+|\beta|)^2.$$

Without loss of generality we may assume that $$\Re(a\bar d)>0,\ \ \ \Re(b\bar c)>0,$$ for if for example $\Re(a\bar d)\leq 0$ then by $$4\Re(a\bar d+b\bar c)\leq 4\Re(b\bar c)
     \leq 4\sqrt{\lambda_1\lambda_2\lambda_3\lambda_4}
     \leq(\sqrt{\lambda_1\lambda_4}
        +\sqrt{\lambda_2\lambda_3})^2.$$ Set $u:=\Re(a\bar d)$ and $\zeta:=-r\bar d/|d|$, where $r>0$ will be determined later.  Then the right-hand side of we can write as follows $$\begin{aligned}
\big(\lambda_1-&\frac{2ur}{|d|}+\lambda_2r^2\big)
  \big(\lambda_3-2r|d|+\lambda_4r^2\big)\\
 &=\left(\lambda_1+\lambda_2r^2\right)
    \left(\lambda_3+\lambda_4r^2\right)+4ur^2
    -2r\left[\lambda_1|d|+\lambda_3\frac{u}{|d|}
      +r^2\left(\lambda_2|d|+\lambda_4\frac{u}{|d|}\right)\right]\\
 &\leq\left(\lambda_1+\lambda_2r^2\right)
     \left(\lambda_3+\lambda_4r^2\right)+4ur^2
     -4r^2(\sqrt{\lambda_1\lambda_4}+\sqrt{\lambda_2\lambda_3})
            \sqrt u\\
  &=\big(\sqrt{\lambda_1\lambda_4}+\sqrt{\lambda_2\lambda_3}
            -2\sqrt u\big)^2r^2+\big(\sqrt{\lambda_1\lambda_3}
               -\sqrt{\lambda_2\lambda_4}\,r^2\big)^2.\end{aligned}$$ For $r=(\frac{\lambda_1\lambda_3}{\lambda_2\lambda_4})^{1/4}$ from we thus obtain $$\left|\frac br+\frac{\bar d}{|d|}\alpha-\frac{d}{|d|}\beta+cr\right|
     \leq\sqrt{\lambda_1\lambda_4}+\sqrt{\lambda_2\lambda_3}
          -2\sqrt u.$$ We also have $$\left|\frac br+\frac{\bar d}{|d|}(\alpha-\bar\beta)+cr\right|
    \geq \left|\frac br+cr\right|-|\alpha|-|\beta|\geq
    2\sqrt{\Re(b\bar c)}-|\alpha|-|\beta|$$ and therefore $$2\sqrt{\Re(a\bar d)}+2\sqrt{\Re(b\bar c)}
     \leq\sqrt{\lambda_1\lambda_4}+\sqrt{\lambda_2\lambda_3}
       +|\alpha|+|\beta|.$$ To get we can now use the following fact: if $0\leq a_1\leq x$, $0\leq a_2\leq x$ and $a_1+a_2\leq x+y$ then $a_1^2+a_2^2\leq
x^2+y^2$. This can be easily verified: if $a_1+a_2\leq x$ then $a_1^2+a_2^2\leq x^2$ and if $a_1+a_2\geq x$ then $$x^2+y^2\geq x^2+(a_1+a_2-x)^2=a_1^2+a_2^2+2x(x-a_1)(x-a_2).$$

$(p,p)$-forms in $\mathbb C^{2p}$ {#s3}
=================================

In $\mathbb C^{2p}$ we choose the positive volume form $$dV:=idz_1\wedge d\bar z_1\wedge\dots\wedge idz_{2p}
     \wedge d\bar z_{2p}=dz_1\wedge\dots\wedge dz_{2p}
     \wedge d\bar z_1\wedge\dots\wedge d\bar z_{2p}.$$ By $\mathcal I$ we will denote the set of subscripts $J=(j_1,\dots,
j_p)$ such that $1\leq j_1<\dots<j_p\leq 2p$. For every $J\in\mathcal I$ there exists unique $J'\in\mathcal I$ such that $J\cup J'=\{1,\dots,2p\}$. We also denote $dz_J=dz_{j_1}\wedge\dots\wedge dz_{j_p}$ and $\varepsilon_J=\pm 1$ is defined in such a way that $$dz_J\wedge dz_{J'}=\varepsilon_J\,dz_1\wedge\dots\wedge
     dz_{2p}.$$ Note that $$\label{eps}
\varepsilon_J\varepsilon_{J'}=(-1)^p.$$

Every $(p,p)$-form $\alpha$ in $\mathbb C^{2p}$ we can associate with a ${N\times N}$-matrix $(a_{JK})$, where $$N=\sharp\mathcal I=\frac{(2p)!}{(p!)^2},$$ by $$\label{mat}
\alpha=\sum_{J,K}a_{JK}idz_{j_1}\wedge d\bar z_{k_1}
     \wedge\dots\wedge idz_{j_p}\wedge d\bar z_{k_p}
     =i^{p^2}\sum_{J,K}a_{JK}dz_J\wedge d\bar z_K$$ (note that $(-1)^{p(p-1)/2}i^p=i^{p^2}$).  Then $$\label{sq}
\alpha^2=\sum_{J,K}\varepsilon_J\varepsilon_K a_{JK}a_{J'K'}\,dV$$ and for $\gamma_1,\dots,\gamma_p\in(\mathbb C^{2p})^\ast$ $$\alpha\wedge i\gamma_1\wedge\bar\gamma_1\wedge\dots\wedge 
     i\gamma_p\wedge\bar\gamma_p
     =\sum_{J,K}a_{JK}\,\gamma_1\wedge\dots\wedge\gamma_p\wedge dz_J
        \wedge\overline{(\gamma_1\wedge\dots\wedge\gamma_p\wedge dz_K)}.$$ Therefore $(a_{JK})$ has to be positive semi-definite on the image of Plücker embedding $$\label{pl}
((\mathbb C^{2p})^\ast)^p\ni(\gamma_1,\dots,\gamma_p)\longmapsto\left(\frac
     {\gamma_1\wedge\dots\wedge\gamma_p\wedge dz_J}{dz_1\wedge\dots\wedge
          dz_{2p}}\right)_{J\in
     \mathcal I}\in\mathbb C^{N}$$ which is well known to be a variety in $\mathbb C^{N}$ (see e.g. [@H], p. 64). 

We are now ready to prove Theorem \[thmp\]:

First note that it is enough to show it for $p=3$. For if $\alpha$ is a $(3,3)$-form in $\mathbb C^6$ such that $\alpha\geq 0$ and $\alpha^2<0$ then for $p>3$ we set $$\beta:=idz_7\wedge d\bar z_7+\dots+idz_{2p}\wedge d\bar z_{2p}.$$ We now have $\alpha\wedge\beta^{p-3}\geq 0$ but $(\alpha\wedge\beta^{p-3})^2=
\alpha^2\wedge\beta^{2p-6}<0$.

Set $p=3$, so that $N=20$, and order $\mathcal I=\{J_1,\dots,J_{20}\}$ lexicographically.  Then the image of Plücker embedding is in particular contained in the quadric $$\label{quad}
z_1z_{20}-z_{10}z_{11}+z_5z_{16}-z_2z_{19}=0.$$ For positive $a,\lambda,\mu$ to be determined later define $$\alpha:=\lambda i\big(dz_{J_1}\wedge d\bar z_{J_1}+dz_{J_{20}}
      \wedge d\bar z_{J_{20}}\big)
     +\mu i\!\!\!\!\!\sum_{\substack{k\in\{2,5,10,\\ \ \ 11,16,19\}}}
     dz_{J_k}\wedge d\bar z_{J_k}
     +a\big(dz_{J_1}\wedge d\bar z_{J_{20}}+
          dz_{J_{20}}\wedge d\bar z_{J_1}\big).$$ Then, similarly as in the proof of Proposition \[prop\], $$\begin{aligned}
\bar zAz^T&=\lambda(|z_1|^2+|z_{20}|^2)
     +\mu(|z_2|^2+|z_5|^2+|z_{10}|^2+|z_{11}|^2+|z_{16}|^2+|z_{19}|^2)
      +2a\Re(\bar z_1z_{20})\\
  &\geq 2(\lambda-a)|z_1z_{20}|+2\mu|-z_{10}z_{11}+z_5z_{16}-z_2z_{19}|\\
  &=2(\lambda+\mu-a)|z_1z_{20}|\end{aligned}$$ if $z$ satisfies .  Therefore $\alpha\geq 0$ if $a\leq\lambda+\mu$.

On the other hand by and $$\alpha^2=2(\lambda^2+3\mu^2-a^2)dV.$$ We see that if we take $a=\lambda+\mu$ and $\lambda>\mu>0$ then $\alpha\geq 0$ but $\alpha^2<0$.

[3]{}

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, [*Complex Analytic and Differential Geometry*]{}, monograph, 1997, available at [http://www-fourier.ujf-grenoble.fr/$\widetilde{\phantom{a}}$demailly]{}

, [*On positive $\mathbb C_{(2,2)}
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, [*Algebraic Geometry.  A First Course*]{}, Graduate Texts in Math.  133, Springer, 1995

, [*Positive $(p,p)$ forms, Wirtinger’s inequality, and currents*]{}, Value Distribution Theory, R.O. Kujala and A.L. Vitter III (ed.),  Part A, pp.  43–62, Dekker, 1974

[^1]: The second named author was partially supported by the NCN grant 2011/01/D/ST1/04192.