The second fundamental form of the real Kaehler submanifolds

TL;DR

This paper investigates the second fundamental form of real Kaehler submanifolds in Euclidean space, providing evidence supporting a conjecture about their geometric structure in low codimension cases.

Contribution

It proves that the second fundamental form behaves as predicted by the conjecture for low codimension, advancing understanding of Kaehler submanifold geometry.

Findings

Second fundamental form aligns with conjectured behavior in low codimension cases

Supports the conjecture for codimension up to 11

Counterexample indicates limitations for higher codimension

Abstract

Let $f\colon M^{2n}\to\R^{2n+p}$, $2\leq p\leq n-1$, be an isometric immersion of a Kaehler manifold into Euclidean space. Yan and Zheng conjectured in \cite{YZ} that if the codimension is $p\leq 11$ then, along any connected component of an open dense subset of $M^{2n}$, the submanifold is as follows: it is either foliated by holomorphic submanifolds of dimension at least $2n-2p$ with tangent spaces in the kernel of the second fundamental form whose images are open subsets of affine vector subspaces, or it is embedded holomorphically in a Kaehler submanifold of $\R^{2n+p}$ of larger dimension than $2n$. This bold conjecture was proved by Dajczer and Gromoll just for codimension three and then by Yan and Zheng for codimension four. In this paper we prove that the second fundamental form of the submanifold behaves pointwise as expected in case that the conjecture is true. This result is a first fundamental step for a possible classification of the non-holomorphic Kaehler submanifolds lying with low codimension in Euclidean space. A counterexample shows that our proof does not work for higher codimension, indicating that proposing $p=11$ in the conjecture as the largest codimension is appropriate.

Equations

$$\mathcal{N}({\alpha }_L)=\{Y\in T_{x}M:{\alpha }_L(X,Y)=0\text{mbox}foranyX\in T_{x}M\}.$$

$$f=F\circ j:M^{2n}\to {\mathbb{R}}^{2n+p}$$

$$\mathcal{J}{\alpha }_{\Omega }^f(X,Y)={\alpha }_{\Omega }^f(X,JY)\text{mbox}foranyX,Y\in \mathfrak{X}(M)$$

$$Q(x)=\{\eta \in N_1(x):⟨\eta ,\alpha (Z,T)⟩=⟨\bar{\eta },\alpha (Z,JT)⟩\text{mbox}foranyZ,T\in T_{x}M\}$$

$$\mathcal{J}{\alpha }_Q(X,Y)={\alpha }_Q(X,JY)\text{mbox}foranyX,Y\in T_{x_0}M.$$

$$\mathcal{S}(\phi )=\text{mbox}span\{\phi (X,Y)\text{mbox}forallX\in V_1\text{mbox}andY\in V_2\}$$

$$\mathcal{N}(\phi )=\{Y\in V_2:\phi (X,Y)=0\text{mbox}forallX\in V_1\}$$

$$\kappa (\phi )=\underset{X\in V_1}{\max }\{\dim {\phi }_X(V_2)\}$$

$${\phi }_{X}Y=\phi (X,Y).$$

$$\dim N(X)=\dim V_2-\kappa (\phi ).$$

$$\mathcal{U}(X)={\phi }_X(V_2)\cap {\phi }_X(V_2{)}^{\perp },$$

$$R{E}^{\ast }(\phi )=\{X\in RE(\phi ):\phi (X,X)\ne 0\}$$

$$R{E}^{\#}(\phi )=\{X\in RE(\phi ):{\tau }_{\phi }(X)=\tau (\phi )\}$$

$$⟨⟨\phi (X,Y),\phi (Z,T)⟩⟩-⟨⟨\phi (X,T),\phi (Z,Y)⟩⟩=0$$

$$⟨⟨\phi (X,Y),\phi (Z,T)⟩⟩=0$$

$$\mathcal{L}(X)=\mathcal{S}(\phi {|}_{V_1\times N(X)}).$$

$$\mathcal{L}(X)\subset \mathcal{U}(X)$$

$$⟨⟨({\xi }_1,{\xi }_2),({\eta }_1,{\eta }_2)⟩⟩=⟨{\xi }_1,{\eta }_1⟩-⟨{\xi }_2,{\eta }_2⟩.$$

$$\mathcal{T}(\xi ,\eta )=(\eta ,-\xi )$$

$$⟨⟨\mathcal{T}\delta ,\nu ⟩⟩=⟨⟨\delta ,\mathcal{T}\nu ⟩⟩.$$

$$W^{p,p}={\mathcal{U}}^r\oplus {\hat{\mathcal{U}}}^r\oplus {\mathcal{V}}^{p-r,p-r}$$

$$\mathcal{T}\phi (X,Y)=\phi (X,JY)\text{mbox}foranyX\in V_1\text{mbox}andY\in V_2$$

$$⟨⟨\phi (X,Y),\mathcal{T}(\xi ,\bar{\xi })⟩⟩=⟨⟨\mathcal{T}\phi (X,Y),(\xi ,\bar{\xi })⟩⟩=⟨⟨\phi (X,JY),(\xi ,\bar{\xi })⟩⟩=0.$$

$$4\dim \mathcal{S}(\phi )\le \kappa (\phi )(\kappa (\phi )+2).$$

$${\phi }_X(V)=\text{mbox}span\{{\phi }_X{X}_j,{\phi }_{X}J{X}_j,1\le j\le m\}$$

$$\phi (Z,X_q)=\phi (X_1,Y)\text{mbox}and\phi (Z,J{X}_q)=\phi (X_1,JY).$$

$$\phi (X,JY)=\mathcal{T}\phi (X,Y)=\mathcal{T}\phi (Y,X)=\phi (Y,JX)=\phi (JX,Y)$$

$$\mathcal{S}(\phi )=\text{mbox}span\{\phi (X_i,X_j),\phi (X_i,J{X}_j),1\le i\le j\le m\},$$

$$\gamma (X,Y)=(\alpha (X,Y),\alpha (X,JY)).$$

$$\alpha (JX,Y)=\alpha (X,JY)\text{mbox}foranyX,Y\in V^{2n}.$$

Full text

In this paper we prove that the second fundamental form of the submanifold behaves pointwise as expected in case that the conjecture is true. This result is a first fundamental step for a possible classification of the non-holomorphic Kaehler submanifolds lying with low codimension in Euclidean space. A counterexample shows that our proof does not work for higher codimension, indicating that proposing $p=11$ in the conjecture as the largest codimension is appropriate.

††2020 Mathematics Subject Classification: 53B25, 53C42.††Key words: Real Kaehler submanifolds, the index of complex relative nullity.

An isometric immersion $f\colon M^{2n}\to\mathbb{R}^{2n+p}$ is called a real Kaehler submanifold if $(M^{2n},J)$ is a connected Kaehler manifold of complex dimension $n\geq 2$ isometrically immersed into Euclidean space with local substantial codimension $p$. The latter means that the image of $f$ restricted to any open subset of $M^{2n}$ does not lie in a proper affine subspace of $\mathbb{R}^{2n+p}$. Moreover, when $p$ is even, we focus in the case in which $f$ restricted to any open subset of $M^{2n}$ is not holomorphic with respect to any complex structure of the ambient space $\mathbb{R}^{2n+p}$.

Since the pioneering work by Dajczer and Gromoll [8], there has been an increasing interest in the study of the real Kaehler submanifolds. The reason, in good part, it is due because when these submanifolds are minimal then they enjoy several of the feature properties of minimal surfaces. For instance, they admit an associated one-parameter family of non congruent isometric minimal submanifolds all with the same Gauss map. Another one is being the real part of its holomorphic representative. Moreover, the immersions are pluriharmonic maps and, in some cases, they admit a Weierstrass type representation. For a partial account of results on this subject of research, as well as many references, we refer to [12].

There is plenty of knowledge on real Kaehler submanifolds $f\colon M^{2n}\to\mathbb{R}^{2n+p}$ when the codimension is as low as $p=1,2$. For instance, in the hypersurface case, there is the local parametric classification obtained in [8] that can be seen in [12] as Theorem $15.14$. The classification of the metrically complete submanifolds with codimension $p=2$ follows from [9] and [15]. Moreover, for both codimensions, the submanifolds carry a foliation by complex relative nullity leaves of dimension $2n-2p$ as described next.

Let $f\colon M^{2n}\to\mathbb{R}^{2n+p}$ be a real Kaehler submanifold and let $L\subset N_{f}M(x)$ be a normal vector subspace at $x\in M^{2n}$. We denote the $\alpha_{L}\colon TM\times TM\to L$ the $L$-component of its normal vector valued second fundamental form $\alpha\colon TM\times TM\to N_{f}M$ and by $\mathcal{N}(\alpha_{L})\subset T_{x}M$ the tangent vector subspace

[TABLE]

Then $\Delta(x)=\mathcal{N}(\alpha_{N_{f}M(x)})$ is called the relative nullity subspace of $f$ at $x\in M^{2n}$. Its complex part $\Delta_{c}(x)=\Delta(x)\cap J\Delta(x)$ is named the complex relative nullity subspace whose dimension $\nu_{f}^{c}(x)$ is the index of complex relative nullity. It is well-known that the vector subspaces $\Delta_{c}(x)$ form a smooth integrable distribution on any open subset of $M^{2n}$ where $\nu_{f}^{c}(x)$ is constant. Moreover, the totally geodesic leaves are holomorphic submanifolds of $M^{2n}$ as well as open subsets of even dimensional affine vector subspaces of $\mathbb{R}^{2n+p}$.

Real Kaehler submanifolds in codimension at least three can be obtained just by considering holomorphic submanifolds of a given real Kaehler submanifold. More precisely, let $F\colon N^{2n+2m}\to\mathbb{R}^{2n+p}$, $m\geq 1$, be a real Kaehler submanifold, and then let $j\colon M^{2n}\to N^{2n+2m}$ be any holomorphic isometric immersion. Then the composition of isometric immersions $f\colon M^{2n}\to\mathbb{R}^{2n+p}$ given by

[TABLE]

is a real Kaehler submanifold.

It is clearly relevant to establish conditions asserting that a real Kaehler submanifold is locally a composition as in (1). This was achieved for $p=3$ by Dajczer and Gromoll [10] and for $p=4$ by Yan and Zheng [16] under the assumption that the index of complex relative nullity of $f\colon M^{2n}\to\mathbb{R}^{2n+p}$ satisfies $\nu_{f}^{c}(x)<2n-2p$ at any $x\in M^{2n}$. The result in the latter paper was complemented by us in [5].

A bold conjecture by Yan and Zheng in [16] states, under the same assumption as above on the index of complex relative nullity, that any real Kaehler submanifold in codimension $p\leq 11$ is a composition as in (1) along connected components of an open dense subset of $M^{2n}$. The purpose of this paper is to walk a fundamental step in order to treat that rather challenging conjecture. We prove that the second fundamental form of the submanifold behaves pointwise as expected if the conjecture were true. Moreover, we have that our proof fails for $p=12$, indicating that proposing $p=11$ in the conjecture as the largest codimension seems appropriate. For codimension $p\leq 6$ our result was obtained in [1] up to some inconsistencies in the argument; see Remark 18. As for higher codimension, it is shown by this paper that the proof is much more difficult.

Before stating our main theorem we roughly explain why this result turns out to be the one we expected. For this purpose, let $f=F\circ j\colon M^{2n}\to\mathbb{R}^{2n+p}$ be a composition, as in (1) but where $F$ itself is not such a composition. Then the second fundamental form $\alpha^{f}\colon TM\times TM\to N_{f}M$ of $f$ splits as the sum of the second fundamental forms $\alpha^{j}$ of $j$ and $\alpha^{F}$ of $F$ that one restricted to $TM$, and where both components need to satisfy certain conditions now discussed. On one hand, there is a vector bundle isometry ${\cal J}\in\Gamma(Aut(\Omega))$ such that

[TABLE]

where $\Omega=F_{*}N_{j}M$. In fact, being $j$ holomorphic we have that $j_{*}JX=J^{N}j_{*}X$ for any $X\in\mathfrak{X}(M)$. Differentiating once and then taking normal component yields $J^{N}\alpha^{j}(X,Y)=\alpha^{j}(X,JY)$. Then ${\cal J}F_{*}|_{N_{j}M}=F_{*}J^{N}|_{F_{*}N_{j}M}$ satisfies the requirement. On the other hand, since $F$ is not a composition then $\alpha^{F}$ should have a large index of complex relative nullity, and hence the same remains to be the case when it is restricted to $TM$.

Let $N_{1}(x)\subset N_{f}M(x)$ denote the vector subspace spanned at $x\in M^{2n}$ by the second fundamental form of $f$, namely, $N_{1}(x)=\mbox{span\,}\{\alpha(X,Y)\colon X,Y\in T_{x}M\}$. It is usually called the first normal space of $f$ at $x\in M^{2n}$. Then let $Q(x)\subset N_{1}(x)$ the complex vector subspace defined as

[TABLE]

where if $\eta=\sum_{i=1}^{k}\alpha(X_{i},Y_{i})$ then $\bar{\eta}=\sum_{i=1}^{k}\alpha(X_{i},JY_{i})$.

• (i)

If $Q=Q(x_{0})$ then $\dim Q=\ell>0$ and there is an isometry ${\cal J}\in Aut(Q)$ such that

[TABLE]

• (ii)

If $N_{1}(x_{0})=Q\oplus P$ is an orthogonal decomposition then $\nu^{c}(\alpha_{P})\geq 2(n-p+\ell)$.

If the submanifold satisfies $\dim N_{1}(x_{0})=q<p$ then the proof of Theorem 1 gives a stronger result. Indeed, one can replace the assumption $p\leq 11$ by $q\leq 11$ and assume in part $(ii)$ that $\nu_{f}^{c}(x_{0})<2n-2q$.

The extrinsic assumption on index of complex relative nullity in Theorem 1 can be replaced by an intrinsic hypothesis, namely, there is no complex vector subspace $L^{2n-2p}\subset T_{x_{0}}M$ such that the sectional curvature satisfies $K_{M}(P)=0$ for any plane $P^{2}\subset L^{2n-2p}$. Notice that part $(i)$ gives that ${\cal J}$ is a complex structure, that is, that we have ${\cal J}^{2}=-I$. It also yields that $\alpha_{Q}(JX,Y)=\alpha_{Q}(X,JY)$ holds for any $X,Y\in T_{x_{0}}M$. Finally, we observe that the inequality $\nu_{f}^{c}(x)<2n-2p$ holds in a neighborhood of $x_{0}$ in $M^{2n}$.

Although the above result can be seen as a validation of the Yan and Zheng conjecture at the level of the structure of the second fundamental form of the submanifold, it is a distance apart from proving that the conjecture is true. In fact, we believe that for codimensions $p\geq 7$ there is just one other possibility, namely, that we may have complex ruled submanifolds that are not compositions. By being complex ruled we mean that there is a holomorphic foliation of $M^{2n}$ such that the image by $f$ of each leaf is part of an affine vector subspace of $\mathbb{R}^{2n+p}$ but it does not have to be part of the complex relative nullity.

An immediate application of Theorem 1 is the following result under a pinching curvature condition.

For $p\leq n$ and without the assumption on the index of complex relative nullity the weaker estimate $m\geq n-p$ was given as Corollary $15.6$ in [12].

Finally, we observe that if $p$ is even and $f\colon M^{2n}\to\mathbb{R}^{2n+p}$ is holomorphic with respect to some complex structure in the ambient space then $Q(x)=N_{1}(x)$ holds everywhere and hence both results given above are trivial.

1 Preliminaries

This section provides several basic facts used throughout the paper.

Let $\varphi\colon V_{1}\times V_{2}\to W$ denote a bilinear form between real vector spaces of finite dimension. The image of $\varphi$ is the vector subspace of $W$ defined by

[TABLE]

whereas the (right) nullity of $\varphi$ is the vector subspace of $V_{2}$ given by

[TABLE]

whose dimension $\nu(\varphi)$ is the index of nullity of $\varphi$.

A vector $X\in V_{1}$ is called a (left) regular element of $\varphi$ if $\dim\varphi_{X}(V_{2})=\kappa(\varphi)$ where

[TABLE]

and $\varphi_{X}\colon V_{2}\to W$ is the linear map defined by

[TABLE]

Then $RE(\varphi)\subset V_{1}$ denotes the subset of regular elements of $\varphi$. Given $X\in RE(\varphi)$ then the vector subspace $N(X)=\ker\varphi_{X}$ satisfies

[TABLE]

Let $W$ be endowed with an inner product of any signature. Then we denote

[TABLE]

$\tau_{\varphi}(X)=\dim\mathcal{U}(X)$ and $\tau(\varphi)=\min_{X\in RE(\varphi)}\{\tau_{\varphi}(X)\}$.

The following result will be used throughout the paper without further reference.

Proposition 3

**. **The following facts hold:

• (i)

The subset $RE(\varphi)\subset V_{1}$ is open and dense.

• (ii)

If $V_{1}=V_{2}=V$ and $\varphi$ is symmetric then

[TABLE]

is an open dense subset of $V$.

• (iii)

If $W$ is endowed with an inner product then

[TABLE]

is an open dense subset of $V_{1}$.

*Proof: *Part $(i)$ is Proposition $4.4$ in [12] whereas the proof of Lemma $2.1$ in [11] gives part $(iii)$. An easy argument gives part $(ii)$, for instance, see the proof of Lemma $4.5$ in [12].

Let $W$ be endowed with the inner product ${\langle\!\langle}\,,\,{\rangle\!\rangle}$. Then the bilinear form $\varphi$ is said to be flat if

[TABLE]

for any $X,Z\in V_{1}$ and $Y,T\in V_{2}$. It is said that $\varphi$ is null if

[TABLE]

for any $X,Z\in V_{1}$ and $Y,T\in V_{2}$.

Given $X\in RE(\varphi)$ we denote

[TABLE]

Then let $\sigma_{\varphi}(X)=\dim\mathcal{L}(X)$ and $\sigma(\varphi)=\min_{X\in RE(\varphi)}\{\dim\sigma_{\varphi}(X)\}$.

Proposition 4

*. *** If $X\in RE(\varphi)$ then $\mathcal{L}(X)\subset\varphi_{X}(V_{2})$. Moreover, if $\varphi$ is flat then

[TABLE]

and thus $\sigma(\varphi)\leq\sigma_{\varphi}(X)\leq\tau_{\varphi}(X)$.

*Proof: *See Proposition $4.6$ in [12].

Let $U^{p}$ be a $p$-dimensional vector space induced with a positive definite inner product ${\langle}\,,\,{\rangle}$. Set $W^{p,p}=U^{p}\oplus U^{p}$ and let $\pi_{1}\colon W^{p,p}\to U^{p}$ (respectively, $\pi_{2}$) denote taking the first (respectively, second) component of $W^{p,p}$. Then let $W^{p,p}$ be endowed with the inner product ${\langle\!\langle}\,,\,{\rangle\!\rangle}$ of signature $(p,p)$ given by

[TABLE]

Then $\mathcal{T}\in\text{Aut}(W)$ defined by

[TABLE]

is a complex structure which means that $\mathcal{T}^{2}=-I$. Moreover, it holds that

[TABLE]

A vector subspace $L\subset W^{p,p}$ is called degenerate if $L\cap L^{\perp}\neq 0$ and nondegenerate if otherwise. A degenerate vector subspace $L\subset W^{p,p}$ is called isotropic if $L=L\cap L^{\perp}$.

Proposition 5

*. *** Given a vector subspace $L\subset W^{p,p}$ there is a direct sum decomposition

[TABLE]

where $\mathcal{U}^{r}=L\cap L^{\perp}$, the vector subspace $\hat{\mathcal{U}}^{r}$ is isotropic, the vector subspace $\mathcal{U}^{r}\oplus\hat{\mathcal{U}}^{r}$ is nondegenerate and $L\subset\mathcal{U}^{r}\oplus\mathcal{V}^{p-r,p-r}$ where $\mathcal{V}^{p-r,p-r}=(\mathcal{U}^{r}\oplus\hat{\mathcal{U}}^{r})^{\perp}$.

*Proof: *See Sublemma $2.3$ in [3] or Corollary $4.3$ in [12].

Remark 6

**. **In the decomposition (6) only $\mathcal{U}^{r}$ is completely determined by $L$. In fact, if $\hat{\mathcal{U}}^{r}=\mbox{span\,}\{\xi_{1},\ldots,\xi_{r}\}$ then any alternative description is as $\mbox{span\,}\{\xi_{1}+\delta_{1},\ldots,\xi_{r}+\delta_{r}\}$ where $\{\delta_{1},\ldots,\delta_{r}\}$ is any set of vectors belonging to $\mathcal{V}^{p-r,p-r}$ that span an isotropic subspace.

Let the vector space $V_{2}$ carry a complex structure $J\in\text{Aut}(V_{2})$. It is a standard fact that $V_{2}$ is even-dimensional and admits a basis of the form $\{X_{j},JX_{j}\}_{1\leq j\leq n}$. Assume that the bilinear form $\varphi\colon V_{1}\times V_{2}\to W^{p,p}$ satisfies that

[TABLE]

and let $W^{p,p}=\mathcal{U}\oplus\hat{\mathcal{U}}\oplus\mathcal{V}$ be the decomposition given by (6) for $L={\cal S}(\varphi)$. Then we have $\mathcal{T}\mathcal{U}=\mathcal{U}$. In effect, if ${\langle\!\langle}\varphi(X,Y),(\xi,\bar{\xi}){\rangle\!\rangle}=0$ for any $X\in V_{1}$ and $Y\in V_{2}$ then

[TABLE]

Proposition 7

*. *** The following facts hold:

• (i)

$\mathcal{T}|_{{\cal S}(\varphi)}\in\text{Aut}\,\,({\cal S}(\varphi))$* and $\mathcal{T}|_{\mathcal{U}}\in\text{Aut}(\mathcal{U})$ are complex structures.*

• (ii)

The vector subspaces ${\cal S}(\varphi)$ and $\mathcal{U}$ of $W^{p,p}$ have even dimension.

• (iii)

The vector subspace $\mathcal{N}(\varphi)\subset V_{2}$ is $J$-invariant and thus of even dimension.

• (iv)

If $\Omega=\pi_{1}(\mathcal{U})$ then $\dim\mathcal{U}=\dim\Omega$ and if $\varphi_{\Omega}=\pi_{\Omega\times\Omega}\circ\varphi$ then ${\cal S}(\varphi_{\Omega})=\mathcal{U}$.

*Proof: *The considerations given above yield parts $(i)$ to $(iii)$. Being the subspace $\mathcal{U}$ isotropic then $\pi_{1}|_{\mathcal{U}}\colon\mathcal{U}\to\Omega$ is an isomorphism. Since $\mathcal{T}\mathcal{U}=\mathcal{U}$ gives that $\pi_{2}(\mathcal{U})=\Omega$ then part $(iv)$ follows.

Proposition 8

**. **Let the bilinear form $\varphi\colon V^{2n}\times V^{2n}\to W^{p,p}$ be symmetric and satisfy the condition (7). Then

[TABLE]

*Proof: *Since $\mathcal{T}|_{\varphi_{X}(V)}$ is a complex structure then $\kappa(\varphi)=2m$. Fix $X\in RE^{*}(\varphi)$ and let $\{X_{j},JX_{j}\}_{1\leq j\leq n}$ be a basis of $V^{2n}$ with $X_{1}=X$ such that

[TABLE]

and $X_{r},JX_{r}\in\ker\varphi_{X}$ for $r\geq m+1$. Since Proposition 4 yields ${\cal S}(\varphi|_{V\times\ker\varphi_{X}})\subset\varphi_{X}(V)$ then given $Z\in V^{2n}$ and $q\geq m+1$ there is $Y\in\mbox{span\,}\{X_{j},JX_{j},1\leq j\leq m\}$ such that

[TABLE]

Being $\varphi$ symmetric, we have

[TABLE]

for any $X,Y\in V^{2n}$. Hence

[TABLE]

and (8) follows.

2 The proofs

In this section we first give a general result in the theory of flat bilinear forms tailored for our purposes in this paper. After that we prove both results that have been stated in the Introduction.

Let $\alpha\colon V^{2n}\times V^{2n}\to U^{p}$ be a symmetric bilinear form and $J\in\text{Aut}(V)$ a complex structure. Then let $\gamma\colon V^{2n}\times V^{2n}\to W^{p,p}$ be the associated bilinear form defined by

[TABLE]

Then $\gamma$ is symmetric if and only if $\alpha$ is pluriharmonic with the latter meaning that

[TABLE]

If $\mathcal{T}\in\text{Aut}(W)$ is the complex structure given by (5) then

[TABLE]

and thus Proposition 7 applies to $\gamma$.

Let $\beta\colon V^{2n}\times V^{2n}\to W^{p,p}$ be the bilinear form defined by

[TABLE]

By (10) we have that

[TABLE]

and hence Proposition 7 applies to $\beta$. Then part $(iii)$ gives that $\nu(\beta)$ is even. We observe that $\nu(\beta)$ was called in [13] the index of pluriharmonic nullity since it satisfies

[TABLE]

Theorem 9

*. *** Let the bilinear forms $\gamma,\beta\colon V^{2n}\times V^{2n}\to W^{p,p}$, $p\leq n$, be flat and satisfy

[TABLE]

If $p\leq 11$ and $\nu(\gamma)<2n-\dim{\cal S}(\gamma)$ then the vector subspace ${\cal S}(\gamma)$ is degenerate. Moreover, if $\Omega=\pi_{1}({\cal S}(\gamma)\cap{\cal S}(\gamma)^{\perp})$ and $U^{p}=\Omega\oplus P$ is an orthogonal decomposition then the following holds:

• (i)

There is an isometric complex structure ${\cal J}\in\text{End}(\Omega)$ so that $\alpha_{\Omega}=\pi_{\Omega}\circ\alpha$ satisfies

[TABLE]

• (ii)

The bilinear form $\gamma_{P}=\pi_{P\times P}\circ\gamma$ is flat, the vector subspace ${\cal S}(\gamma_{P})$ is nondegenerate and $\nu(\gamma_{P})\geq 2n-\dim{\cal S}(\gamma_{P})$.

The proof of Theorem 9 will require several lemmas.

Lemma 10

*. *** Let the bilinear form $\gamma\colon V^{2n}\times V^{2n}\to W^{p,p}$ be symmetric and flat. If $p\leq 11$ and ${\cal S}(\gamma)=W^{p,p}$ then

[TABLE]

*Proof: *We argue for the most difficult case $p=11$ being the other cases similar but easier as $p$ decreases. The first inequality in (14) just means that

[TABLE]

Thus for what follows we fix $X\in RE(\gamma)$ and prove the latter. Proposition 5 yields

[TABLE]

where $\mathcal{U}^{\tau}(X)=\gamma_{X}(V)\cap\gamma_{X}(V)^{\perp}$, $\gamma_{X}(V)\subset\mathcal{U}^{\tau}(X)\oplus\mathcal{V}^{p-\tau,p-\tau}(X)$ and $\tau=\tau_{\gamma}(X)$ for simplicity. Thus $\kappa(\gamma)\leq 2p-\tau$. Then (4) yields $\kappa(\gamma)+\sigma_{\gamma}(X)\leq\kappa(\gamma)+\tau\leq 2p$. which gives the second inequality in (14).

The vector subspace $\mathcal{U}^{\tau}(X)$ is zero or is by Proposition 7 isotropic of even dimension. It follows from (4) that $0\leq\sigma\leq\tau\leq 10$ where $\sigma=\sigma_{\gamma}(X)$ for simplicity of notation.

If $\sigma=0$, that is, we have $N(X)=\mathcal{N}(\gamma)$ and then (14) is just (3). Hence we assume $\sigma>0$. Moreover, using first part $(iii)$ and then part $(ii)$ of Proposition 7 we obtain that $\sigma$ is even. Thus, henceforth we assume $\sigma\geq 2$.

In view of (4) there is a decomposition

[TABLE]

where $\hat{\mathcal{L}}^{\sigma}(X)\subset\hat{\mathcal{U}}^{\tau}(X)$ is such that the vector subspace $\mathcal{V}_{0}^{\tau-\sigma,\tau-\sigma}=(\mathcal{L}^{\sigma}(X)\oplus\hat{\mathcal{L}}^{\sigma}(X))^{\perp}$ is nondegenerate. We denote $\hat{\gamma}=\pi_{\hat{\mathcal{L}}^{\sigma}(X)}\circ\gamma$ and show that $\mathcal{T}\hat{\gamma}(Y,Z)=\hat{\gamma}(Y,JZ)$, that is, that

[TABLE]

Hence $\mathcal{T}|_{{\cal S}(\hat{\gamma})}$ is a complex structure and $\kappa_{0}=\kappa(\hat{\gamma})$ is even. Part $(iii)$ of Proposition 7 gives that $N(X)$ is $J$-invariant. If $(\xi,\bar{\xi})\in\mathcal{L}^{\sigma}(X)$ then part $(i)$ of Proposition 7 applied to $\varphi=\gamma|_{V\times N(X)}$ yields that $\mathcal{T}(\xi,\bar{\xi})\in\mathcal{L}^{\sigma}(X)$. Using (15) and (16) we have

[TABLE]

for any $(\xi,\bar{\xi})\in\mathcal{L}^{\sigma}(X)$ and this gives (17).

We have that

[TABLE]

In effect, it follows from (3) that the dimension of $N(Y)$ on $RE(\gamma)$ is constant. Then by continuity $\sigma\leq\sigma_{\varphi}(Y)$ in a neighborhood of $X$ in $RE(\gamma)$. On the other hand, we obtain from (4) that $\sigma_{\varphi}(Y)\leq\tau(\gamma)$ for any $Y\in RE^{\#}(\gamma)$ which is open and dense in $V^{2n}$. Then (18) follows.

Claim. Given $Z\in V^{2n}$ then $\dim\gamma_{Z}(N(X))$ is even and

[TABLE]

That $\dim\gamma_{Z}(N(X))$ is even follows from parts $(ii)$ and $(iii)$ of Proposition 7 whereas (18) yields the second inequality in (19).

To prove the first inequality in (19) it suffices to argue for $Z\in RE^{\#}(\gamma)\cap RE(\hat{\gamma})$ since this subset of $V^{2n}$ is open and dense. Let $V_{0}\subset V^{2n}$ be the vector subspace $V_{0}=\gamma_{Z}^{-1}(\mathcal{L}^{\sigma}(X))$ and $s_{0}=\dim\gamma_{Z}(V_{0})$. Since $N(X)\subset V_{0}$ by (4) then $r\leq s_{0}$ where $r=\dim\gamma_{Z}(N(X))$. Because $Z\in RE(\hat{\gamma})$ there is a vector subspace $V_{1}^{\kappa_{0}}\subset V^{2n}$ satisfying $\hat{\gamma}_{Z}(V_{1})=\hat{\gamma}_{Z}(V)$. Since any vector in $\gamma_{Z}(V_{1})$ has a nonzero $\hat{\mathcal{L}}^{\sigma}(X)$-component then

[TABLE]

Let $Y_{0}\in V_{0}$ satisfy $\gamma_{Z}Y_{0}\in\mathcal{U}^{\bar{\tau}}(Z)$ where $\bar{\tau}=\tau(\gamma)$ for simplicity of notation. Since $\gamma_{Z}(V_{0})\subset\mathcal{L}^{\sigma}(X)$ and $\hat{\gamma}_{Z}(V_{1})\subset\hat{\mathcal{L}}^{\sigma}(X)$ then using (16) we have

[TABLE]

Hence

[TABLE]

Let $Y_{1}\in V_{1}^{\kappa_{0}}$ satisfy $\gamma_{Z}Y_{1}\in\mathcal{U}^{\bar{\tau}}(Z)$. Since $\gamma_{Z}(V_{0})\subset\mathcal{L}^{\sigma}(X)$ and $\hat{\gamma}_{Z}(V_{1})\subset\hat{\mathcal{L}}^{\sigma}(X)$ then (16) gives

[TABLE]

Hence $\dim\pi_{\hat{\mathcal{L}}(X)}(\gamma_{Z}(V_{1})\cap\mathcal{U}^{\tilde{\tau}}(Z))\leq\sigma-s_{0}$. But since $V_{1}^{\kappa_{0}}$ has been chosen to satisfy that $\pi_{\hat{\mathcal{L}}(X)}|_{\gamma_{Z}(V_{1})}$ is injective, then

[TABLE]

The decomposition (15) for $Z$ yields $\gamma_{Z}(V)\subset\mathcal{U}^{\bar{\tau}}(Z)\oplus\mathcal{V}^{p-\bar{\tau},p-\bar{\tau}}(Z)$. Let the vector subspace $\mathcal{R}\subset\gamma_{Z}(V)$ be such that $\gamma_{Z}(V)=(\gamma_{Z}(V)\cap\mathcal{U}^{\bar{\tau}}(Z))\oplus\mathcal{R}$. Since any vector in $\mathcal{R}$ has a nonzero $\mathcal{V}^{p-\bar{\tau},p-\bar{\tau}}(Z)$-component then $\pi_{\mathcal{V}(Z)}|_{\mathcal{R}}$ is injective.

Set $\mathcal{S}=\pi_{\mathcal{V}(Z)}(\gamma_{Z}(V_{0})\cap\mathcal{R})$ and $\hat{\mathcal{S}}=\pi_{\mathcal{V}(Z)}(\gamma_{Z}(V_{1})\cap\mathcal{R})$. Since $\dim\gamma_{Z}(V_{0})=s_{0}$ and $\dim\gamma_{Z}(V_{1})=\kappa_{0}$, it follows from (21) and (22) that $\dim\mathcal{S},\dim\hat{\mathcal{S}}\geq\kappa_{0}-\sigma+s_{0}$. Let $\delta\in\mathcal{S}\cap\hat{\mathcal{S}}$. Then $\delta=\pi_{\mathcal{V}(Z)}(\gamma_{Z}Y_{i})$ where $Y_{i}\in V_{i}$ and $\gamma_{Z}Y_{i}\in\mathcal{R}$, $i=0,1$. By (20) and the injectivity of $\pi_{\mathcal{V}(Z)}|_{\mathcal{R}}$ we have that $\gamma_{Z}Y_{1}=\gamma_{Z}Y_{0}=0$. Thus $\delta=0$, and hence

[TABLE]

Since $r\leq s_{0}$, then that

[TABLE]

concludes the proof of the claim.

Since $\mathcal{S}(\gamma)=W^{p,p}$ it holds that

[TABLE]

From (23) and $\sigma\geq 2$ we have $\hat{\gamma}\neq 0$. Since $\alpha$ is pluriharmonic then $\gamma$ symmetric and hence also is $\hat{\gamma}$. Thus (8), (17) and (23) yield that

[TABLE]

Case $\kappa_{0}=\sigma$. This says that $\hat{\gamma}_{Z}(V)=\hat{\mathcal{L}}^{\sigma}(X)$ for any $Z\in RE(\hat{\gamma})$. Given $Z\in RE(\hat{\gamma})$ set $\gamma_{1}=\gamma_{Z}|_{N(X)}\colon N(X)\to\mathcal{L}^{\sigma}(X)$ and $N_{1}=\ker\gamma_{1}$. Then $\dim N_{1}\geq\dim N(X)-\sigma$. On one hand, if $\eta\in N(X)$ and $Y\in V^{2n}$ it follows from (16) that $\gamma_{Y}\eta=0$ if and only if ${\langle\!\langle}\gamma_{Y}\eta,\hat{\gamma}_{Z}(V){\rangle\!\rangle}=0$. On the other hand, from (15), (16) and the flatness of $\gamma$ we obtain

[TABLE]

for any $\eta\in N_{1}$ and $Y\in V^{2n}$. Thus $N_{1}=\mathcal{N}(\gamma)$. Now (3) yields

[TABLE]

and gives (14).

We have seen that $\kappa_{0}$ and $2\leq\sigma\leq 10$ are both even. Since $\kappa_{0}\leq\sigma$ and the case of equality has already been considered then we assume that $\kappa_{0}<\sigma$. Hence, in view of (24) it remains to consider the cases $(\kappa_{0},\sigma)=(4,6),(6,8),(6,10)$ and $(8,10)$.

Cases $(6,8)$ and $(8,10)$. By (17) the vector subspace $\hat{\gamma}_{R}(V)\cap\hat{\gamma}_{S}(V)$ is $\mathcal{T}|_{{\cal S}(\hat{\gamma})}$-invariant for any $R,S\in V^{2n}$ and thus of even dimension . Then by (23) there are $Z_{1},Z_{2}\in RE(\hat{\gamma})$ such that

[TABLE]

If $\eta\in N(X)$ and $Y\in V^{2n}$ it follows from (16) and (26) that $\gamma_{Y}\eta=0$ if and only if ${\langle\!\langle}\gamma_{Y}\eta,\hat{\gamma}_{Z_{j}}(V){\rangle\!\rangle}=0$ for $j=1,2$. Set $\gamma_{1}=\gamma_{Z_{1}}|_{N(X)}\colon N(X)\to\mathcal{L}^{\sigma}(X)$, $N_{1}=\ker\gamma_{1}$, $\gamma_{2}=\gamma_{Z_{2}}|_{N_{1}}\colon N_{1}\to\mathcal{L}^{\sigma}(X)$ and $N_{2}=\ker\gamma_{2}$. Then $N_{2}=\mathcal{N}(\gamma)$ since from (15), (16) and the flatness of $\gamma$ we have

[TABLE]

for any $\eta\in N_{2}$ and $Y\in V^{2n}$. From the Claim above $\dim\gamma_{Z_{j}}(N(X))\leq 4$, $j=1,2$, and

[TABLE]

as wished.

Case $(6,10)$. If we have $Z_{1},Z_{2}\in RE(\hat{\gamma})$ such that (26) holds then a similar argument as in the previous case gives (14). Otherwise, by (23) there are $Z_{1},Z_{2},Z_{3}\in RE(\hat{\gamma})$ such that

[TABLE]

and $\dim(\hat{\gamma}_{Z_{1}}(V)+\hat{\gamma}_{Z_{2}}(V))=8$. Set $\gamma_{1}=\gamma_{Z_{1}}|_{N(X)}\colon N(X)\to\mathcal{L}^{10}(X)$, $N_{1}=\ker\gamma_{1}$, $\gamma_{2}=\gamma_{Z_{2}}|_{N_{1}}\colon N_{1}\to\mathcal{L}^{10}(X)$, $N_{2}=\ker\gamma_{2}$, $\gamma_{3}=\gamma_{Z_{3}}|_{N_{2}}\colon N_{2}\to\mathcal{L}^{10}(X)$ and $N_{3}=\ker\gamma_{3}$. From (15) and (16) and the flatness of $\gamma$ we have

[TABLE]

for $\eta_{2}\in N_{2}$ and $Y\in V^{2n}$. Hence $\dim\gamma_{Z_{3}}(N_{2})\leq 2$. Moreover, as in the previous case we obtain $N_{3}=\mathcal{N}(\gamma)$. From the Claim above we have $\dim\gamma_{Z_{j}}(N(X))\leq 4$, $j=1,2$, and

[TABLE]

as wished.

Case $(4,6)$. Given $Z_{1}\in RE(\hat{\gamma})$ by (23) there is $Z_{2}\in RE(\hat{\gamma})$ such that (26) holds. Suppose that there is $Z_{1}\in RE(\hat{\gamma})$ such that $\dim\gamma_{Z_{1}}(N(X))\leq 4$. Since $\tau(\gamma)$ is even by (19) this always holds if $\tau(\gamma)>6$. Set $\gamma_{1}=\gamma_{Z_{1}}|_{N(X)}\colon N(X)\to\mathcal{L}^{6}(X)$ and $N_{1}=\ker\gamma_{1}$. From (15) and (16) and the flatness of $\gamma$ we obtain

[TABLE]

for any $\eta_{1}\in N_{1}$. Since $\kappa_{0}=4$ then $\dim\gamma_{Z_{2}}(N_{1})\leq 2$.

If $\eta\in N(X)$ and $Y\in V^{2n}$ it follows from (16) and (26) that $\gamma_{Y}\eta=0$ if and only if ${\langle\!\langle}\gamma_{Y}\eta,\hat{\gamma}_{Z_{j}}(V){\rangle\!\rangle}=0$ for $j=1,2$. Set $\gamma_{2}=\gamma_{Z_{2}}|_{N_{1}}\colon N_{1}\to\mathcal{L}^{6}(X)$ and $N_{2}=\ker\gamma_{2}$. As above we obtain that $N_{2}=\mathcal{N}(\gamma)$. Now (3) yields

[TABLE]

as wished.

By the above it remains to consider the case when $\tau(\gamma)=6$ and

[TABLE]

If $Y\in RE(\gamma)$ then $\sigma_{\gamma}(Y)\leq\tau(\gamma)=6$ by (18). Suppose that there is $Y\in RE(\gamma)$ such that $\sigma_{\gamma}(Y)\leq 4$. From (24) we are in case $\kappa_{0}=\sigma$ for $Y$ and thus $\nu(\gamma)\geq 2n-k(\gamma)-\sigma_{\gamma}(Y)$. Since $\sigma_{\gamma}(Y)<6=\sigma$ then (14) also holds for $X$.

In view of the above, we assume further that $\sigma_{\gamma}(Y)=6$ for any $Y\in RE(\gamma)$. Now let $Z_{1}\in RE(\gamma)\cap RE(\hat{\gamma})$ and then let $\tilde{\gamma}\colon V^{2n}\times V^{2n}\to\hat{\mathcal{L}}^{6}(Z_{1})$ stand for taking the $\hat{\mathcal{L}}^{6}(Z_{1})$-component of $\gamma$. Suppose that there is $Z_{2}\in RE(\tilde{\gamma})$ such that $\tilde{\gamma}_{Z_{2}}(V)=\hat{\mathcal{L}}^{6}(Z_{1})$. Under this assumption for $Z_{1}$ we are in the situation analyzed in the Case $\kappa_{0}=\sigma$ and thus (14) holds for $Z_{1}$. Since $\sigma_{\gamma}(Z_{1})=\sigma$ it also holds for $X$.

In view of (24) we now also assume that $\dim\tilde{\gamma}_{Z_{2}}(V)=4$ for any $Z_{2}\in RE(\tilde{\gamma})$. If $\dim\gamma_{Z_{2}}(N(Z_{1}))\leq 4$ for some $Z_{2}\in RE(\tilde{\gamma})$ then the initial part of the proof of this case gives that (14) holds for $Z_{1}$ and then also for $X$ since $\sigma_{\gamma}(Z_{1})=\sigma$. Hence we assume that $\gamma_{Z_{2}}(N(Z_{1}))=\mathcal{L}(Z_{1})$ for any $Z_{2}\in RE(\tilde{\gamma})$.

The remaining case to consider is when there are $Z_{1},Z_{2}\in RE(\gamma)\cap RE(\hat{\gamma})$ and $Z_{2}\in RE(\tilde{\gamma})$ for which (26) holds, $\sigma_{\gamma}(Z_{j})=6$, $j=1,2$, $\gamma_{Z_{2}}(N(Z_{1}))=\mathcal{L}(Z_{1})$ and $\dim\tilde{\gamma}_{Z_{2}}(V)=4$. To conclude the proof we show that this situation is not possible. Hence suppose otherwise. In particular, we have $\mathcal{L}(Z_{1})\subset\gamma_{Z_{2}}(V)$. From (27) we obtain that $\mathcal{L}(X)\subset\gamma_{Z_{j}}(V)$, $j=1,2$. Thus given $\eta_{0}\in\mathcal{L}(X)$ there are $Y_{1},Y_{2}\in V^{2n}$ such that $\eta_{0}=\gamma_{Z_{1}}Y_{1}=\gamma_{Z_{2}}Y_{2}$. Let $\xi_{0}\in\mathcal{L}(X)$ and $\xi_{j}\in\mathcal{L}(Z_{j})$, $j=1,2$. Then

[TABLE]

If $\eta_{1}\in\mathcal{L}(Z_{1})$ then ${\langle\!\langle}\xi_{0},\eta_{1}{\rangle\!\rangle}=0$ since $\mathcal{L}(X)\subset\gamma_{Z_{1}}(V)$ and $\mathcal{L}(Z_{1})\subset\mathcal{U}(Z_{1})$. Let $Y_{3}\in V^{2n}$ be such that $\eta_{1}=\gamma_{Z_{2}}Y_{3}$. Since $\mathcal{L}(Z_{2})\subset\mathcal{U}(Z_{2})$ then

[TABLE]

If $\eta_{2}\in\mathcal{L}(Z_{2})$ then ${\langle\!\langle}\xi_{j},\eta_{2}{\rangle\!\rangle}=0$, $j=0,1$, since $\mathcal{L}(X)\subset\gamma_{Z_{2}}(V)$, $\mathcal{L}(Z_{1})\subset\gamma_{Z_{2}}(V)$ and $\mathcal{L}(Z_{2})\subset\mathcal{U}(Z_{2})$. Thus ${\langle\!\langle}\xi_{0}+\xi_{1}+\xi_{2},\eta_{2}{\rangle\!\rangle}=0$. Hence $\mathcal{L}(X)+\mathcal{L}(Z_{1})+\mathcal{L}(Z_{2})$ is an isotropic vector subspace.

We argue that $\dim\mathcal{L}(X)\cap\mathcal{L}(Z_{j})=\dim\mathcal{L}(Z_{1})\cap\mathcal{L}(Z_{2})=2$. On one hand, we have that $\mathcal{L}(X)\cap\mathcal{L}(Z_{j})\neq 0$ since otherwise the vector subspace $\mathcal{L}(X)\oplus\mathcal{L}(Z_{j})$ would be isotropic of dimension $12$ which is not possible. On the other hand, we have

[TABLE]

for any $\xi\in\mathcal{L}(X)\cap\mathcal{L}(Z_{j})$. Since $\kappa_{0}=4$ it follows from part $(ii)$ of Proposition 7 that $\dim\mathcal{L}(X)\cap\mathcal{L}(Z_{j})=2$. Having that $\mathcal{L}(Z_{1})\oplus\mathcal{L}(Z_{2})\subset\gamma_{Z_{2}}(V)$ is isotropic yields $\mathcal{L}(Z_{1})\cap\mathcal{L}(Z_{2})\neq 0$. If $\xi\in\mathcal{L}(Z_{1})\cap\mathcal{L}(Z_{2})$ then

[TABLE]

where the second equality follows from $\mathcal{L}(Z_{2})\subset\mathcal{U}(Z_{2})$. Since $\dim\tilde{\gamma}_{Z_{2}}(V)=4$ then $\mathcal{L}(Z_{1})\cap\mathcal{L}(Z_{2})=2$. We have shown that $\mathcal{L}(X)+\mathcal{L}(Z_{1})+\mathcal{L}(Z_{2})$ has dimension $12$, but this is a contradiction.

Remark 11

**. **The estimate $\nu(\gamma)\geq 2n-2p$ is Proposition $10$ in [2]. A counterexample constructed in [2] shows that this estimate is false already for $p=12$.

Henceforward $U^{p}=U_{1}^{s}\oplus U_{2}^{p-s}$ is an orthogonal decomposition where

[TABLE]

Lemma 12

*. *** If (13) holds then

[TABLE]

and $\mathcal{N}(\beta)=\mathcal{N}(\gamma_{U_{1}})$ where $\gamma_{U_{1}}=\pi_{U_{1}\times U_{1}}\circ\gamma$.

*Proof: *We have that

[TABLE]

Thus if $(\xi,\eta)=\sum_{k}\beta(X_{k},Y_{k})$ then

[TABLE]

Hence if $(\xi,\eta)\in{\cal S}(\beta)$ then $(\xi,0),(0,\xi),(\eta,0)\in{\cal S}(\beta)$ and thus $\mathcal{S}(\beta)\subset U_{1}\oplus U_{1}$. On the other hand, if $(\xi,\eta)\in U_{1}\oplus U_{1}$ there are $\bar{\xi},\bar{\eta}\in U^{p}$ so that $(\xi,\bar{\xi}),(\eta,\bar{\eta})\in{\cal S}(\beta)$ and thus $(\xi,\eta)\in{\cal S}(\beta)$ which proves (28).

From (13), (28) and $(U_{2}\oplus U_{2})^{\perp}=U_{1}\oplus U_{1}$ we obtain $\mathcal{S}(\gamma|_{V\times\mathcal{N}(\beta)})\subset U_{2}\oplus U_{2}$. Then ${\langle\!\langle}\gamma(X,Y),(\xi,\bar{\xi}){\rangle\!\rangle}=0$ if $X\in V^{2n}$, $Y\in\mathcal{N}(\beta)$ and $\xi,\bar{\xi}\in U_{1}^{s}$. Thus $\mathcal{N}(\beta)\subset\mathcal{N}(\gamma_{U_{1}})$. On the other hand, if $S\in\mathcal{N}(\gamma_{U_{1}})$ then $\beta(X,S)=\gamma_{U_{2}}(X,S)+\gamma_{U_{2}}(JX,JS)$=0 by (28) for any $X\in V^{2n}$.

Lemma 13

*. *** Let the bilinear form $\beta\colon V^{2n}\times V^{2n}\to W^{p,p}$, $p\leq n$, be flat. Then

[TABLE]

Moreover, if $\kappa(\beta)=2p$ there is a basis $\{X_{i},JX_{i}\}_{1\leq i\leq n}$ of $V^{2n}$ such that:

• (i)

$\mathcal{N}(\beta)=\mbox{span\,}\{X_{j},JX_{j},\;p+1\leq j\leq n\}$.

• (ii)

$\beta(Y_{i},Y_{j})=0\;\;\mbox{if}\;\;i\neq j$* and $Y_{k}\in\mbox{span\,}\{X_{k},JX_{k}\}\;\;\mbox{for}\;\;k=i,j$.*

• (iii)

$\{\beta(X_{j},X_{j}),\beta(X_{j},JX_{j})\}_{1\leq j\leq p}$* is an orthonormal basis of $W^{p,p}$.*

*Proof: *Proposition $7$ in [2] gives (29). The remaining of the statement is Proposition $2.6$ in [4] as well as Lemma $7$ in [14].

Let $\theta\colon V^{2n}\times V^{2n}\to W^{p,p}$ be the pluriharmonic symmetric bilinear form defined as

[TABLE]

It follows from (10) that

[TABLE]

If the condition (13) holds then also

[TABLE]

In fact, we have using (10) and (12) that

[TABLE]

If $\gamma$ is flat then also is $\theta$. In effect, using (10) we have

[TABLE]

Since $2\gamma=\beta+\theta$ then $\mathcal{N}(\beta)\cap\mathcal{N}(\theta)\subset\mathcal{N}(\gamma)$ whereas the opposite inclusion follows from (11), (30) and that $\mathcal{N}(\gamma)$ is $J$-invariant. Therefore

[TABLE]

and, in particular, we have that $\nu(\beta)\geq\nu(\gamma)$.

Lemma 14

*. *** Let $\gamma,\beta\colon V^{2n}\times V^{2n}\to W^{p,p}$, $p\leq n$, be flat and satisfy the condition (13). If $\nu(\beta)=2n-2s$ then the bilinear forms $\theta_{j}=\pi_{U_{j}\times U_{j}}\circ\theta$, $j=1,2$, are flat.

*Proof: *By (28) and Lemma 13 there is a basis $\{X_{j},JX_{j}\}_{1\leq j\leq n}$ of $V^{2n}$ satisfying that $\mathcal{N}(\beta)=\mbox{span\,}\{X_{j},JX_{j},\;s+1\leq j\leq n\}$, that

[TABLE]

and that $\{\beta(X_{j},X_{j}),\beta(X_{j},JX_{j})\}_{1\leq j\leq s}$ is an orthonormal basis of $U_{1}^{s}\oplus U_{1}^{s}$.

From (28) we have that (32) is equivalent to

[TABLE]

In particular, we obtain using (34) that

[TABLE]

Moreover, since $\theta_{1}$ is symmetric it follows from (34) and (35) for $k\neq\ell$ that

[TABLE]

[TABLE]

and thus

[TABLE]

We have that

[TABLE]

for any $X,Y\in V^{2n}$. Then

[TABLE]

and hence

[TABLE]

We have shown that $\theta_{1}$ and $\theta$ are flat. Since $\theta=\theta_{1}\oplus\theta_{2}$ then also $\theta_{2}$ is flat.

Lemma 15

**. **Let $\gamma,\beta\colon V^{2n}\times V^{2n}\to W^{p,p}$, $p\leq n$, be flat and satisfy the condition (13). If ${\cal S}(\gamma)=W^{p,p}$ and $p-s\leq 9$ the flat bilinear form $\varphi=\theta|_{V\times\mathcal{N}(\beta)}\colon V^{2n}\times\mathcal{N}(\beta)\to W^{p,p}$ satisfies

[TABLE]

*Proof: *It follows from (28) and (32) that ${\cal S}(\varphi)\subset U_{2}^{p-s}\oplus U_{2}^{p-s}\subset W^{p,p}$. Thus $\varphi$ is seen in the sequel as a map

[TABLE]

To obtain the proof it suffices to show for any $X\in RE(\varphi)$ that we have

[TABLE]

Fix $X\in RE(\varphi)$ and set $\sigma=\sigma_{\varphi}(X)$ for simplicity. Proposition 5 gives a decomposition

[TABLE]

where $\mathcal{U}^{\tau}(X)=\varphi_{X}(\mathcal{N}(\beta))\cap\varphi_{X}(\mathcal{N}(\beta))^{\perp}$ with $\tau=\tau_{\varphi}(X)$ for simplicity of notation and $\varphi_{X}(\mathcal{N}(\beta))\subset\mathcal{U}^{\tau}(X)\oplus\mathcal{V}^{p-s-\tau,p-s-\tau}$. Thus $\kappa(\varphi)\leq 2p-2s-\tau$. From (4) we obtain that $\sigma(\varphi)\leq\sigma\leq\tau$ and the last inequality in (38) follows.

We have that $0\leq\sigma\leq\tau\leq p-s\leq 9$. From (31) and the $J$-invariance of $\mathcal{N}(\beta)$ we obtain that $\mathcal{T}\varphi(Y,Z)=\varphi(Y,JZ)$ for any $Y\in V^{2n}$ and $Z\in\mathcal{N}(\beta)$. Thus part $(iii)$ of Proposition 7 gives that $N(X)=\ker\varphi_{X}$ is $J$-invariant and hence part $(ii)$ that $\sigma$ is even. Therefore, we have that $0\leq\sigma\leq 8$.

Case $\sigma=0$. Since $\theta(Y,N(X))=0$ for any $Y\in V^{2n}$ then $N(X)\subset\mathcal{N}(\beta)\cap\mathcal{N}(\theta)=\mathcal{N}(\gamma)$ from (33). On the other hand, it is a general fact that

[TABLE]

and since $\sigma=0$ then (38) follows from (40).

Henceforward, we assume that $\sigma\geq 2$. As in (16) we have a decomposition

[TABLE]

We claim that the symmetric bilinear form defined by $\hat{\theta}=\pi_{\hat{\mathcal{L}}^{\sigma}(X)}\circ\theta$ satisfies

[TABLE]

In fact, if otherwise there is $0\neq\eta=\sum_{j=1}^{r}\varphi(Y_{j},S_{j})\in\mathcal{L}^{\sigma}(X)$ with $Y_{1},\ldots,Y_{r}\in V^{2n}$ and $S_{1},\ldots,S_{r}\in N(X)$ such that

[TABLE]

for any $Z,T\in V^{2n}$. Since $2\gamma=\beta+\theta$ we obtain from (32) and (43) that

[TABLE]

for any $Z,T\in V^{2n}$. Since $0\neq\eta=2\sum_{j=1}^{r}\gamma(Y_{j},S_{j})$ this is a contradiction by the assumption for ${\cal S}(\gamma)$ that proves the claim.

Part $(i)$ of Proposition 7 gives that $\mathcal{T}|_{\mathcal{L}^{\sigma}(X)}\in\text{Aut}(\mathcal{L}^{\sigma}(X))$ is a complex structure. Then from (39) and (41) we obtain for any $(\xi,\bar{\xi})\in\mathcal{L}^{\sigma}(X)$ that

[TABLE]

which says that

[TABLE]

for any $Z,Y\in V^{2n}$. Hence if $Z\in RE(\hat{\theta})$ then $\kappa_{0}=\kappa(\hat{\theta})$ is even. Being $\theta$ symmetric then also is $\hat{\theta}$ and it follows from (8) and (42) that

[TABLE]

Therefore, if $\sigma=2,4$ then $\sigma=\kappa_{0}$ and if $\sigma=6,8$ then either $\sigma=\kappa_{0}$ or $\sigma=\kappa_{0}+2$.

Case $\sigma=\kappa_{0}$. Given $Z\in RE(\hat{\theta})$ we have $\hat{\theta}_{Z}(V)=\hat{\mathcal{L}}^{\sigma}(X)$. Since $\varphi(Y,\eta)\in\mathcal{L}^{\sigma}(X)$ if $Y\in V^{2n}$ and $\eta\in N(X)$, then

[TABLE]

for any $T\in V^{2n}$. Set $\theta_{1}=\theta_{Z}|_{N(X)}\colon N(X)\to\mathcal{L}^{\sigma}(X)$ and $N_{1}=\ker\theta_{1}$. From (39), (41) and the flatness of $\theta$ we obtain

[TABLE]

for any $\delta\in N_{1}$ and $Y,T\in V^{2n}$. Hence from (46) we have $N_{1}\subset\mathcal{N}(\beta)\cap\mathcal{N}(\theta)=\mathcal{N}(\gamma)$. Then (33) and (40) give

[TABLE]

and (38) follows.

Case $\sigma=\kappa_{0}+2$. Suppose that there is $Z\in RE(\varphi)$ such that $\mathcal{L}(Z)={\cal S}(\varphi|_{V\times N(Z)})$ satisfies $\sigma_{\varphi}(Z)\leq 4$. Then by (45) for such $Z\in RE(\varphi)$ we are in Case $\sigma=\kappa_{0}$. Hence

[TABLE]

and since $\sigma\geq 6$ then (38) holds. Thus henceforward we assume that $\sigma_{\varphi}(Z)\geq 6$ for any $Z\in RE(\varphi)$.

If $Z_{1},Z_{2}\in RE(\hat{\theta})$ then (44) gives that $\hat{\theta}_{Z_{1}}(V)\cap\hat{\theta}_{Z_{2}}(V)$ is $\mathcal{T}$-invariant and therefore of even dimension. Given $Z_{1}\in RE(\hat{\theta})$ then by (42) there is $Z_{2}\in RE(\hat{\theta})$ such that we have $\hat{\mathcal{L}}^{\sigma}(X)=\hat{\theta}_{Z_{1}}(V)+\hat{\theta}_{Z_{2}}(V)$.

Since $\varphi(Y,\eta)\in\mathcal{L}^{\sigma}(X)$ if $\eta\in N(X)$ and $Y\in V^{2n}$, then

[TABLE]

for any $T\in V^{2n}$. If $\theta_{1}=\theta_{Z_{1}}|_{N(X)}\colon N(X)\to\mathcal{L}^{\sigma}(X)$ and $N_{1}=\ker\theta_{1}$, then

[TABLE]

If $\theta_{2}=\theta_{Z_{2}}|_{N_{1}}\colon N_{1}\to\mathcal{L}^{\sigma}(X)$ and $N_{2}=\ker\theta_{2}$ we have from (39) and (41) that

[TABLE]

for any $\delta_{1}\in N_{1}$ and $Y\in V^{2n}$. Thus $\dim\theta_{2}(N_{1})\leq\sigma-\kappa_{0}=2$ and hence

[TABLE]

It follows from (39) and (41) that

[TABLE]

for any $\delta_{2}\in N_{2}$, $Y,T\in V^{2n}$ and $j=1,2$. Then from (47) we have $N_{2}\subset\mathcal{N}(\beta)\cap\mathcal{N}(\theta)$. Hence using (33), (40), (48) and (49) we obtain

[TABLE]

Since $\sigma=\kappa_{0}+2$ then in order from the above to have (38) it is necessary to show that there is $Z\in RE(\hat{\theta})$ such that $\dim\theta_{Z}(N(X))\leq\kappa_{0}$. Arguing by contradiction, assume that $\dim\theta_{Z}(N(X))>\kappa_{0}$ for any $Z\in RE(\hat{\theta})$. Since $N(X)$ is $J$-invariant then $\theta_{Z}(N(X))$ has even dimension and thus the assumption means that $\theta_{Z}(N(X))=\mathcal{L}^{\sigma}(X)$ for any $Z\in RE(\hat{\theta})$. Let us take $Z\in RE(\varphi)\cap RE(\theta)\cap RE(\hat{\theta})$. Then $N(Z)=\ker\varphi_{Z}$ satisfies $N(Z)\subset\ker\theta_{Z}$. From (4) and (37) we have

[TABLE]

and from (39) and (41) that there is a decomposition

[TABLE]

Thus if $\delta\in{\cal R}$ then $\delta=\delta_{1}+\delta_{2}+\delta_{3}$ where $\delta_{1}\in\mathcal{L}^{\sigma}(X)$, $\delta_{2}\in\hat{\mathcal{L}}^{\sigma}(X)$ and $\delta_{3}\in\mathcal{V}_{0}\oplus\mathcal{V}$. Since $\delta\in\theta_{Z}(V)^{\perp}$ and $\theta_{Z}(N(X))=\mathcal{L}^{\sigma}(X)$ then

[TABLE]

and hence $\delta_{2}=0$. Thus if $\delta^{1},\delta^{2}\in{\cal R}$ we have

[TABLE]

Hence if $\pi\colon{\cal R}\to\mathcal{V}_{0}\oplus\mathcal{V}$ is defined by $\pi(\delta)=\delta_{3}$, then the vector subspace $\pi({\cal R})\subset\mathcal{V}_{0}\oplus\mathcal{V}$ is isotropic and thus $\dim\pi({\cal R})\leq p-s-\sigma$. Since $p-s\leq 9$ by assumption and $\sigma\geq 6$ then $\dim\pi({\cal R})\leq 3$. From (50) we have $\dim{\cal R}\geq\sigma_{\varphi}(Z)$, and hence $\dim\ker\pi\geq\sigma_{\varphi}(Z)-3\geq 3$. On the other hand, we have that $\ker\pi\subset\mathcal{L}^{\sigma}(X)\cap{\cal R}$. Since ${\cal R}\subset\theta_{Z}(V)^{\perp}$, we obtain from (39) and (41) that

[TABLE]

for any $\zeta\in\ker\pi$ and $Y\in V^{2n}$. Therefore, that $\ker\pi\subset\mathcal{L}^{\sigma}(X)$, $\hat{\theta}_{Z}(V)\subset\hat{\mathcal{L}}^{\sigma}(X)$ and that $\dim\hat{\theta}_{Z}(V)=\kappa_{0}=\sigma-2$ yield $\dim\ker\pi\leq 2$, and we reached a contradiction.

Lemma 16

*. *** Let $\gamma,\beta\colon V^{2n}\times V^{2n}\to W^{p,p}$, $p\leq n$, be flat and satisfy the condition (13). If $p\leq 11$ and ${\cal S}(\gamma)=W^{p,p}$ then $\nu(\gamma)\geq 2n-2p$.

*Proof: *First assume that $s\geq 2$ in which case (36) holds. Let $\varphi\colon V\times\mathcal{N}(\beta)\to U_{2}^{p-s}\oplus U_{2}^{p-s}$ be given by (37) and let $X\in RE(\varphi)$ satisfy $\sigma_{\varphi}(X)=\sigma(\varphi)$. By (39) we have that

[TABLE]

with $\varphi_{X}(\mathcal{N}(\beta))\subset\mathcal{U}^{\tau}(X)\oplus\mathcal{V}^{p-s-\tau,p-s-\tau}$. Then $\kappa(\varphi)\leq 2p-2s-\tau$. On the other hand, it follows from (28) and (29) that $\nu(\beta)\geq 2n-2s$ whereas by (36) we have that $\nu(\beta)-\nu(\gamma)\leq\kappa(\varphi)+\sigma(\varphi)$. Hence

[TABLE]

Since $\tau\geq\sigma(\varphi)$ by (4) then $\nu(\gamma)\geq 2n-2p$ as we wished.

If $s=0$ then $\beta=0$, that is, $\alpha$ is pluriharmonic and then Lemma 10 gives the result. Thus it remains to consider the case $s=1$ and thus $\beta\neq 0$. Part $(ii)$ of Proposition 7 yields that the vector space $\beta_{X}(V)$ is even dimensional. Thus $\kappa(\beta)=2$ and then (29) gives that $\nu(\beta)=2n-2$. Then by Lemma 14 we have that $\theta=\theta_{1}+\theta_{2}$ where the bilinear form $\theta_{2}\colon V^{2n}\times V^{2n}\to U_{2}^{p-1}\oplus U_{2}^{p-1}$ is flat.

We claim that $\theta_{2}$ satisfies the assumptions of Lemma 10. From (28) and $2\gamma=\beta+\theta$ we have

[TABLE]

Since ${\cal S}(\gamma)=W^{p,p}$ by assumption then ${\cal S}(\theta_{2})=U_{2}^{p-1}\oplus U_{2}^{p-1}$. If follows from (28) that the symmetric bilinear form $\alpha_{U_{2}}$ is pluriharmonic and hence also is $\theta_{2}$.

By Lemma 10 we have $\nu(\theta_{2})\geq 2n-2p+2$. Then (30) yields $\mathcal{N}(\gamma_{U_{1}})\subset\mathcal{N}(\theta_{1})$ whereas Lemma 12 that $\mathcal{N}(\beta)\subset\mathcal{N}(\theta_{1})$. Then (33) gives

[TABLE]

Hence, we have

[TABLE]

and since $\nu(\beta)=2n-2$, we conclude that $\nu(\gamma)\geq 2n-2p$.

Remark 17

**. **The estimate given by Lemma 16 is sharp. For instance, if we take as $\alpha$ in (9) and (11) the second fundamental form of a product of real Kaehler hypersurfaces then the hypotheses of the lemma are satisfied and we have equality in the estimate.

Proof of Theorem 9: By Lemma 16 the vector subspace ${\cal S}(\gamma)$ is degenerate since, if otherwise, then by (10) it is of the form ${\cal S}(\gamma)=W_{1}^{q,q}\subset W^{p,p}$ and then Lemma 16 yields a contradiction. Parts $(ii)$ and $(iv)$ of Proposition 7 give, respectively, that $\dim\Omega\geq 2$ and that $\gamma_{\Omega}=\pi_{\Omega\oplus\Omega}\circ\gamma$ satisfies ${\cal S}(\gamma_{\Omega})={\cal S}(\gamma)\cap{\cal S}(\gamma)^{\perp}$. Hence

[TABLE]

for any $X,Y,Z,T\in V^{2n}$. Thus the complex structure ${\cal J}\in\text{End}({\cal S}(\alpha_{\Omega}))$ defined by ${\cal J}\alpha_{\Omega}(X,Y)=\alpha_{\Omega}(X,JY)$ is an isometry. Part $(iv)$ of Proposition 7 gives that ${\cal S}(\alpha_{\Omega})=\Omega$ and hence ${\cal J}\in\text{Aut}(\Omega)$ is a complex structure. In particular, we have that $\alpha_{\Omega}$ is pluriharmonic. Then ${\cal S}(\beta)\subset P\oplus P$ and hence $\beta=\beta_{P}=\pi_{P\times P}\circ\beta$. Thus if $\gamma_{P}=\pi_{P\times P}\circ\gamma$ and $\gamma=\gamma_{\Omega}\oplus\gamma_{P}$ then

[TABLE]

Hence $\gamma_{P}$ and $\beta_{P}$ satisfy the condition (13). Since $\gamma$ is flat and the bilinear form $\gamma_{\Omega}$ is null then also $\gamma_{P}$ is flat. Then by (10) we have that ${\cal S}(\gamma_{P})=W^{q_{1},q_{1}}$ and the remaining of the proof follows from Lemma 16.

We now prove the results stated in the introduction.

Proof of Theorem 1: Let the bilinear forms $\gamma,\beta\colon T_{x_{0}}M\times T_{x_{0}}M\to N_{1}^{f}(x_{0})\oplus N_{1}^{f}(x_{0})$ be defined by (9) and (11) in terms of the second fundamental form $\alpha$ of $f$ at $x_{0}\in M^{2n}$. We endow $N_{1}^{f}(x_{0})\oplus N_{1}^{f}(x_{0})$ with the inner product defined by

[TABLE]

We claim that $\gamma$ and $\beta$ are flat and that (13) holds. For a Kaehler manifold it is a standard fact that the curvature tensor $x\in M^{2n}$ satisfies $R(X,Y)JZ=JR(X,Y)Z$ for any $X,Y,Z\in T_{x}M$. From this and the Gauss equation for $f$ we obtain

[TABLE]

Since $\gamma$ satisfies (10) then using (5) we have

[TABLE]

Using (13) and since $\beta(JX,Y)=-\beta(X,JY)$ from (11) then

[TABLE]

Then by (11) we have

[TABLE]

and the claim has been proved. The proof now follows from Theorem 9 since we have that $\Delta_{c}(x_{0})=\mathcal{N}(\gamma)$ and $Q(x_{0})=\Omega$.

Remark 18

**. **The proof of Theorem 1 in [1] makes use of Theorem $1$ in [6] but that result does not hold for $p=6$ as was clarified in [7]. Nevertheless, it was also established in [7] that Theorem $1$ in [6] still holds for $p=6$ under a slightly stronger assumption which happens to be satisfied in our case of real Kaehler submanifolds.

Proof of Theorem 2: Theorem 1 and the Gauss equation for $f$ give

[TABLE]

for any $X\in\mathcal{N}(\alpha_{P})$.

Marcos Dajczer is partially supported by the grant PID2021-124157NB-I00 funded by MCIN/AEI/10.13039/501100011033/ “ERDF A way of making Europe”, Spain, and is also supported by Comunidad Autónoma de la Región de Murcia, Spain, within the framework of the Regional Programme in Promotion of the Scientific and Technical Research (Action Plan 2022), by Fundación Séneca, Regional Agency of Science and Technology, REF, 21899/PI/22.