A Conjecture of Trautman

TL;DR

This paper discusses Trautman's conjecture linking the local realizability of 3D CR manifolds to the properties of their canonical bundle, reviewing definitions, context, and prior partial results.

Contribution

It reviews the conjecture, provides relevant background, and outlines a previous partial proof, advancing understanding of the conjecture's validity.

Findings

Reviewed the conjecture and its mathematical context

Outlined an earlier partial proof of the conjecture

Connected the conjecture to physical and geometric concepts

Abstract

In 1998 the physicist Andre Trautman conjectured that a three-dimensional CR manifold is locally realizable if and only if its canonical bundle admits a closed nowhere zero section. We review the relevant definitions, give the physical context, and outline an earlier result which established a weak version of the Conjecture.

Equations

$$J:\mathbf{C}\otimes H\to \mathbf{C}\otimes H.$$

$$H=\{\Re Z:Z\in B\}$$

$$J(Z)=iZ\text{mbox}ifZ\in B$$

$$J(Z)=-iZ\text{mbox}ifZ\in \overline{B}.$$

$$JX=Y\text{mbox}andJY=-X.$$

$$B=T^{1,0}({\mathbf{C}}^2)\cap \mathbf{C}\otimes TM$$

$$\bigg{\{}\frac{\partial}{\partial z_{1}},\ \ \frac{\partial}{\partial z_{2}}\bigg{\}}$$

$$\Omega =\{\omega \in \mathbf{C}\otimes {\Lambda }^2(TM):i_b\omega =0\text{mbox}forallb\in \overline{B}\}.$$

$$(X+iJX)f_k=0$$

$$\displaystyle F:M\to{{\bf C}}^{2}\ \ \ \ \$$

$$x\to (\Re f_1,\Im f_1,\Re f_2,\Im f_2)$$

$$T^{1,0}({\mathbf{C}}^{n+1})\cap \mathbf{C}\otimes TM=B.$$

$$\Omega =\{\omega \in \mathbf{C}\otimes {\Lambda }^{n+1}(TM):i_v\omega =0,\forall v\in T^{0,1}\}.$$

$$d{f}_1\wedge \dots \wedge d{f}_{n+1}\ne 0.$$

$$d{f}_1\wedge \dots \wedge d{f}_{n+1}({\overline{L}}_1,\dots ,{\overline{L}}_n,T)\ne 0.$$

$$d{f}_1\wedge \dots \wedge f_n\wedge d{f}_{n+1}\wedge \overline{d{f}_1}\wedge \dots \wedge \overline{d{f}_n}\ne 0.$$

$$F=(f_1,\dots ,f_{n+1})$$

$$\Im z_{n+1}=f(z_1,\dots ,z_n,\Re z_{n+1})$$

$$L\in B\to [L,\overline{L}]mod\{B\oplus \overline{B}\}$$

$${\pounds }_v{T}^{1,0}=T^{1,0}$$

$$K_p^{\perp }=\{v\in T_{p}M:g(v,k)=0\}.$$

$$g(v_1,v_2)$$

$$g([n],[n])\equiv g(n,n)=0.$$

$$g(k,k)=0$$

$$g(k,n)=0.$$

$$N=\{n\in \mathbf{C}\otimes K^{\perp }:J[n]=-i[n]\}.$$

$$g(n_1,n_2)$$

$$g(n_1,\overline{n_1})=2g(\xi ,\xi )\ne 0.$$

$$N\subset \mathbf{C}\otimes K^{\perp }\subset \mathbf{C}\otimes TM$$

$$\begin{gathered} N\cap \overline{N}=\mathbf{C}\otimes K, \\ N\text{plus}\overline{N}=\mathbf{C}\otimes K^{\perp }. \end{gathered}$$

$${\pounds }_k{K}^{\perp }\subset K^{\perp }.$$

Full text

A Conjecture of Trautman

Howard Jacobowitz

Department of Mathematical Sciences, Rutgers University, Camden New Jersey

jacobowi@rutgers.edu

The present article is based on a talk by the author at the University of Basilicata in Fall 2018. The author would like to express his appreciation of the courtesy shown to him by Professors Barletta and Dragomir during his visit.

1.

In 1998 the physicist Andre Trautman conjectured that a three-dimensional CR manifold is locally realizable if and only if its canonical bundle admits a closed nowhere zero section. First we review the relevant definitions and in the next section give the physical context. In Section 3 we outline the earlier results in [2] which had proved a weak version of the Conjecture.

A CR structure on a three-dimensional manifold $M$ is a two-plane distribution $H\subset TM$ and a fiber preserving anti-involution $J:H\to H$. We denote this structure by $(M,H,J)$. It is often useful to extend $J$ by complex linearity to a map

[TABLE]

Then $J$ is completely determined by the eigenspace corresponding to the eigenvalue $i$ (or to the eigenvalue -$i$).

An equivalent definition of a CR structure on a three-dimensional manifold may be given in terms of a complex line bundle: A CR structures on $M$ is a line bundle $B\subset{{\bf{C}}\otimes TM}$ with the property that $B\cap\overline{B}$ contains only the zero section. Then

[TABLE]

is of rank 2 and $J$ is defined on ${{\bf C}}\otimes H=B\oplus\overline{B}$ by setting

[TABLE]

and

[TABLE]

So for $X-iY\in B$

[TABLE]

**Example ** Let $M^{3}\subset{{\bf C}}^{2}$ be a real hypersurface and let $J$ denote the usual operator on ${\bf R}^{4}$ giving the complex structure. Set $H_{p}=T_{p}M\cap JT_{p}M$ for each $p\in M$. Now $J$ acts on $H$ and (M,H,J) is a CR structure. Or, to use the alternative definition, just take

[TABLE]

where $T^{1,0}$ is the linear span of

[TABLE]

(and $T^{0,1}$ is the span of the conjugates).

So later we write $B=T^{1,0}(M)=T^{1,0}$ and write $T^{0,1}$ for $\overline{B}$.

The canonical bundle is another complex line bundle associated to a CR structure. It is a subbundle of the second exterior product. For a real hypersurface in ${{\bf C}}^{2}$ it is generated by the two-form $dz_{1}\wedge dz_{2}$ restricted to $M$. More generally, if the CR structure is given by a complex line bundle $B$ then

[TABLE]

The interior product $i_{b}\omega$ is given by $i_{b}\omega(X)=\omega(b\wedge X)$.

Definition**.**

(M,H,J) is realizable in a neighborhood of $p$ if there exist complex functions $f_{1}$ and $f_{2}$ such that

[TABLE]

for all $X\in H$ and

[TABLE]

is an embedding.

It follows upon identifying $M$ with its image $F(M)$ that the original structure $(M,H,J)$ coincides with the CR structure induced as in the Example.

We digress briefly to discuss higher-dimensional CR structures and return to this in Section 3.

Definition**.**

$(M^{2n+1},B)$* is a CR manifold if $B\subset{{\bf{C}}\otimes TM}$ is a vector subspace of rank $n$ with $B\cap\overline{B}=\{0\}$ and $[\Gamma B,\Gamma B]\subset\Gamma B$. I.e., the commutator of local sections of $B$ is always in $B$.*

More precisely, we have defined a CR manifold of hypersurface type.

Definition**.**

$(M^{2n+1},B)$* is realizable if there is an embedding $F:M\to{\bf C}^{n+1}$ with, after identifying $M$ with $F(M)$,*

[TABLE]

The canonical bundle is now a complex line bundle in the exterior product $\Lambda^{n+1}({{\bf C}}\otimes TM^{2n+1}).$ Namely,

Definition**.**

The canonical bundle is

[TABLE]

Definition**.**

A function $f:M\to{\bf C}$ is a CR function if $Lf=0$ for all $L\in T^{0,1}$.

Lemma 1.1**.**

$M^{2n+1}$* is realizable in ${{\bf C}}^{n+1}$ if there exist CR functions $f_{1},\ldots,f_{n+1}$ such that*

[TABLE]

Proof.

Let $L_{1},\ldots L_{n}$ be a basis for $T^{0,1}$ and let $T$ be any nonzero vector transverse to $H$. From (1.1) and using that the functions are CR, we have

[TABLE]

So $df_{j}T\neq 0$ for some $j$, say $j=n+1$, which now implies

[TABLE]

Thus

[TABLE]

is a local embedding. Indeed perhaps after multiplying $F$ by $i$, $F(M)$ has the form

[TABLE]

∎

The realizability problem is quite subtle. For instance, most three-dimensional $C^{\infty}$ CR structures are not locally realizable [4], [7].

Most realizability results in higher dimensions concern strictly pseudo-convex CR structures.

Definition**.**

A CR structure $(M,B)$ is strictly pseudo-convex if the quadratic form

[TABLE]

is definite.

Such structures are realizable if $dim\ M\geq 7$. See [1] and [5] for the original proofs and [11] for a variation.

Although, as we said, the general realizability problem is subtle there are two easy results.

Proposition 1**.**

Real analytic CR manifolds are locally realizable.

A proof can be found, for instance, in [3, page 22].

Proposition 2**.**

A CR manifold admiting a vector field $v$ transverse to $H$ and preserving the CR structure is locally realizable.

To preserve the CR structure means that the Lie derivative in the direction of $v$ satisfies

[TABLE]

A generalization of this result is important in Section 3 and will be proved there.

2.

We first wish to explain the observation of [8] that a shear-free congruence of null geodesics on a four-dimensional manifold induces a three-dimensional CR structure on a quotient manifold.

Let $M^{4}$ be a Lorentz manifold with metric $g$ and let $k$ be a null vector field, $g(k,k)=0$. Let $K$ be the real line bundle generated by $k$. Set

[TABLE]

Note that $K\subset K^{\perp}$ and that $K^{\perp}/K$ is an $R^{2}$ bundle on $M$. Following the notation in [10], let $n\in K^{\perp}$. Denote the equivalency class of $n$ in $K^{\perp}/K$ by $[n]$ and use the same notation for $n\in{{\bf C}}\otimes(K^{\perp}/K)={\bf C}\otimes K^{\perp}/{\bf C}\otimes K$.

Lemma 2.1**.**

The metric $g$ induces a well-defined positive definite inner product on $K^{\perp}/K$.

Proof.

Let $[n_{1}]$ and $[n_{2}]$ belong to the fiber of $K^{\perp}/K$ over some point of $M$. Define $g([n_{1}],[n_{2}])$ to be $g(n_{1},n_{2})$. If $v_{1}$ and $v_{2}$ are different choices then $v_{j}=n_{j}+a_{j}k$ and so

[TABLE]

since $k$ is a null vector and $n_{j}\in K^{\perp}$. This shows that $g$ is well-defined.

To see that $g$ is definite, assume that for some $[n]$ we have

[TABLE]

By the definitions of $k$ and $K^{\perp}$ we also have

[TABLE]

and

[TABLE]

So either $n$ is a multiple of $k$ or $g$ vanishes on a two-dimensional plane. The second alternative is not possible for a Lorentz metric. So $n=ak$ and thus $[n]=0$. Hence $g$ is definite, and since it arises from a Lorentz metric it is positive definite.

∎

Fix an orientation for $K^{\perp}/K$ (this is not a problem, as long as we care only about local results) and then let $J:K^{\perp}/K\to K^{\perp}/K$ be the operation of rotation by $\pi/2$ radians with respect to the induced metric and orientation. Finally, set

[TABLE]

Note that $N$ is a two-dimensional complex vector bundle on $M$. Extend the inner product $g$ to $N$ as a complex linear form. For $n_{1}=\xi+iJ\xi$ and $n_{2}=\eta+iJ\eta$ in $N$ we have

[TABLE]

since $J$ is rotation by $\pi/2$ radians. So $N$ is said to be totally null. On the other hand,

[TABLE]

We have

[TABLE]

and

[TABLE]

Now consider the flow generated by the vector field $k$. For small values of the time parameter, the orbit space is a three-dimensional manifold (again, for local results this is clear); call it $M^{\prime}$. Without additional assumptions on $k$ the bundle $N$ does not project to a well-defined subbundle of ${{\bf C}}\otimes TM^{\prime}$. Here is where physics enters.

We temporarily drop the assumption that $k$ is null.

Definition**.**

[8, page 1426]** The vector field $k$ is said to be conformally geodesic if the associated flow preserves $K^{\perp}$ and $g(k,k)$ does not change sign.

Note that this definition depends only on the conformal class of $g$ and also that in Riemannian geometry the condition on the flow and $g(k,k)=c$ imply $\nabla_{k}k=0$.

The flow condition may be rewritten as

[TABLE]

and is equivalent to

[TABLE]

where $g(k)$ is the one-form defined by $g(k)v=g(k,v)$. To see this equivalence, we first note that if $v$ is a vector field satisfying $g(k)v=0$ then also $k(g(k)v)=0$ and so

[TABLE]

We want to derive $g(k)\wedge\pounds_{k}g(k)=0.$ It is enough to show that

[TABLE]

That is, if $g(k)$ and $\pounds_{k}g(k)$ have the same kernel then these one-forms are linearly dependent. So assume $\pounds_{k}K^{\perp}\subset K^{\perp}$ and $g(k)v=0$. We now have

[TABLE]

where the last implication follows from (2.2).

On the other hand, if $g(k)\wedge\pounds_{k}g(k)=0$, then

[TABLE]

We are interested in the case where $k$ is a null vector, $g(k,k)=0$. When $k$ is null the foliation of $M$ by the integral curves of $k$ is called a congruence of null geodesics.

The Lorentz metric $g$ induces a degenerate inner product on $K^{\perp}$ and therefore also a (degenerate) conformal structure.

Definition**.**

A conformally geodesic vector field is shear-free if the associated flow preserves the conformal structure of $K^{\perp}$.

The physical hypothesis that $k$ generates a shear-free congruence of null geodesics also can be formulated in terms of the Lie derivative.

Theorem 2.1**.**

[9]** A vector field $k$ on a manifold $M^{4}$ with Lorentz metric $g$ generates a shear-free congruence of null geodesics if and only if

[TABLE]

where $\lambda$ is a function, $\phi$ is a one-form, $g(k)$ is as defined above, and $\phi\otimes g(k)$ signifies the symmetric product constructed from the one-forms.

Proof.

We first show that (2.4), together with (2.3), implies (2.1) and hence $k$ is a conformally geodesic vector field. We start with the Leibniz rule:

[TABLE]

Setting $u=k$ and rearranging this becomes

[TABLE]

Further

[TABLE]

and so

[TABLE]

Now

[TABLE]

We know that $g(k)\wedge\pounds_{k}g(k)=0$ implies that $K^{\perp}$ is preserved and so $k$ is conformally geodesic.

To show that the conformal class $g$ induces on $K^{\perp}$ is constant along the flow on $M$ induced by $k$, we let $v\in K^{\perp}$ be constant along the flow. So $g(k)v=0$ and $\pounds_{k}v=0$. Thus

[TABLE]

This gives us an ordinary differential equation. If local coordinates $(t,x)$ are introduced with $k=\partial_{t}$ then the equation has the form

[TABLE]

and the solutions are

[TABLE]

for some function $\Lambda$. Thus

[TABLE]

This shows that the conformal class of the metric on $K^{\perp}$ does not change under the flow.

Conversely,we want to show that if $k$ generates a shear-free congruence of null geodesics then there exist a scalar function $\lambda$ and a one-form $\phi$ satisfying (2.4). To see this, we start with a frame invariant along the orbits, labeled $e_{1},e_{2},e_{3},e_{4}$ with $\{e_{1},e_{2},e_{3}\}$ a basis for $K^{\perp}$ and $g(e_{4},e_{4})=0$. Let $0\leq i\leq 3,\ 0\leq j\leq 3$. Note that $g(k)e_{i}=0$ and $g(k)x_{4}\neq 0$. For $p\in M$ parametrize the orbit through $p$ by $t$. Since the conformal class of $g$ on $K^{\perp}$ is constant

[TABLE]

Thus

[TABLE]

Define

[TABLE]

We have for $0\leq i,j\leq 3$

[TABLE]

while for $i\leq 4$ we have

[TABLE]

Thus

[TABLE]

∎

Let $\pi$ denote the map of $M$ to the orbit space

[TABLE]

Lemma 2.2**.**

Under the conditions of the Theorem, $\pi_{*}(N)$ is a complex line bundle $\overline{B}\subset{{\bf C}}\otimes TM^{\prime}$ which satisfies $B\cap\overline{B}=\{0\}.$

Proof.

Since $K^{\perp}$ is itself invariant under the flow, $K^{\perp}/K$ projects to a well-defined two-plane distribution $H$ on $M^{\prime}$ and on $H$ we have a well-defined conformal class of metrics. Thus ${\bf C}\otimes TH$ splits into the eigenspaces of $J$

[TABLE]

with $\pi_{*}N=\overline{B}$. ∎

That is, the physical assumptions lead to a CR structure on the orbit space. Further, as we now show, the same conditions provide a two-form $F$ associated to $N$ which itself also passes down to $M^{\prime}$. The interest in such a two-form comes from considerations of Maxwell’s equations. In classical physics, the components of the magnetic and electrical fields can be used to construct a real two-form $F$, called the Faraday tensor. Then, in the absence of charge, Maxwell’s equations become $dF=0$. Naturally, in relativistic physics the situation is more complicated.

To define $F$ we first find a basis for $N$. Let $\xi\in K^{\perp}$ and $\xi\notin K$.Choose any $\eta\in K^{\perp}$ such that $J[\xi]=[\eta]$. Then $n=\xi+i\eta$ and $k$ form a basis for $N$.

Let $g(k)$, defined above, and $g(n)$, defined in the same way, be one-forms on $M$. Set

[TABLE]

Note that $F$ is nowhere zero since the one-forms $g(n)$ and $g(k)$ are independent. For example, $g(n)\overline{n}\neq 0$ while $g(k)\overline{n}=0$.

The two-form $F$ is associated to $N$ in the following sense:

Lemma 2.3**.**

$N=\{v\in{{{\bf{C}}\otimes TM}}\ :\ i_{v}F=0\}$.

Proof.

We have $g(k,k)=0$ because $k$ is null; $g(k,n)=0$ because $N\subset{\bf C}\otimes K^{\perp}$; and $g(n,n)=0$ because $N$ is totally null. So for our basis $i_{k}F=0$ and $i_{n}F=0$. Thus

[TABLE]

Now let $t\in\{v\in{{{\bf{C}}\otimes TM}}\ :\ i_{v}F=0\}$. So

[TABLE]

The independence of $g(n)$ and $g(k)$ implies $t\in{\bf C}\otimes K^{\perp}$ at some point of $M$. Thus

[TABLE]

for constants $\alpha,\beta,\gamma$. Since $g(n,t)=0$ and $g(n,\overline{n})\neq 0$, we see that $\beta=0$ and thus $t\in N$. ∎

We may use $F$ to define a two-form on $M^{\prime}$: Let $t_{1}$ and $t_{2}$ be vectors in ${\bf{C}\otimes TM^{\prime}}$. Lift $t_{j}$ to a vector $t_{j}+\alpha_{j}k$ in ${{\bf{C}}\otimes TM}$. Then

[TABLE]

So $F$ evaluated on the lift is independent of choices and gives a well-defined two-form on $M^{\prime}$. Call this form $F^{\prime}$. For $t\in\overline{B}={{\bf C}}\otimes T^{0,1}(M^{\prime})$ the natural lift, also called $t$ is in $N$. Thus from the Lemma

[TABLE]

Hence $F^{\prime}$ is section of the canonical bundle of $M^{\prime}$ and is nowhere zero.

In summary, the local quotient of a Lorentzian manifold under a shear-free congruence of null geodesics is a CR manifold which has a nowhere zero section of its canonical bundle. This section being closed is related to Maxwell’s equationa and so is a reasonable hypothesis for physicists. We now repeat Trautman’s conjecture.

Conjecture 2.1**.**

If a CR manifold $M^{3}$ admits a nowhere zero closed section of its canonical bundle, then the CR structure is locally realizable.

As we have seen, the converse is true even globally.

3.

A weak version of the conjecture is true and holds for all dimensions.

Functions satisfying

[TABLE]

are called independent. Functions satisfying

[TABLE]

are called strongly independent.

Example**.**

The hyperquadric $Q^{3}\subset{\bf C}^{2}$ is defined by $\Im z_{2}=\lvert z_{1}\rvert^{2}$. The bundle $T^{0,1}$ is generated by

[TABLE]

where $u=\Re z_{2}$. The CR function $f=z$ is strongly independent; The function $f=u+i\lvert z\rvert^{2}$ is independent, but not strongly independent (at the origin).

The following theorem preceded the formulation of Trautman’s Conjecture and establishes a weak form.

Theorem 3.1**.**

[2]** If the CR structure $M^{2n+1}$ has $n$ strongly independent CR functions near $p$ and if the canonical bundle has a closed nowhere zero section then $M^{2n+1}$ is realizable in a neighborhood of $p$.

The proof depends on the following complex version of Proposition 2

Proposition 3**.**

$M$* is realizable in a neighborhood of $p$ if and only if there exists a complex vector field $Y$ near $p$ such that*

• •

$Y$* is transverse to $T^{1,0}\oplus T^{0,1}$*

• •

$\pounds_{Y}T^{1,0}=T^{1,0}$.

Thus the existence of a real vector field such that $\pounds_{v}T^{1,0}=T^{1,0}$ is very special (since most realizable CR structures do not have such a vector field) but the existence of such a complex vector field characterizes realizability.

Proof.

We first prove the necessity. So assume $M$ is realizable near $p$. Without loss of generality we assume $p=0$ and $M$ is given as

[TABLE]

Define $\overline{Y}$ by

[TABLE]

Note that $\overline{Y}$ (and also $Y$) is transverse to $T^{1,0}\oplus T^{0,1}$. Set

[TABLE]

This is a nowhere zero closed section of the canonical bundle. As a consequence of Cartan’s formula

[TABLE]

we have

[TABLE]

This implies $\pounds_{\overline{Y}}T^{0,1}=T^{0,1}$ and so also

[TABLE]

Conversely, we will assume that $\pounds_{Y}T^{1,0}=T^{1,0}$ with $Y$ transverse to $T^{1,0}\oplus T^{0,1}$, and show that $M$ is locally realizable. This is just a slight modification of a standard proof of Proposition 2. Extend $Y$ and each of the vectors in $T^{1,0}$ to ${\bf C}\otimes T(M\times{\bf R})$ by taking them constant in the ${\bf R}$ direction. Let $Y$ still denote this extension and let $V$ denote the extension of the bundle $T^{1,0}$. Set $Z$ to be the complex line bundle spanned by $Y+i\frac{\partial}{\partial t}$ where $t$ is the natural parameter for ${\bf R}$. Then $W=V\oplus Z$ satisfies

[TABLE]

Finally, as is easily seen, $W$ is closed under the commutation of vector fields,

[TABLE]

Thus $W$ satisfies the conditions of the Newlander-Nirenberg Theorem [6] and so defines a complex structure on $M\times{\bf R}$. Since $W\cap{\bf C}\otimes TM=T^{1,0}(M\times R)$, the CR structure induced on $M$ is the one we started with. ∎

All that is left to do in the proof of Theorem 3.1 is to show that if $f_{1},\ldots,f_{n}$ are CR functions on $M^{n+1}$ with

[TABLE]

and if $\omega$ is a nowhere zero section of the canonical bundle with

[TABLE]

then there is a complex vector field $Y$ with

• •

$Y$ transverse to $T^{1,0}\oplus T^{0,1}$

• •

$\pounds_{Y}T^{1,0}=T^{1,0}$.

We just use the closed section to find a replacement for $dz_{n+1}$ in (3.1). Because we prefer to work with the canonical bundle and not its conjugate, we start, as in the Proposition, by defining a vector field $\zeta$ and then let $Y=\overline{\zeta}$. Towards this end, let $\theta$ be a nowhere zero one-form annihilating $T^{1,0}\oplus\overline{T^{1,0}}$. Then

[TABLE]

is a nowhere zero section of the canonical bundle. This bundle is one dimensional, so

[TABLE]

Define $\zeta$ by

[TABLE]

$\zeta$ can be thought of as a complex version of the Reeb vector field. In particular, it is transverse to $T^{1,0}\oplus T^{0,1}$.

We have

[TABLE]

Lemma 3.1**.**

If $\pounds_{\zeta}\omega=0$ then $\pounds_{\zeta}T^{0,1}=T^{0,1}$.

Proof.

We have for all vector fields $\zeta$ and $v$ and all forms $\omega$

[TABLE]

So, if $v\in T^{0,1}$, hence $i_{v}\omega=0$, and $\pounds_{\zeta}\omega=0$, then

[TABLE]

and so $\pounds_{\zeta}v$ is also in $T^{0,1}$. ∎

This Lemma has a partial converse: If $\pounds_{\zeta}T^{0,1}=T^{0,1}$ then $\pounds\omega=\alpha\omega$ for some function $\alpha$.

Finally, we set $Y={\overline{\zeta}}$. Thus, $Y$ is transverse to $T^{1,0}\oplus T^{0,1}$ and

[TABLE]

and we are done.