Let $(N,J_N)$ and $(M, J_M)$ be two compact almost complex manifolds with non integrable almost complex structures (i.e., the Nijenhuis tensor is non zero for both $J_N$ and $J_M$).

Does it imply that there can not exist any non constant pseudo-holomorphic map $~f:N\rightarrow M$? Pseudo-holomorphic means $$ df\circ J_M = J_N\circ df. $$

If this is true, can someone point out a reference for this fact? I am aware that vanishing of the Nijenhuis tensor is a necessary and sufficient condition for an integrable ACS, but I don't see immediately why its non vanishing implies the non existence of non constant pseudo-holomorphic maps.

Secondly, what if $J_N$ is integrable, but $J_M$ is not and the real dimension of $N$ is greater than two?

Note that in my original question, both $J_N$ and $J_M$ were non integrable. Secondly, non constant pseudo-holomorphic curves exist from $(\Sigma,j) $ to $(M, J_M)$ when $(\Sigma,j)$ is an almost complex manifold of real dimension two (i.e. a Riemann Surface). Hence, I added the condition that real dimension of $N$ is greater than two.


The brief answer to your question is 'no': For example, take $N=M$ and $J_N=J_M$. Then the identity map of $N$ is a nonconstant pseudo-holomorphic map.

What is true is that the nonvanishing of the Nijnhuis tensors of the two manifolds puts nontrivial conditions (beyond merely being complex linear) on the induced map on the tangent bundles. Depending on the algebra of the two Nijnhuis tensors when the dimension of $N$ is greater than $2$, it can indeed happen that these conditions imply that any pseudo-holomorphic mapping from $N$ to $M$ must be constant.

For example, if $N=S^6$ and $J_N$ is the 'standard' $G_2$-invariant almost-complex structure on $S^6$, then there are no nonconstant pseudo-holomorphic functions $f:U\to\mathbb{C}$ for any open subset $U\subset S^6$.

Added in response to the OP's comment:

Here is how you can see how the Nijnhuis tensor induces restrictions on the possible first derivatives of a pseudoholomorphic map: Suppose that $(N,J_N)$ and $(M,J_M)$ are almost-complex manifolds and one wants to study the conditions on a pseudoholomorphic mapping $f:N\to M$ that satisfies $f(y) = x$ for some $y\in N$ and $x\in M$. Choose a basis $\alpha^i\ (1\le i\le m)$ for the $(1,0)$-forms on an $x$-neighborhood $U\subset M$ and a basis $\beta^p\ (1\le p\le n)$ for the $(1,0)$-forms on a $y$-neighborhood $V\subset N$. (Assume that $U$ and $V$ have been chosen so that $V$ is in the domain of $f$ and $f(V)\subset U$.) Then the desired mapping $f$ will satisfy $f^*\alpha^i = F^i_p\ \beta^p$ for some functions $F^i_p$ that are essentially the components of the Jacobian of $f$ relative to the bases $\alpha$ and $\beta$. Now, there will be unique functions $A^i_{\bar k\bar l} = -A^i_{\bar l\bar k}$ on $U$ and $B^p_{\bar q\bar r} = - B^p_{\bar r\bar q}$ on $V$ such that $$ \begin{aligned} \mathrm{d} \alpha^i \ &\equiv \tfrac12A^i_{\bar k\bar l}\ \overline{\alpha^k}\wedge\overline{\alpha^l} \ \mod \alpha^1,\ldots,\alpha^m \\ \mathrm{d} \beta^p \ &\equiv \tfrac12B^p_{\bar q\bar r}\ \overline{\beta^q}\wedge\overline{\beta^r} \ \mod \beta^1,\ldots,\beta^n \end{aligned} $$ (These functions are the components of the Nijnhuis tensors of the two almost-complex structures relative to the chosen bases of $(1,0)$-forms; they vanish identically if and only if $J_N$ and $J_M$ are integrable on $V$ and $U$.) Now, since $f^*$ preserves $(p,q)$-type and commute with the exterior derivative, it follows that, upon taking the exterior derivative of the relation $f^*\alpha^i = F^i_p\ \beta^p$ and comparing the $(0,2)$-components, one has $$ f^*\left(\tfrac12A^i_{\bar k\bar l}\ \overline{\alpha^k}\wedge\overline{\alpha^l}\right) = F^i_p\ \tfrac12B^p_{\bar q\bar r}\ \overline{\beta^q}\wedge\overline{\beta^r}, $$ which implies $$ A^i_{\bar k\bar l}{\circ}f\ \overline{\left(F^k_q\ \beta^q\right)}\wedge \overline{\left(F^l_r\ \beta^r\right)} = F^i_p\ B^p_{\bar q\bar r}\ \overline{\beta^q}\wedge\overline{\beta^r} $$ i.e., one has the algebraic equations $$ \tfrac12\ A^i_{\bar k\bar l}{\circ}f\ \overline{(F^k_qF^l_r-F^k_rF^l_q)} = B^p_{\bar q\bar r}\ F^i_p\ . $$ In particular, since $f(y) = x$, one has $$ \tfrac12\ A^i_{\bar k\bar l}(x)\ \overline{(F^k_q(y)F^l_r(y)-F^k_r(y)F^l_q(y))} = B^p_{\bar q\bar r}(y)\ F^i_p(y)\ . $$ Of course, these relations are trivial if the respective Nijnhuis tensors vanish at $y$ and $x$, but they can easily be quite restrictive. For example, if $J_M$ is integrable, so that the $A$s vanish, then this is a set of linear equations for $F(y)$ that, depending on $B^p_{\bar q\bar r}(y)$, could easily have only $F(y)=0$ as solutions.

In fact, as soon as the dimension of $N$ is at least $3$, the 'generic' Nijnhuis tensor $B$ will have this property, and a conclusion will be, for example, that, for the 'generic' almost-complex manifold $(N,J_N)$ of dimension $3$ or more, there are no nonconstant pseudoholomorphic functions $f:V\to\mathbb{C}$ for any open set $V\subset N$. (The conclusion holds in the case $\dim N = 2$ as well, but you have to differentiate one more time to get this in the $2$-dimensional case.)

  • $\begingroup$ @Bryant: Thank you... I should have thought about the example you just gave with the identity map! Can you give a reference where I can read a bit more about the second thing you said (non vanishing of the Nijnhuis tensor sometimes implying that any pseudo holomorphic map is constant)? $\endgroup$
    – Ritwik
    Jun 17 '14 at 14:27
  • $\begingroup$ Unfortunately, I don't know any specific references where this is discussed in detail, but there's nothing deep about it. I'll add some remarks to my answer above that may be helpful. $\endgroup$ Jun 18 '14 at 8:20

This is intended to compliment Robert Bryant's post, and answers Ritwik's question in his comment.

For $\dim_\mathbb{R} N>2$ this is not a useful notion. Indeed, for a generic $(N,J_N)$ there are no nonconstant $f$ (even locally). This is where the integrability condition (vanishing of the Nijenhuis tensor $\mathcal{N}_{J_M}$) needs to kick in for complex manifolds. Explicitly, the local form of such a map $f$ is given by the ``nonlinear Cauchy-Riemann equations'', an elliptic PDE system that is overdetermined for $\dim_\mathbb{R} N>2$ because (locally) any map $(N,J)\to(\mathbb{C},i)$ has $\dim_\mathbb{R} N$ equations in 2 unknowns (real and imaginary parts of function).

Put differently, $J$-holomorphic submanifolds are rare unless $(M,\omega)$ is complex (then use IFT on holomorphic functions $f:M\to\mathbb{C}^n$) or $\dim_\mathbb{R} M=2$ (the Cauchy-Riemann operator $\bar\partial_J$ is elliptic). If $\dim_\mathbb{R}M>2$ then $\bar\partial_J$ is overdetermined and has Fredholm index $-\infty$.

  • $\begingroup$ But remember that existence/nonexistence of a nonconstant pseudoholomorphic mapping depends on the target as well. As I observed above, the identity map is a nonconstant pseudoholomorphic map for any $(N,J_N)$. $\endgroup$ Jun 17 '14 at 18:33
  • $\begingroup$ Ah I am implicitly fixing the target $(M,J_M)$. I speak of generic $(N,J_N)$ which is consistent with your observation. $\endgroup$ Jun 17 '14 at 18:37
  • $\begingroup$ It's not completely clear what the OP had in mind. As I read it, he was asking whether having $J_N$ and $J_M$ be nonintegrable implied (in general) that all pseudoholomorphic maps from $N$ to $M$ must be constant. Is is really obvious that there does not exist a (probably very high dimensional and necessarily) non-integrable, almost complex manifold $(M,J_M)$ such that every almost complex $4$-manifold $(N,J_N)$ has a locally defined, nonconstant pseudoholomorphic map into $(M,J_M)$? I think that statement might not be as easy to actually prove as it is to believe. $\endgroup$ Jun 17 '14 at 18:47
  • $\begingroup$ @Bryant and Chris: Actually Chris's comment was the real motivation for my question. In Symplectic Geometry the moduli space of J holomorphic maps from one Riemann Surface to a Symplectic Manifold is studied. I was wondering why a higher dimensional analogue of this is not studied; I think Chris's answer explains that. $\endgroup$
    – Ritwik
    Jun 18 '14 at 13:07

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