Assume that $U$ is an open set in the complex plane $\mathbb{C}$ and $A$ is a real $2\times 2$ matrix.

We define

$$\mathcal{S}_{A}=\{f:U\to \mathbb{C}\mid Df.A=A.Df \}$$ where $Df$ is the $2\times 2$ Jacobian matrix of smooth function $f:U\to \mathbb{R}^{2}\simeq \mathbb{C}$.

Every function with this property is called an $A$-holomorphic function.

Example: When $J=\begin{pmatrix} 0&-1\\1&0\end{pmatrix}$, $\mathcal{S}_{J}$ is the space of (usual) holomorphic functions.

Assume that this vector space $\mathcal{S}_{A}$ is closed under "uniform convergence on compact subsets of $U$". That is: Assume that $f_{n}$ is a sequence in $\mathcal{S}_{A}$ and $f_{n}$ converges to $f$, uniformly on compact subsets of $U$. Then $f$ must be in $\mathcal{S}_{A}$.

Does this imply that $A$ is in the form $$A=\begin{pmatrix} a&-b\\b&a \end{pmatrix}$$

The question is motivated by the fact that the uniform limit of a sequence of holomorphic functions is a holomorphic function.

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    $\begingroup$ I meant convergent power series whose coefficients are dual numbers and whose argument is a dual number variable: en.wikipedia.org/wiki/Dual_number But nevermind, this alone does not suffice to produce a counterexample since your vector space $S$ is possibly much bigger. $\endgroup$
    – M.G.
    Jul 11, 2016 at 8:29
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    $\begingroup$ @AliTaghavi: Do you know a first-order linear partial differential operator (of two variables) with constant coefficients, whose kernel is closed under compact convergence? :-) BTW, a PDE tag seems highly appropriate. $\endgroup$
    – M.G.
    Jul 11, 2016 at 10:56
  • $\begingroup$ @July With real coefficients?..I do not know. but the CR equation can be read as $f_{x}+if_{y}=0$. Thanks for your suggestion of new tag PDE, I add it:-) $\endgroup$ Jul 11, 2016 at 11:27
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    $\begingroup$ Yes, with real coefficients. If we had one and a bit of luck, one might be able to reconstruct a counter-example $A$. (On the other hand, if $A$ were complex of any dim., things would have been much easier.) $\endgroup$
    – M.G.
    Jul 11, 2016 at 11:36
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    $\begingroup$ The only other thing that comes to my mind is the following: if $C(A)$ denotes the centralizer subalgebra of $A$ in $M_2(\mathbb{R})$, then, by smoothness, $f$ admits Taylor $C(A)$-expansion (not obvious!), meaning that the Taylor polynomial(s) has coefficients and argument in $C(A)$ and the Legendre remainder term has arguments in $C(A)$ (similar to my example above with the dual numbers replaced by the algebra $C(A)$). I've always wondered in which cases (other than over the complex numbers) the remainder term converges to 0 :-) Sorry, I could not be of much help. $\endgroup$
    – M.G.
    Jul 11, 2016 at 12:22

2 Answers 2


Simple counterexample:

Suppose $f(x,y) = u(x,y) + i v(x,y)$ is a holomorphic function. Then it satisfies the Cauchy-Riemann equations \begin{align} u_x = v_y\\ v_x = -u_y \end{align}

Then the function $g(x,y) = \tilde{u}(x,y) + i \tilde{v}(x,y)$ given by \begin{align} \tilde{u}(x,y) = \frac12 u(2x,y) \\ \tilde{v}(x,y) = v(2x,y) \end{align} satisfies \begin{align} \tilde{u}_x = \tilde{v}_y \\ \tilde{v}_x = -4 \tilde{u}_y \end{align} which is the same as $ A.D g = D g.A $ for $$ A = \begin{pmatrix} 0 & -1/4 \\ 1 & 0 \end{pmatrix} $$

This example is obtained geometrically by doing a linear change of variables $(x,y) \mapsto (2x,y)$ on the domain and $(u,v) \mapsto (\frac12 u,v)$ on the co-domain and so all convergence and closure of holomorphic functions carry over unchanged.

More generally, the same procedure shows that for any $A$ of the form

$$ A = c P J P^{-1} $$

where $P\in GL(2)$ and $c\in \mathbb{R}_+$ the closure property asked for in the question holds.

[Edit to answer this question]:

The correct way to think about the question is that the two copies of $A$ in the equation $A (Df) = (Df) A$ are not the same. One is the complex structure on the domain and the other is the complex structure on the codomain. If you conjugate the complex structure on the domain side ONLY what you get is $$ P A P^{-1} (Df) = (Df) A$$ where the second $A$ is unchanged because it is the complex structure on the codomain. To make the equation work as written, we then do the reverse conjugation on the complex structure of the codomain which changes the equation to $$ PA P^{-1} (Df) = (Df) P A P^{-1}$$ The action on the codomain side is the reverse (given by the inverse matrix) simply because covariance versus contravariance in the equation.

  • $\begingroup$ It seems I've misunderstood you the first time. The constant $c$ is a matter of taste, i.e. it is independent of the choice of $P$, right? (I thought at first that it depended on $P$, which baffled me a little :-) ) $\endgroup$
    – M.G.
    Jul 11, 2016 at 16:56
  • $\begingroup$ @July: yes. I included the constant $c$ since in the original post, the matrices don't have to have determinant $\pm 1$. $\endgroup$ Jul 11, 2016 at 17:06
  • $\begingroup$ @WillieWong Thank you very much for your answer. But I do not understand the last part of your answer. Is not $J$ the usual complex structure of Euclidean reagons in $\mathbb{R}^{2n}$? And is not true to say that:" the preserving of this almost complex structure by a map f is equivalent to $DfJ=JDf$?I do not understand that why you think my formulation is not a good way for thinking on this problem?Thanks again for your very interesting answer. $\endgroup$ Jul 12, 2016 at 15:01
  • $\begingroup$ @WillieWong another question: What about odd dimensional case; Is there a $3 \times 3$ matrix $A$ for which $\mathcal{S}_{A}$, the space of A_holomorphic functions from $R^3\to R^3$ is a closed subset of $C(R^3, R^3)$ with the topology of unfiorm convergence on compact subsets. $\endgroup$ Jul 12, 2016 at 15:08
  • $\begingroup$ @AliTaghavi: I have no idea what your comment is trying to say. (1) $J$ is, as you defined, the matrix for the complex structure on $\mathbb{R}^2$. (2) Why must $f$ be a self map? You certainly didn't specify that in your question. In the generality of your question it makes more sense to think about $f$ as a map between complex manifolds $(U,J)$ and $(V,J')$ each with its own complex structure, and think about the relation $(Df)J' = J(Df)$. You can of course restrict to the case where it is a self map, but you don't have to. $\endgroup$ Jul 12, 2016 at 15:10

There is a counterexample:

Let $A=\begin{pmatrix} 0&1\\1&0\end{pmatrix}$,then $\mathcal{S}_{A}=\{f:U\to \mathbb{C}\mid \frac{d}{dx}f = \frac{d}{dy}f\}$.

$\mathcal{S}_{A}$ is closed, but $A$ is not like $\begin{pmatrix} a&-b\\b&a \end{pmatrix}$.

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    $\begingroup$ Why is $\mathcal{S}_{A}$ closed under uniform convergence on compacts? $\endgroup$
    – M.G.
    Jul 11, 2016 at 9:48
  • $\begingroup$ @July I agree with you on this comment. Because in my question the convergence is meant $C^{0}$-convergent. $\endgroup$ Jul 11, 2016 at 10:06
  • $\begingroup$ Also, for this particular $A$, we actually have $\mathcal{S}_{A}=\{\frac{\partial}{\partial x}f_1=\frac{\partial}{\partial y}f_2, \frac{\partial}{\partial x}f_2=\frac{\partial}{\partial y}f_1\}$ (as a PDE system). The centralizer of $A$ is the set of matrices for which the diagonal has same entries and the anti-diagonal has same entries. $\endgroup$
    – M.G.
    Jul 11, 2016 at 10:13

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