Non-standard complex structures on $\Bbb H\times \Bbb H$ so that multiplication is holomorphic

Question

Let $$\mu:\Bbb H\times \Bbb H\to \Bbb H, \qquad (x,y)\mapsto x\cdot_{\Bbb H} y$$ denote the product of two quaternions. With the standard identification $$\Bbb H\cong\Bbb C^2\cong \Bbb R^4, \qquad x_0+ix_1+jx_3+kx_4\equiv (x_0+ix_1, x_3+ix_4)\equiv(x_0,x_1,x_3,x_4)$$ one sees: $$\mu: \Bbb C^4 \to \Bbb C^2,\qquad (z_1,z_2,w_1,w_2)\mapsto (z_1\cdot w_1-z_2\overline{w_2}, z_1w_2+z_2\overline{w_1})$$ and $\mu$ is not holomorphic. My question is whether or one can choose a different complex structure on the domain so that this map does become holomorphic. More precisely:

Does there exist a real orthogonal transformation $A\in O(8)$ so that $$(\mu\circ A): \Bbb C^2\times \Bbb C^2\cong\Bbb R^4\times\Bbb R^4\to \Bbb R^4 \cong \Bbb C^2,\qquad x\mapsto \mu(Ax)$$ is holomorphic?

Answer 1

Its not true, here is a sketch of how it can be shown:

Suppose we have $I_D\in M_{8\times 8}(\Bbb R)$ and $I_T\in M_{4\times 4}(\Bbb R)$ two imaginary units on $\Bbb R^8$ and $\Bbb R^4$. Crucially, from the way the question is set up, these do not depend on the point (because the identification $\Bbb R^8\cong \Bbb C^4$ is linear, same with $\Bbb R^4 \cong \Bbb C^2$). Then if $\mu$ were holomorphic one must have for all $x\in \Bbb R^8$:

$$D_x\mu \circ I_D = I_T\circ D_x \mu \tag{1}$$

If we let $H_x\subseteq \Bbb R^8$ denote the horizontal subspace of $D_x\mu$ (ie $\ker(D_x\mu)^\perp$), then one verifies that this is $4$-dimensional everywhere except at $x=0$. We let $i_x:H_x\to \Bbb R^8$ denote the inclusion and $\pi_x: i_x\circ i_x^T$ the orthogonal projection on $\Bbb R^8$ to this subspace. It is elementary to verify that:

$$\pi_x = (D_x\mu)^T \circ (D_x\mu \circ(D_x\mu)^T)^{-1}\circ D_x\mu\tag{2}$$ and then (1) implies: $$I_D \circ \pi_x = (D_x\mu)^T \circ (D_x\mu\circ (D_x\mu)^T)^{-1}\circ I_T \circ D_x\mu\tag{3}.$$ Now if $p= (1,0,0,0,0,0,0,0)$, $q=(0,0,0,0,1,0,0,0)$ then a software evaluates for me: $$\pi_p=\begin{pmatrix}\Bbb 1_{4\times 4} &0_{4\times 4} \\ 0_{4\times 4}&0_{4\times 4}\end{pmatrix},\qquad \pi_q=\begin{pmatrix}0_{4\times 4} &0_{4\times 4} \\ 0_{4\times 4}&\Bbb 1_{4\times 4}\end{pmatrix}.$$ And then from (3): $$I_D = I_D\circ \pi_q+I_D\circ \pi_p\\ =(D_p\mu)^T \circ (D_p\mu\circ (D_p\mu)^T)^{-1}\circ I_T \circ D_p\mu+(D_q\mu)^T \circ (D_q\mu\circ (D_q\mu)^T)^{-1}\circ I_T \circ D_q\mu\\ =\begin{pmatrix}\ I_T &0_{4\times 4} \\ 0_{4\times 4}&I_T\end{pmatrix}$$ Where this calculation is again done by a software.

Now that the form of $I_D$ (depending in $I_T$) is known it is easy to get a contradiction by plugging in $z=(0,1,0,0,0,0,0,0)$ into (1) and noting that the left-hand side and right-hand side differ.

As a final remark:

This proof does not show there are no complex structures on $\Bbb R^8$ and $\Bbb R^4$ compatible with the flat metrics so that $\mu$ becomes holomorphic.

It might be possible to adapt the strategy or reduce to the above case, but for my application I'm only interested in the structures conjugate by an isometry to the standard one, hence I'm not looking any further.

Answer 2

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For a mapping between (complex) manifolds, $F:\mathscr{M}\to\mathscr{N}$, and some charts: $(U_i,\varphi_i)$, $(V_j,\psi_j)$ on the domain/codomain respectively, we may consider coordinate representations of $F$:

$$\color{red}{\widetilde{F}_{ij}} := \psi_j\circ F\circ \varphi_i^{-1}:\varphi_i(U_i\cap F^{-1}(V_j))\to \psi_j\big(F\big(U_i\cap F^{-1}(V_j)\big)$$ $$\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\subseteq\mathbb{C}^m\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\subseteq \mathbb{C}^n$$

as the figure above suggests by following the arrows up over and down from left to right.

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Label the coordinates in each of $\varphi_1,\varphi_2,\psi_1,\psi_2$ respectively as $$z_1,z_2, w_1,w_2.$$ Then keeping track of the point, $\color{red}{p_0}$, in each location, further define:

$$\color{red}{z_{i,0}} := \varphi_i(p_0)$$ $$\color{red}{w_{j,0}} := \psi_j(F(p_0))$$

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To answer the problem, we need to consider Laurent expansions, which are of the form:

$$f(z) := \sum\limits_{a\in \mathbb{Z}}\alpha_{a}\cdot (z-z_0)^a$$

for three players: $\widetilde{F}_{ij}$ and two transition maps that will be described next.

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Let us simplify notation a little bit and consider the two coordinate reps:

$$\widetilde{F}_1 := \psi_1\circ F \circ \varphi_1^{-1}$$

and

$$\widetilde{F}_2:= \psi_2 \circ F \circ \varphi_2^{-1}.$$

Since we know charts are invertible, equating accross $F$ gives:

$$\widetilde{F}_1 = (\psi_1\circ \psi^{-1}_2)\circ \widetilde{F}_2 \circ (\varphi_2\circ \varphi^{-1}_1)$$ $$=: \color{red}{\tau_{21}^{cod}}\circ \widetilde{F}_2 \circ \color{red}{\tau_{12}^{dom}},$$

where the subscript on $\tau_{12}$ reads "from chart $1$ to chart $2$".

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We may now list:

$$\tau_{21}^{cod}(w_2) = \sum\limits_{a\in\mathbb{Z}}\alpha_a(w_2 - w_{2,0})^a$$ $$\widetilde{F}_2(z_2) = \sum\limits_{b\in\mathbb{Z}}\beta_b(z_2 - z_{2,0})^b$$

and

$$\tau_{12}^{dom}(z_1) = \sum\limits_{c\in\mathbb{Z}}\gamma_c(z_1 - z_{1,0})^c$$

Substituting these expansions for the composition gives:

$$\color{blue}{\widetilde{F}_1(z_1) = \sum\limits_{a\in\mathbb{Z}}\alpha_a\bigg[\sum\limits_{b\in\mathbb{Z}}\beta_b\big(\sum\limits_{c\in\mathbb{Z}}\gamma_c(z_1 - z_{1,0})^c - z_{2,0}\big)^b - w_{2,0}\bigg]^a}$$

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Theory insertion: Two complex charts are compatible if their intersection is disjoint or their transition maps are both holomorphic.

A maximal atlas, $\mathscr{A}$, is one such that:

$$[\forall \varphi\in \mathscr{A}: (\varphi'\sim \varphi)]\implies \varphi'\in \mathscr{A}$$

in other words, it contains every chart that is mutually compatible with every other chart in the atlas. Note that this implies distinct maximal atlases have non-compatible charts. In other words if two charts are from different maximal atlases, they necessarily intersect and have non-holomorphic transition maps.

Now, the order of a function at a point can be defined as the minimum power at which a nonzero coefficient appears in the Laurent series for $f$ centered at $p$:

$$\color{red}{ord_p(f)} := \inf\big\{n\in\mathbb{Z}\text{ }|\text{ }\alpha_n \neq 0\big\}.$$

Define the following orders:

$$\color{red}{a'}:= ord_{w_{2,0}}(\tau_{21}^{cod})$$ $$\color{red}{b'}:= ord_{z_{2,0}}(\widetilde{F}_2)$$ $$\color{red}{c'}:= ord_{z_{1,0}}(\tau_{12}^{dom})$$

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Finally then, we have the blue series rewritten:

$$\color{blue}{\widetilde{F}_1 = \alpha^{a'}\beta^{b'}\gamma^{c'} \cdot z_1^{a'+b'+c'}+\text{(other terms)}}$$

If the transition maps are holos at the point, then $a', c' \geq 0$. If these maps are not holos at the point, then $a',c'<0$ and the order decreases.

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From this point of view, assuming I've picked out the lowest term in the series correctly, it is in general possible to change charts (within a maximal atlas) to increase the order (i.e. to make holomorphic) a coordinate representation.

Apply this reasoning with $F:\mathscr{M}\to\mathscr{N}$ replaced by $\mu:\mathbb{H}\times\mathbb{H}\to\mathbb{H}$. $\square$

References: Conway's "Functions of One Complex Variable" and Miranda's "Algebraic Curves and Riemann Surfaces".