Is complex symplectic structure equivalent to a quaternionic structure?

Question

Let $S$ be a vector space over $\mathbb{C}$ of dimension $\operatorname{dim} S =2n$, let me denote by $S_\mathbb{R}$ real form of $S$ i.e. vector space over $\mathbb{R}$ of dimension $4n$. Is it true that endowing $S$ with a complex symplectic form $\omega$ is equivalent to endowing $S_\mathbb{R}$ with a structure of a module over quaternions $\mathbb{H}$?

In one direction this is clear. If $(I,J,K)$ is a quaternionic structure on $S_\mathbb{R}$ we can choose any inner product $g$ (we only need $g$ to be non-degenerate and symmetric) on $S_\mathbb{R}$ s.t. $J$ and $K$ are orthogonal with respect to $g$ and define $$ \omega(v,w)=g(v,Jw)+i g(v,Kw) $$ then $\omega$ is a complex symplectic structure in the complex structure $I$ on $S$.

How one can go in other direction?

Answer 1

First of all, the formula should read $$ ω(v,w) = g(v,Jw) + ig(v,Kw)$$ instead of “$Ig$”, because the value of an inner product is a number, not vector. But it is just a formalistic quibble.

Arguments presented by original poster are flawed bidirectionally, or possibly thoughts are expressed without a careful distinction between a vector space and an inner product (Hilbert) space. I’m going to explain why we must have a fixed inner product form to interchange between complex symplectic form and quaternion multiplication.

Consider a one-dimensional quaternionic space S, i.e. a 1-dimensional ℍ-module. You can multiply its real 4-vectors by quaternions, but you don’t see their (real) magnitudes; we have only |u|  ∕  |v|, where u ∈ S, v ∈ S\{0} as a well-defined real number. Of course, you can choose an inner product with necessary compatibility property, but this choice contains an arbitrary positive (real) factor. Whichever you think about it, resulting symplectic structures are not identical as such. Quaternionic structure, even in 1 dimension, doesn’t induce certain complex symplectic structure, it only has (several) compatible complex symplectic structures. Of course, things are worse in several quaternionic dimensions.

Consider another direction. Let S = ℂ² and $$ ω(v,w) = v^{\mathsf T}\begin{pmatrix} 0 & 1 \\ −1 & 0 \end{pmatrix}w,\ g(v,w)=\operatorname{Re}(v^†\,w).$$ We can then guess that $$ I\begin{pmatrix}w_1\\w_2\end{pmatrix} = \begin{pmatrix}iw_1\\iw_2\end{pmatrix},\ J\begin{pmatrix}w_1\\w_2\end{pmatrix} = \begin{pmatrix}\overline{w_2}\\−\overline{w_1}\end{pmatrix},\ K\begin{pmatrix}w_1\\w_2\end{pmatrix} = \begin{pmatrix}i\,\overline{w_2}\\−i\,\overline{w_1}\end{pmatrix}. $$

In invariant terms of S_ℝ, we should make J of real part of ω by converting 0,2-tensor to 1,1-tensor (raising one tensor index), and K likewise of imaginary part of ω, so both might be unique. The problem is that the index-raising operation depends on the inner product form $g$, whereas we have many inner products compatible with both complex structure and given symplectic form, each producing its quaternionic structure. For example the following real 1-dimensional family of inner products: $$ g_λ(v,w)=\operatorname{Re}(λ\,\overline{v_1}\,w_1 + λ^{−1}\overline{v_2}\,w_2),\ λ>0$$ Like the ℍ-module-to-symplectic-form case, without a fixed inner product structure we have an ambiguity also starting from n = 1.

If you are interested in quaternionic manifolds, then read about “Hyperkähler structure” first.

Answer 2

The other direction goes this way. Note that since $$g(v,Jw) = \mathrm{Re}( \omega(v,w) ),$$ you can recover $J$ provided you have $g$, by requiring that $Jw$ satisfies the above equation. Indeed, for each real linear form $f:S \to \mathbb{R}$ there is exactly one element $z$ such that $$\forall_{v \in S}\ f(v) = g(v,z).$$ Apply this to the linear form $f_w(v) = \mathrm{Re}( \omega(v,w) )$, and you have $Jw$ defined for each $w$.

Having $J$, set $I$ to be the multiplication by $i$ and $K=IJ$.

That this gives a quaternionic structure may not be obvious from the description above. However, we can recover $J$ in another way. Set $$\tilde g(v,w) = g(v,w) + i g(v,iw).$$ This is a Hermitian inner product on $S$, and we have $$\tilde g(v,Jw) = \omega(v,w).$$ By the same sort of argument, you recover $J$ from this formula. You get $\mathbb{C}$-linearity of $J$ for free and it is easy to prove that $IJ=-JI$ and $J^2=-1$ in this setting.

Note that this identification between hermitian structures and $\mathbb{C}$-symplectic forms is well-defined only when you have the inner product $g$ (or $\tilde g$) fixed. Different choices of $g$ will give you different symplectic forms from a given quaternionic structure, and vice versa.