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Let $V$ be a real vector space with a linear complex structure $J$ (see Wikipedia).

Denote $V_J$ the complex vector space induced from $V$ by the complex structure $J$.

Also, define the complexification of $V$ as usual $$ V^{\mathbb{C}} = V \otimes_{\mathbb{R}} \mathbb{C} $$ and let $\tilde{J}$ be the complex structure induced from $V$ to $V^{\mathbb{C}}$ $$ \tilde{J}(v \otimes \lambda) = J(v) \otimes \lambda \qquad v \in V \quad \lambda \in \mathbb{C} . $$

$V^{\mathbb{C}}$ then decomposes as $$ V^{\mathbb{C}} = V^+ \oplus V^- $$ where $V^{\pm}$ are the $\pm i$ eigenspaces of the operator $\tilde{J}$.

$$ V^{\pm} = \{ v \otimes 1 \mp Jv \otimes i : v \in V \} $$

Wikipedia claims the following:

There is a natural complex linear isomorphism between $V_J$ and $V^+$, so these vector spaces can be considered the same, while $V^-$ may be regarded as the complex conjugate of $V_J$.

My questions are:

  1. In what sense is the isomorphism natural? Does it mean the same thing as canonical, i.e. basis-independent?
  2. Is it also possible to construct such a natural/canonical complex-linear isomorphism between $V^-$ and $V_J$? If not, why?
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The isomorphisms are given by

$$ v \mapsto \frac{1}{2} (v - \tilde{i} Jv) \colon (V, J) \rightarrow (V^{+}, \tilde{i}|_{V^{+}}), \\ v \mapsto \frac{1}{2} (v + \tilde{i}Jv) \colon (V,J) \rightarrow (V^{-},\tilde{i}|_{V^{-}}). $$

Both isomorphisms are basis independent. The first is $\mathbb{C}$-linear while the second is $\mathbb{C}$-antilinear which naturally identifies $(V^{-}, \tilde{i}|_{V^{-}})$ with the conjugate complex vector space $(V,-J)$ in a $\mathbb{C}$-linear way. Here, $\tilde{i}$ is the complex structure on $V^{\mathbb{C}}$ induced via the complexification (that is, $\tilde{i} = \operatorname{id}_V \otimes i$ where $i$ is the natural complex structure on $\mathbb{C}$ just like $\tilde{J} = J \otimes \operatorname{id}_{\mathbb{C}}$). It is usually denoted just by $i$, if at all.

Regarding your second question, a natural $\mathbb{C}$-linear isomorphism between $(V,J)$ and $(V^{-}, \tilde{i}|_{V^{-}})$ would give you a natural $\mathbb{C}$-linear isomorphism between $(V,J)$ and $(V,-J)$. This isomorphism should extend to the level of complex vector bundles so you would get that any complex vector bundle $E$ is $\mathbb{C}$-linear isomorphic to the conjugate bundle $\overline{E}$. However, it is known to be false in general (for example, using Chern classes).

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  • $\begingroup$ But, what if I consider first mapping $v \in (V,J)$ to $Av$ where $A$ is some invertible linear operator in $V$ which anticommutes with $J$? Then composing the result with the second map, I will obtain a $\mathbb{C}$-linear map $(V,J) \rightarrow V^-$. Also if $\bar{V}$ is the complex conjugate vector space, then I can find a $\mathbb{C}$-linear isomorphism with $V$. It is just complex conjugation $\bar{zv}=\bar{z} \cdot v \equiv z \ast v$ with $\ast$ the multiplication by complex numbers in $\bar{V}$. $\endgroup$ – label Feb 23 '18 at 8:05
  • $\begingroup$ @Dominik: Of course $\overline{V}$ and $V$ are isomorphic - they are both complex vector spaces of the same dimension. The issue is that there is no natural isomorphism between them. There is no natural operator $A$ on $V$ which anticommutes with $J$ (such as conjugation). $\endgroup$ – levap Feb 23 '18 at 12:22
  • $\begingroup$ I think I got it. $(V,J)$ and $V^+$ are equivalent in the sense of categories. The map $A$ which anticommutes with $J$ is not "natural" because it doesn't preserve the complex structure on $V$, i.e. it is not a morphism (the morphisms are just linear maps which commute with $J$). However, if I also had a real structure $\sigma$ on $V$, $(V,J,\sigma)$, then I could construct a natural complex-linear isomorphism between $V$ and $\overline{V}$ using the real structure $\sigma$. Do you agree with my claims? $\endgroup$ – label Feb 23 '18 at 21:00
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    $\begingroup$ @Dominik: Yes. If $(V,J)$ carries a real structure $\sigma$, then you have a natural complex-linear isomorphism between $(V,J)$ and $\overline{V} = (V,-J)$ given by $\sigma$. Any complex vector space $V$ has a real structure (in fact many) but there isn't any "canonical" real structure which doesn't involve some choices. $\endgroup$ – levap Feb 25 '18 at 10:17

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