$f$ is the complexification of a map if $f$ commutes with almost complex structure and standard conjugation. What if we had anti-commutation instead?

I started studying the book of Daniel Huybrechts, Complex Geometry An Introduction. I tried studying backwards as much as possible, but I have been stuck on the concepts of almost complex structures and complexification. I have studied several books and articles on the matter including ones by Keith Conrad, Jordan Bell, Gregory W. Moore, Steven Roman, Suetin, Kostrikin and Mainin, Gauthier

I have several questions on the concepts of almost complex structures and complexification. Here are some:

Let $V$ be $\mathbb R$-vector space, possibly infinite-dimensional.

Complexification of space definition: Its complexification can be defined as $V^{\mathbb C} := (V^2,J)$ where $J$ is the almost complex structure $J: V^2 \to V^2, J(v,w):=(-w,v)$ which corresponds to the complex structure $s_{(J,V^2)}: \mathbb C \times V^2 \to V^2,$$ s_{(J,V^2)}(a+bi,(v,w))$$:=s_{V^2}(a,(v,w))+s_{V^2}(b,J(v,w))$$=a(v,w)+bJ(v,w)$ where $s_{V^2}$ is the real scalar multiplication on $V^2$ extended to $s_{(J,V^2)}$. In particular, $i(v,w)=(-w,v)$.

Complexification of map definition: See a question I posted previously.

Proposition (Conrad, Bell): Let $f \in End_{\mathbb C}(V^{\mathbb C})$. We have that $f$ is the complexification of a map if and only if $f$ commutes with the standard conjugation map $\chi$ on $V^{\mathbb C}$, $\chi(v,w):=(v,-w)$. In symbols:

If $f \circ J = J \circ f$, then the following are equivalent:

• Condition 1. $f=g^{\mathbb C}$ for some $g \in End_{\mathbb R}(V)$

• Condition 2. $f \circ \chi = \chi \circ f$

  • I think Bell would rewrite Condition 2 as $f = \chi \circ f \circ \chi$ and say $f$ 'equals its own conjugate'.

Questions: Considering the half of the above proposition that says '$f$ commutes with both $J$ and $\chi$ implies $f$ is complexification of a map', what do we get if we instead have the following?

1. commutes with $J$ and anti-commutes with $\chi$ ($f \circ \chi = - \chi \circ f$)

2. anti-commutes with $J$ ($f \circ J = - J \circ f$, i.e. $f$ is $\mathbb C$-anti-linear) and commutes with $\chi$

3. anti-commutes with $J$ and anti-commutes with $\chi$

Motivation: $f=J$ satisfies the case in Question 1, and $f=\chi$ satisfies the case in Question 2.

Guess (for Question 2):

Similar to this (the $K=-J$ part), I kind of had the idea to define something like anti-complexification of a map: for $g \in End_{\mathbb R}(V)$, $g^{anti-\mathbb C}$ is any $\mathbb C$-anti-linear map such that $g^{anti-\mathbb C} \circ cpx = cpx \circ g$, where $cpx: V \to V^{\mathbb C}$ is the complexification map, as Roman (Chapter 1) calls it, or the standard embedding, as Conrad calls it. I think $g^{anti-\mathbb C}$ turns out to always exist uniquely as $g^{anti-\mathbb C}(v,w)=(g(v),-g(w))$.

Then, I think the answer for Question 2 is that $f$ is the anti-complexification of a map. We can strengthen the result to: Let $f$ be $\mathbb C$-anti-linear on $V^{\mathbb C}$, i.e. $f$ anti-commutes with $J$. We have that $f$ is the anti-complexification of a map $g \in End_{\mathbb R}V$, i.e. $f=g^{anti-\mathbb C}$ if and only if $f$ commutes with the standard conjugation map $\chi$, i.e. $f \circ \chi = \chi \circ f$.

In the case of $f=\chi$ for Question 2, $f=\chi = g^{anti-\mathbb C}$ for $g=id_{V}$, the identity map on $V$, which by the way gives us $(id_{V})^{\mathbb C} = id_{V^{\mathbb C}}$

Solution 1:

I have to get things straight in my head every time I do this. Let $V^2$ be a complex vector space. This is equivalent to the data of a real vector space $V^2$, along with an $\mathbb{R}$-linear operator $J: V^2 \to V^2$ satisfying $J^2 = -1$. We will say that an $\mathbb{R}$-linear map $T: V^2 \to V^2$ is $\mathbb{C}$-linear if $TJ = JT$, and $\mathbb{C}$-antilinear if $TJ = -JT$.

You cannot "de-complexify" $(V^2, J)$, or connect it to an original un-complexified space, without a conjugation map $\chi: V^2 \to V^2$, by which we mean an $\mathbb{R}$-linear, $\mathbb{C}$-antilinear operator satisfying $\chi^2 = 1$. Once we have such a $\chi$, we can decompose $V^2$ into a real subspace $V^2_{\mathrm{re}}$ as the 1-eigenspace of $\chi$, and $V^2_{\mathrm{im}}$ as the (-1)-eigenspace of $\chi$. Note that $J$ gives a choice of isomorphism $V_{\mathrm{re}} \to V_{\mathrm{im}}$, and so does $J^{-1} = -J$.

Now consider the whole structure $(V^2, J, \chi)$. Given an $\mathbb{R}$-linear map $g: V^2_{\mathrm{re}} \to V^2_{\mathrm{re}}$, we can complexify it by defining how it acts on "real and imaginary parts", across the direct sum decomposition $V^2 = V_\mathrm{re} \oplus V_\mathrm{im}$. Note that we need to apply $J$ to an imaginary part to make it real, apply $g$, then apply $J^{-1} = -J$ to send it back to the imaginary subspace: $$ g^\mathbb{C}(v_{\mathrm{re}} + v_{\mathrm{im}}) = g(v_{\mathrm{re}}) - J g( J v_{\mathrm{im}}).$$ Let's quickly check that this is indeed $\mathbb{C}$-linear: $$ \begin{aligned} g^\mathbb{C} J (v_{\mathrm{re}} + v_{\mathrm{im}}) &= g^\mathbb{C}(J v_\mathrm{im} + J v_{\mathrm{re}}) \\ &= g(J v_\mathrm{im}) - J g(J^2 v_{\mathrm{re}}) \\ &= J (g(v_\mathrm{re}) - J g (J v_\mathrm{im})) \\ &= J g^\mathbb{C}(v_\mathrm{re} + v_\mathrm{im}). \end{aligned}$$ It's quite simple to check the commutation property of the conjugation $\chi$ with a complexification $g^\mathbb{C}$: since $\chi$ acts by $1$ on $V^2_\mathrm{re}$, $-1$ on $V^2_\mathrm{im}$ and anticommutes with $J$, we have $$ \chi g^\mathbb{C}(v_{\mathrm{re}} + v_{\mathrm{im}}) = g(v_{\mathrm{re}}) + J g(J v_\mathrm{im}) = g^\mathbb{C}(v_{\mathrm{re}} - v_{\mathrm{im}}) = g^\mathbb{C} \chi (v_{\mathrm{re}} + v_{\mathrm{im}}).$$ You can also see that if we wanted to define the anticomplexification of $g$, we could just swap out the $+$ sign for a $-$ sign in the formula for $g^\mathbb{C}$, which would make it anticommute with $J$.

Now consider our $(V^2, J, \chi)$ and consider any $\mathbb{R}$-linear map $f: V^2 \to V^2$ and its potential $\chi$-commutation properties:

1. $\chi f = f \chi$ is equivalent to $f(V^2_\mathrm{re}) \subseteq V^2_\mathrm{re}$ and $f(V^2_\mathrm{im}) \subseteq V^2_\mathrm{im}$.
2. $\chi f = - f \chi$ is equivalent to $f(V^2_\mathrm{re}) \subseteq V^2_\mathrm{im}$ and $f(V^2_\mathrm{im}) \subseteq V^2_\mathrm{re}$.
3. Note that replacing $f$ by $Jf$ swaps 1. and 2.

So the four classes of maps you are considering are complexifications, anticomplexifications, and $J$ multiplied by the first two.

Note that commutation with $\chi$ is really about doing something to the real and imaginary subspaces: either preserving them or swapping them. However, the action on $f$ on each of these subspaces could be wildly different, for example the action on $V^2_\mathrm{re}$ could be the identity, while on $V^2_\mathrm{im}$ it might be zero. Commutation with $J$ will ensure the actions are similar, in the sense that we can conjugate one action into another via the identity $f = - J f J$. This is the kind of intuition I have: $\chi$ is a choice of real and imaginary subspaces, and $J$ is the "rotation" which identifies them.