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

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 $$(id_{V})^{\mathbb C} = id_{V^{\mathbb C}}$$

Based on Joppy's answer here, this is an answer to both of the following questions

Here, I will derive a formula for general complexification and present generalised versions of both Conrad Theorem 2.6 and Conrad Theorem 4.16 (but for simplicity I focus only on endomorphisms of a space rather than homomorphisms between two spaces).

Part 0. Assumptions:

Let $$V$$ be an $$\mathbb R$$-vector space. Let $$A$$ be an $$\mathbb R$$-subspace of $$V^2$$ such that $$A \cong V$$. Let $$cpx: V \to V^2$$ be any injective $$\mathbb R$$-linear map with $$image(cpx)=A$$. (I guess for any $$\mathbb R$$-isomorphism $$\gamma: V \to A$$, we can choose $$cpx = \iota \circ \gamma$$, where $$\iota: A \to V^2$$ is inclusion.) Let $$K \in Aut_{\mathbb R}(V^2)$$ be any almost complex structure on $$V^2$$ (i.e. $$K$$ is anti-involutory, i.e. $$K \circ K = -id_{V^2}$$, i.e. $$K^{-1} = -K$$). Let $$f \in End_{\mathbb R}(V)$$. Let $$g \in End_{\mathbb R}(V^2)$$.

• 0.1. Intuition on $$A$$: $$A$$ is the subspace of $$V^2$$ that we use to identify $$V$$ with. Originally, this is $$A=V \times 0$$ and then $$cpx$$ is something like $$cpx(v):=(v,0)$$. However, I think $$cpx(v):=(7v,0)$$ will also work.

Part I. On $$\sigma_{A,K}$$ and on $$K(A)$$ the image of $$A$$ under $$K$$:

1. $$K \circ cpx: V \to V^2$$ is an injective $$\mathbb R$$-linear map with $$image(K \circ cpx) = K(A)$$.

2. $$A \cong K(A)$$

3. $$K(A)$$ is an $$\mathbb R$$-subspace of $$V^2$$ such that $$K(A) \cong V$$.

4. There exists a unique map $$\sigma_{A,K} \in Aut_{\mathbb R}(V^2)$$ such that

• 4.1. $$\sigma_{A,K}$$ is involutory, i.e. $$\sigma_{A,K} \circ \sigma_{A,K} = id_{V^2}$$, i.e. $$\sigma_{A,K}^{-1} = \sigma_{A,K}$$,

• 4.2. $$\sigma_{A,K}$$ anti-commutes with $$K$$, i.e. $$\sigma_{A,K} \circ K = - K \circ \sigma_{A,K}$$, and

• 4.3. The set of fixed points of $$\sigma_{A,K}$$ is equal to $$A$$.

1. By (I.4.1), $$\sigma_{A,K}$$ has exactly 2 eigenvalues $$\pm 1$$.

2. $$A$$ is also the eigenspace for the eigenvalue $$1$$.

3. $$K(A)$$ is both the eigenspace for the eigenvalue $$-1$$ of $$\sigma_{A,K}$$, and the set of fixed points of $$-\sigma_{A,K}$$.

4. $$A + K(A) = V^2$$ and $$A \cap K(A) = \{0_{V^2}\}$$, i.e. we have a literal internal direct sum $$A \bigoplus K(A) = V^2$$.

Part II. On real and imaginary parts when we have commutation with $$\sigma_{A,K}$$:

1. If $$g$$ commutes or anti-commutes with $$K$$, we have that $$image(g \circ cpx) \subseteq image(cpx)$$ if and only if $$image(g \circ K \circ cpx) \subseteq image(K \circ cpx)$$.

2. $$image(g \circ cpx) \subseteq image(cpx)$$ and $$image(g \circ K \circ cpx) \subseteq image(K \circ cpx)$$ if and only if $$g$$ commutes with $$\sigma_{A,K}$$.

3. $$image(g \circ cpx) \subseteq image(K \circ cpx)$$ and $$image(g \circ K \circ cpx) \subseteq image(cpx)$$ if and only if $$g$$ anti-commutes with $$\sigma_{A,K}$$.

4. $$image(g \circ cpx) \subseteq image(cpx)$$ if and only if $$g \circ cpx = cpx \circ G$$, for some $$G \in End_{\mathbb R}(V)$$.

• II.4.1. $$G$$ turns out to be uniquely $$G = cpx^{-1} \circ g \circ cpx$$.
1. $$image(g \circ K \circ cpx) \subseteq image(K \circ cpx)$$ if and only if $$g \circ K \circ cpx = K \circ cpx \circ H$$, for some $$H \in End_{\mathbb R}(V)$$.
• II.5.1. $$H$$ turns out to be uniquely $$H = cpx^{-1} \circ K^{-1} \circ g \circ K \circ cpx$$.
1. $$image(g \circ cpx) \subseteq image(cpx)$$ and $$image(g \circ K \circ cpx) \subseteq image(K \circ cpx)$$ if and only if for some $$G, H \in End_{\mathbb R}(V)$$, we can write $$g(a \oplus K(b)) = cpx \circ G \circ cpx^{-1}(a) \oplus K \circ cpx \circ H \circ cpx^{-1} \circ K^{-1} (K(b)),$$ where $$a,b \in A = image(cpx)$$.
• II.6.1. $$g$$ commutes with $$K$$ if and only if $$G=H$$.

• II.6.2. $$g$$ anti-commutes with $$K$$ if and only if $$G=-H$$.

• II.6.3. $$G$$ and $$H$$ turns out to be uniquely as given in (II.4.1) and (II.5.1).

• II.6.4. I don't believe there's any relation between $$G$$ and $$H$$ if we don't know any further information on $$g$$ (e.g. commutes or anti-commutes with $$K$$).

Part III. For generalising Conrad Theorem 2.6:

1. Just as with Conrad Theorem 2.6, there exists a unique map $$f_1 \in End_{\mathbb R}(V^2)$$ such that $$f_1$$ commutes with $$K$$ and $$f_1 \circ cpx = cpx \circ f$$.

2. Observe that there also exists a unique map $$f_2 \in End_{\mathbb R}(V^2)$$ such that $$f_2$$ commutes with $$K$$ and $$f_2 \circ K \circ cpx = K \circ cpx \circ f$$.

3. By (II.6.1), $$f_1=f_2$$. Define $$(f^\mathbb C)_{\mathbb R}:=f_1=f_2$$. Equivalently, $$f^\mathbb C:=f_1^K=f_2^K$$.

• III.3.1. Meaning: The original definition of complexification is based on $$cpx$$. If we have another definition of complexification $$K \circ cpx$$ instead of $$cpx$$, then this definition will be equivalent to the original.
1. The formula for $$(f^\mathbb C)_{\mathbb R}$$ actually turns out to be $$(f^\mathbb C)_{\mathbb R}(a \oplus K(b)) = cpx \circ f \circ cpx^{-1}(a) \oplus K \circ cpx \circ f \circ cpx^{-1} \circ K^{-1} (K(b)),$$ where $$a,b \in A = image(cpx)$$. We can derive this similarly to the derivation in the first part of the proof of Conrad Theorem 2.6.

2. (I'm not sure if I use this fact anywhere in this post.) The map that yields a complexification unique: $$f=h$$ if and only if $$(f^\mathbb C)_{\mathbb R} = (h^\mathbb C)_{\mathbb R}$$.

Part IV. For generalising Conrad Theorem 4.16:

1. We can see that this formula for $$(f^\mathbb C)_{\mathbb R}$$ also allows a generalisation of Conrad Theorem 4.16: $$g=(f^\mathbb C)_{\mathbb R}$$ for some (unique) $$f$$ if and only if $$g$$ commutes with $$K$$ and $$g$$ commutes with $$\sigma_{A,K}$$.
• IV.1.1. By the way, I think Conrad Theorem 4.16 is better stated as 'commutes with both $$J$$ and $$\chi$$ iff complexification' instead of 'If commutes with $$J$$, then we have commutes with $$\chi$$ iff complexification' since, in the latter case, the 'if' direction doesn't use the 'commutes with $$J$$' assumption. It might be wrong to talk about complexification if we don't assume 'commutes with $$J$$', so in this case, we could say like '$$g=f \oplus f$$' instead of '$$g$$ is the complexification of some (unique) $$f$$')

• IV.1.2. Equivalently, $$g=(f^\mathbb C)_{\mathbb R}$$ if and only if $$g$$ commutes with $$K$$ and $$image(g \circ cpx) \subseteq image(cpx)$$

• IV.1.3. Equivalently, $$g=(f^\mathbb C)_{\mathbb R}$$ if and only if $$g$$ commutes with $$K$$ and $$image(g \circ K \circ cpx) \subseteq image(K \circ cpx)$$

Part V. For the analogue of Conrad Theorem 2.6 for anti-complexification (anti-commuting with $$K$$ but still commuting with $$\sigma_{A,K}$$):

1. Just as with Conrad Theorem 2.6, there exists a unique map $$f_1 \in End_{\mathbb R}(V^2)$$ such that $$f_1$$ anti-commutes with $$K$$ and $$f_1 \circ cpx = cpx \circ f$$.

2. There exists a unique map $$f_2 \in End_{\mathbb R}(V^2)$$ such that $$f_2$$ anti-commutes with $$K$$ and $$f_2 \circ K \circ cpx = K \circ cpx \circ f$$.

3. However, by (II.6.2), $$f_1=-f_2$$.

• V.3.1. Meaning: Hence, $$f_1 \ne -f_2$$, unlike with the case of complexification, where we had $$f_1=f_2$$. Therefore, we have two unequivalent definitions of anti-complexification.

• V.3.2. However, observe that if we define $$f^{anti-\mathbb C}:=f_1$$, then $$(-f)^{anti-\mathbb C}=f_2$$. This way, even though $$f_2$$ isn't the anti-complexification of $$f$$, $$f_2$$ is still the anti-complexification of something, namely of $$-f$$.

• V.3.3. Same as V.3.2, but interchange $$f_1$$ and $$f_2$$.

1. The formula for $$(f^{anti-\mathbb C})_{\mathbb R}$$ actually turns out to be (I use the $$f_1$$ definition) $$f_1(a \oplus K(b)) = cpx \circ f \circ cpx^{-1}(a) \oplus K \circ cpx \circ -f \circ cpx^{-1} \circ K^{-1} (K(b)),$$ where $$a,b \in A = image(cpx)$$. We can derive this similarly to the derivation in the first part of the proof of Conrad Theorem 2.6.

2. (I'm not sure if I use this fact anywhere in this post.) The map that yields an anti-complexification is unique (as with complexification): $$f=h$$ if and only if $$(f^{anti-\mathbb C})_{\mathbb R} = (h^{anti-\mathbb C})_{\mathbb R}$$.

Part VI. For the analogue of Conrad Theorem 4.16 for anti-complexification (anti-commuting with $$K$$ but still commuting with $$\sigma_{A,K}$$):

1. The analogue of Conrad Theorem 4.16 for generalised anti-complexification is that: $$g=f^{anti-\mathbb C}$$ if and only if $$g$$ anti-commutes with $$K$$ and $$g$$ commutes with $$\sigma_{A,K}$$.
• VI.1.1. Equivalently, $$g=(f^{anti-\mathbb C})_{\mathbb R}$$ if and only if $$g$$ anti-commutes with $$K$$ and $$image(g \circ cpx) \subseteq image(cpx)$$.

• VI.1.1.1. However, $$cpx^{-1} \circ g \circ cpx$$ may be either of $$\pm f$$, depending on the choice of definition.
• VI.1.2. Equivalently, $$g=(f^{anti-\mathbb C})_{\mathbb R}$$ if and only if $$g$$ anti-commutes with $$K$$ and $$image(g \circ K \circ cpx) \subseteq image(K \circ cpx)$$.

• VI.1.2.1. However, $$cpx^{-1} \circ K^{-1} \circ g \circ K \circ cpx$$ may be either of $$\pm f$$, depending on the choice of definition.
• VI.1.3. Regardless of the definition, $$cpx^{-1} \circ K^{-1} \circ g \circ K \circ cpx = - cpx^{-1} \circ g \circ cpx$$.

Part VII. On real and imaginary parts when we have anti-commutation with $$\sigma_{A,K}$$:

1. $$image(g \circ cpx) \subseteq image(K \circ cpx)$$ if and only if $$g \circ cpx = K \circ cpx \circ G$$, for some $$G \in End_{\mathbb R}(V)$$.
• VII.1.1. $$G$$ turns out to be uniquely $$G = cpx^{-1} \circ K^{-1} \circ g \circ cpx$$.
1. $$image(g \circ K \circ cpx) \subseteq image(cpx)$$ if and only if $$g \circ K \circ cpx = cpx \circ H$$, for some $$H \in End_{\mathbb R}(V)$$.
• VII.2.1. $$H$$ turns out to be uniquely $$H = cpx^{-1} \circ g \circ K \circ cpx$$.
1. $$image(g \circ cpx) \subseteq image(K \circ cpx)$$ and $$image(g \circ K \circ cpx) \subseteq image(cpx)$$ if and only if for some $$G, H \in End_{\mathbb R}(V)$$, we can write $$g(a \oplus K(b)) = K \circ cpx \circ G \circ cpx^{-1}(a) \oplus cpx \circ H \circ cpx^{-1} \circ K^{-1} (K(b)),$$ where $$a,b \in A = image(cpx)$$.
• VII.3.1. Observe that both $$\pm K \circ g$$ commute with $$K$$ if and only if $$g$$ commutes with $$K$$ (if and only if both $$g \circ \pm K$$ commute with $$K$$).

• VII.3.2. Same as (VII.3.1), but 'anti-commute/s' instead of 'commute/s'.

• VII.3.3. $$G$$ and $$H$$ turns out to be uniquely as given in (VII.1.1) and (VII.2.1).

• VII.3.4. I don't believe there's any relation between $$G$$ and $$H$$ if we don't know any further information on $$g$$.

• VII.3.5. By (VII.3.1), apply (II.6.1) to $$K^{-1} \circ g$$: $$K^{-1} \circ g = (G^\mathbb C)_{\mathbb R}$$ if and only if $$G=H$$ if and only if $$K^{-1} \circ g$$ commutes with $$K$$ if and only if $$g$$ commutes with $$K$$.

• VII.3.6. By (VII.3.2), apply (II.6.2) to $$K^{-1} \circ g$$: $$K^{-1} \circ g = (G^{anti-\mathbb C})_{\mathbb R}$$ or $$((-G)^{anti-\mathbb C})_{\mathbb R}$$ (depending on definition) if and only if $$G=-H$$ if and only if $$K^{-1} \circ g$$ anti-commutes with $$K$$ if and only if $$g$$ anti-commutes with $$K$$.

1. $$g$$ anti-commutes with $$\sigma_{A,K}$$ if and only if $$g=K \circ h$$, for some $$h \in End_{\mathbb R}(V)$$ that commutes with $$\sigma_{A,K}$$.
• VIII.1.1. This $$h$$ is uniquely $$h = K^{-1} \circ g$$
1. $$g$$ commutes with $$\sigma_{A,K}$$ if and only if $$g=K^{-1} \circ j$$, for some $$j \in End_{\mathbb R}(V)$$ that anti-commutes with $$\sigma_{A,K}$$.
• VIII.2.1. This $$j$$ is uniquely $$j = K \circ g$$