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

Question

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

Answer 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 $J$.

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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.

Answer 2

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

• Complexification of a map under nonstandard complexifications of vector spaces

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

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).

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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 $A$.

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

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

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

8. $A + K(A) = V^2$ 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 $G = cpx^{-1} \circ g \circ cpx$.

5. $image(g \circ K \circ cpx) \subseteq image(K \circ cpx)$ 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 $H = cpx^{-1} \circ K^{-1} \circ g \circ K \circ cpx$.

6. $image(g \circ cpx) \subseteq image(cpx)$ 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.

4. 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.

5. (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 $f_2$.

4. 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.

5. (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 $G = cpx^{-1} \circ K^{-1} \circ g \circ cpx$.

2. $image(g \circ K \circ cpx) \subseteq image(cpx)$ 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 $H = cpx^{-1} \circ g \circ K \circ cpx$.

3. $image(g \circ cpx) \subseteq image(K \circ cpx)$ 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 $K$.

Part VIII. Additional remarks:

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 $h = K^{-1} \circ g$

2. $g$ 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 $j = K \circ g$