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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:

Definitions, Assumptions, Notations

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, $i(v,w)=(-w,v)$.

Note on Complexification of space definition: The above definition however depends on $J$, so to denote this dependence, we may write $V^{(\mathbb C,J)}=V^{\mathbb C}$. We could have another definition replacing $J$ with any other almost complex structure $K$ which necessarily relates to $J$ by $K = S \circ J \circ S^{-1}$ for some $S \in Aut_{\mathbb R}(V^2)$. For example with $K = - J$ (I think $S$ would be $S(v,w):=(v,-w)$, which is $\mathbb C$-antilinear with respect to $J$, and even to $K=-J$ I think), we get $i(v,w)=(w,-v)$.

Complexification of map definition: Based on Conrad, Bell, Suetin, Kostrikin and Mainin (12.10-11 of Part I) and Roman (Chapter 2), it looks like we can define the complexification (with respect to $J$) $f^{\mathbb C}: V^{\mathbb C} \to V^{\mathbb C}$ of $f: V \to V$, $f \in End_{\mathbb R}V$ as any of the following equivalent, I think, ways (Note: we could actually have different vector spaces such that $f: V \to U$, but I'll just talk about the case where $V=U$)

Definition 1. $f^{\mathbb C}(v,w):=(f(v),f(w))$

  • I think '$\mathbb C$-linear (with respect to $J$)' isn't part of this definition but is deduced anyway.

Definition 2. $f^{\mathbb C}$ the unique $\mathbb C$-linear (with respect to $J$) map such that $f^{\mathbb C} \circ cpx = cpx \circ f$, 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. (Note: I think $cpx$ doesn't depend on $J$.)

Definition 3. $f^{\mathbb C}$ the unique $\mathbb C$-linear (with respect to $J$) map such that $(f^{\mathbb C})_{\mathbb R} = f \oplus f$

Definition 4. $f^{\mathbb C} := (f \oplus f)^J$ and again '$\mathbb C$-linear (with respect to $J$)' isn't part of this definition but is deduced anyway. Here, the notation $(\cdot)^I$ is:

  • Complex structure on map definition: The operator '$(\cdot)^I$' is supposed to be something like an inverse of the realification functor $(\cdot)_{\mathbb R}$ (see Jordan Bell and Suetin, Kostrikin and Mainin). If $(\cdot)^I$ is some kind of functor, then $W^I := (W,I)$.

    • I couldn't find any book that uses this kind of notation, but the point of this '$g^I$' is mainly to be specific and allow shortcuts. Example: The statement '$g$ is $\mathbb C$-linear with respect to $I$' becomes just '$g^I$ is $\mathbb C$-linear'. Another example: For any almost complex structure $K$ on $W$, $K^K$ is $\mathbb C$-linear, but $I^K$ and $K^I$ are not necessarily $\mathbb C$-linear. However, with $-I$ as another almost complex structure on $W$, I think $I^{-I}$and ${-I}^{I}$ are $\mathbb C$-linear.

    • Proposition: $g^I$ is $\mathbb C$-linear if and only if $g$ is $\mathbb R$-linear and $g$ 'commutes with scalar multiplication by i (with respect to $I$)', meaning $g \circ I = I \circ g$.

    • We can also extend to defining maps like $g^{(I,H)}: (W,I) \to (U,H)$ and saying $g^{(I,H)}$ is $\mathbb C$-linear if and only if $g$ is $\mathbb R$-linear and $g \circ I = H \circ g$. In this notation and for the case of $W=U$, $g^{(I,I)}=g^I$.

Regardless of the definition, we end up with the formula given in Definition 1 (Even if the definitions aren't equivalent, whichever definitions are correct, I think will give this formula in Definition 1).

Note on Complexification of map definition: The above definition/s however depends on $J$, so to denote this dependence, we may write $f^{(\mathbb C,J)}=f^{\mathbb C}$.

Questions:

Question 1: What is the formula for $f^{(\mathbb C,K)}$ for any almost complex structure $K$ on $V^2$, assuming it exists, whether uniquely or not?

  • Note: I actually didn't think $f^{(\mathbb C,K)}$ wouldn't be unique or even exist until mid way through typing this (so I added 2 more questions below), so there might be kind of a definition issue here, but I guess it's ok to define $f^{(\mathbb C,K)}$ as any $\mathbb C$-linear (with respect to $K$) map such that $f^{(\mathbb C,K)} \circ cpx = cpx \circ f$

  • Example: For $K=-J$, I think we get still $f^{(\mathbb C,-J)}(v,w)=(f(v),f(w))$ (I derived this in a similar way that Conrad derived the formula for $K=J$).

  • Example: Suppose $V$ in turn has an almost complex structure $k$. Then $k \oplus k$ is an almost complex structure on $V^2$. For $K=k \oplus k$, I don't know how to get the formula for $f^{(\mathbb C,k \oplus k)}(v,w)$, similar to the cases of $K= \pm J$. Maybe it doesn't exist.

Question 2: Does $f^{(\mathbb C,K)}$ always exist even if not uniquely?

Question 3: Whenever $f^{(\mathbb C,K)}$ exists, is $f^{(\mathbb C,K)}$ unique?

Note: This question might be answered by the answer, that I'm still analysing, to another question I posted.

More thoughts based on these:

It appears that:

  1. complexification relies not only on an almost complex structure $K$ on $V^2$ but also on a choice of subspace $A$ of $V^2$, where $A$ is not $V^2$ or $0$. This $A$ is what we use to identify $V$ as an embedded $\mathbb R$-subspace of $V^2$

  2. For any subspace $A$ of $V^2$, except $V^2$ and $0$, and for any almost complex structure $K$ on $V^2$, there exists a unique involutive $\mathbb R$-linear map $\sigma_{A,K}$, on $V^2$, such that $\sigma_{A,K}$ anti-commutes with $K$ and the set of fixed points of $\sigma_{A,K}$ is equal to $A$.

    • 2.1. For example, $\sigma_{V \times 0,J} = \chi$, where $\chi(v,w):=(v,-w)$

  3. Therefore, I should ask about $f^{(\mathbb C,K,A)}$, not $f^{(\mathbb C,K)}$.

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1 Answers1

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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 $A$.

  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 $j = K \circ g$
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