# $f$ is the complexification of a map if $f$ commutes with structure $J$ and conjugation $\chi$. What is the relationship between $J$ and $\chi$? [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 1 (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 $$V^{\mathbb C}$$, $$\chi: V^2 \to V^2$$, $$\chi(v,w):=(v,-w)$$ (Or $$\chi^J: (V^2,J)=V^{\mathbb C} \to V^{\mathbb C}$$, $$\chi^J(v,w):=(v,-w)$$, where $$\chi^J$$ is $$\chi$$ but viewed as map on $$\mathbb C$$-vector space $$V^{\mathbb C}$$ instead of a map on $$\mathbb R$$-vector space $$V^2$$. See the bullet after 'Definition 4' here). 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'.

Proposition 2: $$\chi \circ J = - J \circ \chi$$, i.e. $$\chi: V^2 \to V^2$$ is $$\mathbb C$$-anti-linear with respect to $$J$$, i.e. $$\chi^J: (V^2,J)=V^{\mathbb C} \to V^{\mathbb C}$$ is $$\mathbb C$$-anti-linear, i.e. $$J$$ anti-commutes with $$\chi$$, i.e. $$J$$ is the negative of 'its own conjugate'.

Question 1: What exactly is the relationship between the (seemingly standard) almost complex structure $$J$$ and the standard conjugation $$\chi$$ that tells us that if $$f$$ commutes both with $$J$$ and with $$\chi$$, then $$f$$ is the complexification of a map?

• Well, $$f$$ commutes with $$J$$ if and only if $$f$$ commutes with $$-J$$. Similarly, $$f$$ commutes with $$\chi$$ if and only if $$f$$ commutes with $$-\chi$$, so $$f$$ is the complexification of a map if $$f$$ commutes both-(with $$J$$ or, equivalently, with $$-J$$)-and-(with $$\chi$$ or, equivalently, with $$-\chi$$)

• Proposition 2 obviously gives a way that $$\chi$$ and $$J$$ are related, but I think Proposition 2 does not tell us much because we can replace $$\chi$$ not only with $$-\chi$$ and not only with any conjugation on $$V^{\mathbb C}$$ but also with any $$\mathbb C$$-anti-linear map on $$V^{\mathbb C}$$.

Motivation:

1. From almost complex structure to conjugation: I'm thinking that of what '$$\chi$$' (or $$\chi$$'s) would be if we used a nonstandard definition of complexification. If we had $$V^{(\mathbb C, K)} = (V^2,K)$$ for some almost complex structure $$K$$ on $$V^2$$ (such as anything besides $$\pm J$$), then we might say, for any $$f \in End_{\mathbb R}(V^2)$$ with $$f \circ K = K \circ f$$, that $$f=g^{(\mathbb C,K)}$$ if and only if $$f \circ$$ '$$\chi$$' = '$$\chi$$' $$\circ f$$ assuming '$$g^{(\mathbb C,K)}$$' is defined (see here).
• 1.1. (Added on February 3, 2020) Since the set of fixed points of the original $$\chi$$ (for the original $$K=J$$) is equal to the image of the complexification map $$cpx: V \to V^{\mathbb C}$$, $$cpx(v):=(v,0_V)$$ (see Chapter 1 of Roman; Conrad calls this the standard embedding), I guess we will have to change our notion of 'complexification map'. Maybe $$V \times 0$$ will not be the 'standard' (see here) $$\mathbb R$$-subspace of $$(V^2,K)$$ as it was for $$K=J$$ (because somehow $$\chi$$ is the standard conjugation for $$K=J$$).
1. From conjugation to almost complex structure: I am really not sure what is the correct question to ask here which is why I was reading as many references as possible, but it's kind of a headache to even formulate the question here, especially considering that calling a map a 'conjugation' depends on the almost complex structure in the first place. I think Suetin, Kostrikin and Mainin (specifically 12.9b of Part I) could be helpful.

Question 2: Besides Propositions 1 and 2 and whatever answer/s is/are given for Question 1, what are some relationships between the (seemingly standard) almost complex structure $$J$$ and the standard conjugation $$\chi$$?

(Later added) More thoughts on the above:

Based on the equivalent condition of $$f \circ \chi = \chi \circ f$$ given in an answer here (I'm still analysing this answer) and based on Conrad's proof of Conrad's Theorem 4.16, I make the following observations:

1. For any $$f \in End_{\mathbb R+0i}(V^{\mathbb C})$$, whether or not $$f \in End_{\mathbb C}(V^{\mathbb C})$$, we have that $$f \circ \chi = \chi \circ f$$, we have that there exist unique $$g,h \in End_{\mathbb R}(V)$$ such that $$f = (g \oplus g)^J$$ on $$V \times 0$$ and $$f = (h \oplus h)^J$$ on $$0 \times V = J(0 \times V)$$. Hence, (on all of $$V^{\mathbb C}$$) $$f = (g \oplus h)^J$$, i.e. $$f_{\mathbb R} = g \oplus h$$

2. From Chapter 1 of Roman, we have the complexification map $$cpx: V \to V^{\mathbb C}$$ (see ), $$cpx(v):=(v,0_V)$$. Conrad calls this the standard embedding.

• 2.1. The set of fixed points of $$\chi$$ is equal to the image of $$cpx$$.
1. We can similarly define what I like to call the anti-complexification map $$anticpx: V \to V^{\mathbb C}$$, $$anticpx(v):=(0_V,v)$$.
• 3.1. The fixed points of $$-\chi$$ is equal to the image of $$anticpx$$.
1. Because $$f \in End_{\mathbb R+0i}(V^{\mathbb C})$$, $$f$$ commutes with scalar multiplication by $$-1$$ and so '$$f \circ \chi = \chi \circ f$$' is equivalent to '$$f \circ (-\chi) = (-\chi) \circ f$$'.

2. I like to think that:

• 5a. Observation 2.1 and $$f \circ \chi = \chi \circ f$$ are what give us the $$g$$ as $$g:= cpx^{-1} \circ f \circ cpx$$: In this case, $$f \circ \chi = \chi \circ f$$ for $$V \times 0 = image(cpx)$$ gives us $$image(f \circ cpx) \subseteq image(cpx)$$.

• 5b. $$f \circ \chi = \chi \circ f$$ and Observation 3.1. don't directly give us $$h$$, in the sense that it's $$f \circ (-\chi) = (-\chi) \circ f$$ and Observation 3.1 that (directly) give us $$h:=anticpx^{-1} \circ f \circ anticpx$$: In this case, $$f \circ (-\chi) = (-\chi) \circ f$$ for $$0 \times V = image(anticpx)$$ gives us $$image(f \circ anticpx) \subseteq image(anticpx)$$.

1. We can view Conrad's Theorem 4.16 as saying that if $$f \in End_{\mathbb R+0i}(V^{\mathbb C})$$ and if $$f \circ J = J \circ f$$, then '$$f \circ \chi = \chi \circ f$$' is equivalent to '$$f=(g \oplus g)^J$$ for some $$g \in End_{\mathbb R}(V)$$'.
• 6.1. (I guess we need not say $$g$$ is unique since I guess we have that for any $$g,h \in End_{\mathbb R}(V)$$, $$g \oplus g = h \oplus h$$ on all of $$V^2$$ if and only if $$g=h$$).
1. However, it seems now that we can view Conrad's Theorem 4.16 as saying that if $$f \circ \chi = \chi \circ f$$, or equivalently, that $$f$$ decomposes into $$f=(g \oplus h)^J$$ as described in Observation 1, then '$$f \circ J = J \circ f$$' if and only if '$$g=h$$' proved as follows:
• Proof: (If) Suppose $$g=h$$. Then $$f \circ J = J \circ f$$ because for any $$g \in End_{\mathbb R}(V)$$, $$(g \oplus g)^J$$ is $$\mathbb C$$-linear. (Only if) Suppose $$f \circ J = J \circ f$$. Then $$(0_V,h(v))=f(0_V,v)=(f \circ J)(v,0_V)=(J \circ f)(v,0_V)=J(g(v),0_V)=(0_V,g(v))$$ for all $$v \in V$$. QED
1. I just realised after typing all of Observations 1 - 7 that I think Observations 1 - 7 are more for Motivation 2 than for Motivation 1.
• 8.1. For Motivation 1, I think we can think of, for any $$K$$, finding $$\chi_K$$ such that '$$f: (V^2,K) \to (V^2,K)$$ is the complexification (with respect to $$K$$) of a map' if and only if $$f$$ commutes with $$\chi_K$$.

• 8.2. For Motivation 2, I think we can think of, for any $$\gamma: V^2 \to V^2$$ such that '$$f: V^2 \to V^2$$ commutes with $$\gamma$$' is equivalent to '$$f$$ decomposes into $$f=g \oplus h$$', finding $$K_{\gamma}$$ such that '$$f$$ commutes with $$K_{\gamma}$$' is equivalent to some condition $$P(g,h)$$ on $$g$$ and $$h$$ that is equivalent to saying that '$$f^{K_{\gamma}}$$ is $$\mathbb C$$-linear'.

• 8.2.1. For example: with $$\gamma=\chi$$ and $$K=J$$, we have $$P(g,h)=$$'$$g=h$$'. With $$\gamma=\chi$$ and $$K=-J$$, I think we have $$P(g,h)=$$'$$g=-h$$'

• 8.2.2. I guess '$$\gamma: W \to W$$ such that '$$f: W \to W$$ commutes with $$\gamma$$' is equivalent to '$$f$$ decomposes into $$f=g \oplus h$$' is the definition of a 'conjugation' on an $$\mathbb R$$-vector space $$W$$ that isn't odd-dimensional if it were finite-dimensional or at least is equal to the external direct sum $$W = U \bigoplus U$$ for some $$\mathbb R$$-vector space $$U$$.

• Why are you posting this here? MathOverflow is not a blog. – Mike Miller Nov 13 '20 at 15:59
• @MikeMiller 1 - to get confirmation 2 - stackoverflow.blog/2011/07/01/… – John Smith Kyon Nov 13 '20 at 16:00
• @MikeMiller cunningham's law - if no one disagrees with answer, then i'll just assume answer is correct (at least essentially. some details could be wrong) – John Smith Kyon Nov 20 '20 at 16:27

I believe $$\chi$$ and $$J$$ are related by $$V \times 0$$.

Part I of explanation:

For the two choices of

1. $$J(v,w):=(-w,v)$$ as the almost complex structure on $$V^2$$ that we use to define complexification of both $$V$$ and $$\mathbb R$$-endomorphisms $$f$$ of $$V$$ and

2. $$V \times 0$$ as the $$\mathbb R$$-subspace of $$V^2$$ that we use to identify $$V$$,

we will uniquely get $$\chi(v,w):=(v,-w)$$ as the unique involutive $$\mathbb R$$-linear map on $$V^2$$ such that $$\chi^J$$ is $$\mathbb C$$-anti-linear and the set of fixed points of $$\chi$$ is equal to $$V \times 0$$.

In other words:

If we were to try solve for the possible $$\sigma$$'s, $$\sigma \in End_{\mathbb R} (V^2)$$, such that

1. $$\sigma \circ J = - J \circ \sigma$$,

2. $$\sigma \circ \sigma = id_{V^2}$$

3. The set of fixed points of $$\sigma$$ is equal to $$V \times 0$$, then

we would get that the unique solution to the above system of 3 equations (2 matrix equations and 1 set equation) is $$\sigma = \chi$$.

Part II of explanation:

Let $$V$$ be an $$\mathbb R$$-vector space. Define $$K \in Aut_{\mathbb R} (V^2)$$ as anti-involutive if $$K^2 = -id_{V^2}$$. Observe that $$K$$ is anti-involutive on $$V^2$$ if and only if $$K$$ is an almost complex structure on $$V^2$$. Let $$\Gamma(V^2)$$ be the $$\mathbb R$$-subspaces of $$V^2$$ that are isomorphic to $$V$$ (i.e. $$\mathbb R$$-subspaces of $$V^2$$ except for $$V^2$$ and $$0$$). Let $$AI(V^2)$$ and $$I(V^2)$$ be, respectively, the anti-involutive and involutive maps on $$V^2$$.

Conrad's Theorem 4.11 without reference to complex numbers seems to be able to be restated as:

Let $$V$$ be $$\mathbb R$$-vector space. Let $$J(v,w):=(-w,v)$$. There exists a bijection between $$\Gamma(V^2)$$ and involutive $$\mathbb R$$-linear maps that anti-commute with $$J$$.

Let $$V$$ be an $$\mathbb R$$-vector space. Let $$K \in AI(V^2)$$. There exists a bijection between $$\Gamma(V^2)$$ and involutive $$\mathbb R$$-linear maps $$\sigma$$ that anti-commute with $$K$$.
In relation to the answer in the other question (which I've started to analyse), it appears we have that $$V \times 0$$ is the '$$V^2_{re}$$' (I believe '$$V^2_{re}$$' represents an arbitrary element of $$\Gamma(V^2)$$) that we use to identify $$V$$ as an embedded $$\mathbb R$$-subspace of $$V^2$$.