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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:

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 $U$.

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    $\begingroup$ Why are you posting this here? MathOverflow is not a blog. $\endgroup$ – Mike Miller Nov 13 '20 at 15:59
  • $\begingroup$ @MikeMiller 1 - to get confirmation 2 - stackoverflow.blog/2011/07/01/… $\endgroup$ – John Smith Kyon Nov 13 '20 at 16:00
  • $\begingroup$ @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) $\endgroup$ – John Smith Kyon Nov 20 '20 at 16:27
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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 $J$.

And then possibly (I ask about this here) generalised to:

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 $K$.

Part III of explanation:

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 $V^2$.

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  • $\begingroup$ Note for other readers: this self-answer of the OP's original question appears to be a verbatim copy of content already posted months ago as a self-answer at math.stackexchange.com/a/3537806 $\endgroup$ – Yemon Choi Nov 25 '20 at 14:03
  • $\begingroup$ I do not allege any attempt by the OP to mislead, since it seems this was done in the belief/ethos that "StackExchange sites in general believe that self-answering one's questions ca create good content for the site". Discussion has already been taking place on meta.MO regarding the appropriateness or otherwise of such action $\endgroup$ – Yemon Choi Nov 25 '20 at 14:04

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