2
$\begingroup$

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 $K=J$).
  2. 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 $cpx$.
  3. 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 $anticpx$.
  4. 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 '$f \circ (-\chi) = (-\chi) \circ f$'.

  5. 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 $image(f \circ anticpx) \subseteq image(anticpx)$.

  6. 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 $g=h$).
  7. 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
  8. 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 $U$.

$\endgroup$
1
$\begingroup$

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 $V^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.