Restriction of complexification is almost complex but not conversely

abstract-algebraalmost-complexcomplex-analysiscomplex-geometrylinear algebra

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:


Notations: Let $V$ be a $\mathbb C$-vector space. Let $V_{\mathbb R}$ be the realification of $V$. For any almost complex structure $I$ on $V_{\mathbb R}$, denote by $(V_{\mathbb R},I)$ as the unique $\mathbb C$-vector space whose complex structure is given $(a+bi) \cdot v := av + bI(v)$.

  • Let $W$ be an $\mathbb R$-vector space. Let $W^{\mathbb C}$ denote the complexification of $W$ given by $W^{\mathbb C} := (W^2,J)$, where $J$ is the canonical almost complex structure on $W^2$ given by $J(v,w):=(-w,v)$.

  • For any map $f: V_{\mathbb R} \to V_{\mathbb R}$ and for any almost complex structure $I$ on $V_{\mathbb R}$, denote by $f^I$ as the unique map $f^I: (V_{\mathbb R}, I) \to (V_{\mathbb R}, I)$ such that $(f^I)_{\mathbb R} = f$. With this notation, the conditions '$f$ is $\mathbb C$-linear with respect to $I$' and '$f$ is $\mathbb C$-anti-linear with respect to $I$' are shortened to, respectively, '$f^I$ is $\mathbb C$-linear' and '$f^I$ is $\mathbb C$-anti-linear'.

  • The complexification, under $J$, of any $g \in End_{\mathbb R}W$ is $g^{\mathbb C} := (g \oplus g)^J$, i.e. the unique $\mathbb C$-linear map on $W^{\mathbb C}$ such that $(g^{\mathbb C})_{\mathbb R} = g \oplus g$.

Assumptions: Let $H$ be an almost complex structure on $V_{\mathbb R}^2$. Suppose $H^J$ is $\mathbb C$-linear.


Questions:

  1. Suppose $H^J = f^{\mathbb C} \in End_{\mathbb C}(V_{\mathbb R})^{\mathbb C}$, for some $f \in End_{\mathbb R}V_{\mathbb R}$. Restrict domain of $H$ to $V_{\mathbb R} \times 0$ to get $H|_{V_{\mathbb R} \times 0}: V_{\mathbb R} \times 0 \to (V_{\mathbb R})^2$. Observe that $image(H|_{V_{\mathbb R} \times 0}) \subseteq V_{\mathbb R} \times 0$ such that we can define $(H|_{V_{\mathbb R} \times 0})^{\tilde{}}: V_{\mathbb R} \times 0 \to V_{\mathbb R} \times 0$.

Is $(H|_{V_{\mathbb R} \times 0})^{\tilde{}}$ an almost complex structure on $V_{\mathbb R} \times 0$?

  • Guess: I think yes because: $f$ is an almost complex structure on $V_{\mathbb R}$ by this. Then (or even equivalently), $f \oplus 0$ is an almost complex structure on $V_{\mathbb R} \times 0$. Finally, $(H|_{V_{\mathbb R} \times 0})^{\tilde{}} = f \oplus 0$.
  1. Suppose $image(H|_{V_{\mathbb R} \times 0}) \subseteq V_{\mathbb R} \times 0$ and $image(H|_{0 \times V_{\mathbb R}}) \subseteq 0 \times V_{\mathbb R}$ such that we can define $(H|_{V_{\mathbb R} \times 0})^{\tilde{}}$ and $(H|_{0 \times V_{\mathbb R}})^{\tilde{}}$. Suppose $(H|_{V_{\mathbb R} \times 0})^{\tilde{}}$ and $(H|_{0 \times V_{\mathbb R}})^{\tilde{}}$ are almost complex structures on, respectively, $V_{\mathbb R} \times 0$ and $0 \times V_{\mathbb R}$. Is $H^J$ not necessarily the complexification of some $f \in End_{\mathbb R}V_{\mathbb R}$, i.e. $H^J=f^{\mathbb C}$ is not necessarily true?
  • Guess: Based on this, I think we could have $H=f \oplus g$ and $f \ne g$. Therefore, I think yes because even though $(H|_{V_{\mathbb R} \times 0})^{\tilde{}} = f \oplus 0$ and $(H|_{0 \times V_{\mathbb R}})^{\tilde{}} = 0 \oplus g$ for some $f,g \in End_{\mathbb R}V_{\mathbb R}$, we don't know that $f=g$.

Best Answer

I think my guess to Question 1 is correct, but my guess to Question 2 is incorrect.

For Question 2: I think the answer is no i.e. $H^J = f^{\mathbb C}$ for some $f$: We indeed have $H=f \oplus g$ but then because $H^J$ is $\mathbb C$-linear, we'll get that $f=g$ by II.6.1 here. Note that $f$ is the unique map such that $(H|_{V_{\mathbb R} \times 0})^{\tilde{}} = f \oplus 0$.