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:

I notice that the standard definitions of complexification of an $\mathbb R-$ vector space are as follows:

  1. In terms of direct sums, $V^{\mathbb C, sum} := (V^2,J)$ 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, $i(v,w)=(-w,v)$

  2. In terms of tensor products $V^{\mathbb C, tensor} := V \bigotimes \mathbb C$. Here, $\mathbb C$ scalar multiplication is as follows on elementary tensors $z(v \otimes \alpha) := v \otimes (z\alpha)$, for $v \in V$ and $z, \alpha \in \mathbb C$.

I notice we can have a different definition for sum $V^{\mathbb C, sum, -J} := (V^2,-J)$, where $\mathbb C$ scalar multiplication is now $i(v,w)=(-J)(v,w) := -J(v,w) := (w,-v)$.

  • Note: In this notation, $V^{\mathbb C, sum, J} = V^{\mathbb C, sum}$.

Question 1: Does $V^{\mathbb C, sum, -J}$ somehow correspond to $V^{\mathbb C, tensor, f(z)=\overline z} := (V \bigotimes \mathbb C, f(z)=\overline z)$, where $\mathbb C$ scalar multiplication is as follows on elementary tensors $z(v \otimes \alpha) := v \otimes (f(z)\alpha)$ $ = v \otimes (\overline z \alpha)$, for $v \in V$ and $z, \alpha \in \mathbb C$?

  • Note: In this notation, $V^{\mathbb C, tensor, f(z)=id_{\mathbb C}(z)} = (V \bigotimes \mathbb C)$

    • Note: Any general correspondence between almost complex structures $K$ on $V^2$ and the $f$'s on $V \bigotimes \mathbb C$ may be reserved for Question 2. For Question 1, I'm interested to see if $V^{\mathbb C, sum, -J}$ and $V^{\mathbb C, tensor, f(z)=\overline z}$ are 'more isomorphic' than $V^{\mathbb C, sum, -J}$ and $V^{\mathbb C, sum, J}$ (I think Gauthier would say they are not $\mathbb C$-isomorphic by the identity map or something) in the sense that $V^{\mathbb C, sum, -J}$ and $V^{\mathbb C, tensor, f(z)=\overline z}$ are not just $\mathbb C$-isomorphic, but $\mathbb C$-isomorphic is some unique way, I guess, like Theorem 3.1 of Keith Conrad or here.

Question 2: What are these mysterious $f$'s on $V \bigotimes \mathbb C$, and what is the (or 'a' instead of 'the') correspondence, if any, with the almost complex structures $K$ on $V^2 = (V^{\mathbb C, sum})_{\mathbb R}$ (for tensor product, I think $V \bigotimes \mathbb R^2 = (V^{\mathbb C, tensor})_{\mathbb R}$) ?

  • Update based on Yai0Phah's answer: It looks like these mysterious $f$'s are '$\mathbb R$-algebra endormophisms' of $\mathbb C$. Then each '$\mathbb R$-algebra endormophism' $f$ gives almost complex structure $J_f(v) := f(i)v$ on $V^2$. However, it might be the case that not every almost complex structure $K$ on $V^2$ comes from some '$\mathbb R$-algebra endormophisms' $f$.
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    $\begingroup$ If I am not mistaken, then first one should be correct (note that $\mathbb C\cong\mathbb R\oplus i\mathbb R$ as real vector spaces). The second one might not be as easy as you thought. The problem is that, $V\oplus V$ is no more special than any real vector spaces of even dimension, and there are pretty many almost complex structure on that. For example, let $E$ be a real vector space of even dimension, and $u,v\in E$ two linearly independent vectors, then there exists an almost complex structure $K\colon E\to E$ such that $K(u)=v$ (and consequently $K(v)=-u$). $\endgroup$ – Yai0Phah Jan 24 '20 at 12:48
  • $\begingroup$ @Yai0Phah Thanks. 1. For the first question, what then is the relationship between $f(z) = \overline z$ and $-J$? 2. What exactly is the relevance of $\mathbb C_{\mathbb R} = \mathbb R + 0i \bigoplus 0+\mathbb Ri$ here? 3. For the second question, do you mean that it might be hard to find $f$ since there are a lot of $K$'s in the sense that any 2 linearly independent vectors give rise to some (probably non-unique) $K$? 4. In your opinion, do you think this could/should be on-topic on overflow? $\endgroup$ – John Smith Kyon Jan 25 '20 at 10:14
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    $\begingroup$ By the way, it is clearly improper for MO. $\endgroup$ – Yai0Phah Jan 26 '20 at 16:08
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    $\begingroup$ Ignore my comment about $\mathbb C\cong\mathbb R\oplus i\mathbb R$. This was devoted to the "universal" case of $\mathbb C$, but seemingly more difficult to explain. $\endgroup$ – Yai0Phah Jan 26 '20 at 16:20
  • $\begingroup$ @Yai0Phah Ok, thanks. I guess we'll refer to your answer now. $\endgroup$ – John Smith Kyon Jan 27 '20 at 6:33

Let me try to clarify your questions: in fact, essentially your questions have nothing to do with complexifications. The first thing is that: all complex vectors spaces $E$ are real vector spaces (called the underlying real vector space) with an almost complex structure $J\colon E\to E$ being a map of real vector spaces given by multiplication by $i$, that is, $v\mapsto iv$. On the other hand, given a real vector space $E$, almost complex structures $J\colon E\to E$ give rise to complex vector spaces $E_J$ given by $(a+bi)v=av+bJ(v)$. The data of whether the complex vector space $E$ comes from complexification or "how" it comes from is irrelevant.

Second, given a $\mathbb C$-vector space $E$ and an $\mathbb R$-algebra morphism $f\colon\mathbb C\to\mathbb C$, we have a $\mathbb C$-vector space, denoted by $f_*E$, of which the underlying real vector space is the real vector space $E$ with the multiplication $\mathbb C\times f_*E\to f_*E$ given by $(c,v)\mapsto f(c)v$ where the multiplication $f(c)v$ is taken in the complex vector space $E$.

Given these, I would like to rephrase your two questions as follows:

  1. Let $E$ be a complex vector space which corresponds to an almost complex structure $J\colon E\to E$, and let $f\colon\mathbb C\to\mathbb C$ be the complex conjugate (which is, of course, an $\mathbb R$-algebra endomorphism). Then the almost complex structure corresponds to the complex vector space $f_*E$ is given by $-J\colon E\to E$.
  2. Let $E$ be a complex vector space. Does all almost complex structures come from those correspond to complex vector spaces $f_*E$ where $f\colon\mathbb C\to\mathbb C$ runs through all $\mathbb R$-algebra endomorphisms (Exercise: there are only two $\mathbb R$-algebra endomorphisms on $\mathbb C$)?

Then the first statement is clearly true and the second is false if $E\neq0$. As explained in the comment, it could be seen from the following proposition:

Let $E$ be a real vector space of even dimension and let $u,v\in E$ be two $\mathbb R$-linearly independent vectors, then there exists an almost complex structure $J\colon E\to E$ such that $J(u)=v$ and $J(v)=-u$.

  • $\begingroup$ Thanks Yai0Phah. 1. The $f$'s in question have to be $\mathbb R$-algebra endomorphisms? 2. Do the $f$'s at least embed in the space of almost complex structures? It looks like for every $f$, we can have some corresponding almost complex structure. For example, on $E=(E_{\mathbb R}, J)$, the identity on $\mathbb C$ corresponds to $J$ on $E_{\mathbb R}$ and conjugation on $\mathbb C$ corresponds to $-J$ on $E_{\mathbb R}$ $\endgroup$ – John Smith Kyon Jan 27 '20 at 6:17
  • $\begingroup$ Maybe we cannot find an $f$ for every $J$, but I think we can find a $J$ for every $f$: It appears that the complex structure $\mu_f: \mathbb C \times E_{\mathbb R} \to E_{\mathbb R},$ $\mu_f(z,v) := f(z)v$ uniquely corresponds (based on correspondence in Huybrechts) to the almost complex structure $J_f(v) := (\mu_f)_i(v) := \mu_f(i,v)$ such that $(a+ib) \cdot v := av + bJ_f(v) := av + b\mu_f(i,v) := av + bf(i)v$. $\endgroup$ – John Smith Kyon Jan 27 '20 at 6:22
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    $\begingroup$ @JohnSmithKyon It is not clear in the question what $f$ is supposed to be and I guess from the context that $f$ is an $\mathbb R$-algebra endomorphism. You need to clarify the notations and the definitions in the question. The clearer the question is, the more people can understand and help. $\endgroup$ – Yai0Phah Jan 27 '20 at 17:25
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    $\begingroup$ @JohnSmithKyon By the way, it seems to me that you are not fully prepared for the prerequisites of Huybrechts' book. In particular, it seems that you are not acquainted with rings (especially, commutative ones) and modules, which will be used in Huybrechts' book. The classical textbook is Atiyah-Macdonald's Commutative Algebra, and it might be helpful to read Reid's Undergraduate Commutative Algebra (with more illustrations about geometric intuitions) and look occasionally at Eisenbud's Commutative Algebra (voluminous, but more intuitive than A-M). $\endgroup$ – Yai0Phah Jan 27 '20 at 17:34
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    $\begingroup$ @JohnSmithKyon If you don't know the classification of finitely generated modules over a PID, you also need to learn this topic from, say, Artin's Algebra, or Jacobson's Basic Algebra (just go to the related chapter directly). Sorry, I don't have much time to address all questions. $\endgroup$ – Yai0Phah Jan 27 '20 at 17:35

Complexification is a functor from the category of $\mathbb{R}$-vector spaces to the category of $\mathbb{C}$-vector spaces. To specify a functor we need to specify:

  1. For each $\mathbb{R}$-vector space $V$, a way to produce a complex vector space $V^\mathbb{C}$.
  2. For each $\mathbb{R}$-linear map $g: V \to W$, a way to produce a $\mathbb{C}$-linear map $g^\mathbb{C}: V^\mathbb{C} \to W^\mathbb{C}$.

To be truly functorial, the identity map on $V$ needs to complexify to the identity map on $V^\mathbb{C}$, and composition of maps must complexify nicely: $(g \circ f)^\mathbb{C} = g^\mathbb{C} \circ f^\mathbb{C}$.

Don't forget to define how to complexify a linear map, not just the vector space.

Here are four different ways to define a complexification:

  1. By a tensor product (this is called extension of scalars): $V^\mathbb{C} = V \otimes_\mathbb{R} \mathbb{C}$ where $i \cdot (v \otimes w) = v \otimes iw$ for $z \in \mathbb{C}$. The linear map $g: V \to W$ complexifies to $g^\mathbb{C} = g \otimes 1_\mathbb{C}$.
  2. We could do the same as above, but conjugate things. So define $V^\mathbb{C} = V \otimes_\mathbb{R} \mathbb{C}$ as a real vector space, and define $i(v \otimes w) = - v \otimes iw$. We still set $g^\mathbb{C} = g \otimes 1_\mathbb{C}$.
  3. By a direct sum: $V^\mathbb{C} = V \oplus V$, where $i(v_1, v_2) = (-v_2, v_1)$. The linear map $g: V \to W$ complexifies to $g^\mathbb{C}(v_1, v_2) = (g(v_1), g(v_2))$.
  4. Doing the conjugate of the thing above: $V^\mathbb{C} = V \oplus V$, where $i(v_1, v_2) = (v_2, -v_1)$. We still set $g^\mathbb{C}(v_1, v_2) = (g(v_1), g(v_2))$.

What is the relationship between these methods of complexification, as functors? The answer is that they are all isomorphic functors, meaning that for any two of them there exists a natural transformation such that each component of the natural transformation is an isomorphism.

Consider 3 and 4, which we will differentiate by writing $V^{3 \mathbb{C}}$ and $V^{4 \mathbb{C}}$. We can define a natural transformation $\eta: (-)^{3 \mathbb{C}} \to (-)^{4 \mathbb{C}}$ by setting $$ \eta_V: V^{3 \mathbb{C}} \to V^{4 \mathbb{C}}, \quad \eta(v_1, v_2) = (v_1, -v_2).$$ We need to check that $\eta_V$ is $\mathbb{C}$-linear for each $V$: $$ \eta_V i (v_1, v_2) = \eta_V(-v_2, v_1) = (-v_2, -v_1) = i(v_1, -v_2) = i \eta_V (v_1, v_2).$$ Hence the components $\eta_V$ are all $\mathbb{C}$-linear (they lie in the correct category), and are clearly isomorphisms. We now need to verify the other condition on being a natural transformation, which is that for each map $g: V \to W$ of $\mathbb{R}$-vector spaces, we have $\eta_W \circ g^{3 \mathbb{C}} = g^{4 \mathbb{C}} \circ \eta_V$. Indeed, $$\eta_W g^{3 \mathbb{C}}(v_1, v_2) = \eta_W(g(v_1), g(v_2)) = (g(v_1), -g(v_2))$$ and $$ g^{4 \mathbb{C}} \eta_V(v_1, v_2) = g^{4 \mathbb{C}}(v_1, -v_2) = (g(v_1), -g(v_2)).$$

So indeed $\eta$ gives a natural isomorphism between the third and fourth methods of complexification. We can give a natural isomorphism from the first to the fourth method, where the natural transformation $\mu: (-)^{1 \mathbb{C}} \to (-)^{3 \mathbb{C}}$ will have components $$ \mu_V: V \otimes_\mathbb{R} \mathbb{C} \to V \oplus V, \quad \mu_V(v_1 \otimes 1 + v_2 \otimes i) = (v_1, v_2),$$ where we have used the fact that every tensor in $V \otimes_\mathbb{R} \mathbb{C}$ uniquely decomposes into the form $v_1 \otimes 1 + v_2 \otimes i$. I think this should address most of your questions about the relationships between these.

There is something else going on here: the first and second methods are "conjugates" of each other, and the third and fourth methods are "conjugates" of each other. We can formalise this as follows.

There is a (yet another) functor $\mathbb{C}$-vect to $\mathbb{C}$-vect, the complex conjugate space functor. For a complex vector space $V$, its conjugate vector space is $\overline{V}$, where $\overline{V} = V$ as sets, but with the new scalar multiplication $z \cdot v = \overline{z} v$ for $v \in \overline{V}$. Given a $\mathbb{C}$-linear map $g: V \to W$, the conjugate map $\overline{g}: \overline{V} \to \overline{W}$ is defined to be the same map of sets as $g$. (A pleasant exercise: even though $\overline{g}$ is the same map of sets as $g$, if you choose bases and write out a matrix for $g$, the corresponding matrix for $\overline{g}$ will have every entry conjugated).

The complexification methods 1 and 2 differ by composition with the conjugate functor, as do 3 and 4.

Last note: if we instead wrote every complex vector space as a pair $(V, J)$ of a real vector space $V$ and a $\mathbb{R}$-linear map $J: V \to V$ satisfying $J^2 = -1$, then the complex conjugation functor is simply $\overline{(V, J)} = (V, -J)$. From this viewpoint, a $\mathbb{C}$-linear map is just an $\mathbb{R}$-linear map commuting with $J$, and such a map also commutes with $-J$.

  • $\begingroup$ Thanks, Joppy. 1. To answer Question 1, one should prove isomorphism between second and fourth methods right? 2. Do you have answer for Question 2 please? 3. Based on this and this, why do we still have the same complexifications of maps in Methods 2 and 4 as, respectively, Methods 1 and 3? I'm thinking we don't really need to so we could even have additional Methods 2A and 4A where the complexifications of maps are now different from, respectively, Methods 1 and 3. $\endgroup$ – John Smith Kyon Jan 27 '20 at 6:02
  • $\begingroup$ Note: I slightly edited post. $\endgroup$ – John Smith Kyon Jan 27 '20 at 6:33
  • $\begingroup$ 4. Joppy, actually also, $V^2 = (V^{\mathbb C, sum})_{\mathbb R}$ and then $V \bigotimes \mathbb R^2 = (V^{\mathbb C, tensor})_{\mathbb R}$ ? I understand in the latter case that the '$\bigotimes$' in $V \bigotimes \mathbb R^2$ has a different definition from the '$\bigotimes$' in $V \bigotimes \mathbb R^2$. However, I think $V \bigotimes \mathbb R^2 = (V^{\mathbb C, tensor})_{\mathbb R}$ because $V \bigotimes \mathbb R^2 \cong V \bigotimes (\mathbb R \bigoplus \mathbb R)$ $ \cong (V \bigotimes \mathbb R) \bigoplus (V \bigotimes \mathbb R)$ $ \cong (V) \bigoplus (V) = V^2$ $\endgroup$ – John Smith Kyon Jan 27 '20 at 6:45
  • $\begingroup$ Joppy, never mind my follow-up question 3 ('Based on this and this...'). I started to analyse your answer and so I get why $(f^{\mathbb C, J})_{\mathbb R} = (f^{\mathbb C, -J})_{\mathbb R}$ $\endgroup$ – John Smith Kyon Feb 5 '20 at 3:26

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