Nonstandard definitions of complexifications

Question

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:

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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, $V^{\mathbb C, sum, J} = V^{\mathbb C, sum}$.

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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' $f$.

Answer 1

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 $J(v)=-u$.

Answer 2

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.

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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.

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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 $-J$.