Nonstandard definitions of complexifications

Solution 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 $J(v)=-u$.

Solution 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.

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 $-J$.