Is there a notion that $0 + \mathbb Ri$ or $\mathbb R + 0i$ has an 'internal complexification' equal to 'external complexification' of $\mathbb R$?

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 is one:

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Note: For the question stated next, I mean 'internal' and 'external' as something like internal vs external direct sum. What I mean by 'internal' and 'external' complexification is not the same as in these articles: Article 1, Article 2, Article 3

In the articles, we have 'external complexification' as complexification like $\mathbb R^2$ to $\mathbb C^2$ and 'internal complexification' as adding almost complex structures/complex structures like $\mathbb R^2$ to $\mathbb C$.

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Question: In Conrad, is there some kind of notion of 'internal' vs 'external' complexification, such that, for example, the 'internal' complexification of the $\mathbb R$-subspace $\mathbb R + 0i$ of $\mathbb C$ is actually 'literally' the same (and not merely $\mathbb C$-isomorphic) as the 'external' complexification of $\mathbb R$?

• Here, I mean 'literally' as with the following examples

  • Example 1: $GL(\mathbb R,n)$ is a set of matrices while $Aut(\mathbb R^n)$ is a set of maps so they are not 'literally' equal since they do not have the same underlying set.

  • Example 2: I consider $\mathbb R^{\mathbb C}$, $(\mathbb R^2,J)$ (see below) and $\mathbb C$ as literally equal to each other and literally unequal to $(\mathbb R^2,-J)$, even though they are all $\mathbb C$-isomorphic to each other.

• Below, I try to use the symbol '=' to denote what I believe are literal equalities (under certain conventions) and use the symbol '$\cong$' for isomorphisms.

What I understand:

1. It seems in Conrad that the 'complexification' of $\mathbb R$ is, not merely $\mathbb C$-isomorphic, but actually literally the same as both of the 'complexifications' of the $\mathbb R$-subspaces $\mathbb R + 0i$ and $0 + \mathbb Ri$ of $\mathbb C$. (I kind of don't see $\mathbb R + 0i$ as literally the same thing as $\mathbb R$, but if you do, then you may focus on $0 + \mathbb Ri = \mathbb Ri$ instead of $\mathbb R + 0i$).

2. The complexification of $\mathbb R$ to be $\mathbb R^{\mathbb C} = \mathbb C = (\mathbb R^2,J)$, with $J:\mathbb R^2 \to \mathbb R^2$, $J(u,v) := (-v,u)$, for $u,v \in \mathbb R$. Here, we have $\mathbb R^{\mathbb C}$'s underlying set ($\mathbb R^2$) to be a subset of the underlying set of $\mathbb C$ (which is also $\mathbb R^2$).

3. $(0 + \mathbb Ri)^{\mathbb C} = ((0 + \mathbb Ri)^2,K)$, with $K((0,vi),(0,wi)) := ((0,-wi),(0,vi))$. Here, we have $(0 + \mathbb Ri)^{\mathbb C}$'s underlying set as a subset of the underlying set of $\mathbb C^2$ (which is $\mathbb R^2 \times \mathbb R^2$ or $\mathbb R^4$, depending on convention).

Here's my guess: I kind of think the definition of internal complexification as that for $U$ an $\mathbb R$-subspace (or $(\mathbb R+0i)$-subspace) of a $\mathbb C$-vector space $W$, we have $U^{\text{internal}-\mathbb C} = W$ if and only if any of the following, which I think are equivalent

1. any $\mathbb R$-basis (or $(\mathbb R+0i)$-basis) of $U$ is a $\mathbb C$-basis of $W$

2. (for finite dimensions) $\dim_{\mathbb R} U = \dim_{\mathbb C} W$ and $\mathbb C$-span $U$ = $W$

   • I think of condition (1) as the generalisation of condition (2), similar to this question I just posted.

3. $iU := \{iu | u \in U\}$ is an $\mathbb R$-subspace of $W$ such that $U \cap iU = \{0_W\}$. Then $W_\mathbb R$ can be written as an internal direct sum $W_\mathbb R = U \bigoplus iU$, where $W_\mathbb R$ is $W$ treated as an $\mathbb R$-vector space.

I was also thinking something like if $U^{\text{internal}-\mathbb C} \ne W$, then at least $U^{\text{internal}-\mathbb C}$ is the unique $\mathbb C$-subspace of $W$ such that internally, $(U^{\text{internal}-\mathbb C})_{\mathbb R} = U \bigoplus iU$.

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Additional note based on the comments of reuns: I think that Suetin, Kostrikin and Mainin, like Conrad, also have some notion of internal complexification.

In Suetin, Kostrikin and Mainin, specifically 12.15 of Part I, the authors seem to be talking about how complexification is or can be seen as (by some isomorphism I guess) a specific case of the more general notion of extension of scalars, as Wikipedia does. (Note: The authors don't introduce tensor products until 3 chapters later.)

The definition is that for $\mathcal K$ a subfield of a field $K$ and for a $\mathcal K$-vector space $L$, $L$ has extension $L^{K}$, a $K$-vector space given by formal linear combinations. The definition they gave is for finite $L$, but I believe the same idea works for infinite $L$. I believe the intended definition extended to allow for infinte-dimensional $L$ is as follows:

For $L$ with basis $E=\{e_{\alpha}\}_{\alpha \in A}$

$$L = \{\sum_{j=0}^{n} b_j e_j | \text{for unique} \ b_j \in \mathcal K, e_j \in E, n \ge 0, j \in A \}$$

$$L^{K} := \{\sum_{j=0}^{n} a_j e_j | \text{for unique} \ a_j \in K, e_j \in E, n \ge 0, j \in A \}$$

Applied to $\mathcal K = \mathbb R$ and $K = \mathbb C$ (treating $\mathbb R$ as identical to $\mathbb R + 0i$), it seems then that this generalisation is what one (or maybe just 'I' instead of 'one') might call internal rather than external complexification.

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Reason this could be important and is not some nitpicking that is resolved with 'up to isomorphism':

If both $\overline {\mathbb C}$ and $\mathbb C$ have $\mathbb R$-form $0 + i\mathbb R$, but $0 + i\mathbb R$'s (internal) complexification is $\mathbb C$, then it seems $\overline {\mathbb C}$ is never the literal (internal or external) complexification of any $\mathbb R$-vector space. (Of course, this may depend on your definition of complexification, and I do kind of assume that the '$0 + i\mathbb R$' $\subseteq \mathbb C$ is the same as the '$0 + i\mathbb R$' $\subseteq \overline{\mathbb C}$.) However, that $\overline{\mathbb C}$ has conjugations but has no $\mathbb R$-forms seems to contradict Conrad Theorem 4.11.

Answer 1

Two things:

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First thing:

Conrad actually gives the definition, of $\mathbb C$-vector space $V$ as the internal complexification of $\mathbb R$-subspace $W$ of $W$, just before Theorem 3.2 as $V=W + iW$ and $W \cap iW = 0$, which I think is equivalent to saying $V_{\mathbb R} = W \bigoplus iW$ I just wish 2 things.

1. That this definition was more explicit. Perhaps as something like

Definition 3.2 We say that $V$ is the internal complexification, or just complexification if the context is clear, of $W$ if $V=W + iW$ and $W \cap iW = 0$

2. That we could define internal complexifications of any $\mathbb R$-subspace $W$ of $V$ as something like $W \bigoplus iW$ or like $\mathbb C$-span$(W)$. In this case of $W \bigoplus iW$, I think we would have to restrict to considering $\mathbb R$-subspaces such that $W \cap iW = 0$, thus excluding $W = V_{\mathbb R}$

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Second thing: The unresolved issue is whether or not Suetin, Kostrikin and Mainin has implicit notion for internal complexification such that it is internal and not external complexification that gets generalised by 12.15 of Part I (and 12.4 of Part I).