In what way is $(L_{\mathbb R})^{\mathbb C}$ more like $L \bigoplus \overline L$ than like $L \bigoplus L = L^2$

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 is one:

Question:

Let $L$ be $\mathbb C$-vector space, possibly infinite-dimensional. From Suetin, Kostrikin and Mainin (see 12.13 of Part I), Wikipedia and (implicitly) Daniel Huybrechts, Complex Geometry An Introduction (see Chapter 1.2), we get that $(L_{\mathbb R})^{\mathbb C}$ is $\mathbb C$-isomorphic to an external direct sum: $(L_{\mathbb R})^{\mathbb C} \cong L \ \text{external-}\bigoplus \ \overline L$ in a 'canonical' way.

How exactly does this 'canonical' $\mathbb C$-isomorphism make us think of $(L_{\mathbb R})^{\mathbb C}$ as more like $L \bigoplus \overline L$ than like $L \bigoplus L = L^2$ ? I think of something like 'unique' isomorphisms as asked about in this post. I might be confusing the terms 'canonical' and 'unique'. Also, this post might be relevant.

My understanding of this:

We have the literal (not just isomorphism) internal direct sum $$(L_{\mathbb R})^{\mathbb C} = (L_{\mathbb R}^2,J) = (L^{1,0},J^{1,0}) \ \text{internal-} \ \bigoplus (L^{0,1},J^{0,1})$$

where

  • $J$ is the almost complex structure on $L_{\mathbb R}^2$, $J(l,m):=(-m,l)$,

  • $J^{1,0}$ is $J$ with domain and range restricted to $L^{1,0}$ (we can check that $J(L^{1,0}) \subseteq L^{1,0}$) such that $J^{1,0}$ is an almost complex structure on $L^{1,0}$

  • and $J^{0,1}$ is $J$ with domain and range restricted to $L^{0,1}$ (we can check that $J(L^{0,1}) \subseteq L^{0,1}$) such that $J^{0,1}$ is an almost complex structure on $L^{0,1}$.

Then

  1. Let $\hat i$ be the unique almost complex structure on $L_{\mathbb R}$ such that $L=(L_{\mathbb R}, \hat i)$. We have that $L$ and $(L^{1,0},J^{1,0})$ are $\mathbb C$-isomorphic by $\gamma_L(l)=(l,-\hat i(l))$.

  2. $\overline L=(L_{\mathbb R}, -\hat i)$ and $(L^{0,1},J^{0,1})$ are $\mathbb C$-isomorphic by $\gamma_{\overline L}(l)=(l,\hat i(l))$

  3. Finally, the isomorphism is $f = (\varphi \circ (\gamma_L \ \text{external-}\oplus \ \gamma_{\overline L}))^{-1}$, where $\varphi$ is the standard $\mathbb C$-isomorphism between internal and external direct sums: $\varphi: (L^{1,0},J^{1,0}) \ \text{external-} \ \bigoplus (L^{0,1},J^{0,1}) \to (L^{1,0},J^{1,0}) \ \text{internal-} \ \bigoplus (L^{0,1},J^{0,1})$.

Guess: Based on this post and this post (and 3 of my other posts: Post 1, Post 2, Post 3), I guess canonical/natural just means basis-free, i.e. we don't need axiom of choice, instead of saying that $(L_{\mathbb R})^{\mathbb C}$ is 'more like' $L \bigoplus \overline L$ than like $L \bigoplus L = L^2$ unless an isomorphism constructed without axiom of choice is 'more' of an isomorphism than one constructed with the axiom of choice. I don't really bother to think of 'unique' isomorphism anymore. I think only of 'canonical'/'natural' isomorphism as in basis-free i.e. no axiom of choice.

Best Answer

Let $L$ be a complex vector space, with $L_\mathbb{R}$ its underlying real vector space. We can then complexify the underlying real vector space to get $K = L_\mathbb{R} \otimes_\mathbb{R} \mathbb{C}$. The question is: why is $K$ naturally isomorphic to $L \oplus \overline{L}$?

First lets try to figure out how to split up $K$ into two canonically defined $\mathbb{C}$-subspaces. Consider $J: L_\mathbb{R} \to L_\mathbb{R}$, the complex structure on $L_\mathbb{R}$ coming from multiplication by $i$ in $L$. What we can do is complexify $J$ to get a $\mathbb{C}$-linear map $J^\mathbb{C}: K \to K$. Since $(J^\mathbb{C})^2 = -1$, the complex vector space $K$ decomposes into the $(+i)$ and $(-i)$ eigenspaces of $J^\mathbb{C}$, lets call these $K = K_i \oplus K_{-i}$.

Consider the $\mathbb{R}$-linear map $p_i = (L \to L_\mathbb{R} \to K \to K_i)$, where the last map is projection along the eigendecomposition $K = K_i \oplus K_{-i}$. Then this map is in fact $\mathbb{C}$-linear, since $p_i(Jl) = i p_i(l)$ for all $l \in L$. Furthermore, $p_i$ is an isomorphism (why? consider the kernel of the last projection...) and so we've found a $\mathbb{C}$-linear isomorphism $L \to K_i$. If you do the same thing for $K_{-i}$, you get a $\mathbb{C}$-antilinear isomorphism $p_{-i}: L \to K_{-i}$, and hence $K_{-i}$ looks like the complex conjugate vector space $\overline{L}$.

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