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Suppose $(V, J)$ is a real vector space of dimension $2n$ equipped with an almost complex structure $J$ such that $J^2 = -Id$. $V$ can be realized as a complex vector space by setting $iv = J(v)$ for any $v \in V$. On the other hand one can consider the complexification of $V$, $V \otimes_{\mathbb R} \mathbb C$ and the complexification can be viewed as a complex vector space naturally. There are two almost complex structures on the complexification, one is the $\mathbb C-$linear extension of $J$ and the other is multiplication by $i$. What is the motivation behind considering the complexification in terms of doing complex geometry? Why can't we just consider the original tangent space?

I understand the complexification can be decomposed into eigenspaces of $J$, but what confuses me a bit is what is the benefit of doing so.

efron
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  • Note that with $(V,J)$, $V$ refers to the entirety of the complexified space. On the other hand, with the complexification $V \otimes_R \Bbb C$, $V$ denotes the "real" subspace of the complexification. The point, perhaps, is that even in the first case where we are not given the "real" subspace, we can build it using the eigenspace structure of $J$. – Ben Grossmann Nov 29 '21 at 16:33
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    Thanks for the clarification, but I still do not see what is the point here. For the tangent space of a complex manifold, why do we need to consider its complexification? – efron Nov 29 '21 at 17:18
  • Take this with a grain of salt because my primary experience with complexifications does not come from geometry. I would guess, however, that the point is that when one considers a complex manifold, there already exists a built-in definition of multiplication by $i$ whereas for $V \otimes_{\Bbb R} \Bbb C$, one constructs elements in order to make multiplication by $i$ make sense. In other words, the first approach is compatible with an existing complex-multiplication structure. – Ben Grossmann Nov 29 '21 at 18:44
  • I don't see why you would need to consider complexifications if your manifold comes with a tangent space that already permits multiplication by complex numbers (i.e. is a complex vector space). – Ben Grossmann Nov 29 '21 at 18:45
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    The reason one may want to complexify the real tangent bundle of an almost complex manifold $(M,J)$ is that the eigenbundles $T^{1,0}M$ and $T^{0,1}M$ of $J$ contain useful information about $J$. For example integrability of the $J$ can be determined on these bundles or their dual. If your manifold is already complex then the complexified tangent space is isomorphic to the tangent space of the complexification of the underlying real analytic manifold. This also sees application every now and then. – Thomas Kurbach Nov 30 '21 at 16:38

2 Answers2

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Say $V$ is a real vector space of dimension $2n$, and $J \colon V \to V$ is an operator squaring to $-1$. Then (as you know) the pair $(V, J)$ defines a complex vector space of dimension $n$, with $i \cdot v := Jv$ giving the action of $i$. This means that in any basis, the matrix of $J$ is an $n \times n$ diagonal matrix, with every diagonal entry equal to $i$. Stated in another way, $J$ now satisfies a linear polynomial $J - i = 0$, rather than a quadratic $J^2 + 1 = 0$.

On the other hand, the complexification $V_\mathbb{C} := V \otimes \mathbb{C}$ is also a complex vector space, this time of dimension $2n$, and the action of $i$ is given by $i \cdot (v \otimes z) = v \otimes iz$. The complexification $J_{\mathbb{C}} := J \otimes \operatorname{id}_\mathbb{C}$ is also an operator on this space, sending an element $v \otimes z$ to $Jv \otimes z$. $J_\mathbb{C}$ still satisfies the quadratic $J_\mathbb{C}^2 + 1 = 0$, and hence splits $V_\mathbb{C}$ into two subspaces: the $+i$ eigenspace and the $-i$ eigenspace of $J_\mathbb{C}$. We can see that $$ J_\mathbb{C}(v \otimes 1 - Jv \otimes i) = Jv \otimes 1 + v \otimes i = i(v \otimes 1 - Jv \otimes i),$$ $$ J_\mathbb{C}(v \otimes 1 + Jv \otimes i) = Jv \otimes 1 - v \otimes i = -i(v \otimes 1 + Jv \otimes i),$$ and by further playing around with these eigenspaces you can see that they are each $n$-dimensional. Therefore the matrix of $J_\mathbb{C}$ in an appropriate basis here is diagonal, with $n$ entries equal to $i$, and $n$ entries equal to $-i$.

So in the case of the complex structure $(V, J)$, $J$ simply acts as the $n \times n$ multiplication-by-$i$ scalar operator, while in the complexification $V_\mathbb{C}$, the operator $J_{\mathbb{C}}$ is a particular $2n \times 2n$ diagonalisable operator with eigenvalues $\pm i$ (in particular, it is nothing like a complex structure, but still an operator of a special form).

As for why one wants to consider complex tangent bundles by occasionally forgetting their complex structure and then remembering them again, I would draw an analogy with taking real and complex parts of a complex number (although this analogy is a bit dubious, as a "real structure" on a complex vector space is a different thing...), it allows one to work with all the usual tools one has for real vector spaces or manifolds, then try to use the complex structure to deduce things for the complex manifold.

Joppy
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In fact, if $(V, J)$ is a real vector space equipped with a complex structure $J$, there are two ways to endow $V$ with the structure of a complex vector space:

  1. $i \cdot v := Jv$, denote this complex vector space by $V^+$;
  2. $i \cdot v := -Jv$, denote this complex vector space by $V^-$.

Which of these complex vector space structures is more canonical?

Now, what is the complexification $V_{\mathbb{C}}$ of $V$, it is the direct sum of two complex vector space:$$ V_{\mathbb{C}}=V^+\oplus V^- $$

Yang
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