# Confusion on the definition of a complex structure

I am reading the notes of Moroianu's Lectures on Kahler Geometry: https://moroianu.perso.math.cnrs.fr/tex/kg.pdf. On page 30 we want to prove that the Levi-Civita connection and Chern connection coincide iff the manifold is Kahler. It was first note that $$TM$$ can be viewed as a complex vector bundle, as the complex structure $$J$$ allow us to identify $$iX:= JX$$.

In the first line of the proof of Lemma 5.7, we have $$\bar{\partial}f(X) = \frac{1}{2}(X+iJX)(f)$$. My question is that, is this not just $$0$$ using the identification we made?

I believe that I have probably understood something wrongly here, but I cannot warp my head around what is going on.

EDIT: Thank you all for the response below, which was very helpful. In the book version of this notes the author clarified further what he meant in that case:

In other words, a product $$iX$$ for some $$X\in TM$$ is identified with $$JX$$. Since this point is particularly confusing, we insist a little more on it: we don’t say that $$iX = JX$$ on $$TM$$ (this actually would make no sense because $$TM$$ is a real bundle), we just say that the complex structure on $$TM$$ (which is usually denoted by $$i$$ on vector bundles) is, in this case, given by the tensor $$J$$.

This smells like the usual confusion of complex and real tangent bundles. Let me try to clarify the situation, especially where the identification $$J = i$$ arises. I hope this clarifies something, though I admit that this also confused me for a while.
If $$M$$ is a complex manifold, we have the real tangent bundle $$TM$$, which comes with an endomorphism $$J: TM \to TM$$, determined by the complex structure. We may also consider the complexified tangent bundle $$TM_{\mathbb C} = TM \otimes_{\mathbb R} \mathbb C$$. The endomorphism $$J$$ uniquely extends to a $$\mathbb C$$-linear endomorphism $$J_{\mathbb C}: TM_{\mathbb C} \to TM_{\mathbb C}$$. But we also have the standard complex multiplication on $$TM_{\mathbb C}$$, so for each $$v \in TM_{\mathbb C}$$ we might consider $$i \cdot v \in TM_{\mathbb C}$$. Note however that this is not the same as applying $$J_{\mathbb C}$$, which usually leads to the definitions of the complex subbundles of $$TM_{\mathbb C}$$ $$T^{1,0} = \{v \in TM_{\mathbb C}: J_{\mathbb C}(v) = i \cdot v \} \quad \text{and} \quad T^{0,1} = \{v \in TM_{\mathbb C}: J_{\mathbb C}(v) = -i \cdot v\}.$$ So $$T^{1,0}$$ is the eigenspace to the eigenvalue $$i$$, and $$T^{0,1}$$ is the eigenspace to the eigenvalue $$-i$$. The minimal polynomial of $$J_{\mathbb C}$$ is $$X^2 + 1 = (X - i)(X + i)$$, so those bundles give a direct sum decomposition $$TM = T^{1,0} \oplus T^{0,1}.$$ Also note that the $$\mathbb R$$-linear conjugation $$v = u + iw \mapsto u - iw = \overline{v}$$ for $$u, w \in TM$$ maps $$T^{1,0}$$ to $$T^{0,1}$$, because the $$\mathbb C$$-linearity of $$J_{\mathbb C}$$ gives for $$v \in T^{1,0}$$ $$J_{\mathbb C}(\overline v) = \overline{J_{\mathbb C}(v)} = \overline{iv} = -i \overline v.$$ So we have $$\overline{T^{1,0}} = T^{0,1}$$. This actually proves my claim that the application of $$J_{\mathbb C}$$ is not the same as multiplication by $$i$$ on $$TM_{\mathbb C}$$, since $$\dim T^{1,0} = \dim T^{0,1}$$.
However, the application of the real part $$\Re$$ to $$TM_{\mathbb C}$$ you will give you a complex isomorphism $$\Re: T^{1,0} \to TM$$, and in this way we might think of $$TM$$ as the subbundle $$T^{1,0}$$ sitting inside $$TM_{\mathbb C}$$.
So I guess in your notes you should consider $$0 \neq \frac 1 2 (X + iJ X) \in T^{0,1} \subset TM_{\mathbb C}.$$
• In other words, $JX = iX$ only when $X \in T^{1,0}$. Sep 27, 2021 at 8:42
• @ArcticChar Yeah for $X \in TM_{\mathbb C}$ you can express it in that way. Though I guess that the confusion comes from defining $iX := JX$ on $TM$ (which is not even wrong), without realising that one works in $TM_{\mathbb C}$, where the two operations differ. At least that is where my confusion came from when I learned that stuff. Sep 27, 2021 at 11:08