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 $J$.


1 Answer 1


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 $$0 \neq \frac 1 2 (X + iJ X) \in T^{0,1} \subset TM_{\mathbb C}.$$

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    $\begingroup$ In other words, $JX = iX$ only when $X \in T^{1,0}$. $\endgroup$ Sep 27, 2021 at 8:42
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    $\begingroup$ @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. $\endgroup$ Sep 27, 2021 at 11:08

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