Let $(M, J, h)$ be a complex Hermitian manifold with Levi-Civita connection $\nabla$. If we see $\nabla$ as an operator $\nabla: \Gamma(TM) \rightarrow \Gamma(\Omega^1(M)) \otimes \Gamma(TM)$, and $\pi: \Gamma(\Omega^1(M) ) \otimes \Gamma(TM) \rightarrow \Gamma(\Omega^{0,1}(M)) \otimes \Gamma(TM)$ is the projection on the first factor, then $\nabla^{0,1}:= \pi \circ \nabla$ by definition.
I'm trying to see that $\nabla^{0,1}_X = 1/2 (\nabla_X + J \nabla_{JX} ) $
(where $\nabla^{0,1}_X (Y) := \nabla^{0,1}(Y)(X)$; in other words, if $\nabla (Y) = \alpha \otimes Z$, then $\nabla^{0,1}_X (Y) = \alpha^{0,1} (X) \otimes Z$)
My attempt:
First a comment: $TM$ becomes a complex vector bundle with multiplication by $i$ given by $i X := JX \in \Gamma(TM)$. From what I know, this has absolutely nothing to do with multiplying by $i$ in the complexification $TM^\mathbb{C}:= TM \oplus (iTM)$, where multiplying by $i$ simply moves the vector $X \in TM$ to the vector $iX \in iTM$. This will become relevant as we can express any $X \in TM$ as $1/2(X - iJX) + 1/2(X + iJX)$ where $i$ is the second multiplication, i.e. not the one induced by $J$ on $TM$.
Now for the attempt, say $\nabla(Y) = \alpha \otimes Z$, so $\nabla_X Y = \alpha(X) \otimes Z$. By definition $\nabla_X^{0,1} Y = (\alpha^{0,1})(X) \otimes Z$.
But because $\alpha^{0,1}(X) = \alpha^{0,1}(X_{1,0} + X_{0,1}) = \alpha^{0,1} (X_{0,1})$ and $\alpha(X_{0,1}) = \alpha^{1,0}(X_{0,1}) + \alpha^{0,1}(X_{0,1}) = \alpha^{0,1}(X_{0,1})$, we have: $$ \nabla_X^{0,1} Y = \alpha(X_{0,1}) \otimes Z = \nabla_{X_{0,1}} (Y) = \nabla_{1/2(X+iJX)} ( Y ) $$
The rightmost term is $1/2(\nabla_X Y + i\nabla_{JX} Y) )$. So it seems that I need to show that $i\nabla_{JX} Y = J \nabla_{JX} Y$. I don't know why this is true, given that the $i$ appearing in the equation is not the action of $J$ but it is multiplication by $i$ in the complexified bundle $TM \oplus (iTM)$, as I mentioned in the comment. In fact, if $i \nabla_{JX} Y = J \nabla_{JX} Y$, doesn't this mean precisely that $\nabla_{JX} Y$ is a $(1,0)$-vector field? Why would this be guaranteed?
(In the original post, I asked the question in the context where $h$ is Kähler, this is the context I'm interested in, but I don't see how being Kähler would make a difference)