Why is every complex manifold not Kähler? I'm playing around with what happens when we have a short exact sequence
$$
0 \longrightarrow T_X \stackrel{j}{\longrightarrow} E \stackrel{g}{\longrightarrow} Q \longrightarrow 0
$$
of holomorphic vector bundles over a complex manifold $X$. The question I'd like to answer is: Suppose we put a Hermitian metric on the middle bundle $E$. When is the induced metric on $X$ Kähler?
This can happen; submanifolds of Kähler manifolds are Kähler, and the induced metric arises in this way. This can also not happen; take a non-Kähler manifold and consider it as a submanifold in itself times a curve. No matter what metric we put on $E$, we won't get a Kahler metric on $X$.
Luckily this situation has been studied before, in particular by Demailly in his book (Theorem 14.3, page 273). Using that reference, we argue as follows:
The torsion tensor $\tau$ of the induced metric on $X$ is the exterior covariant derivative of the identity morphism on $T_X$. We have $\operatorname{id}_{T_X} = j^\dagger \circ j$, where $j^\dagger : E \to T_X$ is the adjoint to $j$. Then
$$
\tau
= D_{\operatorname{End}(T_X)} \operatorname{id}
= D_{\operatorname{Hom}(E,T_X)}j^\dagger \circ j
+ j^\dagger \circ D_{\operatorname{Hom}(T_X,E)}j.
$$
By Demailly, we have
$$
D_{\operatorname{Hom}(E,T_X)}j^\dagger
= \beta^\dagger \circ g,
$$
where $\beta$ is the second fundamental form of $T_X$, and
$$
D_{\operatorname{Hom}(T_X,E)}j
= g^\dagger \circ \beta.
$$
This gives
$$
\tau
= \beta^\dagger \circ g \circ j + j^\dagger \circ g^\dagger \circ \beta
= 0
$$
because $g \circ j = 0$ and $j^\dagger \circ g^\dagger = 0$.
Thus the torsion tensor of the Chern connection of the induced metric on $X$ is zero, so that metric is Kähler. As the tangent bundle of any manifold trivially fits into a short exact sequence
$$
0 \to T_X \to T_X \oplus \mathcal O_X \to \mathcal O_X \to 0
$$
we conclude that all complex manifolds are Kähler. Hurray!
Clearly this is not true, but I can't find the mistake in this argument. The induced connection is really the Chern connection, and as far as I can tell this calculates its torsion. What is the mistake here?
 A: I think I've figured out where this goes wrong (nothing like asking people in public and then sleeping on the problem).
The problem with this "proof" is that what we actually calculate is the covariant derivative of the identity, and not the covariant exterior derivative. If $D$ is a connection on a vector bundle $E$, we have
$$
D_E s
= D_E(\operatorname{id}_{E} s)
= (D_{\operatorname{End} E}\operatorname{id}_E)(s) + \operatorname{id}_E(D_Es)
= (D_{\operatorname{End} E}\operatorname{id}_E)(s) + D_Es
$$
for any section $s$ of $E$, so $D_{\operatorname{End} E}\operatorname{id}_E = 0$.
Now, the torsion tensor is the covariant exterior derivative $d^D$ of the identity morphism considered as a one-form with values in $E = T_X$, and the two notions are not the same. If $f \in \operatorname{End} E$ then $D_{\operatorname{End} E}f \in \mathcal C^{\infty}(X, T_X^* \otimes \operatorname{End} E)$. For $T_X$, this is $\mathcal C^\infty(X,T_X^* \otimes T_X^* \otimes T_X)$. Meanwhile the torsion tensor $\tau = d^D \operatorname{id}_{T_X}$ is antisymmetric, and thus an element of $\mathcal C^\infty(X,\bigwedge^2 T_X^* \otimes T_X)$. What the above argument establishes at some length is that $D_{\operatorname{End} T_X} \operatorname{id}_{T_X} = 0$, which is true for any connection as we saw above.
