Does the natural exact sequence of the holomorphic jet bundle spilt? Let $X$ be a complex manifold. Let us consider the jet bundle of the trivial line bundle on $X$. We denote it as $J_{1}(\mathbb{C})$. We have the short exact sequence:
$$0\rightarrow\Omega_{X}\rightarrow J_{1}(\mathbb{C})\rightarrow \mathbb{C}\rightarrow 0$$
I think by using the same method for the smooth case we could conclude that $J_{1}(\mathbb{C})\cong \Omega_{X}\oplus \mathbb{C}$ and the short exact sequence above actually splits
. The method for the smooth case is mentioned for example in wiki: https://en.wikipedia.org/wiki/Jet_bundle
On the other hand, since $J_{1}(L)\cong J_{1}(\mathbb{C})\otimes L$ for any holomorphic line bundle. If the short exact sequence above splits, we actually know that
$$0\rightarrow\Omega_{X}\otimes L\rightarrow J_{1}(L)\rightarrow L\rightarrow 0$$
also splits. But this is too weird since we know that one version of the Euler sequence on the projective space is:
$$0\rightarrow \Omega_{\mathbb{P}^{n}}(1)\rightarrow J_{1}(\mathcal{O}(1))\rightarrow \mathcal{O}(1)\rightarrow 0$$
see for example page 34 of the book https://www.google.ca/books/edition/On_the_Geometry_of_Some_Special_Projecti/_OKKCwAAQBAJ?hl=fr&gbpv=1&dq=euler+sequence+jet+bundle&pg=PA34&printsec=frontcover
However, Euler sequence should not split in the holomorphic category.
 A: Question: "On the other hand, since J1(L)≅J1(C)⊗L for any holomorphic line bundle. If the short exact sequence above splits, we actually know that 0→ΩX⊗L→J1(L)→L→0 also splits."
Answer: If $X\subseteq \mathbb{P}^n_k$ where $k$ is the complex number field and $X$ is a projective manifold, it follows $X$ is an algebraic variety. If $L$ is a holomorphic line bundle on $X$, it follows $L$ is algebraic. The sequence
$$A1.\text{  }0 \rightarrow \Omega^1_X(L) \rightarrow J^1(L) \rightarrow L \rightarrow 0$$
splits iff $L$ has an algebraic (or holomorphic) connection and $L$ does not have such a connection in general. The splitting of your first sequence does not imply the splitting of the sequence A1. The sequence $A1$ defines a class
$$a(L) \in Ext^1_{\mathcal{O}_X}(L , \Omega^1_X(L)).$$
You may "view" the sequence $A1$ as a sequence of real $C^{\infty}$ vector bundles and as such it splits.
Your comment: "We have the short exact sequence: 0→ΩX→J1(C)→C→0"
Answer: It is unclear what you mean when you write $\mathbb{C}$ in the above sequence. Usually this sequence involves the structure sheaf $\mathcal{O}_X$ of the complex manifold $X$: If $U \subseteq X$ is an open subset it follows $\mathcal{O}_X(U)$ is the ring of holomorphic functions on $U$.
Note: If $A$ is a commutative $k$-algebra and $E$ is a left $A$-module let $\Omega:=\Omega^1_{A/k}$. Let
$$J^1(E):=\Omega\otimes_A E \oplus E$$
with the following left $A$-module structure:
$$a(\omega \otimes u,v):=((a\omega)\otimes u +d(a)\otimes v,av)$$
where $d:A \rightarrow \Omega$ is the universal derivation. It follows
$$A1.\text{  }0 \rightarrow \Omega \otimes E \rightarrow J^1(E) \rightarrow E \rightarrow 0$$
is an exact sequence of left $A$-modules and $A1$ is split iff there is a connection
$$\nabla: E \rightarrow \Omega \otimes E.$$ If $E$ is a projective $A$-module it follows $A1$ is split, hence any projective $A$-module has a connection.
Why is it "split by a connection"? If $s: E \rightarrow J^1(E)$ is a left $A$-linear splitting it follows $s(e):=(\nabla(e),e)$. It follows
$$s(ae)=(\nabla(ae),ae)=as(e):=a(\nabla(e),e):=(a\nabla(e)+da\otimes e,ae)$$
hence the map $\nabla$ is a connection on $E$.
Example: If $X$ is a complex projective manifold it follows $X$ is algebraic and has an open affine cover $U_i:=Spec(A_i)$. If $E^*$ is a locally trivial sheaf on $X$ it follows $E_i:=E^*(U_i)$ is a projective $A_i$-module and by the above construction it follows $E_i$ has a connection. There is in general no "global" holomorphic/algebraic connection on $E^*$.
