Does the sheaf $\mathcal{O}^*$ on a complex manifold have an acyclic cover?

Question

Let $X$ be a complex manifold and let $\mathcal{O}^*$ be the sheaf nonvanishing holomorphic functions on it. Does it have an acyclic cover? That is, a cover for which all open sets and all intersections are acyclic for this sheaf.

According to this related MathSE question, the sheaf $\mathcal{O}$ does since any complex manifold can be covered by Stein manifolds.

Therefore, the result would follow if the open sets and all intersections of this cover by Stein manifolds could be taken to be contractible, due of the long exact sequence in cohomology associated to the short exact sequence

$0\to \mathbb{Z}\to \mathcal{O}\to \mathcal{O}^*\to 0\,.$

But I don't know if this can be done.

Answer 1

The answer is Yes for complex manifolds of dimension one.
Indeed for any open subset $U\subset X$ the long exact cohomology sequence associated to the exponential sequence you mention yields the fragment $$\cdots\to H^i(U,\mathcal O_X)\to H^i(U,\mathcal O_X^*)\to H^{i+1}(U,\mathbb Z)\to \cdots$$ Now $U$, like any non-compact Riemann surface, is Stein and thus $H^i(U,\mathcal O_X)=0$ for $i\geq 1$.
And $ H^{i+1}(U,\mathbb Z)=0$ for $i\geq 1$: for dimension reason if $i\geq 2$ and because $U$ is non-compact $i=1$.
In conclusion $H^i(U,\mathcal O_X^*)=0$ ($i\geq 1$) for $U$ non-compact of dimension one, which of course shows that any covering is acyclic for $\mathcal O^*_X$.

Remarks
a) On a non-compact Riemann surface all holomorphic vector bundles, of whatever rank, are trivial !
This amazing result is Theorem 30.4, page 229, of Forster's wonderful book Lectures on Riemann Surfaces.

b) The acyclicity result in the answer is completely false for a smooth algebraic curve $Y$ of dimension one over $\mathbb C$ of dimension one, for example for the algebraic curve underlying a Riemann surface.
Indeed for any open (in the Zariski topology!) subset $V\subset Y $, the group $H^1(V, \mathcal O_{Y,\operatorname {alg}}^*)=\operatorname {Pic}_{alg}(V)$ is non denumerable. This means that there are more than denumerably many non-isomorphic algebraic line bundles on $V$ which are all holomorphically trivial!

c) See also this answer.