Why is $H^1(X, \mathcal{O}) \neq 0$ for $X = (B(0, 1)\times B(0, 2)) \cup (B(0, 2)\times B(0, 1))$? In this MathOverflow answer, David Speyer says that 
\begin{align*}
X &= \{ (z,w) \in \mathbb{C}^2 : (|z|, |w|) \in [0,1) \times [0,2) \cup [0,2) \times [0,1) \}\\
&= (B(0, 1)\times B(0, 2)) \cup (B(0, 2)\times B(0, 1))
\end{align*}
is an example of a contractible manifold such that $H^1(X, \mathcal{O}) \neq 0$. I can see that $X$ is contractible, but I don't know how to show that $H^1(X, \mathcal{O}) \neq 0$. Can we explicitly write down a $\bar{\partial}$-closed $(0, 1)$-form on $X$ which is not exact or do we need to use some sheaf cohomology?
 A: $\def\CC{\mathbb{C}}\def\cO{\mathcal{O}}$Let $U = \{ (z,w) : |z| < 1,\ |w| < 2 \}$ and $V = \{ (z,w) : |z|<2, |w|<1 \}$. So $X = U \cup V$. Note that
$$\cO(U) = \left\{ \sum a_{ij} z^i w^j : |a_{ij}| = O((1+\delta)^i (1/2+\epsilon)^j)  \ \forall \delta, \epsilon>0 \right\}$$
$$\cO(V) = \left\{ \sum b_{ij} z^i w^j : |b_{ij}| = O((1/2+\delta)^i (1+\epsilon)^j)  \ \forall \delta, \epsilon>0 \right\}$$
$$\cO(U \cap V) = \left\{ \sum c_{ij} z^i w^j : |c_{ij}| = O((1+\delta)^i (1+\epsilon)^j)  \ \forall \delta, \epsilon>0 \right\}.$$
Since $U$, $V$ and $U \cap V$ are polydiscs, they are Stein and we can compute $H^1(X, \cO)$ with respect to the Cech cover by $U$ and $V$. This tells us that $H^1(X, \cO)$ is the cokernel of $\cO(U) \oplus \cO(V) \longrightarrow \cO(U \cap V)$. If $\sum a_{ij} z^i w^j$ and $\sum b_{ij} z^i w^j$ are holomorphic on $U$ and $V$ respectively, then $|a_{ij} + b_{ij}| = O((1/2+\delta)^{\min(i,j)})$. So, for example, $\sum z^i w^i$ is holomorphic on $U \cap V$ but cannot be expressed as the sum of such a $\sum a_{ij} z^i w^j$ and $\sum b_{ij} z^i w^j$.
Turning this into an explicit $\bar{\partial}$-closed non-$\bar{\partial}$-exact $(0,1)$-form is doable but not illuminating. Let $f+g=1$ be a partition of unity with $f$ supported on $U$ and $g$ on $V$, and let $c \in \cO(U \cap V)$ represent a nontrivial Cech-cocycle. Then $\bar{\partial}(f c) + \bar{\partial}(gc) = \bar{\partial}(c) = 0$ on $U \cap V$. Define a $1$-form on $X$ to be $\bar{\partial}(f c)$ on $U$ and $-\bar{\partial}(f c)$ on $V$; this is the $\bar{\partial}$-closed non-$\bar{\partial}$-exact given by the isomorphism between sheaf and Dolbeault cohomologies. 
The usual way to think about $X$ is in terms of holomorphically convex hulls: Let $h(z,w) = \sum d_{ij} z^i w^j \in \cO(X)$ and let $(z_0, w_0) \in \CC^2$ with $|z_0 w_0| < 2$. Taking a weighted geometric mean of the bounds $|d_{ij}| = O((1+\delta)^i (1/2+\epsilon)^j)$ and $|d_{ij}| = O((1/2+\delta)^i (1+\epsilon)^j)$, we obtain $|d_{ij}| = O((|z_0|+\delta)^{-i} (|w_0|+\epsilon)^{-j})$, so $h$ extends to $(z_0, w_0)$. If we choose $(z_0, w_0)$ not to lie in $X$ (for example, $(1.2, 1.2)$), then we have given a point in the holomorphically convex hull of $X$ but not in $X$. (And, in fact, the holomorphically convex hull of $X$ is $\{ (z,w) : |z|<2, |w|<2, |zw|<2 \}$.)
A non-holomorphically convex set can never have vanishing $H^1(\cO)$. Consider the Koszul sequence $0 \to \cO \to \cO^2 \to \cO \to \CC_{(z_0, w_0)} \to 0$ where the last term is the skyscraper sheaf at $(z_0, w_0)$. Restricting to $X$, we obtain an exact sequence of sheaves $0 \to \cO \to \cO^2 \to \cO \to 0$. But the corresponding map of global sections $\cO(X)^2 \to \cO(X)$ is not surjective, because any function in the image of this map can be evaluated at $(z_0, w_0)$ and must vanish there. So there must be an $H^1(X, \cO)$ term to recieve the cokernel of $\cO(X)^2 \to \cO(X)$.
