What happens with negative plurigenus? It is a well known result that for a smooth, projective k-variety, the dimension of the global section $H^0(X,K_X^j)$ of $j$-powers of the canonical bundle. Also called plurigenus, are birational invariants when $j\geq 0$. I was wondering what goes wrong in the proof for negative powers of the canonical bundle.The proof that I read via a series of exercises goes like this: 


*

*Fix a dense open subset of definition of the birational map with
$U \subset X$ with $X\setminus U$ of codimension at least 2. 

*Since $K_X^j$ is locally free, and $X\setminus U$ has codim 2 or more, every regular section of $H^0(K_X^j, U)$ can be extended to a section of $H^0(K_X^j, X)$. 
-If $V$ is the dense open subset isomorphic to $U$ on $Y$.
We have $H^0(K_Y^j, Y)$ again equal to $H^0(K_Y^j, V)$. But $H^0(K_Y^j, V)$ is equal to $H^0(K_X^j,U)$.

The proof that is sketched here does not quite work. 
In the first bullet point, $U$ is chosen to be an open set such that $X \setminus U$ has codimension at least 2. In the last bullet point, it is claimed that there should be an open subset $V \subset Y$ isomorphic to $U$. But there is no reason that both of these properties should be attainable at the same time. Indeed, if $X$ and $Y$ are normal surfaces then it is impossible to find such a $U$, unless $X$ and $Y$ are isomorphic. 
The correct proof goes like this. Assume $X \dashrightarrow Y$ is a birational map, defined on an open subset $U$ with $X \setminus U$ of codimension at least 2. Then pullback of differential forms gives a map $H^0(Y,K_Y^j) \rightarrow H^0(U,K_U^j)$. Then your extension argument shows this is actually a map $H^0(Y,K_Y^j) \rightarrow H^0(X,K_X^j)$. It's trivial to see this is injective. Now apply the same argument to the inverse map.
Now it is evident where this fails for negative powers of the canonical bundle: 
there is no pullback map for sections of $K^j$ when $j<0$. 
