Is there a version of the adjunction formula for subvarieties of $\mathbb{P}^n\times\mathbb{P}^m$ I want to try and find an example of a projective variety which does not have an ample canonical or anti-canonical bundle. I think I should try and look at a subvariety of $\mathbb{P}^n\times\mathbb{P}^m$ which has a canonical bundle of the form $\mathcal{O}(k,-l)$, which is not ample/anti-ample. Does there exist an adjunction formula for bihomogeneous smooth complete intersections inside $\mathbb{P}^n\times\mathbb{P}^m$, or more generally, for any $\prod^k_i \mathbb{P}^{n_i}$? 
 A: By the way, the canonical bundle of $\mathbb P^n \times \mathbb P^m$ is just $\mathcal O(- n - 1, - m - 1)$. If $X \subset \mathbb P^n \times \mathbb P^m$ is the vanishing locus of a homogenous polynomial of bidegree $(d, e)$, the canonical bundle of $X$ is $\mathcal O (d - n - 1, e - m - 1)$. A similar result applies for complete intersections.
For example, the complete intersection of four bilinear polynomials in $\mathbb P^3 \times \mathbb P^3 $ is a K3 surface (if smooth), and this has a trivial canonical bundle. 
But from the way you phrased your question, I guess you are actually looking for a variety whose canonical bundle is not ample, anti-ample or trivial. Take the previous example, and chose your four bilinear polynomials to be symmetric under the $\mathbb Z_2$ involution exchanging the two $\mathbb P^3$'s. Then quotient by this $\mathbb Z_2$ involution. As long as you choose the polynomials generically enough that the K3 surface doesn't intersect the diagonal in $\mathbb P^3 \times \mathbb P^3 $, the surface that you get after quotienting is smooth. This quotient surface is an Enriques surface. Its canonical bundle is non-trivial. However, the square of its canonical bundle is trivial, which means its canonical bundle cannot be ample or anti-ample either.
Edit: If you want a variety whose canonical bundle is neither ample nor anti-ample and such that no power of the canonical bundle is trivial, you could take $\mathbb P^2$ blown up at nine points. The self-intersection of the canonical bundle of the blow-up $X$ is $K_X^2 = K_{\mathbb P^2}^2 - 9 = 0$, which immediately implies that the canonical bundle of the blow-up cannot be ample or anti-ample. If some multiple of $K_X$ is trivial, then we would have $K_X.E = 0$ for any exceptional divisor $E$ on $X$, so the genus formula gives $g(E) = 1 + \frac 1 2 (K_X.E + E^2) = 1 + \frac 1 2 (0 -1) = \frac 1 2$, which is a contradiction since $g(E)$ is in fact $0$.
