(I have almost no background in algebraic geometry, so excuse the naivete of my question.)
Some ideas I've been thinking about recently led to the question of whether there is a notion of 'cobordism' for algebraic curves-- specifically, an explicitly geometric kind that would e.g. allow one to get some analogue of a 'cobordism category' with smooth manifolds being replaced by algebraic curves. I have stumbled across the page https://ncatlab.org/nlab/show/algebraic+cobordism, and found this statement about algebraic cobordism:
The (2n,n) -graded part has a geometric description via cobordism classes, at least over fields of characteristic zero.
So, my question is just whether 'algebraic cobordism' allows one to define constructions something like what I described above with cobordism classes coming from this '(2n,n) graded part' of algebraic cobordism, and how similar geometrically this is to the smooth manifold case; if there is an intuitive description with little AG technicalities that would be nice, but I do not expect one.
Actually there is. In fact there are several, but if you want the most explicit one it is given by "double degeneration relations" of Levine and Pandharipande.
The basic idea is the following, you want to consider fibers of projection to $\mathbb{P}^1$ to be "cobordant", but you need to be cautious here because if you want a functorial theory, you need to take into considerations not only varieties over $k$, but also what one may call relative varieties, which are simply projective morphisms $X\to Y$ with $X$ being smooth, because those may impose extra relations. Also you want to be a bit stringent on which fibers you allow.
You need to know when you want two such relative cycles to be cobordant, if $V$ is a (quasi projective integral) variety and $V\to \mathbb{P}^1$ a dominant morphism then all fibers are equidimensional and if two happen to be smooth you can declare them to be cobordant, more precisely the relative cycles $V_0 \to V$ and $V_\infty \to V$ are declared cobordant if the fibers in questions are above $0$ and $\infty$ respectively.
But that's not sufficient you need to add up another type of relation coming from fibers which are not smooth but are the transverse intersection of two smooth effective cartier divisors. Since the intersection is transverse, the local equation defining the two divisors form a regular sequence and we have $N_{Y/E}\otimes N_{Y/D} \simeq \mathcal{O}_Y$ with $Y$ being the smooth intersection of the two (possibly non connected), here $N=N_{Y/E}$ is the normal bundle of $Y$ in $E$.
This tells you that the projective bundle $\mathbb{P}(1\oplus N)$ does not depend on whether you choose $N=N_{Y/E}$ or $N=N_{Y/D}$. The extra relation that you impose is that in the previous context with $f:V \to \mathbb{P}^1$ being given, if you have a smooth fiber (say $V_0$) and a fiber of the preceding type, you impose $[V_0 \to V]-[E\to V]-[D\to V]+ [\mathbb{P}(1\oplus N)\to V]$ to vanish.
This relationship exactly comes from the deformation to the normal cone (for a regular immersion), this is typically the exceptional fiber of the deformation to the normal cone that has such a behavior, two smooth Cartier divisors intersecting transversely.