A [recent MO question about non-rigorous reasoning](https://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics) reminded me of something I've wondered about for some time.
The genus–degree formula says that genus $g$ of a nonsingular projective plane curve of degree $d$ is given by the formula $g=(d−1)(d−2)/2$. Here is a heuristic argument for the formula that someone once told me. Take $d$ lines in general position in the plane; collectively these form a (singular) degree-$d$ curve. There are $d\choose 2$ points of intersection. Now think in terms of complex numbers and visualize each line as a Riemannian sphere. If you start with $d$ disjoint spheres and then bring them together so that every one touches every other one (deforming when necessary) then you expect the genus of the resulting surface to be ${d\choose 2}−(d−1)=(d−1)(d−2)/2$, because after you connect them together in a line with $d−1$ connections, each subsequent connection increases the genus by one.



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> Is there any rigorous proof of the genus–degree formula that closely follows the above line of argumentation?
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A standard proof of the genus–degree formula proceeds by way of the adjunction formula. This doesn't seem to me to answer my question, but perhaps I just don't understand the adjunction formula properly?



Rigorous version of heuristic argument for genus-degree formula?
Yes, this argument can be made rigorous. One needs three steps.


**Step 1.** Show that there is *at least one* smooth plane curve of degree $d$ with the expected genus. Essentially, the proof is given by your heuristic topological argument (deform the union of $d$ lines in general position).


**Step 2.** Show that if one slightly perturbs the coefficients of a homogeneous polynomial defining a smooth curve, the genus remain unchanged. This is basically a continuity argument.


**Step 3.** Show that the space $\mathbb{C}^{\rm nonsing}[x,\,y,\,z]\_d$ of homogeneous polynomials of degree $d$ in three variables defining smooth curves is path-connected. This is because the complement of $\mathbb{C}^{\rm nonsing}[x,\,y,\,z]\_d$ in $\mathbb{C}[x,\,y,\,z]\_d$ (the so-called "discriminant locus") has real codimension $2$.


Putting these three steps together one easily obtains the desired result. For further details you can look at Chapter 4 of Kirvan's book *Plane algebraic curves*. 


