Theorem: Definitions of Genera are Equivalent

In appropriate circumstances, the following definitions of genus are equivalent:
=== Genus of Manifold ===
{{:Definition:Genus of Manifold}}
=== Genus of Riemann Surface ===
{{:Definition:Genus of Riemann Surface}}
=== Genus of Plane Algebraic Curve ===
{{:Definition:Genus of Plane Algebraic Curve}}

Proof:
{{tidy|''please'' make an attempt to comply with house style and rules. It's good manners.}}
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{{ProofWanted}}
'''Theorem:''' Let $C$ be a non-singular projective algebraic curve of degree $n$.
Then the genus of $C$ as a Riemann surface is given by
:$g = \dfrac {\paren {n - 1} \paren {n - 2} } 2$
'''Proof:'''
Take the map $f = \frac y x$ from $C$ to $\mathbf P^1$.
By Ramification points of y/x, the ramification points occur where the tangents pass through $\sqbrk {0, 0, 1}$.
{{explain|what is the meaning of $\sqbrk {0, 0, 1}$?}}
These are therefore given by the equation:
:$\dfrac {\partial P} {\partial z} = 0$
The multiplicity is bigger than $2$ only if $\map {I_p} {C, T_p} > 2$.
That is, if $p$ is an inflection point, but there are only finitely many of these.
So, by a projective transformation, we can assume that $\sqbrk {0, 0, 1}$ does not lie on the tangent to any one of them.
This means that each $m_k$ in the Riemann-Hurwitz Formula is $2$
It remains to calculate the number of ramification points.
This is the number of points of intersection of $P = 0$, the curve $C$ of degree $n$, and $\dfrac {\partial P} {\partial z} = 0$, a curve $D$ which is of degree $n-1$.
Since $C$ is nonsingular it is irreducible.
Hence $C$ and $D$ can have no common component.
We will use Bézout's Theorem.
Thus we need to check that:
:$\sqbrk a = \sqbrk {a_0, a_1, a_2} \in C \cap D$ is a nonsingular point of $D$ and that the tangent lines are distinct.
{{explain|the meaning of $\sqbrk a$ and $\sqbrk {a_0, a_1, a_2}$}}
Now $\tuple {P_{z x}, P_{z y}, P_{z z} }$ is not identically zero at $\sqbrk {a_0, a_1, a_2}$, because this would make the Hessian of $C$ vanish, and we know that $\sqbrk {a_0, a_1, a_2}$ is not an inflection point.
This shows that $D$ is nonsingular here.
Suppose that the tangents of $C$ and $D$ coincide then $\tuple {P_{z x}, P_{z y}, P_{z z} }$ is a multiple of $\paren {P_x, P_y, P_z}$.
{{mistake|Presumably the above is a run-on sentence that does not even include a comma. If this is not the case, then please rewrite it to say what it is meant to say/}}
As in our discussion of inflection points, we use the symmetric bilinear form $B$ defined by the matrix of partial derivatives $P_{i j}$.
Then $\map B {a, a} = 0 = \map B {a, \alpha}$ where the tangent line joins $\sqbrk a$ and $\sqbrk \alpha$.
Put $v = \tuple {0, 0, 1}$.
By Euler's Homogeneous Function Theorem:
:$a_0 P_{z x} + a_1 P_{z y} + a_2 P_{z z} = \paren {n - 1} P_z = 0$
since $\map {P_z} a = 0$.
This gives:
:$\map B {a, v} = 0$
Moreover since:
:$\map {P_{z z} } a = \lambda \map {P_z} a = 0$
we have:
:$\map B {v, v} = 0$
Since $\sqbrk a $ is not an inflection point:
:$\map \det B \ne 0$
So, from:
:$0 = \map B {a, a} = \map B {a, \alpha} = \map B {a, v}$
we deduce:
:$v = \mu a + \nu \alpha$
But then:
:$0 = \map B {v, v} = \nu^2 \map B {\alpha, \alpha}$
and, by Quadric containing a line is singular, this gives:
:$\map \det B = 0$
unless $\nu = 0$.
But then:
:$\sqbrk a = \sqbrk {0, 0, 1}$
which we have specifically excluded.
We conclude that the tangents are distinct
It follows that the conditions for Bézout's Theorem hold.
So the number of ramification points is exactly $n \paren {n - 1}$.
From the Riemann-Hurwitz Formula we obtain:
:$2 - 2 g = 2 n - n \paren {n - 1}$
and so:
:$g = \dfrac 1 2 \paren {n - 1} \paren {n - 2}$
{{qed}}