# More Kirwan Excerpts

In this installment, a little lemma about tangent lines for curves in $\mathbb{P}^2$, resultants and a curves intersecting, and recovering a compact riemann surface from an equation of an algebraic curve.

### Tangent Line Lemma

I give a quick definition of tangent lines for curves in affine space, and for curves in projective space.  The lemma shows they definitions behave well with one another.  Let $f \in k[x,y,z]$ be homogeneous, so that it defines a curve in $\mathbb{P}^2$.  For a point $P$ on the curve, the tangent line is the projective line given by the vanishing of

$\frac{\partial f}{\partial x}|_P \cdot x + \frac{\partial f}{\partial y}|_P \cdot y + \frac{\partial f}{\partial z}|_P \cdot z$

This is motivated by the fact tangent vectors should be orthogonal to the vector of partials $(f_x, f_y, f_z)$.  Also note that being perpendicular to this vector of partial uniquely determines a projective line, and (by the Euler relation) this line goes through the original point $P$.

For the affine case, we could copy the above and say the tangent line at a point is given by the vanishing of

$\frac{\partial f}{\partial x}|_P \cdot x + \frac{\partial f}{\partial y}|_P \cdot y$

This gives a unique line through the origin, but this line will NOT necessarily go through our original point $P$ unless $P = (0,0)$.  But the above condition determines the slope of the line, so forcing it to go through the point $P$ uniquely determines the tangent line to be

$\frac{\partial f}{\partial x}|_P \cdot (x - x(P)) + \frac{\partial f}{\partial y}|_P \cdot (y - y(P))$

Lemma: If $f$ defines a projective plane curve, and $l$ is the tangent line at a point on the curve, then $l$ restricted to the usual affine patch is the affine tangent line.

proof: I assume $P = [a:b:1]$.  Write $f_{x,P}$ for $\frac{\partial f}{\partial x}|_P$ etc.  The projective tangent line is

$xf_{x,P}+ yf_{y,P} + zf_{z,P}$

Note that as $f(x,y,z)$ is homogeneous, so are all its partials.  This is because it is a sum of monomials, and when differentiated each monomial either drops out completely or goes down one in total degree.  It follows that $\frac{\partial f(x,y,z)}{\partial x}|_{[a:b:1]} = \frac{\partial f(x,y,1)}{\partial x}|_{(a,b)}$  Also from the Euler relation we have $f_{z,P} = -af_{x,P} - bf_{y,P}$.  So plugging we get in the affine patch the line

$xf_{x,P} + yf_{y,P} + -af_{x,P} - bf_{y,P}$

$= f_{x,P}(x-a) + f_{y,P}(y - b)$

QED.

### Resultants and Intersections.

The reason to care about resultants is if $p(x),r(x)$ are polynomials then they have a common root iff the resultant $R(p,q) = 0$.  Write $p(x) = \sum_i a_i x^i$ and $r(x) = \sum_j b_j x^j$.  The resolvent is defined as the determinant of some matrix constructed from the $a_i,b_j$.

The way to prove this property of the resultant is as follows.  To have a common factor means $p(x) = r(x)s(x)$ and $q(x) = r(x)s(x)$ and this is equivalent to t(x)p(x) = s(x)q(x)$. Now write everything as a sum of monomials and you get a bunch of conditions on the coefficients, and being able to obtain solutions is equivalent to the previous determinant vanishing. Prop: Any two curves in $\mathbb{P}^2$ intersect. proof: Let the equations be $f(x,y,z), g(x,y,z)$ Consider them as polynomials in $x$ with coefficients in $k[y,z]$. Calculate the resultant. It is a polynomial in $y,z$, over an algebraically closed field it splits as a sum of linear factors, so there are numbers $b,c$ such that $R(f(x,b,c), g(x,b,c)) = 0$, so there is a common root $a$ and then $f(a,b,c) = g(a,b,c) = 0$ as desired. QED. ### Recovering a compact curve from its equations. (See around pg. 90) You might say that the projective closure of $f(x,y)$ sitting in $\mathbb{P}^2$ is already a compact Riemann surface, but there is something else you can do. Thinking of $f(x,y)$ as an implicit function for$late y$, it is not well defined, you need to make various branch cuts, but then you can glue these things together to get something that looks more like a sphere with $g$ handles. For example $y^2 - x$ is not well defined. You need to make a branch cut. Take two copies of $\mathbb{C}$ minus the negative reals. Then you can glue these guys so that you get a well defined holomoprhic function $\sqrt x$ on this glued together peice. You do this in general for more complicated equations$laetx f(x,y)\$ but the identifications can get tricky (see page 90).  But the moral of the story is you can play around with them and eventually glue it into something that looks like a sphere with $g$ handles.

In general to show that any compact connected Riemann surface is a sphere with $g$ handles, you triangulate it so that you can flatted it out and cleverly cut and paste it around (all the while preserving a triangulation of the same abstract space) so that you eventually get it to look like some $4n$-gon where all vertices are identified and you get the thing you want.  This is the appendix of this Kirwan book.

Also a good proof of the implicit function theorem.