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Take the conic $X: z^2-xy=0$ in $\mathbb P^2$. On the patch with $z=1$ and coordinates $x, y$, $X$ is cut out by $1-xy=0$. On the patch with $y=1$ and coordinates $u, v$, $X$ is cut out by $v^2-u=0$.

For an affine plane curve $C$ cut out by $f(x, y) = 0$ in $\mathbb A^2$, we have $$ \frac{dx}{f_y} = -\frac{dy}{f_x} $$ (because $f$ is the zero function on $C$ so $f_x \, dx + f_y \, dy =0$).

So in our situation, on the patch $\mathbb A^2_{x, y}$, we have $$ \frac{dx}{-x} = \frac{dy}{y} $$ and on the patch $\mathbb A^2_{u, v}$, we have $$ \frac{du}{2v} = \frac{dv}{1} $$. The coordinate changes from $\mathbb A^2_{u, v}$ to $\mathbb A^2_{x, y}$ is given by $x=u/v$ and $y=1/v$. I am trying to show that a Kähler differential on the projective curve are Kähler differentials on the affine curves that paste correctly. Namely, the Kähler differentials above should agree. We have $$ \frac{dy}{y} = \frac{d(1/v)}{1/v} = v \cdot \frac{-1}{v^2} dv = -\frac{1}{v} dv $$ but this should agree with $dv$. Certainly I'm making some very dumb mistake. There is an example of this type of calculation here where it does work, on the second page with a cubic. Please save me from my misery.

C.F.G
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hwong557
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1 Answers1

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Ok here's the problem. I believe holomorphic 1-forms on $X$ look like

$$ p(x, y) \frac{dx}{f_y} $$ where $p$ is a polynomial with degree at most 3 less than the polynomial cutting out $X$ (Show that a smooth plane quartic is never hyperelliptic). And in this situation there are no such polynomials $p$.

Another way to see there are no Kahler differentials on $X (=\mathbb P^1)$. The sheaf of meromorphic 1-forms is given by a canonical divisor $K$ of degree $2g-2$, where $g$ is the genus of $X$. If $g=0$ (as it is in my example), then the degree of $K$ is negative so the sheaf $\mathcal O(K)$ has no global sections. The genus of a plane curve of degree $d$ is $\frac{1}{2}(d-1)(d-2)$ so we need to work with curves of degree at least $3$ to get a positive genus.

hwong557
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