k-differentials and their residues

I posted a question earlier. This question is smaller, more to the point and more basic I hope. Yet it is at the heart of my problems, I think.

In one of my references, degree k meromorphic differentials with poles of order $k$ at $n$ points $z_i$ on the Riemann sphere are mentioned: $\phi_k(z)=F(z)dz^k$. Here $F(z)$ is a rational function on the complex plane.

If we want to know the behaviour of $\phi_k(z)$ at $\infty$ we change coordinates to $z'=1/z$ under which the k-differential transforms as $\phi_k(z')=F(1/z')(-dz'/(z')^2)^k$. Furthermore, these k-differentials are required to have fixed residues $\alpha_i$ on each $z_i$.

Then it is stated that we thus allow: $$\phi_k(z) \sim \frac{\alpha_i}{(z-z_i)^k} dz^k +...$$

1) May I also think of these k-differentials as $$\phi_k(z)=\frac{\prod^m_{i=1}c(z-u_i)}{\prod^n_{i=1}(z-z_i)^k}dz^k$$ Here, $c$ is a constant, $m=n-2$ in order that the degree of the divisor of this k-differential is $-2k$ as required on a Riemann surface of genus $g=0$. Furthermore, some of the $u_i$ and/or $c$ may depend on the fixed residues $\alpha_i$. If this is wrong, what is meant by the above stated form of the k-differential?

2) What does it mean for a k-differential to have a residue? For instance, how does Cauchy's formula work for a quadratic differential? Does a second order pole behave for a quadratic differential as a simple pole behaves for a 1-differential? If so, I don't think the residues of a quadratic differential sum to zero.

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The meaning of residue in this context is that this quantity is a conformal invariant - and even a formal invariant - and furthermore the unique such invariant. All this is to say that any k-differential with a pole of order (at worst) k can be expressed a scalar multiple of the standard differential, meaning $c\cdot\frac{dz^k}{z^k}$, in an appropriate coordinate $z$. This much works the same way for any $k\in\mathbb{Z}$. For $k=1$ the residue has an additivity property which allows for a global residue theorem, but this won't be the case more generally: for example, $\frac{dz^2}{z^2}$ has precisely two poles of equal nonzero residue.
Thanks for your answer. For 3 or more second order poles of a quadratic residue, $\frac{c(z-u_1)(z-u_2)}{(z-z_1)^2(z-z_2)^2(z-z_3)^2}dz^2$, there do not seem to be any relations anymore between the residues at the separate points $z_i$. Does this mean that when we fix the residues of a quadratic differential, we lose one more freely chooseable zero than in the case with a 1-differential (from which the residues do sum to zero and therefore we have one less constraint)? –  sam Oct 11 '13 at 11:54
I am asking about these 'free parameters' or freely chooseable zeroes because in the end I need to check that a certain one form, $\lambda = xdz$ with $x=\frac{G(F_1(z),F_2(z),...,F_k(z)))}{\prod (z-z_i)}$, differentiated wrt to one of the moduli is a holomorphic one-form on a curve defined by: $$x^k+F_1(z)x^{k-1}+...+F_k(z)=0$$. Hence, I want some sort of feeling or better yet concrete vision of how the $F_i$ depend on their moduli (and ofcourse how many moduli there are for given k and n). –  sam Oct 11 '13 at 13:23
I'n not entirely sure what you are asking. But, for example, if you specify a finite set of $n\geq 2$ points then there will exists a quadratic differential which has a double pole with 'residue' 1 at each of those points, and which is holomorphic elsewhere. You can write down such a quadratic differential as a sum of Mbbius translates of $\frac{dz^2}{z^2}$,at least for $n$ even, and a modified recipe works for $n$ odd. Moreover, the space of such quadratic differentials will have dimension $\max(n-3,0)$. –  Adam Epstein Oct 11 '13 at 18:27
I don't really see why this space is n-3-dimensional. Doesn't there exist a quadratic differential for $n=2$ right? This would be the one you've mentioned, $dz^2/z^2$ which has a double pole at $z=0$ and a double pole at $z=\infty$. If not, could you explain for $n=4,5$ what would be the beasis vectors of the space? –  sam Oct 14 '13 at 13:27
By the way, Adam, isn't the dimension of quadratic differentials with simple poles equal to $n-3$. Does this also hold for quadratics with double poles? –  sam Oct 14 '13 at 17:04