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I'm a humble physicist, looking for a reference that will explain the Dolbeault cohomology of holomorphic functions of homogeneity $-2$ on the Riemann sphere. In particular, my geometry is sufficiently rusty that I can't work out what the general form of the functions in this space are. Could someone point me to a (very pedestrian) reference, crucially with explicit representatives of the cohomology?

Just so you know where I am at present, I'm fairly sure that $$ c \frac{ [\bar \lambda d \bar \lambda] } { [\lambda \bar \lambda]^2 }$$

is one option, where $c$ is a constant. Is it true that I can use any homogeneous function of $\lambda$ as my $c$ and this will give me the full cohomology? Or does the most compact general formula involve further $\bar \lambda$ terms that cancel out in the $\bar \partial$ derivative, as happens for the measure?

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    This isn't really a matter of geometry, but of manipulating analytic functions/power series. What do you mean by Dolbeault cohomology of ... ? In math language, you're looking at the line bundle $L$ whose (local) sections are holomorphic functions of homogeneity $-2$, and you're trying to compute $H^{0,1}(\Bbb P^1,L)$, i.e., $\bar\partial$-closed $1$-forms with values in $L$ modulo $\bar\partial$-exact such $1$-forms? The Dolbeault isomorphism will allow us to compute this as a Čech cohomology $H^1(\Bbb P^1,E)$, where we work with the usual easy 2-element open cover of $\Bbb P^1$. – 2017-02-07
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    At any rate, if I was right about which cohomology you want, the correct answer is that the vector space is $1$-dimensional. One can write down the generator either as a twisted $(0,1)$-form (which is I guess what you did) or, via the Dolbeault isomorphism, as a $(1,1)$-form (so the area 2-form). – 2017-02-07
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    @TedShifrin - thanks for your response. Just so I'm clear, can my function $c$ be any homogeneous function of the $\lambda$ then? And if so, does that capture all possibly representatives of the cohomology? – 2017-02-08

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I guess that the notion "holomorphic functions of homogeneity $−2$" means sections of the line bundle $\mathcal{O}(-2)$, which is isomorphic to the line bundle of holomorphic $1$-forms $\Omega^1$ (look at its degree$\ldots$). So the Dolbeault resolution of $\mathcal{O}(-2)$ is the Dolbeault resolution of $\Omega^1$, which consists in degree $0$ of ($C^{\infty}$ sections of) $(1,0)$-forms and in degree $1$ of $(1,1)$-forms, which are just the $2$-forms. The Dolbeault operator between them coincides with the Cartan differential and its image are the exact $2$-forms (since $\text{H}^{(0,1)}$ is vanishing, every $1$-form is the sum of a $(1,0)$-form and of an exact $1$-form).

So a $2$-form represents a nontrivial cohomology class in $\text{H}^1(\mathbb{P}^1, \Omega^1)$ if and only if its integral over the projective line is nonvanishing. In Čech cohomology you can represent it by the $1$-form $\text{d}z/z$ on $\mathbb{P}^1 \setminus \{0,\infty\}$, the intersection of the disks around $0$ and $\infty$ (it is not a boundary, because it has nonvanishing residues at $0$ and $\infty$).