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I fear I'm missing something important here, so forgive me if my question is stupid.

Consider $\mathcal{M}_g$ the moduli space of Riemann surfaces of genus $g>2$ and $\mathcal{H}_g$ the moduli space of abelian differentials on Riemann surfaces of genus $g$. By that I mean $\mathcal{H}_g:=\{(C,\omega)\}/_\simeq$ with $C$ a Riemann surface and $\omega$ an abelian differential on $C$. The moduli space $\mathcal{H}_g$ can be considered as an holomorphic fibration on $\mathcal{M}_g$ of rank $g$. From this fact we get that, since $\mathcal{M}_g$ has complex dimension $3g-3$, that $\mathcal{H}_g$ has complex dimension $4g-3$.

An abelian differential on $C$ of genus $g$ has degree $2g-2$. For every partition $\underline{k}$ of $2g-2$ we can consider the stratum $\mathcal{H}_g(\underline{k})\subset\mathcal{H}_g$ of abelian differentials with degrees for the zeroes prescribed by $\underline{k}$. It is known that each stratum $\mathcal{H}_g(\underline{k})$ has complex dimension $2g+k-1$.

From what I wrote follows that the dimensions of the strata should add up to the dimension of $\mathcal{H}_g$.

But this doesn't follow from my computations..

Consider the easiest case of $g=2$, then $dim_{\mathbb{C}}\mathcal{H}_2=5$ and $dim_{\mathbb{C}}\mathcal{H}_2(2,0)=4$, $dim_{\mathbb{C}}\mathcal{H}_2(1,1)=5$, but $9\neq 5$.

What am I getting wrong?

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    $\begingroup$ For a given partition $\underline{k} = (k_1,\dots,k_l)$ of $2g-2$ with $1\leq k_1 \leq \dots \leq k_l$ and $k_1 + \dots + k_l = 2g-2,$ the expected dimension of $\mathcal{H}_g(\underline{k})$ equals $2g-1+l$. In particular, the stratum of $\underline{k} = (1,\dots,1) = (1^{2g-2})$ is open in $\mathcal{H}_g$. $\endgroup$ – Jason Starr Jun 12 '16 at 16:25
  • $\begingroup$ When you have a stratification of an irreducible algebraic variety into a union of locally closed subsets, there is a unique open stratum that has dimension equal to the dimension of the variety. It is (typically) not the case that the sum of the dimensions of the strata equals the dimension of the variety. $\endgroup$ – Jason Starr Jun 12 '16 at 16:36
  • $\begingroup$ yes that's what I was missing, thank you! $\endgroup$ – user0029 Jun 12 '16 at 16:47

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