Topological dimension and derham cohomological dimension

Question

If G is a compact complex manifold then does the topological dimension bound the deRham cohomological dimension below?

By derham cohomological dimension, I mean the largest extended natrual number $n$ such that $H_{dr}^n(G)\not\cong 0$

Answer 1

Let $M$ be a compact connected complex manifold of real dimension $n$. Then $M$ is orientable, hence $H_\mathrm{dR}^n (M, \mathbb{R}) \cong \mathbb{R}$ by Poincaré duality. (Alternatively, the Stokes theorem implies $H_\mathrm{dR}^n (M, \mathbb{R}) \ne 0$.) Moreover, $H^i_\mathrm{dR} (M, \mathbb{R}) = 0$ for all $i > n$. Thus, the topological dimension is equal to the de Rham cohomological dimension.

Answer 2

De Rham cohomology is meaningless for general topological spaces. One can, however, use other cohomology theories to define cohomology and cohomological dimension. One can define Chech cohomology groups of $X$ with sheaf coefficients, $H^*(X; F)$, where $F$ is a sheaf on $X$. Then one defines the cohomological dimension of $X$ (say, over the integers) as $$ cd(X)=\sup \{n: \exists F \ \hbox{such that}\ H^n(X; F)\ne 0\}, $$ where $F$ are sheafs of abelian groups on $X$. (One can also define cohomological dimension over fields in a similar fashion, by considering sheafs of vector spaces.) By taking the constant sheaf (where the abelian group is ${\mathbb Z}$) one obtains the trivial inequality $$ cd(X)\ge \dim_{const}(X):=\sup\{n: H^n(X; {\mathbb Z})\ne 0\}. $$ One more ingredient one needs to know is that $$ \dim(X)\ge cd(X) $$ where $dim$ is the covering dimension. This inequality is essentially a tautology once you get thorough the definitions. Thus, you get $$ \dim_{const}(X)\le \dim(X). $$

Check "Dimension theory" by Engelking for (many more) details.