All Questions
9 questions
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Existence of covering space with trivial pullback map on $H^1$
I have seen somewhere the following claim (but can't remember where): let $M$ be a connected orientable closed smooth manifold with $b_1(M)=1$, then there exists a connected covering space $p:\tilde{M}...
5
votes
0
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248
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Algebraic de Rham cohomology with torus coefficients
Let $X$ be a smooth projective variety over $\mathbb{C}.$
On page 3 in this preprint of Simpson, it is stated that
Notice first of all that the algebraic de Rham theory is not going to work well in ...
2
votes
0
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241
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Monodromy group action on de Rham cohomology
Let $f : Y \longrightarrow X := \mathbb{P}^1\setminus\{0,1,\infty\}$ be the smooth proper morphism associated to the Legendre family, which is an elliptic fibration of the punctured line, with fibre ...
9
votes
2
answers
1k
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Hodge dual of de Rham cohomology and singular cohomology
We know that the de Rham cohomology is isomorphic to the singular cohomology, does the Hodge dual of differential forms induce a dual operation on de Rham cohomology, hence also on singular cohomology?...
17
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1
answer
1k
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Direct proof that Chern-Weil theory yields integral classes
Suppose $E$ is a complex vector bundle of rank $n$ on a compact oriented manifold (both assumed smooth). Let $h$ be a Hermitian metric on $E$, and let $A$ be a Hermitian connection on $E$ and $F_A$ ...
2
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0
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152
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When are automorphisms of the cohomology ring realized by isometries?
Let $(M,g)$ be a closed smooth Riemannian manifold, and denote by $G$ a closed subgroup of its isometry group. By considering the maps $g^*$ induced by elements $g\in G$ in the (de Rham) cohomology $H^...
6
votes
1
answer
2k
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Integration currents vs Poincaré dual
Let $M$ be a complex manifold of dimension $n$ and $S \subset M$ a closed complex submanifold of complex codimension $r$. Let $[S] \in H_{2r}(S)$ be the fundamental class of $S$.
We have the ...
13
votes
1
answer
2k
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de rham model for relative cohomology
In GTM82, I read a model for the relative cohomology of (M,N) with N a submanifold of M.
And in the page:
Relative De Rham cohomologies,
I got to know that there is another model for relative ...
38
votes
4
answers
8k
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Relative De Rham cohomologies
as far as I know, there are two main ways to have a relative version of De Rham Cohomology for a pair (M,N), where M and N are smooth manifolds and N is a closed (as a topological subspace) ...