12 votes

The convex hull of a manifold whose cobordism class is trivial

Implicit in the other responses is the fact that if $M$ bounds a convex manifold $W$, then $W$ is contractible and so M has the homology of a sphere. So any null-cobordant manifold that is not a ...
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10 votes
Accepted

Can we prescribe the $L^2$ norm of the scalar curvature on a four-manifold?

This is not always possible. Let $M$ be a compact smooth manifold of dimension $n$. Consider the Einstein-Hilbert functional $\mathcal{E}$ given by $$\mathcal{E}(g) = \dfrac{\displaystyle\int_Ms_g d\...
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8 votes

The convex hull of a manifold whose cobordism class is trivial

There are exotic spheres (which are null cobordant) which do not bound a parallelisable manifold. Since the convex hull is contractible, it would be parallelisable if it were a manifold, so these guys ...
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  • 6,185
7 votes
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How wild can an open topological 3-manifold be if it has a compact quotient?

The possible universal covers of closed 3-manifolds are $S^3-C$, where $|C|=0, 1, 2$ or $C$ is a tame Cantor set, corresponding to the space of ends of the fundamental group as you suspect. This ...
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  • 61.2k
5 votes

Relation between de Rham cohomology group of Lie group as a manifold and group cohomology of Lie group

The answer to the question is: not really. Consider $G=\mathbb{R}^n\,.$ This has nontrivial group cohomology in all degrees $\le n$ (by the van Est isomorphism theorem, for example), while the ...
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5 votes

Manifolds whose tangent spaces have a special behavior

The answer to the first question is No. The assumption you made is equivalent to stating that for every $q\in M$ that the vector $q\in T_qM$. This is satisfied whenever $M$ is a portion of a cone, ...
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  • 30.7k
3 votes

Lie algebroid in algebraic geometry

I suggest having a look at Beĭlinson, A.; Bernstein, J. A proof of Jantzen conjectures. MR: Matches for: MR=1237825 §1.2 . https://people.math.harvard.edu/~gaitsgde/grad_2009/BB%20-%20Jantzen.pdf
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  • 3,733
3 votes
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A question about complex Laplacian on compact Hermitian manifolds

For your first question, note that (let $\omega$ be the Hermitian form) $$\int_M\Delta_c(f) \omega^n=\int_M(\mathrm{tr}_\omega \sqrt{-1}\partial \overline{\partial}f)\omega^n=n\int_M \sqrt{-1}\partial ...
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2 votes

Isometry between Minkowski space and Tangent space in an article by Stefan Waldmann

I think you are confused by two different uses for the word "isometry". There is first the notion of a linear isometry between vector spaces equipped with a non-degenerate bilinear form. ...
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  • 7,513
2 votes
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Concrete descriptions of $S^1$-bundles over smooth manifold $Y$ underying a K3 surface

You can say a fair amount about the topology of the total spaces of the different bundles, although I suspect none of them is a particularly well-known manifold that has a `name'. (Except of course ...
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2 votes
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Expressing a vector valued function in terms of its derivatives

$\newcommand{\pa}{\partial}\newcommand{\R}{\mathbb R}$The answer is no. Indeed, suppose the contrary: that for each polynomial $f$ there are functions $a_j,b,c_j$ such that \begin{equation} f(x_1,\...
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