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 ...
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
$$\mathcal{E}(g) = \dfrac{\displaystyle\int_Ms_g d\...
8
votes
The convex hull of a manifold whose cobordism class is trivial
There are exotic spheres (which are null cobordant) which do not bound a parallelisable manifold. Since the convex hull is contractible, it would be parallelisable if it were a manifold, so these guys ...
7
votes
Accepted
How wild can an open topological 3-manifold be if it has a compact quotient?
The possible universal covers of closed 3-manifolds are $S^3-C$, where $|C|=0, 1, 2$ or $C$ is a tame Cantor set, corresponding to the space of ends of the fundamental group as you suspect. This ...
5
votes
Relation between de Rham cohomology group of Lie group as a manifold and group cohomology of Lie group
The answer to the question is: not really.
Consider $G=\mathbb{R}^n\,.$ This has nontrivial group cohomology in all degrees $\le n$ (by the van Est isomorphism theorem, for example), while the ...
5
votes
Manifolds whose tangent spaces have a special behavior
The answer to the first question is No.
The assumption you made is equivalent to stating that for every $q\in M$ that the vector $q\in T_qM$.
This is satisfied whenever $M$ is a portion of a cone, ...
3
votes
Lie algebroid in algebraic geometry
I suggest having a look at
Beĭlinson, A.; Bernstein, J. A proof of Jantzen conjectures.
MR: Matches for: MR=1237825
§1.2 .
https://people.math.harvard.edu/~gaitsgde/grad_2009/BB%20-%20Jantzen.pdf
3
votes
Accepted
A question about complex Laplacian on compact Hermitian manifolds
For your first question, note that (let $\omega$ be the Hermitian form)
$$\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
...
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. ...
2
votes
Accepted
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 ...
2
votes
Accepted
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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