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2 votes

$2\mathrm{d}$ area minimizing short embeddings

If think yes, because the assumption $f^*h \leq g$ gives pointwise bounds on the derivatives of $f$ : in local charts, $|\partial_i f|^2 \leq g_{ii}$. So up to translation, you can put the image of $f$...
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10 votes
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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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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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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
2 votes
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Closed almost geodesics in a Riemannian manifold

Any curve $\gamma:[a,b]\to M$ parametrized by arc length is an $\varepsilon$-geodesic for any $\varepsilon>0$. The inequality $(1-\varepsilon)dist_M(\gamma(x),\gamma(y))\leq length[\gamma(x),\gamma(...
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1 vote

Explicit computation of the vertical and horizontal vector bundles

I am just answering to (1) and (2): The horizontal bundle is defined as follows. Consider a point $v\in T_pX\subset TX,$ and $w\in T_pm$. Consider a curve $\gamma$ through $p$ with $\gamma'(0)=w.$ ...
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  • 6,290
1 vote
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Cross product of two infinitesimal bendings

Let $M$ be a rotationally symmetric cylinder, $Z_1$ be a vector field tangent to M generating the isometric translation along the cylinder and $Z_2$ be a vector field also tangent to M generating the ...
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1 vote

What are good Morse Theory lecture notes and books?

also in the M. W. Hirsch, Differential topology ,Chapter 6 : Morse Theory.
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10 votes
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Does the first Laplacian eigenfunction on a homogeneous space have a unique maximum?

The flat torus $\mathbb{T} = \mathbb{R}^2/\Lambda$ gives a counterexample: The first nontrivial eigenvalue is of the form $\lambda_1 = \xi_1^2+\xi_2^2$, where $\xi = (\xi_1,\xi_2)$ is a nonzero ...
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1 vote
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Convergence of extremal subsets in Alexandrov spaces

The limit of extremal subsets is an extremal subset, see Lemma 4.1.3 in Petrunin's Semiconcave functions in Alexandrov geometry. The non-collapsing assumption is not needed.
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0 votes

Area of a deformation of a closed surface

The answer to the second question is still negative. Let $M^3$ be the cylinder $\mathbb{S}^2 \times \mathbb{R}$. Then $\Sigma = \mathbb{S}^2\times \{0\}$ is totally geodesic in $M$, and so for any ...
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2 votes
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A problem arising from reading a lecture on the Yamabe problem of how the Hölder inequality is used

Application of Holder's inequality Notice that the estimate on the Green's function means that $$ \int |G(P,Q)|^\alpha ~dQ $$ is bounded whenever $\alpha < \frac{n}{n-2}$ (and the bound can be ...
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1 vote
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Weitzenböck formula and comparison of norms

I am not an expert in all the delicate points of clifford algebras and spin structures, but I think the following shows the constant can't be 1 in general: let $(M,g)$ be a Riemannian manifold of ...
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  • 336
5 votes

Harmonic functions on complete Riemannian manifolds

For the $n$-dimensional hyperbolic space it is already the space of bounded harmonic functions that is infinitely dimensional, which follows from the integral Poisson formula.
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1 vote

Decomposition of tensors

If what @AMath91 meant by a codazzi tensor is a symmetric tensor $\lambda \in \Gamma(S^{2}T^{*}M)$ satisfying $d^{\nabla}\lambda=0$ then I think the decompostion in the question cannot be true. Take ...
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  • 336
5 votes
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Why are we interested in spectral gaps for Laplacian operators

A spectral gap gives information on geometry of the manifold via Cheeger's inequality, https://en.wikipedia.org/wiki/Cheeger_constant See also Buser's inequality discussed there. More directly, a ...
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  • 11.3k
2 votes

Is it known whether a closed simply-connected manifold of non-negative curvature admits positive Ricci?

No. Every simply connected compact manifold of nonnegative sectional curvature admits positive Ricci curvature. This has been proven by Boehm--Wilking: https://link.springer.com/article/10.1007/s00039-...
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