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Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.
7
votes
Can anyone give an example of Ricci flat Riemannian or Lorentzian Manifold that is not flat?
Your claim that "all solutions from general relativity are geodesically incomplete" is not true. The classical Schwarzschild/Kerr black hole solutions are geodesically incomplete, and along with Minko …
21
votes
Geometric picture of scalar curvature
I'll add a few things which I've found helpful to get intuition to what Renato's already written.
Scalar curvature has the very simple geometric interpretation as the volume defect of small balls (as …
6
votes
Ricci flow descending from an universal cover
I was going to post this as a comment, but it got too long. I'm not exactly sure how to answer the question as written, but the following might be enlightening:
A crucial piece of information for your …
4
votes
Accepted
Minimal surface on $R^3$ with with non Euclidean metric
Below I sketch the proof of the following theorem:
Theorem: Suppose that $\Sigma^n\subset (M^{n+1},g_0)$ is smooth and uniquely area-minimizing relative to it's boundary $\Gamma : = \partial\Sigma$ a …
3
votes
Accepted
Isometries of manifolds with non-positive sectional curvature
Bochner's theorem extends to nonpositive Ricci to give:
If $(M,g)$ is compact and has $\textrm{Ric}\leq 0$ then any Killing vector $X$ is parallel and $\textrm{Ric}(X,X) = 0$.
See Petersen (3rd ed) …
4
votes
Do minimal submanifolds minimize area locally?
Just a comment to supplement both nice answers. Robert Bryant alludes to the commonly held idea that since minimal surfaces are strictly stable on small scales (this is easy to prove) then the result …
4
votes
Accepted
Difference between parallel transport and ambient projection
This is false as stated. Take a surface of revolution generated by $(r(t),z(t))$. I claim I can choose the curve so that there are pieces that look like $(e^{-j},t)$ for $j$ large. The point is that i …
3
votes
Accepted
Volume of balls in 3-dimensional manifolds with nonpositive Ricci tensor
If you are OK with considering large balls, there are easy counterexamples. For example $T^2 \times \mathbb{R}$. Alternatively, there is a metric of negative Ricci curvature on $S^3$ (I think original …
6
votes
1
answer
388
views
Do manifolds with no Ricci lower bounds for any metric exist?
Is there a smooth (noncompact) manifold $M$ such that for any Riemannian metric $g$ on $M$ there are $p_i$ and unit tangent vectors $v_i \in T_{p_i}M$ such that $Ricc(g)|_{p_i}(v_i,v_i) \leq -i$? This …
6
votes
Accepted
What is the current status on bad tangent cones at isolated singularities?
(i) This used to be a wide open area, but recently there has been some progress: Gabor Székelyhidi has constructed an example of an isolated singularity with a cylindrical tangent cone here: https://a …
6
votes
Accepted
The negative gradient flow of a Morse-Bott function on a compact manifold converges to a cri...
I will assume that "converges in the critical set of $f$" is asking that if $f$ is MB then $\phi_t(y)$ (the flowlines) converge as $t\to\infty$ to $y_\infty$ a critical point (of course, depending on …
19
votes
1
answer
505
views
Smooth curve in $\mathbb{R}^3$ not contained in real analytic surface?
Is there a $C^\infty$-smooth embedding $\gamma : I \to \mathbb{R}^3$ so that there is no real analytic $2$-dimensional submanifold $M \subset \mathbb{R}^3$ with $\gamma(I)\subset M$?
3
votes
Complete stable minimal hypersurface in positively curved manifolds
You can construct a positively curved $(M^n,g)$ for any $n\geq 4$ that admits a stable minimal hypersurface. This is described in Example 1.2 here. (That paper also contains some non-existence results …
5
votes
Accepted
Finding vector fields on $S^2$ with equal divergence
I think that this is not possible:
Per my comment on Divergence of conformal Killing vector fields on $S^2$ and the spherical harmonics you want to solve
$$
\textrm{div} (Y) = -2a\cdot x
$$
for $Y$ or …
2
votes
Accepted
Minimal surfaces with increasing area but bounded Morse index
Positive scalar curvature implies that if $\textrm{index}(\Sigma_j)\leq I$ then $\Sigma_j$ have bounded area and genus. This is proven here https://arxiv.org/pdf/1509.06724.pdf (Theorem 1.3). That pap …