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Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.
3
votes
1
answer
656
views
harmonic curvature versus Einstein metric
A metric is said to have harmonic curvature if the exterior derivative of the Ricci tensor vanishes. It is known that there exists manifolds with dRic=0 that are not Einstein. My question is whether o …
4
votes
1
answer
288
views
upper bound on first non-zero eigenvalue of the Laplacian
I'm looking for an upper bound on the first non-zero eigenvalue of the Laplace-Beltrami operator on compact manifolds of dimension greater than four that have constant negative curvature. In particula …
4
votes
1
answer
283
views
non-negativity to positivity
Let $(M,g)$ be a closed Riemannian manifold of dimension $n>7$. In this setting I have been able to prove that the Green's function of a positive Paneitz-Branson operator is non-negative. Furthermore, …
12
votes
2
answers
1k
views
Obstructions to Einstein metrics in high dimensions
It is well known that there exists three and four manifolds that do not admit an Einstein metric, but I wonder if this question is still open for manifolds of dimension higher than four. That is, does …
1
vote
Higher order Sobolev inequality
The desired inequality is correct. It is easily generalized from Aubin's proof (which is given in Lee&Parker's "Yamabe Problem") of the classical Sobolev inequality for Riemannian manifolds. One simpl …
4
votes
1
answer
627
views
Higher order Sobolev inequality
Let $(M,g)$ be a closed, Riemannian manifold of dimension $n>4$. Let $K$ be the best constant for the Sobolev inequality
$||u||^2_p \leq K \int_{{\Bbb{R}}^n} (\Delta u)^2 dx,$
where $p=\frac{2n}{n-4 …
10
votes
2
answers
1k
views
Einstein metrics and conformal geometry
I recall reading somewhere that if a conformal class contains an Einstein metric then that metric is the unique metric with constant scalar curvature in its conformal class, with the exception of the …
3
votes
Some questions about scalar curvature
Also: when one is talking about positive scalar curvature, the Yamabe invariant is important. The Yamabe invariant is the supremum over all conformal classes of the Yamabe constants of a manifold. The …
4
votes
0
answers
343
views
Constant scalar curvature+Constant $\sigma_2(C_g)$ curvature = ?
Let $(M,g)$ be a closed, smooth, Riemannian manifold of dimension $n>4$. Suppose both the scalar curvature and norm of the Ricci tensor are constant. In addition suppose that $g$ satisfies the followi …
5
votes
PDE on manifolds
The tour-de-force of elliptic pde on manifolds is the Yamabe problem. There the pde is a second-order, elliptic, and semilinear with a Sobolev critical exponent. The analysis can become incredibly dif …
7
votes
Riemannian manifolds that are scalar flat but not Ricci flat
To generalize Anton's comment a little, I should add that with the appropriate choice of $l$ and $k$, the product manifold $S^l \times N^k$ will have the property that you are looking for, where $N^k$ …
3
votes
0
answers
234
views
Negative Paneitz constant on $n$-sphere
Let $\Pi$ be the Riemannian functional defined on the space of Riemannian metrics on $S^n$, $n>4$, as follows:
$$
\Pi(g) = \int_M \frac{(n-4)(n^3-4n^2+16n-16)}{16(n-1)^2(n-2)^2} R_g^2 - \frac{2(n-4)}{ …
27
votes
0
answers
3k
views
Ricci flat metric on $n$-sphere?
Can you put a Ricci flat metric on the $n$-sphere, $n>4$?