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Results tagged with riemannian-geometry
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user 121820
Riemannian Geometry is a subfield of Differential Geometry, which specifically studies "Riemannian Manifolds", manifolds with "Riemannian Metrics", which means that they are equipped with continuous inner products.
2
votes
Accepted
Tensor component calculation
Start by writing
$$ \tag{1}\label{1} B_{nn} = \nabla^\mu \nabla^\nu C_{\mu n n \nu} - \frac{1}{2}C_{\mu n n \nu}R^{\mu\nu} . $$
Here I abuse notation to write $\nabla^\mu\nabla^\nu C_{\mu n n \nu} = n …
- 1,713
11
votes
Locally conformally flat
The connected sum of two locally conformally flat manifolds can be equipped with a metrics which is locally conformally flat. To see this, recall that the product metric on $\mathbb{R} \times S^{n-1} …
- 1,713
2
votes
Accepted
Expansion of metric near boundary of 3 dimensional Poincaré-Einstein/hyperbolic manifolds
This is only an answer to the request for references about the expansion and an interpretation of $h_2$ from the standpoint of the metric $x^2g$.
Theorem 7.4 in the book The Ambient Metric proves that …
- 1,713
2
votes
Accepted
Does any warped product metric admit a function with hessian proportional to the metric?
Yes. (With the caveat that if there is a function such that $\nabla^2\varphi = \psi g$, then necessarily it is a warped product over a one-dimensional base, so your question should really require $\d …
- 1,713
4
votes
Accepted
Lee-Parker Yamabe problem proposition 4.6
You are forgetting that, as part of the renormalization, the functions $\psi_s$ all satisfy $\psi_s(-P)=1$ (here $-P$ is the south pole). In particular, since $\psi$ is $C^2$ away from the north pole …
- 1,713
5
votes
Finding a hyperbolic metric with geodesic boundary on a given Riemann surface
See Theorem 1(b) in the following paper; it is enough that the boundary is smooth.
Osgood, B.; Phillips, R.; Sarnak, P. Extremals of determinants of Laplacians. J. Funct. Anal. 80 (1988), no. 1, 148-- …
- 1,713
10
votes
Accepted
Definition of the conformal metric
Let $(M,[g])$ be a conformal manifold;
i.e. $(M,g)$ is a Riemannian manifold and $[g] = \{ u^2g \mathrel{}:\mathrel{} u \in C^\infty(M), u>0 \}$ is the set of Riemannian metrics conformal to $g$.
It i …
- 1,713
4
votes
What are some explicit examples of nontrivial gradient almost Ricci solitons with harmonic c...
The Riemannian product $\mathbb{R}^m \times S^n$ is always of this type, where $\mathbb{R}^m$ is given the flat metric and $S^n$ the round metric of constant sectional curvature one. In this case $\l …
- 1,713
1
vote
Accepted
How to calculate the infimum of Yamabe functional on upper hemisphere
The first observation is that $Q_g(\phi)$ is conformally invariant.
Define
\begin{align*}
L_gu & := -\Delta u + \frac{n-2}{4(n-1)}Ru , \\
B_gu & := \partial_\nu u + \frac{n-2}{2}Hu ,
\end{align*}
wh …
- 1,713
3
votes
Accepted
Yamabe operator, conformal transformations and square of the Dirac operator
The discrepancy comes from the fact that the square of the Dirac operator is in general not conformally covariant. There are ways to modify powers of the Dirac operator to get a conformally covariant …
- 1,713
5
votes
Trace and divergence of the Bach tensor
The Bach tensor is only divergence-free in dimension four. In general dimension, it holds that $\nabla^j B_{ij} = (n-4)P^{jk} ( \nabla_i P_{jk} - \nabla_j P_{ik} )$. I pulled this formula from the b …
- 1,713
5
votes
Accepted
Example of shrinking Ricci soliton
Yes: Take the shrinking Gaussian $(\mathbb{R}^n, dx^2)$ with $X=\rho\nabla\rho$, where $\rho$ denotes the distance to the origin. This space is Ricci flat, so your inequality holds.
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18
votes
Accepted
Geometric interpretation of the Weyl tensor?
There is such an interpretation, with a few caveats. Essentially, there is a canonical connection on a certain vector bundle for which the "principal part" of the curvature is the Weyl tensor in dime …
- 12.9k
6
votes
On equation $\Delta \circ \partial/\partial X=\partial/\partial X \circ \Delta$ on a Riemann...
This is an expansion on Igor's comment and fixes my previous mistake (see also Willie's answer). A direct computation gives
$$ D(f) = 2f^{ab}\nabla_{(a}X_{b)} + X_a(\nabla^b\nabla_b\nabla^a-\nabla^a\ …
- 1,713
3
votes
Einstein warped product manifold Ricci flat
If you also assume that $(N,\ddot g)$ is complete, then your assumptions imply that $(N,\ddot g)$ is Ricci flat. I proved this in my article The nonexistence of quasi-Einstein metrics.
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