All Questions
7 questions
6
votes
1
answer
766
views
Does every ‘curvature’ tensor induce a metric? [duplicate]
So we know that given a Riemannian manifold $(M,g)$ we can calculate its Riemannian curvature $R$. This covariant $4$-tensor field then satisfies some important symmetries
\begin{gather*}
R_{ijkl} = - ...
5
votes
2
answers
340
views
Example of a curvature with no associated metric
Is there a concrete example of a $4$ tensor $R_{ijkl}$ with the same symmetries as the Riemannian curvature tensor, i.e.
\begin{gather*}
R_{ijkl} = - R_{ijlk},\quad R_{ijkl} = R_{jikl},\quad R_{ijkl} =...
5
votes
0
answers
101
views
How is this product of tensors defined?
I am reading the paper “ The first eigenvalue of a small geodesic ball in a riemannian manifold”, by Karp and Pinsky, from where I took the following:
Here, $\Delta_{-2}$ denotes the usual Laplacian ...
4
votes
1
answer
747
views
The Laplacian of an expression involving the Ricci tensor
While doing some computations on a compact Riemannian manifold I have reached the following expression:
$$ \Delta_y \big( Ric_y (\exp_y ^{-1} x, \exp_y ^{-1} x) \big) (x)$$
where $\Delta_y$ is the ...
3
votes
0
answers
367
views
Obtaining the metric from the mixed Ricci tensor $R^i{}_j$
In chapter 5 of the book "Einstein Manifolds", Arthur Besse discusses the possibility to find the metric $g$ when knowing the Ricci curvature tensor $Ric(g)$ ($=R_{ij}$).
But what do we know about ...
2
votes
2
answers
515
views
A kind of "Curvature tensor" for higher dimensional tensors
I begin my question with a multilinear question then I will consider two local smooth analogies:
Assume that $\alpha$ is a real valued symmetric $k$-tensor, that is a $k$-linear map $\alpha:\...
2
votes
2
answers
5k
views
Calculating the Riemann Christoffel tensor for a diagonal metric
I am trying to calculate the entries of the Riemann curvature tensor $R^m_{\phantom{m}ijk}$ for the metric $g_{ij}$.
The Riemann-Christoffel tensor is given as
\begin{align}
R^m_{\phantom{m}ijk} = \...