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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.
3
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kernel of the conformal Laplacian
Let $M$ be a smooth, closed manifold of dimension $n>2$. Let $L_g$ be the conformal Laplacian of the metric $g$. That is, $L_g=-\Delta_g + \frac{n-2}{4(n-1)}R_g$, where $R_g$ is the scalar curvature o …
1
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0
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271
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Bound on $Q$ implies bound on $R$?
Let $(M,g)$ be a smooth, closed Riemannian manifold with dimension $n>4$. Define the $Q$-curvature through the formula
$Q = \Delta R + \frac{n^3-4n^2+16n-16}{(n-1)(n-2)^2} R^2 - \frac{8(n-1)}{(n-2)^2 …
0
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Estimate on the size of flat balls where the Weyl tensor vanishes
I believe that the answer to the first question is the injectivity radius if the manifold is locally conformally flat. The injectivity radius provides us with an estimate on the size of a chart that c …
0
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2
answers
237
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Estimate on the size of flat balls where the Weyl tensor vanishes
Let $(M,g)$ be a Riemannian manifold and suppose that the Weyl tensor of $g$ vanishes at a point $p \in M$. Can one estimate the size of the largest geodesic ball around $p$ that we can make $g$ flat …
12
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2
answers
1k
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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 …
15
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1
answer
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Geometric Interpretation of $Q$-curvature
Let $(M,g)$ be a Riemannian manifold of dimension $n>2$. Thanks to the late T.Branson we have the following definition of the so-called $Q$-curvature:
$Q= \Delta R + \frac{n^3-4n^2+16n-16}{4(n-1)(n-2 …
10
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2
answers
1k
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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
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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
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0
answers
343
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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 …
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Positivity of Second-Order Elliptic Differential Operators
I believe the answer is no if $n>2$. Let $g$ be a metric with a negative Yamabe constant. There will be a metric $h$ in the conformal class of $g$ such that $\int_M R_h dv_g> 0$. Let $L_h$ be the conf …
2
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2
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506
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Positivity of Second-Order Elliptic Differential Operators
Let $(M,g)$ be a closed, smooth Riemannian manifold. Let $\Delta = -div\nabla$ be the Laplace-Beltrami operator. Let $h$ be a smooth function on $M$. Is there a condition on $h$ weaker than non-negati …
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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
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234
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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
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Ricci flat metric on $n$-sphere?
Can you put a Ricci flat metric on the $n$-sphere, $n>4$?