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Results tagged with riemannian-geometry
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user 90655
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.
8
votes
0
answers
405
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What specifically is the gap in Aubin's argument about positive Ricci curvature that Paul Eh...
In his paper [2], Paul Ehrlich write
In [1], Aubin stated a theorem which implied as a corollary that if a manifold
$M$ admits a Riemannian metric with nonnegative Ricci curvature and
all Ricci curva …
- 4,195
1
vote
Positive sectional curvature does not imply positive definite curvature operator?
This argument is due to Burkhard Wilking. The other answers, answer to the OP question very well but I want to give a sharp example. I.e. a manifold with positive sectional curvature and curvature op …
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3
votes
Accepted
Closed manifolds of nonnegative curvature operator are symmetric spaces
As Igor Belegradek commented, the correct statement is as follows:
Theorem (classification of closed simply connected manifold with nonnegative curvature operator): A closed simply connected manifold …
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3
votes
1
answer
358
views
Closed manifolds of nonnegative curvature operator are symmetric spaces
In an online webinar, I heard (not directly) the statement that (closed) manifolds of nonnegative curvature operator $\mathcal{R}\geq 0$ are symmetric spaces. Is this a valid theorem? Any reference t …
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4
votes
1
answer
309
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Existence parallel vector fields and its effect on the topology of manifolds (Karp's Thesis)
It seems that there is no digital copy of Leon Karp's Ph.D. thesis
L. Karp, Vector fields on manifolds, Thesis, New York Univ., 1976.
on internet and his paper excerpted from his thesis is very brief …
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4
votes
1
answer
208
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Existence non-trivial parallel $p$-form implies non-triviality of $p$-th cohomology group us...
Cross-post from MSE.
Suppose $(M,g)$ be a closed Riemannian manifold. Because every parallel (nontrivial) $p$-form $\omega$ is harmonic so the $p$-th Betti number should be positive i.e. $b_p\geq 1$. …
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1
vote
1
answer
239
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Reference for non-parallel harmonic $k$-forms
I want to get some deep understanding on closed orientable Riemannian manifolds admitting $k$-forms ($k\geq 2$) $\omega$ that satisfices the following conditions:
$$\nabla \omega\neq 0,\quad \Delta\om …
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2
votes
1
answer
578
views
Why trace is more natural than (preferred to) determinant for smooth map $f:M\to N$?
Cross-post from MSE.
For a continuous map $f:(M,g)\to (N,h)$, between Riemannian manifolds $(M,g)$ and $(N,h)$ we can pullback $h$ by $f$. Most experts take the trace from this new tensor and work wit …
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14
votes
Accepted
Information about Milnor conjecture
According to David Roberts comment and the following paper it is open for dimensions $n\geq 4$.
Pan, Jiayin, A proof of Milnor conjecture in dimension 3, J. Reine Angew. Math. 758, 253-260 (2020). ZBL …
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3
votes
1
answer
181
views
Open neighbourhood of a point of space of Riemannian metrics
Let $M$ be a finite-dimensional compact smooth manifold and
$$\mathcal{M}et(M) = \{ g : g \text{ is a Riemannian metric on }M \}.$$
Q1-a: What metrics $g$ are very close to the given metric $g_0$? I.e …
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6
votes
1
answer
227
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Does $\pi_k(M)\neq 0$ implies $\operatorname{ind}(\gamma) < k$?
Cross post from MSE. and sorry if this is an obvious question.
Here is a line of proof of Theorem 1.15 from
Brendle, Simon, Ricci flow and the sphere theorem, Graduate Studies in Mathematics 111. Prov …
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18
votes
1
answer
908
views
Consequences of Gromov's Conjecture
In Peter Petersen words, Gromov Betti number estimate is considered one of the deepest and most beautiful results in Riemannian geometry; which asserts that
Theorem (Gromov 1981). There is a constant …
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-3
votes
1
answer
374
views
Is the 2-dimensional Gauss-Bonnet theorem applicable in higher dimensions? [closed]
This is a cross-post of this MSE post that users commented that it is appropriate for MO.
I want to know
Question: Is the 2-dimensional Gauss-Bonnet theorem applicable (any topological or geometrica …
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2
votes
1
answer
270
views
Is there any Riemannian manifold of zero dimensional isometry group such that
Sorry if this question is belongs to MSE. I have no idea about it.
Question: Is there any Riemannian manifold of zero dimensional isometry group which its Ricci curvature is positive (or maybe zer …
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6
votes
1
answer
252
views
A metric has positive sectional curvature if and only if ${\rm Ric}_{ij} < \frac{r}{2}g_{ij}$
This is a cross-post from my question on MSE.
It is well known that
In dimension three a metric has positive sectional curvature if
and only if ${\rm Ric}_{ij} < \frac{r}{2}g_{ij}$. where $r$ …
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