2022
DOI: 10.2140/agt.2022.22.325
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Moduli space of nonnegatively curved metrics on manifolds of dimension 4k + 1

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Cited by 4 publications
(6 citation statements)
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“…The manifold β however is not oriented nullbordant, since the α-invariant of a spin manifold in these dimensions is equal to a multiple of its Â-genus, which is a Pontryagin number, hence it is not possible to use these rigidity results. 7 However, when working with concordance classes instead, we have the following result pointing towards an affirmative answer of Question 4.1. Theorem 4.2 Let M be a d-dimensional, closed, oriented, totally nonspin manifold with D ⊂ M an embedded disk and let g ∈ R + (M \ int(D)) h • .…”
Section: Question 41 Do the Metrics (G N ) N∈n Become Isotopic After ...mentioning
confidence: 77%
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“…The manifold β however is not oriented nullbordant, since the α-invariant of a spin manifold in these dimensions is equal to a multiple of its Â-genus, which is a Pontryagin number, hence it is not possible to use these rigidity results. 7 However, when working with concordance classes instead, we have the following result pointing towards an affirmative answer of Question 4.1. Theorem 4.2 Let M be a d-dimensional, closed, oriented, totally nonspin manifold with D ⊂ M an embedded disk and let g ∈ R + (M \ int(D)) h • .…”
Section: Question 41 Do the Metrics (G N ) N∈n Become Isotopic After ...mentioning
confidence: 77%
“…In the case where M is nonspin, but its universal cover is Spin, there are cases where R + (M) is not connected or has even infinitely many path components, see [2,7,8,15,16,23,25]. For totally nonspin manifolds, Kastenholz-Reinhold give an example of a closed, totally nonspin manifold of dimension 6 whose space of psc-metrics has infinitely many components in [21].…”
Section: Remark 12 (State Of the Art)mentioning
confidence: 99%
“…The codimension 3 case requires a bit more work. Recall that if M is the total space of a linear S 3 -bundle over S 4 , then M admits a metric of K 0 [23], and moreover, if the bundle has nonzero Euler number, then M K 0 (M ) has infinitely many connected components [12,21]. We prove: Theorem 1.4.…”
Section: Motivation and Resultsmentioning
confidence: 95%
“…Examples of closed manifolds S for which M K 0 (S) has infinitely many connected components can be found in [12,13,14,15,21,25]. These metrics on S have K 0 and scal > 0, and the connected components are distinguished by index-theoretic invariants that are constant on paths of scal > 0.…”
Section: Motivation and Resultsmentioning
confidence: 99%
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