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Quarto Bendir's user avatar
Quarto Bendir's user avatar
Quarto Bendir's user avatar
Quarto Bendir
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Defining a probability distribution on each tangent space of a manifold?
(I think I mixed up Yau and Gaffney; Gaffney proved the conservation of unit mass)
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Defining a probability distribution on each tangent space of a manifold?
It isn't clear to me that your definition is well defined. Is it?
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Defining a probability distribution on each tangent space of a manifold?
The heat kernel on Euclidean space is the family of Gaussian probability measures at a given center. S.-T. Yau proved that, on complete Riemannan manifold with Ricci curvature bounded below, the heat kernel consists of probability measures. So this could be what you want. Together with result of M. Gaffney, the heat kernel under the same conditions even defines a Feller process
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Defining a probability distribution on each tangent space of a manifold?
I believe you might as well just take the pushforward of the Gaussian on the given tangent space by the exponential map, no need for any conditional assumptions.
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How much results in Calabi-Yau manifolds and mirror symmetry depends on the existence of a ricci-flat metric?
I know little about this, but as I understand, the particular CY metric is fundamental in formulating the notion of special Lagrangian, and so the Ricci aspect is crucial for the SYZ conjecture in mirror symmetry.
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Breakthroughs in mathematics in 2021
It's definitely possible to learn something from the answers to this kind of question, but it's worth bearing in mind that answers are pretty much limited to various papers in one or two prestige journals, and/or those papers which Quanta writers chose to pay attention to. So it's hard to get a balanced perspective.
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Breakthroughs in mathematics in 2021
The media hype around the AI work seems highly debatable, see e.g. arxiv.org/abs/2112.04324 (a direct response to the Nature article linked to in the answer)
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Is there a natural connection on $TM$?
It's a good question, many people are not familiar with these things. My understanding is the following. Consider a curve $\gamma:I\to TM$ which projects to $c:I\to M$; $\gamma'(0)$ represents a general element of $T(T(M))$ and $\gamma$ can be interpreted as a section of $c^\ast TM$, i.e. as a vector field along $c$. A connection on $M$ defines a connection on $c^\ast TM$, and so you can consider $\nabla_{\partial/\partial t}\gamma$, which will be an element of $T_{c(0)}M$. And $c'(0)$ is another element of $T_{c(0)}M$. This pair of elements is the splitting of $\gamma'(0)$.
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Schoen and Yau's proof of the higher dimensional positive mass theorem
There are certainly some papers citing SY's paper for this result, e.g. by Coley and Ellis, Sakovich and Sormani, and Borghini and Mazzieri. Anyway, Schoen and Yau's approach seems obviously cogent and credible, and and so maybe the best answer to this question would clarify if there are particular parts of their work that are particularly technically demanding (or even possibly unclear). It is disappointing that their paper, along with Lohkamp's, have been out for four years without any (to my knowledge) detailed expository accounts or commentary (whether positive or negative).
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Shing-Tung Yau's doubts about Perelman's proof
I am inclined to agree that this is a bad question for mathoverflow, and that both answers at present (including my own) are unsatisfactory. If mathoverflow is intended to be about providing expert opinion and knowledge and not just about finding and quoting reliable sources, then I think the collection of comments and answers given here marks a failure. The only voice here which has any particular expertise in the relevant field of mathematics seems to be Yau himself.
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Riemannian vector bundle
Ok. I think the two explanations have identical length. Anyway it's good to have facility and to know how the argument looks both ways
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Riemannian vector bundle
In my view it's exactly the same as what I said in the comments, just put in a different language. I think it's very clean, but I know opinions vary on use of coordinates.
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Riemannian vector bundle
Your formula cannot be correct since the RHS is tensorial in X,Y and the LHS is not. Metric compatibility of connection D means X<s,t> = <D_Xs,t> + <s,D_Xt>. Under this condition one has the skew-symmetry <R(X,Y)s,t>+<s,R(X,Y)t> = 0. Under the rank-one condition, this skew-symmetry implies R(X,Y)=0. (When I was learning this material the most useful textbook I found was Morita "Geometry of Differential Forms.")
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Riemannian vector bundle
1) You can put a bundle metric on any vector bundle using a partition of unity. 2) The curvature of a metric-compatible connection is skew-symmetric and a skew-symmetric map of a one-dimensional space is zero.
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