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OT: Math Problem, an early post led me to this math problem. Can someone explain this to me in easy

Mr. Potter

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Dec 27, 2004
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analogy. I'm not a math guy: Need help with the following:

soul conjecture states:

Suppose (M, g) is complete, connected and non-compact with sectional curvature K ≥ 0, and there exists a point in M where the sectional curvature (in all sectional directions) is strictly positive. Then the soul of M is a point; equivalently M is diffeomorphic to Rn.
Perelman proved the conjecture by establishing that in the general case K ≥ 0, Sharafutdinov's retraction P : M → S is a submersion.
 
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Usually folks say this in a mocking, condescending tone, but not me: If you don't understand what that means, then I can't help you!
 
I've never even heard of soul conjecture. I've heard of Soul Caliber, though.
 
analogy. I'm not a math guy: Need help with the following:

soul conjecture states:

Suppose (M, g) is complete, connected and non-compact with sectional curvature K ≥ 0, and there exists a point in M where the sectional curvature (in all sectional directions) is strictly positive. Then the soul of M is a point; equivalently M is diffeomorphic to Rn.
Perelman proved the conjecture by establishing that in the general case K ≥ 0, Sharafutdinov's retraction P : M → S is a submersion.

 
You got me. The best mathematicians, like Perelman, are so Goddamn smart, it's astonishing. I have a PhD in mathematics but this stuff is way beyond my ken, no apologies for that.

What, you didn't have fun doing differential geometry?
 
What, you didn't have fun doing differential geometry?
People do. I've never taken a class in it. But you should know that we mathematicians are never happier than when we're doing math. We manage to find it exciting, every time. My wife screams at the TV when she's watching political shows and I'm sitting there doing math, as happy as a clam. It's a definite luxury.
 
See, the sad thing about a guy like you is, in 50 years you're gonna start doin' some thinkin' on your own and you're going to come up with the fact that there are two certainties in life: one, don't do that, and two, you dropped 150 grand on a ****in' education you could have got for a dollar fifty in late charges at the public library!
 
You got me. The best mathematicians, like Perelman, are so Goddamn smart, it's astonishing. I have a PhD in mathematics but this stuff is way beyond my ken, no apologies for that.


Props to you sir. I have my undergrad in applied mathematics, but anything past undergrad for math is impressive. You can fake undergrad, but actually have to know what you’re doing at the graduate level.
 
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People do. I've never taken a class in it. But you should know that we mathematicians are never happier than when we're doing math. We manage to find it exciting, every time. My wife screams at the TV when she's watching political shows and I'm sitting there doing math, as happy as a clam. It's a definite luxury.
Bravo!
 
analogy. I'm not a math guy: Need help with the following:

soul conjecture states:

Suppose (M, g) is complete, connected and non-compact with sectional curvature K ≥ 0, and there exists a point in M where the sectional curvature (in all sectional directions) is strictly positive. Then the soul of M is a point; equivalently M is diffeomorphic to Rn.
Perelman proved the conjecture by establishing that in the general case K ≥ 0, Sharafutdinov's retraction P : M → S is a submersion.
 
analogy. I'm not a math guy: Need help with the following:

soul conjecture states:

Suppose (M, g) is complete, connected and non-compact with sectional curvature K ≥ 0, and there exists a point in M where the sectional curvature (in all sectional directions) is strictly positive. Then the soul of M is a point; equivalently M is diffeomorphic to Rn.
Perelman proved the conjecture by establishing that in the general case K ≥ 0, Sharafutdinov's retraction P : M → S is a submersion.

giphy.gif
 
analogy. I'm not a math guy: Need help with the following:

soul conjecture states:

Suppose (M, g) is complete, connected and non-compact with sectional curvature K ≥ 0, and there exists a point in M where the sectional curvature (in all sectional directions) is strictly positive. Then the soul of M is a point; equivalently M is diffeomorphic to Rn.
Perelman proved the conjecture by establishing that in the general case K ≥ 0, Sharafutdinov's retraction P : M → S is a submersion.

Dude...who gives a shit! If this diffeomorphic could help our offensive line out I would be for it! Stop messing with our heads on a football board. We are way to busy figgering out who we are gonna be playing on Jan 1!!
 
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People do. I've never taken a class in it. But you should know that we mathematicians are never happier than when we're doing math. We manage to find it exciting, every time. My wife screams at the TV when she's watching political shows and I'm sitting there doing math, as happy as a clam. It's a definite luxury.
my math....

 
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analogy. I'm not a math guy: Need help with the following:

soul conjecture states:

Suppose (M, g) is complete, connected and non-compact with sectional curvature K ≥ 0, and there exists a point in M where the sectional curvature (in all sectional directions) is strictly positive. Then the soul of M is a point; equivalently M is diffeomorphic to Rn.
Perelman proved the conjecture by establishing that in the general case K ≥ 0, Sharafutdinov's retraction P : M → S is a submersion.
Well, duh! That is so obvious.
The best analogy that I can come up with is to think of one of ro's sig pics. When looking at her sectional curvature, at some (albeit early) point, you'll consider it strictly positive down in your soul. As Perelman imagined, if you can't find anything positive about the curvature (think Triponey), even Sharafutdinov had a, well, "retraction". In most cases, as with ro's sig pics, you'll need a [cold water] submersion.
 
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Is this about 9 point circles? Or what about this math? I always show this to my calc classes.

 
J. DIFFERENTIAL GEOMETRY 40(1994)209-212
PROOF OF THE SOUL CONJECTURE OF CHEEGER AND GROMOLL
G. PERELMAN
In this note we consider complete noncompact Riemannian manifolds M of nonnegative sectional curvature. The structure of such manifolds was discovered by Cheeger and Gromoll [2]: M contains a (not neces- sarily unique) totally convex and totally geodesic submanifold S without boundary, 0 < dimS < d i m M , such that M is diffeomorphic to the total space of the normal bundle of *S in ¥ . (S is called a soul of M.) In particular, if S is a single point, then M is diffeomorphic to a Euclidean space. This is the case if all sectional curvatures of M are posi- tive, according to an earlier result of Gromoll and Meyer [3]. Cheeger and Gromoll conjectured that the same conclusion can be obtained under the weaker assumption that M contains a point where all sectional curvatures are positive. A contrapositive version of this conjecture expresses certain rigidity of manifolds with souls of positive dimension. It was verified in [2] in the cases dim S = 1 and codimS = 1, and by Marenich, Walschap, and Strake in the case codimS = 2. Recently Marenich [4] published an argument for analytic manifolds without dimensional restrictions. (We were unable to get through that argument, containing over 50 pages of computations.)
In this note we present a short proof of the Soul Conjecture in full generality. Our argument makes use of two basic results: the Berger's version of Rauch comparison theorem [1] and the existence of distance nonincreasing retraction of M onto S due to Sharafutdinov [5].
Theorem. Let M be a complete noncompact Riemannian manifold of nonnegative sectional curvature, let S be a soul of M, and let P: M —> S be a distance nonincreasing retraction.
(A)ForanyxeS, veSN(S) wehave P{expχ{tυ)) = x for all t>0,
where SN(S) denotes the unit normal bundle of S in M.
Received January 10, 1994. The author is a Miller Fellow at the University of California at Berkeley.

210 G. PERELMAN
(B) ForanygeodesicγcS andanyvectorfieldveΓ(SN(S)) paral- lel along y, the "horizontal curves γt, γt(u) = exp ^(fi/), are geodesies,
filling aflat totallygeodesicstrip (t>0).Moreover,if γ[uQ,ux] ismini- mizing, then all yt[u0, uχ] are also minimizing.
(C) P is a Riemannian submersion of class C 1 . Moreover, the eigen- values of the second fundamental forms of the fibers of P are bounded above,inbarriersense,by injrad(5')~1.
The Soul Conjecture is an immediate consequence of(B) since the normal exponential map N(S)—•M issurjective.
Proof We prove (A) and (B) first. Clearly it is sufficient to check that if (A) and (B)hold for 0 < t < I for some / > 0, then they continue to hold for 0 < t < I + ε(l), for some ε(l) > 0 . In particular, we can start from / = 0 , in which case some of the details of the argument below are redundant.
Suppose that (A) and (B)hold for 0 < t < I. For small r > 0 consider
a function f{r) =max{\xP(expχ((l+r)ι>))\\x eS, v e SNχ(S)} .Clearly / is a Lipschitz nonnegative function, and /(0)= 0. We are going to prove that / = 0 (thereby establishing (A) for 0 < t < I + ε(/)) by showing that the upper left derivative of / is nowhere positive.
Fix r>0.Letf(r)=\x0-xo\wherex0 =P(expx ((/+r)uQ). Since r is small and P is distance decreasing, we can assume that \XQX0\ < injrad(S). Pick a point xχ e S so that xQ lies on a minimizing geodesic between x0 and xχ\ let x0 = γ(u0), xχ = y{μx). Let v e Γ(SN(S)) be a parallel vector field along y, v\χ = v0. Then, according to our assumption, the curves γt(u) = exp /ttj(ίi/), 0 < t < I, are minimizing geodesies of constant length filling a flat totally geodesic rectangle. In particular, the tangent vectors to the normal geodesies σu(t) = expγ^(tv) form a parallel vector field along y{. Therefore, according to Berger's comparison theorem, the arcs of yι+r are no longer than corresponding arcs of γι, with equality only if γt, I < t < I + r, are geodesies filling a flat totally geodesic rectangle.
Now consider the point ~xχ = P(σu (l+r)). Using the distance decreas- ing property of P and the above observation we get
(1)
On
(2)
Taking into account that by construction
\χΰχχ\<|σWo(/+r)σWi(/+r)|<1^(0^(/)| =|*0*il the other hand,
\xχxx\<f{r) = \xoxo\.
^ 1*1*11+ 1*0*11>

PROOF OF THE SOUL CONJECTURE OF CHEEGER AND GROMOLL 211
we see that (1) and (2) must be equalities, and therefore
(3) VtlUotUj, l<t<l + r,
are minimizing geodesies filling a flat totally geodesic rectangle. N o w f o r δ —• 0 , w e o b t a i n
f(r-δ)>\xιP(σUι(Ur-δ))\>\xoxι\-\xQP(σUι(Ur--δ))\
= \xox{\ - \xox{\ - O(δ2) = \xoxo\ - O(δ2) = f(r) - O(δ2),
where we have used the definition o f x 0 and distance nonincreasing prop- erty o f P i n the third inequality, and (3) i n the fourth one.
Thusf[r)=0for0<r<ε(l), and(A)isprovedfor0</</+ε(l). To prove (B) for such t one can repeat a part o f the argument above, up to assertion (3),takingintoaccountthat (JC0,v0),γ,xχ cannowbechosen arbitrarily, and x 0 = x 0 , 'xι=xι.
Assertion (C) i s an easy corollary o f (A), (B) and the distance decreasing property o f P . Indeed, let x be an interior point o f a minimizing geodesic γ c S, σ bea normal geodesic starting at x. Then, according to (B), we can construct a flattotally geodesic strip spanned by γ andσ, and,for any point y on σ, say y = σ(t),wecan define a lift γy of γ through y asahorizontalgeodesic γt ofthatstrip. Thislift isindependentof σ: if incidentally y = σ(t')9 then the corresponding lift y must coincide with yy because otherwise |j^(tto)y (Wj)| < \y(u^)y(uλ)\, and this would contradict(A)andthedistancedecreasingpropertyofP.
Thus we have correctly defined continuous horizontal distribution. Sim- ilar arguments show that P has a correctly defined differential—a linear map which is isometric on horizontal distribution and identically zero on its orthogonal complement. For example, suppose two geodesies γ ι , γ 2 c
S are orthogonal a t their intersection point x . Then their lifts yι orthogonal at y , because otherwise we would have |j^(wo)z| < \yι{u0)P(z)\ forsomepointzonγ2 closetoy.
The estimate on the second fundamental form o f the fiber P~ι(x) a t y followsfromtheinequality \P~ι(x)yy(u0)\ >|*y(tto)|,validforallmin- imizing geodesies γ c S passing through x, and from the standard esti- mate o f the second fundamental form o f a metric sphere i n nonnegatively curved manifold.
y
, γ 2 a r e

212 G. PERELMAN
Remarks. (1) The fibers of the submersion P are not necessarily iso- metric to each other, and not necessarily totally geodesic (see [6]).
(2) Existence of a Riemannian submersion of M onto S was conjec- tured some time ago by D. Gromoll.
(3) It would be interesting to find a version of our theorem for Alexan- drov spaces (which may occur, for instance, as Gromov-Hausdorff limits of blowups of Riemannian manifolds, collapsing with lower bound on sec- tional curvature). We hope to address this and other rigidity problems for Alexandrov spaces elsewhere.
References
[1] M. Berger, An extension ofRauch's metric comparison theorem and some applications, Illinois J. Math. 6 (1962) 700-712.
[2] J. Cheeger & D. Gromoll, On the structure of complete manifolds of nonnegative curva- ture, Ann. of Math. 96 (1972) 413-443.
[3] D. Gromoll & W. T. Meyer, On complete open manifolds of positive curvature, Ann. of Math. 90 (1969) 75-90.
[4] V. Marenich, Structure of open manifolds of nonnegativecurvatureI, II. Siberian Ad- vances in Math. 2 (1992), no. 4, 104-146; 3 (1993) no. 1, 129-151.
n
[5] V. Sharafutdinov, Pogorelov-Klingenberg theoremfor manifolds homeomorphic to R ,
Sibirsk. Math. Zh. 18 (1977) 915-925.
[6] G. Walschap, Nonnegatively curved manifolds with souls of codimension 2, J. Differen-
tial Geometry 27 (1988) 525-537.
ST. PETERSBURG BRANCH OF STEKLOV MATHEMATICAL INSTITUTE UNIVERSITY OF CALIFORNIA, BERKELEY

Thank you, I’ll show myself out!
 
Well, duh! That is so obvious.
The best analogy that I can come up with is to think of one of ro's sig pics. When looking at her sectional curvature, at some (albeit early) point, you'll consider it strictly positive down in your soul. As Perelman imagined, if you can't find anything positive about the curvature (think Triponey), even Sharafutdinov had a, well, "retraction". In most cases, as with ro's sig pics, you'll need a [cold water] submersion.
And in the few other cases (where cold water submersion isn’t necessary), there’s a good chance you’ll need a topical ointment — or an antibiotic.
 
analogy. I'm not a math guy: Need help with the following:

soul conjecture states:

Suppose (M, g) is complete, connected and non-compact with sectional curvature K ≥ 0, and there exists a point in M where the sectional curvature (in all sectional directions) is strictly positive. Then the soul of M is a point; equivalently M is diffeomorphic to Rn.
Perelman proved the conjecture by establishing that in the general case K ≥ 0, Sharafutdinov's retraction P : M → S is a submersion.

My father is who is a freakin genius in math and mechanical engineering would say. Their are no problems, only equations to be solved.
 
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