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LongTheta
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### A Differential Geometric Question for all of you geometers out there.

Nonius,have you noticed that you start each day with a post addressing Newton? What a life!LT.

Nonius
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### A Differential Geometric Question for all of you geometers out there.

QuoteOriginally posted by: LongThetaNonius,have you noticed that you start each day with a post addressing Newton? What a life!LT.I also start the day by taking out the trash.....

Nonius
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### A Differential Geometric Question for all of you geometers out there.

QuoteOriginally posted by: chiral3Are you back on exp[Ut] if N<KI never left it. exp in the geodesic sense, yes.

Nonius
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### A Differential Geometric Question for all of you geometers out there.

anyway, the construction is pretty easy in fact......If G is a Lie Group and G/H is a homogeneous space, consider f:G/H->R smooth. Then, G acts on G/H on the left through isometries. Let Lam be an element of the Lie Algebra of G. Consider the curve f(Exp(Lam t)m) where m is in G/H. Differentiation and setting t = yields, d/dtf(Exp(Lam t)m). This defines a map from G/H to linear functionals defined on the Lie Algebra of G. Using the metric, there is a natural identification between linear functionals and elements of the Lie Algebra. Define GradNonius(f) to be that element of the Lie Algebra such that <GradNonius(f)(m),Lam>=d/dt(t=0)f(Exp(Lam t)m).I claim that GradNonius(f) is orthogonal to the space that represents a vertical fiber over m in the bundle G->H. A vertical fiber over m is by definition the elements V in the tangent space of TG at any g such that p(g)=m such that dp(g)V=0. This tangent space is naturally identified with the Lie Algebra via the adjoint rep. If dp(g)V=0, then Exp(tV) fixes m, which means Exp(tV) is in the fixed group of m. Thus, d/dt(t=0)f(Exp(V t)m)=0. Since GradNonius lives in a space that is orthogonal to the vertical fibers, it is in a horizontal lift, which means that it lives in a space that is naturally identified with the tangent space of m. It also means that Exp(GradNonius(f)(m)t)m is a geodesic in G/H passing through m.Anyway, all of this means that grad descent is relatively easy.

N
Posts: 2808
Joined: May 9th, 2003, 8:26 pm

### A Differential Geometric Question for all of you geometers out there.

anyway, the construction is pretty easy in fact......If G is a Lie Group and G/H is a homogeneous space, consider f:G/H->R smooth. Then, G acts on G/H on the left through isometries. Let Lam be an element of the Lie Algebra of G. Consider the curve f(Exp(Lam t)m) where m is in G/H. Differentiation and setting t = yields, d/dtf(Exp(Lam t)m). This defines a map from G/H to linear functionals defined on the Lie Algebra of G. Using the metric, there is a natural identification between linear functionals and elements of the Lie Algebra. Define GradNonius(f) to be that element of the Lie Algebra such that <GradNonius(f)(m),Lam>=d/dt(t=0)f(Exp(Lam t)m).I claim that GradNonius(f) is orthogonal to the space that represents a vertical fiber over m in the bundle G->H. A vertical fiber over m is by definition the elements V in the tangent space of TG at any g such that p(g)=m such that dp(g)V=0. This tangent space is naturally identified with the Lie Algebra via the adjoint rep. If dp(g)V=0, then Exp(tV) fixes m, which means Exp(tV) is in the fixed group of m. Thus, d/dt(t=0)f(Exp(V t)m)=0. Since GradNonius lives in a space that is orthogonal to the vertical fibers, it is in a horizontal lift, which means that it lives in a space that is naturally identified with the tangent space of m. It also means that Exp(GradNonius(f)(m)t)m is a geodesic in G/H passing through m.Anyway, all of this means that grad descent is relatively easy. Nonius,What a bunch of crap. Your manifold ain't smooth.Rocky

Nonius
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Joined: January 22nd, 2003, 6:48 am

### A Differential Geometric Question for all of you geometers out there.

QuoteOriginally posted by: Nanyway, the construction is pretty easy in fact......If G is a Lie Group and G/H is a homogeneous space, consider f:G/H->R smooth. Then, G acts on G/H on the left through isometries. Let Lam be an element of the Lie Algebra of G. Consider the curve f(Exp(Lam t)m) where m is in G/H. Differentiation and setting t = yields, d/dtf(Exp(Lam t)m). This defines a map from G/H to linear functionals defined on the Lie Algebra of G. Using the metric, there is a natural identification between linear functionals and elements of the Lie Algebra. Define GradNonius(f) to be that element of the Lie Algebra such that <GradNonius(f)(m),Lam>=d/dt(t=0)f(Exp(Lam t)m).I claim that GradNonius(f) is orthogonal to the space that represents a vertical fiber over m in the bundle G->H. A vertical fiber over m is by definition the elements V in the tangent space of TG at any g such that p(g)=m such that dp(g)V=0. This tangent space is naturally identified with the Lie Algebra via the adjoint rep. If dp(g)V=0, then Exp(tV) fixes m, which means Exp(tV) is in the fixed group of m. Thus, d/dt(t=0)f(Exp(V t)m)=0. Since GradNonius lives in a space that is orthogonal to the vertical fibers, it is in a horizontal lift, which means that it lives in a space that is naturally identified with the tangent space of m. It also means that Exp(GradNonius(f)(m)t)m is a geodesic in G/H passing through m.Anyway, all of this means that grad descent is relatively easy. Nonius,What a bunch of crap. Your manifold ain't smooth.RockyLifeless Rock,why do you communicate to a living organism this information? Yes, I agree, rocks are never smooth. Rocks, like you, have sharp edges n shit. I never ever ever met a rock that understood math. Are you communicating to me that you, Rock, actually understand the math behind this? Is there a seperate realm of math that is applicable to the universe of rocks? Perhaps I am naive. I have an open mind though. Teach me about the way rocks, stones, dirt, and such understand the logic and beauty of mathematics.

frolloos
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### Re: A Differential Geometric Question for all of you geometers out there.

So in two dimensions the Ricci flow equation is the heat equation for the metric. That's quite interesting isn't it?

Cuchulainn
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### Re: A Differential Geometric Question for all of you geometers out there.

So in two dimensions the Ricci flow equation is the heat equation for the metric. That's quite interesting isn't it?
Do you have a concrete example, as Paul Halmos would say?
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frolloos
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### Re: A Differential Geometric Question for all of you geometers out there.

Do you have a concrete example, as Paul Halmos would say?
Nope, because I had meant to write "So is in two dimensions the Ricci flow equation a heat equation for the metric?". Missed the "?" sorry.

Actually for any Einstein manifold by definition the Ricci tensor satisfies $R_{ab} = k g_{ab}$, examples in two dimensions are the sphere and hyperbolic plane (so I should have wrote in two dimensions for surfaces of constant curvature).

The Ricci flow equation is $\partial_t g_{ab} = -2 R_{ab}$. For Einstein manifolds that becomes $\partial_t g_{ab} = -2k g_{ab}$. I am just wondering if there is a metric such that the action of the Laplacian on the metric gives the constant curvature $k$, i.e. $\Delta g = k g$ which looks like an eigenvalue equation. Anyway, probably doesn't make sense But the formal similarity between ricci flow and heat equation is interesting.

FaridMoussaoui
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### Re: A Differential Geometric Question for all of you geometers out there.

Rcci flow -> heat equation was the main ingredient of Richard Hamilton "program" to advance the studies on Poincaré conjecture. Perelman completed Hamilton program to prove the conjecture.

frolloos
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### Re: A Differential Geometric Question for all of you geometers out there.

Rcci flow -> heat equation was the main ingredient of Richard Hamilton "program" to advance the studies on Poincaré conjecture. Perelman completed Hamilton program to prove the conjecture.
But what does Ricci flow *mean*? Specifically, is it a process that occurs in nature, or is it purely a construct/strategy to solve certain problems?

Cuchulainn
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### Re: A Differential Geometric Question for all of you geometers out there.

But the formal similarity between ricci flow and heat equation is interesting.

Glancing at Wiki, it states that it smooths the metric in the same that the solution of the heat equation is smoothed from its initial conditions. So, the Riemannian satisfies a heat equation! Why not?

Seems to used in computer graphics, which is not surprising I suppose.

https://www.researchgate.net/profile/Fe ... ion_detail

From the article they state that discrete RF is the negative gradient of the discrete Ricci energy and the metric minimises this energy. See the pic for the heat equation .. energy is getting less and less.

Doesn't sound out of this world. That was my take.
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Cuchulainn
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### Re: A Differential Geometric Question for all of you geometers out there.

This looks nice because it explains Ricci flow in the context of parabolic PDEs (generalisation of heat equaton)
Attachments
Parabolic PDEs and Ricci Flow.pdf
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frolloos
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### Re: A Differential Geometric Question for all of you geometers out there.

Thanks for the notes Cuch, does look like a good set of notes.

Quickly read it and lost it towards the end: Ricci flow is weakly parabolic and then using a trick it becomes strongly parabolic (??). Need to spend more time on it. I am actually reading a book on heat kernel expansions and, well, things seem to be interconnected

Through Feynman-Kac you can link parabolic PDEs to SDEs. Letting my imagination run (too) wild the question I also have is if there is a Feynman-Kac for generalised heat equations such as Ricci flow and what does it mean to have a SDE for a metric.

Cuchulainn
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### Re: A Differential Geometric Question for all of you geometers out there.

Quickly read it and lost it towards the end: Ricci flow is weakly parabolic and then using a trick it becomes strongly parabolic (??). Need to spend more time on it. I am actually reading a book on heat kernel expansions and, well, things seem to be interconnected

DD see section 2.1, Weak ==> a_ij is only positive semi-definite. A transformation e.g. x = logS can make it positive definite.
It's like elliptic operators. I suspect they make the distinction for boundaries where no boundary conditions are needed/allowed.(?)

Through Feynman-Kac you can link parabolic PDEs to SDEs. Letting my imagination run (too) wild the question I also have is if there is a Feynman-Kac for generalised heat equations such as Ricci flow and what does it mean to have a SDE for a metric.

DD Alan and myself were discussing the SDE-PDE connection a few days ago when the manifold is n-dim Euclidean space.

https://forum.wilmott.com/viewtopic.php?f=34&t=67488

IMO all this stuff is about minimisation.
Last edited by Cuchulainn on February 12th, 2020, 11:43 am, edited 1 time in total.
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