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Reference request: Recovering a Riemannian metric from the distance function
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Let $M = (M, g)$ be a Riemannian manifold, and let $p in M$.
Writing $d$ for the geodesic distance in $M$, there is a function
$$
d(-, p)^2 : M to mathbbR.
$$
This function is smooth near $p$. Hence for each point $x in M$ sufficiently close to $p$, we have the Hessian
$$
textHess_x(d(-, p)^2)
$$
(defined using the Levi-Civita connection), which is a bilinear form on $T_x M$. In particular, we can take $x$ to be equal to $p$ itself, giving a bilinear form
$$
textHess_p(d(-, p)^2)
$$
on $T_p M$. But of course, we already have another bilinear form on $T_p M$, namely, the Riemannian metric $g_p$ itself. And the fact is that up to a constant factor, these two forms are equal:
$$
g_p = frac12 textHess_p(d(-, p)^2).
$$
I'm looking for a reference for this fact. For the purposes of what I'm writing, it would ideally be a reference that states this fact in the same simple direct terms as above, without involving any other differential-geometric concepts (e.g. normal coordinates).
I understand that this is a basic fact of Riemannian geometry, so I've already looked for it in various introductions to the subject, including those by do Carmo, Jost, Lee, and Petersen. But I haven't found it stated in any of those sources (which isn't to say it's not there). I have found more sophisticated stuff about $textHess_x(d(-, p)^2)$ for points $x$ different from $p$, but not the simple fact I'm looking for.
Requests for references often result in people giving their favourite proofs rather than a reference. While that doesn't do any harm (and can be quite interesting), I emphasize that it's a reference I'm looking for, not a proof.
reference-request dg.differential-geometry riemannian-geometry
asked Aug 9 at 17:56
Tom Leinster
18.9k474125
 |Â
show 2 more comments
up vote
9
down vote
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Let $M = (M, g)$ be a Riemannian manifold, and let $p in M$.
Writing $d$ for the geodesic distance in $M$, there is a function
$$
d(-, p)^2 : M to mathbbR.
$$
This function is smooth near $p$. Hence for each point $x in M$ sufficiently close to $p$, we have the Hessian
$$
textHess_x(d(-, p)^2)
$$
(defined using the Levi-Civita connection), which is a bilinear form on $T_x M$. In particular, we can take $x$ to be equal to $p$ itself, giving a bilinear form
$$
textHess_p(d(-, p)^2)
$$
on $T_p M$. But of course, we already have another bilinear form on $T_p M$, namely, the Riemannian metric $g_p$ itself. And the fact is that up to a constant factor, these two forms are equal:
$$
g_p = frac12 textHess_p(d(-, p)^2).
$$
I'm looking for a reference for this fact. For the purposes of what I'm writing, it would ideally be a reference that states this fact in the same simple direct terms as above, without involving any other differential-geometric concepts (e.g. normal coordinates).
I understand that this is a basic fact of Riemannian geometry, so I've already looked for it in various introductions to the subject, including those by do Carmo, Jost, Lee, and Petersen. But I haven't found it stated in any of those sources (which isn't to say it's not there). I have found more sophisticated stuff about $textHess_x(d(-, p)^2)$ for points $x$ different from $p$, but not the simple fact I'm looking for.
Requests for references often result in people giving their favourite proofs rather than a reference. While that doesn't do any harm (and can be quite interesting), I emphasize that it's a reference I'm looking for, not a proof.
reference-request dg.differential-geometry riemannian-geometry
asked Aug 9 at 17:56
Tom Leinster
18.9k474125
1
I asked a similar question a while ago on MSE, see math.stackexchange.com/questions/1161589/…. My answer there seems to be related to what you are looking for. In addition, there is another answer that I haven't been able to make sense of.
– S.Surace
Aug 9 at 20:47
@S.Surace: thanks, I hadn't seen that MSE question. Nothing there answers my question (i.e. provides a reference to the stated equation), but it seems that you're interested in this stuff for similar reasons to me. In particular, I'd seen the some of that literature on contrast functions that you mention in your MSE answer, which seems to take as its starting point the result that I want a reference for.
– Tom Leinster
Aug 9 at 21:28
@S.Surace: It seems that you have read only the title of this post, but not its content. Indeed, the title suggests a completely diffferent question - the one that you have asked.
– Alex M.
Aug 9 at 21:58
@AlexM. Am I sure it's true? I believe it's true because someone whose expertise I trust tells me that it is. For a proof, they pointed me to p.4-5 of the paper "Hessian of the Riemannian squared distance" by Pennec: www-sop.inria.fr/members/Xavier.Pennec/… . But the fact I'm interested in isn't stated directly there; you have to do a bit of work to dig it out. I'm looking for a reference where it's stated directly.
– Tom Leinster
Aug 9 at 22:09
It seems to me that the answer is stated explicitly in equation (5) in the Pennec paper, if you take into account the displayed equation after (2). I'm not sure you'll get anything more explicit than that, My preferred approach to this is the equation of Villani stated in the last sentence before section 2.2.
– Deane Yang
Aug 10 at 0:48
 |Â
show 2 more comments
up vote
9
down vote
favorite
up vote
9
down vote
favorite
Let $M = (M, g)$ be a Riemannian manifold, and let $p in M$.
Writing $d$ for the geodesic distance in $M$, there is a function
$$
d(-, p)^2 : M to mathbbR.
$$
This function is smooth near $p$. Hence for each point $x in M$ sufficiently close to $p$, we have the Hessian
$$
textHess_x(d(-, p)^2)
$$
(defined using the Levi-Civita connection), which is a bilinear form on $T_x M$. In particular, we can take $x$ to be equal to $p$ itself, giving a bilinear form
$$
textHess_p(d(-, p)^2)
$$
on $T_p M$. But of course, we already have another bilinear form on $T_p M$, namely, the Riemannian metric $g_p$ itself. And the fact is that up to a constant factor, these two forms are equal:
$$
g_p = frac12 textHess_p(d(-, p)^2).
$$
I'm looking for a reference for this fact. For the purposes of what I'm writing, it would ideally be a reference that states this fact in the same simple direct terms as above, without involving any other differential-geometric concepts (e.g. normal coordinates).
I understand that this is a basic fact of Riemannian geometry, so I've already looked for it in various introductions to the subject, including those by do Carmo, Jost, Lee, and Petersen. But I haven't found it stated in any of those sources (which isn't to say it's not there). I have found more sophisticated stuff about $textHess_x(d(-, p)^2)$ for points $x$ different from $p$, but not the simple fact I'm looking for.
Requests for references often result in people giving their favourite proofs rather than a reference. While that doesn't do any harm (and can be quite interesting), I emphasize that it's a reference I'm looking for, not a proof.
reference-request dg.differential-geometry riemannian-geometry
asked Aug 9 at 17:56
Tom Leinster
18.9k474125
Let $M = (M, g)$ be a Riemannian manifold, and let $p in M$.
Writing $d$ for the geodesic distance in $M$, there is a function
$$
d(-, p)^2 : M to mathbbR.
$$
This function is smooth near $p$. Hence for each point $x in M$ sufficiently close to $p$, we have the Hessian
$$
textHess_x(d(-, p)^2)
$$
(defined using the Levi-Civita connection), which is a bilinear form on $T_x M$. In particular, we can take $x$ to be equal to $p$ itself, giving a bilinear form
$$
textHess_p(d(-, p)^2)
$$
on $T_p M$. But of course, we already have another bilinear form on $T_p M$, namely, the Riemannian metric $g_p$ itself. And the fact is that up to a constant factor, these two forms are equal:
$$
g_p = frac12 textHess_p(d(-, p)^2).
$$
I'm looking for a reference for this fact. For the purposes of what I'm writing, it would ideally be a reference that states this fact in the same simple direct terms as above, without involving any other differential-geometric concepts (e.g. normal coordinates).
I understand that this is a basic fact of Riemannian geometry, so I've already looked for it in various introductions to the subject, including those by do Carmo, Jost, Lee, and Petersen. But I haven't found it stated in any of those sources (which isn't to say it's not there). I have found more sophisticated stuff about $textHess_x(d(-, p)^2)$ for points $x$ different from $p$, but not the simple fact I'm looking for.
Requests for references often result in people giving their favourite proofs rather than a reference. While that doesn't do any harm (and can be quite interesting), I emphasize that it's a reference I'm looking for, not a proof.
reference-request dg.differential-geometry riemannian-geometry
asked Aug 9 at 17:56
Tom Leinster
18.9k474125
edited Aug 9 at 21:42
asked Aug 9 at 17:56
Tom Leinster
18.9k474125
asked Aug 9 at 17:56
Tom Leinster
18.9k474125
asked Aug 9 at 17:56
Tom Leinster
18.9k474125
18.9k474125
1
I asked a similar question a while ago on MSE, see math.stackexchange.com/questions/1161589/…. My answer there seems to be related to what you are looking for. In addition, there is another answer that I haven't been able to make sense of.
– S.Surace
Aug 9 at 20:47
@S.Surace: thanks, I hadn't seen that MSE question. Nothing there answers my question (i.e. provides a reference to the stated equation), but it seems that you're interested in this stuff for similar reasons to me. In particular, I'd seen the some of that literature on contrast functions that you mention in your MSE answer, which seems to take as its starting point the result that I want a reference for.
– Tom Leinster
Aug 9 at 21:28
@S.Surace: It seems that you have read only the title of this post, but not its content. Indeed, the title suggests a completely diffferent question - the one that you have asked.
– Alex M.
Aug 9 at 21:58
@AlexM. Am I sure it's true? I believe it's true because someone whose expertise I trust tells me that it is. For a proof, they pointed me to p.4-5 of the paper "Hessian of the Riemannian squared distance" by Pennec: www-sop.inria.fr/members/Xavier.Pennec/… . But the fact I'm interested in isn't stated directly there; you have to do a bit of work to dig it out. I'm looking for a reference where it's stated directly.
– Tom Leinster
Aug 9 at 22:09
It seems to me that the answer is stated explicitly in equation (5) in the Pennec paper, if you take into account the displayed equation after (2). I'm not sure you'll get anything more explicit than that, My preferred approach to this is the equation of Villani stated in the last sentence before section 2.2.
– Deane Yang
Aug 10 at 0:48
 |Â
show 2 more comments
1
I asked a similar question a while ago on MSE, see math.stackexchange.com/questions/1161589/…. My answer there seems to be related to what you are looking for. In addition, there is another answer that I haven't been able to make sense of.
– S.Surace
Aug 9 at 20:47
@S.Surace: thanks, I hadn't seen that MSE question. Nothing there answers my question (i.e. provides a reference to the stated equation), but it seems that you're interested in this stuff for similar reasons to me. In particular, I'd seen the some of that literature on contrast functions that you mention in your MSE answer, which seems to take as its starting point the result that I want a reference for.
– Tom Leinster
Aug 9 at 21:28
@S.Surace: It seems that you have read only the title of this post, but not its content. Indeed, the title suggests a completely diffferent question - the one that you have asked.
– Alex M.
Aug 9 at 21:58
@AlexM. Am I sure it's true? I believe it's true because someone whose expertise I trust tells me that it is. For a proof, they pointed me to p.4-5 of the paper "Hessian of the Riemannian squared distance" by Pennec: www-sop.inria.fr/members/Xavier.Pennec/… . But the fact I'm interested in isn't stated directly there; you have to do a bit of work to dig it out. I'm looking for a reference where it's stated directly.
– Tom Leinster
Aug 9 at 22:09
It seems to me that the answer is stated explicitly in equation (5) in the Pennec paper, if you take into account the displayed equation after (2). I'm not sure you'll get anything more explicit than that, My preferred approach to this is the equation of Villani stated in the last sentence before section 2.2.
– Deane Yang
Aug 10 at 0:48
1
1
I asked a similar question a while ago on MSE, see math.stackexchange.com/questions/1161589/…. My answer there seems to be related to what you are looking for. In addition, there is another answer that I haven't been able to make sense of.
– S.Surace
Aug 9 at 20:47
I asked a similar question a while ago on MSE, see math.stackexchange.com/questions/1161589/…. My answer there seems to be related to what you are looking for. In addition, there is another answer that I haven't been able to make sense of.
– S.Surace
Aug 9 at 20:47
@S.Surace: thanks, I hadn't seen that MSE question. Nothing there answers my question (i.e. provides a reference to the stated equation), but it seems that you're interested in this stuff for similar reasons to me. In particular, I'd seen the some of that literature on contrast functions that you mention in your MSE answer, which seems to take as its starting point the result that I want a reference for.
– Tom Leinster
Aug 9 at 21:28
@S.Surace: thanks, I hadn't seen that MSE question. Nothing there answers my question (i.e. provides a reference to the stated equation), but it seems that you're interested in this stuff for similar reasons to me. In particular, I'd seen the some of that literature on contrast functions that you mention in your MSE answer, which seems to take as its starting point the result that I want a reference for.
– Tom Leinster
Aug 9 at 21:28
@S.Surace: It seems that you have read only the title of this post, but not its content. Indeed, the title suggests a completely diffferent question - the one that you have asked.
– Alex M.
Aug 9 at 21:58
@S.Surace: It seems that you have read only the title of this post, but not its content. Indeed, the title suggests a completely diffferent question - the one that you have asked.
– Alex M.
Aug 9 at 21:58
@AlexM. Am I sure it's true? I believe it's true because someone whose expertise I trust tells me that it is. For a proof, they pointed me to p.4-5 of the paper "Hessian of the Riemannian squared distance" by Pennec: www-sop.inria.fr/members/Xavier.Pennec/… . But the fact I'm interested in isn't stated directly there; you have to do a bit of work to dig it out. I'm looking for a reference where it's stated directly.
– Tom Leinster
Aug 9 at 22:09
@AlexM. Am I sure it's true? I believe it's true because someone whose expertise I trust tells me that it is. For a proof, they pointed me to p.4-5 of the paper "Hessian of the Riemannian squared distance" by Pennec: www-sop.inria.fr/members/Xavier.Pennec/… . But the fact I'm interested in isn't stated directly there; you have to do a bit of work to dig it out. I'm looking for a reference where it's stated directly.
– Tom Leinster
Aug 9 at 22:09
It seems to me that the answer is stated explicitly in equation (5) in the Pennec paper, if you take into account the displayed equation after (2). I'm not sure you'll get anything more explicit than that, My preferred approach to this is the equation of Villani stated in the last sentence before section 2.2.
– Deane Yang
Aug 10 at 0:48
It seems to me that the answer is stated explicitly in equation (5) in the Pennec paper, if you take into account the displayed equation after (2). I'm not sure you'll get anything more explicit than that, My preferred approach to this is the equation of Villani stated in the last sentence before section 2.2.
– Deane Yang
Aug 10 at 0:48
 |Â
show 2 more comments
5 Answers
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7
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While it does not answer your question, the following direct argument may clarify certain things:
Since the Hessian is a symmetric bilinear form, it suffices to show $frac12Hess_p(d^2(cdot,p))(v,v)=|v|^2$.
If $f$ is a smooth function near $p$ in $mathbbR^n$, then $Hess_p(f)(v,v)=fracd^2dt^2vert_t=0f(gamma(t)) $, where $gamma$ is any
smooth path with $gamma(0)=p$ and $gamma'(0)=v$. This formula continues to hold, if $p$ is a critical point of $f$ on a manifold (in this case the definition of the Hessian does not rely on the choice of a Riemannian metric).
If $gamma$ is the geodesic through $p$ with $gamma'(0)=v$, then, since $gamma$ is locally distance minimizing, $d(gamma(t),p)=|tv|$ for $t$ near $0$. Combined with the above this gives the result.
(If we use a Riemannian metric $g$ and its associated Levi-Civita connection $nabla$ to define $Hess_p^g(f)$ at a noncritical point $p$ of $f$, then the formula $fracd^2dt^2vert_t=0f(gamma(t)) =Hess^g_p(f)(v,v)$ still holds, if $nabla_tgamma'(0)=0$. This is however not used above).
answered Aug 9 at 23:15
user_1789
34636
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This is described in painstaking detail in the paper of Xavier Pennec (2017). (Hessian of he Riemannian Squared Distance).
answered Aug 9 at 23:29
Igor Rivin
77.2k8109289
Thanks. I've seen that paper (and mentioned it in my conversation with Alex M in the comments above), but I don't see the fact I'm concerned with stated explicitly. You can get it by combining the first equation on p.4 with equation (5) on p.5, at least if you're fluent in normal coordinates. But do you see anywhere in this paper where Pennec states it explicitly?
– Tom Leinster
Aug 9 at 23:38
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For the requested reference: I believe it should follow from inequalities (5.6.6) in Jost (2011, p. 235) (plus user_1789’s polarization argument) because $r(x)to0$ as $xto p$.
answered Aug 9 at 23:46
Francois Ziegler
18.8k368112
Funny: I was looking at that exact theorem when your post popped up. What values of $lambda$ and $mu$ are you taking in Jost's theorem?
– Tom Leinster
Aug 9 at 23:59
I think it doesn’t matter, they simplify in a (l’Hospital) limit, no?
– Francois Ziegler
Aug 10 at 0:13
2
I see what you mean! Good, so that's one reference - thanks. Though it does remind me slightly of the famous mathoverflow.net/a/42519.... I mean, I'd love to have a reference where the statement is made directly rather than derived from something much more sophisticated.
– Tom Leinster
Aug 10 at 0:55
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I believe that the reason why you cannot find the result that you are asking about printed anywhere is that it is, after all, a mere exercise in Riemannian computation. First, it is easy to show that if $f$ is smooth around $p$, then $(Hf)_ij = partial^2_ijf - Gamma_ij ^k partial_k f$ in any system of coordinates around $p$. Now, since your $f = d_p^2$ has radial symmetry, it is natural to continue the work in spherical normal coordinates, i.e. you go in $T_pM$ through $exp_p ^-1$ and there you introduce spherical coordinates $r, sigma_1, dots, sigma_n$, with $n = dim M$. Since $Gamma_ij^k (p) = 0$ as a consequence of your coordinates being normal, you will have $(Hf)_ij (p) = (partial^2_ijf) (p) = (partial ^2 _rr r^2) (p) = 2$ (all the other second-order partial derivatives vanish at $p$ because $f=r^2$ does not contain the variables $sigma_1, dots, sigma_n$).
On the other hand, it is known that in normal spherical coordinates the expression of the metric tensor is $g_ij = delta_ij + o(r)$, so that $g_ij (p) = delta_ij (p)$ (the Kronecker symbol), whence it follows that $(Hf)(p) = 2g(p)$ (the metric evaluated at $p$). See p.114 of I. Chavel, "Riemannian Geometry - A Modern Introduction", 2006, or the more general theorem 2.53 of Cartan on p.83 of S. Rosenberg, "The Laplacian on a Riemannian Manifold", 1997, or Petersen's book cited here.
answered Aug 10 at 6:41
Alex M.
2,32121531
Thanks for the references. I'll look up Chavel and Rosenberg (but as I said in the question, I've already tried Petersen).
– Tom Leinster
Aug 10 at 12:42
My guess is that you won't find it anywhere, simply because it's an easy exercise. It doesn't "deserve" to be a theorem. If you want to use it in an article, do like it's often done in differential geometry: prepend the statement by the words "it's well known that".
– Alex M.
Aug 10 at 13:45
2
Hmm. I'm not convinced by the principle of your argument. There are many easy exercises that are found in introductions to the subject concerned - e.g. it's an easy exercise that inverses in a group are unique, and you'll find that stated & proved in every intro to group theory. Personally, I like to give good references when I can. I think the culture of saying "it's well known" without giving a reference is tremendously off-putting to non-expert readers, and contributes to the harmful atomization of mathematics. All that said, you may be right that I won't find what I want.
– Tom Leinster
Aug 10 at 13:57
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If I am not mistaken, in order to define the Hessian you need to fix a connection. I suspect the Riemannian metric you get will depend on this connection, as Finsler metrics also have distance functions.
answered Aug 9 at 20:13
David Hughes
291211
Right; I meant the Hessian with respect to the Levi-Civita connection.
– Tom Leinster
Aug 9 at 20:16
Actually, since $p$ is a critical point of $d(-,p)^2$, one can define the Hessian in $p$ without any choice of connection (just locally as the Hessian of that function in a chart).
– Panagiotis Konstantis
Aug 9 at 20:25
I believe the way to avoid using the Levi-Civita connection (which already determines the metric) is to use the distance function to define geodesics as length minimizing curves and define the Hessian using the second derivative of $d^2$ in the direction of each geodesic. However, there are lots of details to work out. There's a slight chance that this is worked out in the book by Gromov et all, Metric Structure for Riemannian and non-Riemannian Spaces.
– Deane Yang
Aug 9 at 20:38
1
It might appear in a paper that is trying to show that a length space with some additional properties is in fact a Riemannan manifold. Alas, I don’t know of any offhand.
– Deane Yang
Aug 9 at 22:53
1
I suggest looking at the papers by Karcher with his collaborators that are cited in Smith, P. D.; Yang, Deane, "Removing point singularities of Riemannian manifolds", especially those that discuss what they call "almost linear coordinates", which are coordinates constructed using only the distance function. What you want should be at the very least a corollary of something proved in one of these papers.
– Deane Yang
Aug 9 at 23:03
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
While it does not answer your question, the following direct argument may clarify certain things:
Since the Hessian is a symmetric bilinear form, it suffices to show $frac12Hess_p(d^2(cdot,p))(v,v)=|v|^2$.
If $f$ is a smooth function near $p$ in $mathbbR^n$, then $Hess_p(f)(v,v)=fracd^2dt^2vert_t=0f(gamma(t)) $, where $gamma$ is any
smooth path with $gamma(0)=p$ and $gamma'(0)=v$. This formula continues to hold, if $p$ is a critical point of $f$ on a manifold (in this case the definition of the Hessian does not rely on the choice of a Riemannian metric).
If $gamma$ is the geodesic through $p$ with $gamma'(0)=v$, then, since $gamma$ is locally distance minimizing, $d(gamma(t),p)=|tv|$ for $t$ near $0$. Combined with the above this gives the result.
(If we use a Riemannian metric $g$ and its associated Levi-Civita connection $nabla$ to define $Hess_p^g(f)$ at a noncritical point $p$ of $f$, then the formula $fracd^2dt^2vert_t=0f(gamma(t)) =Hess^g_p(f)(v,v)$ still holds, if $nabla_tgamma'(0)=0$. This is however not used above).
answered Aug 9 at 23:15
user_1789
34636
add a comment |Â
up vote
7
down vote
While it does not answer your question, the following direct argument may clarify certain things:
Since the Hessian is a symmetric bilinear form, it suffices to show $frac12Hess_p(d^2(cdot,p))(v,v)=|v|^2$.
If $f$ is a smooth function near $p$ in $mathbbR^n$, then $Hess_p(f)(v,v)=fracd^2dt^2vert_t=0f(gamma(t)) $, where $gamma$ is any
smooth path with $gamma(0)=p$ and $gamma'(0)=v$. This formula continues to hold, if $p$ is a critical point of $f$ on a manifold (in this case the definition of the Hessian does not rely on the choice of a Riemannian metric).
If $gamma$ is the geodesic through $p$ with $gamma'(0)=v$, then, since $gamma$ is locally distance minimizing, $d(gamma(t),p)=|tv|$ for $t$ near $0$. Combined with the above this gives the result.
(If we use a Riemannian metric $g$ and its associated Levi-Civita connection $nabla$ to define $Hess_p^g(f)$ at a noncritical point $p$ of $f$, then the formula $fracd^2dt^2vert_t=0f(gamma(t)) =Hess^g_p(f)(v,v)$ still holds, if $nabla_tgamma'(0)=0$. This is however not used above).
answered Aug 9 at 23:15
user_1789
34636
add a comment |Â
up vote
7
down vote
up vote
7
down vote
While it does not answer your question, the following direct argument may clarify certain things:
Since the Hessian is a symmetric bilinear form, it suffices to show $frac12Hess_p(d^2(cdot,p))(v,v)=|v|^2$.
If $f$ is a smooth function near $p$ in $mathbbR^n$, then $Hess_p(f)(v,v)=fracd^2dt^2vert_t=0f(gamma(t)) $, where $gamma$ is any
smooth path with $gamma(0)=p$ and $gamma'(0)=v$. This formula continues to hold, if $p$ is a critical point of $f$ on a manifold (in this case the definition of the Hessian does not rely on the choice of a Riemannian metric).
If $gamma$ is the geodesic through $p$ with $gamma'(0)=v$, then, since $gamma$ is locally distance minimizing, $d(gamma(t),p)=|tv|$ for $t$ near $0$. Combined with the above this gives the result.
(If we use a Riemannian metric $g$ and its associated Levi-Civita connection $nabla$ to define $Hess_p^g(f)$ at a noncritical point $p$ of $f$, then the formula $fracd^2dt^2vert_t=0f(gamma(t)) =Hess^g_p(f)(v,v)$ still holds, if $nabla_tgamma'(0)=0$. This is however not used above).
answered Aug 9 at 23:15
user_1789
34636
While it does not answer your question, the following direct argument may clarify certain things:
Since the Hessian is a symmetric bilinear form, it suffices to show $frac12Hess_p(d^2(cdot,p))(v,v)=|v|^2$.
If $f$ is a smooth function near $p$ in $mathbbR^n$, then $Hess_p(f)(v,v)=fracd^2dt^2vert_t=0f(gamma(t)) $, where $gamma$ is any
smooth path with $gamma(0)=p$ and $gamma'(0)=v$. This formula continues to hold, if $p$ is a critical point of $f$ on a manifold (in this case the definition of the Hessian does not rely on the choice of a Riemannian metric).
If $gamma$ is the geodesic through $p$ with $gamma'(0)=v$, then, since $gamma$ is locally distance minimizing, $d(gamma(t),p)=|tv|$ for $t$ near $0$. Combined with the above this gives the result.
(If we use a Riemannian metric $g$ and its associated Levi-Civita connection $nabla$ to define $Hess_p^g(f)$ at a noncritical point $p$ of $f$, then the formula $fracd^2dt^2vert_t=0f(gamma(t)) =Hess^g_p(f)(v,v)$ still holds, if $nabla_tgamma'(0)=0$. This is however not used above).
answered Aug 9 at 23:15
user_1789
34636
edited Aug 10 at 7:42
answered Aug 9 at 23:15
user_1789
34636
answered Aug 9 at 23:15
user_1789
34636
answered Aug 9 at 23:15
user_1789
34636
34636
add a comment |Â
add a comment |Â
up vote
3
down vote
This is described in painstaking detail in the paper of Xavier Pennec (2017). (Hessian of he Riemannian Squared Distance).
answered Aug 9 at 23:29
Igor Rivin
77.2k8109289
Thanks. I've seen that paper (and mentioned it in my conversation with Alex M in the comments above), but I don't see the fact I'm concerned with stated explicitly. You can get it by combining the first equation on p.4 with equation (5) on p.5, at least if you're fluent in normal coordinates. But do you see anywhere in this paper where Pennec states it explicitly?
– Tom Leinster
Aug 9 at 23:38
add a comment |Â
up vote
3
down vote
This is described in painstaking detail in the paper of Xavier Pennec (2017). (Hessian of he Riemannian Squared Distance).
answered Aug 9 at 23:29
Igor Rivin
77.2k8109289
Thanks. I've seen that paper (and mentioned it in my conversation with Alex M in the comments above), but I don't see the fact I'm concerned with stated explicitly. You can get it by combining the first equation on p.4 with equation (5) on p.5, at least if you're fluent in normal coordinates. But do you see anywhere in this paper where Pennec states it explicitly?
– Tom Leinster
Aug 9 at 23:38
add a comment |Â
up vote
3
down vote
up vote
3
down vote
This is described in painstaking detail in the paper of Xavier Pennec (2017). (Hessian of he Riemannian Squared Distance).
answered Aug 9 at 23:29
Igor Rivin
77.2k8109289
This is described in painstaking detail in the paper of Xavier Pennec (2017). (Hessian of he Riemannian Squared Distance).
answered Aug 9 at 23:29
Igor Rivin
77.2k8109289
answered Aug 9 at 23:29
Igor Rivin
77.2k8109289
answered Aug 9 at 23:29
Igor Rivin
77.2k8109289
answered Aug 9 at 23:29
Igor Rivin
77.2k8109289
77.2k8109289
Thanks. I've seen that paper (and mentioned it in my conversation with Alex M in the comments above), but I don't see the fact I'm concerned with stated explicitly. You can get it by combining the first equation on p.4 with equation (5) on p.5, at least if you're fluent in normal coordinates. But do you see anywhere in this paper where Pennec states it explicitly?
– Tom Leinster
Aug 9 at 23:38
add a comment |Â
Thanks. I've seen that paper (and mentioned it in my conversation with Alex M in the comments above), but I don't see the fact I'm concerned with stated explicitly. You can get it by combining the first equation on p.4 with equation (5) on p.5, at least if you're fluent in normal coordinates. But do you see anywhere in this paper where Pennec states it explicitly?
– Tom Leinster
Aug 9 at 23:38
Thanks. I've seen that paper (and mentioned it in my conversation with Alex M in the comments above), but I don't see the fact I'm concerned with stated explicitly. You can get it by combining the first equation on p.4 with equation (5) on p.5, at least if you're fluent in normal coordinates. But do you see anywhere in this paper where Pennec states it explicitly?
– Tom Leinster
Aug 9 at 23:38
Thanks. I've seen that paper (and mentioned it in my conversation with Alex M in the comments above), but I don't see the fact I'm concerned with stated explicitly. You can get it by combining the first equation on p.4 with equation (5) on p.5, at least if you're fluent in normal coordinates. But do you see anywhere in this paper where Pennec states it explicitly?
– Tom Leinster
Aug 9 at 23:38
add a comment |Â
up vote
1
down vote
For the requested reference: I believe it should follow from inequalities (5.6.6) in Jost (2011, p. 235) (plus user_1789’s polarization argument) because $r(x)to0$ as $xto p$.
answered Aug 9 at 23:46
Francois Ziegler
18.8k368112
Funny: I was looking at that exact theorem when your post popped up. What values of $lambda$ and $mu$ are you taking in Jost's theorem?
– Tom Leinster
Aug 9 at 23:59
I think it doesn’t matter, they simplify in a (l’Hospital) limit, no?
– Francois Ziegler
Aug 10 at 0:13
2
I see what you mean! Good, so that's one reference - thanks. Though it does remind me slightly of the famous mathoverflow.net/a/42519.... I mean, I'd love to have a reference where the statement is made directly rather than derived from something much more sophisticated.
– Tom Leinster
Aug 10 at 0:55
add a comment |Â
up vote
1
down vote
For the requested reference: I believe it should follow from inequalities (5.6.6) in Jost (2011, p. 235) (plus user_1789’s polarization argument) because $r(x)to0$ as $xto p$.
answered Aug 9 at 23:46
Francois Ziegler
18.8k368112
Funny: I was looking at that exact theorem when your post popped up. What values of $lambda$ and $mu$ are you taking in Jost's theorem?
– Tom Leinster
Aug 9 at 23:59
I think it doesn’t matter, they simplify in a (l’Hospital) limit, no?
– Francois Ziegler
Aug 10 at 0:13
2
I see what you mean! Good, so that's one reference - thanks. Though it does remind me slightly of the famous mathoverflow.net/a/42519.... I mean, I'd love to have a reference where the statement is made directly rather than derived from something much more sophisticated.
– Tom Leinster
Aug 10 at 0:55
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For the requested reference: I believe it should follow from inequalities (5.6.6) in Jost (2011, p. 235) (plus user_1789’s polarization argument) because $r(x)to0$ as $xto p$.
answered Aug 9 at 23:46
Francois Ziegler
18.8k368112
For the requested reference: I believe it should follow from inequalities (5.6.6) in Jost (2011, p. 235) (plus user_1789’s polarization argument) because $r(x)to0$ as $xto p$.
answered Aug 9 at 23:46
Francois Ziegler
18.8k368112
answered Aug 9 at 23:46
Francois Ziegler
18.8k368112
answered Aug 9 at 23:46
Francois Ziegler
18.8k368112
answered Aug 9 at 23:46
Francois Ziegler
18.8k368112
18.8k368112
Funny: I was looking at that exact theorem when your post popped up. What values of $lambda$ and $mu$ are you taking in Jost's theorem?
– Tom Leinster
Aug 9 at 23:59
I think it doesn’t matter, they simplify in a (l’Hospital) limit, no?
– Francois Ziegler
Aug 10 at 0:13
2
I see what you mean! Good, so that's one reference - thanks. Though it does remind me slightly of the famous mathoverflow.net/a/42519.... I mean, I'd love to have a reference where the statement is made directly rather than derived from something much more sophisticated.
– Tom Leinster
Aug 10 at 0:55
add a comment |Â
Funny: I was looking at that exact theorem when your post popped up. What values of $lambda$ and $mu$ are you taking in Jost's theorem?
– Tom Leinster
Aug 9 at 23:59
I think it doesn’t matter, they simplify in a (l’Hospital) limit, no?
– Francois Ziegler
Aug 10 at 0:13
2
I see what you mean! Good, so that's one reference - thanks. Though it does remind me slightly of the famous mathoverflow.net/a/42519.... I mean, I'd love to have a reference where the statement is made directly rather than derived from something much more sophisticated.
– Tom Leinster
Aug 10 at 0:55
Funny: I was looking at that exact theorem when your post popped up. What values of $lambda$ and $mu$ are you taking in Jost's theorem?
– Tom Leinster
Aug 9 at 23:59
Funny: I was looking at that exact theorem when your post popped up. What values of $lambda$ and $mu$ are you taking in Jost's theorem?
– Tom Leinster
Aug 9 at 23:59
I think it doesn’t matter, they simplify in a (l’Hospital) limit, no?
– Francois Ziegler
Aug 10 at 0:13
I think it doesn’t matter, they simplify in a (l’Hospital) limit, no?
– Francois Ziegler
Aug 10 at 0:13
2
2
I see what you mean! Good, so that's one reference - thanks. Though it does remind me slightly of the famous mathoverflow.net/a/42519.... I mean, I'd love to have a reference where the statement is made directly rather than derived from something much more sophisticated.
– Tom Leinster
Aug 10 at 0:55
I see what you mean! Good, so that's one reference - thanks. Though it does remind me slightly of the famous mathoverflow.net/a/42519.... I mean, I'd love to have a reference where the statement is made directly rather than derived from something much more sophisticated.
– Tom Leinster
Aug 10 at 0:55
add a comment |Â
up vote
1
down vote
I believe that the reason why you cannot find the result that you are asking about printed anywhere is that it is, after all, a mere exercise in Riemannian computation. First, it is easy to show that if $f$ is smooth around $p$, then $(Hf)_ij = partial^2_ijf - Gamma_ij ^k partial_k f$ in any system of coordinates around $p$. Now, since your $f = d_p^2$ has radial symmetry, it is natural to continue the work in spherical normal coordinates, i.e. you go in $T_pM$ through $exp_p ^-1$ and there you introduce spherical coordinates $r, sigma_1, dots, sigma_n$, with $n = dim M$. Since $Gamma_ij^k (p) = 0$ as a consequence of your coordinates being normal, you will have $(Hf)_ij (p) = (partial^2_ijf) (p) = (partial ^2 _rr r^2) (p) = 2$ (all the other second-order partial derivatives vanish at $p$ because $f=r^2$ does not contain the variables $sigma_1, dots, sigma_n$).
On the other hand, it is known that in normal spherical coordinates the expression of the metric tensor is $g_ij = delta_ij + o(r)$, so that $g_ij (p) = delta_ij (p)$ (the Kronecker symbol), whence it follows that $(Hf)(p) = 2g(p)$ (the metric evaluated at $p$). See p.114 of I. Chavel, "Riemannian Geometry - A Modern Introduction", 2006, or the more general theorem 2.53 of Cartan on p.83 of S. Rosenberg, "The Laplacian on a Riemannian Manifold", 1997, or Petersen's book cited here.
answered Aug 10 at 6:41
Alex M.
2,32121531
Thanks for the references. I'll look up Chavel and Rosenberg (but as I said in the question, I've already tried Petersen).
– Tom Leinster
Aug 10 at 12:42
My guess is that you won't find it anywhere, simply because it's an easy exercise. It doesn't "deserve" to be a theorem. If you want to use it in an article, do like it's often done in differential geometry: prepend the statement by the words "it's well known that".
– Alex M.
Aug 10 at 13:45
2
Hmm. I'm not convinced by the principle of your argument. There are many easy exercises that are found in introductions to the subject concerned - e.g. it's an easy exercise that inverses in a group are unique, and you'll find that stated & proved in every intro to group theory. Personally, I like to give good references when I can. I think the culture of saying "it's well known" without giving a reference is tremendously off-putting to non-expert readers, and contributes to the harmful atomization of mathematics. All that said, you may be right that I won't find what I want.
– Tom Leinster
Aug 10 at 13:57
add a comment |Â
up vote
1
down vote
I believe that the reason why you cannot find the result that you are asking about printed anywhere is that it is, after all, a mere exercise in Riemannian computation. First, it is easy to show that if $f$ is smooth around $p$, then $(Hf)_ij = partial^2_ijf - Gamma_ij ^k partial_k f$ in any system of coordinates around $p$. Now, since your $f = d_p^2$ has radial symmetry, it is natural to continue the work in spherical normal coordinates, i.e. you go in $T_pM$ through $exp_p ^-1$ and there you introduce spherical coordinates $r, sigma_1, dots, sigma_n$, with $n = dim M$. Since $Gamma_ij^k (p) = 0$ as a consequence of your coordinates being normal, you will have $(Hf)_ij (p) = (partial^2_ijf) (p) = (partial ^2 _rr r^2) (p) = 2$ (all the other second-order partial derivatives vanish at $p$ because $f=r^2$ does not contain the variables $sigma_1, dots, sigma_n$).
On the other hand, it is known that in normal spherical coordinates the expression of the metric tensor is $g_ij = delta_ij + o(r)$, so that $g_ij (p) = delta_ij (p)$ (the Kronecker symbol), whence it follows that $(Hf)(p) = 2g(p)$ (the metric evaluated at $p$). See p.114 of I. Chavel, "Riemannian Geometry - A Modern Introduction", 2006, or the more general theorem 2.53 of Cartan on p.83 of S. Rosenberg, "The Laplacian on a Riemannian Manifold", 1997, or Petersen's book cited here.
answered Aug 10 at 6:41
Alex M.
2,32121531
Thanks for the references. I'll look up Chavel and Rosenberg (but as I said in the question, I've already tried Petersen).
– Tom Leinster
Aug 10 at 12:42
My guess is that you won't find it anywhere, simply because it's an easy exercise. It doesn't "deserve" to be a theorem. If you want to use it in an article, do like it's often done in differential geometry: prepend the statement by the words "it's well known that".
– Alex M.
Aug 10 at 13:45
2
Hmm. I'm not convinced by the principle of your argument. There are many easy exercises that are found in introductions to the subject concerned - e.g. it's an easy exercise that inverses in a group are unique, and you'll find that stated & proved in every intro to group theory. Personally, I like to give good references when I can. I think the culture of saying "it's well known" without giving a reference is tremendously off-putting to non-expert readers, and contributes to the harmful atomization of mathematics. All that said, you may be right that I won't find what I want.
– Tom Leinster
Aug 10 at 13:57
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I believe that the reason why you cannot find the result that you are asking about printed anywhere is that it is, after all, a mere exercise in Riemannian computation. First, it is easy to show that if $f$ is smooth around $p$, then $(Hf)_ij = partial^2_ijf - Gamma_ij ^k partial_k f$ in any system of coordinates around $p$. Now, since your $f = d_p^2$ has radial symmetry, it is natural to continue the work in spherical normal coordinates, i.e. you go in $T_pM$ through $exp_p ^-1$ and there you introduce spherical coordinates $r, sigma_1, dots, sigma_n$, with $n = dim M$. Since $Gamma_ij^k (p) = 0$ as a consequence of your coordinates being normal, you will have $(Hf)_ij (p) = (partial^2_ijf) (p) = (partial ^2 _rr r^2) (p) = 2$ (all the other second-order partial derivatives vanish at $p$ because $f=r^2$ does not contain the variables $sigma_1, dots, sigma_n$).
On the other hand, it is known that in normal spherical coordinates the expression of the metric tensor is $g_ij = delta_ij + o(r)$, so that $g_ij (p) = delta_ij (p)$ (the Kronecker symbol), whence it follows that $(Hf)(p) = 2g(p)$ (the metric evaluated at $p$). See p.114 of I. Chavel, "Riemannian Geometry - A Modern Introduction", 2006, or the more general theorem 2.53 of Cartan on p.83 of S. Rosenberg, "The Laplacian on a Riemannian Manifold", 1997, or Petersen's book cited here.
answered Aug 10 at 6:41
Alex M.
2,32121531
I believe that the reason why you cannot find the result that you are asking about printed anywhere is that it is, after all, a mere exercise in Riemannian computation. First, it is easy to show that if $f$ is smooth around $p$, then $(Hf)_ij = partial^2_ijf - Gamma_ij ^k partial_k f$ in any system of coordinates around $p$. Now, since your $f = d_p^2$ has radial symmetry, it is natural to continue the work in spherical normal coordinates, i.e. you go in $T_pM$ through $exp_p ^-1$ and there you introduce spherical coordinates $r, sigma_1, dots, sigma_n$, with $n = dim M$. Since $Gamma_ij^k (p) = 0$ as a consequence of your coordinates being normal, you will have $(Hf)_ij (p) = (partial^2_ijf) (p) = (partial ^2 _rr r^2) (p) = 2$ (all the other second-order partial derivatives vanish at $p$ because $f=r^2$ does not contain the variables $sigma_1, dots, sigma_n$).
On the other hand, it is known that in normal spherical coordinates the expression of the metric tensor is $g_ij = delta_ij + o(r)$, so that $g_ij (p) = delta_ij (p)$ (the Kronecker symbol), whence it follows that $(Hf)(p) = 2g(p)$ (the metric evaluated at $p$). See p.114 of I. Chavel, "Riemannian Geometry - A Modern Introduction", 2006, or the more general theorem 2.53 of Cartan on p.83 of S. Rosenberg, "The Laplacian on a Riemannian Manifold", 1997, or Petersen's book cited here.
answered Aug 10 at 6:41
Alex M.
2,32121531
answered Aug 10 at 6:41
Alex M.
2,32121531
answered Aug 10 at 6:41
Alex M.
2,32121531
answered Aug 10 at 6:41
Alex M.
2,32121531
2,32121531
Thanks for the references. I'll look up Chavel and Rosenberg (but as I said in the question, I've already tried Petersen).
– Tom Leinster
Aug 10 at 12:42
My guess is that you won't find it anywhere, simply because it's an easy exercise. It doesn't "deserve" to be a theorem. If you want to use it in an article, do like it's often done in differential geometry: prepend the statement by the words "it's well known that".
– Alex M.
Aug 10 at 13:45
2
Hmm. I'm not convinced by the principle of your argument. There are many easy exercises that are found in introductions to the subject concerned - e.g. it's an easy exercise that inverses in a group are unique, and you'll find that stated & proved in every intro to group theory. Personally, I like to give good references when I can. I think the culture of saying "it's well known" without giving a reference is tremendously off-putting to non-expert readers, and contributes to the harmful atomization of mathematics. All that said, you may be right that I won't find what I want.
– Tom Leinster
Aug 10 at 13:57
add a comment |Â
Thanks for the references. I'll look up Chavel and Rosenberg (but as I said in the question, I've already tried Petersen).
– Tom Leinster
Aug 10 at 12:42
My guess is that you won't find it anywhere, simply because it's an easy exercise. It doesn't "deserve" to be a theorem. If you want to use it in an article, do like it's often done in differential geometry: prepend the statement by the words "it's well known that".
– Alex M.
Aug 10 at 13:45
2
Hmm. I'm not convinced by the principle of your argument. There are many easy exercises that are found in introductions to the subject concerned - e.g. it's an easy exercise that inverses in a group are unique, and you'll find that stated & proved in every intro to group theory. Personally, I like to give good references when I can. I think the culture of saying "it's well known" without giving a reference is tremendously off-putting to non-expert readers, and contributes to the harmful atomization of mathematics. All that said, you may be right that I won't find what I want.
– Tom Leinster
Aug 10 at 13:57
Thanks for the references. I'll look up Chavel and Rosenberg (but as I said in the question, I've already tried Petersen).
– Tom Leinster
Aug 10 at 12:42
Thanks for the references. I'll look up Chavel and Rosenberg (but as I said in the question, I've already tried Petersen).
– Tom Leinster
Aug 10 at 12:42
My guess is that you won't find it anywhere, simply because it's an easy exercise. It doesn't "deserve" to be a theorem. If you want to use it in an article, do like it's often done in differential geometry: prepend the statement by the words "it's well known that".
– Alex M.
Aug 10 at 13:45
My guess is that you won't find it anywhere, simply because it's an easy exercise. It doesn't "deserve" to be a theorem. If you want to use it in an article, do like it's often done in differential geometry: prepend the statement by the words "it's well known that".
– Alex M.
Aug 10 at 13:45
2
2
Hmm. I'm not convinced by the principle of your argument. There are many easy exercises that are found in introductions to the subject concerned - e.g. it's an easy exercise that inverses in a group are unique, and you'll find that stated & proved in every intro to group theory. Personally, I like to give good references when I can. I think the culture of saying "it's well known" without giving a reference is tremendously off-putting to non-expert readers, and contributes to the harmful atomization of mathematics. All that said, you may be right that I won't find what I want.
– Tom Leinster
Aug 10 at 13:57
Hmm. I'm not convinced by the principle of your argument. There are many easy exercises that are found in introductions to the subject concerned - e.g. it's an easy exercise that inverses in a group are unique, and you'll find that stated & proved in every intro to group theory. Personally, I like to give good references when I can. I think the culture of saying "it's well known" without giving a reference is tremendously off-putting to non-expert readers, and contributes to the harmful atomization of mathematics. All that said, you may be right that I won't find what I want.
– Tom Leinster
Aug 10 at 13:57
add a comment |Â
up vote
0
down vote
If I am not mistaken, in order to define the Hessian you need to fix a connection. I suspect the Riemannian metric you get will depend on this connection, as Finsler metrics also have distance functions.
answered Aug 9 at 20:13
David Hughes
291211
Right; I meant the Hessian with respect to the Levi-Civita connection.
– Tom Leinster
Aug 9 at 20:16
Actually, since $p$ is a critical point of $d(-,p)^2$, one can define the Hessian in $p$ without any choice of connection (just locally as the Hessian of that function in a chart).
– Panagiotis Konstantis
Aug 9 at 20:25
I believe the way to avoid using the Levi-Civita connection (which already determines the metric) is to use the distance function to define geodesics as length minimizing curves and define the Hessian using the second derivative of $d^2$ in the direction of each geodesic. However, there are lots of details to work out. There's a slight chance that this is worked out in the book by Gromov et all, Metric Structure for Riemannian and non-Riemannian Spaces.
– Deane Yang
Aug 9 at 20:38
1
It might appear in a paper that is trying to show that a length space with some additional properties is in fact a Riemannan manifold. Alas, I don’t know of any offhand.
– Deane Yang
Aug 9 at 22:53
1
I suggest looking at the papers by Karcher with his collaborators that are cited in Smith, P. D.; Yang, Deane, "Removing point singularities of Riemannian manifolds", especially those that discuss what they call "almost linear coordinates", which are coordinates constructed using only the distance function. What you want should be at the very least a corollary of something proved in one of these papers.
– Deane Yang
Aug 9 at 23:03
 |Â
show 1 more comment
up vote
0
down vote
If I am not mistaken, in order to define the Hessian you need to fix a connection. I suspect the Riemannian metric you get will depend on this connection, as Finsler metrics also have distance functions.
answered Aug 9 at 20:13
David Hughes
291211
Right; I meant the Hessian with respect to the Levi-Civita connection.
– Tom Leinster
Aug 9 at 20:16
Actually, since $p$ is a critical point of $d(-,p)^2$, one can define the Hessian in $p$ without any choice of connection (just locally as the Hessian of that function in a chart).
– Panagiotis Konstantis
Aug 9 at 20:25
I believe the way to avoid using the Levi-Civita connection (which already determines the metric) is to use the distance function to define geodesics as length minimizing curves and define the Hessian using the second derivative of $d^2$ in the direction of each geodesic. However, there are lots of details to work out. There's a slight chance that this is worked out in the book by Gromov et all, Metric Structure for Riemannian and non-Riemannian Spaces.
– Deane Yang
Aug 9 at 20:38
1
It might appear in a paper that is trying to show that a length space with some additional properties is in fact a Riemannan manifold. Alas, I don’t know of any offhand.
– Deane Yang
Aug 9 at 22:53
1
I suggest looking at the papers by Karcher with his collaborators that are cited in Smith, P. D.; Yang, Deane, "Removing point singularities of Riemannian manifolds", especially those that discuss what they call "almost linear coordinates", which are coordinates constructed using only the distance function. What you want should be at the very least a corollary of something proved in one of these papers.
– Deane Yang
Aug 9 at 23:03
 |Â
show 1 more comment
up vote
0
down vote
up vote
0
down vote
If I am not mistaken, in order to define the Hessian you need to fix a connection. I suspect the Riemannian metric you get will depend on this connection, as Finsler metrics also have distance functions.
answered Aug 9 at 20:13
David Hughes
291211
If I am not mistaken, in order to define the Hessian you need to fix a connection. I suspect the Riemannian metric you get will depend on this connection, as Finsler metrics also have distance functions.
answered Aug 9 at 20:13
David Hughes
291211
answered Aug 9 at 20:13
David Hughes
291211
answered Aug 9 at 20:13
David Hughes
291211
answered Aug 9 at 20:13
David Hughes
291211
291211
Right; I meant the Hessian with respect to the Levi-Civita connection.
– Tom Leinster
Aug 9 at 20:16
Actually, since $p$ is a critical point of $d(-,p)^2$, one can define the Hessian in $p$ without any choice of connection (just locally as the Hessian of that function in a chart).
– Panagiotis Konstantis
Aug 9 at 20:25
I believe the way to avoid using the Levi-Civita connection (which already determines the metric) is to use the distance function to define geodesics as length minimizing curves and define the Hessian using the second derivative of $d^2$ in the direction of each geodesic. However, there are lots of details to work out. There's a slight chance that this is worked out in the book by Gromov et all, Metric Structure for Riemannian and non-Riemannian Spaces.
– Deane Yang
Aug 9 at 20:38
1
It might appear in a paper that is trying to show that a length space with some additional properties is in fact a Riemannan manifold. Alas, I don’t know of any offhand.
– Deane Yang
Aug 9 at 22:53
1
I suggest looking at the papers by Karcher with his collaborators that are cited in Smith, P. D.; Yang, Deane, "Removing point singularities of Riemannian manifolds", especially those that discuss what they call "almost linear coordinates", which are coordinates constructed using only the distance function. What you want should be at the very least a corollary of something proved in one of these papers.
– Deane Yang
Aug 9 at 23:03
 |Â
show 1 more comment
Right; I meant the Hessian with respect to the Levi-Civita connection.
– Tom Leinster
Aug 9 at 20:16
Actually, since $p$ is a critical point of $d(-,p)^2$, one can define the Hessian in $p$ without any choice of connection (just locally as the Hessian of that function in a chart).
– Panagiotis Konstantis
Aug 9 at 20:25
I believe the way to avoid using the Levi-Civita connection (which already determines the metric) is to use the distance function to define geodesics as length minimizing curves and define the Hessian using the second derivative of $d^2$ in the direction of each geodesic. However, there are lots of details to work out. There's a slight chance that this is worked out in the book by Gromov et all, Metric Structure for Riemannian and non-Riemannian Spaces.
– Deane Yang
Aug 9 at 20:38
1
It might appear in a paper that is trying to show that a length space with some additional properties is in fact a Riemannan manifold. Alas, I don’t know of any offhand.
– Deane Yang
Aug 9 at 22:53
1
I suggest looking at the papers by Karcher with his collaborators that are cited in Smith, P. D.; Yang, Deane, "Removing point singularities of Riemannian manifolds", especially those that discuss what they call "almost linear coordinates", which are coordinates constructed using only the distance function. What you want should be at the very least a corollary of something proved in one of these papers.
– Deane Yang
Aug 9 at 23:03
Right; I meant the Hessian with respect to the Levi-Civita connection.
– Tom Leinster
Aug 9 at 20:16
Right; I meant the Hessian with respect to the Levi-Civita connection.
– Tom Leinster
Aug 9 at 20:16
Actually, since $p$ is a critical point of $d(-,p)^2$, one can define the Hessian in $p$ without any choice of connection (just locally as the Hessian of that function in a chart).
– Panagiotis Konstantis
Aug 9 at 20:25
Actually, since $p$ is a critical point of $d(-,p)^2$, one can define the Hessian in $p$ without any choice of connection (just locally as the Hessian of that function in a chart).
– Panagiotis Konstantis
Aug 9 at 20:25
I believe the way to avoid using the Levi-Civita connection (which already determines the metric) is to use the distance function to define geodesics as length minimizing curves and define the Hessian using the second derivative of $d^2$ in the direction of each geodesic. However, there are lots of details to work out. There's a slight chance that this is worked out in the book by Gromov et all, Metric Structure for Riemannian and non-Riemannian Spaces.
– Deane Yang
Aug 9 at 20:38
I believe the way to avoid using the Levi-Civita connection (which already determines the metric) is to use the distance function to define geodesics as length minimizing curves and define the Hessian using the second derivative of $d^2$ in the direction of each geodesic. However, there are lots of details to work out. There's a slight chance that this is worked out in the book by Gromov et all, Metric Structure for Riemannian and non-Riemannian Spaces.
– Deane Yang
Aug 9 at 20:38
1
1
It might appear in a paper that is trying to show that a length space with some additional properties is in fact a Riemannan manifold. Alas, I don’t know of any offhand.
– Deane Yang
Aug 9 at 22:53
It might appear in a paper that is trying to show that a length space with some additional properties is in fact a Riemannan manifold. Alas, I don’t know of any offhand.
– Deane Yang
Aug 9 at 22:53
1
1
I suggest looking at the papers by Karcher with his collaborators that are cited in Smith, P. D.; Yang, Deane, "Removing point singularities of Riemannian manifolds", especially those that discuss what they call "almost linear coordinates", which are coordinates constructed using only the distance function. What you want should be at the very least a corollary of something proved in one of these papers.
– Deane Yang
Aug 9 at 23:03
I suggest looking at the papers by Karcher with his collaborators that are cited in Smith, P. D.; Yang, Deane, "Removing point singularities of Riemannian manifolds", especially those that discuss what they call "almost linear coordinates", which are coordinates constructed using only the distance function. What you want should be at the very least a corollary of something proved in one of these papers.
– Deane Yang
Aug 9 at 23:03
 |Â
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1
I asked a similar question a while ago on MSE, see math.stackexchange.com/questions/1161589/…. My answer there seems to be related to what you are looking for. In addition, there is another answer that I haven't been able to make sense of.
– S.Surace
Aug 9 at 20:47
@S.Surace: thanks, I hadn't seen that MSE question. Nothing there answers my question (i.e. provides a reference to the stated equation), but it seems that you're interested in this stuff for similar reasons to me. In particular, I'd seen the some of that literature on contrast functions that you mention in your MSE answer, which seems to take as its starting point the result that I want a reference for.
– Tom Leinster
Aug 9 at 21:28
@S.Surace: It seems that you have read only the title of this post, but not its content. Indeed, the title suggests a completely diffferent question - the one that you have asked.
– Alex M.
Aug 9 at 21:58
@AlexM. Am I sure it's true? I believe it's true because someone whose expertise I trust tells me that it is. For a proof, they pointed me to p.4-5 of the paper "Hessian of the Riemannian squared distance" by Pennec: www-sop.inria.fr/members/Xavier.Pennec/… . But the fact I'm interested in isn't stated directly there; you have to do a bit of work to dig it out. I'm looking for a reference where it's stated directly.
– Tom Leinster
Aug 9 at 22:09
It seems to me that the answer is stated explicitly in equation (5) in the Pennec paper, if you take into account the displayed equation after (2). I'm not sure you'll get anything more explicit than that, My preferred approach to this is the equation of Villani stated in the last sentence before section 2.2.
– Deane Yang
Aug 10 at 0:48