Are the exp and log maps of Riemannian geometry conformalComplex manifolds in which the exponential map is holomorphicRiemannian Geometry How metric is Riemannian geometryThe necessary and sufficient condition for $textbfglobal$ conformal flatness of a n-dim (pseudo-)Riemannian manifold$2-$conformal vector fields on Riemannian manifoldsConformal Transformations that are Ricci Positive InvariantConformal harmonic maps in high dimensions are scaled isometriesIs this expression for the Laplacian of conformal maps between Riemannian manifolds known?Conformal $L^p$ rigidity of Riemannian manifoldsConformal factors and light raysAre conformal maps between Riemannian manifolds real-analytic?
Are the exp and log maps of Riemannian geometry conformal
Complex manifolds in which the exponential map is holomorphicRiemannian Geometry How metric is Riemannian geometryThe necessary and sufficient condition for $textbfglobal$ conformal flatness of a n-dim (pseudo-)Riemannian manifold$2-$conformal vector fields on Riemannian manifoldsConformal Transformations that are Ricci Positive InvariantConformal harmonic maps in high dimensions are scaled isometriesIs this expression for the Laplacian of conformal maps between Riemannian manifolds known?Conformal $L^p$ rigidity of Riemannian manifoldsConformal factors and light raysAre conformal maps between Riemannian manifolds real-analytic?
$begingroup$
For any Riemannian manifold are the exp and log maps (from a predetermined base) conformal? If not, are there some manifolds where they are and others where they aren't?
riemannian-geometry
edited Aug 8 at 15:05
Yemon Choi
17.1k5 gold badges49 silver badges110 bronze badges
asked Aug 8 at 9:19
Justin DieterJustin Dieter
384 bronze badges
$endgroup$
add a comment |
$begingroup$
For any Riemannian manifold are the exp and log maps (from a predetermined base) conformal? If not, are there some manifolds where they are and others where they aren't?
riemannian-geometry
edited Aug 8 at 15:05
Yemon Choi
17.1k5 gold badges49 silver badges110 bronze badges
asked Aug 8 at 9:19
Justin DieterJustin Dieter
384 bronze badges
$endgroup$
$begingroup$
They are for Euclidean space, clearly. I think otherwise conformality is rare.
$endgroup$
– Ben McKay
Aug 8 at 9:56
$begingroup$
Have you checked whether this is true for the standard unit sphere?
$endgroup$
– Deane Yang
Aug 8 at 12:41
$begingroup$
The Taylor series expansion of the exponential map shows that, at the very least, the curvature has to vanish at the base point.
$endgroup$
– Deane Yang
Aug 8 at 12:52
1
$begingroup$
I have rolled back an edit which seemed needlessly pedantic
$endgroup$
– Yemon Choi
Aug 8 at 15:07
$begingroup$
@Yemon Choi:A question is asked, not a topic is explained. The omitted ? makes a grammar mistake.
$endgroup$
– user64494
Aug 8 at 16:16
add a comment |
$begingroup$
For any Riemannian manifold are the exp and log maps (from a predetermined base) conformal? If not, are there some manifolds where they are and others where they aren't?
riemannian-geometry
edited Aug 8 at 15:05
Yemon Choi
17.1k5 gold badges49 silver badges110 bronze badges
asked Aug 8 at 9:19
Justin DieterJustin Dieter
384 bronze badges
$endgroup$
For any Riemannian manifold are the exp and log maps (from a predetermined base) conformal? If not, are there some manifolds where they are and others where they aren't?
riemannian-geometry
riemannian-geometry
edited Aug 8 at 15:05
Yemon Choi
17.1k5 gold badges49 silver badges110 bronze badges
asked Aug 8 at 9:19
Justin DieterJustin Dieter
384 bronze badges
edited Aug 8 at 15:05
Yemon Choi
17.1k5 gold badges49 silver badges110 bronze badges
asked Aug 8 at 9:19
Justin DieterJustin Dieter
384 bronze badges
edited Aug 8 at 15:05
Yemon Choi
17.1k5 gold badges49 silver badges110 bronze badges
edited Aug 8 at 15:05
Yemon Choi
17.1k5 gold badges49 silver badges110 bronze badges
edited Aug 8 at 15:05
Yemon Choi
17.1k5 gold badges49 silver badges110 bronze badges
17.1k5 gold badges49 silver badges110 bronze badges
asked Aug 8 at 9:19
Justin DieterJustin Dieter
384 bronze badges
asked Aug 8 at 9:19
Justin DieterJustin Dieter
384 bronze badges
asked Aug 8 at 9:19
Justin DieterJustin Dieter
384 bronze badges
384 bronze badges
$begingroup$
They are for Euclidean space, clearly. I think otherwise conformality is rare.
$endgroup$
– Ben McKay
Aug 8 at 9:56
$begingroup$
Have you checked whether this is true for the standard unit sphere?
$endgroup$
– Deane Yang
Aug 8 at 12:41
$begingroup$
The Taylor series expansion of the exponential map shows that, at the very least, the curvature has to vanish at the base point.
$endgroup$
– Deane Yang
Aug 8 at 12:52
1
$begingroup$
I have rolled back an edit which seemed needlessly pedantic
$endgroup$
– Yemon Choi
Aug 8 at 15:07
$begingroup$
@Yemon Choi:A question is asked, not a topic is explained. The omitted ? makes a grammar mistake.
$endgroup$
– user64494
Aug 8 at 16:16
add a comment |
$begingroup$
They are for Euclidean space, clearly. I think otherwise conformality is rare.
$endgroup$
– Ben McKay
Aug 8 at 9:56
$begingroup$
Have you checked whether this is true for the standard unit sphere?
$endgroup$
– Deane Yang
Aug 8 at 12:41
$begingroup$
The Taylor series expansion of the exponential map shows that, at the very least, the curvature has to vanish at the base point.
$endgroup$
– Deane Yang
Aug 8 at 12:52
1
$begingroup$
I have rolled back an edit which seemed needlessly pedantic
$endgroup$
– Yemon Choi
Aug 8 at 15:07
$begingroup$
@Yemon Choi:A question is asked, not a topic is explained. The omitted ? makes a grammar mistake.
$endgroup$
– user64494
Aug 8 at 16:16
$begingroup$
They are for Euclidean space, clearly. I think otherwise conformality is rare.
$endgroup$
– Ben McKay
Aug 8 at 9:56
$begingroup$
They are for Euclidean space, clearly. I think otherwise conformality is rare.
$endgroup$
– Ben McKay
Aug 8 at 9:56
$begingroup$
Have you checked whether this is true for the standard unit sphere?
$endgroup$
– Deane Yang
Aug 8 at 12:41
$begingroup$
Have you checked whether this is true for the standard unit sphere?
$endgroup$
– Deane Yang
Aug 8 at 12:41
$begingroup$
The Taylor series expansion of the exponential map shows that, at the very least, the curvature has to vanish at the base point.
$endgroup$
– Deane Yang
Aug 8 at 12:52
$begingroup$
The Taylor series expansion of the exponential map shows that, at the very least, the curvature has to vanish at the base point.
$endgroup$
– Deane Yang
Aug 8 at 12:52
1
1
$begingroup$
I have rolled back an edit which seemed needlessly pedantic
$endgroup$
– Yemon Choi
Aug 8 at 15:07
$begingroup$
I have rolled back an edit which seemed needlessly pedantic
$endgroup$
– Yemon Choi
Aug 8 at 15:07
$begingroup$
@Yemon Choi:A question is asked, not a topic is explained. The omitted ? makes a grammar mistake.
$endgroup$
– user64494
Aug 8 at 16:16
$begingroup$
@Yemon Choi:A question is asked, not a topic is explained. The omitted ? makes a grammar mistake.
$endgroup$
– user64494
Aug 8 at 16:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
See Robert Bryant's answer to Complex manifolds in which the exponential map is holomorphic in which he proves that the surfaces for which the exponential map is conformal are precisely the flat ones.
If the exponential map is conformal on some Riemannian manifold of dimension $ge 3$, it is a conformal map on the intersection $P cap B$ of any 2-plane $Psubset T_m M$ in any tangent space with the ball $Bsubset T_m M$ of radius equal to the injectivity radius. So the curvature vanishes on $e^Pcap B$, i.e. the sectional curvature vanishes, so the Riemannian manifold is flat.
answered Aug 8 at 9:59
Ben McKayBen McKay
15.9k2 gold badges32 silver badges64 bronze badges
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f337890%2fare-the-exp-and-log-maps-of-riemannian-geometry-conformal%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
See Robert Bryant's answer to Complex manifolds in which the exponential map is holomorphic in which he proves that the surfaces for which the exponential map is conformal are precisely the flat ones.
If the exponential map is conformal on some Riemannian manifold of dimension $ge 3$, it is a conformal map on the intersection $P cap B$ of any 2-plane $Psubset T_m M$ in any tangent space with the ball $Bsubset T_m M$ of radius equal to the injectivity radius. So the curvature vanishes on $e^Pcap B$, i.e. the sectional curvature vanishes, so the Riemannian manifold is flat.
answered Aug 8 at 9:59
Ben McKayBen McKay
15.9k2 gold badges32 silver badges64 bronze badges
$endgroup$
add a comment |
$begingroup$
See Robert Bryant's answer to Complex manifolds in which the exponential map is holomorphic in which he proves that the surfaces for which the exponential map is conformal are precisely the flat ones.
If the exponential map is conformal on some Riemannian manifold of dimension $ge 3$, it is a conformal map on the intersection $P cap B$ of any 2-plane $Psubset T_m M$ in any tangent space with the ball $Bsubset T_m M$ of radius equal to the injectivity radius. So the curvature vanishes on $e^Pcap B$, i.e. the sectional curvature vanishes, so the Riemannian manifold is flat.
answered Aug 8 at 9:59
Ben McKayBen McKay
15.9k2 gold badges32 silver badges64 bronze badges
$endgroup$
add a comment |
$begingroup$
See Robert Bryant's answer to Complex manifolds in which the exponential map is holomorphic in which he proves that the surfaces for which the exponential map is conformal are precisely the flat ones.
If the exponential map is conformal on some Riemannian manifold of dimension $ge 3$, it is a conformal map on the intersection $P cap B$ of any 2-plane $Psubset T_m M$ in any tangent space with the ball $Bsubset T_m M$ of radius equal to the injectivity radius. So the curvature vanishes on $e^Pcap B$, i.e. the sectional curvature vanishes, so the Riemannian manifold is flat.
answered Aug 8 at 9:59
Ben McKayBen McKay
15.9k2 gold badges32 silver badges64 bronze badges
$endgroup$
See Robert Bryant's answer to Complex manifolds in which the exponential map is holomorphic in which he proves that the surfaces for which the exponential map is conformal are precisely the flat ones.
If the exponential map is conformal on some Riemannian manifold of dimension $ge 3$, it is a conformal map on the intersection $P cap B$ of any 2-plane $Psubset T_m M$ in any tangent space with the ball $Bsubset T_m M$ of radius equal to the injectivity radius. So the curvature vanishes on $e^Pcap B$, i.e. the sectional curvature vanishes, so the Riemannian manifold is flat.
answered Aug 8 at 9:59
Ben McKayBen McKay
15.9k2 gold badges32 silver badges64 bronze badges
edited Aug 8 at 13:50
answered Aug 8 at 9:59
Ben McKayBen McKay
15.9k2 gold badges32 silver badges64 bronze badges
answered Aug 8 at 9:59
Ben McKayBen McKay
15.9k2 gold badges32 silver badges64 bronze badges
answered Aug 8 at 9:59
Ben McKayBen McKay
15.9k2 gold badges32 silver badges64 bronze badges
15.9k2 gold badges32 silver badges64 bronze badges
add a comment |
add a comment |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f337890%2fare-the-exp-and-log-maps-of-riemannian-geometry-conformal%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
They are for Euclidean space, clearly. I think otherwise conformality is rare.
$endgroup$
– Ben McKay
Aug 8 at 9:56
$begingroup$
Have you checked whether this is true for the standard unit sphere?
$endgroup$
– Deane Yang
Aug 8 at 12:41
$begingroup$
The Taylor series expansion of the exponential map shows that, at the very least, the curvature has to vanish at the base point.
$endgroup$
– Deane Yang
Aug 8 at 12:52
1
$begingroup$
I have rolled back an edit which seemed needlessly pedantic
$endgroup$
– Yemon Choi
Aug 8 at 15:07
$begingroup$
@Yemon Choi:A question is asked, not a topic is explained. The omitted ? makes a grammar mistake.
$endgroup$
– user64494
Aug 8 at 16:16