Is the differential of the exponential map orthogonal?
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Let $M$ be a Riemannian manifold, and let $exp_x$ denote the exponential map at $x in M$. Assume the differential of the exponential map $D exp_x(v), vin T_x M,$ is well defined. Is $D exp_x(v)^* D exp_x(v) = Id$? If not, is there an explicit expression for $D exp_x(v)^* D exp_x(v)$?
differential-geometry riemannian-geometry
asked Dec 29 '18 at 19:23
DoggyyDoggyy
211
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Let $M$ be a Riemannian manifold, and let $exp_x$ denote the exponential map at $x in M$. Assume the differential of the exponential map $D exp_x(v), vin T_x M,$ is well defined. Is $D exp_x(v)^* D exp_x(v) = Id$? If not, is there an explicit expression for $D exp_x(v)^* D exp_x(v)$?
differential-geometry riemannian-geometry
asked Dec 29 '18 at 19:23
DoggyyDoggyy
211
$endgroup$
add a comment |
$begingroup$
Let $M$ be a Riemannian manifold, and let $exp_x$ denote the exponential map at $x in M$. Assume the differential of the exponential map $D exp_x(v), vin T_x M,$ is well defined. Is $D exp_x(v)^* D exp_x(v) = Id$? If not, is there an explicit expression for $D exp_x(v)^* D exp_x(v)$?
differential-geometry riemannian-geometry
asked Dec 29 '18 at 19:23
DoggyyDoggyy
211
$endgroup$
Let $M$ be a Riemannian manifold, and let $exp_x$ denote the exponential map at $x in M$. Assume the differential of the exponential map $D exp_x(v), vin T_x M,$ is well defined. Is $D exp_x(v)^* D exp_x(v) = Id$? If not, is there an explicit expression for $D exp_x(v)^* D exp_x(v)$?
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
asked Dec 29 '18 at 19:23
DoggyyDoggyy
211
asked Dec 29 '18 at 19:23
DoggyyDoggyy
211
asked Dec 29 '18 at 19:23
DoggyyDoggyy
211
asked Dec 29 '18 at 19:23
DoggyyDoggyy
211
asked Dec 29 '18 at 19:23
DoggyyDoggyy
211
211
add a comment |
add a comment |
1 Answer
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No, this is true only for flat metrics. Indeed, $T_xM$ is flat (it is just a vector space with the Riemannian metric associated to a certain inner product on the vector space), so if $Dexp_x$ were orthogonal on some open subset of $T_xM$, then $exp_x$ would be locally an isometry on that open subset and so the image of that open subset in $M$ would be flat.
answered Dec 29 '18 at 19:31
Eric WofseyEric Wofsey
181k12208336
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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$begingroup$
No, this is true only for flat metrics. Indeed, $T_xM$ is flat (it is just a vector space with the Riemannian metric associated to a certain inner product on the vector space), so if $Dexp_x$ were orthogonal on some open subset of $T_xM$, then $exp_x$ would be locally an isometry on that open subset and so the image of that open subset in $M$ would be flat.
answered Dec 29 '18 at 19:31
Eric WofseyEric Wofsey
181k12208336
$endgroup$
add a comment |
$begingroup$
No, this is true only for flat metrics. Indeed, $T_xM$ is flat (it is just a vector space with the Riemannian metric associated to a certain inner product on the vector space), so if $Dexp_x$ were orthogonal on some open subset of $T_xM$, then $exp_x$ would be locally an isometry on that open subset and so the image of that open subset in $M$ would be flat.
answered Dec 29 '18 at 19:31
Eric WofseyEric Wofsey
181k12208336
$endgroup$
add a comment |
$begingroup$
No, this is true only for flat metrics. Indeed, $T_xM$ is flat (it is just a vector space with the Riemannian metric associated to a certain inner product on the vector space), so if $Dexp_x$ were orthogonal on some open subset of $T_xM$, then $exp_x$ would be locally an isometry on that open subset and so the image of that open subset in $M$ would be flat.
answered Dec 29 '18 at 19:31
Eric WofseyEric Wofsey
181k12208336
$endgroup$
No, this is true only for flat metrics. Indeed, $T_xM$ is flat (it is just a vector space with the Riemannian metric associated to a certain inner product on the vector space), so if $Dexp_x$ were orthogonal on some open subset of $T_xM$, then $exp_x$ would be locally an isometry on that open subset and so the image of that open subset in $M$ would be flat.
answered Dec 29 '18 at 19:31
Eric WofseyEric Wofsey
181k12208336
answered Dec 29 '18 at 19:31
Eric WofseyEric Wofsey
181k12208336
answered Dec 29 '18 at 19:31
Eric WofseyEric Wofsey
181k12208336
answered Dec 29 '18 at 19:31
Eric WofseyEric Wofsey
181k12208336
181k12208336
add a comment |
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