Meaning of commutativity of a local flow of a vector field and left-translations
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In DoCarmo’s Riemannian Geometry book, it is writen that if $x_t$ is the flow of a left invariant vector field $X$ of a lie group $G$, then $L_ycirc x_t=x_t circ L_y$. But the domains of the left and right side do not necessarily coincide since the domain of $L_y$ is the whole $G$ while the domain of $x_t$ is in general a neighborhood of $G$, so does the author simply mean that for all $u$ in the domain of $x_t$ we have $L_ycirc x_t(u)=x_t(u)$, or what? That expression makes sense only if the domain of $x_t$ is also the whole $G$, but is it really so?
differential-geometry riemannian-geometry smooth-manifolds vector-fields
asked Nov 21 at 12:23
Selflearner
377214
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In DoCarmo’s Riemannian Geometry book, it is writen that if $x_t$ is the flow of a left invariant vector field $X$ of a lie group $G$, then $L_ycirc x_t=x_t circ L_y$. But the domains of the left and right side do not necessarily coincide since the domain of $L_y$ is the whole $G$ while the domain of $x_t$ is in general a neighborhood of $G$, so does the author simply mean that for all $u$ in the domain of $x_t$ we have $L_ycirc x_t(u)=x_t(u)$, or what? That expression makes sense only if the domain of $x_t$ is also the whole $G$, but is it really so?
differential-geometry riemannian-geometry smooth-manifolds vector-fields
asked Nov 21 at 12:23
Selflearner
377214
This isn't true in general. Consider, for example, $G = SO(2)$, $y neq 1$, and a vector field $X$ that vanishes at $1$ but nowhere else. Then, $L_y(x_t(1)) = L_y(1) = y$, but $x_t(L_y(1)) = x_t(y)$, which (since $X_y neq 0$) cannot be $y$ for small nonzero $t$.
– Travis
Nov 21 at 12:43
I managed to confuse myself in a standard way: The flow $x_t$ of a left-invariant vector field $X$ is right multiplication, so that $x_t(h) = h exp(t X)$, and left- and right-multiplication maps commute. A little more explicitly: $L_g(x_t(h)) = L_g(h exp (t X)) = g h exp(tX) = x_t(gh) = x_t(L_g(h))$.
– Travis
Nov 21 at 13:46
Thanks, I haven’t yet reached the section on exponential maps but once I read that section your last comment will solve my problem
– Selflearner
Nov 21 at 14:24
add a comment |
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up vote
0
down vote
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In DoCarmo’s Riemannian Geometry book, it is writen that if $x_t$ is the flow of a left invariant vector field $X$ of a lie group $G$, then $L_ycirc x_t=x_t circ L_y$. But the domains of the left and right side do not necessarily coincide since the domain of $L_y$ is the whole $G$ while the domain of $x_t$ is in general a neighborhood of $G$, so does the author simply mean that for all $u$ in the domain of $x_t$ we have $L_ycirc x_t(u)=x_t(u)$, or what? That expression makes sense only if the domain of $x_t$ is also the whole $G$, but is it really so?
differential-geometry riemannian-geometry smooth-manifolds vector-fields
asked Nov 21 at 12:23
Selflearner
377214
In DoCarmo’s Riemannian Geometry book, it is writen that if $x_t$ is the flow of a left invariant vector field $X$ of a lie group $G$, then $L_ycirc x_t=x_t circ L_y$. But the domains of the left and right side do not necessarily coincide since the domain of $L_y$ is the whole $G$ while the domain of $x_t$ is in general a neighborhood of $G$, so does the author simply mean that for all $u$ in the domain of $x_t$ we have $L_ycirc x_t(u)=x_t(u)$, or what? That expression makes sense only if the domain of $x_t$ is also the whole $G$, but is it really so?
differential-geometry riemannian-geometry smooth-manifolds vector-fields
differential-geometry riemannian-geometry smooth-manifolds vector-fields
asked Nov 21 at 12:23
Selflearner
377214
asked Nov 21 at 12:23
Selflearner
377214
edited Nov 21 at 12:49
asked Nov 21 at 12:23
Selflearner
377214
asked Nov 21 at 12:23
Selflearner
377214
asked Nov 21 at 12:23
Selflearner
377214
377214
This isn't true in general. Consider, for example, $G = SO(2)$, $y neq 1$, and a vector field $X$ that vanishes at $1$ but nowhere else. Then, $L_y(x_t(1)) = L_y(1) = y$, but $x_t(L_y(1)) = x_t(y)$, which (since $X_y neq 0$) cannot be $y$ for small nonzero $t$.
– Travis
Nov 21 at 12:43
I managed to confuse myself in a standard way: The flow $x_t$ of a left-invariant vector field $X$ is right multiplication, so that $x_t(h) = h exp(t X)$, and left- and right-multiplication maps commute. A little more explicitly: $L_g(x_t(h)) = L_g(h exp (t X)) = g h exp(tX) = x_t(gh) = x_t(L_g(h))$.
– Travis
Nov 21 at 13:46
Thanks, I haven’t yet reached the section on exponential maps but once I read that section your last comment will solve my problem
– Selflearner
Nov 21 at 14:24
add a comment |
This isn't true in general. Consider, for example, $G = SO(2)$, $y neq 1$, and a vector field $X$ that vanishes at $1$ but nowhere else. Then, $L_y(x_t(1)) = L_y(1) = y$, but $x_t(L_y(1)) = x_t(y)$, which (since $X_y neq 0$) cannot be $y$ for small nonzero $t$.
– Travis
Nov 21 at 12:43
I managed to confuse myself in a standard way: The flow $x_t$ of a left-invariant vector field $X$ is right multiplication, so that $x_t(h) = h exp(t X)$, and left- and right-multiplication maps commute. A little more explicitly: $L_g(x_t(h)) = L_g(h exp (t X)) = g h exp(tX) = x_t(gh) = x_t(L_g(h))$.
– Travis
Nov 21 at 13:46
Thanks, I haven’t yet reached the section on exponential maps but once I read that section your last comment will solve my problem
– Selflearner
Nov 21 at 14:24
This isn't true in general. Consider, for example, $G = SO(2)$, $y neq 1$, and a vector field $X$ that vanishes at $1$ but nowhere else. Then, $L_y(x_t(1)) = L_y(1) = y$, but $x_t(L_y(1)) = x_t(y)$, which (since $X_y neq 0$) cannot be $y$ for small nonzero $t$.
– Travis
Nov 21 at 12:43
This isn't true in general. Consider, for example, $G = SO(2)$, $y neq 1$, and a vector field $X$ that vanishes at $1$ but nowhere else. Then, $L_y(x_t(1)) = L_y(1) = y$, but $x_t(L_y(1)) = x_t(y)$, which (since $X_y neq 0$) cannot be $y$ for small nonzero $t$.
– Travis
Nov 21 at 12:43
I managed to confuse myself in a standard way: The flow $x_t$ of a left-invariant vector field $X$ is right multiplication, so that $x_t(h) = h exp(t X)$, and left- and right-multiplication maps commute. A little more explicitly: $L_g(x_t(h)) = L_g(h exp (t X)) = g h exp(tX) = x_t(gh) = x_t(L_g(h))$.
– Travis
Nov 21 at 13:46
I managed to confuse myself in a standard way: The flow $x_t$ of a left-invariant vector field $X$ is right multiplication, so that $x_t(h) = h exp(t X)$, and left- and right-multiplication maps commute. A little more explicitly: $L_g(x_t(h)) = L_g(h exp (t X)) = g h exp(tX) = x_t(gh) = x_t(L_g(h))$.
– Travis
Nov 21 at 13:46
Thanks, I haven’t yet reached the section on exponential maps but once I read that section your last comment will solve my problem
– Selflearner
Nov 21 at 14:24
Thanks, I haven’t yet reached the section on exponential maps but once I read that section your last comment will solve my problem
– Selflearner
Nov 21 at 14:24
add a comment |
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This isn't true in general. Consider, for example, $G = SO(2)$, $y neq 1$, and a vector field $X$ that vanishes at $1$ but nowhere else. Then, $L_y(x_t(1)) = L_y(1) = y$, but $x_t(L_y(1)) = x_t(y)$, which (since $X_y neq 0$) cannot be $y$ for small nonzero $t$.
– Travis
Nov 21 at 12:43
I managed to confuse myself in a standard way: The flow $x_t$ of a left-invariant vector field $X$ is right multiplication, so that $x_t(h) = h exp(t X)$, and left- and right-multiplication maps commute. A little more explicitly: $L_g(x_t(h)) = L_g(h exp (t X)) = g h exp(tX) = x_t(gh) = x_t(L_g(h))$.
– Travis
Nov 21 at 13:46
Thanks, I haven’t yet reached the section on exponential maps but once I read that section your last comment will solve my problem
– Selflearner
Nov 21 at 14:24