'Constant' vector field on a Lie group is smooth
$begingroup$
I have a really basic question which I am struggling to articulate formally in differential geometry. I have a Lie group $G$, and a tangent vector $vin T_{1_G}(G)$. I want to claim that the vector field $gmapsto(g,v)$ is a smooth vector field, but I am unsure as to whether I can say formally that $vin T_g(G)$ for all $gin G$.
With $G$ being a topological group, it is homogeneous, and given a chart $(U,varphi)$ of the identity, $(gU, g^{-1} cdot varphi)$ is a chart around any $g$. But I'm still unsure as whether that means that $vin T_g(G)$.
If it is trivially true that $vin T_g(G)$ for $g$ in the same component as $1_G$, or if it true even without the topological group sturcture, I would be very thankful if someone were to help my overcome this conceptual hurdle.
lie-groups smooth-manifolds
asked Dec 9 '18 at 16:41
Keen-ameteurKeen-ameteur
1,413316
$endgroup$
add a comment |
$begingroup$
I have a really basic question which I am struggling to articulate formally in differential geometry. I have a Lie group $G$, and a tangent vector $vin T_{1_G}(G)$. I want to claim that the vector field $gmapsto(g,v)$ is a smooth vector field, but I am unsure as to whether I can say formally that $vin T_g(G)$ for all $gin G$.
With $G$ being a topological group, it is homogeneous, and given a chart $(U,varphi)$ of the identity, $(gU, g^{-1} cdot varphi)$ is a chart around any $g$. But I'm still unsure as whether that means that $vin T_g(G)$.
If it is trivially true that $vin T_g(G)$ for $g$ in the same component as $1_G$, or if it true even without the topological group sturcture, I would be very thankful if someone were to help my overcome this conceptual hurdle.
lie-groups smooth-manifolds
asked Dec 9 '18 at 16:41
Keen-ameteurKeen-ameteur
1,413316
$endgroup$
$begingroup$
What do you mean by $(g, v)$? Is it a translation?
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 16:44
$begingroup$
Elements of the tangent bundles are $(g,w)$ where $gin G$ and $win T_g(G)$, or am I wrong?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:03
$begingroup$
This does not make sense in general, and this is a fundamental fact of differential geometry. Precisely, you cannot translate vectors like that; this is why the standard notation is $w_g$ instead of $(g, w)$. Imagine a vector tangent to a circle. Move that vector elsewhere on the circle; it is not tangent anymore. In general, translation is a subtle matter.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:08
add a comment |
$begingroup$
I have a really basic question which I am struggling to articulate formally in differential geometry. I have a Lie group $G$, and a tangent vector $vin T_{1_G}(G)$. I want to claim that the vector field $gmapsto(g,v)$ is a smooth vector field, but I am unsure as to whether I can say formally that $vin T_g(G)$ for all $gin G$.
With $G$ being a topological group, it is homogeneous, and given a chart $(U,varphi)$ of the identity, $(gU, g^{-1} cdot varphi)$ is a chart around any $g$. But I'm still unsure as whether that means that $vin T_g(G)$.
If it is trivially true that $vin T_g(G)$ for $g$ in the same component as $1_G$, or if it true even without the topological group sturcture, I would be very thankful if someone were to help my overcome this conceptual hurdle.
lie-groups smooth-manifolds
asked Dec 9 '18 at 16:41
Keen-ameteurKeen-ameteur
1,413316
$endgroup$
I have a really basic question which I am struggling to articulate formally in differential geometry. I have a Lie group $G$, and a tangent vector $vin T_{1_G}(G)$. I want to claim that the vector field $gmapsto(g,v)$ is a smooth vector field, but I am unsure as to whether I can say formally that $vin T_g(G)$ for all $gin G$.
With $G$ being a topological group, it is homogeneous, and given a chart $(U,varphi)$ of the identity, $(gU, g^{-1} cdot varphi)$ is a chart around any $g$. But I'm still unsure as whether that means that $vin T_g(G)$.
If it is trivially true that $vin T_g(G)$ for $g$ in the same component as $1_G$, or if it true even without the topological group sturcture, I would be very thankful if someone were to help my overcome this conceptual hurdle.
lie-groups smooth-manifolds
lie-groups smooth-manifolds
asked Dec 9 '18 at 16:41
Keen-ameteurKeen-ameteur
1,413316
asked Dec 9 '18 at 16:41
Keen-ameteurKeen-ameteur
1,413316
asked Dec 9 '18 at 16:41
Keen-ameteurKeen-ameteur
1,413316
asked Dec 9 '18 at 16:41
Keen-ameteurKeen-ameteur
1,413316
asked Dec 9 '18 at 16:41
Keen-ameteurKeen-ameteur
1,413316
1,413316
$begingroup$
What do you mean by $(g, v)$? Is it a translation?
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 16:44
$begingroup$
Elements of the tangent bundles are $(g,w)$ where $gin G$ and $win T_g(G)$, or am I wrong?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:03
$begingroup$
This does not make sense in general, and this is a fundamental fact of differential geometry. Precisely, you cannot translate vectors like that; this is why the standard notation is $w_g$ instead of $(g, w)$. Imagine a vector tangent to a circle. Move that vector elsewhere on the circle; it is not tangent anymore. In general, translation is a subtle matter.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:08
add a comment |
$begingroup$
What do you mean by $(g, v)$? Is it a translation?
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 16:44
$begingroup$
Elements of the tangent bundles are $(g,w)$ where $gin G$ and $win T_g(G)$, or am I wrong?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:03
$begingroup$
This does not make sense in general, and this is a fundamental fact of differential geometry. Precisely, you cannot translate vectors like that; this is why the standard notation is $w_g$ instead of $(g, w)$. Imagine a vector tangent to a circle. Move that vector elsewhere on the circle; it is not tangent anymore. In general, translation is a subtle matter.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:08
$begingroup$
What do you mean by $(g, v)$? Is it a translation?
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 16:44
$begingroup$
What do you mean by $(g, v)$? Is it a translation?
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 16:44
$begingroup$
Elements of the tangent bundles are $(g,w)$ where $gin G$ and $win T_g(G)$, or am I wrong?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:03
$begingroup$
Elements of the tangent bundles are $(g,w)$ where $gin G$ and $win T_g(G)$, or am I wrong?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:03
$begingroup$
This does not make sense in general, and this is a fundamental fact of differential geometry. Precisely, you cannot translate vectors like that; this is why the standard notation is $w_g$ instead of $(g, w)$. Imagine a vector tangent to a circle. Move that vector elsewhere on the circle; it is not tangent anymore. In general, translation is a subtle matter.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:08
$begingroup$
This does not make sense in general, and this is a fundamental fact of differential geometry. Precisely, you cannot translate vectors like that; this is why the standard notation is $w_g$ instead of $(g, w)$. Imagine a vector tangent to a circle. Move that vector elsewhere on the circle; it is not tangent anymore. In general, translation is a subtle matter.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The concept you're looking for here is left-invariant vector field. For Lie group $G$ and any $v in T_1 G$, we can define an (a priori, rough) vector field $$tilde v in Gamma(TG), qquad tilde v_g := T_1 L_g cdot v .$$ Here, $L_g : G to G$ is the left multiplication map $h mapsto gh$. Then, unwinding definitions shows that $tilde v$ is invariant under left multiplication, that is that
$$T_1 L_g cdot tilde v_h = tilde v_{gh}$$
for all $g, h in G$.
Conversely, there is canonical map $$omega : TG to T_1 G , qquad omega : w_g mapsto T L_{g^{-1}} cdot w_g ,$$ and unwinding definitions gives that $$omega(tilde v_g) = v.$$ If we identify $T_1 G$ with the Lie algebra $mathfrak g$ of $G$, $omega$ is precisely the Maurer-Cartan form on $G$. This in turn defines a canonical bundle isomorphism $$Phi : TG to G times T_1 G, qquad Phi : w_g mapsto (g, omega(w_g)) .$$ This bundle isomorphism formalizes the notion in the original question: Unwinding definitions shows that $Phi^{-1}$ identifies $(g, v) in G times T_1 G$ with $tilde v_g in T_g G$.
With this in hand, we can check smoothness directly by verifying that $tilde v cdot f$ is smooth whenever $f$ is, and this again amounts to unwinding definitions and recognizing that an expression for $tilde v cdot f$ is manifestly smooth.
answered Dec 9 '18 at 17:03
TravisTravis
60.3k767147
$endgroup$
$begingroup$
I am actually trying to show that there exists a unique left invariant field. I am trying to define a 'simple' left invariant vector field, and later showing that it is unique.
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:14
$begingroup$
The left-invariant vector field $tilde v$ is unique more or less by definition: In order for $tilde v$ to be left-invariant, then specializing the equation $T_1 L_g cdot tilde v_h = tilde v_{gh}$ to $h = e$ gives $tilde v_g = T_1 L_g cdot v$.
$endgroup$
– Travis
Dec 9 '18 at 17:21
$begingroup$
Just to be sure, when working with a definition not by derivations, what exactly does $L_gcdot v$ means?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:23
$begingroup$
The notation I used is, e.g., $T_1 L_g cdot v$. Here, $T_1 L_g : T_1 G to T_g G$ is the pushforward of the smooth map $L_g$ at $1 in G$.
$endgroup$
– Travis
Dec 9 '18 at 17:26
$begingroup$
So is it the differential of $L_g$ at $1_G$, right?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:32
|
show 1 more comment
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The concept you're looking for here is left-invariant vector field. For Lie group $G$ and any $v in T_1 G$, we can define an (a priori, rough) vector field $$tilde v in Gamma(TG), qquad tilde v_g := T_1 L_g cdot v .$$ Here, $L_g : G to G$ is the left multiplication map $h mapsto gh$. Then, unwinding definitions shows that $tilde v$ is invariant under left multiplication, that is that
$$T_1 L_g cdot tilde v_h = tilde v_{gh}$$
for all $g, h in G$.
Conversely, there is canonical map $$omega : TG to T_1 G , qquad omega : w_g mapsto T L_{g^{-1}} cdot w_g ,$$ and unwinding definitions gives that $$omega(tilde v_g) = v.$$ If we identify $T_1 G$ with the Lie algebra $mathfrak g$ of $G$, $omega$ is precisely the Maurer-Cartan form on $G$. This in turn defines a canonical bundle isomorphism $$Phi : TG to G times T_1 G, qquad Phi : w_g mapsto (g, omega(w_g)) .$$ This bundle isomorphism formalizes the notion in the original question: Unwinding definitions shows that $Phi^{-1}$ identifies $(g, v) in G times T_1 G$ with $tilde v_g in T_g G$.
With this in hand, we can check smoothness directly by verifying that $tilde v cdot f$ is smooth whenever $f$ is, and this again amounts to unwinding definitions and recognizing that an expression for $tilde v cdot f$ is manifestly smooth.
answered Dec 9 '18 at 17:03
TravisTravis
60.3k767147
$endgroup$
$begingroup$
I am actually trying to show that there exists a unique left invariant field. I am trying to define a 'simple' left invariant vector field, and later showing that it is unique.
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:14
$begingroup$
The left-invariant vector field $tilde v$ is unique more or less by definition: In order for $tilde v$ to be left-invariant, then specializing the equation $T_1 L_g cdot tilde v_h = tilde v_{gh}$ to $h = e$ gives $tilde v_g = T_1 L_g cdot v$.
$endgroup$
– Travis
Dec 9 '18 at 17:21
$begingroup$
Just to be sure, when working with a definition not by derivations, what exactly does $L_gcdot v$ means?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:23
$begingroup$
The notation I used is, e.g., $T_1 L_g cdot v$. Here, $T_1 L_g : T_1 G to T_g G$ is the pushforward of the smooth map $L_g$ at $1 in G$.
$endgroup$
– Travis
Dec 9 '18 at 17:26
$begingroup$
So is it the differential of $L_g$ at $1_G$, right?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:32
|
show 1 more comment
$begingroup$
The concept you're looking for here is left-invariant vector field. For Lie group $G$ and any $v in T_1 G$, we can define an (a priori, rough) vector field $$tilde v in Gamma(TG), qquad tilde v_g := T_1 L_g cdot v .$$ Here, $L_g : G to G$ is the left multiplication map $h mapsto gh$. Then, unwinding definitions shows that $tilde v$ is invariant under left multiplication, that is that
$$T_1 L_g cdot tilde v_h = tilde v_{gh}$$
for all $g, h in G$.
Conversely, there is canonical map $$omega : TG to T_1 G , qquad omega : w_g mapsto T L_{g^{-1}} cdot w_g ,$$ and unwinding definitions gives that $$omega(tilde v_g) = v.$$ If we identify $T_1 G$ with the Lie algebra $mathfrak g$ of $G$, $omega$ is precisely the Maurer-Cartan form on $G$. This in turn defines a canonical bundle isomorphism $$Phi : TG to G times T_1 G, qquad Phi : w_g mapsto (g, omega(w_g)) .$$ This bundle isomorphism formalizes the notion in the original question: Unwinding definitions shows that $Phi^{-1}$ identifies $(g, v) in G times T_1 G$ with $tilde v_g in T_g G$.
With this in hand, we can check smoothness directly by verifying that $tilde v cdot f$ is smooth whenever $f$ is, and this again amounts to unwinding definitions and recognizing that an expression for $tilde v cdot f$ is manifestly smooth.
answered Dec 9 '18 at 17:03
TravisTravis
60.3k767147
$endgroup$
$begingroup$
I am actually trying to show that there exists a unique left invariant field. I am trying to define a 'simple' left invariant vector field, and later showing that it is unique.
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:14
$begingroup$
The left-invariant vector field $tilde v$ is unique more or less by definition: In order for $tilde v$ to be left-invariant, then specializing the equation $T_1 L_g cdot tilde v_h = tilde v_{gh}$ to $h = e$ gives $tilde v_g = T_1 L_g cdot v$.
$endgroup$
– Travis
Dec 9 '18 at 17:21
$begingroup$
Just to be sure, when working with a definition not by derivations, what exactly does $L_gcdot v$ means?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:23
$begingroup$
The notation I used is, e.g., $T_1 L_g cdot v$. Here, $T_1 L_g : T_1 G to T_g G$ is the pushforward of the smooth map $L_g$ at $1 in G$.
$endgroup$
– Travis
Dec 9 '18 at 17:26
$begingroup$
So is it the differential of $L_g$ at $1_G$, right?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:32
|
show 1 more comment
$begingroup$
The concept you're looking for here is left-invariant vector field. For Lie group $G$ and any $v in T_1 G$, we can define an (a priori, rough) vector field $$tilde v in Gamma(TG), qquad tilde v_g := T_1 L_g cdot v .$$ Here, $L_g : G to G$ is the left multiplication map $h mapsto gh$. Then, unwinding definitions shows that $tilde v$ is invariant under left multiplication, that is that
$$T_1 L_g cdot tilde v_h = tilde v_{gh}$$
for all $g, h in G$.
Conversely, there is canonical map $$omega : TG to T_1 G , qquad omega : w_g mapsto T L_{g^{-1}} cdot w_g ,$$ and unwinding definitions gives that $$omega(tilde v_g) = v.$$ If we identify $T_1 G$ with the Lie algebra $mathfrak g$ of $G$, $omega$ is precisely the Maurer-Cartan form on $G$. This in turn defines a canonical bundle isomorphism $$Phi : TG to G times T_1 G, qquad Phi : w_g mapsto (g, omega(w_g)) .$$ This bundle isomorphism formalizes the notion in the original question: Unwinding definitions shows that $Phi^{-1}$ identifies $(g, v) in G times T_1 G$ with $tilde v_g in T_g G$.
With this in hand, we can check smoothness directly by verifying that $tilde v cdot f$ is smooth whenever $f$ is, and this again amounts to unwinding definitions and recognizing that an expression for $tilde v cdot f$ is manifestly smooth.
answered Dec 9 '18 at 17:03
TravisTravis
60.3k767147
$endgroup$
The concept you're looking for here is left-invariant vector field. For Lie group $G$ and any $v in T_1 G$, we can define an (a priori, rough) vector field $$tilde v in Gamma(TG), qquad tilde v_g := T_1 L_g cdot v .$$ Here, $L_g : G to G$ is the left multiplication map $h mapsto gh$. Then, unwinding definitions shows that $tilde v$ is invariant under left multiplication, that is that
$$T_1 L_g cdot tilde v_h = tilde v_{gh}$$
for all $g, h in G$.
Conversely, there is canonical map $$omega : TG to T_1 G , qquad omega : w_g mapsto T L_{g^{-1}} cdot w_g ,$$ and unwinding definitions gives that $$omega(tilde v_g) = v.$$ If we identify $T_1 G$ with the Lie algebra $mathfrak g$ of $G$, $omega$ is precisely the Maurer-Cartan form on $G$. This in turn defines a canonical bundle isomorphism $$Phi : TG to G times T_1 G, qquad Phi : w_g mapsto (g, omega(w_g)) .$$ This bundle isomorphism formalizes the notion in the original question: Unwinding definitions shows that $Phi^{-1}$ identifies $(g, v) in G times T_1 G$ with $tilde v_g in T_g G$.
With this in hand, we can check smoothness directly by verifying that $tilde v cdot f$ is smooth whenever $f$ is, and this again amounts to unwinding definitions and recognizing that an expression for $tilde v cdot f$ is manifestly smooth.
answered Dec 9 '18 at 17:03
TravisTravis
60.3k767147
edited Dec 9 '18 at 17:19
answered Dec 9 '18 at 17:03
TravisTravis
60.3k767147
answered Dec 9 '18 at 17:03
TravisTravis
60.3k767147
answered Dec 9 '18 at 17:03
TravisTravis
60.3k767147
60.3k767147
$begingroup$
I am actually trying to show that there exists a unique left invariant field. I am trying to define a 'simple' left invariant vector field, and later showing that it is unique.
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:14
$begingroup$
The left-invariant vector field $tilde v$ is unique more or less by definition: In order for $tilde v$ to be left-invariant, then specializing the equation $T_1 L_g cdot tilde v_h = tilde v_{gh}$ to $h = e$ gives $tilde v_g = T_1 L_g cdot v$.
$endgroup$
– Travis
Dec 9 '18 at 17:21
$begingroup$
Just to be sure, when working with a definition not by derivations, what exactly does $L_gcdot v$ means?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:23
$begingroup$
The notation I used is, e.g., $T_1 L_g cdot v$. Here, $T_1 L_g : T_1 G to T_g G$ is the pushforward of the smooth map $L_g$ at $1 in G$.
$endgroup$
– Travis
Dec 9 '18 at 17:26
$begingroup$
So is it the differential of $L_g$ at $1_G$, right?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:32
|
show 1 more comment
$begingroup$
I am actually trying to show that there exists a unique left invariant field. I am trying to define a 'simple' left invariant vector field, and later showing that it is unique.
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:14
$begingroup$
The left-invariant vector field $tilde v$ is unique more or less by definition: In order for $tilde v$ to be left-invariant, then specializing the equation $T_1 L_g cdot tilde v_h = tilde v_{gh}$ to $h = e$ gives $tilde v_g = T_1 L_g cdot v$.
$endgroup$
– Travis
Dec 9 '18 at 17:21
$begingroup$
Just to be sure, when working with a definition not by derivations, what exactly does $L_gcdot v$ means?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:23
$begingroup$
The notation I used is, e.g., $T_1 L_g cdot v$. Here, $T_1 L_g : T_1 G to T_g G$ is the pushforward of the smooth map $L_g$ at $1 in G$.
$endgroup$
– Travis
Dec 9 '18 at 17:26
$begingroup$
So is it the differential of $L_g$ at $1_G$, right?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:32
$begingroup$
I am actually trying to show that there exists a unique left invariant field. I am trying to define a 'simple' left invariant vector field, and later showing that it is unique.
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:14
$begingroup$
I am actually trying to show that there exists a unique left invariant field. I am trying to define a 'simple' left invariant vector field, and later showing that it is unique.
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:14
$begingroup$
The left-invariant vector field $tilde v$ is unique more or less by definition: In order for $tilde v$ to be left-invariant, then specializing the equation $T_1 L_g cdot tilde v_h = tilde v_{gh}$ to $h = e$ gives $tilde v_g = T_1 L_g cdot v$.
$endgroup$
– Travis
Dec 9 '18 at 17:21
$begingroup$
The left-invariant vector field $tilde v$ is unique more or less by definition: In order for $tilde v$ to be left-invariant, then specializing the equation $T_1 L_g cdot tilde v_h = tilde v_{gh}$ to $h = e$ gives $tilde v_g = T_1 L_g cdot v$.
$endgroup$
– Travis
Dec 9 '18 at 17:21
$begingroup$
Just to be sure, when working with a definition not by derivations, what exactly does $L_gcdot v$ means?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:23
$begingroup$
Just to be sure, when working with a definition not by derivations, what exactly does $L_gcdot v$ means?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:23
$begingroup$
The notation I used is, e.g., $T_1 L_g cdot v$. Here, $T_1 L_g : T_1 G to T_g G$ is the pushforward of the smooth map $L_g$ at $1 in G$.
$endgroup$
– Travis
Dec 9 '18 at 17:26
$begingroup$
The notation I used is, e.g., $T_1 L_g cdot v$. Here, $T_1 L_g : T_1 G to T_g G$ is the pushforward of the smooth map $L_g$ at $1 in G$.
$endgroup$
– Travis
Dec 9 '18 at 17:26
$begingroup$
So is it the differential of $L_g$ at $1_G$, right?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:32
$begingroup$
So is it the differential of $L_g$ at $1_G$, right?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:32
|
show 1 more comment
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$begingroup$
What do you mean by $(g, v)$? Is it a translation?
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 16:44
$begingroup$
Elements of the tangent bundles are $(g,w)$ where $gin G$ and $win T_g(G)$, or am I wrong?
$endgroup$
– Keen-ameteur
Dec 9 '18 at 17:03
$begingroup$
This does not make sense in general, and this is a fundamental fact of differential geometry. Precisely, you cannot translate vectors like that; this is why the standard notation is $w_g$ instead of $(g, w)$. Imagine a vector tangent to a circle. Move that vector elsewhere on the circle; it is not tangent anymore. In general, translation is a subtle matter.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:08