Mixing
Computing the differential of a Lie group action
$begingroup$
(Ex. 27.4 page 252 Loring Tu) (The differential of an action). Let $mu: P times G rightarrow P$. For $g in G$, the tangent space $T_gG$ may be identified with $l_{g*} mathfrak{g} $, where $l_g:G rightarrow G$ is left multiplication by $g in G$ and $mathfrak g = T_eG$ is the Lie algebra of $G$. An element of the tangent space $T_{(p,g)}P times G$ is of the form
$$(X_P, l_{g*} A)$$
for $X_p in T_pP$ and $A in mathfrak g$. The differential is given by
$$ mu_*= mu_{*,(p,g)} :T_{(p,g)}(P times G) rightarrow T_{pg} P$$
is given by
$$ mu_*(X_p, l_{g*} A) = r_{g*} (X_p) + underline{A}_{pg} $$
Definition: $underline{A}$ is the fundamental vector field on $P$ associated to $A in mathfrak{g}$,
$$ underline{A}_p = frac{d}{dt}Big|_{t=0} p cdot e^{tA} in T_pP$$
I am struggling in writing out the proof rigorously and neatly. I would be greatful if someone may spell this out.
differential-geometry lie-groups lie-algebras differential principal-bundles
asked Feb 3 at 9:54
CL.CL.
2,3123925
$endgroup$
add a comment |
$begingroup$
(Ex. 27.4 page 252 Loring Tu) (The differential of an action). Let $mu: P times G rightarrow P$. For $g in G$, the tangent space $T_gG$ may be identified with $l_{g*} mathfrak{g} $, where $l_g:G rightarrow G$ is left multiplication by $g in G$ and $mathfrak g = T_eG$ is the Lie algebra of $G$. An element of the tangent space $T_{(p,g)}P times G$ is of the form
$$(X_P, l_{g*} A)$$
for $X_p in T_pP$ and $A in mathfrak g$. The differential is given by
$$ mu_*= mu_{*,(p,g)} :T_{(p,g)}(P times G) rightarrow T_{pg} P$$
is given by
$$ mu_*(X_p, l_{g*} A) = r_{g*} (X_p) + underline{A}_{pg} $$
Definition: $underline{A}$ is the fundamental vector field on $P$ associated to $A in mathfrak{g}$,
$$ underline{A}_p = frac{d}{dt}Big|_{t=0} p cdot e^{tA} in T_pP$$
I am struggling in writing out the proof rigorously and neatly. I would be greatful if someone may spell this out.
differential-geometry lie-groups lie-algebras differential principal-bundles
asked Feb 3 at 9:54
CL.CL.
2,3123925
$endgroup$
add a comment |
$begingroup$
(Ex. 27.4 page 252 Loring Tu) (The differential of an action). Let $mu: P times G rightarrow P$. For $g in G$, the tangent space $T_gG$ may be identified with $l_{g*} mathfrak{g} $, where $l_g:G rightarrow G$ is left multiplication by $g in G$ and $mathfrak g = T_eG$ is the Lie algebra of $G$. An element of the tangent space $T_{(p,g)}P times G$ is of the form
$$(X_P, l_{g*} A)$$
for $X_p in T_pP$ and $A in mathfrak g$. The differential is given by
$$ mu_*= mu_{*,(p,g)} :T_{(p,g)}(P times G) rightarrow T_{pg} P$$
is given by
$$ mu_*(X_p, l_{g*} A) = r_{g*} (X_p) + underline{A}_{pg} $$
Definition: $underline{A}$ is the fundamental vector field on $P$ associated to $A in mathfrak{g}$,
$$ underline{A}_p = frac{d}{dt}Big|_{t=0} p cdot e^{tA} in T_pP$$
I am struggling in writing out the proof rigorously and neatly. I would be greatful if someone may spell this out.
differential-geometry lie-groups lie-algebras differential principal-bundles
asked Feb 3 at 9:54
CL.CL.
2,3123925
$endgroup$
(Ex. 27.4 page 252 Loring Tu) (The differential of an action). Let $mu: P times G rightarrow P$. For $g in G$, the tangent space $T_gG$ may be identified with $l_{g*} mathfrak{g} $, where $l_g:G rightarrow G$ is left multiplication by $g in G$ and $mathfrak g = T_eG$ is the Lie algebra of $G$. An element of the tangent space $T_{(p,g)}P times G$ is of the form
$$(X_P, l_{g*} A)$$
for $X_p in T_pP$ and $A in mathfrak g$. The differential is given by
$$ mu_*= mu_{*,(p,g)} :T_{(p,g)}(P times G) rightarrow T_{pg} P$$
is given by
$$ mu_*(X_p, l_{g*} A) = r_{g*} (X_p) + underline{A}_{pg} $$
Definition: $underline{A}$ is the fundamental vector field on $P$ associated to $A in mathfrak{g}$,
$$ underline{A}_p = frac{d}{dt}Big|_{t=0} p cdot e^{tA} in T_pP$$
I am struggling in writing out the proof rigorously and neatly. I would be greatful if someone may spell this out.
differential-geometry lie-groups lie-algebras differential principal-bundles
differential-geometry lie-groups lie-algebras differential principal-bundles
asked Feb 3 at 9:54
CL.CL.
2,3123925
asked Feb 3 at 9:54
CL.CL.
2,3123925
edited Feb 3 at 11:45
CL.
asked Feb 3 at 9:54
CL.CL.
2,3123925
asked Feb 3 at 9:54
CL.CL.
2,3123925
asked Feb 3 at 9:54
CL.CL.
2,3123925
2,3123925
add a comment |
add a comment |
1 Answer
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$begingroup$
So with my edits above I think I have came up with the answer:
$$ cdots cdots cdots $$
Let us identify
$$T_{(p,g)}(P times G) cong T_pP oplus T_gG$$
hence by linearity of differential, it suffices to compute each component.
$$ cdots cdots cdots $$
Let $(X_p, 0) in T_pP oplus T_gG$. $varphi(t):(-varepsilon, varepsilon) rightarrow P$ be a smooth curve, $varphi(0) = p, varphi'(0)=X_p$. Define $$gamma(t):(-varepsilon, varepsilon) rightarrow P times G, quad t mapsto (varphi(t), g)$$
Then
begin{align*}
mu_*(X_p,0) &= mu_* gamma_* (frac{d}{dt} Big|_{t=0}) \
& =(r_g circ varphi)_* (frac{d}{dt} Big|_{t=0}) \
&= (r_g)_* X_p
end{align*}
$$ cdots cdots cdots $$
To compute $(0,(l_g)_*A)$ we consider the curve,
$$ gamma(t): (-varepsilon, varepsilon) rightarrow P times G, (p, g cdot e^{tA}) $$
Then we have
begin{align*}
mu_* gamma_* (frac{d}{dt} Big|_{t=0}) &= (c_{pg})_*(frac{d}{dt} Big|_{t=0}) \
&=: underline{A}_{pg} end{align*}
where $c_{pg}(t):Bbb R rightarrow P$ is the map given by $pg cdot e^{tA}$.
answered Feb 3 at 11:44
CL.CL.
2,3123925
$endgroup$
$begingroup$
Addendum: I have just found out the answer is also included in the end of the book.
$endgroup$
– CL.
Feb 4 at 9:45
add a comment |
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$begingroup$
So with my edits above I think I have came up with the answer:
$$ cdots cdots cdots $$
Let us identify
$$T_{(p,g)}(P times G) cong T_pP oplus T_gG$$
hence by linearity of differential, it suffices to compute each component.
$$ cdots cdots cdots $$
Let $(X_p, 0) in T_pP oplus T_gG$. $varphi(t):(-varepsilon, varepsilon) rightarrow P$ be a smooth curve, $varphi(0) = p, varphi'(0)=X_p$. Define $$gamma(t):(-varepsilon, varepsilon) rightarrow P times G, quad t mapsto (varphi(t), g)$$
Then
begin{align*}
mu_*(X_p,0) &= mu_* gamma_* (frac{d}{dt} Big|_{t=0}) \
& =(r_g circ varphi)_* (frac{d}{dt} Big|_{t=0}) \
&= (r_g)_* X_p
end{align*}
$$ cdots cdots cdots $$
To compute $(0,(l_g)_*A)$ we consider the curve,
$$ gamma(t): (-varepsilon, varepsilon) rightarrow P times G, (p, g cdot e^{tA}) $$
Then we have
begin{align*}
mu_* gamma_* (frac{d}{dt} Big|_{t=0}) &= (c_{pg})_*(frac{d}{dt} Big|_{t=0}) \
&=: underline{A}_{pg} end{align*}
where $c_{pg}(t):Bbb R rightarrow P$ is the map given by $pg cdot e^{tA}$.
answered Feb 3 at 11:44
CL.CL.
2,3123925
$endgroup$
$begingroup$
Addendum: I have just found out the answer is also included in the end of the book.
$endgroup$
– CL.
Feb 4 at 9:45
add a comment |
$begingroup$
So with my edits above I think I have came up with the answer:
$$ cdots cdots cdots $$
Let us identify
$$T_{(p,g)}(P times G) cong T_pP oplus T_gG$$
hence by linearity of differential, it suffices to compute each component.
$$ cdots cdots cdots $$
Let $(X_p, 0) in T_pP oplus T_gG$. $varphi(t):(-varepsilon, varepsilon) rightarrow P$ be a smooth curve, $varphi(0) = p, varphi'(0)=X_p$. Define $$gamma(t):(-varepsilon, varepsilon) rightarrow P times G, quad t mapsto (varphi(t), g)$$
Then
begin{align*}
mu_*(X_p,0) &= mu_* gamma_* (frac{d}{dt} Big|_{t=0}) \
& =(r_g circ varphi)_* (frac{d}{dt} Big|_{t=0}) \
&= (r_g)_* X_p
end{align*}
$$ cdots cdots cdots $$
To compute $(0,(l_g)_*A)$ we consider the curve,
$$ gamma(t): (-varepsilon, varepsilon) rightarrow P times G, (p, g cdot e^{tA}) $$
Then we have
begin{align*}
mu_* gamma_* (frac{d}{dt} Big|_{t=0}) &= (c_{pg})_*(frac{d}{dt} Big|_{t=0}) \
&=: underline{A}_{pg} end{align*}
where $c_{pg}(t):Bbb R rightarrow P$ is the map given by $pg cdot e^{tA}$.
answered Feb 3 at 11:44
CL.CL.
2,3123925
$endgroup$
$begingroup$
Addendum: I have just found out the answer is also included in the end of the book.
$endgroup$
– CL.
Feb 4 at 9:45
add a comment |
$begingroup$
So with my edits above I think I have came up with the answer:
$$ cdots cdots cdots $$
Let us identify
$$T_{(p,g)}(P times G) cong T_pP oplus T_gG$$
hence by linearity of differential, it suffices to compute each component.
$$ cdots cdots cdots $$
Let $(X_p, 0) in T_pP oplus T_gG$. $varphi(t):(-varepsilon, varepsilon) rightarrow P$ be a smooth curve, $varphi(0) = p, varphi'(0)=X_p$. Define $$gamma(t):(-varepsilon, varepsilon) rightarrow P times G, quad t mapsto (varphi(t), g)$$
Then
begin{align*}
mu_*(X_p,0) &= mu_* gamma_* (frac{d}{dt} Big|_{t=0}) \
& =(r_g circ varphi)_* (frac{d}{dt} Big|_{t=0}) \
&= (r_g)_* X_p
end{align*}
$$ cdots cdots cdots $$
To compute $(0,(l_g)_*A)$ we consider the curve,
$$ gamma(t): (-varepsilon, varepsilon) rightarrow P times G, (p, g cdot e^{tA}) $$
Then we have
begin{align*}
mu_* gamma_* (frac{d}{dt} Big|_{t=0}) &= (c_{pg})_*(frac{d}{dt} Big|_{t=0}) \
&=: underline{A}_{pg} end{align*}
where $c_{pg}(t):Bbb R rightarrow P$ is the map given by $pg cdot e^{tA}$.
answered Feb 3 at 11:44
CL.CL.
2,3123925
$endgroup$
So with my edits above I think I have came up with the answer:
$$ cdots cdots cdots $$
Let us identify
$$T_{(p,g)}(P times G) cong T_pP oplus T_gG$$
hence by linearity of differential, it suffices to compute each component.
$$ cdots cdots cdots $$
Let $(X_p, 0) in T_pP oplus T_gG$. $varphi(t):(-varepsilon, varepsilon) rightarrow P$ be a smooth curve, $varphi(0) = p, varphi'(0)=X_p$. Define $$gamma(t):(-varepsilon, varepsilon) rightarrow P times G, quad t mapsto (varphi(t), g)$$
Then
begin{align*}
mu_*(X_p,0) &= mu_* gamma_* (frac{d}{dt} Big|_{t=0}) \
& =(r_g circ varphi)_* (frac{d}{dt} Big|_{t=0}) \
&= (r_g)_* X_p
end{align*}
$$ cdots cdots cdots $$
To compute $(0,(l_g)_*A)$ we consider the curve,
$$ gamma(t): (-varepsilon, varepsilon) rightarrow P times G, (p, g cdot e^{tA}) $$
Then we have
begin{align*}
mu_* gamma_* (frac{d}{dt} Big|_{t=0}) &= (c_{pg})_*(frac{d}{dt} Big|_{t=0}) \
&=: underline{A}_{pg} end{align*}
where $c_{pg}(t):Bbb R rightarrow P$ is the map given by $pg cdot e^{tA}$.
answered Feb 3 at 11:44
CL.CL.
2,3123925
answered Feb 3 at 11:44
CL.CL.
2,3123925
answered Feb 3 at 11:44
CL.CL.
2,3123925
answered Feb 3 at 11:44
CL.CL.
2,3123925
2,3123925
$begingroup$
Addendum: I have just found out the answer is also included in the end of the book.
$endgroup$
– CL.
Feb 4 at 9:45
add a comment |
$begingroup$
Addendum: I have just found out the answer is also included in the end of the book.
$endgroup$
– CL.
Feb 4 at 9:45
$begingroup$
Addendum: I have just found out the answer is also included in the end of the book.
$endgroup$
– CL.
Feb 4 at 9:45
$begingroup$
Addendum: I have just found out the answer is also included in the end of the book.
$endgroup$
– CL.
Feb 4 at 9:45
add a comment |
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