Computing the differential of a Lie group action












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(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







share|cite|improve this question











$endgroup$

















    1












    $begingroup$



    (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







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$



      (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







      share|cite|improve this question











      $endgroup$





      (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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 3 at 11:45







      CL.

















      asked Feb 3 at 9:54









      CL.CL.

      2,3123925




      2,3123925






















          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 $pg cdot e^{tA}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Addendum: I have just found out the answer is also included in the end of the book.
            $endgroup$
            – CL.
            Feb 4 at 9:45












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          1












          $begingroup$

          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 $pg cdot e^{tA}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Addendum: I have just found out the answer is also included in the end of the book.
            $endgroup$
            – CL.
            Feb 4 at 9:45
















          1












          $begingroup$

          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 $pg cdot e^{tA}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Addendum: I have just found out the answer is also included in the end of the book.
            $endgroup$
            – CL.
            Feb 4 at 9:45














          1












          1








          1





          $begingroup$

          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 $pg cdot e^{tA}$.






          share|cite|improve this answer









          $endgroup$



          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 $pg cdot e^{tA}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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


















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