Worked examples of Lie derivatives












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I'm trying to find the Lie derivative of a 2-form $sin(theta)dtheta wedge dphi$ with respect to a vector field given in a differential basis $a partial/ partial phi$ and I think the way to go here is to use Cartan's formula but I'm not sure how to do that. I can't find any worked examples in any differential textbook or online, everything seems to focus on proving identities not performing actual calculations. Can somebody point me to a useful resource with some exercises or step-by-step examples?






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    I'm trying to find the Lie derivative of a 2-form $sin(theta)dtheta wedge dphi$ with respect to a vector field given in a differential basis $a partial/ partial phi$ and I think the way to go here is to use Cartan's formula but I'm not sure how to do that. I can't find any worked examples in any differential textbook or online, everything seems to focus on proving identities not performing actual calculations. Can somebody point me to a useful resource with some exercises or step-by-step examples?






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      I'm trying to find the Lie derivative of a 2-form $sin(theta)dtheta wedge dphi$ with respect to a vector field given in a differential basis $a partial/ partial phi$ and I think the way to go here is to use Cartan's formula but I'm not sure how to do that. I can't find any worked examples in any differential textbook or online, everything seems to focus on proving identities not performing actual calculations. Can somebody point me to a useful resource with some exercises or step-by-step examples?






      differential-geometry differential-forms lie-derivative







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      I'm trying to find the Lie derivative of a 2-form $sin(theta)dtheta wedge dphi$ with respect to a vector field given in a differential basis $a partial/ partial phi$ and I think the way to go here is to use Cartan's formula but I'm not sure how to do that. I can't find any worked examples in any differential textbook or online, everything seems to focus on proving identities not performing actual calculations. Can somebody point me to a useful resource with some exercises or step-by-step examples?






      differential-geometry differential-forms lie-derivative




      differential-geometry differential-forms lie-derivative






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      edited Jan 21 at 6:17









      Ivo Terek

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      asked Jan 20 at 23:10









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          I understand that you want to compute $$mathcal{L}_{a partial_theta}(sintheta,{rm d}thetawedge {rm d}phi).$$In this case it is easy enough to use the definition via flows. The flow of the vector field $apartial_theta$ is just $$Phi_{t,apartial_theta}(theta,phi) = (theta+at, phi),$$so $$begin{align}mathcal{L}_{apartial_theta}(sintheta,{rm d}theta wedge {rm d}theta) &= frac{rm d}{{rm d}t}bigg|_{t=0} (Phi_{t,apartial_theta})^* (sintheta,{rm d}theta wedge {rm d}phi) \ &= frac{rm d}{{rm d}t}bigg|_{t=0} sin(theta+at),{rm d}(theta+at)wedge{rm d}phi \ &= frac{{rm d}}{{rm d}t}bigg|_{t=0} sin(theta+at),{rm d}thetawedge {rm d}phi \ &= acos(theta+at)bigg|_{t=0},{rm d}thetawedge{rm d}phi \ &= a costheta,{rm d}thetawedge{rm d}phi.end{align}$$If you want to double-check using Cartan's Homotopy Formula $mathcal{L}_{apartial_theta} = iota_{apartial_theta}circ {rm d}+{rm d}circ iota_{apartial_theta}$, you can just use that $${rm d}(sintheta,{rm d}thetawedge {rm d}phi) = costheta ,{rm d}thetawedge{rm d}thetawedge {rm d}phi = 0$$because ${rm d}theta$ repeats, and compute $$begin{align} iota_{apartial_theta}(sintheta,{rm d}thetawedge{rm d}phi) &= sintheta,({rm d}theta wedge {rm d}phi)(apartial_theta, cdot) \ &= sintheta begin{vmatrix} {rm d}theta(apartial_theta) & {rm d}theta \ {rm d}phi(apartial_theta) & {rm d}phiend{vmatrix} \ &= sin theta begin{vmatrix} a & {rm d}theta \ 0 & {rm d}phiend{vmatrix} \ &= a sin {rm d}phi,end{align}$$leading again to $$mathcal{L}_{apartial_theta} = {rm d}(asintheta,{rm d}phi) = acostheta,{rm d}theta wedge {rm d}phi.$$Anyway, you can look for more examples like this in the book Analysis and Algebra in Differentiable Manifolds -
          A Workbook for Students and Teachers           by Gadea, et al.






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            $L_X=i_Xd+di_X$, you have $d(sin(theta)dthetawedge dphi)=cos(theta)dthetawedge dthetawedge dphi=0$



            Let $(u,v)$ be a vector of $T_{theta,phi}mathbb{R}^2$ $i_{a{partial}over{partialphi}}sin(theta)dthetawedge dphi=sin(theta)detpmatrix{0&ucr a&v}=-asin(theta)u=-asin(theta)dtheta(u,v)$. This implies that $d(i_{a{partial}over{partialphi}}sin(theta)dthetawedge dphi)=0$






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              $begingroup$

              I understand that you want to compute $$mathcal{L}_{a partial_theta}(sintheta,{rm d}thetawedge {rm d}phi).$$In this case it is easy enough to use the definition via flows. The flow of the vector field $apartial_theta$ is just $$Phi_{t,apartial_theta}(theta,phi) = (theta+at, phi),$$so $$begin{align}mathcal{L}_{apartial_theta}(sintheta,{rm d}theta wedge {rm d}theta) &= frac{rm d}{{rm d}t}bigg|_{t=0} (Phi_{t,apartial_theta})^* (sintheta,{rm d}theta wedge {rm d}phi) \ &= frac{rm d}{{rm d}t}bigg|_{t=0} sin(theta+at),{rm d}(theta+at)wedge{rm d}phi \ &= frac{{rm d}}{{rm d}t}bigg|_{t=0} sin(theta+at),{rm d}thetawedge {rm d}phi \ &= acos(theta+at)bigg|_{t=0},{rm d}thetawedge{rm d}phi \ &= a costheta,{rm d}thetawedge{rm d}phi.end{align}$$If you want to double-check using Cartan's Homotopy Formula $mathcal{L}_{apartial_theta} = iota_{apartial_theta}circ {rm d}+{rm d}circ iota_{apartial_theta}$, you can just use that $${rm d}(sintheta,{rm d}thetawedge {rm d}phi) = costheta ,{rm d}thetawedge{rm d}thetawedge {rm d}phi = 0$$because ${rm d}theta$ repeats, and compute $$begin{align} iota_{apartial_theta}(sintheta,{rm d}thetawedge{rm d}phi) &= sintheta,({rm d}theta wedge {rm d}phi)(apartial_theta, cdot) \ &= sintheta begin{vmatrix} {rm d}theta(apartial_theta) & {rm d}theta \ {rm d}phi(apartial_theta) & {rm d}phiend{vmatrix} \ &= sin theta begin{vmatrix} a & {rm d}theta \ 0 & {rm d}phiend{vmatrix} \ &= a sin {rm d}phi,end{align}$$leading again to $$mathcal{L}_{apartial_theta} = {rm d}(asintheta,{rm d}phi) = acostheta,{rm d}theta wedge {rm d}phi.$$Anyway, you can look for more examples like this in the book Analysis and Algebra in Differentiable Manifolds -
              A Workbook for Students and Teachers               by Gadea, et al.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                I understand that you want to compute $$mathcal{L}_{a partial_theta}(sintheta,{rm d}thetawedge {rm d}phi).$$In this case it is easy enough to use the definition via flows. The flow of the vector field $apartial_theta$ is just $$Phi_{t,apartial_theta}(theta,phi) = (theta+at, phi),$$so $$begin{align}mathcal{L}_{apartial_theta}(sintheta,{rm d}theta wedge {rm d}theta) &= frac{rm d}{{rm d}t}bigg|_{t=0} (Phi_{t,apartial_theta})^* (sintheta,{rm d}theta wedge {rm d}phi) \ &= frac{rm d}{{rm d}t}bigg|_{t=0} sin(theta+at),{rm d}(theta+at)wedge{rm d}phi \ &= frac{{rm d}}{{rm d}t}bigg|_{t=0} sin(theta+at),{rm d}thetawedge {rm d}phi \ &= acos(theta+at)bigg|_{t=0},{rm d}thetawedge{rm d}phi \ &= a costheta,{rm d}thetawedge{rm d}phi.end{align}$$If you want to double-check using Cartan's Homotopy Formula $mathcal{L}_{apartial_theta} = iota_{apartial_theta}circ {rm d}+{rm d}circ iota_{apartial_theta}$, you can just use that $${rm d}(sintheta,{rm d}thetawedge {rm d}phi) = costheta ,{rm d}thetawedge{rm d}thetawedge {rm d}phi = 0$$because ${rm d}theta$ repeats, and compute $$begin{align} iota_{apartial_theta}(sintheta,{rm d}thetawedge{rm d}phi) &= sintheta,({rm d}theta wedge {rm d}phi)(apartial_theta, cdot) \ &= sintheta begin{vmatrix} {rm d}theta(apartial_theta) & {rm d}theta \ {rm d}phi(apartial_theta) & {rm d}phiend{vmatrix} \ &= sin theta begin{vmatrix} a & {rm d}theta \ 0 & {rm d}phiend{vmatrix} \ &= a sin {rm d}phi,end{align}$$leading again to $$mathcal{L}_{apartial_theta} = {rm d}(asintheta,{rm d}phi) = acostheta,{rm d}theta wedge {rm d}phi.$$Anyway, you can look for more examples like this in the book Analysis and Algebra in Differentiable Manifolds -
                A Workbook for Students and Teachers                 by Gadea, et al.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I understand that you want to compute $$mathcal{L}_{a partial_theta}(sintheta,{rm d}thetawedge {rm d}phi).$$In this case it is easy enough to use the definition via flows. The flow of the vector field $apartial_theta$ is just $$Phi_{t,apartial_theta}(theta,phi) = (theta+at, phi),$$so $$begin{align}mathcal{L}_{apartial_theta}(sintheta,{rm d}theta wedge {rm d}theta) &= frac{rm d}{{rm d}t}bigg|_{t=0} (Phi_{t,apartial_theta})^* (sintheta,{rm d}theta wedge {rm d}phi) \ &= frac{rm d}{{rm d}t}bigg|_{t=0} sin(theta+at),{rm d}(theta+at)wedge{rm d}phi \ &= frac{{rm d}}{{rm d}t}bigg|_{t=0} sin(theta+at),{rm d}thetawedge {rm d}phi \ &= acos(theta+at)bigg|_{t=0},{rm d}thetawedge{rm d}phi \ &= a costheta,{rm d}thetawedge{rm d}phi.end{align}$$If you want to double-check using Cartan's Homotopy Formula $mathcal{L}_{apartial_theta} = iota_{apartial_theta}circ {rm d}+{rm d}circ iota_{apartial_theta}$, you can just use that $${rm d}(sintheta,{rm d}thetawedge {rm d}phi) = costheta ,{rm d}thetawedge{rm d}thetawedge {rm d}phi = 0$$because ${rm d}theta$ repeats, and compute $$begin{align} iota_{apartial_theta}(sintheta,{rm d}thetawedge{rm d}phi) &= sintheta,({rm d}theta wedge {rm d}phi)(apartial_theta, cdot) \ &= sintheta begin{vmatrix} {rm d}theta(apartial_theta) & {rm d}theta \ {rm d}phi(apartial_theta) & {rm d}phiend{vmatrix} \ &= sin theta begin{vmatrix} a & {rm d}theta \ 0 & {rm d}phiend{vmatrix} \ &= a sin {rm d}phi,end{align}$$leading again to $$mathcal{L}_{apartial_theta} = {rm d}(asintheta,{rm d}phi) = acostheta,{rm d}theta wedge {rm d}phi.$$Anyway, you can look for more examples like this in the book Analysis and Algebra in Differentiable Manifolds -
                  A Workbook for Students and Teachers                   by Gadea, et al.






                  share|cite|improve this answer









                  $endgroup$



                  I understand that you want to compute $$mathcal{L}_{a partial_theta}(sintheta,{rm d}thetawedge {rm d}phi).$$In this case it is easy enough to use the definition via flows. The flow of the vector field $apartial_theta$ is just $$Phi_{t,apartial_theta}(theta,phi) = (theta+at, phi),$$so $$begin{align}mathcal{L}_{apartial_theta}(sintheta,{rm d}theta wedge {rm d}theta) &= frac{rm d}{{rm d}t}bigg|_{t=0} (Phi_{t,apartial_theta})^* (sintheta,{rm d}theta wedge {rm d}phi) \ &= frac{rm d}{{rm d}t}bigg|_{t=0} sin(theta+at),{rm d}(theta+at)wedge{rm d}phi \ &= frac{{rm d}}{{rm d}t}bigg|_{t=0} sin(theta+at),{rm d}thetawedge {rm d}phi \ &= acos(theta+at)bigg|_{t=0},{rm d}thetawedge{rm d}phi \ &= a costheta,{rm d}thetawedge{rm d}phi.end{align}$$If you want to double-check using Cartan's Homotopy Formula $mathcal{L}_{apartial_theta} = iota_{apartial_theta}circ {rm d}+{rm d}circ iota_{apartial_theta}$, you can just use that $${rm d}(sintheta,{rm d}thetawedge {rm d}phi) = costheta ,{rm d}thetawedge{rm d}thetawedge {rm d}phi = 0$$because ${rm d}theta$ repeats, and compute $$begin{align} iota_{apartial_theta}(sintheta,{rm d}thetawedge{rm d}phi) &= sintheta,({rm d}theta wedge {rm d}phi)(apartial_theta, cdot) \ &= sintheta begin{vmatrix} {rm d}theta(apartial_theta) & {rm d}theta \ {rm d}phi(apartial_theta) & {rm d}phiend{vmatrix} \ &= sin theta begin{vmatrix} a & {rm d}theta \ 0 & {rm d}phiend{vmatrix} \ &= a sin {rm d}phi,end{align}$$leading again to $$mathcal{L}_{apartial_theta} = {rm d}(asintheta,{rm d}phi) = acostheta,{rm d}theta wedge {rm d}phi.$$Anyway, you can look for more examples like this in the book Analysis and Algebra in Differentiable Manifolds -
                  A Workbook for Students and Teachers                   by Gadea, et al.







                  share|cite|improve this answer












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                  share|cite|improve this answer










                  answered Jan 21 at 6:15









                  Ivo TerekIvo Terek

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                      0












                      $begingroup$

                      $L_X=i_Xd+di_X$, you have $d(sin(theta)dthetawedge dphi)=cos(theta)dthetawedge dthetawedge dphi=0$



                      Let $(u,v)$ be a vector of $T_{theta,phi}mathbb{R}^2$ $i_{a{partial}over{partialphi}}sin(theta)dthetawedge dphi=sin(theta)detpmatrix{0&ucr a&v}=-asin(theta)u=-asin(theta)dtheta(u,v)$. This implies that $d(i_{a{partial}over{partialphi}}sin(theta)dthetawedge dphi)=0$






                      share|cite|improve this answer











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                        0












                        $begingroup$

                        $L_X=i_Xd+di_X$, you have $d(sin(theta)dthetawedge dphi)=cos(theta)dthetawedge dthetawedge dphi=0$



                        Let $(u,v)$ be a vector of $T_{theta,phi}mathbb{R}^2$ $i_{a{partial}over{partialphi}}sin(theta)dthetawedge dphi=sin(theta)detpmatrix{0&ucr a&v}=-asin(theta)u=-asin(theta)dtheta(u,v)$. This implies that $d(i_{a{partial}over{partialphi}}sin(theta)dthetawedge dphi)=0$






                        share|cite|improve this answer











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                          0












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                          0





                          $begingroup$

                          $L_X=i_Xd+di_X$, you have $d(sin(theta)dthetawedge dphi)=cos(theta)dthetawedge dthetawedge dphi=0$



                          Let $(u,v)$ be a vector of $T_{theta,phi}mathbb{R}^2$ $i_{a{partial}over{partialphi}}sin(theta)dthetawedge dphi=sin(theta)detpmatrix{0&ucr a&v}=-asin(theta)u=-asin(theta)dtheta(u,v)$. This implies that $d(i_{a{partial}over{partialphi}}sin(theta)dthetawedge dphi)=0$






                          share|cite|improve this answer











                          $endgroup$



                          $L_X=i_Xd+di_X$, you have $d(sin(theta)dthetawedge dphi)=cos(theta)dthetawedge dthetawedge dphi=0$



                          Let $(u,v)$ be a vector of $T_{theta,phi}mathbb{R}^2$ $i_{a{partial}over{partialphi}}sin(theta)dthetawedge dphi=sin(theta)detpmatrix{0&ucr a&v}=-asin(theta)u=-asin(theta)dtheta(u,v)$. This implies that $d(i_{a{partial}over{partialphi}}sin(theta)dthetawedge dphi)=0$







                          share|cite|improve this answer














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                          share|cite|improve this answer








                          edited Jan 20 at 23:58

























                          answered Jan 20 at 23:21









                          Tsemo AristideTsemo Aristide

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