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Why is this formula wrong ? lie derivative tensor
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I have derived some formula for the lie derivative of a covariant tensor which seems wrong....
Let $X$ is vector field inducing a flow $phi_t$ and $T$ a rank $2$ covariant tensor, I claim that
$$(mathcalL_XT)(Y,Z)=T(mathcalL_XY,Z)+T(Y,mathcalL_XZ) $$
By definition we have that
$$(mathcalL_XT)(Y,Z)=fracddt(phi_-t)_*T_phi_t(p)(Y,Z) $$
Now by bi-linearity of $T$ we have:
beginalign*
(phi_-t)_*T_phi_t(p)(Y,Z) &= T((phi_-t)*Y_phi_t(p),(phi_-t)*Z_phi_t(p)) \
&=T((phi_-t)*Y_phi_t(p)-Y,(phi_-t)*Z_phi_t(p))+T(Y,(phi_-t)*Z_phi_t(p))
endalign*
Hence
$$frac(phi_-t)_*T_phi_t(p)(Y,Z)-T_p(Y,Z) t= T(frac(phi_-t)*Y_phi_t(p)-Yt,(phi_-t)_*Z_phi_t(p)) +T(Y_p,frac(phi_-t)*Z_phi_t(p)-Z_pt) $$
We conclude by taking the limit as $t$ goes to $0$ using the continuity of $T$.
differential-geometry lie-derivative
asked Jul 19 at 3:21
PLP
162
add a comment |Â
up vote
0
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I have derived some formula for the lie derivative of a covariant tensor which seems wrong....
Let $X$ is vector field inducing a flow $phi_t$ and $T$ a rank $2$ covariant tensor, I claim that
$$(mathcalL_XT)(Y,Z)=T(mathcalL_XY,Z)+T(Y,mathcalL_XZ) $$
By definition we have that
$$(mathcalL_XT)(Y,Z)=fracddt(phi_-t)_*T_phi_t(p)(Y,Z) $$
Now by bi-linearity of $T$ we have:
beginalign*
(phi_-t)_*T_phi_t(p)(Y,Z) &= T((phi_-t)*Y_phi_t(p),(phi_-t)*Z_phi_t(p)) \
&=T((phi_-t)*Y_phi_t(p)-Y,(phi_-t)*Z_phi_t(p))+T(Y,(phi_-t)*Z_phi_t(p))
endalign*
Hence
$$frac(phi_-t)_*T_phi_t(p)(Y,Z)-T_p(Y,Z) t= T(frac(phi_-t)*Y_phi_t(p)-Yt,(phi_-t)_*Z_phi_t(p)) +T(Y_p,frac(phi_-t)*Z_phi_t(p)-Z_pt) $$
We conclude by taking the limit as $t$ goes to $0$ using the continuity of $T$.
differential-geometry lie-derivative
asked Jul 19 at 3:21
PLP
162
1
I think there are three variables $Y,Z$ and the coefficients in $T$, which will change as the transformation of $X$. For example, $T(Y,Z)|_p=a(p)Y(p)bigotimes Z(p)$. The result can be obtained from the Leibniz rule.
– W. mu
Jul 19 at 3:46
I see, I forgot to put the proper subscript $T_p$ or $T_phi_t(p)$ at some places making the third coefficient $a(p)$ a constant. Thanks a lot !
– PLP
Jul 19 at 3:55
2
For the record, the correct formula is $$(mathcalL_XT)(Y,Z) = X(T(Y,Z)) - T(mathcalL_XY,Z) - T(Y,mathcalL_XZ).$$
– Ivo Terek
Jul 19 at 4:11
I derived the wrong formula trying to prove this one. It seems the correct way is to use coordinates or to use the fact that Lie derivative commutes with contraction
– PLP
Jul 19 at 4:27
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have derived some formula for the lie derivative of a covariant tensor which seems wrong....
Let $X$ is vector field inducing a flow $phi_t$ and $T$ a rank $2$ covariant tensor, I claim that
$$(mathcalL_XT)(Y,Z)=T(mathcalL_XY,Z)+T(Y,mathcalL_XZ) $$
By definition we have that
$$(mathcalL_XT)(Y,Z)=fracddt(phi_-t)_*T_phi_t(p)(Y,Z) $$
Now by bi-linearity of $T$ we have:
beginalign*
(phi_-t)_*T_phi_t(p)(Y,Z) &= T((phi_-t)*Y_phi_t(p),(phi_-t)*Z_phi_t(p)) \
&=T((phi_-t)*Y_phi_t(p)-Y,(phi_-t)*Z_phi_t(p))+T(Y,(phi_-t)*Z_phi_t(p))
endalign*
Hence
$$frac(phi_-t)_*T_phi_t(p)(Y,Z)-T_p(Y,Z) t= T(frac(phi_-t)*Y_phi_t(p)-Yt,(phi_-t)_*Z_phi_t(p)) +T(Y_p,frac(phi_-t)*Z_phi_t(p)-Z_pt) $$
We conclude by taking the limit as $t$ goes to $0$ using the continuity of $T$.
differential-geometry lie-derivative
asked Jul 19 at 3:21
PLP
162
I have derived some formula for the lie derivative of a covariant tensor which seems wrong....
Let $X$ is vector field inducing a flow $phi_t$ and $T$ a rank $2$ covariant tensor, I claim that
$$(mathcalL_XT)(Y,Z)=T(mathcalL_XY,Z)+T(Y,mathcalL_XZ) $$
By definition we have that
$$(mathcalL_XT)(Y,Z)=fracddt(phi_-t)_*T_phi_t(p)(Y,Z) $$
Now by bi-linearity of $T$ we have:
beginalign*
(phi_-t)_*T_phi_t(p)(Y,Z) &= T((phi_-t)*Y_phi_t(p),(phi_-t)*Z_phi_t(p)) \
&=T((phi_-t)*Y_phi_t(p)-Y,(phi_-t)*Z_phi_t(p))+T(Y,(phi_-t)*Z_phi_t(p))
endalign*
Hence
$$frac(phi_-t)_*T_phi_t(p)(Y,Z)-T_p(Y,Z) t= T(frac(phi_-t)*Y_phi_t(p)-Yt,(phi_-t)_*Z_phi_t(p)) +T(Y_p,frac(phi_-t)*Z_phi_t(p)-Z_pt) $$
We conclude by taking the limit as $t$ goes to $0$ using the continuity of $T$.
differential-geometry lie-derivative
asked Jul 19 at 3:21
PLP
162
edited Jul 19 at 3:39
asked Jul 19 at 3:21
PLP
162
asked Jul 19 at 3:21
PLP
162
asked Jul 19 at 3:21
PLP
162
162
1
I think there are three variables $Y,Z$ and the coefficients in $T$, which will change as the transformation of $X$. For example, $T(Y,Z)|_p=a(p)Y(p)bigotimes Z(p)$. The result can be obtained from the Leibniz rule.
– W. mu
Jul 19 at 3:46
I see, I forgot to put the proper subscript $T_p$ or $T_phi_t(p)$ at some places making the third coefficient $a(p)$ a constant. Thanks a lot !
– PLP
Jul 19 at 3:55
2
For the record, the correct formula is $$(mathcalL_XT)(Y,Z) = X(T(Y,Z)) - T(mathcalL_XY,Z) - T(Y,mathcalL_XZ).$$
– Ivo Terek
Jul 19 at 4:11
I derived the wrong formula trying to prove this one. It seems the correct way is to use coordinates or to use the fact that Lie derivative commutes with contraction
– PLP
Jul 19 at 4:27
add a comment |Â
1
I think there are three variables $Y,Z$ and the coefficients in $T$, which will change as the transformation of $X$. For example, $T(Y,Z)|_p=a(p)Y(p)bigotimes Z(p)$. The result can be obtained from the Leibniz rule.
– W. mu
Jul 19 at 3:46
I see, I forgot to put the proper subscript $T_p$ or $T_phi_t(p)$ at some places making the third coefficient $a(p)$ a constant. Thanks a lot !
– PLP
Jul 19 at 3:55
2
For the record, the correct formula is $$(mathcalL_XT)(Y,Z) = X(T(Y,Z)) - T(mathcalL_XY,Z) - T(Y,mathcalL_XZ).$$
– Ivo Terek
Jul 19 at 4:11
I derived the wrong formula trying to prove this one. It seems the correct way is to use coordinates or to use the fact that Lie derivative commutes with contraction
– PLP
Jul 19 at 4:27
1
1
I think there are three variables $Y,Z$ and the coefficients in $T$, which will change as the transformation of $X$. For example, $T(Y,Z)|_p=a(p)Y(p)bigotimes Z(p)$. The result can be obtained from the Leibniz rule.
– W. mu
Jul 19 at 3:46
I think there are three variables $Y,Z$ and the coefficients in $T$, which will change as the transformation of $X$. For example, $T(Y,Z)|_p=a(p)Y(p)bigotimes Z(p)$. The result can be obtained from the Leibniz rule.
– W. mu
Jul 19 at 3:46
I see, I forgot to put the proper subscript $T_p$ or $T_phi_t(p)$ at some places making the third coefficient $a(p)$ a constant. Thanks a lot !
– PLP
Jul 19 at 3:55
I see, I forgot to put the proper subscript $T_p$ or $T_phi_t(p)$ at some places making the third coefficient $a(p)$ a constant. Thanks a lot !
– PLP
Jul 19 at 3:55
2
2
For the record, the correct formula is $$(mathcalL_XT)(Y,Z) = X(T(Y,Z)) - T(mathcalL_XY,Z) - T(Y,mathcalL_XZ).$$
– Ivo Terek
Jul 19 at 4:11
For the record, the correct formula is $$(mathcalL_XT)(Y,Z) = X(T(Y,Z)) - T(mathcalL_XY,Z) - T(Y,mathcalL_XZ).$$
– Ivo Terek
Jul 19 at 4:11
I derived the wrong formula trying to prove this one. It seems the correct way is to use coordinates or to use the fact that Lie derivative commutes with contraction
– PLP
Jul 19 at 4:27
I derived the wrong formula trying to prove this one. It seems the correct way is to use coordinates or to use the fact that Lie derivative commutes with contraction
– PLP
Jul 19 at 4:27
add a comment |Â
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1
I think there are three variables $Y,Z$ and the coefficients in $T$, which will change as the transformation of $X$. For example, $T(Y,Z)|_p=a(p)Y(p)bigotimes Z(p)$. The result can be obtained from the Leibniz rule.
– W. mu
Jul 19 at 3:46
I see, I forgot to put the proper subscript $T_p$ or $T_phi_t(p)$ at some places making the third coefficient $a(p)$ a constant. Thanks a lot !
– PLP
Jul 19 at 3:55
2
For the record, the correct formula is $$(mathcalL_XT)(Y,Z) = X(T(Y,Z)) - T(mathcalL_XY,Z) - T(Y,mathcalL_XZ).$$
– Ivo Terek
Jul 19 at 4:11
I derived the wrong formula trying to prove this one. It seems the correct way is to use coordinates or to use the fact that Lie derivative commutes with contraction
– PLP
Jul 19 at 4:27