Identity Involving Lie Derivative and Local Flows












2














I'm trying to show,



$$ frac{d}{dt} varphi_t^* omega = varphi_t^* left( mathcal{L}_{X_t} omega right)$$



but I have another question as well. Every case in which the lie derivative is mentioned, that notation $mathcal{L}_{X_t}$ has never been given interpretation and so I would like to understand this notation first.



Given $frac{d}{dt}bigr|_{t=0} varphi_t = X_p = X_{varphi_0(p)}$ and so $frac{d}{dt} varphi_t = X_{varphi_t(p)}$ which we can call $X_t$. And so,



$$mathcal{L}_{X_t} = mathcal{L}_{frac{d}{dt} varphi_t}$$



Is this correct?



Update: attempt to prove identity










share|cite|improve this question









New contributor




Topology Guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • When $(varphi_t)_t$ is the local flow of a time-dependent vector field, then $varphi_{t+s}neqvarphi_tcircvarphi_s$.
    – C. Falcon
    Jan 4 at 20:34










  • I'm sorry, I wasn't clear but, ${varphi_t}$ are induced by $X in mathcal{X}(M)$ and so by ODE theory, we do have this relation.
    – Topology Guy
    Jan 4 at 20:54










  • Oh, then the identity is really straightforward, see ktoi's answer. However, this is strange to write $X_t$ for a time-independant vector field...
    – C. Falcon
    Jan 4 at 20:59












  • I updated my "attempt to prove the identity" from some months ago and I came to the understanding that pull-backs commuted with limits. I never checked this off, but you two have helped tremendously and now I can move on.
    – Topology Guy
    Jan 4 at 21:01


















2














I'm trying to show,



$$ frac{d}{dt} varphi_t^* omega = varphi_t^* left( mathcal{L}_{X_t} omega right)$$



but I have another question as well. Every case in which the lie derivative is mentioned, that notation $mathcal{L}_{X_t}$ has never been given interpretation and so I would like to understand this notation first.



Given $frac{d}{dt}bigr|_{t=0} varphi_t = X_p = X_{varphi_0(p)}$ and so $frac{d}{dt} varphi_t = X_{varphi_t(p)}$ which we can call $X_t$. And so,



$$mathcal{L}_{X_t} = mathcal{L}_{frac{d}{dt} varphi_t}$$



Is this correct?



Update: attempt to prove identity










share|cite|improve this question









New contributor




Topology Guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • When $(varphi_t)_t$ is the local flow of a time-dependent vector field, then $varphi_{t+s}neqvarphi_tcircvarphi_s$.
    – C. Falcon
    Jan 4 at 20:34










  • I'm sorry, I wasn't clear but, ${varphi_t}$ are induced by $X in mathcal{X}(M)$ and so by ODE theory, we do have this relation.
    – Topology Guy
    Jan 4 at 20:54










  • Oh, then the identity is really straightforward, see ktoi's answer. However, this is strange to write $X_t$ for a time-independant vector field...
    – C. Falcon
    Jan 4 at 20:59












  • I updated my "attempt to prove the identity" from some months ago and I came to the understanding that pull-backs commuted with limits. I never checked this off, but you two have helped tremendously and now I can move on.
    – Topology Guy
    Jan 4 at 21:01
















2












2








2







I'm trying to show,



$$ frac{d}{dt} varphi_t^* omega = varphi_t^* left( mathcal{L}_{X_t} omega right)$$



but I have another question as well. Every case in which the lie derivative is mentioned, that notation $mathcal{L}_{X_t}$ has never been given interpretation and so I would like to understand this notation first.



Given $frac{d}{dt}bigr|_{t=0} varphi_t = X_p = X_{varphi_0(p)}$ and so $frac{d}{dt} varphi_t = X_{varphi_t(p)}$ which we can call $X_t$. And so,



$$mathcal{L}_{X_t} = mathcal{L}_{frac{d}{dt} varphi_t}$$



Is this correct?



Update: attempt to prove identity










share|cite|improve this question









New contributor




Topology Guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I'm trying to show,



$$ frac{d}{dt} varphi_t^* omega = varphi_t^* left( mathcal{L}_{X_t} omega right)$$



but I have another question as well. Every case in which the lie derivative is mentioned, that notation $mathcal{L}_{X_t}$ has never been given interpretation and so I would like to understand this notation first.



Given $frac{d}{dt}bigr|_{t=0} varphi_t = X_p = X_{varphi_0(p)}$ and so $frac{d}{dt} varphi_t = X_{varphi_t(p)}$ which we can call $X_t$. And so,



$$mathcal{L}_{X_t} = mathcal{L}_{frac{d}{dt} varphi_t}$$



Is this correct?



Update: attempt to prove identity







differential-geometry differential-topology symplectic-geometry






share|cite|improve this question









New contributor




Topology Guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Topology Guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 20:33







Topology Guy













New contributor




Topology Guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Jan 4 at 19:27









Topology GuyTopology Guy

133




133




New contributor




Topology Guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Topology Guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Topology Guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • When $(varphi_t)_t$ is the local flow of a time-dependent vector field, then $varphi_{t+s}neqvarphi_tcircvarphi_s$.
    – C. Falcon
    Jan 4 at 20:34










  • I'm sorry, I wasn't clear but, ${varphi_t}$ are induced by $X in mathcal{X}(M)$ and so by ODE theory, we do have this relation.
    – Topology Guy
    Jan 4 at 20:54










  • Oh, then the identity is really straightforward, see ktoi's answer. However, this is strange to write $X_t$ for a time-independant vector field...
    – C. Falcon
    Jan 4 at 20:59












  • I updated my "attempt to prove the identity" from some months ago and I came to the understanding that pull-backs commuted with limits. I never checked this off, but you two have helped tremendously and now I can move on.
    – Topology Guy
    Jan 4 at 21:01




















  • When $(varphi_t)_t$ is the local flow of a time-dependent vector field, then $varphi_{t+s}neqvarphi_tcircvarphi_s$.
    – C. Falcon
    Jan 4 at 20:34










  • I'm sorry, I wasn't clear but, ${varphi_t}$ are induced by $X in mathcal{X}(M)$ and so by ODE theory, we do have this relation.
    – Topology Guy
    Jan 4 at 20:54










  • Oh, then the identity is really straightforward, see ktoi's answer. However, this is strange to write $X_t$ for a time-independant vector field...
    – C. Falcon
    Jan 4 at 20:59












  • I updated my "attempt to prove the identity" from some months ago and I came to the understanding that pull-backs commuted with limits. I never checked this off, but you two have helped tremendously and now I can move on.
    – Topology Guy
    Jan 4 at 21:01


















When $(varphi_t)_t$ is the local flow of a time-dependent vector field, then $varphi_{t+s}neqvarphi_tcircvarphi_s$.
– C. Falcon
Jan 4 at 20:34




When $(varphi_t)_t$ is the local flow of a time-dependent vector field, then $varphi_{t+s}neqvarphi_tcircvarphi_s$.
– C. Falcon
Jan 4 at 20:34












I'm sorry, I wasn't clear but, ${varphi_t}$ are induced by $X in mathcal{X}(M)$ and so by ODE theory, we do have this relation.
– Topology Guy
Jan 4 at 20:54




I'm sorry, I wasn't clear but, ${varphi_t}$ are induced by $X in mathcal{X}(M)$ and so by ODE theory, we do have this relation.
– Topology Guy
Jan 4 at 20:54












Oh, then the identity is really straightforward, see ktoi's answer. However, this is strange to write $X_t$ for a time-independant vector field...
– C. Falcon
Jan 4 at 20:59






Oh, then the identity is really straightforward, see ktoi's answer. However, this is strange to write $X_t$ for a time-independant vector field...
– C. Falcon
Jan 4 at 20:59














I updated my "attempt to prove the identity" from some months ago and I came to the understanding that pull-backs commuted with limits. I never checked this off, but you two have helped tremendously and now I can move on.
– Topology Guy
Jan 4 at 21:01






I updated my "attempt to prove the identity" from some months ago and I came to the understanding that pull-backs commuted with limits. I never checked this off, but you two have helped tremendously and now I can move on.
– Topology Guy
Jan 4 at 21:01












2 Answers
2






active

oldest

votes


















0














Edit: The following answer assumes $varphi_t$ is the associated flow of $X,$ but for a general flow it is incorrect.



Edit 2: It seems like OP was asking about this case, in which case my answer would apply.





If $omega$ is some tensor field on $M$ and $X$ is a vector field, the Lie derivative is usually defined as,
$$ mathcal{L}_Xomega|_p = left.frac{mathrm{d}}{mathrm{d}t}right|_{t=0} varphi_t^* omega|_p = lim_{t rightarrow 0} frac1tleft(varphi_t^*(omega_{varphi(p)}) - omega_p right), $$
where $varphi_t$ is the associated flow of $X.$



From this, the identity follows essentially by definition, noting you can commute pullbacks with limits. Indeed assuming the flow exists up to some time $t_0,$ we have,



$$left.frac{mathrm{d}}{mathrm{d}t}right|_{t=t_0} varphi_t^*omega = left.frac{mathrm{d}}{mathrm{d}t}right|_{t=0} varphi_{t_0}^* varphi_t^*omega = left.varphi_{t_0}^* frac{mathrm{d}}{mathrm{d}t}right|_{t=0} varphi_t^*omega = varphi_{t_0}^* mathcal{L}_{X}omega. $$
I'll leave you to check the details on a pointwise level.





Now what I wrote isn't the same as what you asked, since I don't have $X_t$ appearing in my expression. I think this is because of how I interpret the pullback, namely I define,
$$ varphi^*omega|_p = varphi^*|_{varphi(p)} omega_{varphi(p)} in E_p,$$
where $E$ is whatever bundle $omega$ is a section of. If you define the pullback differently however, then you may find you'll need to differentiate with respect to $X_t$ when checking the above pointwise.



In general there's nothing special about $X_t = varphi_{-t}^*X$ however, all you're really doing is shifting the points over so everything lands where they should.






share|cite|improve this answer























  • You are working with time-independent vector fields, while the question is for time-dependent for vector fields. In the latter case, your definition for the Lie derivative is no longer the right one and furtherfore $(varphi_t)_t$ is no longer a one-parameter subgroup.
    – C. Falcon
    Jan 4 at 20:26










  • Ah, thanks a lot. I wrote this down some time ago, could you check the edit in my question? Is that the limit thing you are referring to?
    – Topology Guy
    Jan 4 at 20:27










  • @C.Falcon I think you're right, I assumed $varphi_t$ was the associated flow of a vector field $X$ but reading the question that doesn't seem to be the case (though it's quite unclear what exactly is being asked). I'll edit my answer noting that.
    – ktoi
    Jan 4 at 20:41










  • @ktoi I'm sorry for the confusion. I made it clear now that $varphi_t$ were induced from $X$. I apologize, as I've been working in the field for the past few months, I assumed a lot of things when asking my question. Thank you both for your help.
    – Topology Guy
    Jan 4 at 20:57






  • 1




    @TopologyGuy I see, in that case your attempt seems fine (it is more or less what I posted after all). Since you tagged this question with the sympletic topology tag however, you may also need the time-dependent version at some point (e.g. to prove the Darboux theorem), so knowing about the general case would also be useful.
    – ktoi
    Jan 4 at 21:03



















0














In order to prove the identity:
$$frac{mathrm{d}}{mathrm{d}t}{varphi_t}^*omega={varphi_t}^*(mathcal{L}_{X_t}omega),$$




  1. Prove that it holds for functions ($0$-forms), it is just the chain rule,


  2. Prove that if it holds for $omega$, then it holds for $mathrm{d}omega$, use that the exterior derivative commutes with the pullback and the Lie derivative,


  3. Prove that if it holds for $omega$ and $omega'$, then it holds for $omegawedgeomega'$, use that the pullback of a wedge product is the wedge product of their pullbacks,


  4. Conclude using that the algebra of differential forms is locally generated as an algebra by functions and differentials of functions.



Full computations can be found in Appendix B of An introduction to Contact Topology by H. Geiges.






share|cite|improve this answer























  • Thanks for that. I'm fine with proving the identity that way since you do a similar thing for cartan's homotopy formula $mathcal{L}_X = d iota_X + iota_X d$. I was thinking there was some geometric intuition or "obvious" way to remember the identity, but it seems as though it is one just like the homotopy formula i.e just show that it works for all $k$-forms. Thanks again.
    – Topology Guy
    Jan 4 at 20:20










  • As a follow up, is my interpretation for $mathcal{L}_{X_t}$ correct?
    – Topology Guy
    Jan 4 at 20:21











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});






Topology Guy is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062001%2fidentity-involving-lie-derivative-and-local-flows%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














Edit: The following answer assumes $varphi_t$ is the associated flow of $X,$ but for a general flow it is incorrect.



Edit 2: It seems like OP was asking about this case, in which case my answer would apply.





If $omega$ is some tensor field on $M$ and $X$ is a vector field, the Lie derivative is usually defined as,
$$ mathcal{L}_Xomega|_p = left.frac{mathrm{d}}{mathrm{d}t}right|_{t=0} varphi_t^* omega|_p = lim_{t rightarrow 0} frac1tleft(varphi_t^*(omega_{varphi(p)}) - omega_p right), $$
where $varphi_t$ is the associated flow of $X.$



From this, the identity follows essentially by definition, noting you can commute pullbacks with limits. Indeed assuming the flow exists up to some time $t_0,$ we have,



$$left.frac{mathrm{d}}{mathrm{d}t}right|_{t=t_0} varphi_t^*omega = left.frac{mathrm{d}}{mathrm{d}t}right|_{t=0} varphi_{t_0}^* varphi_t^*omega = left.varphi_{t_0}^* frac{mathrm{d}}{mathrm{d}t}right|_{t=0} varphi_t^*omega = varphi_{t_0}^* mathcal{L}_{X}omega. $$
I'll leave you to check the details on a pointwise level.





Now what I wrote isn't the same as what you asked, since I don't have $X_t$ appearing in my expression. I think this is because of how I interpret the pullback, namely I define,
$$ varphi^*omega|_p = varphi^*|_{varphi(p)} omega_{varphi(p)} in E_p,$$
where $E$ is whatever bundle $omega$ is a section of. If you define the pullback differently however, then you may find you'll need to differentiate with respect to $X_t$ when checking the above pointwise.



In general there's nothing special about $X_t = varphi_{-t}^*X$ however, all you're really doing is shifting the points over so everything lands where they should.






share|cite|improve this answer























  • You are working with time-independent vector fields, while the question is for time-dependent for vector fields. In the latter case, your definition for the Lie derivative is no longer the right one and furtherfore $(varphi_t)_t$ is no longer a one-parameter subgroup.
    – C. Falcon
    Jan 4 at 20:26










  • Ah, thanks a lot. I wrote this down some time ago, could you check the edit in my question? Is that the limit thing you are referring to?
    – Topology Guy
    Jan 4 at 20:27










  • @C.Falcon I think you're right, I assumed $varphi_t$ was the associated flow of a vector field $X$ but reading the question that doesn't seem to be the case (though it's quite unclear what exactly is being asked). I'll edit my answer noting that.
    – ktoi
    Jan 4 at 20:41










  • @ktoi I'm sorry for the confusion. I made it clear now that $varphi_t$ were induced from $X$. I apologize, as I've been working in the field for the past few months, I assumed a lot of things when asking my question. Thank you both for your help.
    – Topology Guy
    Jan 4 at 20:57






  • 1




    @TopologyGuy I see, in that case your attempt seems fine (it is more or less what I posted after all). Since you tagged this question with the sympletic topology tag however, you may also need the time-dependent version at some point (e.g. to prove the Darboux theorem), so knowing about the general case would also be useful.
    – ktoi
    Jan 4 at 21:03
















0














Edit: The following answer assumes $varphi_t$ is the associated flow of $X,$ but for a general flow it is incorrect.



Edit 2: It seems like OP was asking about this case, in which case my answer would apply.





If $omega$ is some tensor field on $M$ and $X$ is a vector field, the Lie derivative is usually defined as,
$$ mathcal{L}_Xomega|_p = left.frac{mathrm{d}}{mathrm{d}t}right|_{t=0} varphi_t^* omega|_p = lim_{t rightarrow 0} frac1tleft(varphi_t^*(omega_{varphi(p)}) - omega_p right), $$
where $varphi_t$ is the associated flow of $X.$



From this, the identity follows essentially by definition, noting you can commute pullbacks with limits. Indeed assuming the flow exists up to some time $t_0,$ we have,



$$left.frac{mathrm{d}}{mathrm{d}t}right|_{t=t_0} varphi_t^*omega = left.frac{mathrm{d}}{mathrm{d}t}right|_{t=0} varphi_{t_0}^* varphi_t^*omega = left.varphi_{t_0}^* frac{mathrm{d}}{mathrm{d}t}right|_{t=0} varphi_t^*omega = varphi_{t_0}^* mathcal{L}_{X}omega. $$
I'll leave you to check the details on a pointwise level.





Now what I wrote isn't the same as what you asked, since I don't have $X_t$ appearing in my expression. I think this is because of how I interpret the pullback, namely I define,
$$ varphi^*omega|_p = varphi^*|_{varphi(p)} omega_{varphi(p)} in E_p,$$
where $E$ is whatever bundle $omega$ is a section of. If you define the pullback differently however, then you may find you'll need to differentiate with respect to $X_t$ when checking the above pointwise.



In general there's nothing special about $X_t = varphi_{-t}^*X$ however, all you're really doing is shifting the points over so everything lands where they should.






share|cite|improve this answer























  • You are working with time-independent vector fields, while the question is for time-dependent for vector fields. In the latter case, your definition for the Lie derivative is no longer the right one and furtherfore $(varphi_t)_t$ is no longer a one-parameter subgroup.
    – C. Falcon
    Jan 4 at 20:26










  • Ah, thanks a lot. I wrote this down some time ago, could you check the edit in my question? Is that the limit thing you are referring to?
    – Topology Guy
    Jan 4 at 20:27










  • @C.Falcon I think you're right, I assumed $varphi_t$ was the associated flow of a vector field $X$ but reading the question that doesn't seem to be the case (though it's quite unclear what exactly is being asked). I'll edit my answer noting that.
    – ktoi
    Jan 4 at 20:41










  • @ktoi I'm sorry for the confusion. I made it clear now that $varphi_t$ were induced from $X$. I apologize, as I've been working in the field for the past few months, I assumed a lot of things when asking my question. Thank you both for your help.
    – Topology Guy
    Jan 4 at 20:57






  • 1




    @TopologyGuy I see, in that case your attempt seems fine (it is more or less what I posted after all). Since you tagged this question with the sympletic topology tag however, you may also need the time-dependent version at some point (e.g. to prove the Darboux theorem), so knowing about the general case would also be useful.
    – ktoi
    Jan 4 at 21:03














0












0








0






Edit: The following answer assumes $varphi_t$ is the associated flow of $X,$ but for a general flow it is incorrect.



Edit 2: It seems like OP was asking about this case, in which case my answer would apply.





If $omega$ is some tensor field on $M$ and $X$ is a vector field, the Lie derivative is usually defined as,
$$ mathcal{L}_Xomega|_p = left.frac{mathrm{d}}{mathrm{d}t}right|_{t=0} varphi_t^* omega|_p = lim_{t rightarrow 0} frac1tleft(varphi_t^*(omega_{varphi(p)}) - omega_p right), $$
where $varphi_t$ is the associated flow of $X.$



From this, the identity follows essentially by definition, noting you can commute pullbacks with limits. Indeed assuming the flow exists up to some time $t_0,$ we have,



$$left.frac{mathrm{d}}{mathrm{d}t}right|_{t=t_0} varphi_t^*omega = left.frac{mathrm{d}}{mathrm{d}t}right|_{t=0} varphi_{t_0}^* varphi_t^*omega = left.varphi_{t_0}^* frac{mathrm{d}}{mathrm{d}t}right|_{t=0} varphi_t^*omega = varphi_{t_0}^* mathcal{L}_{X}omega. $$
I'll leave you to check the details on a pointwise level.





Now what I wrote isn't the same as what you asked, since I don't have $X_t$ appearing in my expression. I think this is because of how I interpret the pullback, namely I define,
$$ varphi^*omega|_p = varphi^*|_{varphi(p)} omega_{varphi(p)} in E_p,$$
where $E$ is whatever bundle $omega$ is a section of. If you define the pullback differently however, then you may find you'll need to differentiate with respect to $X_t$ when checking the above pointwise.



In general there's nothing special about $X_t = varphi_{-t}^*X$ however, all you're really doing is shifting the points over so everything lands where they should.






share|cite|improve this answer














Edit: The following answer assumes $varphi_t$ is the associated flow of $X,$ but for a general flow it is incorrect.



Edit 2: It seems like OP was asking about this case, in which case my answer would apply.





If $omega$ is some tensor field on $M$ and $X$ is a vector field, the Lie derivative is usually defined as,
$$ mathcal{L}_Xomega|_p = left.frac{mathrm{d}}{mathrm{d}t}right|_{t=0} varphi_t^* omega|_p = lim_{t rightarrow 0} frac1tleft(varphi_t^*(omega_{varphi(p)}) - omega_p right), $$
where $varphi_t$ is the associated flow of $X.$



From this, the identity follows essentially by definition, noting you can commute pullbacks with limits. Indeed assuming the flow exists up to some time $t_0,$ we have,



$$left.frac{mathrm{d}}{mathrm{d}t}right|_{t=t_0} varphi_t^*omega = left.frac{mathrm{d}}{mathrm{d}t}right|_{t=0} varphi_{t_0}^* varphi_t^*omega = left.varphi_{t_0}^* frac{mathrm{d}}{mathrm{d}t}right|_{t=0} varphi_t^*omega = varphi_{t_0}^* mathcal{L}_{X}omega. $$
I'll leave you to check the details on a pointwise level.





Now what I wrote isn't the same as what you asked, since I don't have $X_t$ appearing in my expression. I think this is because of how I interpret the pullback, namely I define,
$$ varphi^*omega|_p = varphi^*|_{varphi(p)} omega_{varphi(p)} in E_p,$$
where $E$ is whatever bundle $omega$ is a section of. If you define the pullback differently however, then you may find you'll need to differentiate with respect to $X_t$ when checking the above pointwise.



In general there's nothing special about $X_t = varphi_{-t}^*X$ however, all you're really doing is shifting the points over so everything lands where they should.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 20:58

























answered Jan 4 at 20:21









ktoiktoi

2,3481616




2,3481616












  • You are working with time-independent vector fields, while the question is for time-dependent for vector fields. In the latter case, your definition for the Lie derivative is no longer the right one and furtherfore $(varphi_t)_t$ is no longer a one-parameter subgroup.
    – C. Falcon
    Jan 4 at 20:26










  • Ah, thanks a lot. I wrote this down some time ago, could you check the edit in my question? Is that the limit thing you are referring to?
    – Topology Guy
    Jan 4 at 20:27










  • @C.Falcon I think you're right, I assumed $varphi_t$ was the associated flow of a vector field $X$ but reading the question that doesn't seem to be the case (though it's quite unclear what exactly is being asked). I'll edit my answer noting that.
    – ktoi
    Jan 4 at 20:41










  • @ktoi I'm sorry for the confusion. I made it clear now that $varphi_t$ were induced from $X$. I apologize, as I've been working in the field for the past few months, I assumed a lot of things when asking my question. Thank you both for your help.
    – Topology Guy
    Jan 4 at 20:57






  • 1




    @TopologyGuy I see, in that case your attempt seems fine (it is more or less what I posted after all). Since you tagged this question with the sympletic topology tag however, you may also need the time-dependent version at some point (e.g. to prove the Darboux theorem), so knowing about the general case would also be useful.
    – ktoi
    Jan 4 at 21:03


















  • You are working with time-independent vector fields, while the question is for time-dependent for vector fields. In the latter case, your definition for the Lie derivative is no longer the right one and furtherfore $(varphi_t)_t$ is no longer a one-parameter subgroup.
    – C. Falcon
    Jan 4 at 20:26










  • Ah, thanks a lot. I wrote this down some time ago, could you check the edit in my question? Is that the limit thing you are referring to?
    – Topology Guy
    Jan 4 at 20:27










  • @C.Falcon I think you're right, I assumed $varphi_t$ was the associated flow of a vector field $X$ but reading the question that doesn't seem to be the case (though it's quite unclear what exactly is being asked). I'll edit my answer noting that.
    – ktoi
    Jan 4 at 20:41










  • @ktoi I'm sorry for the confusion. I made it clear now that $varphi_t$ were induced from $X$. I apologize, as I've been working in the field for the past few months, I assumed a lot of things when asking my question. Thank you both for your help.
    – Topology Guy
    Jan 4 at 20:57






  • 1




    @TopologyGuy I see, in that case your attempt seems fine (it is more or less what I posted after all). Since you tagged this question with the sympletic topology tag however, you may also need the time-dependent version at some point (e.g. to prove the Darboux theorem), so knowing about the general case would also be useful.
    – ktoi
    Jan 4 at 21:03
















You are working with time-independent vector fields, while the question is for time-dependent for vector fields. In the latter case, your definition for the Lie derivative is no longer the right one and furtherfore $(varphi_t)_t$ is no longer a one-parameter subgroup.
– C. Falcon
Jan 4 at 20:26




You are working with time-independent vector fields, while the question is for time-dependent for vector fields. In the latter case, your definition for the Lie derivative is no longer the right one and furtherfore $(varphi_t)_t$ is no longer a one-parameter subgroup.
– C. Falcon
Jan 4 at 20:26












Ah, thanks a lot. I wrote this down some time ago, could you check the edit in my question? Is that the limit thing you are referring to?
– Topology Guy
Jan 4 at 20:27




Ah, thanks a lot. I wrote this down some time ago, could you check the edit in my question? Is that the limit thing you are referring to?
– Topology Guy
Jan 4 at 20:27












@C.Falcon I think you're right, I assumed $varphi_t$ was the associated flow of a vector field $X$ but reading the question that doesn't seem to be the case (though it's quite unclear what exactly is being asked). I'll edit my answer noting that.
– ktoi
Jan 4 at 20:41




@C.Falcon I think you're right, I assumed $varphi_t$ was the associated flow of a vector field $X$ but reading the question that doesn't seem to be the case (though it's quite unclear what exactly is being asked). I'll edit my answer noting that.
– ktoi
Jan 4 at 20:41












@ktoi I'm sorry for the confusion. I made it clear now that $varphi_t$ were induced from $X$. I apologize, as I've been working in the field for the past few months, I assumed a lot of things when asking my question. Thank you both for your help.
– Topology Guy
Jan 4 at 20:57




@ktoi I'm sorry for the confusion. I made it clear now that $varphi_t$ were induced from $X$. I apologize, as I've been working in the field for the past few months, I assumed a lot of things when asking my question. Thank you both for your help.
– Topology Guy
Jan 4 at 20:57




1




1




@TopologyGuy I see, in that case your attempt seems fine (it is more or less what I posted after all). Since you tagged this question with the sympletic topology tag however, you may also need the time-dependent version at some point (e.g. to prove the Darboux theorem), so knowing about the general case would also be useful.
– ktoi
Jan 4 at 21:03




@TopologyGuy I see, in that case your attempt seems fine (it is more or less what I posted after all). Since you tagged this question with the sympletic topology tag however, you may also need the time-dependent version at some point (e.g. to prove the Darboux theorem), so knowing about the general case would also be useful.
– ktoi
Jan 4 at 21:03











0














In order to prove the identity:
$$frac{mathrm{d}}{mathrm{d}t}{varphi_t}^*omega={varphi_t}^*(mathcal{L}_{X_t}omega),$$




  1. Prove that it holds for functions ($0$-forms), it is just the chain rule,


  2. Prove that if it holds for $omega$, then it holds for $mathrm{d}omega$, use that the exterior derivative commutes with the pullback and the Lie derivative,


  3. Prove that if it holds for $omega$ and $omega'$, then it holds for $omegawedgeomega'$, use that the pullback of a wedge product is the wedge product of their pullbacks,


  4. Conclude using that the algebra of differential forms is locally generated as an algebra by functions and differentials of functions.



Full computations can be found in Appendix B of An introduction to Contact Topology by H. Geiges.






share|cite|improve this answer























  • Thanks for that. I'm fine with proving the identity that way since you do a similar thing for cartan's homotopy formula $mathcal{L}_X = d iota_X + iota_X d$. I was thinking there was some geometric intuition or "obvious" way to remember the identity, but it seems as though it is one just like the homotopy formula i.e just show that it works for all $k$-forms. Thanks again.
    – Topology Guy
    Jan 4 at 20:20










  • As a follow up, is my interpretation for $mathcal{L}_{X_t}$ correct?
    – Topology Guy
    Jan 4 at 20:21
















0














In order to prove the identity:
$$frac{mathrm{d}}{mathrm{d}t}{varphi_t}^*omega={varphi_t}^*(mathcal{L}_{X_t}omega),$$




  1. Prove that it holds for functions ($0$-forms), it is just the chain rule,


  2. Prove that if it holds for $omega$, then it holds for $mathrm{d}omega$, use that the exterior derivative commutes with the pullback and the Lie derivative,


  3. Prove that if it holds for $omega$ and $omega'$, then it holds for $omegawedgeomega'$, use that the pullback of a wedge product is the wedge product of their pullbacks,


  4. Conclude using that the algebra of differential forms is locally generated as an algebra by functions and differentials of functions.



Full computations can be found in Appendix B of An introduction to Contact Topology by H. Geiges.






share|cite|improve this answer























  • Thanks for that. I'm fine with proving the identity that way since you do a similar thing for cartan's homotopy formula $mathcal{L}_X = d iota_X + iota_X d$. I was thinking there was some geometric intuition or "obvious" way to remember the identity, but it seems as though it is one just like the homotopy formula i.e just show that it works for all $k$-forms. Thanks again.
    – Topology Guy
    Jan 4 at 20:20










  • As a follow up, is my interpretation for $mathcal{L}_{X_t}$ correct?
    – Topology Guy
    Jan 4 at 20:21














0












0








0






In order to prove the identity:
$$frac{mathrm{d}}{mathrm{d}t}{varphi_t}^*omega={varphi_t}^*(mathcal{L}_{X_t}omega),$$




  1. Prove that it holds for functions ($0$-forms), it is just the chain rule,


  2. Prove that if it holds for $omega$, then it holds for $mathrm{d}omega$, use that the exterior derivative commutes with the pullback and the Lie derivative,


  3. Prove that if it holds for $omega$ and $omega'$, then it holds for $omegawedgeomega'$, use that the pullback of a wedge product is the wedge product of their pullbacks,


  4. Conclude using that the algebra of differential forms is locally generated as an algebra by functions and differentials of functions.



Full computations can be found in Appendix B of An introduction to Contact Topology by H. Geiges.






share|cite|improve this answer














In order to prove the identity:
$$frac{mathrm{d}}{mathrm{d}t}{varphi_t}^*omega={varphi_t}^*(mathcal{L}_{X_t}omega),$$




  1. Prove that it holds for functions ($0$-forms), it is just the chain rule,


  2. Prove that if it holds for $omega$, then it holds for $mathrm{d}omega$, use that the exterior derivative commutes with the pullback and the Lie derivative,


  3. Prove that if it holds for $omega$ and $omega'$, then it holds for $omegawedgeomega'$, use that the pullback of a wedge product is the wedge product of their pullbacks,


  4. Conclude using that the algebra of differential forms is locally generated as an algebra by functions and differentials of functions.



Full computations can be found in Appendix B of An introduction to Contact Topology by H. Geiges.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 20:17

























answered Jan 4 at 19:51









C. FalconC. Falcon

15.1k41950




15.1k41950












  • Thanks for that. I'm fine with proving the identity that way since you do a similar thing for cartan's homotopy formula $mathcal{L}_X = d iota_X + iota_X d$. I was thinking there was some geometric intuition or "obvious" way to remember the identity, but it seems as though it is one just like the homotopy formula i.e just show that it works for all $k$-forms. Thanks again.
    – Topology Guy
    Jan 4 at 20:20










  • As a follow up, is my interpretation for $mathcal{L}_{X_t}$ correct?
    – Topology Guy
    Jan 4 at 20:21


















  • Thanks for that. I'm fine with proving the identity that way since you do a similar thing for cartan's homotopy formula $mathcal{L}_X = d iota_X + iota_X d$. I was thinking there was some geometric intuition or "obvious" way to remember the identity, but it seems as though it is one just like the homotopy formula i.e just show that it works for all $k$-forms. Thanks again.
    – Topology Guy
    Jan 4 at 20:20










  • As a follow up, is my interpretation for $mathcal{L}_{X_t}$ correct?
    – Topology Guy
    Jan 4 at 20:21
















Thanks for that. I'm fine with proving the identity that way since you do a similar thing for cartan's homotopy formula $mathcal{L}_X = d iota_X + iota_X d$. I was thinking there was some geometric intuition or "obvious" way to remember the identity, but it seems as though it is one just like the homotopy formula i.e just show that it works for all $k$-forms. Thanks again.
– Topology Guy
Jan 4 at 20:20




Thanks for that. I'm fine with proving the identity that way since you do a similar thing for cartan's homotopy formula $mathcal{L}_X = d iota_X + iota_X d$. I was thinking there was some geometric intuition or "obvious" way to remember the identity, but it seems as though it is one just like the homotopy formula i.e just show that it works for all $k$-forms. Thanks again.
– Topology Guy
Jan 4 at 20:20












As a follow up, is my interpretation for $mathcal{L}_{X_t}$ correct?
– Topology Guy
Jan 4 at 20:21




As a follow up, is my interpretation for $mathcal{L}_{X_t}$ correct?
– Topology Guy
Jan 4 at 20:21










Topology Guy is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















Topology Guy is a new contributor. Be nice, and check out our Code of Conduct.













Topology Guy is a new contributor. Be nice, and check out our Code of Conduct.












Topology Guy is a new contributor. Be nice, and check out our Code of Conduct.
















Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062001%2fidentity-involving-lie-derivative-and-local-flows%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

1300-talet

1300-talet

Display a custom attribute below product name in the front-end Magento 1.9.3.8