Hodge-$star$ operator computation on a smooth two-dimensional manifold












3












$begingroup$


Let $(x,y)$ be the local coordinates on a Riemannian manifold $M$ with $dim(M) =2$. Let $star$ denote the Hodge-$star$ operator, and let $g = g_{ij}$ denote the Riemannian metric on $M$.



I am attempting to compute the formula for the Laplace--Beltrami acting on 1-forms of $M$. I have been stuck on computing that action of the Hodge-$star$ operator on $dx$ and $dy$, and this is what I would like help with.



We know: $star(1) = sqrt{det(g)} dx wedge dy$ and $star(dx wedge dy) = dfrac{1}{sqrt{det(g)}}$. If the metric is simply $delta_{ij}$, then we know that $star dx$ and $star dy = -dx$, but I am interested in the case when the metric $g_{ij} neq delta_{ij}$.



I would appreciate any help with this, and if you need any further details, please let me know.



I have been aware of the following, but has been of little or no avail:begin{eqnarray*}
dx wedge star dx &=& sqrt{det(g_{ij})} g^{11} dx wedge dy,\
dy wedge star dx &=& sqrt{det(g_{ij})} g^{12} dy wedge dx
end{eqnarray*}



From this, it is claimed that we deduce: $$star dx = sqrt{det(g_{ij})}(g^{11} dy - g^{12}dx).$$



This is not clear.






differential-geometry riemannian-geometry laplacian hodge-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Why not? You know you have some expression $*dx = adx + bdy$ and you know what happens when you wedge with $dx$ and $dy$. Though your second to last displayed line should say $dy wedge dx$.
    $endgroup$
    – Mike Miller
    Dec 7 '18 at 8:26












  • $begingroup$
    @MikeMiller I apologise, but I do find this comment illuminating. Can you please elaborate? Thanks in advance.
    $endgroup$
    – AmorFati
    Dec 8 '18 at 20:11










  • $begingroup$
    I am not sure how: can you be precise about what should be elaborated? (I am glad to write an answer but am not sure what to add.)
    $endgroup$
    – Mike Miller
    Dec 8 '18 at 20:14












  • $begingroup$
    @MikeMiller Actually no, I understand what you have said in your first comment now. I also realise how simple this is, thank you! It just was not clicking for me. Thanks Mike! If you post an answer, I am happy to accept it and upvote it. I appreciate your help :)
    $endgroup$
    – AmorFati
    Dec 8 '18 at 20:56






  • 1




    $begingroup$
    No problem at all. In fact, I suggest you write up the answer yourself - I will gladly upvote it!
    $endgroup$
    – Mike Miller
    Dec 8 '18 at 21:11
















3












$begingroup$


Let $(x,y)$ be the local coordinates on a Riemannian manifold $M$ with $dim(M) =2$. Let $star$ denote the Hodge-$star$ operator, and let $g = g_{ij}$ denote the Riemannian metric on $M$.



I am attempting to compute the formula for the Laplace--Beltrami acting on 1-forms of $M$. I have been stuck on computing that action of the Hodge-$star$ operator on $dx$ and $dy$, and this is what I would like help with.



We know: $star(1) = sqrt{det(g)} dx wedge dy$ and $star(dx wedge dy) = dfrac{1}{sqrt{det(g)}}$. If the metric is simply $delta_{ij}$, then we know that $star dx$ and $star dy = -dx$, but I am interested in the case when the metric $g_{ij} neq delta_{ij}$.



I would appreciate any help with this, and if you need any further details, please let me know.



I have been aware of the following, but has been of little or no avail:begin{eqnarray*}
dx wedge star dx &=& sqrt{det(g_{ij})} g^{11} dx wedge dy,\
dy wedge star dx &=& sqrt{det(g_{ij})} g^{12} dy wedge dx
end{eqnarray*}



From this, it is claimed that we deduce: $$star dx = sqrt{det(g_{ij})}(g^{11} dy - g^{12}dx).$$



This is not clear.






differential-geometry riemannian-geometry laplacian hodge-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Why not? You know you have some expression $*dx = adx + bdy$ and you know what happens when you wedge with $dx$ and $dy$. Though your second to last displayed line should say $dy wedge dx$.
    $endgroup$
    – Mike Miller
    Dec 7 '18 at 8:26












  • $begingroup$
    @MikeMiller I apologise, but I do find this comment illuminating. Can you please elaborate? Thanks in advance.
    $endgroup$
    – AmorFati
    Dec 8 '18 at 20:11










  • $begingroup$
    I am not sure how: can you be precise about what should be elaborated? (I am glad to write an answer but am not sure what to add.)
    $endgroup$
    – Mike Miller
    Dec 8 '18 at 20:14












  • $begingroup$
    @MikeMiller Actually no, I understand what you have said in your first comment now. I also realise how simple this is, thank you! It just was not clicking for me. Thanks Mike! If you post an answer, I am happy to accept it and upvote it. I appreciate your help :)
    $endgroup$
    – AmorFati
    Dec 8 '18 at 20:56






  • 1




    $begingroup$
    No problem at all. In fact, I suggest you write up the answer yourself - I will gladly upvote it!
    $endgroup$
    – Mike Miller
    Dec 8 '18 at 21:11














3












3








3


5



$begingroup$


Let $(x,y)$ be the local coordinates on a Riemannian manifold $M$ with $dim(M) =2$. Let $star$ denote the Hodge-$star$ operator, and let $g = g_{ij}$ denote the Riemannian metric on $M$.



I am attempting to compute the formula for the Laplace--Beltrami acting on 1-forms of $M$. I have been stuck on computing that action of the Hodge-$star$ operator on $dx$ and $dy$, and this is what I would like help with.



We know: $star(1) = sqrt{det(g)} dx wedge dy$ and $star(dx wedge dy) = dfrac{1}{sqrt{det(g)}}$. If the metric is simply $delta_{ij}$, then we know that $star dx$ and $star dy = -dx$, but I am interested in the case when the metric $g_{ij} neq delta_{ij}$.



I would appreciate any help with this, and if you need any further details, please let me know.



I have been aware of the following, but has been of little or no avail:begin{eqnarray*}
dx wedge star dx &=& sqrt{det(g_{ij})} g^{11} dx wedge dy,\
dy wedge star dx &=& sqrt{det(g_{ij})} g^{12} dy wedge dx
end{eqnarray*}



From this, it is claimed that we deduce: $$star dx = sqrt{det(g_{ij})}(g^{11} dy - g^{12}dx).$$



This is not clear.






differential-geometry riemannian-geometry laplacian hodge-theory







share|cite|improve this question











$endgroup$




Let $(x,y)$ be the local coordinates on a Riemannian manifold $M$ with $dim(M) =2$. Let $star$ denote the Hodge-$star$ operator, and let $g = g_{ij}$ denote the Riemannian metric on $M$.



I am attempting to compute the formula for the Laplace--Beltrami acting on 1-forms of $M$. I have been stuck on computing that action of the Hodge-$star$ operator on $dx$ and $dy$, and this is what I would like help with.



We know: $star(1) = sqrt{det(g)} dx wedge dy$ and $star(dx wedge dy) = dfrac{1}{sqrt{det(g)}}$. If the metric is simply $delta_{ij}$, then we know that $star dx$ and $star dy = -dx$, but I am interested in the case when the metric $g_{ij} neq delta_{ij}$.



I would appreciate any help with this, and if you need any further details, please let me know.



I have been aware of the following, but has been of little or no avail:begin{eqnarray*}
dx wedge star dx &=& sqrt{det(g_{ij})} g^{11} dx wedge dy,\
dy wedge star dx &=& sqrt{det(g_{ij})} g^{12} dy wedge dx
end{eqnarray*}



From this, it is claimed that we deduce: $$star dx = sqrt{det(g_{ij})}(g^{11} dy - g^{12}dx).$$



This is not clear.






differential-geometry riemannian-geometry laplacian hodge-theory




differential-geometry riemannian-geometry laplacian hodge-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 22:50







AmorFati

















asked Dec 5 '18 at 22:58









AmorFatiAmorFati

396529




396529








  • 1




    $begingroup$
    Why not? You know you have some expression $*dx = adx + bdy$ and you know what happens when you wedge with $dx$ and $dy$. Though your second to last displayed line should say $dy wedge dx$.
    $endgroup$
    – Mike Miller
    Dec 7 '18 at 8:26












  • $begingroup$
    @MikeMiller I apologise, but I do find this comment illuminating. Can you please elaborate? Thanks in advance.
    $endgroup$
    – AmorFati
    Dec 8 '18 at 20:11










  • $begingroup$
    I am not sure how: can you be precise about what should be elaborated? (I am glad to write an answer but am not sure what to add.)
    $endgroup$
    – Mike Miller
    Dec 8 '18 at 20:14












  • $begingroup$
    @MikeMiller Actually no, I understand what you have said in your first comment now. I also realise how simple this is, thank you! It just was not clicking for me. Thanks Mike! If you post an answer, I am happy to accept it and upvote it. I appreciate your help :)
    $endgroup$
    – AmorFati
    Dec 8 '18 at 20:56






  • 1




    $begingroup$
    No problem at all. In fact, I suggest you write up the answer yourself - I will gladly upvote it!
    $endgroup$
    – Mike Miller
    Dec 8 '18 at 21:11














  • 1




    $begingroup$
    Why not? You know you have some expression $*dx = adx + bdy$ and you know what happens when you wedge with $dx$ and $dy$. Though your second to last displayed line should say $dy wedge dx$.
    $endgroup$
    – Mike Miller
    Dec 7 '18 at 8:26












  • $begingroup$
    @MikeMiller I apologise, but I do find this comment illuminating. Can you please elaborate? Thanks in advance.
    $endgroup$
    – AmorFati
    Dec 8 '18 at 20:11










  • $begingroup$
    I am not sure how: can you be precise about what should be elaborated? (I am glad to write an answer but am not sure what to add.)
    $endgroup$
    – Mike Miller
    Dec 8 '18 at 20:14












  • $begingroup$
    @MikeMiller Actually no, I understand what you have said in your first comment now. I also realise how simple this is, thank you! It just was not clicking for me. Thanks Mike! If you post an answer, I am happy to accept it and upvote it. I appreciate your help :)
    $endgroup$
    – AmorFati
    Dec 8 '18 at 20:56






  • 1




    $begingroup$
    No problem at all. In fact, I suggest you write up the answer yourself - I will gladly upvote it!
    $endgroup$
    – Mike Miller
    Dec 8 '18 at 21:11








1




1




$begingroup$
Why not? You know you have some expression $*dx = adx + bdy$ and you know what happens when you wedge with $dx$ and $dy$. Though your second to last displayed line should say $dy wedge dx$.
$endgroup$
– Mike Miller
Dec 7 '18 at 8:26






$begingroup$
Why not? You know you have some expression $*dx = adx + bdy$ and you know what happens when you wedge with $dx$ and $dy$. Though your second to last displayed line should say $dy wedge dx$.
$endgroup$
– Mike Miller
Dec 7 '18 at 8:26














$begingroup$
@MikeMiller I apologise, but I do find this comment illuminating. Can you please elaborate? Thanks in advance.
$endgroup$
– AmorFati
Dec 8 '18 at 20:11




$begingroup$
@MikeMiller I apologise, but I do find this comment illuminating. Can you please elaborate? Thanks in advance.
$endgroup$
– AmorFati
Dec 8 '18 at 20:11












$begingroup$
I am not sure how: can you be precise about what should be elaborated? (I am glad to write an answer but am not sure what to add.)
$endgroup$
– Mike Miller
Dec 8 '18 at 20:14






$begingroup$
I am not sure how: can you be precise about what should be elaborated? (I am glad to write an answer but am not sure what to add.)
$endgroup$
– Mike Miller
Dec 8 '18 at 20:14














$begingroup$
@MikeMiller Actually no, I understand what you have said in your first comment now. I also realise how simple this is, thank you! It just was not clicking for me. Thanks Mike! If you post an answer, I am happy to accept it and upvote it. I appreciate your help :)
$endgroup$
– AmorFati
Dec 8 '18 at 20:56




$begingroup$
@MikeMiller Actually no, I understand what you have said in your first comment now. I also realise how simple this is, thank you! It just was not clicking for me. Thanks Mike! If you post an answer, I am happy to accept it and upvote it. I appreciate your help :)
$endgroup$
– AmorFati
Dec 8 '18 at 20:56




1




1




$begingroup$
No problem at all. In fact, I suggest you write up the answer yourself - I will gladly upvote it!
$endgroup$
– Mike Miller
Dec 8 '18 at 21:11




$begingroup$
No problem at all. In fact, I suggest you write up the answer yourself - I will gladly upvote it!
$endgroup$
– Mike Miller
Dec 8 '18 at 21:11










1 Answer
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$begingroup$

Write $star dx = P dx + Q dy$ for some $P,Q$ to be determined. We have $$dx wedge star dx = P dx wedge dx + Q dx wedge dy = Q dx wedge dy,$$ and $$dy wedge star dx = P dy wedge dx + Q dy wedge dy = P dy wedge dx.$$



With $dx wedge star dx = sqrt{det(g_{ij})} g^{11} dx wedge dy$, we observe that $Q = sqrt{det(g_{ij})} g^{11}$. Similarly, with $dy wedge star dx = sqrt{det(g_{ij})} g^{12} dx wedge dy$, we have $P = - g^{12} sqrt{det(g_{ij})}$. Hence, $$star dx = sqrt{det(g_{ij})} g^{11} dy - sqrt{det(g_{ij})} g^{12} dx,$$ as claimed.






share|cite|improve this answer









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    $begingroup$

    Write $star dx = P dx + Q dy$ for some $P,Q$ to be determined. We have $$dx wedge star dx = P dx wedge dx + Q dx wedge dy = Q dx wedge dy,$$ and $$dy wedge star dx = P dy wedge dx + Q dy wedge dy = P dy wedge dx.$$



    With $dx wedge star dx = sqrt{det(g_{ij})} g^{11} dx wedge dy$, we observe that $Q = sqrt{det(g_{ij})} g^{11}$. Similarly, with $dy wedge star dx = sqrt{det(g_{ij})} g^{12} dx wedge dy$, we have $P = - g^{12} sqrt{det(g_{ij})}$. Hence, $$star dx = sqrt{det(g_{ij})} g^{11} dy - sqrt{det(g_{ij})} g^{12} dx,$$ as claimed.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Write $star dx = P dx + Q dy$ for some $P,Q$ to be determined. We have $$dx wedge star dx = P dx wedge dx + Q dx wedge dy = Q dx wedge dy,$$ and $$dy wedge star dx = P dy wedge dx + Q dy wedge dy = P dy wedge dx.$$



      With $dx wedge star dx = sqrt{det(g_{ij})} g^{11} dx wedge dy$, we observe that $Q = sqrt{det(g_{ij})} g^{11}$. Similarly, with $dy wedge star dx = sqrt{det(g_{ij})} g^{12} dx wedge dy$, we have $P = - g^{12} sqrt{det(g_{ij})}$. Hence, $$star dx = sqrt{det(g_{ij})} g^{11} dy - sqrt{det(g_{ij})} g^{12} dx,$$ as claimed.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Write $star dx = P dx + Q dy$ for some $P,Q$ to be determined. We have $$dx wedge star dx = P dx wedge dx + Q dx wedge dy = Q dx wedge dy,$$ and $$dy wedge star dx = P dy wedge dx + Q dy wedge dy = P dy wedge dx.$$



        With $dx wedge star dx = sqrt{det(g_{ij})} g^{11} dx wedge dy$, we observe that $Q = sqrt{det(g_{ij})} g^{11}$. Similarly, with $dy wedge star dx = sqrt{det(g_{ij})} g^{12} dx wedge dy$, we have $P = - g^{12} sqrt{det(g_{ij})}$. Hence, $$star dx = sqrt{det(g_{ij})} g^{11} dy - sqrt{det(g_{ij})} g^{12} dx,$$ as claimed.






        share|cite|improve this answer









        $endgroup$



        Write $star dx = P dx + Q dy$ for some $P,Q$ to be determined. We have $$dx wedge star dx = P dx wedge dx + Q dx wedge dy = Q dx wedge dy,$$ and $$dy wedge star dx = P dy wedge dx + Q dy wedge dy = P dy wedge dx.$$



        With $dx wedge star dx = sqrt{det(g_{ij})} g^{11} dx wedge dy$, we observe that $Q = sqrt{det(g_{ij})} g^{11}$. Similarly, with $dy wedge star dx = sqrt{det(g_{ij})} g^{12} dx wedge dy$, we have $P = - g^{12} sqrt{det(g_{ij})}$. Hence, $$star dx = sqrt{det(g_{ij})} g^{11} dy - sqrt{det(g_{ij})} g^{12} dx,$$ as claimed.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 22:55









        AmorFatiAmorFati

        396529




        396529






























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