Covariant Exterior Derivative action on the $mathrm{End}(E)$-valued p-forms












1












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Suppose I define an operator $d_A$ by its action on sections $sin Gamma(E)=Omega_M^0(E)$ of some vector bundle $Pi:Erightarrow M$ in a trivializing neighbourhood $Usubset M$ as
$$ d_As|_U=(ds+Awedge s)|_U$$
for some connection $A=(A^i_j)$ with $A^i_jin Omega^1(U)$. By writing $omegainOmega^r_M(E)$ as $omega=e_iwedge omega^i$ where $omega^iin Omega^r(M)$ and $e_iin Gamma(E)$ are local basis sections, and imposing,
$$d_A(etawedge omega^i)=d_Aetawedge omega^i+(-1)^{mathrm{deg}(eta)}etawedge domega^iqquad forall etain Omega^r_M(E)quad (1)$$
I've shown,
$$d_Aomega=d_A(e_iwedge omega^i)=d_Ae_iwedge omega^i+(-1)^0e_iwedge domega^i\
=de_iwedgeomega^i+Awedge e_iwedge omega^i+e_iwedge domega^i\
=d(omega)-e_iwedge domega^i+Awedgeomega+e_iwedge domega^i\
=domega+Awedge omega
$$
Now suppose I have $sigmain Gamma(mathrm{End}(E))=Omega^0_M(mathrm{End}(E))$, I have been told that one can extend $d_A$ as a map,
$$d_A:Omega^r_M(mathrm{End}(E))rightarrow Omega_M^{r+1}(mathrm{End}(E)) $$
by first imposing,
$$d_A(sigma)s=d_A(sigma s)-sigma d_Asquad (2)$$
to show,
$$d_A(muwedge alpha)=d_Amuwedge alpha+(-1)^{mathrm{deg}(mu)}(muwedge d_Aalpha)qquadforall muin Omega_M^p(mathrm{End}(E)),alphainOmega_M^q(E)$$
then it follows that,
$$d_A(muwedge beta)=d_Amuwedge beta+(-1)^{mathrm{deg}(mu)}(muwedge d_Abeta)qquadforall muin Omega_M^p(mathrm{End}(E)),betainOmega_M^q(mathrm{End}(E))$$
It also apparently follows that if $muin Omega_M^2(mathrm{End}(E))$ then,
$$d_Amu=dmu+Awedge mu-muwedge A$$
Here is my confused attempt, maybe consider them more as ideas:



From (2), if I ASSUME that I can do the following,



$d_A(sigmawedge s)=d(sigmawedge s)+Awedgesigmawedge s
=(dsigma)wedge s+sigmawedge ds+Awedgesigmawedge s
$



Combining with (2) we have,



$(d_Asigma)wedge s=(dsigma)wedge s+sigmawedge ds+Awedgesigmawedge s-sigmawedge ds-sigmawedge Awedge s\
=dsigmawedge s+Awedgesigmawedge s-sigmawedge Awedge s\
=(dsigma+Awedgesigma-sigmawedge A)wedge s\
implies d_Asigma=dsigma+Awedgesigma-sigmawedge A
$



Then, if we write $omegain Omega_M^p(mathrm{End}(E))$ as $omega=f_iwedge omega^i$ where $f^iin Gamma(mathrm{End}(E))=Omega_M^0(mathrm{End}(E))$, and if I also ASSUME that (1) holds with for $etain Omega_M^r(mathrm{End}(E))
$ then,



$d_Aomega=d_A(f_iwedge omega^i)=d_Af_iwedge omega^i+(-1)^0f_iwedge domega^i\
=df_iwedge omega^i+Awedge f^iwedgeomega^i-f^iwedge Awedge omega^i+f_iwedge domega^i \
=d(omega)-f^iwedge domega^i+Awedgeomega-(-1)^{(r)}f^iwedgeomega^iwedge A+f_iwedge domega^i\
=domega+Awedgeomega-(-1)^romegawedge A
$



which gives the $Omega^2_M(mathrm{End}(E))$ as a corollary.



I've had some other thoughts, none of them well enough formulated to write down, any help with the above would be very much appreciated. There is a reference:
https://www.dpmms.cam.ac.uk/~agk22/vb.pdf page 35.






differential-geometry differential-forms vector-bundles exterior-algebra







share|cite|improve this question











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    $begingroup$


    Suppose I define an operator $d_A$ by its action on sections $sin Gamma(E)=Omega_M^0(E)$ of some vector bundle $Pi:Erightarrow M$ in a trivializing neighbourhood $Usubset M$ as
    $$ d_As|_U=(ds+Awedge s)|_U$$
    for some connection $A=(A^i_j)$ with $A^i_jin Omega^1(U)$. By writing $omegainOmega^r_M(E)$ as $omega=e_iwedge omega^i$ where $omega^iin Omega^r(M)$ and $e_iin Gamma(E)$ are local basis sections, and imposing,
    $$d_A(etawedge omega^i)=d_Aetawedge omega^i+(-1)^{mathrm{deg}(eta)}etawedge domega^iqquad forall etain Omega^r_M(E)quad (1)$$
    I've shown,
    $$d_Aomega=d_A(e_iwedge omega^i)=d_Ae_iwedge omega^i+(-1)^0e_iwedge domega^i\
    =de_iwedgeomega^i+Awedge e_iwedge omega^i+e_iwedge domega^i\
    =d(omega)-e_iwedge domega^i+Awedgeomega+e_iwedge domega^i\
    =domega+Awedge omega
    $$
    Now suppose I have $sigmain Gamma(mathrm{End}(E))=Omega^0_M(mathrm{End}(E))$, I have been told that one can extend $d_A$ as a map,
    $$d_A:Omega^r_M(mathrm{End}(E))rightarrow Omega_M^{r+1}(mathrm{End}(E)) $$
    by first imposing,
    $$d_A(sigma)s=d_A(sigma s)-sigma d_Asquad (2)$$
    to show,
    $$d_A(muwedge alpha)=d_Amuwedge alpha+(-1)^{mathrm{deg}(mu)}(muwedge d_Aalpha)qquadforall muin Omega_M^p(mathrm{End}(E)),alphainOmega_M^q(E)$$
    then it follows that,
    $$d_A(muwedge beta)=d_Amuwedge beta+(-1)^{mathrm{deg}(mu)}(muwedge d_Abeta)qquadforall muin Omega_M^p(mathrm{End}(E)),betainOmega_M^q(mathrm{End}(E))$$
    It also apparently follows that if $muin Omega_M^2(mathrm{End}(E))$ then,
    $$d_Amu=dmu+Awedge mu-muwedge A$$
    Here is my confused attempt, maybe consider them more as ideas:



    From (2), if I ASSUME that I can do the following,



    $d_A(sigmawedge s)=d(sigmawedge s)+Awedgesigmawedge s
    =(dsigma)wedge s+sigmawedge ds+Awedgesigmawedge s
    $



    Combining with (2) we have,



    $(d_Asigma)wedge s=(dsigma)wedge s+sigmawedge ds+Awedgesigmawedge s-sigmawedge ds-sigmawedge Awedge s\
    =dsigmawedge s+Awedgesigmawedge s-sigmawedge Awedge s\
    =(dsigma+Awedgesigma-sigmawedge A)wedge s\
    implies d_Asigma=dsigma+Awedgesigma-sigmawedge A
    $



    Then, if we write $omegain Omega_M^p(mathrm{End}(E))$ as $omega=f_iwedge omega^i$ where $f^iin Gamma(mathrm{End}(E))=Omega_M^0(mathrm{End}(E))$, and if I also ASSUME that (1) holds with for $etain Omega_M^r(mathrm{End}(E))
    $     then,



    $d_Aomega=d_A(f_iwedge omega^i)=d_Af_iwedge omega^i+(-1)^0f_iwedge domega^i\
    =df_iwedge omega^i+Awedge f^iwedgeomega^i-f^iwedge Awedge omega^i+f_iwedge domega^i \
    =d(omega)-f^iwedge domega^i+Awedgeomega-(-1)^{(r)}f^iwedgeomega^iwedge A+f_iwedge domega^i\
    =domega+Awedgeomega-(-1)^romegawedge A
    $



    which gives the $Omega^2_M(mathrm{End}(E))$ as a corollary.



    I've had some other thoughts, none of them well enough formulated to write down, any help with the above would be very much appreciated. There is a reference:
    https://www.dpmms.cam.ac.uk/~agk22/vb.pdf page 35.






    differential-geometry differential-forms vector-bundles exterior-algebra







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Suppose I define an operator $d_A$ by its action on sections $sin Gamma(E)=Omega_M^0(E)$ of some vector bundle $Pi:Erightarrow M$ in a trivializing neighbourhood $Usubset M$ as
      $$ d_As|_U=(ds+Awedge s)|_U$$
      for some connection $A=(A^i_j)$ with $A^i_jin Omega^1(U)$. By writing $omegainOmega^r_M(E)$ as $omega=e_iwedge omega^i$ where $omega^iin Omega^r(M)$ and $e_iin Gamma(E)$ are local basis sections, and imposing,
      $$d_A(etawedge omega^i)=d_Aetawedge omega^i+(-1)^{mathrm{deg}(eta)}etawedge domega^iqquad forall etain Omega^r_M(E)quad (1)$$
      I've shown,
      $$d_Aomega=d_A(e_iwedge omega^i)=d_Ae_iwedge omega^i+(-1)^0e_iwedge domega^i\
      =de_iwedgeomega^i+Awedge e_iwedge omega^i+e_iwedge domega^i\
      =d(omega)-e_iwedge domega^i+Awedgeomega+e_iwedge domega^i\
      =domega+Awedge omega
      $$
      Now suppose I have $sigmain Gamma(mathrm{End}(E))=Omega^0_M(mathrm{End}(E))$, I have been told that one can extend $d_A$ as a map,
      $$d_A:Omega^r_M(mathrm{End}(E))rightarrow Omega_M^{r+1}(mathrm{End}(E)) $$
      by first imposing,
      $$d_A(sigma)s=d_A(sigma s)-sigma d_Asquad (2)$$
      to show,
      $$d_A(muwedge alpha)=d_Amuwedge alpha+(-1)^{mathrm{deg}(mu)}(muwedge d_Aalpha)qquadforall muin Omega_M^p(mathrm{End}(E)),alphainOmega_M^q(E)$$
      then it follows that,
      $$d_A(muwedge beta)=d_Amuwedge beta+(-1)^{mathrm{deg}(mu)}(muwedge d_Abeta)qquadforall muin Omega_M^p(mathrm{End}(E)),betainOmega_M^q(mathrm{End}(E))$$
      It also apparently follows that if $muin Omega_M^2(mathrm{End}(E))$ then,
      $$d_Amu=dmu+Awedge mu-muwedge A$$
      Here is my confused attempt, maybe consider them more as ideas:



      From (2), if I ASSUME that I can do the following,



      $d_A(sigmawedge s)=d(sigmawedge s)+Awedgesigmawedge s
      =(dsigma)wedge s+sigmawedge ds+Awedgesigmawedge s
      $



      Combining with (2) we have,



      $(d_Asigma)wedge s=(dsigma)wedge s+sigmawedge ds+Awedgesigmawedge s-sigmawedge ds-sigmawedge Awedge s\
      =dsigmawedge s+Awedgesigmawedge s-sigmawedge Awedge s\
      =(dsigma+Awedgesigma-sigmawedge A)wedge s\
      implies d_Asigma=dsigma+Awedgesigma-sigmawedge A
      $



      Then, if we write $omegain Omega_M^p(mathrm{End}(E))$ as $omega=f_iwedge omega^i$ where $f^iin Gamma(mathrm{End}(E))=Omega_M^0(mathrm{End}(E))$, and if I also ASSUME that (1) holds with for $etain Omega_M^r(mathrm{End}(E))
      $       then,



      $d_Aomega=d_A(f_iwedge omega^i)=d_Af_iwedge omega^i+(-1)^0f_iwedge domega^i\
      =df_iwedge omega^i+Awedge f^iwedgeomega^i-f^iwedge Awedge omega^i+f_iwedge domega^i \
      =d(omega)-f^iwedge domega^i+Awedgeomega-(-1)^{(r)}f^iwedgeomega^iwedge A+f_iwedge domega^i\
      =domega+Awedgeomega-(-1)^romegawedge A
      $



      which gives the $Omega^2_M(mathrm{End}(E))$ as a corollary.



      I've had some other thoughts, none of them well enough formulated to write down, any help with the above would be very much appreciated. There is a reference:
      https://www.dpmms.cam.ac.uk/~agk22/vb.pdf page 35.






      differential-geometry differential-forms vector-bundles exterior-algebra







      share|cite|improve this question











      $endgroup$




      Suppose I define an operator $d_A$ by its action on sections $sin Gamma(E)=Omega_M^0(E)$ of some vector bundle $Pi:Erightarrow M$ in a trivializing neighbourhood $Usubset M$ as
      $$ d_As|_U=(ds+Awedge s)|_U$$
      for some connection $A=(A^i_j)$ with $A^i_jin Omega^1(U)$. By writing $omegainOmega^r_M(E)$ as $omega=e_iwedge omega^i$ where $omega^iin Omega^r(M)$ and $e_iin Gamma(E)$ are local basis sections, and imposing,
      $$d_A(etawedge omega^i)=d_Aetawedge omega^i+(-1)^{mathrm{deg}(eta)}etawedge domega^iqquad forall etain Omega^r_M(E)quad (1)$$
      I've shown,
      $$d_Aomega=d_A(e_iwedge omega^i)=d_Ae_iwedge omega^i+(-1)^0e_iwedge domega^i\
      =de_iwedgeomega^i+Awedge e_iwedge omega^i+e_iwedge domega^i\
      =d(omega)-e_iwedge domega^i+Awedgeomega+e_iwedge domega^i\
      =domega+Awedge omega
      $$
      Now suppose I have $sigmain Gamma(mathrm{End}(E))=Omega^0_M(mathrm{End}(E))$, I have been told that one can extend $d_A$ as a map,
      $$d_A:Omega^r_M(mathrm{End}(E))rightarrow Omega_M^{r+1}(mathrm{End}(E)) $$
      by first imposing,
      $$d_A(sigma)s=d_A(sigma s)-sigma d_Asquad (2)$$
      to show,
      $$d_A(muwedge alpha)=d_Amuwedge alpha+(-1)^{mathrm{deg}(mu)}(muwedge d_Aalpha)qquadforall muin Omega_M^p(mathrm{End}(E)),alphainOmega_M^q(E)$$
      then it follows that,
      $$d_A(muwedge beta)=d_Amuwedge beta+(-1)^{mathrm{deg}(mu)}(muwedge d_Abeta)qquadforall muin Omega_M^p(mathrm{End}(E)),betainOmega_M^q(mathrm{End}(E))$$
      It also apparently follows that if $muin Omega_M^2(mathrm{End}(E))$ then,
      $$d_Amu=dmu+Awedge mu-muwedge A$$
      Here is my confused attempt, maybe consider them more as ideas:



      From (2), if I ASSUME that I can do the following,



      $d_A(sigmawedge s)=d(sigmawedge s)+Awedgesigmawedge s
      =(dsigma)wedge s+sigmawedge ds+Awedgesigmawedge s
      $



      Combining with (2) we have,



      $(d_Asigma)wedge s=(dsigma)wedge s+sigmawedge ds+Awedgesigmawedge s-sigmawedge ds-sigmawedge Awedge s\
      =dsigmawedge s+Awedgesigmawedge s-sigmawedge Awedge s\
      =(dsigma+Awedgesigma-sigmawedge A)wedge s\
      implies d_Asigma=dsigma+Awedgesigma-sigmawedge A
      $



      Then, if we write $omegain Omega_M^p(mathrm{End}(E))$ as $omega=f_iwedge omega^i$ where $f^iin Gamma(mathrm{End}(E))=Omega_M^0(mathrm{End}(E))$, and if I also ASSUME that (1) holds with for $etain Omega_M^r(mathrm{End}(E))
      $       then,



      $d_Aomega=d_A(f_iwedge omega^i)=d_Af_iwedge omega^i+(-1)^0f_iwedge domega^i\
      =df_iwedge omega^i+Awedge f^iwedgeomega^i-f^iwedge Awedge omega^i+f_iwedge domega^i \
      =d(omega)-f^iwedge domega^i+Awedgeomega-(-1)^{(r)}f^iwedgeomega^iwedge A+f_iwedge domega^i\
      =domega+Awedgeomega-(-1)^romegawedge A
      $



      which gives the $Omega^2_M(mathrm{End}(E))$ as a corollary.



      I've had some other thoughts, none of them well enough formulated to write down, any help with the above would be very much appreciated. There is a reference:
      https://www.dpmms.cam.ac.uk/~agk22/vb.pdf page 35.






      differential-geometry differential-forms vector-bundles exterior-algebra




      differential-geometry differential-forms vector-bundles exterior-algebra






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      edited Jan 17 at 11:29







      Sam Collie

















      asked Dec 23 '18 at 18:24









      Sam CollieSam Collie

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