Covariant Exterior Derivative action on the $mathrm{End}(E)$-valued p-forms
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
edited Jan 17 at 11:29
Sam Collie
asked Dec 23 '18 at 18:24
Sam CollieSam Collie
507
507
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