Mixing
Differential forms, exactness, closedness, and the Moyal product
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
This is my attempt to prove something (I'm not even sure if it's true to begin with) using fairly loose heuristic arguments. I present here all the steps and ideas and I would be extremely grateful if someone could confirm or refute it, explaining why I'm wrong (if I'm wrong) or explaining why I'm right (if I'm right for the wrong reasons or if I don't seem to be aware of some important subtleties). Thank you!
Assumptions
- let $mathcalP$ be a $2n$-dimensional phase space, i.e. a symplectic manifold with a linear symplectic form given by a $2n times 2n$ matrix$$omega = beginbmatrix 0 & I_n \ -I_n & 0 endbmatrix$$
functions on $mathcalP$ are equipped with the Moyal $star$-product, explicitly given by $$f(x,p)star g(x,p)=f(x,p)exp left[ fraci2left( overleftarrowpartial_x cdot overrightarrowpartial_p - overrightarrowpartial_x cdot overleftarrowpartial_pright) right] g(x,p)$$
we define framelike functions as functions of the form $f_a(x,p)$ where $a$ is a spacetime (pseudo-Euclidean) index such that, using Einstein summation convention $$f_a(x,p) star g^a(x,p) = eta^ab f_a(x,p) g_b(x,p)$$ where $eta=mathrmdiag(-1,underbrace+1,dots,+1_n-1)$
Statement
$$partial_af_b - partial_b f_a + i [f_a oversetstar, f_b] = 0 Longleftrightarrow exists , varepsilon(x,p) , texts.t. , f_a=partial_a varepsilon$$
($f_a equiv f_a(x,p)$, for notational clarity)
Heuristic (?) arguments
- let's assume that we can write $f_a(x,p)$ as a $1$-form $$mathbff=f_a mathbfdx^a$$
- since Moyal product is isomorphic to the matrix product in $M_infty$, we can view $mathbff$ as an operator-valued $1$-form
- The main statement becomes $$ mathbfd mathbff + i mathbff wedge mathbff = 0 Longleftrightarrow mathbff text is exact$$
$Longleftarrow$ proof
Trivial.$$mathbff , textexact Longleftrightarrow exists varepsilon , : , mathbff = mathbfd varepsilon$$
$$implies mathbfd mathbff = mathbfd^2 varepsilon = 0 ,, & ,, mathbff wedge mathbff = mathbfd varepsilon wedge mathbfd varepsilon = 0$$
$Longrightarrow$ proof (EDIT: this "proof" is wrong, I will correct my post as soon as possible)
$$mathbfd mathbff + i mathbff wedge mathbff = 0$$
$$implies mathbfd left( mathbfd mathbff + i mathbff wedge mathbffright) = 0$$
$$implies mathbfd left( mathbff wedge mathbff right) = 0 $$
$$implies mathbfd mathbff wedge mathbff + mathbff wedge mathbfd mathbff = 0 $$
$$implies (-1)^1 cdot 2 mathbff wedge mathbfd mathbff + mathbff wedge mathbfd mathbff = 0 $$
$$implies mathbff wedge mathbfd mathbff = 0 $$
$$implies mathbfd mathbff = 0 ,,, texti.e. ,, mathbff ,, textis closed$$
Now, if I'm not mistaken, Poincaré lemma should hold here, so $mathbff$ should also be an exact 1-form, which proves the $Longrightarrow$ statement.
Anything wrong with this?
mathematical-physics differential-forms symplectic-geometry noncommutative-geometry
asked Aug 3 at 13:01
Ivan V.
591216
add a comment |Â
up vote
3
down vote
favorite
This is my attempt to prove something (I'm not even sure if it's true to begin with) using fairly loose heuristic arguments. I present here all the steps and ideas and I would be extremely grateful if someone could confirm or refute it, explaining why I'm wrong (if I'm wrong) or explaining why I'm right (if I'm right for the wrong reasons or if I don't seem to be aware of some important subtleties). Thank you!
Assumptions
- let $mathcalP$ be a $2n$-dimensional phase space, i.e. a symplectic manifold with a linear symplectic form given by a $2n times 2n$ matrix$$omega = beginbmatrix 0 & I_n \ -I_n & 0 endbmatrix$$
functions on $mathcalP$ are equipped with the Moyal $star$-product, explicitly given by $$f(x,p)star g(x,p)=f(x,p)exp left[ fraci2left( overleftarrowpartial_x cdot overrightarrowpartial_p - overrightarrowpartial_x cdot overleftarrowpartial_pright) right] g(x,p)$$
we define framelike functions as functions of the form $f_a(x,p)$ where $a$ is a spacetime (pseudo-Euclidean) index such that, using Einstein summation convention $$f_a(x,p) star g^a(x,p) = eta^ab f_a(x,p) g_b(x,p)$$ where $eta=mathrmdiag(-1,underbrace+1,dots,+1_n-1)$
Statement
$$partial_af_b - partial_b f_a + i [f_a oversetstar, f_b] = 0 Longleftrightarrow exists , varepsilon(x,p) , texts.t. , f_a=partial_a varepsilon$$
($f_a equiv f_a(x,p)$, for notational clarity)
Heuristic (?) arguments
- let's assume that we can write $f_a(x,p)$ as a $1$-form $$mathbff=f_a mathbfdx^a$$
- since Moyal product is isomorphic to the matrix product in $M_infty$, we can view $mathbff$ as an operator-valued $1$-form
- The main statement becomes $$ mathbfd mathbff + i mathbff wedge mathbff = 0 Longleftrightarrow mathbff text is exact$$
$Longleftarrow$ proof
Trivial.$$mathbff , textexact Longleftrightarrow exists varepsilon , : , mathbff = mathbfd varepsilon$$
$$implies mathbfd mathbff = mathbfd^2 varepsilon = 0 ,, & ,, mathbff wedge mathbff = mathbfd varepsilon wedge mathbfd varepsilon = 0$$
$Longrightarrow$ proof (EDIT: this "proof" is wrong, I will correct my post as soon as possible)
$$mathbfd mathbff + i mathbff wedge mathbff = 0$$
$$implies mathbfd left( mathbfd mathbff + i mathbff wedge mathbffright) = 0$$
$$implies mathbfd left( mathbff wedge mathbff right) = 0 $$
$$implies mathbfd mathbff wedge mathbff + mathbff wedge mathbfd mathbff = 0 $$
$$implies (-1)^1 cdot 2 mathbff wedge mathbfd mathbff + mathbff wedge mathbfd mathbff = 0 $$
$$implies mathbff wedge mathbfd mathbff = 0 $$
$$implies mathbfd mathbff = 0 ,,, texti.e. ,, mathbff ,, textis closed$$
Now, if I'm not mistaken, Poincaré lemma should hold here, so $mathbff$ should also be an exact 1-form, which proves the $Longrightarrow$ statement.
Anything wrong with this?
mathematical-physics differential-forms symplectic-geometry noncommutative-geometry
asked Aug 3 at 13:01
Ivan V.
591216
i) Are framelike functions possibly something else than 1-forms? ii) Is the product of framelike functions supposed to involve the Moyal product? iii) Could you specify what is the definition of the bracket $[ - oversetstar, - ]$? iv) Could you recall the definition of $wedge$ between operator-valued (OV) forms? v) Why does $depsilon wedge depsilon = 0$ for OV forms? vi) Why does $df wedge f = f wedge df$ for OV forms? vii) Why does $f wedge df = 0$ implies $df=0$? (This is false for the real-valued 1-form $f = xdy - ydx$, hence for some 'diagonal' OV forms too.)
– Jordan Payette
Aug 3 at 14:51
@JordanPayette Reasonable and helpful questions, thank you! I'll append the answers to my post as an edit.
– Ivan V.
Aug 3 at 16:24
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This is my attempt to prove something (I'm not even sure if it's true to begin with) using fairly loose heuristic arguments. I present here all the steps and ideas and I would be extremely grateful if someone could confirm or refute it, explaining why I'm wrong (if I'm wrong) or explaining why I'm right (if I'm right for the wrong reasons or if I don't seem to be aware of some important subtleties). Thank you!
Assumptions
- let $mathcalP$ be a $2n$-dimensional phase space, i.e. a symplectic manifold with a linear symplectic form given by a $2n times 2n$ matrix$$omega = beginbmatrix 0 & I_n \ -I_n & 0 endbmatrix$$
functions on $mathcalP$ are equipped with the Moyal $star$-product, explicitly given by $$f(x,p)star g(x,p)=f(x,p)exp left[ fraci2left( overleftarrowpartial_x cdot overrightarrowpartial_p - overrightarrowpartial_x cdot overleftarrowpartial_pright) right] g(x,p)$$
we define framelike functions as functions of the form $f_a(x,p)$ where $a$ is a spacetime (pseudo-Euclidean) index such that, using Einstein summation convention $$f_a(x,p) star g^a(x,p) = eta^ab f_a(x,p) g_b(x,p)$$ where $eta=mathrmdiag(-1,underbrace+1,dots,+1_n-1)$
Statement
$$partial_af_b - partial_b f_a + i [f_a oversetstar, f_b] = 0 Longleftrightarrow exists , varepsilon(x,p) , texts.t. , f_a=partial_a varepsilon$$
($f_a equiv f_a(x,p)$, for notational clarity)
Heuristic (?) arguments
- let's assume that we can write $f_a(x,p)$ as a $1$-form $$mathbff=f_a mathbfdx^a$$
- since Moyal product is isomorphic to the matrix product in $M_infty$, we can view $mathbff$ as an operator-valued $1$-form
- The main statement becomes $$ mathbfd mathbff + i mathbff wedge mathbff = 0 Longleftrightarrow mathbff text is exact$$
$Longleftarrow$ proof
Trivial.$$mathbff , textexact Longleftrightarrow exists varepsilon , : , mathbff = mathbfd varepsilon$$
$$implies mathbfd mathbff = mathbfd^2 varepsilon = 0 ,, & ,, mathbff wedge mathbff = mathbfd varepsilon wedge mathbfd varepsilon = 0$$
$Longrightarrow$ proof (EDIT: this "proof" is wrong, I will correct my post as soon as possible)
$$mathbfd mathbff + i mathbff wedge mathbff = 0$$
$$implies mathbfd left( mathbfd mathbff + i mathbff wedge mathbffright) = 0$$
$$implies mathbfd left( mathbff wedge mathbff right) = 0 $$
$$implies mathbfd mathbff wedge mathbff + mathbff wedge mathbfd mathbff = 0 $$
$$implies (-1)^1 cdot 2 mathbff wedge mathbfd mathbff + mathbff wedge mathbfd mathbff = 0 $$
$$implies mathbff wedge mathbfd mathbff = 0 $$
$$implies mathbfd mathbff = 0 ,,, texti.e. ,, mathbff ,, textis closed$$
Now, if I'm not mistaken, Poincaré lemma should hold here, so $mathbff$ should also be an exact 1-form, which proves the $Longrightarrow$ statement.
Anything wrong with this?
mathematical-physics differential-forms symplectic-geometry noncommutative-geometry
asked Aug 3 at 13:01
Ivan V.
591216
This is my attempt to prove something (I'm not even sure if it's true to begin with) using fairly loose heuristic arguments. I present here all the steps and ideas and I would be extremely grateful if someone could confirm or refute it, explaining why I'm wrong (if I'm wrong) or explaining why I'm right (if I'm right for the wrong reasons or if I don't seem to be aware of some important subtleties). Thank you!
Assumptions
- let $mathcalP$ be a $2n$-dimensional phase space, i.e. a symplectic manifold with a linear symplectic form given by a $2n times 2n$ matrix$$omega = beginbmatrix 0 & I_n \ -I_n & 0 endbmatrix$$
functions on $mathcalP$ are equipped with the Moyal $star$-product, explicitly given by $$f(x,p)star g(x,p)=f(x,p)exp left[ fraci2left( overleftarrowpartial_x cdot overrightarrowpartial_p - overrightarrowpartial_x cdot overleftarrowpartial_pright) right] g(x,p)$$
we define framelike functions as functions of the form $f_a(x,p)$ where $a$ is a spacetime (pseudo-Euclidean) index such that, using Einstein summation convention $$f_a(x,p) star g^a(x,p) = eta^ab f_a(x,p) g_b(x,p)$$ where $eta=mathrmdiag(-1,underbrace+1,dots,+1_n-1)$
Statement
$$partial_af_b - partial_b f_a + i [f_a oversetstar, f_b] = 0 Longleftrightarrow exists , varepsilon(x,p) , texts.t. , f_a=partial_a varepsilon$$
($f_a equiv f_a(x,p)$, for notational clarity)
Heuristic (?) arguments
- let's assume that we can write $f_a(x,p)$ as a $1$-form $$mathbff=f_a mathbfdx^a$$
- since Moyal product is isomorphic to the matrix product in $M_infty$, we can view $mathbff$ as an operator-valued $1$-form
- The main statement becomes $$ mathbfd mathbff + i mathbff wedge mathbff = 0 Longleftrightarrow mathbff text is exact$$
$Longleftarrow$ proof
Trivial.$$mathbff , textexact Longleftrightarrow exists varepsilon , : , mathbff = mathbfd varepsilon$$
$$implies mathbfd mathbff = mathbfd^2 varepsilon = 0 ,, & ,, mathbff wedge mathbff = mathbfd varepsilon wedge mathbfd varepsilon = 0$$
$Longrightarrow$ proof (EDIT: this "proof" is wrong, I will correct my post as soon as possible)
$$mathbfd mathbff + i mathbff wedge mathbff = 0$$
$$implies mathbfd left( mathbfd mathbff + i mathbff wedge mathbffright) = 0$$
$$implies mathbfd left( mathbff wedge mathbff right) = 0 $$
$$implies mathbfd mathbff wedge mathbff + mathbff wedge mathbfd mathbff = 0 $$
$$implies (-1)^1 cdot 2 mathbff wedge mathbfd mathbff + mathbff wedge mathbfd mathbff = 0 $$
$$implies mathbff wedge mathbfd mathbff = 0 $$
$$implies mathbfd mathbff = 0 ,,, texti.e. ,, mathbff ,, textis closed$$
Now, if I'm not mistaken, Poincaré lemma should hold here, so $mathbff$ should also be an exact 1-form, which proves the $Longrightarrow$ statement.
Anything wrong with this?
mathematical-physics differential-forms symplectic-geometry noncommutative-geometry
asked Aug 3 at 13:01
Ivan V.
591216
edited Aug 3 at 18:07
asked Aug 3 at 13:01
Ivan V.
591216
asked Aug 3 at 13:01
Ivan V.
591216
asked Aug 3 at 13:01
Ivan V.
591216
591216
i) Are framelike functions possibly something else than 1-forms? ii) Is the product of framelike functions supposed to involve the Moyal product? iii) Could you specify what is the definition of the bracket $[ - oversetstar, - ]$? iv) Could you recall the definition of $wedge$ between operator-valued (OV) forms? v) Why does $depsilon wedge depsilon = 0$ for OV forms? vi) Why does $df wedge f = f wedge df$ for OV forms? vii) Why does $f wedge df = 0$ implies $df=0$? (This is false for the real-valued 1-form $f = xdy - ydx$, hence for some 'diagonal' OV forms too.)
– Jordan Payette
Aug 3 at 14:51
@JordanPayette Reasonable and helpful questions, thank you! I'll append the answers to my post as an edit.
– Ivan V.
Aug 3 at 16:24
add a comment |Â
i) Are framelike functions possibly something else than 1-forms? ii) Is the product of framelike functions supposed to involve the Moyal product? iii) Could you specify what is the definition of the bracket $[ - oversetstar, - ]$? iv) Could you recall the definition of $wedge$ between operator-valued (OV) forms? v) Why does $depsilon wedge depsilon = 0$ for OV forms? vi) Why does $df wedge f = f wedge df$ for OV forms? vii) Why does $f wedge df = 0$ implies $df=0$? (This is false for the real-valued 1-form $f = xdy - ydx$, hence for some 'diagonal' OV forms too.)
– Jordan Payette
Aug 3 at 14:51
@JordanPayette Reasonable and helpful questions, thank you! I'll append the answers to my post as an edit.
– Ivan V.
Aug 3 at 16:24
i) Are framelike functions possibly something else than 1-forms? ii) Is the product of framelike functions supposed to involve the Moyal product? iii) Could you specify what is the definition of the bracket $[ - oversetstar, - ]$? iv) Could you recall the definition of $wedge$ between operator-valued (OV) forms? v) Why does $depsilon wedge depsilon = 0$ for OV forms? vi) Why does $df wedge f = f wedge df$ for OV forms? vii) Why does $f wedge df = 0$ implies $df=0$? (This is false for the real-valued 1-form $f = xdy - ydx$, hence for some 'diagonal' OV forms too.)
– Jordan Payette
Aug 3 at 14:51
i) Are framelike functions possibly something else than 1-forms? ii) Is the product of framelike functions supposed to involve the Moyal product? iii) Could you specify what is the definition of the bracket $[ - oversetstar, - ]$? iv) Could you recall the definition of $wedge$ between operator-valued (OV) forms? v) Why does $depsilon wedge depsilon = 0$ for OV forms? vi) Why does $df wedge f = f wedge df$ for OV forms? vii) Why does $f wedge df = 0$ implies $df=0$? (This is false for the real-valued 1-form $f = xdy - ydx$, hence for some 'diagonal' OV forms too.)
– Jordan Payette
Aug 3 at 14:51
@JordanPayette Reasonable and helpful questions, thank you! I'll append the answers to my post as an edit.
– Ivan V.
Aug 3 at 16:24
@JordanPayette Reasonable and helpful questions, thank you! I'll append the answers to my post as an edit.
– Ivan V.
Aug 3 at 16:24
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871039%2fdifferential-forms-exactness-closedness-and-the-moyal-product%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
i) Are framelike functions possibly something else than 1-forms? ii) Is the product of framelike functions supposed to involve the Moyal product? iii) Could you specify what is the definition of the bracket $[ - oversetstar, - ]$? iv) Could you recall the definition of $wedge$ between operator-valued (OV) forms? v) Why does $depsilon wedge depsilon = 0$ for OV forms? vi) Why does $df wedge f = f wedge df$ for OV forms? vii) Why does $f wedge df = 0$ implies $df=0$? (This is false for the real-valued 1-form $f = xdy - ydx$, hence for some 'diagonal' OV forms too.)
– Jordan Payette
Aug 3 at 14:51
@JordanPayette Reasonable and helpful questions, thank you! I'll append the answers to my post as an edit.
– Ivan V.
Aug 3 at 16:24