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Is the exterior covariant derivative an anti-derivation? |
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| Feb16-05, 11:36 AM | #1 |
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Is the exterior covariant derivative an anti-derivation?
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hello,\n\nI am trying to find out if the exterior covariant derivative is an\nanti-derivation when applied to two types of Lie algebra valued forms.\nThe definition I am using of an anti-derivation is that for Lie algebra\nvalued forms a and b, the identity\n\nD([a, b]) = [Da, b] + (-1)^k [a, Db]\n\nis satisfied, where a is a k-form. The two cases I am interested in are:\n\n\n(1) a and b take values in the same Lie algebra as the connection 1-form\nA, so that\n\nDa := da + [A, a]\n\nand the Lie algebra valued form [a, b] is defined so that e.g. if a and\nb are 1-forms we have\n\n[a, b](u, v) := [a(u), b(v)] - [a(v), b(u)].\n\nThus the question is whether\n\nD([a, b]) = d[a, b] + [A, [a, b]]\n\nis equal to\n\n[Da, b] + (-1)^k [a, Db] = ([da, b] + [[A, a], b]) + (-1)^k ([a, db] +\n[a, [A, b]]).\n\n\n(2) a and b take values in the vector space acted on by a representation\nof the connection 1-form A, so that\n\nDa := da + A ^ a\n\nand the vector valued form [a, b] is defined as above, while e.g. for a\n1-form a we have\n\n(A ^ a)(u, v) := A(u)a(v) - A(v)a(u).\n\nThus the question is whether\n\nD([a, b]) = d[a, b] + A ^ [a, b]\n\nis equal to\n\n[Da, b] + (-1)^k [a, Db] = ([da, b] + [A ^ a, b]) + (-1)^k ([a, db] +\n[a, A ^ b]).\n\n\nWhat would help here is some identities for expressions involving\nvector-valued forms like d[a, b].\nThanks very much for any help or especially references.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hello,
I am trying to find out if the exterior covariant derivative is an anti-derivation when applied to two types of Lie algebra valued forms. The definition I am using of an anti-derivation is that for Lie algebra valued forms a and b, the identity D([a, b]) = [Da, [itex]b] + (-1)^k[/itex] [a, Db] is satisfied, where a is a k-form. The two cases I am interested in are: (1) a and b take values in the same Lie algebra as the connection 1-form A, so that [tex]Da := da +[/itex] [A, a] and the Lie algebra valued form [a, b] is defined so that e.g. if a and b are 1-forms we have [a, b](u, [itex]v) := [a(u), b(v)] - [a(v), b(u)][/itex]. Thus the question is whether D([a, b]) = d[a, [itex]b] +[/itex] [A, [a, b]] is equal to [Da, [itex]b] + (-1)^k[/itex] [a, Db] [itex]= ([da, b] + [[A, a], b]) + (-1)^k ([a, db] +[a, [A, b]])[/itex]. (2) a and b take values in the vector space acted on by a representation of the connection 1-form A, so that [itex]Da := da + A ^ a[/tex] and the vector valued form [a, b] is defined as above, while e.g. for a 1-form a we have [itex](A ^ a)(u, v) := A(u)a(v) - A(v)a(u)[/itex]. Thus the question is whether D([a, b]) = d[a, [itex]b] + A ^[/itex] [a, b] is equal to [Da, [itex]b] + (-1)^k[/itex] [a, Db] [itex]= ([da, b] + [A ^ a, b]) + (-1)^k ([a, db] +[a, A ^ b])[/itex]. What would help here is some identities for expressions involving vector-valued forms like d[a, b]. Thanks very much for any help or especially references. |
| Feb18-05, 11:41 AM | #2 |
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<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On Wed, 16 Feb 2005 17:36:38 +0000, Mark Adams wrote:\n\n> Hello,\n>\n> I am trying to find out if the exterior covariant derivative is an\n> anti-derivation when applied to two types of Lie algebra valued forms. The\n> definition I am using of an anti-derivation is that for Lie algebra valued\n> forms a and b, the identity\n>\n> D([a, b]) = [Da, b] + (-1)^k [a, Db]\n>\n> is satisfied, where a is a k-form. [...]\n\nSorry, this is not an answer to your question per se, but a suggestion for\nbetter terminology.\n\nUsually an antiderivation is like an "inverse" of a derivation, like the\nindefinite integral is to a derivative. I think what you want to say here\nis that D is a *graded* derivation. And the identity you want to satisfy\nis the graded Leibniz rule.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Wed, 16 Feb 2005 17:36:38 [itex]+0000,[/itex] Mark Adams wrote:
> Hello, > > I am trying to find out if the exterior covariant derivative is an > anti-derivation when applied to two types of Lie algebra valued forms. The > definition I am using of an anti-derivation is that for Lie algebra valued > forms a and b, the identity > > D([a, b]) = [Da, [itex]b] + (-1)^k[/itex] [a, Db] > > is satisfied, where a is a k-form. [...] Sorry, this is not an answer to your question per se, but a suggestion for better terminology. Usually an antiderivation is like an "inverse" of a derivation, like the indefinite integral is to a derivative. I think what you want to say here is that D is a *graded* derivation. And the identity you want to satisfy is the graded Leibniz rule. Igor |
| Feb20-05, 05:04 PM | #3 |
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<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"Igor Khavkine" <igor.kh@gmail.com> wrote in message\nnews:pan.2005.02.17.05.03.52.172287@gmail.com...\n> I think what you want to say here\n> is that D is a *graded* derivation. And the identity you want to\nsatisfy\n> is the graded Leibniz rule.\n\nYes, the term "antiderivation" as I\'m using it means a graded derivation\nof odd degree. For a and b taking values in a graded algebra (e.g. the\nexterior algebra), a graded derivation satisfies the graded Leibniz rule\n\nD(ab) = (Da)b + (-1)^ck a(Db)\n\nwhere a is a k-form and the derivation is of degree c (e.g. in the\nexterior algebra it takes k-forms to (k+c)-forms). Since the exterior\ncovariant derivative is of odd degree +1, if it satisfies the graded\nLeibniz rule it would be an "antiderivation".\n\nIt\'s weird, most books I\'ve found that treat the exterior covariant\nderivative in detail seem to also introduce the idea of a derivation,\nbut none state whether D is a derivation or not...\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Igor Khavkine" <igor.kh@gmail.com> wrote in message
news:pan.2005.02.17.05.03.52.172287@gmail.com... > I think what you want to say here > is that D is a *graded* derivation. And the identity you want to satisfy > is the graded Leibniz rule. Yes, the term "antiderivation" as I'm using it means a graded derivation of odd degree. For a and b taking values in a graded algebra (e.g. the exterior algebra), a graded derivation satisfies the graded Leibniz rule D(ab) [itex]= (Da)b + (-1)^ck a(Db)[/itex] where a is a k-form and the derivation is of degree c (e.g. in the exterior algebra it takes k-forms to (k+c)-forms). Since the exterior covariant derivative is of odd degree +1, if it satisfies the graded Leibniz rule it would be an "antiderivation". It's weird, most books I've found that treat the exterior covariant derivative in detail seem to also introduce the idea of a derivation, but none state whether D is a derivation or not... |
| Feb23-05, 03:57 AM | #4 |
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Is the exterior covariant derivative an anti-derivation?
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On Wed, 16 Feb 2005 17:36:38 +0000, Mark Adams wrote:\n\n> Hello,\n>\n> I am trying to find out if the exterior covariant derivative is an\n> anti-derivation when applied to two types of Lie algebra valued forms. The\n> definition I am using of an anti-derivation is that for Lie algebra valued\n> forms a and b, the identity\n>\n> D([a, b]) = [Da, b] + (-1)^k [a, Db]\n>\n> is satisfied, where a is a k-form. The two cases I am interested in are:\n>\n>\n> (1) a and b take values in the same Lie algebra as the connection 1-form\n> A, so that\n>\n> Da := da + [A, a]\n>\n> and the Lie algebra valued form [a, b] is defined so that e.g. if a and b\n> are 1-forms we have\n>\n> [a, b](u, v) := [a(u), b(v)] - [a(v), b(u)].\n>\n> Thus the question is whether\n>\n> D([a, b]) = d[a, b] + [A, [a, b]]\n>\n> is equal to\n>\n> [Da, b] + (-1)^k [a, Db] = ([da, b] + [[A, a], b]) + (-1)^k ([a, db] + [a,\n> [A, b]]).\n\n> What would help here is some identities for expressions involving\n> vector-valued forms like d[a, b].\n> Thanks very much for any help or especially references.\n\nThe identity is simply d[a,b] = [da,b] + (-1)^k [a,db]. If a and b are\nexpressed in matrix form, each component is a form and they are multiplied\nusing the wedge product. The exterior derivative will pass through the\nsummations over Lie algebra indices. Then the first part of the identity\nyou want to establish is obvious.\n\nThen you should recall that if A, a, b are Lie algebra elements, then\n[A,.] acts as a derivative with respect to the Lie bracket:\n\n[A,[a,b]] = [[A,a],b] + [a,[A,b]].\n\nThis is the Jacobi identity. What you have left to prove that it acts as a\ngraded derivation when A, a, b are Lie algebra valued forms, A being a\n1-form. Once again, this is easy to verify when writing it out in\ncomponents. There is an extra minus sign because components of a and A get\nswapped in the wedge product:\n\n[A,[a,b]] = [[A,a],b] + (-1)^k [a,[A,b]].\n\nThis is the graded Jacobi identity. So both d and [A,.] are graded\nderivations, hence their sum should be a graded derivation as well.\n\n> (2) a and b take values in the vector space acted on by a representation\n> of the connection 1-form A, so that\n>\n> Da := da + A ^ a\n>\n> and the vector valued form [a, b] is defined as above, while e.g. for a\n> 1-form a we have\n>\n> (A ^ a)(u, v) := A(u)a(v) - A(v)a(u).\n>\n> Thus the question is whether\n>\n> D([a, b]) = d[a, b] + A ^ [a, b]\n>\n> is equal to\n>\n> [Da, b] + (-1)^k [a, Db] = ([da, b] + [A ^ a, b]) + (-1)^k ([a, db] + [a,\n> A ^ b]).\n\nI can\'t quite make sense of your second example. How is it different from\nthe first one? If a and b, as you say, take values in a representation of\nthe Lie algebra where A is valued, what is [a,b]?\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Wed, 16 Feb 2005 17:36:38 [itex]+0000,[/itex] Mark Adams wrote:
> Hello, > > I am trying to find out if the exterior covariant derivative is an > anti-derivation when applied to two types of Lie algebra valued forms. The > definition I am using of an anti-derivation is that for Lie algebra valued > forms a and b, the identity > > D([a, b]) = [Da, [itex]b] + (-1)^k[/itex] [a, Db] > > is satisfied, where a is a k-form. The two cases I am interested in are: > > > (1) a and b take values in the same Lie algebra as the connection 1-form > A, so that > > [itex]Da := da +[/itex] [A, a] > > and the Lie algebra valued form [a, b] is defined so that e.g. if a and b > are 1-forms we have > > [a, b](u, [itex]v) := [a(u), b(v)] - [a(v), b(u)][/itex]. > > Thus the question is whether > > D([a, b]) = d[a, [itex]b] +[/itex] [A, [a, b]] > > is equal to > > [Da, [itex]b] + (-1)^k[/itex] [a, Db] [itex]= ([da, b] + [[A, a], b]) + (-1)^k ([a, db] + [a,[/itex] > [itex][A, b]]).[/itex] > What would help here is some identities for expressions involving > vector-valued forms like d[a, b]. > Thanks very much for any help or especially references. The identity is simply d[a,b] = [da,b] [itex]+ (-1)^k[/itex] [a,db]. If a and b are expressed in matrix form, each component is a form and they are multiplied using the wedge product. The exterior derivative will pass through the summations over Lie algebra indices. Then the first part of the identity you want to establish is obvious. Then you should recall that if A, a, b are Lie algebra elements, then [A,.] acts as a derivative with respect to the Lie bracket: [A,[a,b]] = [[A,a],b] + [a,[A,b]]. This is the Jacobi identity. What you have left to prove that it acts as a graded derivation when A, a, b are Lie algebra valued forms, A being a 1-form. Once again, this is easy to verify when writing it out in components. There is an extra minus sign because components of a and A get swapped in the wedge product: [A,[a,b]] = [[A,a],b] [itex]+ (-1)^k[/itex] [a,[A,b]]. This is the graded Jacobi identity. So both d and [A,.] are graded derivations, hence their sum should be a graded derivation as well. > (2) a and b take values in the vector space acted on by a representation > of the connection 1-form A, so that > > [itex]Da := da + A ^ a[/itex] > > and the vector valued form [a, b] is defined as above, while e.g. for a > 1-form a we have > > [itex](A ^ a)(u, v) := A(u)a(v) - A(v)a(u)[/itex]. > > Thus the question is whether > > D([a, b]) = d[a, [itex]b] + A ^[/itex] [a, b] > > is equal to > > [Da, [itex]b] + (-1)^k[/itex] [a, Db] [itex]= ([da, b] + [A ^ a, b]) + (-1)^k ([a, db] + [a,[/itex] > [itex]A ^ b]).[/itex] I can't quite make sense of your second example. How is it different from the first one? If a and b, as you say, take values in a representation of the Lie algebra where A is valued, what is [a,b]? Igor |
| Mar15-05, 12:21 PM | #5 |
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<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n"Igor Khavkine" <igor.kh@gmail.com> wrote in message\nnews:pan.2005.02.21.04.28.32.472820@gmail.com...\n>\n> This is the graded Jacobi identity. So both d and [A,.] are graded\n> derivations, hence their sum should be a graded derivation as well.\n>\n\nI finally got this to work out (thanks). So:\n\n(1) If a and b take values in the same Lie algebra as the connection\n1-form A,\nD([a, b]) = [Da, b] + (-1)^k [a, Db]\n\nholds and so D is a graded derivation in this sense. I also found a\nreference that confirmed this fact.\n\n\n> I can\'t quite make sense of your second example. How is it different\nfrom\n> the first one? If a and b, as you say, take values in a representation\nof\n> the Lie algebra where A is valued, what is [a,b]?\n\nGood question...! What I had in mind here was the case where a and b\ntake values in the tangent space of the manifold; if they take vector\nvalues that are acted on by a representation of the Lie algebra, their\nvalues do not form vector fields and as you say we cannot define the Lie\ncommutator. So:\n\n(2) If a and b take values in the tangent space of the manifold, we can\nask if\n\nD([a, b]) = [Da, b] + (-1)^k [a, Db].\n\nFor example if a is a vector-valued 0-form we have\n\nDa(v) := Del_v(a)\n\nand so for vector fields a and b, the question is whether\n\nDel_v([a, b]) = [Del_v(a), b] + [a, Del_v(b)].\n\nI calculated that this relation does not hold, and thus D is NOT a\nderivation in this sense. I could not find a reference that confirmed\nthis. However, in the process of looking I did come across some other\nrelated constructions.\n\n\n\n(3) Since Del_v is defined to act on tensor products of vector fields a\nand b via\n\nDel_v(a x b) := (Del_v(a)) x b + a x (Del_v(b)),\n\none can consider forms that take values in the tensor algebra. In this\nsense, D is is an (un-graded) derivation, since\n\nD(a x b) = (Da) x b + a x (Db)\n\nholds (I think).\n\n\n\n(4) One can then also consider forms that take values in the exterior\nalgebra; e.g. if a and b are vector-valued 1-forms,\n\n(a ^ b)(v, w) := a(v) ^ b(w) - a(w) ^ b(v)\n\n= (a(v) x b(w) - b(w) x a(v)) - (a(w) x b(v) - b(v) x a(w))\n\nis an antisymmetric tensor valued 2-form. In this sense, D is a graded\nderivation, since\n\nD(a ^ b) = Da ^ b + (-1)^k a ^ Db\n\nholds. I did find a reference for this fact.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Igor Khavkine" <igor.kh@gmail.com> wrote in message
news:pan.2005.02.21.04.28.32.472820@gmail.com... > > This is the graded Jacobi identity. So both d and [A,.] are graded > derivations, hence their sum should be a graded derivation as well. > I finally got this to work out (thanks). So: (1) If a and b take values in the same Lie algebra as the connection 1-form A, D([a, b]) = [Da, [itex]b] + (-1)^k[/itex] [a, Db] holds and so D is a graded derivation in this sense. I also found a reference that confirmed this fact. > I can't quite make sense of your second example. How is it different from > the first one? If a and b, as you say, take values in a representation of > the Lie algebra where A is valued, what is [a,b]? Good question...! What I had in mind here was the case where a and b take values in the tangent space of the manifold; if they take vector values that are acted on by a representation of the Lie algebra, their values do not form vector fields and as you say we cannot define the Lie commutator. So: (2) If a and b take values in the tangent space of the manifold, we can ask if D([a, b]) = [Da, [itex]b] + (-1)^k[/itex] [a, Db]. For example if a is a vector-valued 0-form we have Da(v) [itex]:= Del_v(a)[/itex] and so for vector fields a and b, the question is whether [tex]Del_v([a, b]) = [Del_v(a), b] +[/itex] [a, [itex]Del_v(b)][/itex]. I calculated that this relation does not hold, and thus D is NOT a derivation in this sense. I could not find a reference that confirmed this. However, in the process of looking I did come across some other related constructions. (3) Since [itex]Del_v[/itex] is defined to act on tensor products of vector fields a and b via [itex]Del_v(a x b) := (Del_v(a)) x b + a x (Del_v(b)),[/tex] one can consider forms that take values in the tensor algebra. In this sense, D is is an (un-graded) derivation, since D(a x [itex]b) = (Da) x b + a x (Db)[/itex] holds (I think). (4) One can then also consider forms that take values in the exterior algebra; e.g. if a and b are vector-valued 1-forms, [tex](a ^ b)(v, w) := a(v) ^ b(w) - a(w) ^ b(v)= (a(v) x b(w) - b(w) x a(v)) - (a(w) x b(v) - b(v) x a(w))[/tex] is an antisymmetric tensor valued 2-form. In this sense, D is a graded derivation, since [tex]D(a ^ b) = Da ^ b + (-1)^k a ^ Db[/tex] holds. I did find a reference for this fact. |
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