<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>\nGreetings--I have a question related to classical mechanics. Can somebody\nexplain to me how vectors in the tangent bundle (i.e. velocity) transform\nunder a change of basis differently from one-forms in the cotangent bundle\n(i.e. momentum)?\n\nI know that velocity vectors transform under the jacobian matrix, this makes\nintuitive sense since velocities are just time derivatives of the\ncoordinates. However, I am told that momenta transform under the transpose\nof the jacobian matrix--why is this? Can direct me to a intuitive and/or\nmathematical proof?\n\nThanks much,\nPhilip Tanedo\nflipt@stanford.edu\n\nPS--on a completely different note, are there any condensed matter folks out\nthere who could point me to accessible (read: undergrad-accessible) material\non spin-gap compounds? Thanks doubly!\n\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>Greetings--I have a question related to classical mechanics. Can somebody
explain to me how vectors in the tangent bundle (i.e. velocity) transform
under a change of basis differently from one-forms in the cotangent bundle
(i.e. momentum)?
I know that velocity vectors transform under the jacobian matrix, this makes
intuitive sense since velocities are just time derivatives of the
coordinates. However, I am told that momenta transform under the transpose
of the jacobian matrix--why is this? Can direct me to a intuitive and/or
mathematical proof?
Thanks much,
Philip Tanedo
flipt@stanford.edu
PS--on a completely different note, are there any condensed matter folks out
there who could point me to accessible (read: undergrad-accessible) material
on spin-gap compounds? Thanks doubly!
Pierre Asselin
Jul11-04, 03:57 AM
<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\nFlip Tomato <flipt@stanford.edu> wrote:\n\n> Greetings--I have a question related to classical mechanics. Can somebody\n> explain to me how vectors in the tangent bundle (i.e. velocity) transform\n> under a change of basis differently from one-forms in the cotangent bundle\n> (i.e. momentum)?\n\n> I know that velocity vectors transform under the jacobian matrix, this makes\n> intuitive sense since velocities are just time derivatives of the\n> coordinates. However, I am told that momenta transform under the transpose\n> of the jacobian matrix--why is this? Can direct me to a intuitive and/or\n> mathematical proof?\n\nThe inverse transpose of the Jacobian matrix. Short answer: p^t v\nhas to be an invariant. So if v\' = Jv, you need p\' = J^{-1}^t p in\norder that p\'^t v\' be equal to p^t v.\n\nSee Leon Brillouin\'s book "Tensors in Mechanics and Elasticity" (I\nthink; this reference is from memory).\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>Flip Tomato <flipt@stanford.edu> wrote:
> Greetings--I have a question related to classical mechanics. Can somebody
> explain to me how vectors in the tangent bundle (i.e. velocity) transform
> under a change of basis differently from one-forms in the cotangent bundle
> (i.e. momentum)?
> I know that velocity vectors transform under the jacobian matrix, this makes
> intuitive sense since velocities are just time derivatives of the
> coordinates. However, I am told that momenta transform under the transpose
> of the jacobian matrix--why is this? Can direct me to a intuitive and/or
> mathematical proof?
The inverse transpose of the Jacobian matrix. Short answer: p^t v
has to be an invariant. So if v' = Jv, you need p' = J^{-1}^t p in
order that p'^t v' be equal to p^t v.
See Leon Brillouin's book "Tensors in Mechanics and Elasticity" (I
think; this reference is from memory).
Eric A. Forgy
Jul12-04, 04:44 AM
<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>\npa@invalid.invalid (Pierre Asselin) wrote:\n> The inverse transpose of the Jacobian matrix. Short answer: p^t v\n> has to be an invariant. So if v\' = Jv, you need p\' = J^{-1}^t p in\n> order that p\'^t v\' be equal to p^t v.\n\nHi Flip,\n\nI am going to give the same answer that Pierre gave, but with a little\nmore words and a slightly different notation.\n\nLet X be a vector field and alpha a dual vector field, i.e. a 1-form\n(a dual vector field is a linear function(al) of vector fields,\nmeaning that you can "evaluate" a dual vector field on a vector field\nresulting in a scalar function, i.e. 0-form).\n\nI will denote functional evaluation using angle brackets so that\n\n<alpha,X> := alpha(X),\n\nwhich is a scalar function (0-form).\n\nIf you have any map T that takes a vector field and returns another\nvector field, you can define a transpose map T^t that takes 1-forms\nand returns 1-forms via\n\n<T^t(alpha),X> = <alpha,T(X)>.\n\nNow if we let X\' = J(X) denote the same vector field expressed in\nterms of a different coordinate system (similarly alpha\' is alpha\nexpressed in terms of the new coordinate system), then clearly we must\nhave (in a coordinate chart)\n\n<alpha,X> = <alpha\',X\'>\n\nwhich is exactly what Pierre said because the evaluation does not care\nwhat coordinates you use to express vector fields or 1-forms. Now we\ncan run through the algebra\n\n<alpha\',X\'> = <alpha\',J(X)> = <J^t(alpha\'),X> = <alpha,X>.\n\nTherefore,\n\nJ^t(alpha\') = alpha\n\nif\n\nX\' = J(X).\n\nIf we want to express alpha\' = "something" then we just move J^t to\nthe other side so that\n\nalpha\' = J^{-t}(alpha)\n\nas Pierre said.\n\nBest regards,\nEric\n"The Department of Redundancy Department"\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>pa@invalid.invalid (Pierre Asselin) wrote:
> The inverse transpose of the Jacobian matrix. Short answer: p^t v
> has to be an invariant. So if v' = Jv, you need p' = J^{-1}^t p in
> order that p'^t v' be equal to p^t v.
Hi Flip,
I am going to give the same answer that Pierre gave, but with a little
more words and a slightly different notation.
Let X be a vector field and \alpha a dual vector field, i.e. a 1-form
(a dual vector field is a linear function(al) of vector fields,
meaning that you can "evaluate" a dual vector field on a vector field
resulting in a scalar function, i.e. 0-form).
I will denote functional evaluation using angle brackets so that
<\alpha,X> := \alpha(X),
which is a scalar function (0-form).
If you have any map T that takes a vector field and returns another
vector field, you can define a transpose map T^t that takes 1-forms
and returns 1-forms via
<T^t(\alpha),X> = <\alpha,T(X)>[/itex].
Now if we let X' = J(X) denote the same vector field expressed in
terms of a different coordinate system (similarly \alpha' is \alpha
expressed in terms of the new coordinate system), then clearly we must
have (in a coordinate chart)
<\alpha,X> = <\alpha',X'>
which is exactly what Pierre said because the evaluation does not care
what coordinates you use to express vector fields or 1-forms. Now we
can run through the algebra
If we want to express \alpha' = "something" then we just move J^t to
the other side so that
[itex]\alpha' = J^{-t}(\alpha)
as Pierre said.
Best regards,
Eric
"The Department of Redundancy Department"
arivero
Jul13-04, 03:37 AM
Hi Flip,
Let me to add, once I saw in a russian book the suggestion of visualising covectors as if it were defined by two parallel planes, so you are interested in the normal vector to the plane. Even if mostly 3D, it is a good metaphor.
Flip Tomato
Jul14-04, 05:23 AM
<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>\nThanks Pierre and Eric for the responses--but I\'m still a little confused\nabout part of Eric\'s explanation, which I cite here:\n\n> If you have any map T that takes a vector field and returns another\n> vector field, you can define a transpose map T^t that takes 1-forms\n> and returns 1-forms via\n>\n> <T^t(alpha),X> = <alpha,T(X)>.\n>\n\nThis is exactly the proposition which confsues me. Pierre\'s explanation\n(inner product must be invariant) is nice intuitively, but I don\'t see how\nyou motivate this step mathematically. What is the property of the inner\nproduct or the nature of the one-form that dictates that one forms tranform\nT^t?\n\nSorry if I\'m being obtuse =)\nFlip\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>Thanks Pierre and Eric for the responses--but I'm still a little confused
about part of Eric's explanation, which I cite here:
> If you have any map T that takes a vector field and returns another
> vector field, you can define a transpose map T^t that takes 1-forms
> and returns 1-forms via
>
> <T^t(\alpha),X> = <\alpha,T(X)>.
>
This is exactly the proposition which confsues me. Pierre's explanation
(inner product must be invariant) is nice intuitively, but I don't see how
you motivate this step mathematically. What is the property of the inner
product or the nature of the one-form that dictates that one forms tranform
T^t?
Sorry if I'm being obtuse =)
Flip
Pierre Asselin
Jul16-04, 09:19 AM
<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\n> Pierre\'s explanation\n> (inner product must be invariant) is nice intuitively, but I don\'t see how\n> you motivate this step mathematically.\n\nUh? The inner product *is* the one-form.\n\nMore precisely: the one-form is a linear function taking vectors\nand returning real numbers, independently of any coordinate systems.\nIf you now choose a coordinate system, you can represent your vectors and\none-forms by buckets of numbers, and evaluation of the one-form becomes\nan inner product. If you have two coordinate systems, the inner product\nhas to be invariant because you\'re talking about *the same* one-form as\nseen from the two systems.\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>Pierre's explanation
> (inner product must be invariant) is nice intuitively, but I don't see how
> you motivate this step mathematically.
Uh? The inner product *is* the one-form.
More precisely: the one-form is a linear function taking vectors
and returning real numbers, independently of any coordinate systems.
If you now choose a coordinate system, you can represent your vectors and
one-forms by buckets of numbers, and evaluation of the one-form becomes
an inner product. If you have two coordinate systems, the inner product
has to be invariant because you're talking about *the same* one-form as
seen from the two systems.
Flip Tomato
Jul19-04, 04:09 AM
<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\nYes, I understand that the inner product must be invariant, but why is it\nthat it must transform this way:\n\n<T^t(alpha),X> = <alpha,T(X)>.\n\ni.e. I have a one form alpha (sorry, I am not LaTeX enabled). I represent\nthe action of this one form on a vector X as <alpha, X>. If I transform X to\nT(X), then why is it that alpha is transformed as T^t(alpha).\n\ni.e. I understand that the inner product with a transformed vector and\nuntransformed one-form should yield the same 0-form as the inner product\nwith an untransformed vector and a transformed one-form. I just do not\nunderstand why the transformed one-form takes the form (T^t(alpha)) that it\ndoes.\n\n-Philip\n\n"Pierre Asselin" <pa@invalid.invalid> wrote in message\nnews:0q97dc.8i1.ln@brick.verano.sba.ca.us...\n>\n>\n> > Pierre\'s explanation\n> > (inner product must be invariant) is nice intuitively, but I don\'t see\nhow\n> > you motivate this step mathematically.\n>\n> Uh? The inner product *is* the one-form.\n>\n> More precisely: the one-form is a linear function taking vectors\n> and returning real numbers, independently of any coordinate systems.\n> If you now choose a coordinate system, you can represent your vectors and\n> one-forms by buckets of numbers, and evaluation of the one-form becomes\n> an inner product. If you have two coordinate systems, the inner product\n> has to be invariant because you\'re talking about *the same* one-form as\n> seen from the two systems.\n\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>Yes, I understand that the inner product must be invariant, but why is it
that it must transform this way:
<T^t(\alpha),X> = <\alpha,T(X)>[/itex].
i.e. I have a one form \alpha (sorry, I am not LaTeX enabled). I represent
the action of this one form on a vector X as <\alpha, X>. If I transform X to
T(X), then why is it that \alpha is transformed as [itex]T^t(\alpha).
i.e. I understand that the inner product with a transformed vector and
untransformed one-form should yield the same 0-form as the inner product
with an untransformed vector and a transformed one-form. I just do not
understand why the transformed one-form takes the form (T^t(\alpha)) that it
does.
-Philip
"Pierre Asselin" <pa@invalid.invalid> wrote in message
news:0q97dc.8i1.ln@brick.verano.sba.ca.us...
>
>
> > Pierre's explanation
> > (inner product must be invariant) is nice intuitively, but I don't see
how
> > you motivate this step mathematically.
>
> Uh? The inner product *is* the one-form.
>
> More precisely: the one-form is a linear function taking vectors
> and returning real numbers, independently of any coordinate systems.
> If you now choose a coordinate system, you can represent your vectors and
> one-forms by buckets of numbers, and evaluation of the one-form becomes
> an inner product. If you have two coordinate systems, the inner product
> has to be invariant because you're talking about *the same* one-form as
> seen from the two systems.
Eric A. Forgy
Jul20-04, 04:25 AM
<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\n"Flip Tomato" <flipt@stanford.edu> wrote:\n> Yes, I understand that the inner product must be invariant, but why is it\n> that it must transform this way:\n>\n> <T^t(alpha),X> = <alpha,T(X)>.\n>\n> i.e. I have a one form alpha (sorry, I am not LaTeX enabled). I represent\n> the action of this one form on a vector X as <alpha, X>. If I transform X to\n> T(X), then why is it that alpha is transformed as T^t(alpha).\n>\n> i.e. I understand that the inner product with a transformed vector and\n> untransformed one-form should yield the same 0-form as the inner product\n> with an untransformed vector and a transformed one-form. I just do not\n> understand why the transformed one-form takes the form (T^t(alpha)) that it\n> does.\n>\n> -Philip\n\nHi Flip,\n\nI wrote the following message on July 14, 2004, but it didn\'t seem to\nmake it through. I will try again today (July 19, 2004).\n\n===================================================\n\nFlip Tomato wrote:\n\n>> If you have any map T that takes a vector field and returns another\n>> vector field, you can define a transpose map T^t that takes 1-forms\n>> and returns 1-forms via\n>>\n>> <T^t(alpha),X> = <alpha,T(X)>.\n>>\n> This is exactly the proposition which confsues me. Pierre\'s explanation\n> (inner product must be invariant) is nice intuitively\n>\nWell, what Pierre was talking about was not the inner product, it was\nthe evaluation map. Pierre might have assumed that you knew that (once\nyou have chosen a basis) a 1-form can (although I am not a fan of the\nidea) be expressed as a row vector, whereas a vector field can be\nexpressed as a column vector. When he wrote\n\np^t v,\n\nthe p^t was meant to denote a 1-form (row vector) and v was a vector\nfield (column vector) and p^t v is matrix multiplication.\n\n> but I don\'t see how\n> you motivate this step mathematically. What is the property of the inner\n> product or the nature of the one-form that dictates that one forms tranform\n> T^t?\n>\nI\'ll try to explain in two different ways what is going on.\n\nFirst way:\n----------\n\nExpress everything in coordinates...\n\nLet e_u be a coordinate basis for vector fields (in a coordinate\nchart) and dx^u be a coordinate basis for 1-forms such that\n\n<dx^u,e_v> = dx^u(e_v) = delta^u_v.\n\nIn this basis, the vector field X can be expressed as\n\nX = X^u e_u\n\nand the 1-form alpha can be expressed as\n\nalpha = alpha_v dx^v\n\nso that the evaluation gives\n\n<alpha,X>\n= <alpha_v dx^v,X^u e_u>\n= alpha_v X^u <dx^v,e_u>\n= alpha_v X^u delta^v_u\n= alpha_u X^u.\n\nNow the map T can also be expressed in coordinates as\n\nT = T^u_v e_u (x) dx^v\n\nso that\n\nT(X)\n= (T^u_v X^w) e_u (x) <dx^v,e_w>\n= (T^u_v X^v) e_u.\n\nIf you look at how the indices are working out, you can probably\nrecognize this looks like a matrix vector multiplication.\n\nNow we can compute the evaluation\n\n<alpha,T(X)>\n= alpha_w (T^u_v X^v) <dx^w,e_u>\n= alpha_u (T^u_v X^v).\n\nYou can define some map S on 1-forms via\n\nS = S_v^u dx^v (x) e_u\n\nso that\n\nS(alpha)\n= S_v^u alpha_w dx^v (x) <e_u,dx^w>\n= (S_v^u alpha_u) dx^v.\n\nComputing the evaluation results in\n\n<S(alpha),X>\n= (S_v^u alpha_u) X^w <dx^v,e_w>\n= (S_v^u alpha_u) X^v.\n\nNOW...\n\nLet\'s say we want to find the special map S of 1-forms such that it\nsatisfies the relation\n\n<S(alpha),X> = <alpha,T(X)>,\n\nwhich when written explicitly in coordinates is\n\n(S_v^u alpha_u) X^v = alpha_u (T^u_v X^v).\n\nComparing both sides we see that in order for the relation to be\nsatisfied, we must have\n\nS_v^u = T^u_v.\n\n[Note: The indices should really be slightly shifted relative to one\nanother, e.g. S_v^u = S_{v }^{ u} and T^u_v = T^{u }_{ v}.]\n\nAny map that relates to T in this way is usually denoted T^t and is\ncalled the transpose of T. The standard way to write this would be\n\n<T^t(alpha),X> = <alpha,T(X)>.\n\nWait!! See, I didn\'t really want to express this in coordinates\nbecause coordinates are nasty. The only reason I went through all of\nthis is because you seem to have thought that\n\n<T^t(alpha),X> = <alpha,T(X)>\n\nwas a "derived" result where you assume we already know what ^t means.\nBut in fact, the above relation is the DEFINITION of transpose. It is\nnot a derived result. It is a definition of T^t. (Redundant enough?\n:))\n\nI should also remark that although it is called the transpose, we are\nnot really dealing with matrices, i.e. square blocks of numbers. I\npersonally prefer this way of dealing with tensors because matrices\nare somewhat limited, e.g. good luck representing the Riemannian\ncurvature as a matrix :)\n\nOnce you understand that\n\n<T^t(alpha),X> = <alpha,T(X)>\n\nis actually the definition of transpose, I think the answer to the\nrest of your question should become clear. If not, let us know! :)\n\nOk. Here is the...\n\nSecond Way:\n-----------\n\nThis way uses matrices and I do not recommend it, but maybe it will\nhelp. Once you choose a coordinate basis, you can (although I don\'t\nrecommend it) think of (as I said) a 1-form alpha as a row vector\n[alpha]^t and a vector field X as a column vector [X]. To make it\nclear which is which, I will explicitly put a {}^t on 1-forms.\nTherefore, I will denote the 1-form alpha above as\n\n[alpha]^t\n\nand the vector field X above as\n\n[X].\n\nThe square brackets denote that the object is a square block of\nnumbers (0-forms actually).\n\nThe evaluation can then be written as Pierre had it\n\n<alpha,X> = [alpha]^t [X].\n\nKeep in mind, this is NOT the inner product (although it looks like it\nand a theorem says that it is related to an inner product with a dual\nelement).\n\nNow let [X] = [T][Y] for some matrix [T] and column vector [Y] so that\n\n<alpha,X>\n= [alpha]^t [X]\n= [alpha]^t ([T][Y])\n= ([alpha]^t [T]) [Y]\n= ([T]^t [alpha])^t [Y]\n\nso that\n\n([T]^t [alpha])^t [Y] = [alpha]^t ([T][Y]).\n\nThis is the matrix version of\n\n<T^t(alpha),X> = <alpha,T(X)>\n\nabove.\n\nNow let X\' be the same vector field X expressed in terms of a new\ncoordinate system with the change of coordinates given by a true\nmatrix [T], i.e.\n\n[X]\' = [T][X].\n\nSimilarly, [alpha]\'^t is the row vector representation of the same\n1-form in a new basis. We all agree that the evaluation doesn\'t depend\non which coordinates we use so that\n\n[alpha]^t [X]\n= [alpha]\'^t [X]\'\n= [alpha]\'^t [T][X]\n= ([T]^t [alpha]\') [X]\n\nso that\n\n[alpha] = [T]^t [alpha]\'\n\nor\n\n[alpha]\' = [T]^{-t} [alpha].\n\nConclusion:\n-----------\n\nThe only thing you really need to understand in order to answer your\noriginal question is that\n\n<T^t(alpha),X> = <alpha,T(X)>\n\nis not a derived result, but is actually the definition of T^t. From\nthis, everything Pierre and I said follows automatically.\n\nI hope this helps!\n\n> Sorry if I\'m being obtuse =)\n>\nNo worries. Believe me. I\'ve been there :)\n\nBest wishes,\nEric\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>"Flip Tomato" <flipt@stanford.edu> wrote:
> Yes, I understand that the inner product must be invariant, but why is it
> that it must transform this way:
>
> <T^t(\alpha),X> = <\alpha,T(X)>.
>
> i.e. I have a one form \alpha (sorry, I am not LaTeX enabled). I represent
> the action of this one form on a vector X as <\alpha, X>. If I transform X to
> T(X), then why is it that \alpha is transformed as T^t(\alpha).
>
> i.e. I understand that the inner product with a transformed vector and
> untransformed one-form should yield the same 0-form as the inner product
> with an untransformed vector and a transformed one-form. I just do not
> understand why the transformed one-form takes the form (T^t(\alpha)) that it
> does.
>
> -Philip
Hi Flip,
I wrote the following message on July 14, 2004, but it didn't seem to
make it through. I will try again today (July 19, 2004).
>> If you have any map T that takes a vector field and returns another
>> vector field, you can define a transpose map T^t that takes 1-forms
>> and returns 1-forms via
>>
>> <T^t(\alpha),X> = <\alpha,T(X)>.
>>
> This is exactly the proposition which confsues me. Pierre's explanation
> (inner product must be invariant) is nice intuitively
>
Well, what Pierre was talking about was not the inner product, it was
the evaluation map. Pierre might have assumed that you knew that (once
you have chosen a basis) a 1-form can (although I am not a fan of the
idea) be expressed as a row vector, whereas a vector field can be
expressed as a column vector. When he wrote
p^t v,
the p^t was meant to denote a 1-form (row vector) and v was a vector
field (column vector) and p^t v is matrix multiplication.
> but I don't see how
> you motivate this step mathematically. What is the property of the inner
> product or the nature of the one-form that dictates that one forms tranform
> T^t?
>
I'll try to explain in two different ways what is going on.
First way:
----------
Express everything in coordinates...
Let e_u be a coordinate basis for vector fields (in a coordinate
chart) and dx^u be a coordinate basis for 1-forms such that
<dx^u,e_v> = dx^u(e_v) = \delta^u_v[/itex].
In this basis, the vector field X can be expressed as
Let's say we want to find the special map S of 1-forms such that it
satisfies the relation
<S(\alpha),X> = <\alpha,T(X)>,
which when written explicitly in coordinates is
(S_v^u \alpha_u) X^v = \alpha_u (T^{u_v} X^v).
Comparing both sides we see that in order for the relation to be
satisfied, we must have
S_v^u = T^{u_v}.
[Note: The indices should really be slightly shifted relative to one
another, e.g. S_v^u = S_{v }^{ u} and T^{u_v} = T^{u }_{ v}.]
Any map that relates to T in this way is usually denoted T^t and is
called the transpose of T. The standard way to write this would be
<T^t(\alpha),X> = <\alpha,T(X)>.
Wait!! See, I didn't really want to express this in coordinates
because coordinates are nasty. The only reason I went through all of
this is because you seem to have thought that
<T^t(\alpha),X> = <\alpha,T(X)>
was a "derived" result where you assume we already know what ^t means.
But in fact, the above relation is the DEFINITION of transpose. It is
not a derived result. It is a definition of T^t. (Redundant enough?
:))
I should also remark that although it is called the transpose, we are
not really dealing with matrices, i.e. square blocks of numbers. I
personally prefer this way of dealing with tensors because matrices
are somewhat limited, e.g. good luck representing the Riemannian
curvature as a matrix :)
Once you understand that
<T^t(\alpha),X> = <\alpha,T(X)>
is actually the definition of transpose, I think the answer to the
rest of your question should become clear. If not, let us know! :)
Ok. Here is the...
Second Way:
-----------
This way uses matrices and I do not recommend it, but maybe it will
help. Once you choose a coordinate basis, you can (although I don't
recommend it) think of (as I said) a 1-form \alpha as a row vector
[\alpha]^t and a vector field X as a column vector [X]. To make it
clear which is which, I will explicitly put a {}^t on 1-forms.
Therefore, I will denote the 1-form \alpha above as
[\alpha]^t
and the vector field X above as
[X].
The square brackets denote that the object is a square block of
numbers (0-forms actually).
The evaluation can then be written as Pierre had it
<\alpha,X> = [\alpha]^t [X].
Keep in mind, this is NOT the inner product (although it looks like it
and a theorem says that it is related to an inner product with a dual
element).
Now let [X] = [T][Y] for some matrix [T] and column vector [Y] so that
Now let X' be the same vector field X expressed in terms of a new
coordinate system with the change of coordinates given by a true
matrix [T], i.e.
[X]' = [T][X].
Similarly, [\alpha]'^t is the row vector representation of the same
1-form in a new basis. We all agree that the evaluation doesn't depend
on which coordinates we use so that
The only thing you really need to understand in order to answer your
original question is that
[itex]<T^t(\alpha),X> = <\alpha,T(X)>
is not a derived result, but is actually the definition of T^t. From
this, everything Pierre and I said follows automatically.
I hope this helps!
> Sorry if I'm being obtuse =)
>
No worries. Believe me. I've been there :)
Best wishes,
Eric
Pierre Asselin
Jul22-04, 05:16 AM
<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>\nFlip Tomato <flipt@stanford.edu> wrote:\n\n\n> Yes, I understand that the inner product must be invariant, but why is it\n> that it must transform this way:\n\n> <T^t(alpha),X> = <alpha,T(X)>.\n\n> [ ... ]\n> i.e. I understand that the inner product with a transformed vector and\n> untransformed one-form should yield the same 0-form as the inner product\n> with an untransformed vector and a transformed one-form. I just do not\n> understand why the transformed one-form takes the form (T^t(alpha)) that it\n> does.\n\nOh. Well, that would be because, in matrix notation, your <alpha, T(X)>\nbecomes\nalpha^t * (T * X)\n=( alpha^t * T) * X\n=( alpha^T * (T^t)^t ) * X\n= (T^t * alpha)^t * X\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>Flip Tomato <flipt@stanford.edu> wrote:
> Yes, I understand that the inner product must be invariant, but why is it
> that it must transform this way:
> <T^t(\alpha),X> = <\alpha,T(X)>.
> [ ... ]
> i.e. I understand that the inner product with a transformed vector and
> untransformed one-form should yield the same 0-form as the inner product
> with an untransformed vector and a transformed one-form. I just do not
> understand why the transformed one-form takes the form (T^t(\alpha)) that it
> does.
Oh. Well, that would be because, in matrix notation, your <\alpha, T(X)>
becomes
\alpha^t * (T * X)=( \alpha^t * T) * X=( \alpha^T * (T^t)^t ) * X= (T^t * \alpha)^t * X