Gauge Bosons and Metrics [Archive]

lost.and.lonely.physicist@gmail.com

Feb15-05, 11:48 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>Is it possible that there is some underlying metric from which the\nfields of gauge bosons can be derived? I\'m pretty sure somebody has\nthought about such questions, for otherwise why are objects like del_mu\n+ i e A_mu called "covariant" derivatives in quantum field theory?\n\nIs there perhaps some way we could test such a hypothesis\nexperimentally?\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>Is it possible that there is some underlying metric from which the
fields of gauge bosons can be derived? I'm pretty sure somebody has
thought about such questions, for otherwise why are objects like del_mu+ i e A_{mu} called "covariant" derivatives in quantum field theory?

Is there perhaps some way we could test such a hypothesis
experimentally?

whopkins@csd.uwm.edu

Feb16-05, 11:36 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>lost.and.lonely.physicist@gmail.com wrote:\n> Is it possible that there is some underlying metric from which the\n> fields of gauge bosons can be derived?\n\nThat\'s all old hat and standard textbook material. A metric is:\nh_{mn} = g_{mn}\nh_{mb} = A^a_m k_{ab}\nh_{an} = k_{ab} A^b_n\nh_{ab} = k_{ab}\nwhere k is the (constant) metric associated with the symmetry group;\nA^b_m the components of the gauge field; (a,b,...) the Lie algebra\nindices; (m,n,...) the space-time indices; g the metric for the base\nspace.\n\nA (non-holonomic) basis is given by:\ne_m = d/dx^m - A^a_m Y_a\ne_a = Y_a\nwhere the Y\'s are the basis of the Lie algebra (extended to the lie\ngroup, e.g. as left-invariant or right-invariant fields). Since this\nis non-holonomic, one has non-zero commutators:\n[Y_a, Y_b] = f^c_{ab} Y_c\nwith the f\'s being the structure constants. The metric k is adjoint\ninvariant if it lowers the upper index to make the structure constants\nfully anti-symmetric:\nf_{abc} = k_{ad} f^a_{bc}\nwith f_{abc} = -f_{bac}.\nUnder this condition, the geodesics coincide with the paths given by\nthe combined action of gravity and the Wong equations. The Wong\nequations are the generalization of the Lorentz force law to general\n(non-abelian) symmetry groups; and include a set of Lorentz type force\nequations coupling a particle\'s charge vector e = (e_a) to the field\nF^a_{mn} = d(A^a_m)/dx^n - d(A^a_n)/dx^m + f^a_{bc} A^b_m A^c_n\nForce = e_a F^a_{mn} = e.F_{mn} (as Lie vectors)\nplus an additional equation describing the rotation of the charge\nvector\nde/ds proportional to [e, A_m dx^m/ds]\nwhere A_m is the Lie-vector (A^a_m). The total square charge is\nconserved:\n|e|^2 = k^{ab} e_a e_b.\n\nThe curvature scalar R for the metric h decomposes exactly into a sum\nof the curvature scalar R for the metric g plus an additional term\nproportional to the square of the field F which exactly replicates the\nstress tensor of the gauge fields -- plus a third term proportional to\nthe square of the structure constants which corresponds to a\nCosmological Constant:\nR_h = R_g - 1/4 F^a_{mn} F_a^{mn} + 1/4 f^a_{bc} f_a^{bc}\nindices raised and lowers with the g and k metrics, as needed.\n\n> Is there perhaps some way we could test such a hypothesis\n> experimentally?\n\nThe transformation is mathematical, so it has little physical content\n-- other than the assertion that the fields in question actually be\nYang-Mills fields in the first place, that the metric be adjoint\ninvariant ... which then means that the total square charge of whatever\nfundamental particles are involve should be a constant of motion!\n\nThe fields in the Standard Model are not purely Yang-Mills fields.\nThey\'re a combination of Yang-Mills and Higgs fields. So, the\ngeometric picture presented is incomplete. Completing it with the\naddition of something to account for the Higgs fields (or whatever\nmechanism is to explain mass) is non-trivial.\n\nAlso, given the Yang-Mills part alone, the particles do not have a\nsquare charge invariant. In fact, the old argument Maxwell used for\nthe Displacement Current suddenly pops up here, again, in new guise and\nnew context.\n\nIn order to preserve charge-squared magnitude, one needs to add in an\nextra set of particles for right-handed neutrinos (and left-handed\nantineutrinos); to allow for non-zero neutrino masses (though this is\nnot necessary); to allow for an additional force which couples to the\nquantum number G = (Baryon - Lepton)/2. Then two square invariants\narise:\nI^2 + 3 IR^2 = 3/4; L^2 + 6 G^2 = 3/2\nwhere I^2 is the total isospin for the SU(2) sector (equal to 3/4 for\nleft-handed particles & right-handed anti-particles; 0 for right-handed\nparticles and left-handed anti-particles); IR is Y - G where Y is the\nweak isospin. It\'s the right-handed version of isospin and is equal to\n+/- 1/2 for the I = 0 particles; and 0 otherwise. L^2 is the total\nSU(3) charge and is 4/3 for quarks, 0 for leptons. The Baryon number\nis 1/3 for quarks so G is 1/6. It is 1/2 for leptons.\n\nSo, the closest thing to a bona fide experimental consequence, is the\nprediction of the additional sector for right-neutrinos (left\nanti-neutrinos); the additional force coupling to the Baryon-Lepton\nnumber; and possibly a right-handed analogue of SU(2), which IR is the\nz-component of. Of note is that symmetry breaking in the standard\nmodel has a nice symmetric combination of the key quantum numbers\ncorresponding to the massless photon field:\ncharge (Q) = I3 + IR + G.\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>lost.and.lonely.physicist@gmail.com wrote:
> Is it possible that there is some underlying metric from which the
> fields of gauge bosons can be derived?

That's all old hat and standard textbook material. A metric is:
h_{mn} = g_{mn}h_{mb} = A^{a_m} k_{ab}h_{an} = k_{ab} A^{b_n}h_{ab} = k_{ab}
where k is the (constant) metric associated with the symmetry group;
A^{b_m} the components of the gauge field; (a,b,...) the Lie algebra
indices; (m,n,...) the space-time indices; g the metric for the base
space.

A (non-holonomic) basis is given by:
e_m = d/dx^m - A^{a_m} Y_ae_a = Y_a
where the Y's are the basis of the Lie algebra (extended to the lie
group, e.g. as left-invariant or right-invariant fields). Since this
is non-holonomic, one has non-zero commutators:
[Y_a, Y_b] = f^{c_}{ab} Y_c
with the f's being the structure constants. The metric k is adjoint
invariant if it lowers the upper index to make the structure constants
fully anti-symmetric:
f_{abc} = k_{ad} f^{a_}{bc}
with f_{abc} = -f_{bac}.
Under this condition, the geodesics coincide with the paths given by
the combined action of gravity and the Wong equations. The Wong
equations are the generalization of the Lorentz force law to general
(non-abelian) symmetry groups; and include a set of Lorentz type force
equations coupling a particle's charge vector e = (e_a) to the field
F^{a_}{mn} = d(A^{a_m})/dx^n - d(A^{a_n})/dx^m + f^{a_}{bc} A^{b_m} A^{c_n}
Force = e_a F^{a_}{mn} = e.F_{mn} (as Lie vectors)
plus an additional equation describing the rotation of the charge
vector
de/ds proportional to [e, A_m dx^m/ds]
where A_m is the Lie-vector (A^{a_m}). The total square charge is
conserved:
|e|^2 = k^{ab} e_a e_b.

The curvature scalar R for the metric h decomposes exactly into a sum
of the curvature scalar R for the metric g plus an additional term
proportional to the square of the field F which exactly replicates the
stress tensor of the gauge fields -- plus a third term proportional to
the square of the structure constants which corresponds to a
Cosmological Constant:
R_h = R_g - 1/4 F^{a_}{mn} F_a^{mn} + 1/4 f^{a_}{bc} f_a^{bc}
indices raised and lowers with the g and k metrics, as needed.

> Is there perhaps some way we could test such a hypothesis
> experimentally?

The transformation is mathematical, so it has little physical content
-- other than the assertion that the fields in question actually be
Yang-Mills fields in the first place, that the metric be adjoint
invariant ... which then means that the total square charge of whatever
fundamental particles are involve should be a constant of motion!

The fields in the Standard Model are not purely Yang-Mills fields.
They're a combination of Yang-Mills and Higgs fields. So, the
geometric picture presented is incomplete. Completing it with the
addition of something to account for the Higgs fields (or whatever
mechanism is to explain mass) is non-trivial.

Also, given the Yang-Mills part alone, the particles do not have a
square charge invariant. In fact, the old argument Maxwell used for
the Displacement Current suddenly pops up here, again, in new guise and
new context.

In order to preserve charge-squared magnitude, one needs to add in an
extra set of particles for right-handed neutrinos (and left-handed
antineutrinos); to allow for non-zero neutrino masses (though this is
not necessary); to allow for an additional force which couples to the
quantum number G = (Baryon - Lepton)/2. Then two square invariants
arise:
I^2 + 3 IR^2 = 3/4; L^2 + 6 G^2 = 3/2
where I^2 is the total isospin for the SU(2) sector (equal to 3/4 for
left-handed particles & right-handed anti-particles; for right-handed
particles and left-handed anti-particles); IR is Y - G where Y is the
weak isospin. It's the right-handed version of isospin and is equal to
+/- 1/2 for the I = particles; and otherwise. L^2 is the total
SU(3) charge and is 4/3 for quarks, for leptons. The Baryon number
is 1/3 for quarks so G is 1/6. It is 1/2 for leptons.

So, the closest thing to a bona fide experimental consequence, is the
prediction of the additional sector for right-neutrinos (left
anti-neutrinos); the additional force coupling to the Baryon-Lepton
number; and possibly a right-handed analogue of SU(2), which IR is the
z-component of. Of note is that symmetry breaking in the standard
model has a nice symmetric combination of the key quantum numbers
corresponding to the massless photon field:
charge (Q) = I3 + IR + G.

bjflanagan

Feb16-05, 11:36 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>Have you already looked at Kaluza-Klein theory?\n\nIn his very nice introductory text on \'QFT\' (Cambridge), Ryder provides\na table in the section on the "geometry of gauge fields" which\nhighlights the parallels between gauge theory and GR. I can well\nremember coming across this material years ago and having a very\npleasant "aha" moment.\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>Have you already looked at Kaluza-Klein theory?

In his very nice introductory text on 'QFT' (Cambridge), Ryder provides
a table in the section on the "geometry of gauge fields" which
highlights the parallels between gauge theory and GR. I can well
remember coming across this material years ago and having a very
pleasant "aha" moment.

lost.and.lonely.physicist@gmail.com

Feb16-05, 11:36 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>I understand that the gauge field for U[1] has only 1 index, whereas\nthe Christoffel symbol in GR has 3. The gauge field for SU[3] and SU[2]\nhas 2 indices, one for mu, one for summing over the generators, so they\ndon\'t seem to have enough either.\n\nI guess I should have put "metric" in quotes. What I\'m wondering is\nwhether there is some sort of underlying geometric object from which we\ncan derive the gauge fields. There is some reference to geometry in\nquantum field theory texts, such as the commutator of the covariant\nderivatives giving rise to the field strength. But is the field\nstrength synonymous with some sort of curvature? The latter would\nrequire a metric to intepret what exactly is curving, right?\n\nOn the other hand, in GR the presence of an EM field i.e. U[1] gauge\nfield, does modify the underlying spacetime metric. Here I am recalling\nthe Kerr-Newman metric for a charged rotating black hole. Do these\nnotions - the presence of a "covariant" derivative in qft and the\nmodification of geometry due to the presence of gauge fields - have\nsome sort of relationship?\n\nHave we found any examples where weak or strong interactions have\nmodified the metric of spacetime? If say there is a overall surplus of\nneutrinos over antineutrinos in the universe left over from the Big\nBang, that\'d make our universe weak non-neutral - would that change the\nRobertson-Walker metric?\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>I understand that the gauge field for U[1] has only 1 index, whereas
the Christoffel symbol in GR has 3. The gauge field for SU[3] and SU[2]
has 2 indices, one for \mu, one for summing over the generators, so they
don't seem to have enough either.

I guess I should have put "metric" in quotes. What I'm wondering is
whether there is some sort of underlying geometric object from which we
can derive the gauge fields. There is some reference to geometry in
quantum field theory texts, such as the commutator of the covariant
derivatives giving rise to the field strength. But is the field
strength synonymous with some sort of curvature? The latter would
require a metric to intepret what exactly is curving, right?

On the other hand, in GR the presence of an EM field i.e. U[1] gauge
field, does modify the underlying spacetime metric. Here I am recalling
the Kerr-Newman metric for a charged rotating black hole. Do these
notions - the presence of a "covariant" derivative in qft and the
modification of geometry due to the presence of gauge fields - have
some sort of relationship?

Have we found any examples where weak or strong interactions have
modified the metric of spacetime? If say there is a overall surplus of
neutrinos over antineutrinos in the universe left over from the Big
Bang, that'd make our universe weak non-neutral - would that change the
Robertson-Walker metric?

whopkins@csd.uwm.edu

Feb19-05, 02:48 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>lost.and.lonely.physicist@gmail.com wrote:\n> I understand that the gauge field for U[1] has only 1 index, whereas\n> the Christoffel symbol in GR has 3.\n\nActually, the gauge field has 2 indices, but the second only takes on 1\nvalue. If, in contrast, reverting to a type of "charge with\ncomplexion", where the charge is a vector e=(e1,e2,...) and the\npotential as well V=(V1,V2,...) with an energy term E = e.V (something,\nin fact, not too far away from what Maxwell, himself, described early\non in his treatise when trying to account for how many types of charge\nthere are); then you\'d see the index explicitly.\n\nIf the gauge field is A^a_m, it forms part of a differential form\nA^a = A^a_m dx^m\nand, in turn, is part of a Lie-vector valued differential form:\nA = A^a Y_a\nwhere (Y_1,Y_2,...) are suitable basis vectors in the Lie algebra.\n\nSo the a index is a Lie index; the m index is a spacetime index.\n\nFor the GR coefficient Gamma^m_{np}, the n is the spacetime index; and\nthe (^m_p) is a GL(4) index, corresponding to the basis vectors Y_m^p\nof the transformation group GL(4).\n\nThat\'s why it\'s "3 indices". It\'s really only 2, if you group the m\nand p together.\n\n> I guess I should have put "metric" in quotes. What I\'m wondering is\n> whether there is some sort of underlying geometric object from which\nwe\n> can derive the gauge fields.\n\nAs pointed out in a previous article, those would be directly related\nto the metric for the fibre bundle that combines the gauge group and\nspacetime as the a-m components of the total metric, via:\nh_{am} = k_{ab} A^b_m\nwhere k_{ab} is the a-b component of the total metric, and the metric\nfor the Lie algebra.\n\nAdjoint invariance, as described there, means structure constants\nlowered with the k metric should become totally anti-symmetric. For\nU(1), this imposes no constraint, since the Lie algebra is abelian and\nthe structure constants are 0. For SU(2), the commutators are:\n[Y1,Y2] = Y3; [Y2,Y3]=Y1; [Y3,Y1]=Y2\nso the structure constants would be:\nf^3_{12} = f^1_{23} = f^2_{31} = 1\nf^3_{21} = f^1_{32} = f^2_{13} = -1\nf^a_{bc} = 0 else.\nThe only metric which has the required compatibility with this set is\none of the form:\nk_{ab} = k delta_{ab}.\nThe metrics for any compact simple Lie algebra (e.g. SU(2), SU(3),\netc.) are, likewise, all uniquely determined by this condition up to a\nconstant multiple. For combinations U(1) x SU(2) x SU(3), the metric,\nin general, would be of the form:\nk = g1 k1 + g2 k2 + g3 k3\nfor constants g1, g2, g3; where k1, k2, k3 are the metrics for U(1),\nSU(2), SU(3).\n\n> There is some reference to geometry in quantum field theory texts\n\nFibre Bundles are part of differential geometry. Cocquereaux has an\non-line book which covers the topic thoroughly, for instance. (A web\nsearch should readily find it).\n\n> But is the field\n> strength synonymous with some sort of curvature? The latter would\n> require a metric to intepret what exactly is curving, right?\n\nIn the fibre bundle, the field strength would actually appear in the\nconnection coefficients. So, for instance, the field component\nG^a_{mn} would be related to the a-m-n connection coefficient\nGamma^a_{mn}.\n\nThe gauge theory which has the christoffel coefficients as their field\nstrengths would be a GL(4) gauge theory or -- if adopting the Cartan\nmoving frame basis -- a SO(3,1) gauge theory. That\'s largely separate\nfrom GR.\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>lost.and.lonely.physicist@gmail.com wrote:
> I understand that the gauge field for U[1] has only 1 index, whereas
> the Christoffel symbol in GR has 3.

Actually, the gauge field has 2 indices, but the second only takes on 1
value. If, in contrast, reverting to a type of "charge with
complexion", where the charge is a vector e=(e1,e2,...) and the
potential as well V=(V1,V2,...) with an energy term E = e.V (something,
in fact, not too far away from what Maxwell, himself, described early
on in his treatise when trying to account for how many types of charge
there are); then you'd see the index explicitly.

If the gauge field is A^{a_m}, it forms part of a differential form
A^a = A^{a_m} dx^m
and, in turn, is part of a Lie-vector valued differential form:
A = A^a Y_a
where (Y_1,Y_2,...) are suitable basis vectors in the Lie algebra.

So the a index is a Lie index; the m index is a spacetime index.

For the GR coefficient \Gamma^m_{np}, the n is the spacetime index; and
the (^m_p) is a GL(4) index, corresponding to the basis vectors Y_m^p
of the transformation group GL(4).

That's why it's "3 indices". It's really only 2, if you group the m
and p together.

> I guess I should have put "metric" in quotes. What I'm wondering is
> whether there is some sort of underlying geometric object from which
we
> can derive the gauge fields.

As pointed out in a previous article, those would be directly related
to the metric for the fibre bundle that combines the gauge group and
spacetime as the a-m components of the total metric, via:
h_{am} = k_{ab} A^{b_m}
where k_{ab} is the a-b component of the total metric, and the metric
for the Lie algebra.

Adjoint invariance, as described there, means structure constants
lowered with the k metric should become totally anti-symmetric. For
U(1), this imposes no constraint, since the Lie algebra is abelian and
the structure constants are . For SU(2), the commutators are:
[Y1,Y2] = Y3; [Y2,Y3]=Y1; [Y3,Y1]=Y2
so the structure constants would be:
f^{3_}{12} = f^{1_}{23} = f^{2_}{31} = 1f^{3_}{21} = f^{1_}{32} = f^{2_}{13} = -1f^{a_}{bc} = else.
The only metric which has the required compatibility with this set is
one of the form:
k_{ab} = k \delta_{ab}.
The metrics for any compact simple Lie algebra (e.g. SU(2), SU(3),
etc.) are, likewise, all uniquely determined by this condition up to a
constant multiple. For combinations U(1) x SU(2) x SU(3), the metric,
in general, would be of the form:
k = g1 k1 + g2 k2 + g3 k3
for constants g1, g2, g3; where k1, k2, k3 are the metrics for U(1),
SU(2), SU(3).

> There is some reference to geometry in quantum field theory texts

Fibre Bundles are part of differential geometry. Cocquereaux has an
on-line book which covers the topic thoroughly, for instance. (A web
search should readily find it).

> But is the field
> strength synonymous with some sort of curvature? The latter would
> require a metric to intepret what exactly is curving, right?

In the fibre bundle, the field strength would actually appear in the
connection coefficients. So, for instance, the field component
G^{a_}{mn} would be related to the a-m-n connection coefficient
\Gamma^a_{mn}.

The gauge theory which has the christoffel coefficients as their field
strengths would be a GL(4) gauge theory or -- if adopting the Cartan
moving frame basis -- a SO(3,1) gauge theory. That's largely separate
from GR.

lost.and.lonely.physicist@gmail.com

Mar7-05, 03:48 PM

<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>whopkins@csd.uwm.edu wrote:\n> Fibre Bundles are part of differential geometry. Cocquereaux has an\n> on-line book which covers the topic thoroughly, for instance. (A web\n> search should readily find it).\n\nIs it this?\n\nhttp://www.cpt.univ-mrs.fr/~coque/linkxtoxbook.html\n\nIt seems to be in French, which I do not know, unfortunately.\n\nThis and your follow-up post seems quite fascinating, but I know too\nlittle about differential geometry and group theory to understand it.\nIs there some book that discusses what you\'ve written here? I have T.\nFrankel\'s Geometry of Physics - is that any good?\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>whopkins@csd.uwm.edu wrote:
> Fibre Bundles are part of differential geometry. Cocquereaux has an
> on-line book which covers the topic thoroughly, for instance. (A web
> search should readily find it).

Is it this?

http://www.cpt.univ-mrs.fr/~coque/linkxtoxbook.html

It seems to be in French, which I do not know, unfortunately.

This and your follow-up post seems quite fascinating, but I know too
little about differential geometry and group theory to understand it.
Is there some book that discusses what you've written here? I have T.
Frankel's Geometry of Physics - is that any good?

Aaron Bergman

Mar8-05, 12:49 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>In article <1110161623.965529.247020@l41g2000cwc.googlegroups .com>,\n"lost.and.lonely.physicist@gmail.com"\n<lo st.and.lonely.physicist@gmail.com> wrote:\n\n> This and your follow-up post seems quite fascinating, but I know too\n> little about differential geometry and group theory to understand it.\n> Is there some book that discusses what you\'ve written here? I have T.\n> Frankel\'s Geometry of Physics - is that any good?\n\nI\'m not particularly fond of it. I don\'t remember if it covers what\nyou\'re looking for. I\'d try M. Nakahara _Geometry, Topology and Physics_\nor, possibly, a book by C. Nash and S. Sen _Topology and Geometry for\nPhysicists_.\n\nAaron\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>In article <1110161623.965529.247020@l41g2000cwc.googlegroups. com>,
"lost.and.lonely.physicist@gmail.com"
<lost.and.lonely.physicist@gmail.com> wrote:

> This and your follow-up post seems quite fascinating, but I know too
> little about differential geometry and group theory to understand it.
> Is there some book that discusses what you've written here? I have T.
> Frankel's Geometry of Physics - is that any good?

I'm not particularly fond of it. I don't remember if it covers what
you're looking for. I'd try M. Nakahara _Geometry, Topology and Physics_
or, possibly, a book by C. Nash and S. Sen _Topology and Geometry for
Physicists_.

Aaron