[SOLVED] Group Theory For Dummies


by climbhi
Tags: dummies, solved, theory
Hurkyl
Hurkyl is offline
#91
Jun25-03, 04:27 PM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,101
More talking with him indicates he may have been simplifying quite a bit when he brought up Maxwell EM. I'll let someone else explain what "gauge theory" means in general; I'm presuming I'll understand the ramifications after I work through the EM exercise, but I haven't done that yet. [:)]
chroot
chroot is offline
#92
Jun25-03, 05:03 PM
Emeritus
Sci Advisor
PF Gold
chroot's Avatar
P: 10,424
Just to help motivate the thread, I'll find su(n).

[size=large]Lie algebra of U(n)[/size]

First, as a reminder, we know that U(n) is the unitary group of n x n matrices. You should program the word 'unitary' into your head so it reminds you of these conditions:

1) Multiplication by unitary matrices preserves the complex inner product: <Ax, Ay> = <x, y> = [sum]i xi* yi, where A is any member of U(n), x and y are any complex vectors, and * connotes complex conjugation.

2) A* = A-1

3) A* A = I

4) |det A| = 1

Now, to find u(n), the Lie algebra of the Lie group U(n), I'm going to follow Brian Hall's work on page 43 of http://arxiv.org/math-ph/0005032

Recall that we can represent any1 member of a matrix Lie group G by an exponentiation of a member of its Lie algebra g. In other words, for all U in U(n), there is a u in u(n) such that:

exp(tu) = U

where exp is the exponential mapping defined above. Thus exp(tu) is a member of U(n) when u is a member of u(n), and t is any real number.

Now, given that U* = U-1 for member of U(n), we can assert that

(exp(tu))* = (exp(tu))-1

Both sides of this equation can be simplified. The left side's conjugation operator can be shown to "fall through" the exponential, and the left side is equivalent to exp(tu*). Similarly, the -1 on the right side falls through, and the right side is equivalent to exp(-tu). (Exercise: it's easy and educational to show that the * and -1 work this way.) We thus have a simple relation:

exp(tu*) = exp(-tu)

As Hall says, if you differentiate this expression with respect to t at t=0, you immediately arrive at the conclusion that

u* = -u

Matrices which have this quality are called "anti-Hermitian." (the "anti" comes from the minus sign.) The set of n x n matrices {u} such that u* = -u is the Lie algebra of U(n).

Now how about su(n)?

[size=large]Lie algebra of SU(n)[/size]

SU(n) is a subgroup of U(n) such that all its members have determinant 1. How does this affect the Lie algebra su(n)?

We only need to invoke one fact, which has been proven above. The fact is:

det(exp(X)) = exp(trace(X))

If X is a member of a Lie algebra, exp(X) is a member of the corresponding Lie group. The determinant of the group member must be the same as e raised to the trace of the Lie algebra member.

In this case, we know that all of the members of SU(n) have det 1, which means that exp(trace(X)) must be 1, which means trace(X) must be zero!

You can probably see now how su(n) must be. Like u(n), su(n) is the set of n x n anti-Hermitian matrices -- but with one additional stipulation: members of su(n) are also traceless.

1You can't represent all group members this way in some groups, as has been pointed out -- but it's true for all the groups studied here.

- Warren

edit: A few very amateurish mistakes. Thanks, lethe, for your help.
marcus
marcus is online now
#93
Jun26-03, 10:00 AM
Astronomy
Sci Advisor
PF Gold
marcus's Avatar
P: 22,797
The weather's been pretty hot and chroot's derivation of su(n) is really neat and clear so I'm thinking I will just be shamelessly lazy and quote Warren with modifications to get sl(n, C).

I see that he goes along with Brian Hall and others in using lower case to stand for the Lie Algebra of a group written in upper case. So su(n) is the L.A. that belongs to SU(n).

In accord with that notation, sl(n,C) is the L.A. that goes with the group SL(n,C), which is just the n x n complex matrices with det = 1. Unless I am overlooking something, all I have to do is just a trivial change in what Warren already did:

Originally posted by chroot, with minor change for SL(n, C)


[size=large]Lie algebra of SL(n, C)[/size]

SL(n, C) is a subgroup of GL(n, C) such that all its members have determinant 1. How does this affect the Lie algebra sl(n, C)?

We only need to invoke one fact, which has been proven above. The fact is:

det(exp(X)) = exp(trace(X))

If X is a member of a Lie algebra, exp(X) is a member of the corresponding Lie group. The determinant of the group member must be the same as e raised to the trace of the Lie algebra member.

In this case, we know that all of the members of SL(n, C) have det 1, which means that exp(trace(X)) must be 1, which means trace(X) must be zero!
...sl(n, C) is the set of n x n complex matrices but with one additional stipulation: members of sl(n, C) are...traceless.
That didnt seem like any work at all. Even in this heat-wave.
Hurkyl said to give the L.A. of SO(3,1) so maybe i should do that to so as not to look like a slacker. Really like the clarity of both Hurkyl and Chroot style.

I guess Lethe must have raised the "topologically connected" issue. For a rough and ready treatment, I feel like glossing over manifolds and that but it is nice to picture how the det = 0 "surface" slices the GL group into two chunks...

Because "det = 0" matrices, being non-invertible, are not in the group!


...so that only those with det > 0 are in the "connected component of the identity". The one-dimensional subgroups generated by elements of the L.A. are like curves radiating from the identity and they cannot leap the "det = 0" chasm and reach the negative determinant chunk.

Now that I think of it, Lethe is here and he might step in and do SO(3,1) before I attend to it!
marcus
marcus is online now
#94
Jun26-03, 12:27 PM
Astronomy
Sci Advisor
PF Gold
marcus's Avatar
P: 22,797
Hurkyl has a notion of where to go. I want to follow the hints taking shape here:
***********
....But I did talk to one of my coworkers and got a three hour introductory lecture on Lie Groups / Algebras in various contexts, and I think going down the differential geometry route would be productive (and it allows us to keep the representation theory in the representation theory thread!)... I think we are almost at the point where we can derive Maxwellean Electrodynamics as a U(1) gauge theory (which will motivate some differential geometry notions in the process), but I wanted to work out most of the details before introducing that.

Anyways, my coworker did suggest some things to do in the meanwhile; we should finish deriving the Lie algebras for the other standard Lie groups, such as SU(2), SL(n; C), SO(3, 1)... so I assign that as a homework problem for you guys to do in this thread!
***********
the suggestion is----discuss SO(3,1) and so(3,1). Then back to Hurkyl for an idea about the next step. Lets go with that.

Originally posted by chroot, changed to be about SO(3,1)


[size=large]Lie algebra of SO(3,1)[/size]

SO(3,1) is just the group of Special Relativity that gets you to the moving observer's coordinates---it contains 4x4 real matrices that preserve a special "metric" dx2 + dy2 + dz2 - dt2

to keep the space and time units the same, distance is measured in light-seconds----or anyway time and distance units are made compatible so that c = 1 and I dont have to write ct everywhere and can just write t.

This "metric" is great because light-like vectors have norm zero. So the definition that a matrix in this group takes any vector to one of the same norm means that light-like stays light-like!

All observers, even those in relative motion, agree about what is light-like---the world line of something going that speed. (Another way of saying the grandfather axiom of SR that all agree about the speed of light.)

the (3,1) indicates the 3 plus signs followed by the 1 minus sign in the "metric".

So we implement the grand old axiom of SR by having this special INNER PRODUCT* in our 4D vector

1) Multiplication by SO(3,1) matrices preserves the special inner product: <Ax, Ay> = <x, y> = [sum]*i xiyi, where A is any member of SO(3,1), x and y are any real 4D vectors, and * is a reminder that the last term in the sum gets a minus sign.

2) This asterisk notation is a bit clumsy and what Brian Hall does instead is define a matrix g which is diag(1,1,1,-1).

g looks like the 4 x 4 identity except for one minus sign
BTW notice that g-1 = g
and also that gt = g

and he expresses the condition 1) by saying

At g A = g

[[[[to think about.....express <x,y> as xtg y
express <Ax, Ay> as xt At g A y]]]]

3) Then he manipulates 2) to give
g-1 At g = A-1

.....multiply both sides of 2) on the left by g-1
to give
g-1 At g A= I

then multiply both sides on the right by A-1.....

4) then---ahhhh! the exponential map at last----he writes 3) using a matrix A = expX, and solves for a condition on X


g-1 At g = g-1 exp(Xt) g = exp(g-1 Xt g ) = exp(- X) = A-1

the only way this will happen is if X satisfies the condition
g-1 Xt g = -X

it is something like what we saw before with SO(n) except gussied up with g, so it is not a plain transpose or a simple skew symmetric condition. also the condition is the same as

g Xt g = -X

because g is equal to its inverse.
Better post this and proofread later.

.
BTW multiplying by g on right and left like that does not change trace, so as an additional check

trace(X) = trace(g Xt g) = trace( -X) = - trace (X)

showing that trace (X) = 0

so now we know what matrices comprise so(3,1)

they are the ones that satisfy

g Xt g = -X
Lonewolf
Lonewolf is offline
#95
Jun26-03, 03:13 PM
P: 333
Not sure how relevant this is to where the thread is going, but I didnít want people to think Iíd given up on it.

The Heisenberg Group

The set of all upper triangular 3x3 matrices with determinant 1 coupled with matrix multiplication forms a group known as the Heisenberg Group, which will be denoted H. The matrices A in H are of the form

(1 a b)
(0 1 c)
(0 0 1)
where a,b,c are real numbers.

If A is in the form above, the inverse of A can be computed directly to be

(1 -a ac-b)
(0  1 -c    )
(0  0  1    )
H is thus a subgroup of GL(3:R)

The limit of all matrices in the form of A is again in the form of A. (This bit wasnít as clear to me as the text indicated. Can someone help?)

The Lie Algebra of the Heisenberg Group

Consider a matrix X such that X is of the form

(0  d  e)
(0  0  f)
(0  0  0)
then exp(X) is a member of H.

If W is any matrix such that exp(tW) is of the form of matrix A, then all of the entries of W=d(exp(tW))/dt at t=0 which are on or below the diagonal must be 0, so W is of the form X.

Apologies for the possible lack of clarity. I kinda rushed it.
Hurkyl
Hurkyl is offline
#96
Jun29-03, 08:21 PM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,101
I don't think I'll have time over the next week or so to prepare anything, so it'd be great if someone else can introduce something (or pose some questions) for a little while!
marcus
marcus is online now
#97
Jul1-03, 12:18 AM
Astronomy
Sci Advisor
PF Gold
marcus's Avatar
P: 22,797
Originally posted by Hurkyl
I don't think I'll have time over the next week or so to prepare anything, so it'd be great if someone else can introduce something (or pose some questions) for a little while!
Hey Warren, any ideas?
Maybe we should hunker down and wait till
Hurkyl gets back because he seemed to give the
thread some direction. But on the other hand
we dont want to depend on his initiative to the
point that it is a burden! What should we do?

I am thinking about the Lorentz group, or that thing SO(3,1)
I discussed briefly a few days ago.
Lonewolf is our only audience. (in part a fiction, but one must
imagine some listener or reader)
Maybe we should show him explicit forms of matrices implementing the Lorentz
and Poincare groups.

It could be messy but on the other hand these are so
basic to relal speciativity. Do we not owe it to ourselves
to investigate them?

Any particular interests or thoughts about what to do?
marcus
marcus is online now
#98
Jul1-03, 01:01 AM
Astronomy
Sci Advisor
PF Gold
marcus's Avatar
P: 22,797
If we were Trekies we might call it "the Spock algebra of the Klingon group" or if we were on firstname basis with Sophus Lie and Hendrik Lorentz we would be talking about
"the Sophus algebra of the Hendrik group"
such solemn name droppers... Cant avoid it.

Anyway I just did some scribbling and here it is. Pick any 6 numbers a,b,c,d,e, f
This is a generic matrix in the Lie algebra of SO(3;1):

0   a  b  c
-a  0  d  e
-b -d  0  f
c   e  f  0
what I did was take a line from preceding post (also copied below)
g-1 Xt g = -X

remember that g is a special diagonal matrix diag(1,1,1,-1)

and multiply on both sides by g to get
Xt g = -gX

that says that X transpose with ritemost colum negged
equals -1 times the original X with its bottom row negged.

This should be really easy to see so I want to make it that way.
Is this enough explanation for our reader? Probably it is.

But if not, let's look at the original X with its bottom row negged


0   a  b  c
-a  0  d  e
-b -d  0  f
-c -e  -f  0
And let's look at the transpose with its ritemost column negged


0  -a  -b  -c
a   0  -d  -e
b   d   0  -f
c   e   f   0
And just inspect to see if the first is -1 times the second.
It does seem to be the case.

Multiplying by g on left or right does things either to the
bottom row or the rightmost column, I should have said at the beginning---and otherwise doesnt change the matrix.

Ahah! I see that what I have just done is a homework problem in Brian hall's book. It is exercise #7 on page 51, "write out explicitly the general form of a 4x4 real matrix in so(3;1)



Originally a chroot post but changed to be about SO(3;1)


[size=large]Lie algebra of SO(3;1)[/size]

SO(3;1) is just the group of Special Relativity that gets you to the moving observer's coordinates---it contains 4x4 real matrices that preserve a special "metric" dx2 + dy2 + dz2 - dt2

to keep the space and time units the same, distance is measured in light-seconds----or anyway time and distance units are made compatible so that c = 1 and I dont have to write ct everywhere and can just write t.

1) Multiplication by SO(3;1) matrices preserves the special inner product: <Ax, Ay> = <x, y> = [sum]*i xiyi, where A is any member of SO(3,1), x and y are any real 4D vectors, and * is a reminder that the last term in the sum gets a minus sign.

2) This asterisk notation is a bit clumsy and what Brian Hall does instead is define a matrix g which is diag(1,1,1,-1).

g looks like the 4 x 4 identity except for one minus sign
BTW notice that g-1 = g
and also that gt = g

and he expresses the condition 1) by saying

At g A = g

3) Then he manipulates 2) to give
g-1 At g = A-1

4) then---ahhhh! the exponential map at last----he writes 3) using a matrix A = expX, and solves for a condition on X

g-1 At g = g-1 exp(Xt) g = exp(g-1 Xt g ) = exp(- X) = A-1

the only way this will happen is if X satisfies the condition
g-1 Xt g = -X

it is something like what we saw before with SO(n) except gussied up with g, so it is not a plain transpose or a simple skew symmetric condition. also the condition is the same as

g Xt g = -X

Hurkyl
Hurkyl is offline
#99
Jul6-03, 08:44 PM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,101
I've been trying to devise a good way to introduce differential manifolds...

(by that I mean that I hate the definition to which I was introduced and I was looking for something that made more intuitive sense!)

I think I have a way to go about it, but it dawned on me that I might be spending a lot of effort over nothing, I should have asked if everyone invovled is comfortable with terms like "differentiable manifold" and "tangent bundle".
marcus
marcus is online now
#100
Jul6-03, 10:19 PM
Astronomy
Sci Advisor
PF Gold
marcus's Avatar
P: 22,797
Originally posted by Hurkyl
I've been trying to devise a good way to introduce differential manifolds...

(by that I mean that I hate the definition to which I was introduced and I was looking for something that made more intuitive sense!)

I think I have a way to go about it, but it dawned on me that I might be spending a lot of effort over nothing, I should have asked if everyone invovled is comfortable with terms like "differentiable manifold" and "tangent bundle".
I like Marsden's chapter 4 very much
"Manifolds, Vector Fields, and Differential Forms"
pp 121-145 in his book----25 pages
His chapter 9 covers Lie groups and algebras, not too
differently from Brian Hall that we have been using.
So Marsden is describing only the essentials.
I will get the link so you can see if you like it.

Lonewolf and I started reading Marsden's chapter 9 before
we realized Brian Hall was even better. So at least two of us
have some acquaintance with the Marsden book.

We could just ask if anybody had any questions about
Marsden chapter 4----those 25 pages----and if not simply
move on.

On the other hand if you have thought up a better way
to present differential geometry and want listeners, go for it!
Here is the url for Marsden.

http://www.cds.caltech.edu/~marsden/bib_src/ms/Book/
marcus
marcus is online now
#101
Jul6-03, 11:46 PM
Astronomy
Sci Advisor
PF Gold
marcus's Avatar
P: 22,797
H., I had another look at Marsden.
His chapter 9 is too hard and the book as a whole is
too hard. It is a graduate textbook.
But maybe his short chapter 4 on manifolds, vector
fields and differential forms is not too hard.
a short basic summary. It seems to me OK.
If you agree then perhaps this is a solution.
We dont have to give the definitions because
they are all summarized for us.

We should proceed only where it will give us pleasure,
and at our own pace, being under no obligation to anyone. If Lonewolf is still around we can provide whatever explanations
he asks for so he can keep up with the party. If we decide
it is time to stop we will stop (substantial ground has already
been covered). I shall be happy with whatever you decide.

I am interested to know if there are any matrix group, lie group,
lie algebra, repr. theory topics that you would like to hit.
E.g. sections or chapters of Brian Hall (or propose some other online text).

I am currently struggling to understand a little about spin foams
but can find no direct connection there to this thread.
Baez has a introductory paper gr-qc/9905087
Hurkyl
Hurkyl is offline
#102
Jul8-03, 08:35 PM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,101
I've been thinking more about my idea of trying to derive Maxwell's equations from the geometry of M4*U(1) (M4=Minowski space)... the way the idea was presented to me, I got the impression it would be an interesting application of lie groups requiring just a minimal amount of differential geometry... but as I've been mulling over what we'd have to do to get there I'm thinking it might actually be an interesting application of differential geometry requiring just a minimal amount of lie groups. [:(]

So basically, I don't know where to go from here!


The way I usually like to learn is to delve a little bit into a subject, then figure out a (possibly almost trivial) concrete example of how the subject can be used to describe "real world" things, and then continue studying deeper into the subject. The problem is I just don't know what "real world" thing we can get to early on. I guess the solution is to just delve deeper into the math before looking back at the real world.
marcus
marcus is online now
#103
Jul9-03, 12:34 AM
Astronomy
Sci Advisor
PF Gold
marcus's Avatar
P: 22,797
Originally posted by Hurkyl

The way I usually like to learn is to delve a little bit into a subject, then figure out a (possibly almost trivial) concrete example of how the subject can be used to describe "real world" things, and then continue studying deeper into the subject. The problem is I just don't know what "real world" thing we can get to early on. I guess the solution is to just delve deeper into the math before looking back at the real world.
I just happened onto a 3 page online account of
"Representation Theory of SL(2,C)"

It is an appendix in an 8 page paper by Perez Rovelli
"Spin Foam Model for Lorentzian General Relativity"

They lifted it from W. Ruhl (1970) "The Lorentz Group and Harmonic Analysis" and some other classical sources like that.

Baez also reviews SL(2,C) rep theory on page 4 of what I think is a great paper he wrote with Barrett, gr-qc/0101107.
That paper Baez and Barrett "Integrability for Relativistic Spin
Networks" is 22 pages but there is already a good bit of grist for the mill in just the first 4 or 5 pages.

If you have other directions in mind, drop a few hints and I will try to come up with source material.

Oh! we had better not forget to go over the irreps of SU(2)
Do you happen to have an online source? That is easier.
What was I thinking of! irreps of SU(2) naturally come well
before one tries SL(2,C).

think of something nice and simple, my brain is fried from spin foams and 10j symbols
Lonewolf
Lonewolf is offline
#104
Jul9-03, 03:46 PM
P: 333
I'm still around. Can someone explain tangent bundles, please. Marsden defines them as the disjoint union of tangent vectors to a manifold M at the points m in M. Am I right in thinking that this gives us a set containing every tangent vector to the manifold, or did I miss something?
lethe
lethe is offline
#105
Jul9-03, 05:08 PM
lethe's Avatar
P: 657
Originally posted by Lonewolf
I'm still around. Can someone explain tangent bundles, please. Marsden defines them as the disjoint union of tangent vectors to a manifold M at the points m in M. Am I right in thinking that this gives us a set containing every tangent vector to the manifold, or did I miss something?
that s almost right. the tangent bundle is every tangent vector at any point of the manifold, along with the manifold itself.

the tangent bundle is itself given the structure of a manifold.
marcus
marcus is online now
#106
Jul9-03, 05:15 PM
Astronomy
Sci Advisor
PF Gold
marcus's Avatar
P: 22,797
Originally posted by Lonewolf
I'm still around. Can someone explain tangent bundles, please. Marsden defines them as the disjoint union of tangent vectors to a manifold M at the points m in M. Am I right in thinking that this gives us a set containing every tangent vector to the manifold, or did I miss something?
Such a good question! Thread would die without a questioner like that---Hurkyl and I, chroot and/or Lethe etc. wouldnt like to just talk to selves. I want to let Hurkyl answer this because will do it in clear orderly reliable manner.

But look, it is a necessary and basic construction! The tangent vectors on a manifold are the most important vectorspacetype things in sight!
And yet each tangentspace at each separate point is different. So at the outset all one has is a flaky disjoint collection of vectorspaces. One HAS to glue it all together into a coherent structure and give it a topology and, if possible, what is even better namely a differential structure.

Imagine a surface with a postagestampsized tangent plane at every point but all totally unrelated to each other. How flaky and awful! But now imagine that with a little thought you can merge all those tangentplanes into a coherent thing-----a dualnatured thing because it is both a linearspace (in each "fiber") and a manifold. Now I am imagining each tangentspace as a one-dimensional token (in reality n-dimensional) and sort of like a hair growing out of the point x in the manifold. All the hairs together making a sort of mat.

And these things generalize----not just tangentspace bundles but higher tensor bundles and Lie algebra bundles. Socalled fiber bundles (general idea). It is a great machine.

A vectorfield is a "section" of the tangent bundle. The graph of a function is a geometrical object and the "graph" of a vectorfield lives in the tangent bundle. A choice of one vector "over" each point x in the manifold. Great way to visualize things.

The problem is how to be rigorous about it!!!! Hurkyl is good at this. You get this great picture but how do you objectify gluing and interrelating all the tangent spaces into a coherent bundle and giving them usable structure.

It turns out to be ridiculously easy. To make a differentiable manifold out of anything you merely need to specify local coordinate charts. The vectorspace has obvious coordinates and around every point in the manifold you have a coordinatized patch so if it is, like, 3D, you have 3 coordinates for the manifold and 3 for the vectorspace. So you have 6 coordinates of a local chart in the bundle.

Charts have to fit together at the overlaps and the teacher wastes some time and chalk showing that the 6D charts for the bundle are smoothly transformable one to the other on overlapping territory------why? because, surprise, the original 3D manifold charts were smoothly compatible on overlaps.

You will see a bit of magic. An innocent looking LOCAL condition
taking almost no time to mention will unexpectedly suffice to make it all coherent. All that is needed is that, at every point, right around that point, the tangent bundle looks like a cartesian product of a patch of manifold and a fixed vectorspace.

what should I do, edit this? delete it? it is attitudinal prep before someone else writes out the definition (if they ever do) of a fiber bundle-----tangent bundle just a special case of fiber bundle
once that is done, can erase this----I dont want to bother editing it since just provisional. Glad yr still around LW
marcus
marcus is online now
#107
Jul9-03, 05:55 PM
Astronomy
Sci Advisor
PF Gold
marcus's Avatar
P: 22,797
Lethe, chroot, Hurkyl
does everybody know the dodge used to represent the lorentz group on infinite dimensional function spaces?

I sort of suspect you all do.

Let's identify at least the proper lorentz group, the connected component of the identity or whatever, with SL(2,C)
and just look at SL(2,C)

to transform a function f(x) using a 2x2 complex matrix all you need to do is process the x with the matrix first before you feed it to the function

in a generalized sense it is SL(2,C) "acting by translation".

x ----> (ax + b)/(cx + d)

f[x] ----> f[(ax + b)/(cx + d)]

Lately when I see representations of SL(2,C) they mostly involve this action on functions by generalized "translation". So I assume its familiar to y'all.

And the vectorspace of the action is infinite dimensional. There isn't just a discrete set of reps labeled by integers or halfintegers, rather a whole slew labeled by the real line (and maybe another parameter as well).

something to prove here, namely that composing
two maps of the form
x ----> (ax + b)/(cx + d)
results in one of that form
and if you do things in the right order it really gives a group representation
maybe someone should state the definition so we could check details like that?
Hurkyl
Hurkyl is offline
#108
Jul9-03, 09:49 PM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,101


Doh, and I thought I had weaseled out of explaining fiber bundles!

Disclaimer: the terminology "fiber bundle" is somewhat new to me... although I've known the idea behind them for quite a while. It is possible I have some subtle detail wrong about them. (really, studying this whole subject with rigor is new to me... I just had the good fortune to have gotten partway through Brian Hall's text and another one on differential geometry before this discussion started! It's fortunate that much or even all of differential geometry is ideas we've had since our calculus days... just phrased in nifty ways that make them precise, easy to use, and generalizable)


Let's start with something mundane. Everyone remember studying calculus of a single variable? [:)]

In this context we were primarily concerned with functions in the domain of the real numbers and range in the real numbers. Real numbers have some very nice properties (i.e. they form a differentiable manifold), and we were primarily concerned with how the nice properties of the domain and range interacted with our single variable functions. For example, a continuous function is (essentially) one that preserves the property of nearness in the domain and range. A differentiable function relates the differential structure of the domain to that of the range.

So, our studies began by considering certain functions that map R to R. The "total space" of functions was simply R*R; the set that contains all possible graphs of functions.


Eventually our studies became more sophisticated. We no longer considered x an independant variable and y a dependant variable, but we treated x and y both as fully fledged variables. It became interesting to study the structure of R*R as a differentiable manifold in its own right (though, of course, we didn't call it that!).


These same ideas can be applied to any 2 differentiable manifolds; all of our ideas from calculus still apply (but may be a little tricker) when we're considering functions from, say, R2 to SO(3)! And just like in the single variable case, it pays to also be able to consider the total space as a structure on its own merit.


A fiber bundle is the abstraction that covers the above notions. We have a manifold M (analogous to a domain) and a fiber F (analogous to a range). We consider a "total space" E which looks locally like M*F. More precisely, that means we have a projection mapping &pi; that projects E onto M... so that for any neighborhood U of M, &pi;-1(U) is isomorphic (or isometric or diffeomorphic or whatever) to U*F. That is, the set of all points in E that map onto U must have the same structure as U*F.

Why do we only ask this to hold locally? Why not just take the total space to be equal to U*F? (that would be called a trivial bundle... and incidentally, the local isomorphisms from E to U*F are called trivializations) Well, it helps to consider what appears to be everyone's favorite first nontrivial fiber bundle... the mobius strip!

Recall the procedure for making a mobius strip: you take a long thin rectangle of paper, you loop it around into a cylinder, but then before pasting the ends together so it really is a cylinder, you twist the strip one time, and you get a mobius strip.

But what does this have to do with fiber bundles?

Well, we can describe the strip as being:

S = [0, 10]*[-1, 1]

"Looping" the strip means topologically identifying the two ends of the strip {0}*[-1, 1] and {10}*[-1, 1]. The "domain" (aka base manifold) here is [0, 10] with 0 and 10 identified; this is just the circle S1. The fiber is [-1, 1]. This pasting gives us the cylinder S1*[-1, 1]... a trivial fiber bundle.

But wait! There are two (topologically distinct) ways we can map [-1, 1] onto [-1, 1]; instead of using the identity map f(x)=x, use the map f(x)=-x. This corresponds to the "twist" we use when making a mobius strip. This fiber bundle is (obviously) different from the cylinder. Describing the mobius strip as S1*[-1, 1] does not work anymore...

Presume it does work. Then define f from S1 to [-1, 1] as:

f(x) = 1

This is a constant function, and clearly continuous... but it is not a continuous function on the mobius strip! (please ignore the minor details that would take too much effort to patch this into a perfectly rigorous demonstration)


A "section" is simply a fancy name for a graph of a function! But recall that the total space E cannot always be decomposed into M*F... so "section" only coincides with "graph" locally. More precisely, a section S is simply a surface in the total space E such that for every point x on the manifold M, there is a unique point y on S such that &pi;(y) = x.



For a n-dimensional manifold M, the tangent bundle is a special fiber bundle where the fiber is Rn; for any point x on M, &pi;-1(x) = TMx. In other words, the total space E is simply the collection of the tangent spaces for every point on the manifold. The associated projection map &pi; takes each tangent space onto the point on the manifold to which it is tangent. (alternatively, &pi;-1 inverse takes each point on the manifold to its associated tangent space)


Question: is the tangent bundle a trivial bundle? In other words, is the tangent bundle TM diffeomorphic to M*Rn? (I don't know the answer to this one... I presume it is yes, but I haven't tried to prove or disprove it yet)



Anyways, to summarize:

A fiber bundle is a generalization of the cross product; it permits the resulting structure to "twist" as it goes around the base manifold so that the net result does not globally have the same structure as M*F (though it does locally). A section S is the corresponding generalization of a graph of a function; to each point x on the manifold corresponds a unique point y of S, and the projection onto M of y is simply x. A tangent bundle is simply one where the fibers are the tangent spaces of M.

Lonewolf, you were correct (I believe) in your summary of what a tangent bundle is... but the important thing is that bundles also has properties related to the structures of M and F, be it merely topological, differential, metric, or even a linear structure in the case of vector spaces.


Sorry this isn't as clear as my other ones, I hadn't prepared any nice examples and demonstrations of the concepts. [:(]


Register to reply

Related Discussions
[SOLVED] group theory problem Calculus & Beyond Homework 13
[SOLVED] group theory question Calculus & Beyond Homework 6
[SOLVED] group theory question Calculus & Beyond Homework 2
[SOLVED] group theory Calculus & Beyond Homework 4
[SOLVED] group theory Calculus & Beyond Homework 0