So, this was the transformation we worked
out by demanding, so this came out of the
requirement 
that the covariant derivative of phi which
we defined to be D mu of phi minus i A mu
of phi should transform nicely. So by that
what did we mean; we meant D mu prime of phi
prime should be 
and this sort of gave us something very nice
transformation for A this is how this has
to this. So, and the key point to remember
is I am using script A to remind you that
this is a matrix valued object, so that is
one thing and this can be rewritten in slightly
different forms. For instance I could try
to put the d mu on top of this.
And the way you do that is by just taking
equal to identity which I write as 1 and take
d mu of this. So, what this tells you is that
d mu of g dot g inverse plus g dot 
because identity is a constant matrix. I think
I wrote dagger here; I will keep switching
back and forth, but in unity representations
g and g inverse and g dagger are the same.
So, this tells you that you can see that this
is exactly this quantity but this can be rewritten.
So, you can go back and forth between these
two things but there is something nice about
writing it in the second form which you will
see pretty soon. If I do that it will just
correspond to putting the d mu on this and
changing the sign; that is all that it corresponds
to but you can something nice happening that
you see that A mu prime is equal to g plus
I; it is kind of something like this, maybe
it is useful pneumonic to remember things.
Otherwise, I mean there is nothing special
about this over this. Now coming back to this
object the thing is suppose we looked at things
under global transformations.
So let us assume what happens. So, the question
is how does A transform? The nice guess would
be I mean you would have said oh, it should
be invariant. So, what do the global imply?
g is derivatives so g should vanish but this
term will not go away. So, it is very important
to realize that A mu prime is not invariant
but covariant. Unlike the case of the abelian
or u one case what happened there it was invariant.
t a mu prime was equal to a mu and that you
can see happened when things are abelian or
communicating, g can go through and cancel
the g inverse; everything commutes so then
it is invariant. So, this is the natural generalization.
So, what one says is that this transforms
covalently. This is generalizing what we know
for normal transformations on space and time;
we say if something that transforms nicely,
we say it transforms covalently. It does not
have to be ah covariant vector, even a contravariant
vector transforms nicely; you do not say contract
variantly or whatever, you say covalently
for everything; that means it transforms nicely
under global SU 2 transformations.
In fact the key thing as I mentioned last
time is that will you write this only thinking
keeping taking g to be SU 2 in mind but actually
if you look through none of the calculations
I really needed anything very deeply I mean
which made use a particular properties of
SU 2. So, all these things actually go through.
You do not have to believe me; just sit quietly
in your room or wherever and convince yourself
that that is indeed true, just go back through
the alternates. So, in fact 
one can show that actually this kind 
of transformation 
is that of the adjoint representation; we
haven’t defined this but I am just mentioning
it of the Lie group. So, the adjointable representation
has one property; it is dimension is the same
as the dimension of the lie group, not the
rank, so the dimension of the lie group.
So, another point which may not have been
clear is the following thing is this was a
claim and I leave it as a claim because I
know I have given enough quasi proof but not
a real proof. Claim is that A mu is Lie algebra
value; by that what do I mean? Remember Lie
algebra is a vector space and the generators
of that thing provide a basis. So, what that
says that A mu can be written as. Since we
are doing we are discussing SU 2, it will
be. So, coming back to this then we have to
look at something like this and ask is this
also Lie algebra value. Of course, it follows
that if this is true this has to be true.
It is like saying that let us look at this
transformation; this is Lie algebra value
this is also Lie algebra valued so this should
be. But the thing is but this is something
you can check explicitly and I sort of indicated
last lecture how this is done to first order
and said it in word but I think it is worth
seeing of what happens to second order and
so let us do that.
And so first thing is we already worked out
what g mu would be g mu d mu of e power i
and we expand this and we get first term is
if you notice I have put t a instead of half
sigma a because again it 
is not so important. So, this is what we get
to second order keeping terms up to second
order and so now next thing I need to do is
to multiply by g inverse. Right now this T
a T b is not Lie algebra value or anything
but we should see; what we should see is it
would almost be that if this sign were different
one of them because I can always re-enable
things and make it look like this and that
is exactly what we will see will happen. Again
it is useful to put this lambda kind of book
keeping parameter out here and the key point
here is that d mu of g is actually order lambda
and if you want to keep terms order theta
square it is sufficient for me to look at
g inverse only to first order.
So, whatever I have written at the end of
the board times minus. So this will be equal
to, so now I will be going back and forth
from this end of the board to the other end.
So, this term will give you the first term
into one, not a problem. Now so this is multiplying
by theta c from the right side. So, coming
back here that would give you a term with
the d mu theta here with theta on the other
side and that is exactly like this term but
has a minus sign. So, the plus half minus
one will make it a minus half. So, I can just
put everything together minus half mu or there
will be i’s; I am just awful with these
i’s no more terms. Now I have to do some
juggling which basically corresponds to saying
let us go ahead and write, call this guy,
call this b and call this a and then you can
see that this is equal to i.
So, it is the magical combination which is
the commutator comes out and this is now again
in the Lie algebra. So you can see at least,
so this implies that all these terms are in
the Lie algebra. In fact you can also see
that this is the complicated function, so
this would be some function of theta d mu
of theta theta some a and t a and there will
be only one derivative but it can be nonlinear.
So, the next term will have one d mu of theta
and two thetas and you can see it and I can
replace this by the structure constants and
you can massage it look and make it look like
something like that. So, this is in some ways
just like the analog of the b c h formula
which again you can only check order by order
and of course the thing is that you know that
there exist a full formula and the same thing
holds out here but there is something very
geometric about this thing.
So, what this actually tells you is that we
know that g is an element of the group and
if you take SU 2 it is like every point, every
group element is like a point on the S 3 and
what this is doing is looking at small variations.
So, a derivative is like doing a delta g.
So, think of this as your S 3; I pick any
point it does not matter and then sort of
look at a neighborhood of that point epsilon
neighborhood which is all you need to do a
first derivative and you sort of linearize
things, it is called the tangent space. So,
this is like looking at the tangent space
at that particular point and so Lie algebra
is always related is valued in the tangent
space of the group manifold; that is the statement
that this is actually valued in the Lie algebra.
It is a very, very important point.
So, what it says is something it need not
have been if you just did some delta g dot
g inverse for any delta g so by delta g, I
mean so point here is given by some value
for thetas and you change your thetas by take
theta a to theta a plus delta theta a but
this is variably taken to be small and so
you work out what delta g would be in that
for that kind. So, then this object would
be valued in the Lie algebra; that is what
it says. But you can check I mean if you feel
up to it you can check that the third order
piece also works out. I do not know, often
I do not know of a neat way of algebraic way
that is to show that it is always Lie algebra
the algebra valued. So, we have actually accomplished
the goal which we set out which was to start
with a Lagrangian which was globally invariant
and we made it locally invariant.
So, the rule now is very clear, you just do
this and this takes care of it. And in fact
if you have several fields again for each
field every time there is a covariant derivative
they are suitable depending on how that transforms
you write the corresponding this thing a covariant
derivative. It is similar to what we did even
in the u one case, if the charges were different
you put a different q but in the non-abelian
cases what you would do is you would put.
So, in the non-abelian case one chooses what
plays the role of q, any idea? q determines
the representation how it transforms; what
determines in the non-abelian case, what is
it I have to change? So, the thing is that
suppose you have two fields in the u one case
let us say with charge q 1 and charge q 2,
the covariant derivatives depended on the
charge. My question to you is in the non-abelian
case what is the analog of charge? No, thetas
are the parameter even there it is thetas
is the parameters. So, it is just a representation.
It is a representation one chooses the T a‘s
in a suitable representation. So, suppose
so let us go back to the SU 2 and yesterday
we looked at the case where phi was phi 1
and phi 2 in the two dimension representation
and then we wrote the gauge field A mu to
be A mu half sigma a. So, this is in the two
dimensional representation or spin half representation
but I could have chosen a different one. I
could have chosen phi to be in the three dimensional
representation; this is a different phi, let
me call it psi just to in the three dimensional
representation. And then the A mu’s it will
be the these fields are the same but I have
to change the t’s because these are two
by two matrices I need to write the corresponding
three by three guys. We know what that is,
it is a minus I epsilon a b c or whatever.
So then so we would write A mu, I use the
same symbol but really you will see what is
different on this side by T a of b c. I think
it is minus i epsilon a b c. So, what will
remain unchanged is this is the gauge field
that these guy’s the three guys here and
three here they are the same. So, it is similar
to if you remember what happened we wrote
q a mu except q gets replaced really by the
generators of the Lie algebra. The reason
is it is just a number in u 1 is because all
represent an irreducible representations of
u 1 or one dimensional. So, you would use
a one by one matrix which we will not call
a matrix; we usually call it a number or the
charge. So the point here is that if you go
back to the covariant derivative which I will
write again so you can compare.
So in the u 1 case we wrote D mu of phi which
was in some q dimensional representing and
charge q we wrote as d mu minus i q A mu;
this is what we wrote. Now for SU 2 some arbitrary
representation phi is in some arbitrary representation
R, then you would write D mu of phi R where
I am putting this superscript R to indicate
that I have to choose, whatever is the matrix.
So, I gave you two examples but there can
be many many more examples. Are there any
questions? The thing is that what I am saying
here the point here is that for u 1, there
is only one T number one and the Lie algebra
is very trivial t t is zero. So the only way
and it is one dimensional. It is a one by
one matrix which is a number. So, instead
of calling q I could have called it just the
t; I could replace it by t if you wish but
I am just calling the number so that I stick
to my notation.
So, it is just to make the parallel. So, really
how this reduces to this is more what I am
getting at rather than the other way. So,
now we are ready to discuss what the field
strength would be and we just repeat what
we did. What did we do in the u 1 case in
the abelian case? We worked out the commutator.
So let us see, so the analog of field strength.
So, for that we need to compute something
like this and what does this become, what
is this equal to? This is I just need the
factor of i. How did we define that? Was it
i times f mu nu or minus i times. So, I will
define again some matrix valued object dot
phi. Was it minus; there was a q that is okay
but the q is replaced by a t here. So, I am
writing this, yeah t inside the f, so that
is perfect. So, again this analogy is now
useful. So, we just have to compute this and
the calculation is quite simple but there
are some the factor things do not commute
give you extra terms.
So, let us look at this. So, this would be
d mu. I will do things all are quantum mechanics
now and rewrite things because these are matrices.
So I can rewrite this as that is this, last
thing is minus, minus i into minus i is minus
one. First term obviously vanishes because
derivatives commute once acting on smooth
functions so this becomes zero but what does
this give you. So, let us look at this 
d mu of A mu of phi minus A mu of d mu of
phi; this cancels with this only when it acts
on this. So, these are all script A’s. I
am going to erase it in a moment but in your
note books you can. So this gets this and
this is just minus of the other thing. So
what we end up getting, so this term is equal
to.
Now I have to pull out n s, so this implies
that f mu nu. So, I am pulling out a minus
I so that should get a minus. There are several
ways to see how phi transforms; one is to
go back, is this clear? Yes or no; you can
say no also it is fine, I can try to explain.
See there was something wrong out there. So,
the hard way of getting the transformation
of f mu nu which I would heartily recommend
that you do is to we know how A transforms;
go and plug and charge but I will do it in
two seconds by using this definition. We know
how phi transforms and these things act nicely.
So, phi transforms with a g out here and so
what is it we have to write, so I start like
this put primes on all these guys and so this
will be equal to this is just g phi. This
is what I get but you look here and since
these things transform nicely what this will
tell you are that this whole object on this
side transforms like g times.
So, putting things back together this implies
A transforms nicely. Again so, this is exactly
how the gauge field transformed under global
transformations but this is true even under
local transformations. So, again unlike the
u 1 case the field strength is not invariant
but rather it is covariant. Is this clear
and the most important part here is that it
also has this kind of a piece A square piece
it is not just a derivative piece. So, if
you try I mean so in some ways this definition
is a much nicer definition of the field strength
because if you try to say let me try to start
with some gauge field and try to construct
something which is invariant. You can do it
but, then it will be messy figuring out this
thing; you can I mean I am not saying it cannot
done. So, now that we have field strength
the idea would be to go ahead and try to write
an action for the field strength and so for
this.
So, what about a kinetic energy for a non-abelian
gauge field, any suggestions? f mu nu square
you would say. So, we have to you would write
something like this but we need to, so but
this is a matrix valued thing, so it says
we take a trace. So, now while the thing is
that so again this is also Lie algebra valued,
so you could write this as trace. So, these
are the only matrix part. So, I can pull these
things out 
but now this what number you get; what you
get here depends on the representation. You
can show that it is always proportional to
these things but they will differ by some
factor. So, then we have to fix the normalization,
so you need to choose some representation.
So, what is usually done is that you choose
to define it in what is called the fundamental
representation. So, if you take SU 2 the fundamental
representation is the two dimensional representation,
for s u n it is the n.
So, all the continuous groups which we discussed
earlier we actually it was like chicken and
egg; if you remember we defined the group
through a particular realization of it; that
is the fundamental representation for all
the groups the basic groups that we looked
at, the orthogonal, the unitary and the syntactic.
That is this in some sense they are also the
smallest representation that you have for
this thing. So, for SU 2 the smallest representation
is two for SO 3 it is three. So, what one
does is to, say, choose to fix normalizations
you choose things to be in the fundamental
representation and in most cases the normalizations
are such that trace of these things you call
this some this is called the cutter metric;
it is symmetric of course from the property
of the traces and this is the definition of
this thing and if you just take the case of
SU 2 you will find that h a b is delta a b.
In fact for SU n you did the Gellman matrices
also, for there also it was delta a b. It
was two delta a b, thank you. So, this is
two delta a b, yeah; probably it is true even
for the other thing for the probable matrices,
all of them it is the same. So, up to this
normalization factor you can see that this
is the definition which you would have for
the kinetic energy but now comes if you look
at this thing it is f square and coming back
to what we have out here, if you take f square
it has terms like this with square and these
look like kinetic energy terms but it has
it has cubic and quadratic pieces because
when I square this I get one term which would
be d A A square which is cubic, then there
is a quadratic piece which is A square and
this is no interactions nothing; it is just
a non-abelian gauge field. It is highly nontrivial.
It is equation of motion is nonlinear unlike
Maxwell’s or the abelian case where it was
linear. There were no interactions, nothing.
So, this already shows you that non-abelian
gauge theories are very different in characteristics
and so the thing is that it is not there is
no analog of a free theory but what you do
in quantum mechanics or quantum field theory.
It is that you break this up and treat this
as interactions and do them perturbatively
but in terms of gauge invariance or if you
take only this combination this does not quite
transform nicely. So, you have to be quite
clever about how you go about doing things
and maintaining gauge invariance, etc because
this combination is not invariant nor is this
combination; it is only this particular combination
which transforms nicely. I should not gauge
invariant gauge covariant. So, that that itself
shows. So, for instance q c d is a theory
where the group is SU 3 and so SU 3 the dimension
of it is 8; it is 3 square minus 1 which is
8.
So, that will tell you that it has to have
eight gauge fields and in quantum mechanics
for every particular gauge field you will
expect one spin one particle so that would
predict that you should see eight spin one
particles. They are called gluons and except
that we do not see any gluons in nature free
gluons in nature. What happens is that quantum
mechanically this theory is like I said there
are interactions are they interact such that
they are confining that you never see objects
which actually carry this charge. So, there
are quarks also which transforms like which
are analogous to this; they transform in something.
You could write some Lagrangian but that cannot
describe low energy physics as we see it.
But there are high energy regimes where you
go to accelerators, etc where you break things
apart and then available to see I mean evidence
of existence of this object.
And so this is even classically it is a complicated
theory but even quantum mechanically obviously
it is much much more complicated. So, non-abelian
gauge theories are also called Yang-Mills
theories. This is after two persons Yang and
Mills who independently kind of came up with
this proposal. I do not know when this was
may be late fifty’s or early sixty’s and
another point here is that you cannot write
any mass term for this. You might think I
mean that the mass term would be a quadric
term. There you cannot write a gauge invariant
mass term. So, exactly like so the thing is
that, yeah, you cannot write a mass term because
there is not gauge invariant and so the predication
would be if you took some gauge group; let
us say you took SU 2 you would say that you
should see three massless spin one particles.
So, in the early sixty’s when Glashow, Weinberg
and Salem, they were the first people to actually
use this setup and they actually said that
there exist some SU 2 cross U 1 whatever that
is; we will see in more details of it later
but then the predication was like I said for
every vector field we should see the thing
but nobody saw any massless spin one particles.
The only massless spin one particle we have
seen is a photon and so to be honest I think
it must have been very courageous of those
people to actually propose things and actually
push it and they may have been a laughing
stalk of people at that point in time because
they are saying look these guys are writing
out these theories and then we have not seen
any spin one particles and so it lead to this
issue of how do we understand this. And today
we know that these particles are seen but
they have masses and so the question is, how
do they get masses and so I will just discuss
that in the next ten minutes because we already
have the structure for it and we already have
seen how to get a massive photon.
How did we get a massive photon? Higgs mechanism,
so what happened in that case? We started
out with the situation where you had a global
symmetry which was under the vacuum spontaneous
broke the symmetry to some subgroup h, u 1
broken down to nothing; that is what we looked
at and so there was one Goldstone boson but
when we went to the local theory that goldstone
boson disappeared and what happened? We ended
up with a massive photon and as I explained
to you a massive particle has one degree of
freedom more than a massless particle; excuse
me, sorry. So, the framework is very clear
to us. We should look for a situation where
we have global symmetry which is spontaneously
broken. So, let us go back to our SU 2 model
and ask these questions and see what we get.
So, we will be very very specific here; we
will take SU 2 the group to be SU 2 and phi
will be exactly the example we have been looking
at in the two dimensional representation and
what is it we have to do? We have to choose
a u of phi with such that the ground state
has a nontrivial value. So let us choose u
of phi. As we all know this is the only potential
I seem to know. So, the ground state 
would correspond to, say, remember phi 1 and
phi 2 are complex fields. So this is like
a point on a three dimensional sphere and
so let us say that we go ahead and do the
following. We choose a particular solution;
so let us choose something like this, choose
phi such that it is zero for this and a out
here. Now the question I want to ask you is
what is the unbroken symmetry, what is the
analog of h, what is the symmetry of this
solution, is there any subgroup which is preserved?
Yeah but that is not a subgroup of this. So
yeah, so there is a good point. So, somebody
is saying there is a u 1 I could have just
changed the phase of A for instance but that
is not an element of this group.
So but I could either think of it as I could
do in two ways to add that u 1 into the story;
one of which is to say that it is just some
cross a second u 1 which would act in some
phases on these guys or go to u 2 if I choose
the global symmetry is also u 2 it could be.
U 2 would be a special case of u 1. So, you
can get SU 2 cross u 1 and SU 2 cross u 2
but right now I am not looking at that part.
Eventually we will in the Weinberg-Salam model
you will see that it is an SU 2 cross u 1
will have this same thing. This is called
a Higgs doublet or whatever; does not matter
but right now since that is not part of it,
there is no so it is not so complicated. What
you have to do is you just look at that you
know t 1 t 2 t 3 or sigma 1 sigma 2 to sigma
3 we will see which of them kills this; that
means it is invariant. So, there is nothing.
So H is phi; it is broken down to nothing.
So, what does Goldstone’s theorem tell us?
We should expect three Goldstone bosons. So,
implies three Goldstone bosons. So, what we
want to do next is to gauge it and what do
we do by gauging? We have to follow the same
procedure that we did a little while ago and
convert this global SU 2 into local SU 2.
So, coming back to this generalization you
see that if I want to convert these guys also
into Su 2 cross u 1 or u 2 into a local these
things then I have to add one more gauge field.
I would have four gauge fields but in our
setup here we have only three gauge fields
because I am focusing on the s u 2. But one
thing you will agree is that even if I go
to these groups the Goldstone boson theory
number would be the same because the number
of generators will go, this will increase
by one, this will increase by one. So, the
count of goldstone bosons will not change.
So now, we go to the locally invariant theory;
and actually coming back to this you can see
that these Goldstone bosons are exactly the
flag directions on this three sphere which
is going along the sphere three sphere. So
d mu of phi, so this is what we were looking
at and so we look at what we did; what we
can do is to just expand things to quadratic
order and so we just look at phi, and write
out plus I can put an eta here. So, what I
am here is parameterizing fluctuations and
like I said what we should do is write the
broken generators out here. We have already
seen how to do this. And the key point here
is this theta guys can be completely eaten
up by just making a gauge transformation and
a local gauge transformation. So, all I need
to really do is in this not even worry about
the thetas; assume that there are no thetas
in this thing and just write out take phi
by a gauge transformation.
So, for practical purposes I can just write
phi equal to zero a plus eta and that is it.
So, now I can just go there and plug these
things in here and expand to quadratic order
in etas and a’s, etcetera. So, this is not
as hard as it looks. So, d mu of this is just
d mu of eta, so I will get one term which
is d mu of eta whole square. But here eta
is a real field and what else, yes and then
but now the point here is that there are terms
this thing d mu has also a mu piece. So, there
will be one piece which will be, I am not
writing all the pieces so there will be one
piece which will look like a mu zero a. If
you put all these things together you will
start seeing, so this term will look like
a square A mu trace A mu square. So, what
does this tell you that this mechanism is
exactly like what happened in the abelian
case except 
that this now has all the guys all the three
it is there and you see that they all getting
masses which are again exactly like what happened
to the photon.
The photon mass was proportional to a and
it is exactly what we see here also the photon
mass is indeed proportional to a. So, there
will be n square so not the photon these three
guys. So, this is called the Higgs mechanism.
This is not the Higgs mechanism which is used
in the standard model; it is an example of
a non-abelian Higgs mechanism. After some
about four, five lectures down the road we
will actually discuss the standard model where
we will look at SU 2 cross u 1 structure with
a similar doublet and there will be some unbroken
guy; that unbroken guy will be identified
with the electromagnetism. The H that will
be your u 1; so it will not be this u 1 nor
will it be thing, it will be some linear combination
of them.
So, this u 1 charge is called hypercharge.
So, one usually writes that with a u 1 y not
q. So, hopefully you are able to see the structure
of these things and so you can see that you
can get masses through this mechanism. Just
one more bit, it is sort of interesting; if
you take SU 2 and you take a doublet you can
see that you can never you can never get a
u 1 unbroken just with plain SU 2, because
it is like saying if you pick a direction
in two dimensional space there is what remains
is nothing; roughly speaking but now so the
question is suppose I want to get a unbroken
u 1, what should I do? Any ideas; I want to
start with SU 2 but I would like to get an
unbroken u 1. I will let you think about it;
may be you can back and tell me; that is a
very simple answer to that. So, I will stop
here.
