I'm confused, how does one explicitly calculate those matrices. I need a systematic way to find them. I can see they satisfy the commutation relations, etc as required, but I don't know how to actually calculate the matrices.

other quick question, its late and my mind ceased to function many hours ago..

Does anyone recognize the following operator

(x^2 + y^2)(d^2/dx^2 + d^2/dy^2)

The last part is the laplace operator. But the first part (which looks like r^2 in polar) messes everything up.

ZM

matt grime
Homework Helper
A lie algebra is just a vector space. You have a basis. And you know how each element acts on that basis ad(x)y = [x,y], so you can just write out the matrix. Although it is dressed upion fancy language, it is just 1st year linear algebra at this point.

Do you have the structure constants for the Lie algebra? If $$\big[ X_{a},X_{b} \big] = f_{abc}X_{c}$$ then your adjoint representation is $$[T_{a}]_{bc} = -if_{abc}$$ (factors choosen for hermiticity).

So if you're given the $$f_{abc}$$ terms, you automatically have the adjoint matrix terms.

nrqed
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Do you have the structure constants for the Lie algebra? If $$\big[ X_{a},X_{b} \big] = f_{abc}X_{c}$$ then your adjoint representation is $$[T_{a}]_{bc} = -if_{abc}$$ (factors choosen for hermiticity).

So if you're given the $$f_{abc}$$ terms, you automatically have the adjoint matrix terms.
I have a very stupid question concerning the adjoint representation.

The way I read the definition, we have
$ad(X) Y = [X,Y]$. Which I interpret as meaning :If we apply X given in the adjoint representation to the element Y, we get the same result as evaluating the commutator between X and Y. On mathworld (see the link on the first post of this thread) the show an dexample of the 4 basis (2 by 2) matrices of gl2 and they give their expressions in the adjoint representation (which are 4 by 4). I know the trick of using the tsructure constants to build the adjoint representation. But I still don't understand the line $ad(X) Y = [X,Y]$. For example, consider X=Y. Then we get zero. But it's clear that multiplying the matrices in the adjoint representation given on mathworld by themselves, we don't get zero. So I clearly completely misunderstand the meaning of that relation. I would appreciate some help in understanding this!

Patrick

matt grime
Homework Helper
You do not multiply the matrices of the representations of X and Y when you apply ad(X) to Y. ad(X) is a 4x4 matrix. Y is a 4x1 vector.

Multiplication of matrices corresponds to composition of linear maps ad(X) and ad(Y) (i.e. ad(X)ad(Y) is the map that sends Z to [X[Y,Z]] which in general is not zero as you have noticed).

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nrqed
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You do not multiply the matrices of the representations of X and Y when you apply ad(X) to Y. ad(X) is a 4x4 matrix. Y is a 4x1 vector.

Multiplication of matrices corresponds to composition of linear maps ad(X) and ad(Y) (i.e. ad(X)ad(Y) is the map that sends Z to [X[Y,Z]] which in general is not zero as you have noticed).
Ok. Thanks for the help. But X and Y here are meant to be elemenst of the group, right? So now there are some elements (here labeled arbitrarily as "Y") which are representend by 4 components column vectors and some other elements (X) represented by 4 x 4 matrices. Does that mean that there are two representations for each element of the group, one as a 4 component vector and one as a 4x4 matrix? And when I want the "commutator" of two elements [X,Y], I must apply the 4x4 representation of X on the column vector representation of Y?
I know this is all very basic and I have been using the adjoint representation for years but it's one of those things that when you go back to it and really try to understand it from scartch is not that clear.

nrqed, I'd really recommend the textbook 'Lie Algebras in Particle Physics' by Georgi for this kind of stuff. It covers groups and algebras (and then their applications to physics) in a very friendly way, more in understanding and working than the indepth generality which a maths textbook would often do. Georgi explains this stuff in a very nice way.

I personally find it easier to think of the X and Y as not vectors, because then you get confused by the whole [X,Y] thing. Thinking of them as non-communting differential operators is nicer. It also shows why you need [X,Y] to generate a new element of the group and NOT XY or YX, such as having the group of tangent vectors $$X = X^{a}\frac{\partial}{\partial x^{a}}$$. XY is NOT a first order differential operator, but XY-YX is since partial derivatives commute and you end up with something within the original group, while it's individual parts aren't.

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matt grime
Homework Helper
I think you're confusing things.

A representation of a lie algebra L is a homomorphism from L to the space End(V) of linear homs of a vector space.

In the adjoint rep we're just letting the vector space by L, and the hom is written as ad(?).

So there are two copies of L lying around. The elements of L just as elements in V, and then as elements in End(V). Strictly speaking, the second copy is not a copy, but a homomorphic image - there is no reason why it should be an isomorphism and in general it won't be.

When you want to evaluate ad(X), the element of End(V) to an element Y in V, you just remember that X and Y are elements of L and write down [X,Y].

And when I want the "commutator" of two elements [X,Y], I must apply the 4x4 representation of X on the column vector representation of Y?
This doesn't make sense. The commutator of two elements only makes sense when you're thinking of them as elements of a matrix ring.

You're misusing the word representation. Or rather using it in two incompatible ways. Remember a representation is a homomorphism into End(V) for some vector space (thus the 4x4 matrices are a representation of X in the correct sense). Saying that the 4x1 vector is a representation of Y is using an entirely different meaning of the word representation (its more traditional simplistic literal sense).

If you want to evaluate the linear map ad(X) as an element of End(V), then it sends Y to [X,Y]. Nothing else more complicated.

nrqed
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Thanks for the help.

I think you're confusing things.
Yes, I realize that I am .
A representation of a lie algebra L is a homomorphism from L to the space End(V) of linear homs of a vector space.
Is this the same as saying that a representation of a lie algebra is a set of matrices acting on some vector space?
In the adjoint rep we're just letting the vector space by L, and the hom is written as ad(?).
Is the first sentence meant obe be "letting the vector space be L"? In that case, what does that mean in concrete terms? I thought you mean that the elements of the Lie algebra are represented (or "associated" if the word "represented" is bad) to column vectors on which the matrices are acting?

So there are two copies of L lying around. The elements of L just as elements in V, and then as elements in End(V).
Ok, but isn't that saying that we are also representing elements of the Lie algebra by column vectors? To be concrete, if I have a certain element of SU(2), how do I find its corresponding element of the vector space V on which the adjoint representation is acting?

Thanks fo ryour patience. I realize that I am missing the point but I am hoping that if I keep asking enough questions, I will gather enough tidbits of information to put it together.

matt grime
Homework Helper
Let's stop using the word representing entirely and only use representation in its correct meaning of a homomorphism from L to End(V).

In a representation the elements of the Lie algebra are sent to elements of End(V). They are *not* associated to elements of the underlying vector space V. ad is a homomorphism from L to End(L), thus it is a representation of L.

If it really helps, let's write V for the underlying elements of the vector space of L. The adjoint representation is then a map from L the lie algebra to End(V).

So, elements of V are just elements of L. (And SU(2) is not a lie algebra).

So, consider the lie algebra sl_2(C). it has the obvious/standard basis e,f,h with the relations that [ef]=0, [eh]=2e, [fh]=-2f.

So you can write down the adjoint rep from this information.

If you want to make as(x) act on any element k in L, the ad(x) sends it to [kx].

sl_2(C)=C^3 as a vector space. There is an obvious action of sl_2 on this vector space.

I really am just repeating myself once more.

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matt grime
Homework Helper
Ok, but isn't that saying that we are also representing elements of the Lie algebra by column vectors? To be concrete, if I have a certain element of SU(2), how do I find its corresponding element of the vector space V on which the adjoint representation is acting?
.
That is a different use of the word representing than is meant by representation. SU(2) is not a lie algebra, but *any* lie algebra is a vector space. You don't have to 'associate' elements of L to anything in a vector space in the adjoint representation - they already are a a vector space. What the corresponding column vector is depends entirely on the choice of basis you made. There is then a Lie Algebra representation of L acting on this vector spacce, the adjoint representation.

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matt grime
Homework Helper
Let's try a third way.

What is a representation of a lie algebra space? Let V be a vector space. A rep is a map from LxV to V satisfying certain rules. I hope you're used to this notation. Since L is itself a vector space there is nothing to stop us using that as V, and the map sends (x,y) to ad(x)y=[x,y].

nrqed
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nrqed, I'd really recommend the textbook 'Lie Algebras in Particle Physics' by Georgi for this kind of stuff. It covers groups and algebras (and then their applications to physics) in a very friendly way, more in understanding and working than the indepth generality which a maths textbook would often do. Georgi explains this stuff in a very nice way.

I personally find it easier to think of the X and Y as not vectors, because then you get confused by the whole [X,Y] thing. Thinking of them as non-communting differential operators is nicer. It also shows why you need [X,Y] to generate a new element of the group and NOT XY or YX, such as having the group of tangent vectors $$X = X^{a}\frac{\partial}{\partial x^{a}}$$. XY is NOT a first order differential operator, but XY-YX is since partial derivatives commute and you end up with something within the original group, while it's individual parts aren't.
Very nice example. And thanks for the recommendation.

nrqed
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Gold Member
Let's try a third way.

What is a representation of a lie algebra space? Let V be a vector space. A rep is a map from LxV to V satisfying certain rules. I hope you're used to this notation. Since L is itself a vector space there is nothing to stop us using that as V, and the map sends (x,y) to ad(x)y=[x,y].
I am learning, so thanks.
(btw, I meant su(2), not SU(2))

My problem I think arises because I am trying to translate what you are saying to an explicit representation in terms of matrices (I don't know when to correctly use the word "representation" anymore) . I do see what you are saying: there is a vector space which we take as being the Lie algebra itself. A rep is a map from L x V to V. Since [x,y] does send an element of V to V, we can take this as being a possible rep, and it's called the adjoint rep.
This makes complete sense to me as long as I leave it formal.

Now, my problem is if I try to translate this to the language of explicit matrices. They say that one must use the structure constants to build the explicit adjoint representation. These wil represent the map LxV -> V.

For sl_2(C) you gave for example [eh]=2e and so on.

All I am saying is that to me it seems that the explicit adjoint representation of that algebra will be 3x3 matrices. And now, to reproduce the different commutation relations of the group, we will have to represent
the elements e,f, h by the column vectors (1,0,0), (0,1,0), (0,0,1).
Then the map (x,y) = [x,y] wil be reproduced in that explicit representation by simply multiplying the 3x3 matrix representing x by the column vector associated to the element y.

This is what I meant in my post #8. Isn't that correct?

Patrick

matt grime
Homework Helper
All I am saying is that to me it seems that the explicit adjoint representation of that algebra will be 3x3 matrices. And now, to reproduce the different commutation relations of the group, we will have to represent
the elements e,f, h by the column vectors (1,0,0), (0,1,0), (0,0,1).
Then the map (x,y) = [x,y]
I'd prefer it if you wrote ad(x).y, so it is clear that we're thinking of the action of x in the adjoint representation on the element of the vector space y.

wil be reproduced in that explicit representation by simply multiplying the 3x3 matrix representing x by the column vector associated to the element y.

This is what I meant in my post #8. Isn't that correct?
Yes. That is correct.

matt grime
Homework Helper
For sl_2(C) you gave for example [eh]=2e and so on.
Let's do it (and I'll get it right this time - never do commutator relations in your head)

[he]=2e
[hf]=-2f
[hh]=0

So ad(h) sends e to 2e, f to -2f and h to 0, so appealling to our first linear algebra course, the matrix of ad(h) is just diag(2,-2,0) in this representation (ordering the basis as e,f,h).

nrqed
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Gold Member
Let's do it (and I'll get it right this time - never do commutator relations in your head)

[he]=2e
[hf]=-2f
[hh]=0

So ad(h) sends e to 2e, f to -2f and h to 0, so appealling to our first linear algebra course, the matrix of ad(h) is just diag(2,-2,0) in this representation (ordering the basis as e,f,h).
yes, it's crystal clear now. Thank you for your help. It is very much appreciated.

nrqed
Homework Helper
Gold Member
Let's do it (and I'll get it right this time - never do commutator relations in your head)

[he]=2e
[hf]=-2f
[hh]=0

So ad(h) sends e to 2e, f to -2f and h to 0, so appealling to our first linear algebra course, the matrix of ad(h) is just diag(2,-2,0) in this representation (ordering the basis as e,f,h).
Since this is clear now, I'd like to take advantage of the situation to clarify something else. What about the fundamental representation? How is it defined, formally? Now we want a rep of LxV -> V but how is the action defined? For the adjoint representation, the structure of L itself is used to build the rep, but what is used to define the fundamental rep?

Thanks again

matt grime
Homework Helper
You use the classical theory of reps of lie algebras. You know what a root space is? Cartan subalgebra? It's a beautiful, classical, theory. You should look it up.

Dox
In this cases I prefer the Quantum Mechanics notation, so from

$Ad(Y)\left. |X\right>=\left. |[Y,X]\right>,$​

It is trivially to find the matrix element''

$\left<Z|Ad(Y)|X\right>= \left<Z|[Y,X]\right>.$​

So, since the commutators of the basis elements are known, the problem is solved.

Enjoy!​