SU(3) Gauge Group: QCD & SM Invariance Explained

[CAF123]

The lagrangian of a non interacting quark is made to be invariant under local SU(3) transformations by introduction of a new field, the gauge field, giving rise to the gluon. This gives us a locally gauge invariant lagrangian for the quark field and together with the construction of a locally gauge invariant kinetic term gives us the classical QCD lagrangian (ignoring the redundancy in the large degeneracy of gauge transformations resulting in additional ghost terms etc)

This gauging of the local symmetry results in the transformation law of the gauge field to be $A_{\mu} \rightarrow U A_{\mu}U^{-1} - i/g (\partial_{\mu} U) U^{-1}$. It is said that the gluon field transforms in the adjoint representation of the gauge group SU(3) which contains all group elements U such that $U = \exp(i \theta^a t^a)$ with $\theta^a$ constant.

My question is, even though QCD is invariant under local gauge transformations, that is transformations of the form $\exp(i \theta^a(x) t^a)$ with $\theta^a$ now a function of x, we write the gauge group of the SM as $SU(3)_c \otimes SU(2)_L \otimes U(1)_Y$. But SU(3)_c only contains transformations where theta is restricted to be a constant. So why are the local transformations not contained in the gauge group of the SM? Is it because the charge of the gauged SU(3) symmetry is colour and since this is an internal degree of freedom, we restrict to global transformations independent of spacetime?

Thanks!

 

[MathematicalPhysicist]

Have you read Srednicki cover to cover?
I remember some of your notation from his book, even if he doesn't discuss your particular problem, you can check his references.

I can't help more than this, I cannot remember everything I read from it.

 

[CAF123]

I have used Srednicki in the past but I didn't like his presentation of a few things so sought other books. I don't have access to his book at the moment. I'll see if any QFT experts on the forum can provide any insight - certainly in the resources I have been using I couldn't find an answer to my question specifically.

 

[MathematicalPhysicist]

The answer to your questions should appear on chapters 84-86 of Srednicki, or so I think.

Check it out in case no one replies, my 2 cents.

 

[CAF123]

I found his book on the net but couldn't see anything relevant in those chapters - I checked his chapter on QCD too and on group representations but found nothing to help. I've done courses on this stuff before but more thinking on the matter is leading to more questions haha - I'll wait to see if anyone has an idea.

 

[Dr.AbeNikIanEdL]

Could you maybe state again what the actual question is?

 

[CAF123]

Sure. Basically the question stemmed from the transformation law for the gluon field - it's said everywhere that the quarks transform in the fundamental rep of the gauge group SU(3) and the gluons in the adjoint rep of the gauge group SU(3). The latter statement is only clear for global transformations, i.e make U independent of x then the $\partial_{\mu} U$ term disappears and we have the usual $A_{\mu} \rightarrow UA_{\mu}U^{-1}$ transformation law for objects transforming in the adjoint rep. The gauge group is also called $SU(3)_c$ because the degree of freedom that is mixed under these transformations is colour. This gauge group is what appears in the SM gauge group. But why? The full symmetry of QCD contains local transformations too (by construction) so why is this not inherent in the SM gauge group?

it feels like a misnomer to call $SU(3)_c$ a gauge group because it is restricted to global transformations and we gauged a local symmetry.

 

[Orodruin]

SU(3) is not restricted to global symmetries. If it was there would be no gluons. You would just have a global symmetry and say "so what?"

 

[CAF123]

Ok, so why is it then said that the gluons transform under the adjoint representation of SU(3)? Under local SU(3) transformations (ie the gauged symmetry), $A_{\mu} \rightarrow UA_{\mu}U^{-1} - i/g (\partial_{\mu}U)U^{-1}$ which is not the transformation law for something transforming in the adjoint representation.

Thanks!

 

[Orodruin]

The field tensor (which contains the physical degrees of freedom) transforms under the adjoint representation. It is not clear what you really mean by saying "gluons transform".

 

[CAF123]

I also made a discussion here
https://www.physicsforums.com/threads/adjoint-transformation-of-gluon.899607/

The physical degrees of freedom are just the $A_{\mu}^a$ no? In that post, I do a small computation showing $A_{\mu}^a t^a$ transforms as $U A_{\mu}^a t^a U^{-1}$ which infinitesimally looks like $A_{\mu}^c \rightarrow A_{\mu}^c + \theta^a f^{acb}A_{\mu}^b$ which is just the transformation law for the components of a 8x1 real vector (c.f infinitesimal cross product rule in 3D Euclidean with $\epsilon_{ijk}$) It's just the reconcilation of the result of my computation and the transformation law $A_{\mu} \rightarrow U A_{\mu}U^{-1} - i/g (\partial_{\mu}U)U^{-1}$ which I'm confused about.

Thanks :)

 

[Orodruin]

The physical degrees of freedom are just the A_{\mu}^a no?

No. There are unphysical degrees of freedom that you need to fix using gauge conditions. Just like you need to fix a gauge in normal electrodynamics. If $A^\mu$ was completely physical you would not need to bother with gauge fixings and ghosts.

 

[samalkhaiat]

I also made a discussion here
https://www.physicsforums.com/threads/adjoint-transformation-of-gluon.899607/

The physical degrees of freedom are just the $A_{\mu}^a$ no? In that post, I do a small computation showing $A_{\mu}^a t^a$ transforms as $U A_{\mu}^a t^a U^{-1}$ which infinitesimally looks like $A_{\mu}^c \rightarrow A_{\mu}^c + \theta^a f^{acb}A_{\mu}^b$ which is just the transformation law for the components of a 8x1 real vector (c.f infinitesimal cross product rule in 3D Euclidean with $\epsilon_{ijk}$) It's just the reconcilation of the result of my computation and the transformation law $A_{\mu} \rightarrow U A_{\mu}U^{-1} - i/g (\partial_{\mu}U)U^{-1}$ which I'm confused about.

Thanks :)

You are dealing with two different groups here. The global compact group $G$, and the infinite-dimensional local gauge group $G(M)$ ($M$ being a contractible space-time manifold). With respect to $G$, the gauge field behaves like a genuine tensor transforming by the adjoint representation: $$\mathbb{A}^{g}_{\mu}(x) = g^{-1}\mathbb{A}_{\mu}(x)g ,$$ where $\mathbb{A}_{\mu} (x) = A^{a}_{\mu}(x) T_{a}$. Locally, i.e., with respect to $G(M)$, and exactly because of the inhomogeneous (second) term in its transformation law: $$\mathbb{A}^{g}_{\mu}(x) = g^{-1}(x) (\mathbb{A}_{\mu}(x) + \partial_{\mu}) g(x) ,$$ the gauge field is, therefore, interpreted as a $\mathcal{L}(G)$-valued connection, $\Gamma(x) = (A^{a}_{\mu}(x) dx^{\mu})T_{a}$, on the trivial principal bundle $\pi (G,M)$.

We have similar story in GR. The Christoffel symbol $$\Gamma (x) = \left( \Gamma^{\rho}_{\mu\nu}(x) dx^{\mu} \right) T^{\nu}{}_{\rho} ,$$$$[T^{\mu}{}_{\nu} , T^{\rho}{}_{\sigma}] = \delta^{\rho}_{\nu} T^{\mu}{}_{\sigma} - \delta^{\mu}_{\sigma} T^{\rho}{}_{\nu} ,$$ is interpreted as a $gl(n,R)$-valued connection on $\mathscr{F}_{M}\left(GL(n,R),M\right)$, the frame bundle of $M$. The $T^{\mu}{}_{\nu}$ is the natural basis of the Lie algebra $gl(n,R) \sim T_{e}\left(G(n,R) \right)$. Now let $g$ be the transition function mappings $g \ : \ U \cap \bar{U} \to GL(n,R)$, where $U, \bar{U} \subset M$ such that $U \cap \bar{U}$ is not the empty set . Then you have $$\bar{\Gamma}(x) = g^{-1}(x) \left( \Gamma (x) + d \right) g(x) .$$ Again this transformation law consists of two part, the first one is tensorial, but the second one is not.

 

[CAF123]

You are dealing with two different groups here. The global compact group $G$, and the infinite-dimensional local gauge group $G(M)$ ($M$ being a contractible space-time manifold).

I see, so just to check, it is the former that is called SU(3) and since the global transformations therein mix the colour charge of QCD, it is denoted as SU(3)_c? The latter are then transformations in the image of the map $\mathbb{R}^4 \simeq \mathcal{M} \rightarrow \mathrm{SU}(3), x \mapsto \exp(i \theta^a(x) t^a)$, where $\mathcal M$ denotes Minkowski space. (I.e all local SU(3) transformations are in the latter group and the global transformations in the former)...That all fine?

With respect to $G$, the gauge field behaves like a genuine tensor transforming by the adjoint representation: $$\mathbb{A}^{g}_{\mu}(x) = g^{-1}\mathbb{A}_{\mu}(x)g ,$$ where $\mathbb{A}_{\mu} (x) = A^{a}_{\mu}(x) T_{a}$. Locally, i.e., with respect to $G(M)$, and exactly because of the inhomogeneous (second) term in its transformation law: $$\mathbb{A}^{g}_{\mu}(x) = g^{-1}(x) (\mathbb{A}_{\mu}(x) + \partial_{\mu}) g(x) ,$$ the gauge field is, therefore, interpreted as a $\mathcal{L}(G)$-valued connection, $\Gamma(x) = (A^{a}_{\mu}(x) dx^{\mu})T_{a}$, on the trivial principal bundle $\pi (G,M)$.

I didn't understand the abstract mathematical terminology in terms of bundles but would be interested to read more about it all from this abstract setting - certainly I have never come across a treatment done in this way in any standard QFT text - could you refer me to a reference?

Why is it then that in the SM gauge group it is $\mathrm{SU}(3)_c$ that is appearing and not this infinite dimensional local gauge group? I thought (as pointed out in OP) that it maybe due to colour being an internal quantum number so transforms under transformations independent of spacetime but maybe there is another argument.

Then you have $$\bar{\Gamma}(x) = g^{-1}(x) \left( \Gamma (x) + d \right) g(x) .$$ Again this transformation law consists of two part, the first one is tensorial, but the second one is not.

I see. So all of this is the underlying reason why we say the Christoffel symbols don't transform like tensors?

Thanks!

 

[samalkhaiat]

It is said that the gluon field transforms in the adjoint representation of the gauge group SU(3)

To make that statement correct, you must add the phrase “up to inhomogeneous term”. Or, as I explained before, another correct statement is “it transforms in the adjoint representation of the global group $SU(3)$”. But the absolutely correct statement is of course “under the local (gauge) group $SU(3)$ the gluon field transforms like a connection or compensating field”
By the way, when we write $SU(3)_{c}$, we mean that this (local and/or global) group acts on the colour space of each quark-flavour, for example, $d = (d_{1}, d_{2} , d_{3})^{t}$. And we do that just to distinguish it from the flavour group $SU(3)_{f}$ which acts on the flavour space of quarks $q = (u,d,s)^{t}$.

we write the gauge group of the SM as $SU(3)_c \otimes SU(2)_L \otimes U(1)_Y$.

No, we don’t. It is meaningless to use the tensor product sign $\otimes$ between groups.

But SU(3)_c only contains transformations where theta is restricted to be a constant.

Who said that? The gauge group of the SM is $$G_{SM} = SU(3)_{c} \times \left( SU(2) \times U(1) \right)_{ew} ,$$ and all factor groups in $G_{SM}$ are understood to be local gauge groups.

So why are the local transformations not contained in the gauge group of the SM?

One can not give an answer to a wrong question: In QCD, the group $SU(3)_{c}$ is the local gauge group.

Is it because the charge of the gauged SU(3) symmetry is colour and since this is an internal degree of freedom, we restrict to global transformations independent of spacetime?

Answering a wrong question, almost always, leads to gibberish: What do you mean by “the charge of the gauged SU(3) symmetry is colour”? In field theories, the symmetry charges are the observables of the theory. So, don’t confuse the symmetry charges with the internal index space of the fields.

 

[CAF123]

@samalkhaiat Thanks! You answered all my questions...Just a few follow up comments if that's ok:

1) So, as Orodruin already mentioned, the field tensor transforms under an adjoint representation too - is this now the adjoint representation of the local gauge group rather than the gluon which is the adjoint representation for the global gauge group (= subgroup of local gauge group with theta restricted to be constant)

2) In my small computation I mentioned in the other thread which I linked from this thread (given again below), I derived the adjoint transformation law $A_{\mu} \rightarrow U A_{\mu}U^{-1}$ - where did I use the fact that I was working in the global gauge group in this derivation?
$(U t_a U^{-1})_{ij} = D_{ab}(t_b)_{ij}$. The U's and the t_a can be in any representation carrying the indices i,j but I don't see where I am using the fact I'm restricting to the global gauge group to obtain the adjoint rep transformation law.

https://www.physicsforums.com/threads/adjoint-transformation-of-gluon.899607/

Thanks!

 

[samalkhaiat]

@samalkhaiat Thanks! You answered all my questions...Just a few follow up comments if that's ok:

1) So, as Orodruin already mentioned, the field tensor transforms under an adjoint representation too - is this now the adjoint representation of the local gauge group

Again, you are asking a wrong question. The word representation always refers to the global (i.e., finite-dimensional) groups, i.e., not to the local (i.e., infinite-dimensional) groups. This is because we (mathematicians) know very little about the representation theory of infinite-dimensional groups on space-time of dimension >2.

2) In my small computation I mentioned in the other thread which I linked from this thread (given again below), I derived the adjoint transformation law $A_{\mu} \rightarrow U A_{\mu}U^{-1}$ - where did I use the fact that I was working in the global gauge group in this derivation?
$(U t_a U^{-1})_{ij} = D_{ab}(t_b)_{ij}$.

Thanks!

The equation $$\mbox{Ad}_{g} X_{a} \equiv g X_{a} g^{-1} = D_{a}{}^{b}(g) X_{b} ,$$ defines the finite-dimensional (matrix) representation, $D(gh) = D(g)D(h)$, of the finite-dimensional (global) group.

 

[CAF123]

Ok thanks! It's clear now - I wanted to ask a few final things about the mappings $\text{ad}$ and $\rho$ that are usually used in the literature:

The adjoint map at the level of the lie group is such that $$\text{ad}: \mathfrak{g} \rightarrow \mathfrak{gl}(\mathfrak g),$$ taking $\lambda_a \mapsto \text{ad}_{\lambda_a}$ where $$\text{ad}_{\lambda_a}: \mathfrak{g} \rightarrow \mathfrak{g}\,\,\,\,\,\text{with}\,\,\, \lambda_b \mapsto [\lambda_a, \lambda_b].$$

So, this means that $$\lambda_b \rightarrow \text{ad}_{\lambda_a} \lambda_b = [\lambda_a, \lambda_b] = if_{abc}\lambda_c.$$

Equivalently, $\rho: \mathfrak{g} \rightarrow \mathfrak{gl}(\mathfrak{g})$ is also defined as the map such that $\rho(\lambda_a)_{bc}\lambda_c = [\lambda_a, \lambda_b] = if_{abc}\lambda_c$. I'm just trying to get all the indices to make sense so what is the relationship between $\text{ad}$ and the $\rho?$

Thanks!

 

[samalkhaiat]

Ok thanks! It's clear now - I wanted to ask a few final things about the mappings $\text{ad}$ and $\rho$ that are usually used in the literature:

The adjoint map at the level of the lie group is such that $$\text{ad}: \mathfrak{g} \rightarrow \mathfrak{gl}(\mathfrak g),$$ taking $\lambda_a \mapsto \text{ad}_{\lambda_a}$ where $$\text{ad}_{\lambda_a}: \mathfrak{g} \rightarrow \mathfrak{g}\,\,\,\,\,\text{with}\,\,\, \lambda_b \mapsto [\lambda_a, \lambda_b].$$

So, this means that $$\lambda_b \rightarrow \text{ad}_{\lambda_a} \lambda_b = [\lambda_a, \lambda_b] = if_{abc}\lambda_c.$$

Equivalently, $\rho: \mathfrak{g} \rightarrow \mathfrak{gl}(\mathfrak{g})$ is also defined as the map such that $\rho(\lambda_a)_{bc}\lambda_c = [\lambda_a, \lambda_b] = if_{abc}\lambda_c$. I'm just trying to get all the indices to make sense so what is the relationship between $\text{ad}$ and the $\rho?$

Thanks!

For matrix Lie groups, they are the same thing. Both define the adjoint representation of (matrix) Lie algebra (or Lie algebra homomorphism) on $gl(\mathscr{L})$, the Lie algebra of the general linear group over the “vector space” $\mathscr{L}(G)$.

I believe, I went through this with you before. The representation of the group $G$ on its Lie algebra $\mathscr{L}(G)$, i.e., the pair $\left(\mbox{Ad} , \mathscr{L}(G) \right)$ is obtained by differentiating the conjugation map (automorphism) $\varphi_{g}(h) = ghg^{-1}$ at the identity, $$\mbox{Ad}_{g} = \mbox{d} (\varphi_{g})|_{e} : \ \mathscr{L}(G) \to \mathscr{L}(G) ,$$ So, by writing $h = e^{tX}$ for $X \in \mathscr{L}(G)$, you get $$\mbox{Ad}_{g}(X) = g \ \left( \frac{d}{dt} e^{tX} \right)_{t = 0} \ g^{-1} = g \ X \ g^{-1} .$$ Similarly, $\mbox{ad}_{X}$ is obtained by differentiating $\mbox{Ad}_{g}$ at the identity: $$\mbox{ad}_{X} = \mbox{d} (\mbox{Ad}_{g})|_{e} : \ \mathscr{L}(G) \to \mathscr{L}(G) .$$ Then, for $X,Y \in \mathscr{L}(G)$, write $g=e^{tX}$ and get $$\mbox{ad}_{X} (Y) = \frac{d}{dt}\left( \mbox{Ad}_{(e^{tX})} (Y)\right)_{t = 0} = \frac{d}{dt}\left( e^{tX}Ye^{-tX}\right)_{t=0} = [X , Y] .$$

 

[CAF123]

Thanks! Indeed it was mentioned before but I want to check the indices on the $\rho$. E.g for the mapping $\text{ad}_X(Y) = [X,Y],$ if we let $X \rightarrow \lambda^a, Y \rightarrow \lambda^b$ then $\text{ad}_{\lambda^a}(\lambda^b) = [\lambda^a, \lambda^b]$. Or we can formulate this as $$\text{ad}_X: \mathfrak{g} \rightarrow \mathfrak{g}\,\,\,\,\text{with}\,\,\,\,Y \mapsto [X,Y].$$

Now what are the equivalent statements for $\rho$? $\rho$ was defined as $\rho (\lambda_a)_{bc}(\lambda_c) = [\lambda_a,\lambda_b]$ but here in this case, in the notation of X and Y used above, the X would be $\lambda_a$ and the Y would be $\lambda_c$ yes? So this seems to suggest I would write something like $$\rho_{(\lambda_a)_{bc}} : \mathfrak{g} \rightarrow \mathfrak{g}\,\,\,\,\,\text{with}\,\,\,\, \lambda_c \mapsto [\lambda_a, \lambda_b]$$

So in one case the index $b$ is used to specify an entry in the matrix representation $\rho$ (as in $\rho(\lambda_a)_{\mathbf{b}c}$) and in the other case (first case with the map $\text{ad}$) it is an index on the $\lambda^b$. Is this the right way to think about it?

Thanks again!

EDIT: Well actually, I've also seen the equation $\rho(\lambda_a)(\lambda_b)= [\lambda_a, \lambda_b]$ used so I think all the equation $\rho(\lambda_a)_{bc} \lambda_c = [\lambda_a, \lambda_b]$ is saying is that the basis vectors {\lambda} of the lie algebra mix between each other under the matrix representation of the adjoint action.

 

[samalkhaiat]

Thanks! Indeed it was mentioned before but I want to check the indices on the $\rho$. E.g for the mapping $\text{ad}_X(Y) = [X,Y],$ if we let $X \rightarrow \lambda^a, Y \rightarrow \lambda^b$ then $\text{ad}_{\lambda^a}(\lambda^b) = [\lambda^a, \lambda^b]$. Or we can formulate this as $$\text{ad}_X: \mathfrak{g} \rightarrow \mathfrak{g}\,\,\,\,\text{with}\,\,\,\,Y \mapsto [X,Y].$$

Now what are the equivalent statements for $\rho$? $\rho$ was defined as $\rho (\lambda_a)_{bc}(\lambda_c) = [\lambda_a,\lambda_b]$ but here in this case, in the notation of X and Y used above, the X would be $\lambda_a$ and the Y would be $\lambda_c$ yes? So this seems to suggest I would write something like $$\rho_{(\lambda_a)_{bc}} : \mathfrak{g} \rightarrow \mathfrak{g}\,\,\,\,\,\text{with}\,\,\,\, \lambda_c \mapsto [\lambda_a, \lambda_b]$$

So in one case the index $b$ is used to specify an entry in the matrix representation $\rho$ (as in $\rho(\lambda_a)_{\mathbf{b}c}$) and in the other case (first case with the map $\text{ad}$) it is an index on the $\lambda^b$. Is this the right way to think about it?

Thanks again!

EDIT: Well actually, I've also seen the equation $\rho(\lambda_a)(\lambda_b)= [\lambda_a, \lambda_b]$ used so I think all the equation $\rho(\lambda_a)_{bc} \lambda_c = [\lambda_a, \lambda_b]$ is saying is that the basis vectors {\lambda} of the lie algebra mix between each other under the matrix representation of the adjoint action.

Okay, if the followings will not make sense to you, then you seriously need to consult a textbook that you do understand.

By definition, a (p-dimensional) representation of a (n-dimensional) Lie algebra $\mathscr{L}^{n}$ is a pair, $(\mathcal{V}^{p} , \rho)$, consisting of a p-dimensional vector space $\mathcal{V}^{p}$, the representation space, and a map $\rho : X_{a} \to \rho(X_{a})$, from $\mathscr{L}^{n}$ onto the algebra of linear operators in $\mathcal{V}^{p}$, satisfying $$\rho (\alpha X_{a} + \beta X_{b}) = \alpha \rho (X_{a}) + \beta \rho (X_{b}) , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ $$\rho \left( [ X_{a} , X_{b} ]\right) =[ \rho (X_{a}) , \rho (X_{b}) ] = C_{abc} \rho (X_{c}) . \ \ \ \ \ \ (2)$$

So, given a complete orthonormal set $\{|\varphi_{i}\rangle \}$ of basis vectors in $\mathcal{V}^{p}$, the action of the generators $X_{a}$ on $|\varphi_{j}\rangle$ is determined by $$X_{a}|\varphi_{j}\rangle = \left(\rho(X_{a}) \right)_{ij} |\varphi_{i}\rangle , \ \ i = 1, \cdots , p. \ \ a = 1, \cdots , n \ , \ \ (3)$$ the $p \times p$ matrix representation of the generators $X_{a}$ $$\left(\rho(X_{a}) \right)_{ij} = \langle \varphi_{i}| X_{a} |\varphi_{j} \rangle .$$ As you can easily check, this matrix satisfies the conditions (1) & (2).

Now, one possible representation space $\mathscr{V}$, on which the generators $X_{a}$ can act, is that with basis vectors $|X_{b}\rangle$. That is, $\mbox{dim}(\mathscr{V}) = n$, the dimension of the Lie algebra $\mathscr{L}^{n}$. So, to turn the Lie algebra $$[X_{a} , X_{b}] = C_{abc}X_{c} ,$$ into representation space $\mathscr{L}^{n} \cong \mathscr{V}^{n}$, we define the inner product $\langle X_{a}|X_{b}\rangle = \delta_{ab} ,$ and the action $$X_{a}|X_{b}\rangle = C_{abc}|X_{c}\rangle . \ \ \ \ \ \ \ \ (4)$$ We now have everything as in the general case discussed above. Comparing (4) with (3), we find $$\left( \rho (X_{a}) \right)_{cb} = C_{abc} .$$ So, in the representation space $\mathscr{L}^{n}$, the Lie algebra relation $$\mbox{ad}_{X_{a}}(X_{b}) = [X_{a} , X_{b}] = C_{abc}X_{c} ,$$ reads $$|\mbox{ad}_{X_{a}}(X_{b}) \rangle = |[X_{a} , X_{b}] \rangle = C_{abc} |X_{c} \rangle ,$$ or $$|\mbox{ad}_{X_{a}}(X_{b}) \rangle = \left( \rho (X_{a})\right)_{cb} |X_{c}\rangle .$$

 

[MathematicalPhysicist]

I suggest for Lie groups and algebras using the textbook by Brian Hall, there are more advanced books (which are also mentioned in this book, like Varadarjan's (not sure for spelling)).

What you call $\rho$ a representation in your definition, is called in Hall's book: "Lie algebra homomorphism".
So if I understand, a representation on a Lie algebra is a Lie algebra homomorphism between the Lie algebra of p-dimensional vector spaces and its linear operators acting on a p-dimensional vector space.

 

[samalkhaiat]

What you call $\rho$ a representation in your definition, is called in Hall's book: "Lie algebra homomorphism".

Linear representation is a linear map (condition 1 above), that preserves the Lie algebra structure, i.e., Lie algebra homomorphism (condition 2 above).

 

[MathematicalPhysicist]

BTW @samalkhaiat I reackon you know more than me in Lie algebras and groups.

Do you happen to know how to solve my question that I posed from Hall's textbook, here:

http://math.stackexchange.com/quest...ion-5-8-in-the-book-of-brian-halls-lie-groups

Appreciate your help.
It's purely mathematical question, but I assume that physicists also know their maths.

 

[samalkhaiat]

BTW @samalkhaiat I reackon you know more than me in Lie algebras and groups.

Do you happen to know how to solve my question that I posed from Hall's textbook, here:

http://math.stackexchange.com/quest...ion-5-8-in-the-book-of-brian-halls-lie-groups

Appreciate your help.
It's purely mathematical question, but I assume that physicists also know their maths.

You can repeat the calculations with $\pi(X_{a}) \to \pi^{*}(X_{a}) = - X_{a}^{T}$. Or, you can use the following theorems to reach the same conclusion:
1) Every irreducible representation $\pi$ of $sl(3,C)$ has a unique highest weight given by a pair of positive integers $\mu (\pi) = (n_{1} , n_{2})$.
2) Two irreducible representations, $\pi_{1}$ and $\pi_{2}$ with $\mu (\pi_{1}) = \mu (\pi_{2})$ are equivalent, i.e., there exists a non-singular matrix $D$ such that $\pi_{2} = D^{-1}\pi_{1}D$.
3) For $n_{1} \neq n_{2}$, a representation $\pi$ and it’s conjugate $\pi^{*}$ are in-equivalent irreducible representations of the same dimension:
$$\mbox{Dim}(\pi) = \mbox{Dim}(\pi^{*}) = \frac{1}{2}(n_{1} + 1)(n_{2} + 1)(n_{1} + n_{2} + 2) .$$

So, if $\mu(\pi) = (n_{1}, n_{2})$, the above theorems tell you that the highest weight of $\pi^{*}$ must be $\mu(\pi^{*}) = (n_{2} , n_{1})$.

 

[CAF123]

_{oi @MathematicalPhysicist , taken over my thread :P ...just kidding :)}

So, given a complete orthonormal set $\{|\varphi_{i}\rangle \}$ of basis vectors in $\mathcal{V}^{p}$, the action of the generators $X_{a}$ on $|\varphi_{j}\rangle$ is determined by $$X_{a}|\varphi_{j}\rangle = \left(\rho(X_{a}) \right)_{ij} |\varphi_{i}\rangle , \ \ i = 1, \cdots , p. \ \ a = 1, \cdots , n \ , \ \ (3)$$ the $p \times p$ matrix representation of the generators $X_{a}$ $$\left(\rho(X_{a}) \right)_{ij} = \langle \varphi_{i}| X_{a} |\varphi_{j} \rangle .$$ As you can easily check, this matrix satisfies the conditions (1) & (2).

Ok many thanks for the detailed explanations! - my comments are:

1) Being treated as a matrix/vector product, should we not write $X_{a}|\varphi_{j}\rangle = \left(\rho(X_{a}) \right)_{ji} |\varphi_{i}\rangle$ with the indices on the rho flipped? Then conclude that $\rho(\lambda_a)_{bc} = C_{abc}$?

2) I'd interpret the rhs of that equation as mixing around the $\varphi$'s with j and i labelling what $\varphi$ we are talking about in the set $\left\{|\varphi \rangle \right\}$. What is wrong with writing the transformation as $$X_{a}|\varphi_{j}\rangle_{\ell} = \left(\rho(X_{a}) \right)_{\ell k} |\varphi_{j}\rangle_k?$$ It does not allow for an easy identification of what rho is and I mean e.g $|\varphi_i\rangle_j$ to stand for the jth component of the ith $p \times 1$ basis vector.

 

[samalkhaiat]

1) Being treated as a matrix/vector product, should we not write $X_{a}|\varphi_{j}\rangle = \left(\rho(X_{a}) \right)_{ji} |\varphi_{i}\rangle$ with the indices on the rho flipped?

No, because we define the matrix element of an operator $X_{a}$ in the basis $\{|\varphi_{i}\rangle\}$ as $(X_{a})_{kj} \equiv \langle \varphi_{k}| X_{a}|\varphi_{j}\rangle$. So, if you multiply Eq(3) above by $\langle \varphi_{k}|$, you get $$(X_{a})_{kj} \equiv \langle \varphi_{k}| X_{a}|\varphi_{j}\rangle = \rho_{ij} \langle \varphi_{k}|\varphi_{i}\rangle = \rho_{kj} .$$ For an arbitrary vector $|\Psi \rangle = \sum^{p} C_{j}|\varphi_{j}\rangle$ in the representation space $\mathcal{V}^{p}$, you have $$X_{a}|\Psi\rangle = \rho_{ij}C_{j}|\varphi_{i}\rangle .$$ This leads to the following transformation law for the components $C_{i}$ of the vector $|\Psi\rangle$ along the basis $|\varphi_{i}\rangle$ $$C^{’}_{k} = \langle \varphi_{k}|X_{a}|\Psi\rangle = \rho_{ij}C_{j}\delta_{ik} = \rho_{kj}C_{j} .$$

Then conclude that $\rho(\lambda_a)_{bc} = C_{abc}$?

No, because of the above mentioned reason as well as the fact that my generators $X_{a}$ are anti-hermitian, i.e., there is no $i$ on the RHS of the algebra.

2) I'd interpret the rhs of that equation as mixing around the $\varphi$'s with j and i labelling what $\varphi$ we are talking about in the set $\left\{|\varphi \rangle \right\}$.

This is what any representation matrix does to the indices of the elements of the representation space. The representation space itself is called the index space.

What is wrong with writing the transformation as $$X_{a}|\varphi_{j}\rangle_{\ell} = \left(\rho(X_{a}) \right)_{\ell k} |\varphi_{j}\rangle_k?$$

What is correct about it? Who (apart from you) wrote down such thing? Vectors carry one and only one index.

I mean e.g $|\varphi_i\rangle_j$ to stand for the jth component of the ith $p \times 1$ basis vector.

Component with respect to what? The jth component of the ith basis vector is simply $\delta_{ij}$. As I just told you, the indices characterize the representation space. So, when you stick an index on some object, you must specify the way it transforms. Don’t feel free to invent meaningless objects.

 

[CAF123]

What is correct about it? Who (apart from you) wrote down such thing? Vectors carry one and only one index.

Component with respect to what? The jth component of the ith basis vector is simply $\delta_{ij}$. As I just told you, the indices characterize the representation space. So, when you stick an index on some object, you must specify the way it transforms. Don’t feel free to invent meaningless objects.

I was trying to think about it in terms of ordinary matrix/vector multiplication, e.g of the form $\mathbf v' = M \mathbf v$ or in component form $v_i' = M_{ij}v_j$ and $\mathbf v = v_i e_i$ so $v_i$ are the components of $\mathbf v$ in the basis $\left\{e_j \right\}$. So, with the problem at hand, say we fix a and b. This means $X_b$ is just some vector in the representation space. The action of another element $X_a$ on this element is through the Lie bracket $[X_a, X_b]$.

I'm saying perhaps this can be written as $\rho(X_a)_{cd} (X_b)_d$, where $\rho(X_a)$ would be equivalent to the $M$ above and the $(X_b)$ would be equivalent to $\mathbf v$. Now, as you said, $(X_i)_j = \delta_{ij}$ so $$\rho(X_a)_{cd} (X_b)_d = \rho(X_a)_{cd} \delta_{bd} = \rho(X_a)_{cb} \overset{!}{=} [X_a, X_b]_c =C_{abd} (X_d)_c = C_{abd} \delta_{dc} = C_{abc}$$ and therefore $\rho(X_a)_{cb} = C_{abc}$?

I realize vectors carry only one index but the i index in $(X_i)_j$ is just a label, yes? So $\rho(X_a)_{bc} X_c$ is like a reshuffling of the X's - but by reshuffling the components of the X's as I think $\rho(X_a)_{cd}(X_b)_d$ means, we are also in effect mixing the X's because rho is a linear operator on the representation space (=linear vector space, so action of operator on a vector produces another vector in the space, by definition). So, in essence, I am saying this way of thinking about it in terms of components is equivalent.

Basically, I agree with everything you wrote but I just want to fully understand why my way thinking in terms of components is not correct - at least unless the calculation above is erroneous, I do get the correct expression for rho. Thanks for your help!