Are there two different adjoint representations used in the SM?

[victorvmotti]

We know from Lie representation theory that the Lie algebra is a vector space. Therefore a representation of the Lie group can be transformation of this vector space itself which we call the Adjoint Representation. An element of this vector space, is itself represented by a matrix. For example, in the case of $SU(3)$, to show the adjoint representation, we write $V\psi(x)V^{-1}$, where $V$ is a $3*3$ special unitary matrix, $\psi(x)$ is also a $3*3$ (traceless) matrix, i.e. an element in the Lie algebra vector space. So we are multiplying three $3*3$ matrices. A new vector, or $3*3$ matrix, is produced, which belongs to or is an element in the Lie algebra vector space.

But in a SM textbook I see another definition/use of the Adjoint Representation. This time we consider the (abstract) commutation relations among the generators of the $SU(3)$ and interpret them as a vector acts on the another to produce a new vector. In other words, use the structure constants as elements to build a $8*8$ matrix to represent the Lie algebra vectors: $L_{1},...L_{8}$. Now if we consider an octet, or an 8 component column object, $\psi(x)=(\psi_{1}, ..., \psi_{8})^T$ each of $\psi_{i}$ a fermion or Dirac spinor, with four complex components, i.e. spinor indices suppressed, we can write for example $L_{1}\psi(x)$ to make or show a transformation of the fermion fields.

Now my question is that in the example of $SU(3)$ are these two different Adjoint Representations related somehow, as one of them involves the operation of $3*3$ matrices and the another one involves the operation of $8*8$ matrices?

My first impression is that in the first case, the 3 dimensional one, we are using the adjoint representation of the group using the Lie algebra as a vector space whereas in the second case, the 8 dimensional one, we are using the adjoint representation of the Lie algebra using the Lie algebra itself as a vector space. And both of them are useful in writing the Lagrangian density of the SM. Is this correct?

 

[fresh_42]

Only $\operatorname{SU}(3)$ and $\mathfrak{su}(3)$ are $3\times 3$ matrices which therefore operate on a three dimensional complex vector space.

There is an adjoint representation of the Lie group:
$$
\operatorname{Ad}\, : \,\operatorname{SU}(3) \longrightarrow \operatorname{GL}(\mathfrak{su}(3)) \, , \,x\longmapsto \left(A\longmapsto xAx^{-1}\right)
$$
on its Lie algebra, and an adjoint representation of the Lie algebra on itself:
$$
\operatorname{ad}\, : \,\mathfrak{su}(3) \longrightarrow \mathfrak{gl}(\mathfrak{su}(3)) \, , \,X\longmapsto \left(A\longmapsto [X,A]=XA-AX\right)
$$
Both adjoint representations - one in group theory, the other one in algebra theory - are related via
$$
\operatorname{Ad}(\exp(A)) = \exp(\operatorname{ad}(A))
$$
Hence the representation spaces is the eight dimensional Lie algebra $\mathfrak{su}(3)$ in both cases, such that we have $8\times 8$ matrices as operation. The representations itself are either a group homomorphism, or a Lie algebra homomorphism, so they are two completely different things. Only the elements of $\operatorname{SU}(3)$ and $\mathfrak{su}(3)$ are $3\times 3$ matrices.

 

[victorvmotti]

Both adjoint representations - one in group theory, the other one in algebra theory - are related via
$$
\operatorname{Ad}(\exp(A)) = \exp(\operatorname{ad}(A))
$$

Thanks. Part of my question answered, that these two adjoint representations are related.

But

1/ shouldn't we use instead of equality, the group homomorphism:

$$
\operatorname{Ad}(\exp(A)) \cong \exp(\operatorname{ad}(A))
$$

Because, assuming that $A$ is a $3 \times 3$ matrix $\in \mathfrak{su}(3)$ then using the maps you defined, we have
$$
\operatorname{Ad}(\exp(A)) \in \operatorname{GL}(\mathfrak{su}(3))
$$
and
$$
\exp(\operatorname{ad}(A)) \in \operatorname{SU}(3)
$$

2/How about instead of using a $3 \times 3$ matrix $A \in \mathfrak{su}(3)$, based on the structure constants for the Lie algebra elements, we use $8 \times 8$ matrix $A \in \mathfrak{su}(3)$ as explained in this lecture note, equation (19)? Will the relationship $\operatorname{Ad}(\exp(A)) \cong \exp(\operatorname{ad}(A))$ still be valid? In this case $\exp(\operatorname{ad}(A))$ and $\exp(A)$ are $8\times 8$ matrices $\in \operatorname{SU}(3)$, right?

 

[fresh_42]

The structure constants are the coefficients of the Lie multiplication. Given a basis $\{L_1,\ldots,L_8\}$ of $\mathfrak{su}(3)$ then
$$
(\operatorname{ad}(L_i))(L_j)=[L_i,L_j]=L_iL_j-L_jL_i=\sum_{k=1}^8 c_{ij}^kL_k
$$

Will the relationship $\operatorname{Ad}(\exp(A)) = \exp(\operatorname{ad}(A))$ still be valid?

It is always valid, modulo technical details. It represents the connection of differentiation and integration: the Lie algebra is the tangent space of the $1$-connection component of the Lie group at $1$. That tangent space attached at group $1$ has its origin $0$ at the same point. $\exp$ is quasi the integration from the Lie algebra as tangent space into the Lie group as manifold. This is the same principle as with solving differential equations, where we also often make an Ansatz with the exponential function. It is only a more generalized context here.

In this case $\exp(\operatorname{ad}(A))$ and $\exp(A)$ are $8\times 8$ matrices $\in \operatorname{SU}(3)$, right?

No. The exponential function doesn't change the dimensions of the matrix it is applied to. Since $A$ is a $3\times 3$ matrix, so is $\exp(A)$. But $\operatorname{ad}$ maps $\mathfrak{su}(3)$ matrices to $\mathfrak{su}(3)$ matrices, i.e. it maps an eight dimensional vector to another eight dimensional vector, which is a $8\times 8$ matrix.

The confusion may come from the fact that the matrices (generators) in $\mathfrak{su}(3)$ are considered vectors in the representation space of the adjoint representation. We have $8$ linear independent $3\times 3$ matrices which build an $8$ dimensional vector space $V=\mathfrak{su}(3)$. Lie group $\operatorname{SU}(3)$ and Lie algebra $\mathfrak{su}(3)$ both operate on $V$ via $8\times 8$ matrices by the corresponding adjoint representations. So in a way, those $3\times 3$ matrices are represented by $8\times 8$ matrices, which is their action on the $8$ dimensional vector space $V$.

https://www.physicsforums.com/insights/representations-precision-important/
has the Gell-Mann matrices as example.

 

[victorvmotti]

So in a way, those $3\times 3$ matrices are represented by $8\times 8$ matrices, which is their action on the $8$ dimensional vector space $V$.

The case of $3 \times 3$ matrix make sense to me as an $8$ dimensional vector space.

But, did you see the lecture note I just shared, eq (19). There $A$ is not a $3 \times 3$ matrix but $A$ is a $8 \times 8$ matrix $\in \mathfrak{su}(3)$. We then map it by adjoint to another $8 \times 8$ matrix in $\mathfrak{su}(3)$ . And then use the exp map to arrive at a $8 \times 8$ matrix in $SU(3)$.

 

[fresh_42]

The case of $3 \times 3$ matrix make sense to me as an $8$ dimensional vector space.

But, did you see the lecture note I just shared, eq (19). There $A$ is not a $3 \times 3$ matrix but $A$ is a $8 \times 8$ matrix $\in \mathfrak{su}(3)$. We then map it by adjoint to another $8 \times 8$ matrix in $\mathfrak{su}(3)$ . And then use the exp map to arrive at a $8 \times 8$ matrix in $SU(3)$.

$\mathfrak{su}(3)$ is an eight dimensional vector space which consists of $3\times 3$ matrices.

However, the adjoint representation is first a Lie algebra homomorphism, because it is a representation. This means that the kernel of this homomorphism is an ideal in $\mathfrak{su}(3)$. But $\mathfrak{su}(3)$ has no proper ideals, because it is a simple Lie algebra. Thus the kernel is either $\{0\}$ or the entire Lie algebra. The latter is not possible, since this would imply that the adjoint representation maps everything onto $0$, which it does not. Hence the adjoint representation has only a trivial kernel, i.e. it is injective. If we now consider the mapping $\mathfrak{su}(3) \longrightarrow \operatorname{im}_{\operatorname{ad}\mathfrak{su}(3)}=\operatorname{ad}(\mathfrak{su}(3)) \subseteq \mathfrak{gl}(\mathfrak{su}(3))$ onto its image under the adjoint representation, which is a subspace of all $8\times 8$ matrices in $\mathfrak{gl}(\mathfrak{su}(3))$, then we have an isomorphism. With $\mathfrak{su}(3) \cong_{\operatorname{ad}} \operatorname{ad}(\mathfrak{su}(3))$ we have identified the $3\times 3$ matrices of $\mathfrak{su}(3)$ with the $8 \times 8$ matrices of their image under the adjoint representation in $\mathfrak{gl}(\mathfrak{su}(3))$.

That is we have a $1:1$ correspondence $L_i \longleftrightarrow \operatorname{ad}(L_i)$ between the $3\times 3$ matrices $L_i$ with the $8\times 8$ matrices $\operatorname{ad}(L_i)$. This means especially that the structure constants are the same.

 

[victorvmotti]

Many thanks, now all clear!

 

[samalkhaiat]

.. my question is that in the example of $SU(3)$ are these two different Adjoint Representations related somehow, as one of them involves the operation of $3*3$ matrices and the another one involves the operation of $8*8$ matrices?

Yes, and it is a simple exercise to find the connection. Recall that the adjoint representation $[8]$ is the traceless part of the tensor product of the two fundamental representations $[\bar{3}] \otimes [3]$. So, as you said, the (Hermitian traceless) matrix field $\Psi (x)$ transforms according to $$\Psi (x) \to U^{\dagger} \Psi (x) U , \ \ \ U = e^{- \frac{i}{2} \alpha_{a} \lambda^{a}} , \ \ \ \ (1)$$ where $\lambda^{a}, \ a = 1, \cdots , 8$ are the $3 \times 3$ (traceless Hermitian) Gell-Mann matrices of the fundamental representation. Since any $3 \times 3$ traceless Hermitian matrix can be expanded in terms of the $\lambda^{a}$’s, we can write $$\Psi (x) = \psi^{a}(x) \lambda^{a} ,$$ where $\psi^{a}(x) = \frac{1}{2} \mbox{Tr} \left( \lambda^{a} \Psi (x)\right)$ are 8 real fields (the components of a vector in 8-dimensional vector space). Now, you can finish the exercise by finding the transformation law of the 8-vector $\psi^{a}$ from (1): $$\lambda^{c}\psi^{c}(x) \to U^{\dagger}\lambda^{a}U \ \psi^{a}(x) .$$ Expand and use the algebra $[\lambda^{a} , \lambda^{b}] = 2if^{abc}\lambda^{c}$: $$U^{\dagger} \lambda^{a} U = \left( 1 + (- i \alpha \cdot t) + \frac{1}{2!} (- i \alpha \cdot t)^{2} + \cdots \right)^{ab} \ \lambda^{b} = \left( e^{- i \alpha \cdot t}\right)^{ab} \ \lambda^{b} ,$$ where $t^{c}$ are eight $8 \times 8$ matrices defined by $(t^{c})^{ab} = - i f^{cab}$ and, therefore, satisfy the Lie algebra of $\mbox{SU}(3)$. So Eq(1) becomes $$\lambda^{c}\psi^{c}(x) \to \psi^{a}(x) \left( e^{- i \alpha \cdot t }\right)^{ab} \lambda^{b} .$$ Thus, using $\mbox{Tr}(\lambda^{a}\lambda^{b}) = 2 \delta^{ab}$, you find the transformation law for the 8 real fields: $$\psi^{b}(x) \to \left( e^{- i \alpha \cdot t}\right)^{ab} \psi^{a}(x) . \ \ \ \ \ (2)$$ So, you have two alternative ways to describe the transformation of fields in the adjoint representation: Eq(1) for the Hermitian traceless (matrix) field $\Psi (x)$ and Eq(2) for the real 8-component field $\psi^{a}(x) = \frac{1}{2} \mbox{Tr} \left(\lambda^{a} \Psi (x) \right)$.

 

[victorvmotti]

Very helpful exercise indeed, thank you!

Any hints on this related question posted here about suppressed and unsuppressed Dirac indices in the matrix field adjoint representation?

 

[samalkhaiat]

Any hints on this related question posted here about suppressed and unsuppressed Dirac indices in the matrix field adjoint representation?

The fields in the above exercise were generic. I supressed all space-time indices because the members of the octet $[8]$(i.e., the matrix elements of $\Psi (x)$) can be Lorentz scalars, vectors, spinnors or tensors of any rank. The three famous examples are: 1) The pseudo-scalar ($J^{p} = 0^{-}$) meson octet $$M[8] = \begin{pmatrix} \frac{\pi^{0}}{\sqrt{2}} + \frac{\eta_{8}}{\sqrt{6}} & \pi^{+} & K^{+} \\ \pi^{-} & - \frac{\pi^{0}}{\sqrt{2}} + \frac{\eta_{8}}{\sqrt{6}} & K^{0} \\ K^{-} & \bar{K}^{0} & - \frac{2 \eta_{8}}{\sqrt{6}} \end{pmatrix},$$ 2) The vector meson ($J^{p} = 1^{-}$) octet $$V_{\mu}[8] = \begin{pmatrix} \frac{\rho_{\mu}^{0}}{\sqrt{2}} + \frac{\omega_{\mu}}{\sqrt{6}} & \rho_{\mu}^{+} & K_{\mu}^{\ast +} \\ \rho_{\mu}^{-} & - \frac{\rho_{\mu}^{0}}{\sqrt{2}} + \frac{\omega_{\mu}}{\sqrt{6}} & K_{\mu}^{\ast 0} \\ K_{\mu}^{\ast -} & \bar{K}_{\mu}^{ \ast 0} & - \frac{2 \omega_{\mu}}{\sqrt{6}} \end{pmatrix}$$ And 3) The baryon ($J^{p} = (1/2)^{-}$) octet $$B[8] = \begin{pmatrix} \frac{\Sigma^{0}}{\sqrt{2}} + \frac{\Lambda}{\sqrt{6}} & \Sigma^{+} & p \\ \Sigma^{-} & - \frac{\Sigma^{0}}{\sqrt{2}} + \frac{\Lambda}{\sqrt{6}} & n \\ \Xi^{-} & \Xi^{0} & - \frac{2 \Lambda}{\sqrt{6}} \end{pmatrix}$$ Here, of course, all the members of $B[8]$ carry Dirac’s indices. Try to obtain the real fields $b^{a} = \frac{1}{2}\mbox{Tr}(\lambda^{a}B[8])$.