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I'm trying to solve the following problem:

Find the matrix elements $\langle u|T_a|v \rangle$ where $T_a$ are the $SU(3)$ generators and $|u\rangle$ and $|v\rangle$ are tensors in the adjoint representation of SU(3) with components $u^i_j$ and $v^i_j$. Write the results in terms of the tensor components and the $\lambda_a$ matrices.

First order of business, I don't think I understand exactly what's meant by the adjoint representation here. The text is Georgi's on Lie algebras and Lie groups. Is this the adjoint representation of the Lie algebra or the Lie group? Can it only be one for some reason?

Also, what exactly are the objects $|u\rangle$ and $|v\rangle$? Are they vectors in the adjoint representation, so linear combinations of the gell-mann matrices? For the lie algebra, to get the hexagonal weight decomposition, it requires 6 ladder operators and 2 Cartan generators. Are these the actual basis of the adjoint representation of the Lie algebra? If so, why?

Depending on the answer to the first question will help me understand how to put matrix elements here in terms of the gell-mann matrices. Can both $|u\rangle$ and $|v\rangle$ be written in terms of the gell-mann matrices (via some basis I guess?).

Lastly, will $T_a$ act on tensors $|u\rangle$ and $|v\rangle$ via commutation since we're working in the adjoint representation? If not, how exactly do indices contract here? Does $T_a$ have mixed, solely upper, or solely lower indices?

Is $T_a |v\rangle$ supposed to live in 3 $\otimes$ 8?

Edit: Georgi's text defines the action of the SU(3) generators on an arbitrary tensor $|v\rangle$ in the representation (n,m) as

$$ T_a v^{j_1 ... j_n}_{i_1 ... i_m} = \sum_{l=1}^n [T_a]^{j_l}_k v^{j_1 ... k ... j_n}_{i_1 ... i_m} - \sum_{l=1}^m [T_a]^{k}_{i_l } v^{j_1 ... j_n}_{i_1 ... k ... i_m} $$

but I don't understand why or where this would come from.

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I don't really understand the $v_j^i$ notation but I suppose $\vert v\rangle$ and $\vert u\rangle$ are basis for the carrier space of the adjoint irrep (1,1) or $\boldsymbol{8}$ as you write it.

Denote by $T_a $ the $8\times 8$ realization of $\lambda_a$. Then you need $T_a\vert v\rangle$, which is the action of $\lambda_a$ on the basis vector $\vert v\rangle$. Because we are dealing with the adjoint, where the action of the algebra on itself is the commutator, this is by definition equivalent to $$ [\lambda_a,\lambda_{v}] $$ where $\lambda_v$ is the Gell-Mann matrix corresponding to $\vert v\rangle$. Thus, your matrix elements are basically the structure constants.

If fact, $T_a\vert v\rangle$ lives in $\boldsymbol{8}\otimes \boldsymbol{8}$ but the action of the generators does not change the irrep so the result is also in $\boldsymbol{8}$.

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    $\begingroup$ $T_a$ acts in an 8-dimensional space as there are 8 basis elements: one per generator. $T_a$ transforms $\lambda_a$ (or $\frac12 \lambda_a$if you prefer; it’s a linear operation so the $\frac12$ is just a nomalization) into a linear combination of the 8 other generators, so it will act by $8\times 8$ matrices. Maybe the sticky point is to think of the $8$ $3\times 3$ matrices as basis vectors. $\endgroup$ Commented Jan 1 at 11:32
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    $\begingroup$ Editorial comment: if you’re trying learn this stuff from George good luck. You may be interested by this and in particular Tung or Cornwell. There’s also the review “Group Theory for unified model building” by Slansky $\endgroup$ Commented Jan 1 at 11:37
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    $\begingroup$ The tensor in the adjoint will have one upper and one lower index (as per the Dynkin notation (1,1).). Again, there are 8 of those tensors so your $T_a$ will be $8\times 8$. From the notation the $T_a$ will mix $3$ upper indices and $3$ lower indices but the tensor is trace less so that $8$ independent components. $\endgroup$ Commented Jan 1 at 11:44
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    $\begingroup$ you have to be careful here. If $W$ is a generic $8$ then in fact you can get anything in $8\otimes 8$ and I dunno this funny notation well enough but you get (2,2) (which I think is 27), 2 copies of (1,1) (which is I think 8 and $8^*$), (3,0) and (0,3) which I think are 10 and $10^*$, and a scalar (which I call $(0,0)$). However by definition the generators cannot change the irrep so if you act on $8$ you end up in $8$ and not in $8^*$, and nothing but $8$. $\endgroup$ Commented Jan 1 at 22:31
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    $\begingroup$ I'm not good with contracting indices but I can guarantee that my reasoning is correct (unless I misunderstood your question). Rutwig and Michel do tensor stuff in Campoamor-Stursberg, Rutwig, and Michel Rausch De Traubenberg. Group theory in physics: a practitioner’s guide. 2019, but to me I just get lost in this alphabet of indices and the rules on contraction so I never use this method. $\endgroup$ Commented Jan 1 at 22:34

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