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Consider the Yang-Mills action $S = -\frac{1}{2g^{2}} \int d^{x}\ \mathrm{Tr}\left( F_{\mu \nu} F^{\mu \nu} \right)$, where $F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\mu} A_{\nu} - i [A_{\mu},A_{\nu}]$ is the field strength tensor. Let the gauge group be $U(N)$, so that $A_{\mu}$ belongs to the adjoint representation of $U(N)$.

I'm trying to do a problem which says the following: the adjoint representation of $U(N)$ can be thought of as products of fundamental and anti-fundamental representations - so $\bar{N} \times N$. They say, adopt the following way to write the group index of $A_{\mu}^{a}$, such that $a = (\bar{j}, k)$. In this way, the above means the following in terms of matrix elements: $$ A^{a}_{\mu} \ \to \ A^{(\bar{j},k)}_{\mu} = \left[ A_{\mu} \right]_{jk} $$

I'm asked to compute the structure constants $f^{(\bar{m},n)}_{(\bar{j},k),(\bar{p},q)}$ which are defined by the expression: $$ [ A_{\mu}, A_{\nu} ]^{(\bar{m}, n)} = i f^{(\bar{m},n)}_{(\bar{j},k),(\bar{p},q)} A^{(\bar{j},k)}_{\mu} A^{(\bar{p},q)}_{\nu} $$

What exactly am I trying to do here?

My Attempt: My thinking is that I pick some set of generators $\{ T^{a} \}$ for $U(N)$, which defines a set of structure constants $\{f_{abc}\}$ (relative to my choice of generators).

The generators of the adjoint representation $\{T_{\mathrm{AD}}^{a}\}$, have elements determined by $[T_{\mathrm{AD}}^{a}]_{bc} = i f_{abc}$. Then I think I can expand $A_{\mu}$ as: $$ A_{\mu} = \sum _{a} A_{\mu}^{a} T_{\mathrm{AD}}^{a} $$

But from here I don't have any idea what i am doing....am I supposed to write the structure constants $f^{(\bar{m},n)}_{(\bar{j},k),(\bar{p},q)}$ in terms of the $f_{abc}$?

(Eventually this problem is to lead me towards the t'Hooft double line formalism.)

EDIT: It occurred to me that maybe I need to write the $f^{(\bar{m},n)}_{(\bar{j},k),(\bar{p},q)}$ in terms of the structure constants of the fundamental and anti-fundamental respresentations?

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    $\begingroup$ Thanks for your reply! I'm ripping my hair out over this problem. I'm sorry, I'm not following. Which expression did I write for the generators of the fundamental representation? $\endgroup$ – Greg.Paul Dec 11 '16 at 1:32
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    $\begingroup$ The structure constants of the Lie algebra are identical in all representations. But your edit is on the right track... You are asked to look at NxN matrices, not (N²-1) x (N²-1) matrices. Look at the NxN matrix expression defining these f's, your second equation. Write out the l.h.side explicitly. Compare to the rhside: f is a tensor with two free indices, and 4 saturated ones. So it starts life out as a δδδδ-δδδδ ... are any of these contracted? $\endgroup$ – Cosmas Zachos Dec 11 '16 at 1:48
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    $\begingroup$ i can work out the following off the top of my head: $\mathrm{LHS}=[A_{\mu},A_{\nu}]^{(\bar{m},n)} = [A_{\mu}A_{nu}]^{(\bar{m},n)}-[A_{\mu}A_{\nu}]^{(\bar{m},n)} = A_{\mu}^{(\bar{m},s)}A_{\nu}^{(\bar{s}, n)} - A_{\nu}^{(\bar{m},s)}A_{\mu}^{(\bar{s}, n)}$. Where would the kronecker $\delta$'s come from? $\endgroup$ – Greg.Paul Dec 11 '16 at 1:54
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    $\begingroup$ Now compare to the right hand side, where the As are in tensor product. So, m is j, k is s, ... etc... $\endgroup$ – Cosmas Zachos Dec 11 '16 at 1:56
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    $\begingroup$ Absolutely! That's the point. I apologize for my glibness above: for U(N) the adjoint has dimension N², not this -1. This is not uncommon language for O(N) and, significantly, the Lorentz group, as well. $\endgroup$ – Cosmas Zachos Dec 11 '16 at 14:53
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Everything happens in the Lie algebra so consider $u$ to be the relevant Lie algebra (I'm tired of typing mathfrak...).

The adjoint representation is the representation $\rho$ given by the Lie algebra acting on itself by the Lie bracket:

$\rho:u\longrightarrow M(u),\quad \rho(X)\longmapsto [X,.].$

It is a representation because the Lie bracket is linear and $u$ itself is a vector space, hence the action can be represented as a matrix. Remember that this map is an algebra homomorphism. Thus algebraic relations are preserved.

Now, suppose the basis $\{T^i\}_{i\in I}$ of $u$ with $[T^i,T^j]=f_{k}^{ij} T^k$. What we want is to express this last algebraic relation in terms of objects in the adjoint representation. Let us first express these matrices as their matrix elements:

For $X\in u$, $X=X_i T^i \implies \rho(X)_{i}^j=[X,T^j]_i=X_k f_i^{kj}$.

In particular, for $X=T^k$, we have that

$\rho(T^k)_{i}^j=f_i^{kj}.$

So we can write

$\rho([T^a,T^b])_{i}^j=f_{c}^{ab}\rho(T^c)_{i}^j=f_c^{ab}f_i^{cj}$

And knowing that $(\rho(T^a)\rho(T^b))_i^j=f_i^{ak}f_k^{bj}$, we have that

$\rho([X_n T^n,Y_m T^m])_i^j=X_n Y_m\, \rho([T^n,T^m])_i^j=-X_n Y_m f_i^{jk} f_k^{nm}=X_n Y_m (f_i^{mk}f_k^{nj}-f_i^{nk}f_k^{mj})$, using Jacobi Identity.

So with $\rho(X)_a^b\, \rho(Y)_c^{d} F_{(ac)i}^{(bd)j}=X_n Y_m f_a^{nb}f_c^{m d}F_{(ac)i}^{(bd)j}\implies F_{(bd)i}^{(ac)j}=(\delta_{d}^a\delta_{b}^j\delta_{i}^c-\delta_{b}^c\delta_{d}^j\delta_{i}^a)$

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  • $\begingroup$ One question: What does $M(u)$ mean in your notation? I'm guess the space of linear mappings $u \to u$? $\endgroup$ – Greg.Paul Dec 11 '16 at 7:14
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    $\begingroup$ Exactly, it is the space of linear operators, also known as matrices. $\endgroup$ – G. Bergeron Dec 11 '16 at 7:30
  • $\begingroup$ One more question: $T^{i}$ are the generators of $U(N)$ right? So does this mean that $T^{i}_{\mathrm{Adj}}:=\rho(T^{i})$ are the generators of the adjoint representation? $\endgroup$ – Greg.Paul Dec 11 '16 at 7:38
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    $\begingroup$ They are basis, not all possible generators, which would be the entire Lie algebra. Your second point is stretching definitions. They are called generators because they generate, when exponentiated, a group element. the fact that you consider them in the abstract or in a representation doesn't change anything. $\endgroup$ – G. Bergeron Dec 11 '16 at 7:45
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    $\begingroup$ Yes, but my point was that you want a basis of generator. They are generators of $U(N)$. The other part about the adjoint representation is that algebraic objects are abstract structures. A representation is a realization of that structure with concrete mathematical objects. So in a sense yes they are "the same generators", but in a concrete realization instead of as an abstract object. Formally speaking, $U(N)$ are not the unitary matrices, only there algebraic structure. Those matrices are the realization called the fundamental representation. $\endgroup$ – G. Bergeron Dec 11 '16 at 10:25
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I think that I have figured it out thanks to Mr. Zachos. Given the equation defining the structure constants: $$ \left[ A_{\mu}, A_{\nu} \right]^{ ( \bar{m}, n ) } = i f^{( \bar{m}, n )}_{ ( \bar{j},k )(\bar{p},q) } A_{\mu}^{(\bar{j},k)} A_{\nu}^{(\bar{p},q)} $$

We can expand the left hand side such that:

$=A_{\mu}^{(\bar{m}, s)} A_{\nu}^{(\bar{s}, n)} - A_{\mu}^{(\bar{s}, n)} A_{\nu}^{(\bar{m}, s)} \\ = \left[ \delta_{mj} \delta_{sk} \delta_{sp} \delta_{nq} - \delta_{sj} \delta_{nk} \delta_{mp} \delta_{sq} \right] A_{\nu}^{(\bar{p}, q)} A_{\mu}^{(\bar{j}, k)} \\ = \left[ \delta_{mj} \delta_{kp} \delta_{nq} - \delta_{jq} \delta_{nk} \delta_{mp} \right] A_{\nu}^{(\bar{p}, q)} A_{\mu}^{(\bar{j}, k)} \\ = i \left[ i \big( \delta_{jq} \delta_{kn} \delta_{pm} - \delta_{jm} \delta_{kp} \delta_{qn} \big) \right] A_{\nu}^{(\bar{p}, q)} A_{\mu}^{(\bar{j}, k)} $

Which means that the structure constants have the form: $$ f^{( \bar{m}, n )}_{ ( \bar{j},k )(\bar{p},q) } = i \big( \delta_{jq} \delta_{kn} \delta_{pm} - \delta_{jm} \delta_{kp} \delta_{qn} \big) $$

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