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I'm studying abelian and non-abelian gauge theories, using the geometric approach. I have defined the generalized potential (that is the connection 1-form of the principal bundle $P(M,G)$ where $M$ is the Minkowski 4-dimentional space-time and $G$ is the structure group of the gauge theory and it has the same role in the principal bundle)

$$A\equiv iA_\mu^k(x)dx^\mu T_k$$

where $T_k$ are the generators of the structure group, so this is a Lie algebra-valued 1-form. Then we can define the curvature 2-form using the Cartan's Structure equation $F=dA+A\wedge A$, so

$$ F=\dfrac{1}{2}iF_{\mu\nu}^k dx^\mu\wedge dx^\nu T_k $$ which is a Lie algebra-valued 2-form. At this stage I'm still able to recognise the abelian Maxwell theory, which has $U(1)$ as structure group (so we have null structure constants and only one generator $\mathbb{I}$).

Then I'd like to say which is the Lagrangian density of the free-field theory (without external sources or matter): we know that we need a Lorentz-invariant and a gauge-invariant Lagrangian density, so we wish we could contract the indices of the group (latin indices). After some consideration and several steps, we conclude that we can use the adjoint representation of our structure group and use trace to contract the indices

$$ \operatorname{Tr}(F\wedge * F)=-\frac{1}{2}F_{\mu\nu}^kF^{\mu\nu\ j}\operatorname{Tr}(T_kT_j)\Omega $$ where $\Omega\equiv dx^0\wedge dx^1\wedge dx^3\wedge dx^3$ is the volume form and $\operatorname{Tr}(T_kT_j)$ is $\operatorname{Tr}(\operatorname{Ad}T_k \operatorname{Ad}T_j)$ for short.

My question is: how could I obtain the Maxwell Lagrangian density from this? I'm not well versed in this topic, but it seems to me that the adjoint representation of the generators of an abelian Lie group is the trivial representation, which means that $\operatorname{Ad}T_k$ is the $n\times n$ matrix with all $0$. So is everything zero here?


Edit: The only clue that I have is that, when the theory is abelian, $F$ is automatically gauge-invariant and so it is $F\wedge *F$, without needing the use of the trace. But I sill need some representation of the generators. Which one? I suppose we should not use the adjoint representation, which isn't faithful. I was thinking about the defining representation, but I'm not sure.


Edit 2: I'm aware that, for the Maxwell theory, if I use the defining representation of the only generator $1$, I obtain the desired Lagrangian density. I'd like to understand why for non-abelian gauge theories I have to use the Killing form ($=$ trace of the adjoint representations), while for abelian gauge theories I have to use the defining representation (or the trace of the defining representation? They're the same for $U(1)$ and I'd like to have a generic rule that works for every abelian group and not just for $U(1)$). I don't know if using the trace of the defining representations is the correct thing to do for a generic abelian theory, I just know that it works for $U(1)$.

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    $\begingroup$ I guess this is due to the fact that the Lagrangian density for gauge theories is defined through an invariant polynomial. For semisimple algebras this is the Killing form as you said. However, $\mathfrak{u}(1)$ is not semisimple and hence the Killing form is not a good choice. $\endgroup$ – NDewolf Dec 21 '20 at 19:55
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Doesn't the standard non abelian Lagrangian just reduce to the correct thing in the abelian case? The Lie algebra $\mathfrak{u}(1)$ is just generated by the $1 \times 1$ matrix known as "$1$", so here the only generator is $T^a$ where $a = 1$, and $T^1 = 1$. Then $\mathrm{Tr}(T^1) = 1, \mathrm{Tr}(T^1 T^1) = 1$, so the whole thing just reduces to $F \wedge \star F$ as it should.

Edit: To sum up the discussion in the comments, often times the Killing form is equal to $\mathrm{Tr}(T^a T^b)$. However, for $\mathfrak{u}(1)$ the Killing form is equal to $0$ because it is abelian. However, you can still just use $\mathrm{Tr}(T^a T^b)$ anyway, which gets you the right thing.

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  • $\begingroup$ Do you mean that I should use the defining representation of the generator of $U(1)$? Do I have to use the defining representation instead of the adjoint representation when I have an abelian group in general? $\endgroup$ – Nabla Dec 21 '20 at 20:21
  • $\begingroup$ In physics I think people usually just use $Tr(T^a T^b)$ as the Killing form and not the definition you gave with the adjoint thing. I agree with you that in the case of $\mathfrak{u}(1)$ the adjoint is just $0$. On this page en.wikipedia.org/wiki/Killing_form#Matrix_elements they give what the Killing form is in terms of symmetric polynomials, and for $\mathfrak{so}(2) = \mathfrak{u}(1)$ it is indeed $0$. $\endgroup$ – user1379857 Dec 21 '20 at 20:33
  • $\begingroup$ I'm sorry but isn't the definition of the Killing form "the trace of the composition of the adjoint representation of two generators"? I don't understand what do you mean saying that that trace is the Killing form and not the adjoint thing, aren't they the same thing? $\endgroup$ – Nabla Dec 21 '20 at 20:40
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    $\begingroup$ If there is something to say then it isn't common knowledge amongst physicists. The main reason to use $Tr(XY)$ is because it equals $Tr(g XY g^{-1})$, making the Lagrangian gauge invariant and hence well defined. The only fact I am aware of is that this will be positive definite if the group is compact. If it is non compact, then those directions will be negative. That's important because you need the kinetic term to have the correct sign. Other than that I'm not aware of any other principles for choosing bilinear forms or gauge groups. $\endgroup$ – user1379857 Dec 21 '20 at 21:35
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    $\begingroup$ It can also be shown that for thr Lie algebras of unitary groups ($n>1$) the Killing form just coincides with the usual trace. $\endgroup$ – NDewolf Dec 22 '20 at 8:34

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