Gauge theory formalism

Question

Source:

Chapter 11 of Freedman and Van Proeyen’s supergravity [textbook][1]

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An infinitesimal symmetry transformation is determined by

1. a parameter, call it $\epsilon^A$, and

2. an operation, call it $\delta_\epsilon$.

The operation $\delta_\epsilon$

1. depends linearly on the parameter $\epsilon^A$, and

2. acts on fields, i.e. $\delta_\epsilon(\phi^i)$.

For some global symmetry, $\epsilon^A$ does not depend on the spacetime $x^\mu$.

Another way to say "$\delta_\epsilon$ depends linearly on the parameter $\epsilon^A$,'' is to write $$\delta_\epsilon =\epsilon^A T_A$$

where the $T_A$ are some operations on fields.

Let $\{(t_A)^i{}_j\}$ be the matrix generators of a representation of some Lie algebra.

This Lie algebra (LA) is defined by $[t_A,t_B]=f_{AB}{}^C t_C$.

The action of $T_A$ on the fields is defined with the LA basis elements,

$$ T_A(\phi^i)=-(t_A)^i{}_j \phi^i $$

So then we have

\begin{eqnarray*} \delta_\epsilon(\phi^i) &=& \epsilon^A T_A(\phi^i) \\ &=& -\epsilon^A (t_A)^i{}_j (\phi^j) \end{eqnarray*}

Then the product of two symmetry transformations reads,

\begin{eqnarray*} \delta_{\epsilon_1}\delta_{\epsilon_2}(\phi^i) &=& \epsilon_1{}^A T_A(\epsilon_2{}^B T_B\phi^i) \\ &=& \epsilon_1{}^A T_A(-\epsilon_2{}^B (t_B)^i{}_j \phi^j) \\ &=& -\epsilon_1{}^A \epsilon_2{}^B (t_B)^i{}_j T_A \phi^j \\ &=& -\epsilon_1{}^A \epsilon_2{}^B (t_B)^i{}_j (-(t_A)^j{}_k \phi^k) \\ &=& \epsilon_1{}^A \epsilon_2{}^B (t_B)^i{}_j (t_A)^j{}_k \phi^k \\ \end{eqnarray*}

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The authors then go on to state the commutator, which I am concerned with,

$$ \begin{aligned} \ [\delta(\epsilon_1), \delta(\epsilon_2)] \phi^i & = \epsilon_2^B \epsilon_1^A [T_A, T_B] \phi^i \\ & = \epsilon_2^B \epsilon_1^A ([t_B, t_A]\phi)^i = -\epsilon_2^B \epsilon_1^A {f_{AB}}^C (t_C\phi)^i\\ & = \epsilon_2^B \epsilon_1^A {f_{AB}}^C T_C\phi^i \end{aligned} \tag{11.6} $$

Notably,

$\epsilon_1{}^A$ and $\epsilon_2{}^B$ are numbers, and are so are commutative, and

$(t_B)^i{}_j,$ $(t_A)^j{}_k,$ and $\phi^k$ are matrices, and so are associative.

Thus, how is $[\delta_{\epsilon_1},\delta_{\epsilon_2}](\phi^i) = \delta_{\epsilon_1}\delta_{\epsilon_2}(\phi^i) - \delta_{\epsilon_2}\delta_{\epsilon_1}(\phi^i)$

not equal to zero?

As far as I can tell, $\delta_{\epsilon_1}\delta_{\epsilon_2}(\phi^i)=\delta_{\epsilon_2}\delta_{\epsilon_1}(\phi^i).$

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Disclaimer:

That formula for the product of symmetry transformations on the fields is my work. The authors have,

$$ \begin{aligned} \delta(\epsilon_1) \delta(\epsilon_2) \phi^i & = \epsilon_1^A T_A \epsilon_2^B [-{(t_B)^i}_j \phi^j] \\ & = \epsilon_1^A \epsilon_2^B {(-t_B)^i}_j T_A \phi^j \\ & = \epsilon_1^A \epsilon_2^B {(-t_B)^i}_j {(-t_A)^j}_k \phi^k . \end{aligned} \tag{11.5} $$

which seems the same as mine but I just want to be cautious.

Cheers [1]: https://www.cambridge.org/gb/academic/subjects/physics/theoretical-physics-and-mathematical-physics/supergravity?localeText=United%20Kingdom&locale=en_GB&remember_me=on

Answer 1

I think its mostly a notational confusion. You derivation for the $\delta$ commutator does lead to the same conclusion as the authors.

$$ \begin{eqnarray*} \delta_1\delta_2 &=&\epsilon_1{}^A \epsilon_2{}^B (t_B)^i{}_j (t_A)^j{}_k \phi^k \\ \delta_2\delta_1 &=&\epsilon_2{}^A \epsilon_1{}^B (t_B)^i{}_j (t_A)^j{}_k \phi^k \\ &=& \epsilon_1{}^B \epsilon_2{}^A (t_B)^i{}_j (t_A)^j{}_k \phi^k \\ &=& \epsilon_1{}^A \epsilon_2{}^B (t_A)^i{}_j (t_B)^j{}_k \phi^k \end{eqnarray*} $$ where the last step follows from $A,B$ being dummy vars.

$$ \begin{eqnarray*} [\delta_1,\delta_2] &=& \epsilon_1{}^A \epsilon_2{}^B\left( (t_B)^i{}_j (t_A)^j{}_k-(t_A)^i{}_j (t_B)^j{}_k\right) \phi^k \\ &=& \epsilon_1{}^A \epsilon_2{}^B\left( (t_B t_A)^i{}_k-(t_At_B)^i{}_k\right) \phi^k \\ &=& \epsilon_1{}^A \epsilon_2{}^B([t_B,t_A] )^i{}_k \phi^k\\ &=& \epsilon_1{}^A \epsilon_2{}^B(-f_{AB}{}^Ct_C )^i{}_k \phi^k\\ &=& \epsilon_1{}^A \epsilon_2{}^Bf_{AB}{}^C(-t_C )^i{}_k \phi^k\\ &=& \epsilon_1{}^A \epsilon_2{}^Bf_{AB}{}^CT_C \phi^i\\ \end{eqnarray*} $$

Answer 2

Thank you to @lineage for the prompting!

\begin{eqnarray*} \delta_{\epsilon_1}\delta_{\epsilon_2}(\phi^i) &=& \epsilon_1{}^A \epsilon_2{}^B (t_B)^i{}_j (t_A)^j{}_k \phi^k \\ &\neq& \epsilon_1{}^A \epsilon_2{}^B (t_A)^i{}_j (t_B)^j{}_k \phi^k \end{eqnarray*}

since matrix multiplication is not commutative.

Thus the two terms in the commutator are not identical.