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

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*}

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

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


$\epsilon_1{}^A$ 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, $\delta_{\epsilon_1}\delta_{\epsilon_2}(\phi^i)=\delta_{\epsilon_2}\delta_{\epsilon_1}(\phi^i).$


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


2 Answers 2


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.

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

  • $\begingroup$ I think an error in your answer pointed me to the right answer! Notably, your order of the matrices in the first two lines are the same. This is incorrect. Matrix multiplication is associative, but not commutative. The order of the parameters are irrelevant, but the order of the matrix multiplication is not (so I do not think your dummy index argument is sound either). Thus I realize the commutator will indeed not vanish since the terms are not equal. $\endgroup$
    – Lopey Tall
    Dec 19, 2019 at 16:27
  • 1
    $\begingroup$ not so: note that the rhs of line 1 depends only on $1,2$ so the dummy indices in the rhs don't change with a $1\to2$ switch..I can elaborate if you wish $\endgroup$
    – lineage
    Dec 19, 2019 at 16:35

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.


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