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Lately I am encountering the commutator of variations of the variables and I'm not quite sure about its physical meaning.

Some examples.

1) "The composition of two supersymmetries generates a time translation: \begin{equation} [\delta_S(\epsilon_1), \delta_S(\epsilon_2)]x=\delta_{T}(a)x" \end{equation} where the subscripts stand for the transformation ($S$ for supersymmetry, $T$ for time translation) and the parenthesis contain the infinitesimal parameter.

2) "One requires the nilpotency, i.e. \begin{equation} [\delta_B(\Lambda_1), \delta_B(\Lambda_2)]=0 \end{equation} on all the variables".

I want to stress out that my question is not about supersymmetry, but it concerns the use of the commutators. I would have thought that requiring nilpotency would have translated in the condition $\delta_B(\Lambda_1) \delta_B(\Lambda_2)=0$ and, similarly, that "The composition of two supersymmetries generates a time translation" would have translated in $\delta_S(\epsilon_1) \delta_S(\epsilon_2)x=\delta_{T}(a)x$.

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    $\begingroup$ This is just the physicist's convention of speaking: The composition of two symmetry transformations generates a trivial "sum part", but, significantly, a further commutator part. Think of the beginning of the CBH expansion and the form of the relevant exponent. Your question might equally well apply to ordinary rotations about different axes, and the composition of the relevant generators. $\endgroup$
    – Cosmas Zachos
    Commented Jan 1, 2019 at 22:53

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It comes from the Baker-Campbell-Hausdorff formula:

$$ e^X e^Y= e^{X+Y +\frac{1}{2} [ X, Y] + \cdots } . $$

We are talking about generators, but remember that the actual group elements are the exponentiation of the Lie algebra. So to first order in the infinitesimal parameter, the composition of two group elements is

\begin{align*} e^{\delta_S(\epsilon_1)}e^{\delta_S(\epsilon_2)} &= e^{\delta_S(\epsilon_1) + \delta_S(\epsilon_S) + \frac{1}{2}[\delta_S(\epsilon_1), \delta_S(\epsilon_2)]}\\ &= e^{\delta_S(\epsilon_1) + \delta_S(\epsilon_S) + \frac{1}{2}\delta_T(a) } \end{align*}

and so we see that the composition of two supersymmetries contains a time translation.

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    $\begingroup$ Thanks, you got the point perfectly! So it "contains" a time translation but it's not purely a time translation. $\endgroup$
    – Luthien
    Commented Jan 2, 2019 at 14:58
  • $\begingroup$ @Luthien that’s right $\endgroup$
    – InertialObserver
    Commented Jan 2, 2019 at 20:31
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In general a Lie group $G$ with an associative product $\circ$ induces a Lie algebra $\mathfrak{g}=T_eG$ with a Lie bracket $[\cdot,\cdot]$ that satisfies the Jacobi identity. Similarly, any representation of the Lie group $G$ induces a corresponding representation of the Lie algebra $\mathfrak{g}$.

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  • $\begingroup$ These are the "technical" details for the argument, but the argument itself is missing (although stated in the other answers). To make this answer self-contained, I would add that, when talking about the composition (product) of generators, one has to use the algebraic structure on $T_eG$ (i.e. the Lie bracket) induced by the Lie group structure on $G$, since the generators are the elements of this space. In a representation, a Lie bracket can be represented as a commutator. $\endgroup$
    – Phoenix87
    Commented Jan 2, 2019 at 10:54
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    $\begingroup$ @Phoenix87: I agree. $\endgroup$
    – Qmechanic ♦
    Commented Jan 2, 2019 at 11:03
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    $\begingroup$ Thanks for your answer @Qmechanic, but I was not confused about the concept of the commutator and Lie algebra. I needed the link between transformations and the commutator, which is the CBH expansion, as the other answer pointed out. $\endgroup$
    – Luthien
    Commented Jan 2, 2019 at 15:01

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