Question about Trivial Gauge Transformation

Question

Suppose we have an action $S=S(a,b,c)$ which is a functional of the fields $a,\, b,\,$ and $c$. We denote the variation of $S$ wrt to a given field, say $a$, i.e. $\frac{\delta S}{\delta a}$, by $E_a$.

Then $S$ is gauge invariant when

$$\delta S = \delta a E_a + \delta b E_b +\delta c E_c = 0 \tag{1}$$

This gives

$$\delta c = - (\delta a E_a + \delta b E_b)/E_c \tag{2}$$

From the above well-defined equation $\delta c$ can be obtained for arbitrary $\delta a$ and $\delta b$. It is not necessary to have a relation b/w $\,\delta a$ and $\delta b$. So there is an infinite number of gauge-transformations here. These gauge transformations vanish when the equations of motion hold, for e.g. $\delta c$ in $(2)$ vanishes when $E_a=0\,\&\,E_b=0$. In other words, these gauge transformations vanish on-shell.

As per sec. 3.1.5 of Ref. 1, such gauge transformations are called "trivial gauge transformations". According to theorem 3.1 such gauge transformations can always be expressed as $\delta y^i=\varepsilon^{ij}\frac{\delta S}{\delta y^j}$ where the action $S$ is a functional of the fields $y^i$ and $\varepsilon^{ij}=-\varepsilon^{ji}$.

I am having trouble obtaining the components of the matrix $\varepsilon^{ij}$ for the above example. If $y^1=a,\,y^2=b,$ and $y^3=c\,$ then from $\delta y^1=-(\delta bE_b+\delta cE_c)/E_a\,$ I read $\varepsilon^{12}=(-\delta b)/E_a$. From $\delta y^2=-(\delta aE_a+\delta cE_c)/E_b\,$ I read $\varepsilon^{21}=(-\delta a)/E_b$. But these two components of $\varepsilon^{ij}$ are not simply (-1) times each other as they should be. What's going wrong?

References:

1. M. Henneaux & C. Teitelboim, Quantization of Gauge Systems, 1994; subsection 3.1.5.

Answer 1

Answer to the question (v5):

1. First all, a gauge transformation that vanish on-shell $$ \delta y^i~ \approx~ 0\quad \wedge \quad \delta y^i\frac{\delta S}{\delta y^i}~=~0 \tag{3.11}$$ is a so-called trivial gauge transformation $$\exists \mu^{ij}=-\mu^{ji}:~~ \delta y^i~=~\mu^{ij}\frac{\delta S}{\delta y^j},\tag{3.7}$$ cf. Thm 3.1 in Ref. 1. The proof of Thm 3.1 is rather technical and involves (among other things) a Koszul-Tate resolution.

2. OP's transformation (2) is singular at the EL equations $E_d=0$ for the $d$-field. Away from the singular locus $E_d\neq 0$, OP's transformation (2) is of the form (3.11) where Thm 3.1 applies.

References:

1. M. Henneaux & C. Teitelboim, Quantization of Gauge Systems, 1994; subsection 3.1.5.

Answer 2

I found a way to express the trivial gauge transformations in a way such that they have anti-symmetric $\varepsilon^{ij}$ as coefficients!

$$\delta a = \frac{-(\delta b E_b + \delta c E_c)}{E_a}=\Big(\frac{\delta a}{E_b}-\frac{\delta b}{E_a}\Big)\frac{E_b}{3}+\Big(\frac{\delta a}{E_c}-\frac{\delta c}{E_a}\Big)\frac{E_c}{3} \tag{a}$$ $$\delta b = \frac{-(\delta c E_c+\delta a E_a)}{E_b}=\Big(\frac{\delta b}{E_c}-\frac{\delta c}{E_b}\Big)\frac{E_c}{3}+\Big(\frac{\delta b}{E_a}-\frac{\delta a}{E_b}\Big)\frac{E_a}{3} \tag{b}$$ $$\delta c = - \frac{(\delta a E_a+\delta b E_b)}{E_c}=\Big(\frac{\delta c}{E_a}-\frac{\delta a}{E_c}\Big)\frac{E_a}{3}+\Big(\frac{\delta c}{E_b}-\frac{\delta b}{E_c}\Big)\frac{E_b}{3} \tag{c}$$

We can read off from eq. (a) that $\varepsilon^{ab}=\frac{1}{3}\Big(\frac{\delta a}{E_b}-\frac{\delta b}{E_a}\Big)$. And we can read off from eq. (b) that $\varepsilon^{ba}=\frac{1}{3}\Big(\frac{\delta b}{E_a}-\frac{\delta a}{E_b}\Big)$. Now we can easily see that $\varepsilon^{ab}=-\varepsilon^{ba}$.