Two days ago I posted a post that discusses a very generic gauge transformation. I repeat it here. 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 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 $\delta b$. This implies that there is an infinite number of gauge-transformations here. At first I was very puzzled by this property of infinite number of gauge transformations. But then a kind PSE responder directed me to look into "trivial gauge transformations" in Ref. 1.
I looked there and I deduced that the gauge transformation discussed in (1) and (2) above is a trivial gauge transformation. I even showed in an answer to my earlier question how the above gauge transformation can be expressed with an antisymmetric matrix as stated by Theorem 3.1 in Ref. 1.
But I think I made a mistake in my earlier post where I said that $\delta c$ vanishes on-shell. When the equations of motion hold, i.e. when $E_a=E_b=E_c=0$, $\delta c$ in (2) becomes undefined; it does not vanish. Does that mean that (2) is not a trivial gauge transformation? If so, how can one explain the infinite number of gauge transformations in (2)?
Equations (1) and (2) hold true for all gauge transformations. If (2) is indeed a trivial gauge transformation then doesn't that imply that all gauge transformations are trivial? But every gauge transformation cannot be trivial. How to distinguish a trivial gauge transformation from a non-trivial one?
References:
1: M. Henneaux & C. Teitelboim, Quantization of Gauge Systems, 1994; subsection 3.1.5.