This is about the KZG Polynomial Commitment Scheme
In Section 2, it's written
We use the notation $e : \mathbb G \times \mathbb G \mapsto \mathbb G_T$ to denote a symmetric (type 1) bilinear pairing.The choice of type 1 pairings was made to simplify presentation, however, our constructions can easily be modified to work with pairings of types 2 and 3 as well.
The above uses only one Elliptic Curve Group whose generator $\mathbb G$.
In the appendix where the document describes Bilinear Pairings, this is the description.
For three cyclic groups $\mathbb G,\widehat{\mathbb G}$ and $\mathbb G_T$ (all of which we shall write multiplicatively) of the same prime order $p$, a bilinear pairing $e$ is a map $e : \mathbb G \times \widehat{\mathbb G} \mapsto \mathbb G_T$
Note that here 2 different elliptic curve groups $\mathbb G$ & $\widehat{\mathbb G}$ respectively are used (instead of $\mathbb G$ being used as both the first & second group in the earlier description).
And this is not just for type 1 pairings, but all 3 types of pairings can have different generators.
I am a little confused about how the proof would work if the 2 generators were different.
For e.g. if we have to prove
$Q(x) = \frac {F(x) - c} {x-b}$,
This is how we do it.
$Q(x)(x-b) = F(x) -c$
Evaluating at $x=a$
$Q(a)(a-b) = F(a) - c$
Now we multiply both sides by $G$ (which is the generator of the group $\mathbb G$ & this becomes
$Q(a).G.(a-b) = F(a).G - c$
Now $C_Q = Q(a).G$ (commitment of $Q$) &
$C_F = F(a).G$ (commitment of $F$)
From here, it's easy to prove that if $e(Q(a)G, (a-b)G) \stackrel{?}{=} e((F(a)-c)G, G)$ is true, then the original statement $Q(x)(x-b) = F(x) -c$ is true. (This proof is also given in the KGZ paper in multiplicative notation). However, I am not able to derive the proof for $G_1 \ne G_2$ - how do I derive the proof for the above in case I am using two different groups with generators $G_1$ & $G_2$.
Now this easy converting of a polynomial into it's commitment by multiplying both sides by the single generator is key to using pairings to prove without knowing the polynomial itself.
I am not sure how this translates to the two groups being different $\mathbb G$ & $\widehat{\mathbb G}$ because they would then have different generators & you won't be able to multiply both sides by the same generator to convert the polynomial into it's commitment so as to verify a proof without knowing the polynomial itself (and just knowing the SRS).
So where exactly is the generator of the 2nd Group $\widehat{\mathbb G}$ used & how does one convert the polynomials into their respective commitments in that case.
