linear computations over bilinear pairings

Question

Does this hold in asymetric bilinear pairings?

$e(x_1,x_2)e(x_3,x_4) = e(x_1x_3,x_2x_4)$, where $x_1,x_3 \in \mathbb{G}_1$ and $x_2,x_4 \in \mathbb{G}_2$ for a bilinear pairing $e$

Answer 1

I use multiplicative notation for $\mathbb{G}_1$ and $\mathbb{G_2}$ (as in your question) as this may make it a bit clearer for you. Lets say the groups are of prime order $p$ and and lets consider your elements w.r.t. bases $g_1$ and $g_2$. Lets write your elements $x_1$ and $x_3$ as $g_1^{a_1}$ and $g_1^{a_2}$ and $x_2$ and $x_4$ as $g_2^{b_1}$ and $g_2^{b_2}$ respectively.

Then lets look at your equation $e(x_1,x_2)e(x_3,x_4) = e(x_1x_3,x_2x_4)$.

We just use bilinearity below. You see that on your left hand side you have $$e(g_1^{a_1},g_2^{b_1})e(g_1^{a_2},g_2^{b_2})=e(g_1,g_2)^{a_1b_1+a_2b_2}$$ Lets take a look at the right hand side $$e(g_1^{a_1}g_1^{a_2},g_2^{b_1}g_2^{b_2})=e(g_1^{a_1+a_2},g_2^{b_1+b_2})=e(g_1,g_2)^{(a_1+a_2)(b_1+b_2)}=e(g_1,g_2)^{a_1b_1+a_2b_2+a_1b_2+a_2b_1}$$ Now if your equality should hold you need to have that $a_1b_2+a_2b_1=0$ (in $\mathbb{Z}_p$) or in other words $e(g_1,g_2)^{a_1b_2}e(g_1,g_2)^{a_2b_1}=e(x_1,x_4)e(x_3,x_2)=1$. So as Thomas has already written this is not true in the general case.

Answer 2

For clarity, I will use an additive notation in the groups $\mathbb{G}_1$ and $\mathbb{G}_2$, and denote group elements with the "$x$" letter, not "$e$".

For all $x_1, x_3 \in \mathbb{G}_1$ and $x_2, x_4 \in \mathbb{G}_2$, we have the following:

\begin{eqnarray*} e(x_1 + x_3, x_2 + x_4) &=& e(x_1, x_2 + x_4) e(x_3, x_2 + x_4) \\ &=& e(x_1, x_2) e(x_1, x_4) e(x_3, x_2) e(x_3, x_4) \end{eqnarray*}

(First equation is by linearity of the pairing for its first operand, second equation is by linearity for its second operand.)

Therefore, one can have the property you seek only when $e(x_1,x_4)$ and $e(x_3,x_2)$ happen to be opposite of each other, which is not true in general (unless the pairing is degenerate, in which case it is quite uninteresting).