I am currently working on Bilinear Pairing.To start my work i need to find the mathematically prove of three properties of bilinear pairing.

Let $ G_{1} $ and $ G_{T} $be a cyclic multiplicative group with the same prime order q, that is, $|G_{1}|$ =$ |G_{2}|$ = q. Let g be a generator of G1. An efficient bilinear map $ e $: $G_{1}$ × $G_{1}$ → $G_{T} $, with the following properties:

  1. Bilinear: for all g ∈ $G_{1}$ and a; b ∈ $ Z_{q}^{^{*}}$ , $ e(g^{a},g^{b})=e(g,g)^{^{ab}} $.
  2. Non-degenerate: there exists g ∈ G1 such that e(g, g) ̸= 1.

I need to find some mathematical prove for Bilinear map e. I choose a cyclic multiplicative group $G_{5} $ having generator g=2 and (a=3,b=4) then how to prove the first property ie $ e(2^{3},2^{4})= e(2,2)^{^{3.4}}$

  • $\begingroup$ What map(s) do you need to prove satisfy(ies) those properties? $\;$ $\endgroup$ – user991 Jan 15 '14 at 11:20
  • $\begingroup$ What you give is a definition. There is nothing to prove for a definition :/ If you have a concrete map $e$ then you could ask proving that it satisfies the definition. $\endgroup$ – DrLecter Jan 15 '14 at 11:32
  • $\begingroup$ @DrLecter i did not understand how the bilinear equation is proved. Say we take G1=(0,1,2,3,4) and g=(2) ,a=3 and b=4 then how we will prove e(2^3,2^4)=e(2,2)^3.4 $\endgroup$ – Raginisingh Jan 15 '14 at 11:55
  • $\begingroup$ @Raginisingh Since you ask the question here, I assume that you are looking for pairings for cryptographic use? This also requires that the discrete logarithm problem in $G1$ and $GT$ is hard. Such pairings are only known to exist on certain elliptic curve groups for $G1$ ($G2$) and related multiplicative groups of finite fields (for $GT$). In your example you use the group $G1=(Z_5,+)$ and what would be $e(2,2)^{3\cdot 4}$, since you have not specified $GT$. For a more general (non crypto) treatment of bilinear maps you may look here. $\endgroup$ – DrLecter Jan 15 '14 at 12:21
  • $\begingroup$ @DrLecter sir actually i am unable to understand how $G_{T}$ is calculated .can u help me out taking some real time example with mathematical equations. $\endgroup$ – Raginisingh Jan 15 '14 at 12:31

You are on the wrong track there. You first choose the groups and then want to find a bilinear map, but it does not work this way. Basically, all you can do is choose groups and function and then check IF your construction fulfills the definitions of pairings. Some basic properties have to "fit" (e.g. the orders of the groups need to be compatible).

If you are able to create pairings on just any groups of your liking, you just had a major algebraic breakthrough.

For cryptographic applications, we also need that the target group has an difficult problem, mostly DLOG or Diffie-Hellman. This eliminates any constructions of groups with addition modulo $n$.

In that case, the only pairing friendly groups we know of are based on Weil pairings and Tate pairings.

If we just use your middle question

I need to find some mathematical prove for Bilinear map e.

you should have a closer look at Weil pairings in Wikipedia. The explanation is quite complicated, but there is no easy one.

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  • $\begingroup$ Actually, the group formed by addition modulo $N$ (for both $G_1, G_2$ and $G_T$) is also pairing friendly, and the pairing operation is quite easy. However, it isn't cryptographically interesting... $\endgroup$ – poncho Sep 7 '16 at 13:46
  • $\begingroup$ @poncho You're right, I'll edit the answer. $\endgroup$ – tylo Sep 7 '16 at 14:04

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