# Importance of non-degeneracy property of bilinear map for cryptography

I'm currently looking into pairing-based cryptography and I stumbled upon the definition of the properties bilinearity, computability and non-degeneracy.

Now I have a problem with understanding the non-degeneracy and how it is important to the security of elliptic curve cryptography. I have not found a paper that goes into detail about it, only from a mathematical standpoint which is a little to abstract for me.

Let $$e: G_1 \times G_2 \rightarrow G_T$$ where $$G_1, G_2$$ and $$G_T$$ are of prime order $$p$$.

Non-degeneracy is defined as: $$\forall P \in G_1,P \neq 0, \exists Q \in G_2: \quad e(P,Q) \neq 1$$ $$\forall Q \in G_2,Q \neq 0, \exists P \in G_1: \quad e(Q,P) \neq 1$$

tl;dr Why is the non-degeneracy an important property for pairings in cryptographic applications?

In cryptography the groups in question are typically cyclic and of prime order and non-degeneracy is equivalent to saying that $$e(P_1,P_2)\neq 1$$ where $$P_1$$ and $$P_2$$ are generators for the groups. If this is not the case, then all of our pairing computations will produce the answer 1.
In pairing-based encryption we regularly compute shared secret values that are of the form $$e(P_1,P_2)^{abc}$$ or other pairing outputs that are supposed to be confidential. In pairing-based verification we regularly produce pairing outputs that are only supposed to be constructible by someone in possession of secret knowledge. In both cases an adversary with a degenerate pairing, the adversary can easily construct any pairing output because they would know that it is always 1.
• Does this relate to the complexity of reversing the output of the pairing? E.g. if i know the result of the pairing is 1, I can construct a pairing that solves the equation $e(P_1, P_2) = 1$? I'm trying to make sense of how an attacker could actually exploit a degenerate pairing. Feb 15 at 13:02