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The mass term for the type-I seesaw is given by $$\mathcal{L}_{mass}=-m_D\overline{\nu_L}N_R+M_R\overline{(N_R)^c}N_R+h.c.$$ where the right-chiral fields $N_R$ are electroweak singlets. Since they do not have any Standard model interactions, they are called sterile. On the other hand, $\nu_L$ fields are not sterile.

Now I've found two more terms in the literature. Light neutrinos and active neutrinos. When we diagonalize the $6\times 6$ neutrino mass matrix $M_\nu=\begin{pmatrix}0 & m_D\\m_D^T & M_R\end{pmatrix}$ resulting from the Lagrangian above, we get, 3 light Majorana neutrinos and three heavy Majorana neutrinos (both are linear combinations of $\nu_L$ and $N_R$).

My question is, among $\nu_L$, the 3 light Majorana neutrinos or the 3 heavy Majorana neutrinos what are referred to as active neutrinos? And why?

In wikipedia, $\nu_L$ of the Standard Model are referred to as active which I think is in contradiction with the link above.

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A sterile neutrino is any neutrino which does not interact with Standard Model gauge bosons, and an active neutrino is any neutrino which does interact with Standard Model gauge bosons.

Easiest way to distinguish them would be to ask whether there is a 3-point Feynman diagram you can draw for that neutrino where one leg is a gauge boson.

I think this definition is consistent both with Wikipedia and with the way it's used in the paper you link to. Note that even in the title they refer to the "mixing of sterile and active neutrinos" which agrees with the Wikipedia definition. The sterile $N_R$ neutrinos mix with the active $\nu_L$ neutrinos to form mass eigenstates which are also active. However, if the heavy mass eigenstates have very weak gauge interactions, they will be approximately sterile.

In the paper you link to, they do choose to refer to these heavy mass eigenstates as sterile, even though they are technically very weakly interacting. They state:

We shall call $N_I$ as sterile neutrinos since they possess very suppressed gauge interactions.

To summarize, $N_R$ are the sterile neutrinos, and any other linear combination between that and $\nu_L$, including $\nu_L$ itself, is active. But if you're talking about an eigenstate that is very nearly sterile, it may make sense to refer to it as sterile as well.

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