Today I was trying to reconstruct some PIE roots by myself and I came across the word for '(to) know' in different indo-european languages. Here are some examples:

  • Eng. (to) know
  • It. conoscere
  • Lat. (g)nōscō (I know); nōtus (known)
  • Gr. γιγνώσκω gignōskō
  • Skr. जानाति jānāti (he knows); ज्ञात​ jñātá (known); ज्ञातृ jñātṛ (someone who knows)

In Sanskrit PIE vowels *a (< *h₂e), *e and *o all merged into अ a, nevertheless we could guess the original vowel quality looking at Greek. Since it has an ω ō I thought that PIE could show an alternation of the kind *ǵneh₃- ~ *ǵn̥h₃- and wikipedia seems to confirm my guess (cfr. the root ǵneh₃-).

If we read the reconstruction that wikipedia proposes, we can see that the Sanskrit forms are augmented with a nasal infix (in fact the verbal base belongs to the ninth class) resulting in *ǵn̥-né-h₃-ti ~ *ǵn̥-n-h₃-énti. Now the root has the zero grade. I know that the resonant *n̥ evolves regularly in अ a in Sanskrit, but why is it long in the verb जानाति jānāti? The long a also appears in the past participle and in the nomen agentis formed upon the same root. Why is that?

Finally Latin and Greek forms seem also quite strange to me. According to wikipedia's article, they are based on *ǵn̥h₃-sḱé-ti which is a kind of present with the durative-iterative suffix -sḱé. In Greek this verb shows a reduplication too. My doubt is one more time about the quantity of the radical vowel. Why is it long both in Latin and Greek? If the suffix *-sḱé requires the zero grade, the laryngeal could develop a short *e that is subsequently colored to *o when the laryngeal is lost, but then this vowel should remain short. Only in the e-grade the laryngeal could color the preceding vowel and then lengthen it when lost.

Is there some phonetic law that explains why these forms have a long vowel? Or it's just a case of analogy?

2 Answers 2


In Indo-Iranian both *eh₃ and *n̥h₃ become *ō, which then becomes ā. In Skt jānāti there is an infix *-ne- before the last consonant of the root, in this case the laryngeal. Thus the zero-grade root *ǵn̥h₃- forms the present *ǵn̥-ne-h₃-ti > *janāti (cf. Avestan zanā-) > jānāti (with ā in the first syllable by analogy to forms with *ǵn̥h₃- > jā-).

  • 1
    More precisely *n̥h₃ become nō (and more generally *RH > RV:).
    – TKR
    Dec 13, 2016 at 1:23
  • @TKR. In Indo-Iranian *n̥h₃ becomes ā (not nō as in Greek).
    – fdb
    Dec 13, 2016 at 11:30
  • @TKR I thought PIE *nH > Lat. nā. Can you add more info?
    – Alex B.
    Dec 13, 2016 at 20:41
  • 1
    @AlexB. Sorry, I was wrong -- my comment applies only to Greek.
    – TKR
    Dec 13, 2016 at 20:46

@fdb's answer addresses the Indo-Iranian forms, so this one will address the Greek and Latin ones.

In Greek, there are two relevant sets of sound changes:

  1. PIE *eH > Gk V̄. That is, *e followed by any laryngeal became a long vowel; which long vowel resulted depends on the specific laryngeal: *eh₁ > ē, *eh₂ > ā, *eh₃ > ō. (In Attic-Ionic, most instances of ā later underwent a further sound change to merge with ē.)
  2. PIE *R̥H > Gk RV̄. That is, any syllabic sonorant followed by a laryngeal turned into the non-syllabic version of that sonorant followed by a long vowel, again depending on the laryngeal. In this case, PIE *n̥h₃ > Gk nō.

What this means for γιγνώσκω is that in theory, it could come from either an e-grade or a zero grade: both *n̥h₃ and *neh₃ would end up as Gk nō.

In Latin, rule 1 is the same, but rule 2 is simpler: PIE *R̥H > Lat Rā. That is, whatever the specific laryngeal is, it gives Latin ā. (This rule is shared with Celtic.) This means that Latin (g)nōscō cannot come from a zero grade, but only from an e-grade.

It seems to be generally true that the suffix *-sḱé prefers the zero grade, but there are some other Latin counterexamples: Weiss (Comp. Hist. Gramm. Lat. 407) lists pāscō, suēscō and OL escit "exits" among these. LIV thinks the e-grade of (g)nōscō serves to avoid homophony with nāscor.

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