Re: [tied] IE *-su and the Nostratic "equational" marker *-n :)

I see two problems here.

(1) The plural ending is not *-(e)s but *-es (*k^wones), and the *-e- is not lost even in i/u-stem plurals like *-ej-es. Moreover, the Loc.pl. is based on a nil-grade form: *k^wnsu, not your "*k^wonsu". You could claim that an original *k^wones-ú became *k^wnsú, but it would be an ad hoc explanation inconsistent with the reconstructable relative chronology of vowel reductions. The Nom.pl. ending is not reduced when unstressed.

I'd claim instead that *k^wn-sú is a compound-like postpositional phrase, in which case vowel reduction is OK (cf. *k^wn-gWHén- 'dog-slayer').

(2) There are some adverbs ending in *-u, but I can't recall any actual locatives with this ending. When you say it's attested, what exactly do you mean?

Piotr

----- Original Message -----
From: Glen Gordon

To: cybalist@egroups.com

Sent: Saturday, November 18, 2000 8:07 AM

Subject: Re: [tied] IE *-su and the Nostratic "equational" marker *-n :)

Is there not attestation of *-su itself without a preceding *-i- (like say... Sanskrit /s'vasu/ < *k^won-su)? If so, and if the locative was unmarked by case ending in the singular at one time, as it appears with *k^won-su, we should expect the locative plural to have been *-(e)s like the nominative if anything. Since *-u is also an attested locative, we have a clear solution: *-su is composed of the plural plus a secondarily attached locative ending *-u. Even if it is found as *-i-su from time to time, the fact that we find *-su without *-i in defiance of the singular shows that the latter, less ordered form must be the original form. Since the pattern of *-i [sg] versus *-su [pl] is not immediately clear logically, a seemingly regular plural in *-is (locative sg + plural) would have to be an ending created to _replace_ the original ending *-su. At what point is the above logic flawed?