Re: Gemination in Celtic

From: fournet.arnaud
Message: 56972
Date: 2008-04-07

----- Original Message -----
From: "Anders R. Joergensen" <ollga_loudec@...>

> =========

>Well, these things may be very difficult to determine, especially as
>no convincing examples have been put forward, on which to form an
>opinion.
>Anders
=======
Well, any time I give an example where -k(k)- alternates with -g-,
it's either unclear or not attested here and there, or unconvincing.
Subtratic French is not Celtic, etc.
Discussing with you is an interesting experiment,
as you generally offer cryptically short answers and all kinds of dodgings.
Arnaud
==========
>It seems that there's an awful lot of intuition in this. I hope the
>evidence for the remaining six alternative h2's is stronger than the
>evidence for our present h2 (= ?). I must say that I have yet to see
>Have you entertained the idea that there is no regular correspondence
>Italo-Celtic *k(k) = Eastern *g?
>Anders
=============
I keep thinking that
*bhel-H2-k "beam"
Gaulish *bala:kon
SKrt bhur-i-jau < -H-g-
Greek phalan-g-s
Latin ful-c-io
is clear

So far you have never a single comment on this.

Quite obviously, if you don't look at the possible examples,
It's little wonder you cannot see a convincing example.
Why don't you try to look at this one ?

Arnaud
===========
>You tell me.

It's from kwreyH2-k > creicc-
I suppose this obvious example is not convincing, as usual.

If you think it has to be cut otherwise,
why don't you explain your point ?
instead of answering with riddles.
Arnaud

> ============
>> the -sk- suffix can turn *bhreH2-k- into *bhreH1-k-sk-
>How?

=========
Because sk is unvoiced and devoices H2 into H1.
Arnaud
===========

>> Isn't there a Letton verb blaz- with a ?

>Pokorny lists a Latvian one (with long -a:-). I don't see the
>relevance.
=========
I suppose a: signals H2,
Bla:z < bla:g < bleH2-k
Without extra unvoiced -sk-, *H2-k > *g
But I suppose this is not a clear example, as usual.
What is the problem with this example ?
Arnaud
==========
>
> I consider this root is *bh_s. with three states :
> unaffixed bh_H2
> infixed bh-l-_H2
> infixed bh-r-_H2
> All meaning to shine, to be bright.

>The relevance of this of course depends on your ability to show that
>bre:ks^ti has -h2- > -h1- somehow.
>Anders
=======
See above.
Arnaud
========