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Antisymmetry Kayne, Richard (1995)

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Antisymmetry Kayne, Richard (1995) CAS LX 523 Syntax II (1) A Spring 2001 March 13, 2001 qp Paul Hagstrom Week 7: Antisymmetry BE 33 Kayne, Richard (1995). The antisymmetry of syntax. Cambridge, MA: MIT Press. CDFG 1111 Koopman, Hilda (2000). The spec-head configuration. In Koopman, H., The syntax of cdef specifiers and heads. London: Routledge. (2) A node α ASYMMETRICALLY C-COMMANDS β if α c-commands β and β does not The basic proposals: c-command α. X-bar structures (universally) have a strict order: Spec-head-complement. There is no distinction between adjuncts and specifiers. • B asymmetrically c-commands F and G. There can be only one specifier. • E asymmetrically c-commands C and D. • No other non-terminal nodes asymmetrically c-command any others. But wait!—What about SOV languages? What about multiple adjunction? Answer: We’ve been analyzing these things wrong. (3) d(X) is the image of a non-terminal node X. Now, we have lots of work to do, because lots of previous analyses relied on d(X) is the set of terminal nodes dominated by node X. the availability of “head-final” structures, or multiple adjunction. • d(C) is {c}. Why make our lives so difficult? Wasn’t our old system good enough? • d(B) is {c, d}. Actually, no. • d(F) is {e}. A number of things had to be stipulated in X-bar theory (which we will review); • d(E) is {e, f}. they can all be made to follow from one general principle. • d(A) is {c, d, e, f}. The availability of a head-parameter actually fails to predict the kinds of languages that actually exist. (4) d<X,Y> for non-terminal nodes X and Y is the set of all ordered pairs of terminals A more restrictive grammar is simpler—perhaps slightly closer to plausible in a and b such that a is in d(X) and b is in d(Y). terms of acquisition. • d<B,F> is {<c,e>, <d,e>} Up through the early 1990’s, people generally considered syntactic structures to be kind of • d<B,E> is {<c,e>, <c,f>, <d,e>, <d,f>} like “mobiles”—c-command, dominance, hierarchy was important, but linear order was not important. Hence, Japanese SOV sentences were in all syntactically relevant respects (5) d{<B,F>,<B,E>} is the union of d<B,F> and d<B,E>. just like English SVO sentences. {<c,e>, <d,e>} ∪ {<c,e>, <c,f>, <d,e>, <d,f>} = {<c,e>, <c,f>, <d,e>, <d,f>} Kayne’s proposal says that (asymmetric) c-command and linear order are directly (6) The Linear Correspondence Axiom related—something that comes first asymmetrically c-commands something that comes Let T be the set of terminals in a given phrase marker P. later. No parameter to set. Let A be the maximal set of pairs of non-terminals X and Y in P such that X asymmetrically c-commands Y d(A) is a linear ordering of T. • if <a,b> and <b,c> then <a,c>. • for any two terminals a and b, either <a,b> or <b,a>. • for no terminals a and b does both <a,b> and <b,a>. What does that mean? Linear precedence is the relation at issue, terminal nodes are the (8) * XP d(A) = d{<YP,ZP>, <YP,X>, <X′,Y′>, <X,Z′>} words, and so the LCA says that, as determined by asymmetric c-command in the tree, 3 ′ every word either precedes or follows every other word, but not both. YP X d<YP,ZP> = {<y,z>} “y precedes z” 13 Y′ XZP d<YP,X> = {<y,x>} “y precedes x” ′ ′ It seemed complicated, but the intuition is clear: 11 1 d<X ,Y > = {<x,y>} “x precedes y” • Language requires articulating words in order. Y 1 Z′ d<X,Z′> = {<x,z>} “x precedes z” (not even forced by the language faculty—forced by the real world) 11 1 • Language ensures that this is possible. 11 Z so d(A) = {<y,z>, <y,x>, <x,y>, <x,z>} • The formal version “it is possible to pronounce the words in order” is just 11 1 z follows both x and y. yx z as given above. y precedes both x and z. but x also precedes y…! The cool thing about the LCA is that a number of things we used to have to stipulate about X-bar theory now just come for free. It couldn’t have been any other way. This won’t do; we haven’t even gotten off the ground. Obviously, the one we need to get rid of is <X′,Y′>. Labeling: A head is a nonterminal that dominates no other nonterminals. We need it to turn out that X′ does not asymmetrically c-command Y′. (A nonhead is a nonterminal that dominates at least one other nonterminal) The solution. Suppose that the X-bar structure arises from adjoining YP to X′ (which we There is an immediate problem: (7) is ruled out. might as well call “XP” now), and define c-command such that XP won’t c-command YP. And this is a big problem, since (7) is the only X-bar structure there’s supposed to be. (9) α c-commands β if and only if: (7) * XP • α and β are categories (that is, not a segment unless it is the top segment) 3 α β α β YP X′ • no segment of dominates (“ excludes ”), and 13 • every category that dominates α also dominates β. Y′ XZP 11 1 In English: adjuncts c-command what they adjoin to, but not vice-versa ′ Y 1 Z (below, YP c-commands XP, but XP doesn’t c-command YP). 11 1 11 Z 11 1 yx z (10) XP d(A) = d{<YP,ZP>, <YP,X>, <X,Z′>} rU YP XP A = {<YP,ZP>, <YP, Z′>, <YP, Z>, <YP,X>, <X′, Y′>, <X′, Y>, <X, Z′>, <X,Z>} d<YP,ZP> = {<y,z>} “y precedes z” 13 have the same image same image same image Y′ XZP ′ ′ 11 1 d<X ,Y > = {<x,y>} “x precedes y” so d(A) = d{<YP,ZP>, <YP,X>, <X′,Y′>,<X,Z′>} Y 1 Z′ d<X,Z′> = {<x,z>} “x precedes z” 11 1 The problem lies in the fact that we have both <X′,Y′> and <YP,X> in there. 11 Z so d(A) = {<y,z>, <y,x>, <x,y>, <x,z>} 11 1 z follows both x and y. yx z y precedes both x and z. d(A) is a linear order. Adjoining a head to a head (e.g., head Notice: X c-commands YP— Now, we can “derive X-bar theory”… movement) and X asymmetrically c-commands Y, ZP. (12) TP • Every phrase must have at least one head. • X and YP are categories (not segments). qp • No phrase can have two heads. TYP• No segment of YP dominates X. • The complement of a head cannot be a head (must be a phrase) rU 3 • Every category that dominates X • A phrase can have only one specifier. XTYZP also dominates YP. • A phrase cannot be adjoined to a head. 1111 —T? No, the category T does not • A head cannot be adjoined to a phrase. 111 Z dominate X, only one segment does. 1111 • A specifier cannot be a head (it must be a phrase). xtyz —TP. Yep, and it dominates YP. • “X′” cannot be moved by itself, leaving the specifier behind. • A head to another head will c-command the c-command domain of the So, if X was a head moved from lower in original head (a moved head will c-command its trace). the tree (say, somewhere in ZP), X would c- command its trace—as is proper. With some extra things as well… There are no specifiers, there are only adjuncts—but you can’t adjoin more than one • There are no specifiers at all—there are only adjuncts; phrase to a phrase. Here’s why: there is no distinction between an adjunct and a specifier. • Multiple clitic adjunctions cannot be analyzed as multiple adjunction to, Adjoining two things to the something The problem here is that: say, T—they must be adjoined to one another, then adjoined to T. B c-commands A • “Multiple specifiers” can exist in a similar way, adjoined to one another, (13) T • B and A are categories rU allowing an explanation for the apparent ability for the highest AT • No segment of A dominates B. specifier to “c-command out” (more below). 1rU • Every category that dominates B 1 BT dominates A. We won’t go through all of the cases; they’re worked out in the book, and the logic is 11 1 and always the same. Property X (say, “A phrase can only have one head”) follows because if ab t A c-commands B X were violated, the resulting tree would be “too symmetric” (e.g., the heads would c- • A and B are categories XP command each other and thus neither would precede the other). • No segment of B dominates A. rU YP XP • Every category that dominates A Some highlights… 1rU dominates B. YZP XP so Two-headed phrase a phrase d(A) = d{<B,Z>, <B,X>} 11 1 Neither asymmetrically c-commands with a head for a complement 1 ZX the other. d<B,Z> = {<b,z>} “b precedes z” 11 1 (11) A yz x d<B,X> = {<b,x>} “b precedes x” Same problem between YP and ZP.
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