HW7: Where do case features go?
Here’s what happened in class last time: We looked at some constraints on the allowable shapes of noun phrases and worked up to the conclusion that what we had been calling noun phrases are actually headed by D. So, they’re DPs. Given that we now have DPs, with NPs inside, we have some things to clarify. There were a number of properties that we assumed held of noun phrases, but now we need to figure out which of those properties are properties of D and which are properties of the N inside.
Many of the properties we thought were of N are actually of D. So: it is DPs that get θ-roles (not NPs), DPs that need case (not NPs), DPs that can be moved to SpecTP to satisfy the [uD*] (”EPP”) feature of T. One thing that still belongs to N are the φ-features (person, number, gender): It is the noun staplers that is intrinsically plural in the DP those staplers.
Since it is DPs that need case, the [ucase: ] features go on the D (not on the N).
Notice that it is quite possible for a DP that gets one case to have other DPs inside it that get other cases. Every DP needs its case checked, and everything that can check case (T, v, P, and the special V that appears in double-object constructions) can only check the case feature on a single DP.
To give an example, here’s what my dessert would look like in the sentence My dessert is burning. The subject is my dessert and so the whole DP gets nominative case. But the possessor my gets genitive case. I’ll draw the top of the tree, with the case features (I am leaving out the [uφ: ] features, among others):
TP
/ \
DP T′
/ \ | \
DP D' T ...
pronoun / \ ^----------[T, uD*, ucase:nom, …]
[D, φ:1sg, D NP
ucase:gen] ∅poss dessert [N]
[D,
ucase:gen,
ucase:nom]
(Side note) Ok, so we end up with two [ucase: ] features on the ∅poss. That’s perhaps not the best aspect of this theory, we need to stipulate something that says that the one that determines the pronunciation of case is the one that was valued (or perhaps the last one that was checked). This came about because ∅poss both checks the case of something else (the genitive pronoun) and needs its own case checked. So, this only arises when we have a D that can check the case of something else, and there are only a small number of those (basically, just ∅poss and maybe every in the king’s every whim).
To clarify how the tree above came about, let me build the DP up step by step:
D′
/ \
D NP
∅poss dessert [N]
[D, φ:1sg,
ucase:gen,
ucase: ]
Then, we put in the possessor, a pronoun with the features [D, φ:1sg, ucase: ], and its [ucase: ] feature is valued and checked by the [ucase:gen] feature of ∅poss.
DP
/ \
DP D′
pronoun / \
[D, φ:1sg, D NP
ucase:gen] ∅poss dessert [N]
[D,
ucase:gen,
ucase: ]
Then, the derivation continues: We merge this in with V burn, merge the result with v (unaccusative), merge the result with T, at which point the [ucase:nom] feature of T can see the still-unvalued [ucase: ] feature on ∅poss, and so it checks and values it, and the whole DP my dessert moves to SpecTP to check the strong [uD*] (”EPP”) feature of T. The top of the tree would then look like the one I drew above.