|How to analyze measure 3 of the Adagio in Beethoven's 32nd Sonata?|
Question: Dear Theorist, I'm trying to analyze the glorious 2nd movement of Beethoven's 32nd piano sonata in C, op. 111. In the 3th and 4th full measures of the 3rd variation (C major, 12/32 time), he introduces some altered pre-dominant chords that are to reappear along with others as the movement progresses. The spelling of the chords in the score and my functional analysis of them is as follows (* * dim):****Var. 3, M. 3:**Beats 01-02: E-G-Bb-C# (#vi*4-3)**Beats 03-04: F-G-B-D (V2)**Beats 05-06: D#-G-B (bIII*)**Beats 07-08: E-G-C (I6)**Beats 09-10: C-E-G (I)**Beats 11-12: C-E-G (I)****Var. 3, M. 4:**Beats 01-04: G(pedal)-F#-A-C-Eb (vii*7/V: with V as pedal)**Beat 05-07: G-B-D (V), Beat 08: F-Ab-B-D (vii*4-3)**Beats 09-10: F-Ab-B-D (vii*4-3)**Beat 11: B-D-G (V6), Beat 12: F-G-B-D (V2) Can you tell me if this is right? The #vi°4-3 of I could also be read as the more common #ii°4-3/V. However, the #ii°7 usually resolves to I6, so I thought the less common but still extant #vi°4-3 was best, as it resolves to the dominant.
That's a tough one. Of course there's always a little room for gentlemanly disagreement in analysis of Beethoven; there's not necessarily only one right answer here. But I'd read it slightly differently at the beginning. For all those whose lives permit them the luxury of thinking about this sort of thing:
The chord at the beginning of measure 3 is a dim 7 chord, true. And these are ambiguous, sound-wise, as to root. But it is spelled as if the root were C# - that is, it contains Bb. If I understand you right you've labeled this chord as #vi 4/3, which I guess means you're reading the Bb as an A#, so that the A# can be the root of A#,C#,E,G - that would be very nice, too, since everything moves as expected then - but isn't that cheating? I mean, we need some law here. Or maybe you mean it's a #vi 4/3 of V, with its root being E? But it doesn't spell ideally for that, either, because in that case the C# would be better off as a Db.
I'm not sure it's very useful to analyze it in terms of functional root progression anyway. Personally I'd just see it in contrapuntal terms: this is a leading-tone chord moving to V7; three of its pitches move by half step and the other remains constant. Similarly, the V+ chord can be seen as a decoration on the V. But if forced by threat of violence to name a specific root in Roman numerals, I suppose I'd say #i for the root of the first chord, which could be called "vii° of ii," making that measure into #vii° 6/5 of ii, V7, V+, I. But the ii is skipped and we proceed directly to V7. Then measure 4, not counting the pedal, is vii° 4/2 of V, followed by V, vii° 4/3, V7 - pretty much as you say from there. Here's the original followed by my own reduction:
As an aside, the 12/32 time signature could throw a diligent music student for a loop. The music appears to have three beats in each measure; that is, each beat ought to add up to an 8th note (a quaver, for you Brits out there). Each of the three beats ought to contain the value of four 32nds (demisemiquavers!) in order to add up to 12. Though a modern reader might expect this to be a compound meter with 4 beats of three 32nds, it clearly is not.
Trouble is, if you read the notation literally each beat of measure 3 is five 32nds in the upper staff, totaling fifteen, while the lower part of the same measure has 18 of them. The next measure doesn't work out either. It's because Beethoven has not included the triplet indications that would be necessary to make the beats come out even. That's no wonder; all those triplets would be very busy visually, and the guy was writing with a goose feather. It's something to keep in mind, though, if you want a computer to play this correctly. The solution is that 32nds followed by 32nds are as written; a 32nd followed by a 64th is really a triplet. Do that, and all is well.
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