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This question already has an answer here:

Is the triad G-C-E an inversion of C major or a dominant with a non-chord tone?

marked as duplicate by Richard, Community Dec 28 '18 at 18:42

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    This is an English forum. Please change your question to English. Auch wenn ich die Frage einwandfrei verstehe (sprachlich, nicht semantisch). – MaestroGlanz Dec 28 '18 at 8:20
  • It is the second inversion of C Maj – ggcg Dec 28 '18 at 12:44
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    @ggcg: This is not what the question is about. The fact that it's an inverted C chord doesn't need any discussion. The question is how do you analyze it in a I64-V-I cadence. – Matt L. Dec 28 '18 at 12:50
  • I think there is an assumption you are asking about a cadential 6/4 chord. Can you clarify if that is what you mean? – Michael Curtis Dec 28 '18 at 13:38
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I don't think there is an "answer" so much as a historically informed way to understand why it could be labelled in those two ways. Regardless of labels a cadential 6/4 functions one way in the common practice style.

There was a time when Roman numeral analysis and chord inversions were not part of music theory. This is critical to understand. There was a time when there wasn't a theory of chord roots and inversions! For reference points we could mention composers like Corelli and Bach. Their theory and notation used figured bass where a cadential 6/4 would look like this...

enter image description here

Historically, harmony was the result of counterpoint and chords were intervals above the bass. The thinking would have been something like: the bass goes from G to C and the G takes a chord of the sixth. The 4th over the G is dissonant and therefore must be resolved so it moves down a step to the 3rd. Then the G moves to the C.

From this perspective we do not have a form of tonic chord. It would be a form of dominant chord, because the bass is playing the dominant scale degree. Also, strictly speaking, it probably wouldn't have been called 'non-chord tones' as the 6/4 is the chord, but it's a dissonance requiring resolution. Probably I wouldn't have been called an appoggiatura.

To capture this historical view in modern Roman numerals it would look like...

enter image description here

As long as you mark the dissonant tones as appoggiaturas the V symbol under the cadential 6/4 is OK.

This...

enter image description here

...is the fully modern view labeling it as a second inversion tonic chord.

Personally, I like the older approach, because it allows me to think about it as the past composers did. But I also understand the modern way. The two aren't mutually exclusive.

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Nobody can tell you what it is (in a given context). Obviously, in isolation without any context it's just an inverted C major triad, but that's a rather uninteresting and obvious fact.

A more reasonable question would concern the way it is analyzed, and indeed, both interpretations are possible. And note that the context is important. When used in an authentic cadence I64-V-I, then the I64 chord can be either interpreted as the second inversion of the tonic triad, or it can be analyzed as a dominant with non-chord tones that resolve to a consonant dominant. The first interpretation is more common in traditional texts, whereas the latter is more often found in modern texts.

  • This isn't quite true, "Nobody can tell you what it is". Anyone can absolutely tell what it is. It is a C Maj second inversion. There needs to be enough of the G triad present for the ear to hear it as G. – ggcg Dec 28 '18 at 12:46
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    @ggcg: No, this is all about context, and in the given context you're free to hear/analyze it as an inverted tonic, or as a double appoggiatura of V. Who is right? Nobody can tell you, that is what my first sentence is about. – Matt L. Dec 28 '18 at 12:48
  • Yes, it is not about context as the OP didn't provide any. Just a chord period. – ggcg Dec 28 '18 at 12:49
  • @ggcg: That question can only occur in the context that I assume as implied. How would that question otherwise come about? So you think the OP might as well have asked if E-G-C is some type of E chord? I don't think so. – Matt L. Dec 28 '18 at 12:52
  • I disagree, It is often that students ask questions like this without context when they are learning. Once you are told you can eliminate the 5th, a common practice with the V7 chord (because there are enough frequencies to identify it). All bets are off in the mind of the student. You can assume away. My interpretation of the OP is valid and assuming turns it into another question. – ggcg Dec 28 '18 at 12:58
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The three notes G C and E are 1,3 and 5 of C major - in C, E, G order it's root; in G C E order it's second inversion. Dominant with a non-chord tone? If a note has been changed in a triad chord, it becomes a different chord! It can be understood that it is the dominant triad of key F - that built off the V note - which is C, but any closed or open inversion of a triad still keeps its original root name.

  • Not always, the interpretation I mentioned in my answer is a very common interpretation, and our ears also tell us so :) In pop/jazz terms, that chord could be called Gsus4 (the additional 'e' is just a color tone, not changing the function; same for the missing 5 of G). – Matt L. Dec 28 '18 at 10:39
  • @MattL. - never come across it in jazz. I understand that G with C instead of B could be construed as Gsus4, even without the D, but with E, it changes the taste somewhat. I can't come up with an outrageous 'changed chord', but surely by altering the basic ingredients, the chord itself cannot be Gsus4 - the basic ingredients thereof being GCD. In that case, any 'colour tone' can be added to anything, and given the same name. If that's the case, a sus is usually followed by a 'normal', but this 'sus' is IV of G. What am I missing? – Tim Dec 28 '18 at 11:07
  • In a jazz context you might have come across Fmaj7/G, which functions as a dominant in C. It is basically a G7sus4, and it has an E as a color note. Anyway, the question is about a standard progression mainly used in classical music, and there the two common analyses are the ones I've summarized in my answer: link – Matt L. Dec 28 '18 at 12:13

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