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I was following a question, https://music.stackexchange.com/questions/57733/tips-on-how-to-add-non-diatonic-chords-to-my-music, and one of the answers introduced a chord progression involving different non-diatonic chords, with an audio sample. As I was analyzing the chord progression, I noticed something rather unfamiliar:

In the second measure a second inversion c minor chord with a V 6/4 (uppercase) roman numeration,that goes to a 5/3.

But the notation suggests it is a second inversion c minor going to a root position G major, so i 6/4 to V.

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Is it just a misprint or is there something that I'm missing?

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Jaafar Jumaa
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3 Answers3

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That's my answer; neat!

This is a contentious issue for musicians. You're right that beat 3 is a i chord in second inversion; it has C–E♭–G, obviously a C-minor chord, and that should be i!

But let's imagine that the E♭–G–C on beat 3 didn't exist, and we just had a half note of D–G–B. We'd label that V, right? Well, this notation just shows that the V chord really "begins" on beat 3, we just have some accented non-chord tones up above. The C is a type of 4–3 suspension, the G already fits, and the E♭ is basically an accented passing tone. Especially when we have that G in the bass, our ears are already conceptualizing that beat 3 as a dominant chord. (It also helps that this particular resolution is one of the most common things in common-practice harmony; we call it the "cadential six-four.")

So to show that this is just an ornamented V chord, we go ahead and label it with a V Roman numeral, even though technically that beat 3 has all the notes of a i chord. Roman-numeral analysis tries to show how a chord functions, and since beat 3 is really the start of the dominant, we go ahead and use a V Roman numeral.

The superscript numbers just show how the upper voices are moving. The 6–5 indicates that a sixth above the bass (E♭) moves to a fifth above (D); likewise, the 4–3 shows the movement C–B.

Other methods of analysis would be:

  • I64–V, as you said. This is very common, but it fails to show how beat 3 is already the dominant.
  • I64–V, but with a bracket below, labelled V, showing that both of those chords are really functioning as dominant. This, in my opinion, is the clearest approach.

PS - Something very similar happens in the next-to-last measure.

Richard
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  • @Richard So V 6/4 functions as a dominant even though it is a minor chord? – Jaafar Jumaa Jun 19 '17 at 17:01
  • Yep! This is because the 6/4 really just shows the non-chord tones above the bass. Just like we label a `4–3` suspension as `V4–3`, this is just a "double suspension" of `6–5` and `4–3`, it just *happens* to create what looks like a tonic chord first. But as I said, plenty of people analyze this as `I64–V`, so both analyses are correct in their own respect. – Richard Jun 19 '17 at 17:04
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My harmony lessons taught me that the second inversion of the tonic chord (like in the question) is notated as V 6/4 instead of I 6/4 or i 6/4, as that chord strongly tends towards the dominant. So no, it's not a misprint.

(Like you, I similarly disagree with my harmony lessons on some days.)

Dekkadeci
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A Cadential Six-Four Chord progression is a chord in the second inversion that decorates two important scale degrees. The Dominant and the Tonic.

When it is the Dominant chord being decorated (like it is in this example.) then the preceding chord is the Tonic chord in the second inversion. When it is Tonic chord being decorated then the preceding chord is the Sub-Dominant chord in the second inversion.

It seems like this is just some sort of awkward printing, the better notation would be I6/4 - V5/3. May have been some printing error.

Neil Meyer
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