Trying to harmonically analyze the first 9 measures of this piece I've come across a few ambiguous chords, and I can't identify correctly the progression.

Saying the tonal center is Eb gives a reasonable progression Eb -> Eb7 -> Eb7b -> Eb6(omit 5) -> Ab9, i.e. I -> IV, on the first phrase (bars 1-4) but this means the second phrase (bars 5-8) is a Eb -> Eb7 -> Eb7b -> Eb6(omit 5) -> Bb/F -> Fm7, or I -> V -> ii which is a bit weird to me.

Saying the tonal center is Cm and saying it's substituting Cm by its III (Eb) makes this second progression a lot more simple III -> VII -> iv, but I'm not satisfied by this interpretation either.

What would be the correct way to interpret these progressions?

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    I'm a little bit tired now, but first thing tomorrow I'll go trough my answers as I believe this is a dupe. – Neil Meyer Sep 8 '16 at 17:53
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    Possible duplicates: How to determine which inversion a chord is? and Understanding Inversions – user28 Sep 8 '16 at 19:31
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    You may be overthinking this. Eb -> Eb7 -> Eb7b -> Eb6(omit 5) isn't "wrong", but to me it sounds more like a single chord of Eb plus a descending chromatic melodic line Eb - D - Db - C. Not every "collection of notes played simultaneously" has a harmonic function. – user19146 Sep 8 '16 at 19:36
  • I think it's better to note them as separate chords because that same line is used later to support the melody line of the singer. – Pedro Montoto García Sep 9 '16 at 12:55
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    I'm sure I've seen a video discussing the theory of descending chromatically from the root of chords in jazz comping. When playing jazz, it's essentially a great "cheat" way to create the "illusion" of harmonic movement when you only have one chord. And sometimes (especially in pop music) this modifying one note in a chord is strong enough to be the actual basis for harmony on its own. It's similar to the james bond/cry me a river effect of raising the fifth of a minor chord chromatically up to the aug 5th then the major 6th. Or the same thing in major with "aguarela do brasil" or "Brazil" – Some_Guy Jan 4 '17 at 12:22

Not quite sure why the word 'inversions' is in your subject title, while the details of the question are trying to determine root progressions. Perhaps this statement will help though, I hope: The inversion of any chord is determined by which note in that chord is pitched lowest. It doesn't matter where any of the other notes are, as long as they're above it. A C major chord is in first inversion if the lowest pitched note in it is an E. Second inversion if the lowest pitched note in it is a G. Etc. A G7 chord is in third inversion if the lowest pitched note in it is an F natural.

Bear in mind that sometimes a given chord can be interpreted two ways. For example, a C6 chord, and an A minor7 chord have identical notes, though they may be in a different order. Most theorists say that we should assume root position unless there is other evidence to the contrary. In other words, if the bottom (loweswt pitched) note of that chord is a C, we should assume it's a C6. If the lowest pitched note is an A, we should assume it's an Amin7. So far so good. But what if the lowest pitched note is an E?!?!? Then it's anybody's guess, but if the piece is predominantly in A minor, we should probably call it a second inversion Amin7 chord, while if the piece is predominantly in C major, we should probably call it a first inversion C6. etc.

Hope that helps.

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I would have interpreted the chords from this music as follows:

Bars 1-4: | Eb - Eb(maj7) | Eb7 - Cm/Eb | Abm6 / | Ab7 / |

Bars 5-8: | Eb - Eb(maj7) | Eb7 - Cm/Eb | Bbm/F / | Fm7/Ab / |

Or to follow a numerical notation:

Bars 1-4: | I - I(maj7) | I7 - VI/I | IVm6 / | IV7 / |

Bars 5-8: | I - I(maj7) | I7 - VI/I | Vm/II | IIm7/IV / |

This isn't a very typical chord progression but perhaps that unpredictability makes it more interesting to listen to.

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It's a straightforward Eb major, followed by a chromatic change in the top (root, Eb) note. It drops to D (=Emaj7), then to Db (=Eb7), then C (=Eb6), then to the IV (=Ab).

Second time round instead of IV, it goes to Vsus, presumably to get back to I.

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