A musician friend mentioned this in passing - she talked about the something chord and then explained it was unresolvable, and how she hates it because she cannot stand music that doesn't resolve.

But I can't recall what it was called. I don't know enough to know if it really is a single chord, or a family of chords, or something less specific entirely.

  • 1
    Actually could it be the "Tristan Chord"? en.wikipedia.org/wiki/Tristan_chord but that page doesn't make it clear. If anyone can confirm this and explain in simple terms, that would be a great answer. – Mr. Boy Dec 18 '14 at 15:33
  • Yeah that page is a little confusing. I tried to sum it up a little more clearly and succinctly in my answer. Let me know if you have any questions. – Casey Rule Dec 18 '14 at 15:49
  • 1
    There is, of course, a huge difference between "unresolved" and "unresolvable"...while I'm guessing your friend was talking about the former, I'd be very interested in examples of the latter. – Kyle Strand Dec 18 '14 at 19:30
  • 2
    What happens when an Unresovable Chord meets an Unrestrainable Progression? :-) – Carl Witthoft Dec 18 '14 at 22:29
  • 1
    Yeah I meant in general, we don't use the Tristan Chord much in what we play! – Mr. Boy Dec 19 '14 at 9:09

It sounds like she may have been talking about the Tristan Chord, a famous chord from the opening of Richard Wagner's opera Tristan und Isolde.

enter image description here

While it can be enharmonically written as a half-diminished 7th chord (F-A♭-C♭-E♭), it does not resolve in the way a half-diminished 7th chord would, nor is it written as a half-diminished 7th chord. For this reason, it has become famous for being a rather difficult chord to analyze with traditional Roman numeral analysis. The issue is not that it doesn't resolve, but that it resolves in a way that makes its relationship to the key so ambiguous.

| improve this answer | |
  • Sounds like a C#9, with the root missing... which usually goes to an F# of some sort, but it resolves to an E7... – Tim Dec 18 '14 at 15:42
  • I beat you to it in my comment on the original question by a few seconds! As I said there, the theory is a bit deep, can you expand your answer to explain (or not need) some of the musical terminology? "it can be enharmonically written as a diminished 7th chord" is gibberish to me! – Mr. Boy Dec 18 '14 at 15:51
  • 1
    What that means is that if you rewrite the chord using different note names, (F-A♭-C♭-E♭), then you could call this a half-diminished chord (that's just certain type of chord that resolves in a very specific way in most classical music), however in the context of the piece, these notes don't seem to function like a typical half-diminished chord. It's difficult to explain exactly why this is so unusual without getting into deeper theory, but what it boils down to is that this chord is famous because it is so difficult to analyze. – Casey Rule Dec 18 '14 at 15:57
  • 2
    As an Fm7b5 (half dim) it also resolves happily to an F# chord. Maybe Wagner wrote it with those particular notes to bamboozle future music students... – Tim Dec 18 '14 at 16:35
  • You sneaky Wagner! – Casey Rule Dec 18 '14 at 16:40

I'm not sure I could call any chord "unresolvable", though I'd have to know the context of the conversation.

However, augmented chords (e.g. C-E-G♯), as well as diminished 7th chords (e.g. C-E♭-G♭-B♭♭) -- both of which have been mentioned in other answers -- share a common trait: they are, in some sense, symmetric. Augmented chords are created by stacking three major thirds, and diminshed 7ths are likewise created by stacking four minor thirds, both of which add to an octave. While neither chord is "unresolvable", these two types of symmetric chords could be said to be "ambiguously resolvable", in that they look the same from three or four different keys (respectively), and can therefore be plausibly resolved to any one of several possible distant keys. As a result, romantic composers will often use them as a sort of "turn-stile" to modulate between unrelated keys.

In fact, the Tristan Chord, mentioned in yet another existing answer, is also somewhat symmetric (at least after the G♯ resolves to the A). It consists of two major thirds (F-A and B-D♯), separated by a tritone (a symmetric interval). As such, you could almost envision it resolving to a Bb7 instead of an E7 (if it weren't for those chromatic passing tones that help to indicate a clear direction to the line).

The takeaway point is that, in music, symmetry leads to a certain ambiguity of tonal center, and therefore can weaken the sense of needing any specific resolution. Such a chord still needs a resolution, but it has several options, and so does not necessarily need any specific resolution.

| improve this answer | |

They may have said the Tristan Chord, but I would argue it is resolvable, and in fact not mysterious... as others have pointed out, the analysis works well if you see the G# as a non-harmonic tone, a lower neighbor (in A minor) to the A at the end of the measure.

Then we have the following notes, spelling them out in thirds: (B D# F A). We have F in the bass. This is a textbook French 6th chord in A minor!


Note that the next chord is V7, a dominant chord in A minor, which is exactly what you would expect (perhaps with an intervening I 6/4 but not seen here) following the Fr6.

As I recall Wagner loved augmented sixth chords. I've not done any research on the Tristan Chord specifically before seeing it here, but I'm surprised it has an aura of mystery.

| improve this answer | |
  • Good call on the French augmented 6th chord. I knew I'd forgotten something important. – Caleb Hines Dec 19 '14 at 17:19
  • Can you give me an example then, if we're playing in some regular key like C Major / A Minor or G Major / E Minor, of a)what the notes in this chord would be in that key b)a chord progression using it which resolves? Ideally using the least weird chords possible although maybe the point is it only works with weird chords... – Mr. Boy Dec 19 '14 at 19:51
  • Nothing irregular about E minor. :) The following is typical: I -> Fr6 -> I 6/4 -> V7 -> I In C, this would be (no attempt at voice leading here): (C E G) -> (Ab C D F#) -> (G C E) -> (G B [D] F) -> (C E G) The Fr6 is the only unusual one here. – Daniel Garrison Dec 19 '14 at 20:07

Personally, I find augmented chords (like c e g#) much worse to resolve than diminished chords: diminished chords reduce to a seventh chord by lowering any chord note by a semitone and resolve obviously from there. One can often actually use them functionally instead of a seventh chord in the first place.

Augmented chords don't work in that manner. If you change one of their notes, the result will be a finished chord, but one that is not in useful harmonic relation to the augmented chord and thus does not serve as its resolution or as a satisfactorily meaningful part of a multi-step resolution.

Consequently, if you take a look at resolution master J.S.Bach, you'll find diminished chords all over the place in cadenzas, but mostly draw a blank on augmented chords.

| improve this answer | |
  • I don't know, lots of Romantic composers (Rachmaninoff is a good example) use augmented chords as a transition between I and vi chords (or, equivalently, i and III or V/VI chords), which seems to provide a pretty good harmonic relation and satisfying resolution. Of course, the relationship and the resolution depend on the context leading up to the augmented chord, but that's also true of fully diminished seventh chords. – Kyle Strand Dec 18 '14 at 19:29

While Dom talks about the V chords which in themselves sound unfinished, they are resolvable. I think it may be diminished chords that you have in mind.They use notes which are usually not within the key in question, thus sound a little strange, and can, by their make-up, go to several different places. The V chord, 9 times out of 10, will find resolution in the I chord. Everywhere in Western music.

| improve this answer | |
  • Yeah I misread the question. I thought it was about unresolved chords. – Dom Dec 18 '14 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.