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proof sketch
I don't really grok this sentence:
"||Ƹ||, the norm of the full p-limit error map, must also be minimal among all valid error maps for S*, or the restriction of Ƹ to G would improve on Ɛ."
What can I do to convince myself of this? Or does it need to be rephrased?
- clumma July 31, 2012, 01:05:37 AM UTC-0700
Let's say that V is the space of monzos and V* is the space of vals, and that S is the subspace of smonzos and S* is the dual space of svals. Note that in the article, "S" referred to a temperament, so this is a different notation I'm using here.
If we place a norm on V, this induces a norm on S, and the unit sphere of S is going to be the intersection of the unit sphere of V and the subspace defined by S. Since this is still a norm, we can define a dual norm on S*, which will by definition have all of the properties we like about dual norms (e.g. tells us the max weighted over all intervals for some tuning map). So S* is what we want, and our goal is to find some way to figure out what this dual norm for S* is.
There is a V-map M going from V* -> S*. The kernel of M is going to be the subspace of vals which is "tempered out" or made irrelevant when restricted to the subgroup S. The corollary to Hahn-Banach which is linked to shows that the norm on S* is actually the same as the quotient norm V*/ker(M), where ker(M) is that subspace of "invariant vals" I mentioned before. The quotient norm by definition is the norm of the smallest vector in the coset of vectors all mapping to the same thing.
In short, what this all means is that for any sval s, we can find the norm ||s|| in two steps:
1) Look at all of the full-limit vals in the equivalence class which map to s
2) Find the val v in this equivalence class with lowest norm
and then ||s|| = ||v||, by definition. Hahn-Banach shows that this also works out to be the dual norm to the norm defined on V, so that for some arbitrary subgroup tuning map t, ||t|| also has the added benefit of telling us the max norm-weighted error over all intervals in S. So it's all the same and works out very neatly.
Now then, as for the thing you asked:
Assume that Ɛ is the error map in S* with shortest norm for the subgroup temperament T we care about, and that Ƹ is the full-limit error map in V* with shortest norm in the coset of error maps that get sent to Ɛ under the V-map M. Then, by definition, ||Ɛ|| = ||Ƹ||, which follows from the definition of quotient norm above.
Note that the preimage under M of the set of all error maps supporting T is going to be an even larger set of error maps. Then for any full-limit error map e which is in this preimage, ||e|| >= ||Ƹ||.
Proof by contradiction: Assume some other full-limit error map X exists which is actually the minimal error map, and for which ||X|| <= ||e|| for every e in the preimage, including Ƹ. Then since there's a V-map M sending error maps in V* to error maps in S*, the subgroup error map M(X), given by the matrix multiplication X*M, would have to have a norm ||M(X)||
||X||. This is because ||X|| has to be equal to the norm of the shortest error map that gets mapped to it, which in this case is M(X). Therefore, since ||X|| < ||Ƹ||, and ||M(X)||
||X|| and ||Ɛ|| = ||Ƹ||, it must also be true that ||M(X)|| < ||Ɛ||. But, this contradicts our initial assumption that ||Ɛ|| is the error map in S* with the shortest norm.
TL;DR, if there were a better full-limit error map than Ƹ, it'd have to map to a better subgroup error map than Ɛ, but that's a contradiction.
- mbattaglia1 July 31, 2012, 06:35:22 AM UTC-0700
wtf, I dunno what the hell happened. Here's the last paragraph again:
Proof by contradiction: Assume some other full-limit error map X exists which is actually the minimal error map, and for which ||X|| <= ||e|| for every e in the preimage, including Ƹ. Then since there's a V-map M sending error maps in V* to error maps in S*, the subgroup error map M(X), given by the matrix multiplication X*M, would have to have a norm ||M(X)|| = ||X||. This is because ||X|| has to be equal to the norm of the shortest error map that gets mapped to it, which in this case is M(X).
Therefore, since
||X|| < ||Ƹ||, and
||M(X)|| = ||X||, and
||Ɛ|| = ||Ƹ||,
it must also be true that ||M(X)|| < ||Ɛ||. But, this contradicts our initial assumption that ||Ɛ|| is the error map in S* with the shortest norm.
I guess I have to put it on separate lines, or it thinks things between equals signs are headers, =like this=.
- mbattaglia1 July 31, 2012, 06:37:11 AM UTC-0700
No, I guess not. Well, I dunno wtf went wrong, but there it is.
- mbattaglia1 July 31, 2012, 06:37:32 AM UTC-0700