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| = ARCHIVED WIKISPACES DISCUSSION BELOW =
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| '''All discussion below is archived from the Wikispaces export in its original unaltered form.'''
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| == proof sketch ==
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| I don't really grok this sentence:
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| "||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."
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| What can I do to convince myself of this? Or does it need to be rephrased?
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| - '''clumma''' July 31, 2012, 01:05:37 AM UTC-0700
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| 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.
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| 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.
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| 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.
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| In short, what this all means is that for any sval s, we can find the norm ||s|| in two steps:
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| 1) Look at all of the full-limit vals in the equivalence class which map to s
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| 2) Find the val v in this equivalence class with lowest norm
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| 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.
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| Now then, as for the thing you asked:
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| 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.
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| 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||.
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| 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)|| <h1 id="toc0"> ||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)|| </h1>
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| ||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.
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| 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.
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| - '''mbattaglia1''' July 31, 2012, 06:35:22 AM UTC-0700
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| wtf, I dunno what the hell happened. Here's the last paragraph again:
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| 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).
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| Therefore, since
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| ||X|| < ||Ƹ||, and
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| ||M(X)|| = ||X||, and
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| ||Ɛ|| = ||Ƹ||,
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| 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.
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| I guess I have to put it on separate lines, or it thinks things between equals signs are headers, =like this=.
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| - '''mbattaglia1''' July 31, 2012, 06:37:11 AM UTC-0700
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| No, I guess not. Well, I dunno wtf went wrong, but there it is.
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| - '''mbattaglia1''' July 31, 2012, 06:37:32 AM UTC-0700
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