Minkowski reduction in diemnsion less than five

According to a paper which can be found here:

http://tinyurl.com/2ec73xe

if the norm is Euclidean and the dimension less than five we can take linear combinations of our candidate basis vectors

s1*v1 + ... + sn*vn

where the coefficients si are -1, 0, or 1, and if the linear combination is shorter than one of the basis vectors, we may replace the basis vector with the combination for a new basis. Iterating this proceedure until no more reductions are possible gives a Minkowski reduced basis.

Sadly, it doesn't claim this will be true for, to pick an example wildly at random, Tenney height reduction. However it is so closely allied with the corresponding weighted Euclidean case it seems to me that if you reduce to the Euclidean Minkowski basis you are probably already in the TM basis, and in any case this proceedure really ought to work.

This all started because I was looking at Minkowski reduced bases for rank three lattices, and found that the basis for Portent is [2, 8/7, 12/11] and also found it's a big fat pain to find these the way I did.

Portent, by the way, is the 1029/1024, 385/384 (and 441/440) temperament.

And without the obnoxious PDF locking...

/tuning-math/files/carl/

-C.

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
>
> According to a paper which can be found here:
>
> http://tinyurl.com/2ec73xe
>
> if the norm is Euclidean and the dimension less than five we
> can take linear combinations of our candidate basis vectors
>
> s1*v1 + ... + sn*vn
>
> where the coefficients si are -1, 0, or 1, and if the linear
> combination is shorter than one of the basis vectors, we may
> replace the basis vector with the combination for a new basis.
> Iterating this proceedure until no more reductions are possible
> gives a Minkowski reduced basis.
>
> Sadly, it doesn't claim this will be true for, to pick an
> example wildly at random, Tenney height reduction. However it
> is so closely allied with the corresponding weighted Euclidean
> case it seems to me that if you reduce to the Euclidean Minkowski
> basis you are probably already in the TM basis, and in any case
> this proceedure really ought to work.
>
> This all started because I was looking at Minkowski reduced
> bases for rank three lattices, and found that the basis for
> Portent is [2, 8/7, 12/11] and also found it's a big fat pain
> to find these the way I did.
>
> Portent, by the way, is the 1029/1024, 385/384 (and 441/440)
> temperament.
>

On 20 June 2010 05:06, genewardsmith <genewardsmith@sbcglobal.net> wrote:
> According to a paper which can be found here:
>
> http://tinyurl.com/2ec73xe
>
> if the norm is Euclidean and the dimension less than five we can take linear combinations of our candidate basis vectors
>
> s1*v1 + ... + sn*vn
>
> where the coefficients si are -1, 0, or 1, and if the
> linear combination is shorter than one of the basis vectors,
> we may replace the basis vector with the combination for
> a new basis. Iterating this proceedure until no more
> reductions are possible gives a Minkowski reduced basis.

At the start, it says it works for positive definite quadratic forms,
which would include my parametric badness. Later on, the restriction
is a gauge function, which looks like a general norm.

This is certainly useful because it'll apply to my parametric badness.
The reduced basis in a badness space (what'll be a "notation" from
what I remember of Gene's terminology, or its dual) looks like a very
useful thing to be able to find. But I'm finding it difficult to find
for high limits.

All I have to do now is understand this paper . . .

> Sadly, it doesn't claim this will be true for, to pick an example
> wildly at random, Tenney height reduction. However it is
> so closely allied with the corresponding weighted Euclidean
> case it seems to me that if you reduce to the Euclidean
> Minkowski basis you are probably already in the TM basis,
> and in any case this proceedure really ought to work.

Is this the general method or the 4 dimensional case you gave above?
Tenney height looks like a gauge function.

Graham

--- In tuning-math@yahoogroups.com, Graham Breed <gbreed@...> wrote:

> Is this the general method or the 4 dimensional case you gave above?
> Tenney height looks like a gauge function.

Tenney height is a gauge function. However, the restriction to -1, 0, or +1 coefficents is only claimed to be proven for Euclidean gauge functions.