Re: generators of SL(2,R)

From: Roger L. Bagula (rlbtftn_at_netscape.net)
Date: 06/27/04 

Date: Sun, 27 Jun 2004 19:40:56 GMT

Thank you so much for again pointing out your great knowledge and
erudition. But you were wrong...

David C. Ullrich wrote:
> On Mon, 21 Jun 2004 19:03:24 GMT, "Roger L. Bagula"
> <rlbtftn@netscape.net> wrote:
>
>
>>Dear Robin Chapman :
>>
>>If you know any transfinite math ,
>> you sure don't show it
>>and then, picking fights over it when it isn't at issue.
>>So My "finite" subset is real..
>>Show me a "real" generator of his type that
>>doesn't go complex?
>
>
> That doesn't make much sense to me. What _you_
> need to do is show us an element of SL(2,R) which
> is not in the group generated by those matrices,
> or vice versa.
>
>
>>I'm trying to keep it "real"
>>on more than one level..
>>
>> semantically speaking.
>>
>>Real number domain sets contain many finite sets..
>>We are looking for an overall generator set that remains real.
>>We know that quadratic number fields are infinite but countable,
>>but they do remain in the real domain:
>>R: a+b*Sqrt[Prime[n]]: real numbers a,b
>>his x goes to--> a*x^2+c*x+b ( as a result of the detm=1 domain
>>definition of SL(2,R))
>>part of which is always complex , real numbers a,b,c.
>
>
> Part of _what_ is always complex?
>
>
>>If you say that x is defined as real then you have limited the domain of
>>{a,b,c} so that it is no longer a group under the axioms of groups.
>>So it is "proved" that his generators are not generators of SL(2,R).
>
>
> None of this makes much sense. Which probably has something to do
> with the fact that those matrices _do_ generate SL(2,R).
>
>
>>Suppose we use all the quadratic number fields Q(Sqrt[Prime[n]]) as the
>>basis of the x ( a subset of the reals), but as individual infinite sets
>>with countable n. We, then, have an infinite set of infinite sets of
>>countable elements that
>>approximates the real domain so well that you can't find a number that
>>isn't covered.
>
>
> You can't? Let's see...
>
> Pi is not "covered". (Not that this has anything to do with
> SL(2,R)...)
>
>
>>It is still aleph zero, but you can't really tell it from
>>aleph one in reality... not with a Turing machine in a finite stopping
>>time anyway. As far as I know that is the Chaitin kind of definition of
>>transfinites with is "equivalent" to the Cantor one.
>>
>>
>>Robin Chapman wrote:
>>
>>>Roger L. Bagula TOP-POSTED:
>>>
>>>
>>>
>>>>Here's one that might give a real only basis
>>>>that is based on golden mean number field.
>>>>gm=(1+Sqrt[5])/2
>>>>e1={{0,gm},{-1/gm,0}}
>>>>e2={{gm,x},{0,1/gm}}
>>>>x={gm,1/gm,1/gm^2}
>>>>It is always SL(2,R)
>>>
>>>
>>>These do not generated SL(2,R). SL(2,R) is
>>>an uncountable group. Hence it cannot be generated
>>>by a countable set of generators, let alone a finite
>>>set such as this.
>>>
>>>
>>>
>>>>and I think since the
>>>>f[n]=mod[n*gm,1]=fractionalpart[n*gm]
>>>>is an infinite set, that the result can cover the
>>>>whole real domain, but only at an infinite level
>>>>for n.
>>>
>>>
>>>What you "think" is hardly relevant.
>>>
>>>
>>
>
>
> ************************
>
> David C. Ullrich

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