Re: how to deal with the algebra problem?
- From: rusin@xxxxxxxxxxxxxxxxxxxxx (Dave Rusin)
- Date: 18 Apr 2005 07:21:59 GMT
In article <A11622TF$scimath@xxxxxxxxxxxxxxxxxxxx>,
SU(2) <idyllic.bbs@xxxxxxxxxxxxxxxxxxxx> wrote:
>assume a1*a2*...*an=1
>does there exist b1,b2,...,bn such that:
>a1(b1/b2)=1
>a2(b2/b3)=1
> .
> .
> .
>an(bn/b1)=1
In addition to the responses you've already gotten (which are the ones
you probably wanted) you need to know that you're encountering
cohomology of groups here!
Your a's and b's all lie in the group R*, the multiplicative group of
the nonzero real numbers. THe group M = (R*)^n is also an abelian
group but in addition is a module for the cyclic group G = Z/nZ of
order n , the generator g acting via
g( ( a1, a2, ..., an ) ) = (a2, a3, ..., an, a1)
[I don't really care which generator g of G acts this way; if it
bothers you, take g = 1 + nZ .]
Next fix G as above but, for the moment, let M be any G-module,
and consider the sequence of G-modules and G-module homomorphisms
...--> M --h1--> M --h2--> M --h1--> M --> ...
where h1 = 1 + g + g^2 + ... + g^{n-1} and h2 = 1 - g . These
bits of arithmetic are carried out in the endomorphism ring of M;
the use of the symbols "+" and "-" is traditional for abelian groups
but in your case M is written multiplicatively so these would mean,
respectively, h1(m) = m . g(m) . g(g(m)) . ... . g^{n-1}(m) and
h2(m) = m . {g(m)}^{-1}. You can calculate -- without knowing what
endomorphism g is, in fact -- that h1 o h2 = 0 and h2 o h1 = 0,
which means that the kernel of h1 contains the image of h2 and
vice-versa. It therefore makes sense to define the _cohomology groups_
H^1( G, M ) = ker(h1) / im(h2)
H^2( G, M ) = ker(h2) / im(h1)
These gadgets show up all over the place in mathematics, e.g. in
Hilbert's "Theorem 90" (in number theory) and in many parts of algebraic
topology.
If you do the calculations in your case, you find that
h1( (a1, a2, ..., an) ) = ( A, A, A, ..., A)
where A = a1 a2 a3 ...; also
h2( (a1, a2, ..., an) ) = ( a1/a2, a2/a3, ..., an/a1 ).
Thus, your question asks precisely, "Is something which lies in the kernel
of h1 necessarily [the inverse of something] in the image of h2 ?"
That's exactly the question, "Is H^1(G,M) = 0 ?", and the answer
(as noted in the other responses in this thread) is "Yes."
I'm almost embarassed to admit that this is really the way I think about
the original question.
dave
.
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