Two responses to invariant theory question......DMD
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Subject: Re: question abt invariant theory
From: Amnon Neeman <neeman@maths.anu.edu.au>
Date: Thu, 30 Jun 2005 11:21:44 +1000 (EST)

The ring is relatively well known. The following facts are true:

(1) Over Q the elements given by Ravenel generate.

(2) Over Z they don't. Nagata proved that they fail to generate over Z/2,
I found a trick that works in every charateristic. The general statement
is that if $m>1$ and $n>p$ then they fail to generate over Z/p. (In my
paper, cited below, I prove the weaker estimate that if both $m$ and $n$
are $>p$ then the ring is not generated by these elements).

(3) The ring is most definitely not smooth. Not even over Q.

(4) The relations are understood quite well. Both of my papers cited below
study the relations.

The following are some references:

@article {Nagata55,
      AUTHOR = {Nagata, Masayoshi},
      TITLE = {On the normality of the {C}how variety of positive
              {$0$}-cycles of degree {$m$} in an algebraic variety},
      JOURNAL = {Mem. Coll. Sci. Univ. Kyoto. Ser. A. Math.},
      VOLUME = {29},
      YEAR = {1955},
      PAGES = {165--176},
}

@ARTICLE{Neeman91C,
      AUTHOR={Neeman, Amnon},
      TITLE={Zero cycles in {${\mathbb P}^n$}},
      JOURNAL={Advances in Math.},
      VOLUME={89},
      YEAR=1991,
      PAGES={217--227},
      }


@ARTICLE{Elmore-Hall-Neeman05,
      AUTHOR = {Elmore, Ryan, Hall, Peter and Neeman, Amnon},
      TITLE =  {An application of classical invariant theory to
        identifiability in nonparametric mixtures},
      JOURNAL = {Ann. Inst. Four. Grenoble},
     VOLUME = {55},
      PAGES  = {1--28},
      YEAR  =  {2005},    }

       Yours, Amnon


On Wed, 29 Jun 2005, Don Davis wrote:


>> Subject: Question about invariant theory
>> From: "Douglas C. Ravenel" <doug@math.rochester.edu>
>> Date: Wed, 29 Jun 2005 15:45:59 -0400 (EDT)
>>
>> Here is a question for the listserve about invariant theory.
>>
>> Let $m$ and $n$ be positive integers and consider the ring
>> \begin{displaymath}
>> R=Z[r_{i,j}: 1\le i \le m, 1\le j \le  n],
>> \end{displaymath}
>>
>> \noindent the polynomial ring on $nm$ variables.  I am especially
>> interesed in the case where $n$ is prime.
>>
>>
>> Let the symmetric group $\Sigma_{n}$ act by permuting the second
>> subscript.
>> What is the structure of the invariant subring $S$?
>> It appears to have $\binom{n+m}{m}-1$ generators,
>> namely the coefficients of that number of mononials of degree $\le n$
>> in the $m$ dummy variables $y_{i}$ in the invariant expression
>>
>> \begin{displaymath} \prod_{j=1}^{^{n}}\left(1+\sum_{i=1}^{m}r_{i,j}y_{i}
>> \right)
>> =1+\sum_{K}a_{K}y^{K}.
>> \end{displaymath}
>>
>> \noindent Here $K= (k_{1},\dotsc ,k_{m})$ is a sequence of $m$
>> nonnegative integers (not all zero) whose sum is at most $n$, and
>>
>> \begin{displaymath}
>> y^{K} = y_{1}^{k_{1}}\dotsb y_{n}^{k_{n}}.
>> \end{displaymath}
>>
>> \noindent On the other hand, its Krull dimension is only $nm$
>> (since it is contained in $R$), so it must be some relations. What are they?
>>
>> A related question: Is the spectrum (in the sense of algebraic
>> geometry) of the invariant ring $S$ smooth?
>>
>> Doug
>>
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Subject: RE: question abt invariant theory
From: "Neil Strickland" <N.P.Strickland@sheffield.ac.uk>
Date: Thu, 30 Jun 2005 10:46:01 +0100

In various places I have made use of the rings that Doug mentions.
However, I have always managed to avoid the need for an explicit
presentation by generators and relations, which seems to be hard.
The rings are certainly not smooth, in any case.  You can see this
quite explicitly in the case n=2, after tensoring with the reals:
you get R^m crossed with the affine cone on RP^{m-1}, which is not
a manifold.  More abstractly, the action is not generated by 
reflections unless m=1.  I forget the precise generality of the 
relevant theorem, but the slogan is that you cannot get a smooth
ring of invariants unless the group action is generated by 
reflections.  If you work p-adically or mod p, you have to consider
pseudoreflections instead, where the fixed space has codimension 
one but the nontrivial eigenvalue need not be -1.  There are still 
no such elements in the action of \Sigma_n on k^{nm}, unless m=1.

Neil