What is known about S\_n group invariants in a free associative (noncommutative) algebra k < x\_1, ...x\_n >) ? (S\_n natural acts by permutations on generators).


What is Poincare series ? Is it finitely generated, is it free ? Are the generators as algebra/vector space known ? 


The same question for the commutative algebra - gives algebra of symmetric polynoms,
which is for-ever-young research topic. To what extent non-commutative version is the same rich ? 




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PS


Is it commutative ? Probably no - however, pay attention on the following simple fact:
consider group algebra of C[G], represent it as a factor of k< x\_1, ...., x\_G > ,
then S\_n invariants go into the center of C[G] (since we need to check invariance with respect to conjugaction - but we have much bigger invariance with respect to S\_G action).
This somewhat may be considered as indication that non-commutativity is not that much big...


In particular T\_k = \sum\_i (x\_i)^k will be mapped to higher Frobenius-Schur indicators
see 


[sum\_g g^k, Frobenius-Schur indicators, S\_n-invariants in freeAss(x\_i), center of the group algebra](https://mathoverflow.net/questions/107256/sum-g-gk-frobenius-schur-indicators-center-of-the-group-algebra) 


This question is motivation to ask present.



S_n invariants in a free associative algebra ("noncommutative symmetric polynoms")
The paper <http://arxiv.org/pdf/math/0502082.pdf> shows the invariants are a free associative algebra and give an explicit basis. Hence it is not commutative. This is proved first in M. C. Wolf, Symmetric functions of noncommutative elements, Duke Math. J. 2 (1936), 626–637 without an explicit basis.


