Consider (over $\mathbf C$) the exterior algebra $\tilde A=\Lambda(x,y,z)$ on three generators, i.e. non-commutative polynomials with the relation that $xy=–yx$, and similarly for the others, as well as the symmetric algebra $A=S(x,y,z)=\mathbf C[x,y,z]$. There is an obvious action of the symmetric group $S_3$ on both. I want to understand the respective coinvariant algebraas $\tilde A_{S_3}=\tilde A/(\tilde A^{S_3})_+$, and $A_{S_3}$ similarly..
What I expect to happen: proposition 0.4 in this document makes me expect that $\dim_\mathbf C A_{S_3} = 6$ (although $A$ is not finite dimensional).
What I actually computed: First for the exterior algebra. $ (\tilde A^{S_3})_+=(x+y+z, xyz)$; the missing elementary symmetric polynomial is inverted by a simple reflection and hence is not contained in the invariants. The naive approach to a basis of the quotient is to check which of $1,x,y,z,xy,yz,xz,xyz$ (which are a basis for $\tilde A$) becomes linearly dependent.
- $z = -x-y$
- $xy = -y^2-zy=-zy$ by $y$ times the relation $x+y+z=0$
- $yz = -z^2-xz=-xz$
- $xz = -x^2-xy=-xy$,
- hence $xy,xy$ and $xz$ are linearly dependent.
The quotient thus has a basis $\langle 1,x,y,xy\rangle$.
For the symmetric algebra, $A^{S_3}=(x+y+z, xy+yz+xz, x^2+y^2+z^2, xyz)$. Then $A_{S_3}$ has a basis $\langle 1,x,y,x^2,y^2\rangle$. This can be obtained by observing that the following naive basis vectors are linearly dependent:
- $z = -x - y$
- $z^2 = -x^2 - y^2$
- $z^2 = (x+y)^2 = x^2 + 2xy + y^2$, hence $x^2+y^2=-xy$
- $xz=-x^2-xy=-x^2+x^2+y^2=y^2$
- $yz=x^2$
- $x^3=xyz=0$
and all others of degree 3 similarly.
Question: Have I calculated wrong, or should I expect a different dimension? Is the theorem valid also for infinite-dimensional vector spaces? Is there a similar theorem for the exterior algebra?
What I am actually after: I want to compare $(A_{S_3})^s$ and $(\tilde A_{S_3})^s$ for $s$ the simple reflection that interchanges $x$ and $y$. From the above bases I obtain:
- $(A_{S_3})^s=(x+y) \ni xy$ with $(x+y)^2=xy$
- $(\tilde A_{S_3})^s=(x+y, xy)$ with $(x+y)^2=0$.
What astonishes me is that they are nearly the same: isomorphic as vector spaces (both have basis $\langle 1,x+y,xy\rangle$), but not as algebras. However, I do not trust in my calculations, so: Are both actually isomorphic (as algebras)?
Your calculation in the original question is correct as far as I can see, but you should have expected a different dimension (although I would not have been able to immediately predict it to be 4). The point is: proposition 0.4 from the paper you mention talks about this construction for the symmetric algebra, but you use the exterior algebra, which already has a much smaller number of dimensions to begin with.
In the bonus question you should expect dimension 6 by proposition 0.4 so something is missing in your basis for $A_{S_3}$. I'm not sure what it is. x^3 perhaps? Also it seems that in the symmetric case $x^2 + y^2 + z^2$ deserves a place in your list of generators of $(A^{S_3})$ and perhaps there are others.