To understand what’s really going on here, we need to consider the ring of *coinvariants* for the action of the symmetric group $S_n$ on $\mathbb{C}[x_1,\ldots,x_n]$, where a permutation of $\{1,\ldots,n\}$ acts by permuting the variables accordingly.

An *invariant* is any polynomial fixed under this action, so it’s just another name for a symmetric function here. Recall that the ring of symmetric functions is generated as an algebra by the power sum symmetric functions $p_1,\ldots,p_n$ where $$p_i=x_1^i+x_2^i+\cdots+x_n^i.$$ So, we say that the ring of *coinvariants* is the quotient ring $$\mathbb{C}[x_1,\ldots,x_n]/(p_1,\ldots,p_n)$$ by the ideal generated by the invariants having no constant term.

In the case of two variables $x$ and $y$, the ring of coinvariants is simply $$\mathbb{C}[x,y]/(x+y,x^2+y^2).$$ Notice that this ideal $I=(x+y,x^2+y^2)$ is also generated by the elementary symmetric functions $x+y$ and $xy$. Now, given any monomial $x^ky^j$, it is zero modulo $I$ unless one of $j$ or $k$ is zero, and we also have $x^k=-x^{k-1}y$ modulo I. So every monomial is zero modulo I except for $1$, $x$, and $y$. Since $x+y=0$ we have $x=-y$, so in fact the two polynomials $1$ and $x$ generate the ring of coinvariants as a $\mathbb{C}$-vector space.

Equivalently, $1$ and $x-y$ generate the ring of coinvariants in two variables (sound familiar?), and the submodules they generate are $S_2$-invariant: the former is the trivial representation and the second is the alternating (sign) representation. (See this post for more information on the characters of the symmetric group). Thus the coinvariant ring decomposes as the direct sum of the trivial and sign representations of $S_2$, and we see that it is therefore isomorphic to the *regular representation*, the representation formed by $S_2$ acting on itself by left composition of permutations.

In fact, this holds in general:

The ring of coinvariants $\mathbb{C}[x_1,\ldots,x_n]/(p_1,\ldots,p_n)$ is isomorphic to the regular representation of $S_n$.

For a proof of this, see for instance Stanley’s paper on invariants of finite groups, Proposition 4.9. Rather than going into the proof, however, let’s focus on how this relates to the decomposition on the previous page.

There is a nice fact called *Chevalley’s theorem* that works for finite groups generated by reflections. For instance, $S_n$ is generated by transpositions, which can be realized as reflections across hyperplanes. One form of Chevalley’s theorem states that if such a group $G$ acts on the ring $R=\mathbb{C}[x_1,\ldots,x_n]$, then $R$ is a finitely generated free module over the ring of invariants $R^G$. Moreover, $R$ is generated as an $R^G$-module by any set of coset representatives for $R/(R^G_+)$, where $R^G_+$ (See, for instance, Humphrey’s book, section 3.6. Thanks to Francois Bergeron for pointing it out to me!)

In the case of $S_n$, we can state the above as the following tensor product decomposition:

$$\mathbb{C}[x_1,\ldots,x_n]=\mathbb{C}[p_1,\ldots,p_n]\otimes_{\mathbb{C}[S_n]}\mathbb{C}[x_1,\ldots,x_n]/(p_1,\ldots,p_n)$$

where $\mathbb{C}[S_n]$ denotes the group algebra over $S_n$. Tensoring with the subring of symmetric functions has the effect of “changing basis” to make the ring of coinvariants into a module over the symmetric functions, and Chevalley’s theorem says that this module is exactly the entire ring $\mathbb{C}[x_1,\ldots,x_n]$, considered as a module over its subring $\mathbb{C}[p_1,\ldots,p_n]$.

Going back to the case of two variables, this decomposition becomes $$\mathbb{C}[x,y]=\Lambda_{C}(x,y)\otimes_{\mathbb{C}[S_n]} \mathbb{C}[x,y]/(x+y,xy),$$ where $\Lambda_{C}(x,y)=\mathbb{C}[p_1,p_2]$ denotes the ring of symmetric functions in $x$ and $y$. Since $1$ and $x-y$ form a basis for $\mathbb{C}[x,y]/(x+y,xy)$, this can be interpreted as the fact that every polynomial in $\mathbb{C}[x,y]$ can be written uniquely in the form $f(x,y)\cdot 1+ g(x,y)\cdot (x-y)$ where $f$ and $g$ are symmetric.

So, we now need only to generalize the polynomials $1$ and $x-y$ to higher numbers of variables. Our question becomes:

What is a natural basis for $$\mathbb{C}[x_1,\ldots,x_n]/(p_1,\ldots,p_n)?$$

I just realized I forgot to reference Francois Bergeron’s book for the proof on the last page of this post: see http://www.amazon.com/Algebraic-Combinatorics-Coinvariant-Mathematics-Mathematiques/dp/1568813244, in the chapter on coinvariants.

I’d much rather use the basis of Schubert polynomials!

Thanks for pointing that out Allen – I had forgotten about the connection with the flag variety. Perhaps I’ll post about Schubert polynomials next week or the week after.

Just discovered this blog– very nice!

Expanding on what Allen K. said about Schubert polynomials, I like to think of at least three different well-known bases for the polynomial ring in n variables over the symmetric polynomials which model the q-factorial in 3 ways:

(1) The Schubert polynomials model the q-factorial as the generating function for permutations w by inv(w), the inversion number or Coxeter group length.

(2) Garsia’s “descent monomial basis” model the q-factorial as the generating function for permutations w by maj(w), the major index.

(3) Then there is what Garsia calls “the Artin basis”, that is the set of all monomials in x_1,…,x_n for which the degree of x_i is at most i-1. They appear in E. Artin’s old slim paperback book “Galois theory”, when he proves the fundamental theorem of symmetric functions. And these monomials model the q-factorial as a product of q-numbers, similarly to your partial derivatives of the Vandermonde product.

Thanks for the detailed reply! I’m sorry it took me so long to see that you commented and approve it. You should be able to comment freely from now on. ðŸ™‚

Maria

I think I’m missing something – why are we tensoring with the base ring $\mathbb{C}[S_n]$? For instance, wouldn’t that make $(x+y)(x+2y) = (-1 \cdot (x+y))\otimes (x+2y) = (x+y)\otimes (-1\cdot (x+2y)) = (x+y)(2x+y)$? (where $-1$ is the nonidentity element of $S_2$ considered inside $\mathbb{C}[S_2]$.

By the way, your blog is awesome!

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