linear algebra - Calculate the dimension of a space of operators

This is a homework question.

First, consider the ring of real polynomials in $n$ variables, $P_n=\mathbb{R}[x_1,\ldots,x_n]$ ($n\geq 2$), and let $S_n$ act on $P_n$ by automorphism (of algebra) by permutating the variables: $\sigma(x_i)=x_{\sigma(i)}$

For each $i=1,\ldots,n-1$, let $s_i$ be the transposition $(i\ i+1)$ (so $s_i(x_i)=x_{i+1}$ and $s_i(x_{i+1})=x_i$).

Finally, for each $i$, consider the operator $\Delta_i:P_n\to P_n$ given by
$$\Delta_i(f)=\frac{f-s_i(f)}{x_i-x_{i+1}}$$

(a) Show that $\Delta_i$ is a linear operator on $P_n$ that satisfies the rule
$$\Delta_i(fg)=\Delta_i(f)g+s_i(f)\Delta_i(g)$$
(b) Let $D_n$ be the subalgebra generated by $Id_P,\Delta_1,\ldots,\Delta_{n-1}$. Calculate the dimension of $D_3$.

This is what I have so far: It is easy to check that $\Delta_i$ is a well-defined operator and that the rule above is satisfied. Moreover, in the basis $\left\{x_1^{\alpha_1}\cdots x_n^{\alpha_n}\right\}$, we have the following (let's write $x(\alpha)=x_1^{\alpha_1}\cdots x_n^{\alpha_n}$ for $\alpha=(\alpha_1,\ldots,\alpha_n)$)
$$\Delta_i(x(\alpha))=x_1^{\alpha_1}\cdots x_{i-1}^{\alpha_{i-1}}x_i^{\min(\alpha_i,\alpha_{i+1})}x_{i+1}^{\min(\alpha_i,\alpha_{i+1})}\left(\sum_{k=0}^{|\alpha_i-\alpha_{i+1}|-1}x_i^kx_{i+1}^{|\alpha_i-\alpha_{i+1}|-1-k}\right)x_{i+1}^{\alpha_{i+2}}\cdots x_n^{\alpha_n}$$
(the sum is $0$ if $\alpha_i=\alpha_{i+1}$)

The problem is finding the dimension of $D_3$. I checked that $\Delta_i^2=0$, so $D_3$ is generated (as a vector space) by $Id_P,\Delta_1,\Delta_1\Delta_2,\Delta_1\Delta_2\Delta_1,\ldots,\Delta_2,\Delta_2\Delta_1,\Delta_2\Delta_1\Delta_2,\ldots$, but I couldn`t find any other good relation between $\Delta_1$ and $\Delta_2$ so obtain an upper bound for the dimension of $D_3$ (which I'm guessing is finite).

Answer

My previous answer was incorrect, here is the correct answer obtained by trying to correct the previous one :)

Claim: $\Delta_1 \Delta_2 \Delta_1 = \Delta_2 \Delta_1 \Delta_2$.

Note that the polynomial $h = (x_1-x_2)(x_2-x_3)(x_3-x_1)$ has the property that $s_i(h) = -h$. It follows that $\Delta_i h^2 p = h^2 \Delta_i p$ for any $p$.

A slightly mundane computation shows that
$$\Delta_1 \Delta_2 h(x_1,x_2,x_3) x_1^a x_2^b x_3^c = - x_1^a x_3^b x_2^{c+1}+x_1^a x_2^b x_3^{c+1}+x_1^{a+1} x_3^b x_2^c-\\x_1^{a+1} x_2^b x_3^c+x_3^a x_1^b x_2^{c+1}-x_2^a x_1^b x_3^{c+1}+x_2^a x_1^{b+1} x_3^c-x_3^a x_1^{b+1} x_2^c =: g(x_1,x_2,x_3).$$

Now, grouping terms with the same power of $x_1$ together and applying $\Delta_2$ shows (with some work) that:
$$\Delta_2 g(x_1,x_2,x_3) = x_1^a x_3^b x_2^c-x_1^a x_2^b x_3^c-x_3^a x_1^b x_2^c+x_2^a x_1^b x_3^c+x_3^a x_2^b x_1^c-x_2^a x_3^b x_1^c$$
A fully analogous argument show gives the same answer if we start with $\Delta_2 \Delta_1 h(x_1,x_2,x_3) x_1^a x_2^b x_3^c$. It follows that:
$$\Delta_1 \Delta_2 \Delta_1 h(x_1,x_2,x_3) x_1^a x_2^b x_3^c = \Delta_2 \Delta_1 \Delta_2 h(x_1,x_2,x_3) x_1^a x_2^b x_3^c.$$
It follows that $\Delta_1 \Delta_2 \Delta_1 h p = \Delta_2 \Delta_1 \Delta_2 h p$ for any polynomial $p$. Thus, also $\Delta_1 \Delta_2 \Delta_1 h^2 p = \Delta_2 \Delta_1 \Delta_2 h^2 p$. We can now bring out $h^2$, and conclude that $\Delta_1 \Delta_2 \Delta_1$ and $\Delta_2 \Delta_1 \Delta_2$.

It follows immediately (together with OP's work) that $\Delta_1\Delta_2\Delta_1\Delta_2 = \Delta_2\Delta_1\Delta_2\Delta_1 =0$. It remains to check that $\mathrm{Id},\Delta_1,\Delta_2,\Delta_1 \Delta_2, \Delta_2 \Delta_1, \Delta_1\Delta_2\Delta_1$ are linearly independent. They are, and it is enough to look at some low degree polynomials to check that. So, the sought dimension is $6$.