*This is a long overdue followup post to Garsia-Procesi Modules: Part 1 that I finally got around to editing and posting. Enjoy!*

In this post, I talked about the combinatorial structure of the Garsia-Procesi modules $R_\mu$, the cohomology rings of the type A Springer fibers. Time to dive even further into the combinatorics!

## Tanisaki generators, visually

Recall that, for a partition $\mu$ of $n$, the graded $S_n$-module $R_\mu$ can be constructed (due to Tanisaki) as the quotient ring $$\mathbb{C}[x_1,\ldots,x_n]/I_\mu$$ where $I_\mu$ is generated by certain partial elementary symmetric functions called *Tanisaki generators*.

Define $d_k(\mu)=\mu’_{n-k+1}+\mu’_{n-k+2}+\cdots$ to be the sum of the last $k$ columns of $\mu$, where we pad the conjugate partition $\mu’$ with $0$’s in order to think of it as as a partition of $n$ having $n$ parts. Then the partial elementary symmetric function $e_r(x_{i_1},\ldots,x_{i_k})$ is a Tanisaki generator if and only if $k-d_k(\mu)\lt r\le k$. Tanisaki and Garsia and Procesi both use this notation, but I find $d_k$ hard to remember and compute with, especially since it involves adding zeroes to $\mu$ and adding parts of its transpose in reverse order, and *then* keeping track of an inequality involving it to compute the generators.

An equivalent, and perhaps simpler, definition is as follows. If $n-k\lt \mu_1$, then the quantity $k-d_k(\mu)$ is equal to the number of squares in the first $n-k$ columns of $\mu$, *excluding the first row*. Indeed, we have $$k-d_k(\mu)=n-d_k(\mu)-(n-k),$$ and on the right hand side, $n-d_k(\mu)$ is simply the number of squares in the first $n-k$ columns of $\mu$. Subtracting $n-k$ is then equivalent to removing one square from each column in that count, which can be done by crossing out the first row of $\mu$.

So, define $$s_t(\mu)=\mu’_1+\cdots+\mu’_t-t$$ to be the number of squares in the first $t$ columns, not including the first row. Then we include an elementary symmetric function $e_r$ in $k$ of the variables $x_i$ if $$r\gt s_{n-k}(\mu).$$ (Note that the upper bound condition, $r\le k$, is necessary for the elementary symmetric function to be nonzero, so we technically don’t need to state the upper bound).

For instance, if $n=8$ and $\mu=(4,3,1)$, then the partition diagram looks like:

where the bottom row is x’ed out to remind ourselves not to count it.

For $k=8$, we require $r\gt s_0(\mu)=0$, so all elementary symmetric functions $e_1,\ldots,e_8$ in the $8$ variables $x_1,\ldots,x_8$ are in $I_\mu$.

For $k=7$ we require $r\gt s_1(\mu)=2$, so $e_3,e_4,\ldots,e_7$ on any seven of the variables are generators.

For $k=6$ we require $r \gt s_2(\mu)=3$, so $e_4,e_5,e_6$ on any six of the variables are generators.

For $k=5$ we require $r \gt s_3(\mu)=4$, so only $e_5$ on any five of the variables is a generator.

For $k=4$ we require $r\gt s_4(\mu)=4$, and we have no additional generators.

For $k$ such that $n-k>\mu_1$, we also never get any additional generators, since $k-d_k(\mu)=k$ in this case. Therefore the combinatorial interpretation above, which is only valid for $n-k<\mu_1$, is in fact an equivalent definition for the Tanisaki generators.

It’s not fundamentally all that different, but working with columns of a partition from the left rather than the right can save a bit of mental space during computations.

## A smaller set of generators

It turns out that, for $|S|<n$, it suffices to only include $e_r(S)$ for the smallest possible $r$ in order to generate the Tanisaki ideal. We prove this here. Write $X={x_1,\ldots,x_n}$ throughout.

**Proposition.** The ideal $I_\mu$ is generated by the following elements:

- The elementary symmetric functions $e_1,\ldots,e_n$ in all $n$ variables $x_1,\ldots,x_n$, and
- The partial elementaries $e_r(S)$ where $S\subseteq X$ with $|S|=k<n$ and $$r=s_{n-k}(\mu)+1.$$

**Proof.** Let $I_\mu^0$ be the ideal generated by the above functions. To show $I_\mu^0=I_\mu$, it suffices to show that for every $k s_{n-k}(\mu)$. We show this by nested induction, first on $n-k$ and then on $r$.

For the base case, $n-k=0$, there is nothing to show since $k=n$. Now fix $k$ and assume the claim holds for all smaller values of $n-k$ (for all larger values of $k$). Let $S\subseteq X$ with $|S|=k$, and since $k<n$, choose a missing variable $x_i\in X-S$.

Define $r_0=s_{n-k}(\mu)+1$; we have $e_{r_0}(S)\in I_\mu^0$ by the definition of $I_\mu^0$. Now let $r>r_0$ and assume $e_t(S)\in I_\mu^0$ for $r_0\le ts_{n-(k+1)}(\mu)$ and so $e_r(S\cup {x_i})\in I_\mu^0$ by the induction hypothesis on $n-k$. Therefore $e_r(S)\in I_\mu^0$ as desired. $\square$

## Recursive algorithm for expanding in basis $\mathcal{B}(\mu)$

Garsia and Procesi define a basis $\mathcal{B}(\mu)$ of monomials for $R_\mu$ recursively by $$\mathcal{B}(\mu)=\bigcup_i x_n^{i-1}\mathcal{B}(\mu^{(i)}).$$ (Recall that $\mu^{(i)}$ is the partition of $n-1$ formed by removing one square from the $i$th part and re-sorting the resulting rows in partition order.) They then give a completely elementary inductive proof that every other element of $R_\mu$ can be expressed as a linear combination of basis elements from $\mathcal{B}(\mu)$. We describe their algorithm here, with a few slight modifications.

First, note that it suffices to describe how to express any monomial $x^\alpha$ as a linear combination of the elements of $\mathcal{B}_\mu$, plus an element of $I_\mu$. For $n=0$, the only partition is the empty partition, $R_\emptyset=\mathbb{C}$, and the unique basis element is $1$, so $x^\alpha=x^0=1$ is its entire expansion (since $I_\emptyset=(0)$).

For $n>0$, we use the following recursive algorithm.

- Given $x^\alpha=x_1^{\alpha_1}\cdots x_n^{\alpha_n}$, let $i-1=\alpha_n$ be the exponent of $x_n$.
- Setting $\beta=(\alpha_1,\ldots,\alpha_{n-1})$, we have that $x^{\beta}=x_1^{\alpha_1}\cdots x_{n-1}^{\alpha_{n-1}}$ can be interpreted as a representative of an element in $R_{\mu^{(i)}}$. Use this algorithm recursively to express $x^\beta$ in terms of $\mathcal{B}(\mu^{(i)})$ plus an error term in $I_{\mu^{(i)}}$, giving an expansion $$x^\beta=\sum_{b\in \mathcal{B}(\mu^{(i)})} c_b b(x_1,\ldots,x_{n-1}) +E$$ where $E\in I_{\mu^{(i)}}$.
- Multiplying both sides by $x_n^{i-1}$, we have $$x^\alpha=\sum_{b\in \mathcal{B}(\mu^{(i)})}c_bx_n^{i-1}b(x_1,\ldots,x_{n-1})+x_n^{i-1}E.$$ Each monomial $x_n^{i-1}b(x_1,\ldots,x_{n-1})$ is an element of $\mathcal{B}(\mu)$ by the recursive definition of the basis $\mathcal{B}(\mu)$.
- We next expand $x_n^{i-1}E$ as a linear combination of elements of $\mathcal{B}(\mu)$ plus an element of $I_\mu$. Since $E\in I_{\mu^{(i)}}$, we can write $$E=\mathop{\sum_{r>s_{n-k}(\mu^{(i)})}}_{i_1,\ldots,i_k\in [n-1]} a_{r,\{i_j\}}e_r(x_{i_1},\ldots,x_{i_k}).$$ We will consider each term individually, expressing $x_n^{i-1}e_r(x_{i_1},\ldots,x_{i_k})$ in terms of $I_{\mu}$ plus elements of $x_n^{i}R_\mu$ (hence reducing to a case in which the exponent of $x_n$ is larger, at which point we can repeat the algorithm until the exponent is above $\mu’_1$.)
- If either $n-k\lt \mu_i$, or if $n-k\ge \mu_i$ and $r>s_{n-1-k}(\mu^{(i)})+1=s_{n-1-k}(\mu)$, then $e_r(x_{i_1},\ldots,x_{i_k},x_n)\in I_\mu$ and so the expansion $$x_n^{i-1}e_r(x_{i_1},\ldots,x_{i_k})=x_n^{i-1}e_r(

x_{i_1},\ldots,x_{i_k} ,x_n)-x_n^ie_{r-1}(x_{i_1},\ldots,x_{i_k})$$ is our desired expression in $I_\mu+x^iR_\mu$. - Otherwise, if $n-k\ge \mu_i$ and $r=s_{n-1-k}(\mu^{(i)})+1=s_{n-1-k}(\mu)$, then note that we must have $i>1$. Iteratively using the identity $x_ne_r(x_{i_1},\ldots,x_{i_k})=e_{r+1}(x_{i_1},\ldots,x_{i_k},x_n)-e_{r+1}(x_{i_1},\ldots,x_{i_k})$, we can multiply by $x_n$ exactly $i-1$ times to obtain $$\begin{align*} x_n^{i-1}e_r(x_{i_1},\ldots,x_{i_k})= & \phantom{+}x_n^{i-2}e_{r+1}(x_{i_1},\ldots,x_{i_k},x_n) \\ &-x_n^{i-3}e_{r+2}(x_{i_1},\ldots,x_{i_k},x_n) \\ &+\cdots \\ &+ (-1)^{i-2}e_{r+i-1}(x_{i_1},\ldots,x_{i_k},x_n) \\ &+(-1)^{i-1}e_{r+i-1}(x_{i_1},\ldots,x_{i_k})\end{align*}$$ all of whose terms on the right hand side are in $I_\mu$.
- We now have expressed each term of $E$ in the form $I+x_n^iE_1$ where $I\in I_\mu$ is expressed in terms of the Tanisaki generators, and $E_1\in R_\mu$. We iterate steps 1-6 on each term of $x_n^iE_1$ and continue until we only have monomials having $x_n^h$ as a factor where $h=\mu_1’$ is the height of $\mu$.
- Note that $$\begin{align*}x_n^h= &\phantom{+}x_n^{h-1}e_1(x_1,\ldots,x_n) \\ &-x_n^{h-2}e_2(x_1,\ldots,x_n) \\ &+\cdots \\ &+(-1)^{h-1}e_h(x_1,\ldots,x_n) \\ &-e_h(x_1,\ldots,x_{n-1}). \end{align*}$$ The first $h$ terms above are clearly in $I_\mu$, and the last term $e_h(x_1,\ldots,x_{n-1})$ is in $I_\mu$ as well because $h>s_{n-(n-1)}(\mu)=\mu’_1-1$. The above expansion, as well as the similar relations for $x_i^h$ obtained by acting by an appropriate element of $S_n$, ensures that the process terminates.

We can now use this algorithm to express every monomial of degree at most $n(\mu)=\sum_i (i-1)\mu_i$ (which is the highest nonzero degree of $R_\mu$) in terms of the basis $\mathcal{B}(\mu)$ plus an element of $I_\mu$, expressed explicitly in terms of Tanisaki generators.

In order to minimize recursive calls, one should build this database of expansions starting with partitions of size $1$, then of size $2$, and so on. Further, for a given partition shape $\mu$, one should first expand the monomials of degree at most $n(\mu)$ with the highest possible exponent of $x_n$ (namely $x_n^h$ where $h=\mu_1’$ and then continue with the monomials having $n$th exponent $x_n^{h-1}$, then with $x_n^{h-2}$, and so on. This ensures that steps 4-8 of the algorithm only need to be run once on each summand in the error term $E$ obtained in step 3.

## Expansions up to $|\mu|=3$

Using the above algorithm, I was able to quickly calculate all monomial expansions of degree at most $n(\mu)$ for all partitions $\mu$ of size at most $3$ by hand. In some cases, I worked out some of the expansions of the monomials of degree larger than $n(\mu)$, which lie entirely in $I_\mu$, as a convenience for the calculations in the subsequent cases. (In general, if one were to code this algorithm with the goal of finding the expansions for $R_\mu$, one should calculate the $I_{\lambda}$ expansions for all monomials of degree up to $n(\mu)$ for ALL partitions $\lambda\le \mu$, in order to maximize efficiency at each step.)

For $\mu=(1)$, we have $\mathcal{B}((1))=\{1\}$ and $I_\mu=(e_1(x_1))$. The expansions are:

- $x_1=e_1(x_1)$
- $1=1$

For $\mu=(2)$, we have $\mathcal{B}((1))=\{1\}$ and $I_\mu=(e_1(x_1,x_2),e_2(x_1,x_2),e_1(x_1),e_1(x_2))$. The expansions are:

- $x_2=e_1(x_2)$
- $x_1=e_1(x_1)$
- $1=1$

For $\mu=(1,1)$, we have $\mathcal{B}((1))=\{1,x_2\}$ and $I_\mu=(e_1(x_1,x_2),e_2(x_1,x_2))$. The expansions are:

- $x_2^2=x_2e_1(x_1,x_2)-e_2(x_1,x_2)$
- $x_2x_1=e_2(x_1,x_2)$
- $x_2=x_2$
- $x_1^2=x_1e_1(x_1,x_2)-e_2(x_1,x_2)$
- $x_1=-x_2+e_1(x_1,x_2)$
- $1=1$

For $\mu=(3)$, we have $\mathcal{B}((1))=\{1\}$ and $I_\mu$ is the set of *all* partial elementary symmetric functions in three variables, so $R_\mu=\mathbb{C}$. The expansions are:

- $1=1$

For $\mu=(2,1)$, we have $\mathcal{B}((1))=\{1,x_2,x_3\}$ and $I_\mu=(e_1,e_2,e_3,e_2(x_1,x_2),e_2(x_1,x_3),e_2(x_2,x_3))$, where $e_i$ without any variables indicates the full elementary symmetric functions using all three variables. The expansions are:

- $x_3^2=x_3e_1-e_2+e_2(x_1,x_2)$
- $x_3=x_3$
- $x_2=x_2$
- $x_1=-x_3-x_2+e_1$
- $1=1$

For $\mu=(1,1,1)$, we have $\mathcal{B}((1))=\{1,x_2,x_3,x_3x_2,x_3^2,x_3^2x_2\}$ and $I_\mu=(e_1(x_1,x_2,x_3),e_2(x_1,x_2,x_3),e_3(x_1,x_2,x_3))$. For shorthand in this case we simply write $e_1,e_2,e_3$ since there are no strictly partial $e_r$’s in $I_\mu$. The expansions are:

- $x_3^3=x_3^2e_1-x_3e_2+e_3$
- $x_3^2x_2=x_3^2x_2$
- $x_3^2x_1=-x_3^2x_2+x_3e_2-e_3$
- $x_3^2=x_3^2$
- $x_3x_2^2=-x_3^2x_2+(x_2x_3+x_3^2)e_1-x_3e_2$
- $x_3x_1^2=x_3^2x_2+(x_3^2+x_3x_1)e_1

-2x_3e_2 +e_3$ - $x_3x_2x_1=e_3$
- $x_3x_2=x_3x_2$
- $x_3x_1=-x_3x_2-x_3^2+x_3e_1$
- $x_3=x_3$
- $x_2^3=x_2^2e_1-x_2e_2+e_3$
- $x_2^2x_1=x_3^2x_2-x_2x_3e_1+x_2e_2$
- $x_2^2=-x_3x_2+(x_2+x_3)e_1-e_2$
- $x_2x_1^2=-x_3^2x_2-x_1x_3e_1+(x_1+x_3)e_2-e_3$
- $x_2x_1=x_3^2-x_3e_1+e_2$
- $x_2=x_2$
- $x_1^3=x_1^2e_1-x_1e_2+e_3$
- $x_1^2=x_3x_2+x_3^2-x_1e_1-e_2$
- $x_1=-x_2-x_3+e_1$
- $1=1$