Today is 19/03/2020 and I will post what I learn from 18-19/03/2020.

Symmetric functions - Hall polynomials

I read chapter II of Macdonald’s book Symmetric Functions and Hall Polynomials. Here is a brief summary from that chapter, talking about Hall polynomials:

  • Define finite $\mathfrak{o}$-module $M$, which corresponds uniquely (up to isomorphism) to a partition $\lambda$ (called type of $M$).
  • Define Hall algebra which is $\mathbf{Z}$-generated by basis $(u_{\lambda})$ indexed by partition with multiplication determined by structure constants $G_{\mu \nu}^{\lambda}(\mathfrak{o})$, which counts number of ubmodule $N$ of type $\nu$ and cotype $\mu$ in finite $\mathfrak{o}$-module $M$ of type $\lambda$ (here we suppose $k$ is finite with $|k|=q$ so $M$ has finitely many elements)
  • We give a formula for $G_{\mu \nu}^{\lambda}(\mathfrak{o})$ which is a Hall polynomial $g_{\mu \nu}^{\lambda}$ evaluated at $q$.

Finite $\mathfrak{o}$-module

Let $\mathfrak{o}$ discrete valuation ring with maximal ideal $\mathfrak{p}$ and residue field $k=\mathfrak{o}/\mathfrak{p}$. A finite $\mathfrak{o}$-module $M$ is a finitely generated $\mathfrak{o}$-module such that $\mathfrak{p}^rM=0$ for some $r \ge 0$.

There is a unique correspondence between finite $\mathfrak{o}$-module $M$ and a partition $\lambda$ (up to isomorphism of $M$): Since $\mathfrak{o}$ is principal ideal domain, $M$ as finitely generated $\mathfrak{o}$-module is direct sum of cyclic modules

\[M=\bigoplus_{i=1}^r \mathfrak{o}/\mathfrak{p}^{\lambda_i} \label{eq1}\tag{1}\]

Furthermore, if $\mu_i=\text{dim }_k(\mathfrak{p}^{i-1} M/\mathfrak{p}^i M)$ then $\mu=(\mu_1,\ldots,)$ is conjugate of $\lambda$. We call $\lambda$ type of $M$.

If $\lambda$ is type of $M$ then let $l(M)=|\lambda|= \sum \lambda_i$ be the length of $M$. The length has very useful property (which will be used frequently in the chapter) is that if $N$ submodule of $M$ then $l(M)=l(N)+l(M/N)$. The cotype of $N$ in $M$ is defined to be the type of $M/N$.

Finite $\mathfrak{o}$-module $M$ is cyclic if its type is $(r)$ consisting of single part (very natural definition from \ref{eq1}). $M$ is elementary (i.e. $\mathfrak{p}M=0$) iff its type if $(1^r)$.

Duality

Let $\pi$ generater of maximal ideal $\mathfrak{p}$. If $m\le n$, multiplication by $\pi^{n-m}$ is injective $\mathfrak{o}$-homomorphism from $\mathfrak{o}/\mathfrak{p}^m$ to $\mathfrak{o}/ \mathfrak{p}^n$. Denote $E=\varinjlim \mathfrak{o}/ \mathfrak{p}^n$ then $E$ can be viewed a the smallest injective $\mathfrak{o}$-module containing $k$ as submodule.

The dual of finite $\mathfrak{o}$-module $M$ is denoted $\widehat{M}=\text{Hom}_{\mathfrak{o}}(M,E)$ which is isomorphic to $M$. Since $E$ injective, exact sequence

\[\xymatrix{0 \ar[r] & N \ar[r] & M \ar[r] & M/N \ar[r] & 0}\]

gives rise to exact sequence

\[\xymatrix{0 & \widehat{N} \ar[l] & \widehat{M} \ar[l] & \widehat{M/N} \ar[l] & 0 \ar[l]}\]

where $\widehat{M/N}$ is the annihilator $N^0$ of $N$ in $\widehat{M}$, i.e. set of all $f\in \widehat{M}$ so $f(N)=0$. The natural mapping $M\to \widehat{\widehat{M}}$ is isomorphism for all finite $\mathfrak{o}$-module, and identifies $N$ with $N^{00}$. Hence,

$N \leftrightarrow N^0$ is a one-one correspondence between submodules of $M, M\widehat{M}$, respectively, which maps set of all $N\subset M$ of type $\nu$ and cotype $\mu$ onto set of all $N^0\subset \widehat{M}$ of type $\mu$ and cotype $\nu$.

QUESTION: What is exactly this correspondence? How do one derive the (co)type here? This is part of this chapter that I couldn’t quite understand.

Hall algebra

We assume $k$ to be finite. Then $M$ has finite number of elements.

Here we define Hall algebra based on structure constants $G_{\mu \nu}^{\lambda} (\mathfrak{o})$ which counts number of submodule $N$ of type $\nu$ and cotype $\mu$ in finite $\mathfrak{o}$-module $M$ of type $\lambda$:

Let $H=H(\mathfrak{o})$ be free $\mathbf{Z}$-module on basis $(u_{\lambda})$ indexed by partitions $\lambda$. Define product in $H$ by the rule

\[u_{\mu}u_{\nu}=\sum_{\lambda} G_{\mu \nu}^{\lambda}(\mathfrak{o}) u_{\lambda}\]

Since $l(M)=l(M/N)+l(N)$ so $G_{\mu\nu}^{\lambda}(\mathfrak{o})=0$ unless $|\lambda|=|\mu|+|\nu|$. Hence, the sum in the above right-hand side is finite.

One can show the following

  • $H(\mathfrak{o})$ is commutative and associative ring with identity element, which is called Hall algebra of $\mathfrak{o}$.
  • Ring $H(\mathfrak{o})$ is generated (as $\mathbf{Z}$-algebra) by elements $u_{(1^r)}$ for $r\ge 1$ and they are algebraically independent over $\mathbf{Z}$. This follows that Hall algebra $H(\mathfrak{o})$ isomorphic to ring $\Lambda$ of symmetric functions.

$LR$-sequence of a submodule

Next, we investigate connection between type of $M$ and type of its submodule:

Let $M$ be finite $\mathfrak{o}$-module of type $\lambda$, we have

  • If $N$ submodule of type $\nu$ and cotype $\mu$ in $M$, then $\mu \subset \lambda$ and $\nu \subset \lambda$.
  • Denote $S=\{x\in M: \mathfrak{p}x=0\}$ called socle of $M$, which is the unique largest elementary submodule of $M$. Then type of $M/S$ is $\tilde{\lambda}= (\lambda_1-1,\lambda_2-1,\ldots)$.
  • If $N$ elementary submodule of $M$ of cotype $\mu$ then $\lambda-\nu$ is vertical strip (i.e. $\lambda_i-\mu_i=0$ or $1$).

Proofs for these properties can be obtained by using additive property of length of finite $\mathfrak{o}$-module, i.e. $l(M)-l(M/N)-l(N)=0$.

From these properties, we find that every submodule $N$ of finite $\mathfrak{o}$-module $M$ gives rise to a $LR$-sequence of type $(\mu’,\nu’,\lambda’)$ where $\lambda,\mu,\nu$ are types of $M,M/N, N$, respectively. One should just think $LR$-sequence of type $(\mu,\nu,\lambda)$, i.e. sequence of partitions

\[\mu=\lambda^{(0)}\subset \lambda^{(1)}\subset \cdots \subset \lambda^{(r)}=\lambda\]

as tableau $T$ of shape $\lambda-\mu$, with weight $\nu_i= |\lambda^{(i)}-\lambda^{(i-1)}|$ and numbers on $T$ are determined by above sequence of partitions such that $w(T)$ is a lattice permutation (see Chapter 1, section 9 of the book).

For each $i\ge 0$, let $\lambda^{(i)}$ the cotype of $\mathfrak{p}^iN$. Then the sequence $S(N)= \left( \lambda^{(0)} ‘, \lambda^{(1)} ‘,\ldots, \lambda^{(r)} ‘ \right)$ where $\mathfrak{p}^rN=0$ is an $LR$-sequence of type $(\mu’,\nu’,\lambda’)$.

Hall polynomial

We compute the structure constants $G_{\mu \nu}^{\lambda}(\mathfrak{o})$ of Hall algebra (given $k$ is finite). Let $S$ be $LR$-sequence of type $(\mu’,\nu’,\lambda’)$ and $M$ finite $\mathfrak{o}$-module of type $\lambda$. Denote $G_S(\mathfrak{o})$ the number of submodules $N$ of $M$ whose associated $LR$-sequence $S(N)$ is $S$. Each such $N$ will have type $\nu$ and cotype $\mu$ (from definition of $S(N)$). Then

\[G_{\mu \nu}^{\lambda}(\mathfrak{o})=\sum_S G_S(\mathfrak{o})\]

summed over all $LR$-sequences $S$ of type $(\mu’,\nu’,\lambda’)$. Hence, it suffices to compute $G_S(\mathfrak{o})$ for certain $S$. To do this, we divide into smaller subproblems that is easier to count: Observe that if $N$ submodule of $M$ such that $S(N)=S = (\lambda^{(0)} ‘, \ldots, \lambda^{(r)} ‘)$, then with $N_1=\mathfrak{p}N$, we have $S(N_1)=\left( \lambda^{(1)} ‘, \ldots, \lambda^{(r)} ‘ \right)=S_1$.

Thus, we have $G_S(\mathfrak{o})$ equals to $G_{S_1}(\mathfrak{o})$ (i.e. number of submodules $N_1$ of $M$ so $S(N_1)=S_1$) times (given such $N_1$) number of submodule $N$ of $M$ so $\mathfrak{p}N=N_1$ and $N$ of cotype $\lambda^{(0)}=\mu$ (note these two conditions will imply $S(N)=S$, which is what we originally want to count). The later factor is equivalent to count number of submodule $N/\mathfrak{p}N_1$ of cotype $\lambda^{(0)}$ in $M/\mathfrak{p}N_1$ of type $\lambda^{(2)}$ such that $\mathfrak{p} \left( N/\mathfrak{p}N_1 \right)= N_1/\mathfrak{p}N_1$ (which has cotype $\lambda^{(1)}$). The advantage of rephrasing this way is that $N_1/\mathfrak{p}N_1$ is an elementary submodule. By generalizing this, the new object we want to count is:

Given $P$ elementary submodule of cotype $\beta$ in $M$, we want to count number of submodules $N$ of certain cotype $\alpha$ in $M$ such that $\mathfrak{p}N=P$.

This is the content of lemma (4.7) in the book. If one follows the proof of this lemma, where we try to rephrase the object in question, just as we did previously, (hoping that the new object is easier to count), we reduce to count the following object

Let $N$ elementary submodule of cotype $\alpha$ in $M$. Let $\beta$ partition such that $\alpha\subset \beta \subset \lambda$. We want to count $H_{\alpha\beta\lambda}$, number of submodules $P\subset N$ of cotype $\beta$ in $M$.

This is the content of lemma (4.4) in the book. Let $P$ be submodule of $N$ of cotype $\beta$ in $M$. From what is written there, the general idea is that by observing $P=P_0\supset P_1 \supset P_2 \cdots$ where $P_i=P\cap \mathfrak{p}^iM$ then

\[l(P_{i-1}/P_i)= \lambda_i'-\beta_i'\]

and conversely, if one knows this condition then we know $P$ has cotype $\beta$ in $M$. Hence, in order to construct $P$, one can start construct $P_i$ inductively as $i\to 0$. Note an extra condition we need to guarantee is that $P\subset N$ so for all $i$, we need $P_i$ such that $P_i\subset N_i= N\cap \mathfrak{p}^iM$. This is equivalent to

\[P_{i-1}\cap N_i=P_i\]

Thus, we need to count the following

Given $P_i$, we want to count number of submodules $P_{i-1}$ of $N_{i-1}$ such that $l(P_{i-1}/P_i) =\lambda_i’-\beta_i’=\theta_i$ and $P_{i-1}\cap N_i=P_i$.

This object can be computed directly, where the idea is that from $P_i$, we want to contruct $P_{i-1} \supset P_i$ by adding new $\theta_i’$ elements (viewed as generating elements) in $N_{i-1}$ which are linearly independent modulo $N_i$.

After all of these counting, we conclude that $G_{\mu\nu}^{\lambda}(\mathfrak{o})$ can be expressed as polynomial $g_{\mu\nu}^{\lambda}(t) \in \mathbf{Z}[t]$ evaluated at $q=|k|$. Such polynomial called Hall polynomial.

From these computations, Macdonald also gives explicit formulas for two special cases $\nu=(r)$ and $\nu=(1^r)$.

OK, I just had a talk with my supervisor and it seems the content of this chapter is not as famous/active to current research anymore (at least in area of algebraic combinatorics) so if anyone wants to read Macdonald’s book, a suggestion is to jump right to chapter 3 after reading chapter 1.