Does anyone know the formula for a Hall polynomial $g_{u,v}^{\lambda}(p)$ when $v$ is the type of cyclic subgroup (ie. $v=(v_{1})$ ) . http://en.wikipedia.org/wiki/Hall_algebra

I was hoping this particular case would be simple enough to describe .


Let's say you want to compute the Hall polynomial $g^\lambda_{(r),\mu}(p)$. According to [Dutta and Prasad, Degenerations and orbits in finite abelian groups], the orbits under the automorphism group of a finite abelian group are given by $\{O_I|I\subset J(P_\lambda)\}$, where $J(P_\lambda)$ denotes the lattice of order ideals in a certain poset $P_\lambda$. The same paper also gives a formula for $|O_I|$.

Clearly, the Hall polynomial that you are looking for is $(p^r-p^{r-1})^{-1}\sum_{I} |O_I|$, the sum being over all $I$ for which the order of an element in $O_I$ is $p^r$ and for which the quotient of the group of type $\lambda$ by any element of $O_I$ is of type $\mu$. As pointed out by David Speyer in the comments to an earlier version of this answer, $I$ is uniquely determined by these conditions. So a final form of the answer is obtained by explaining how to obtain $I$ from $\lambda$ and $\mu$.

Given an element $(p^{v_1},p^{v_2},\dotsc)$, the type of the quotient is found by computing the Smith canonical form of the matrix $\begin{pmatrix} p^{\lambda_1} & & &\\ & p^{\lambda_2} & &\\ & & \ddots & \\& & &p^{\lambda_l}\\ p^{v_1} & p^{v_2} & \cdots & p^{v_l} \end{pmatrix}$.

By the characterization of order ideals in $P_\lambda$, we have that $v_i\leq v_{i-1}\leq v_i+(\lambda_{i-1}-\lambda_i)$.

Proposition. Let $I\subset P_\lambda$ be an order ideal. Let $\mu$ be the type of the group obtained by going modulo an element of $O_I$. Then








Proof. The gcd of $i\times i$ minors of the above matrix can be seen to be $v_{l-i+1}+\lambda_{l-i+2}+\dotsb+\lambda_l$ (using the inequalities on $v_i$). Therefore, we get $\mu_{l-i+1}+\dotsb+\mu_l=v_{l-i+1}+\lambda_{l-i+2}+\dotsb+\lambda_l$, from which the above identities follow.QED.

This allows us to recover $(v_1,v_2,\dotsc)$ once we know $\lambda$ and $\mu$, where (this also follows from the above proposition), $\mu$ is a partition obtained from $\lambda$ by removing a horizontal strip of length $r$. In particular, $v$ is uniquely determined by $\lambda$ and $\mu$.

We get

$v_l=\mu_l$, and $v_i=\mu_i-[(\lambda_{i+1}+\dotsb+\lambda_l)-(\mu_{i+1}+\dotsb+\mu_l)]$ for $i<l$.

Let $I\subset P_\lambda$ be the order ideal defined by $(v_1,v_2,\dotsc)$. We get


It easily follows (from the formula for $|O_I|$ in our paper, which says that $|O_I|$ is a monic polynomial in $p$ of degree $\sum_i (\lambda_i-\nu_i)$) that $g^\lambda_{(r)\mu}(p)$ is monic in $p$ of degree $n(\lambda)-n(\mu)$. This could perhaps give another approach to Hall's theorem (in analogy with the proof that MacDonald gave).

  • $\begingroup$ I've only read the first two pages of your paper, but it appears to me to imply that there will precisely one term in each sum. In other words, it appears to me that the orbit of an element of $g$ in $A_{p,\lambda}$ is determined by the isomorphism class of its quotient. (continued) $\endgroup$ – David E Speyer Apr 30 '12 at 16:12
  • $\begingroup$ Fix $\lambda$ and let $A = \bigoplus \mathbb{Z}/(p^{\lambda_i})$. As $g$ ranges through $A$, the possible isomorphism types of $A/\langle g \rangle$ range through all partitions $\mu$ which can be obtained from $\lambda$ by deleting a horizontal strip. Here en.wikipedia.org/wiki/Pieri%27s_formula is an explanation in the context of Littlewood-Richardson coefficients, and the Littlewood-Richardson coefficient is nonzero iff the Hall polynomial is. (See, for example, section 2 of arxiv.org/abs/math.AG/9908012 ). (continued) $\endgroup$ – David E Speyer Apr 30 '12 at 16:18
  • $\begingroup$ So, the number of groups which are obtainable as $A/\langle g \rangle$ is the number of horizontal strips in $\lambda$. This is easily seen to be $\prod (\lambda_{i} - \lambda_{i+1} +1)$, where we tack infinitely zeroes onto the end of $\lambda$ so that the last few terms of the product are $\cdots (\lambda_{t-1} -\lambda_t+1) (\lambda_t+1) \cdot 1 \cdot 1 \cdots$. (continued) $\endgroup$ – David E Speyer Apr 30 '12 at 16:20
  • 1
    $\begingroup$ But equation (1) in your paper says that this is precisely the number of orbits in $A$ under the action of the automorphism group! So it must be that an orbit is distinguished by the isomorphism type of $A/\langle g \rangle$. I don't know whether the bijection between these horizontal strips and your order ideals is already in your paper, but it should be! $\endgroup$ – David E Speyer Apr 30 '12 at 16:22
  • $\begingroup$ Aha! So $\mu$ is completely determined by $I$? This is a wonderful simplification. Thank you for pointing this out. Then the formula becomes very simple. I think I can work out the exact relations between $I$, $\lambda$ and $(v_1,v_2,\dotsc)$. $\endgroup$ – Amritanshu Prasad May 1 '12 at 3:00

I'll put down what I figured out. Example 2.4 of Schiffman's Lectures on Hall Algebras tells how to multiply by the partition $1^r$. It is a generalization of the Pieri rule. If we set $p=1$, there is a symmetry of the ring of symmetric polynomials which sends a partition to its transpose, so we can just transpose this formula. But I don't know if there is an analogous symmetry for Hall polynomials. (The ring of Hall polynomials has an antipode, which morally should do this, but I couldn't find a statement of whether it does in a quick search.)

The formula for multiplying by $1^r$ also appears as Lemma 2.4 in this paper. According to this, the formula occurs on page 341 of Macdonald's book. I don't have the book available, but if I did I would look there to see whether the transpose formula also occurs.

Finally, I share your intuition that this computation should be doable by hand. If I were you, I'd look at Schifman's proof and see if I could adapt it.


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