The Benson - Symonds Invariant for Permutation Modules

Abstract

In a recent paper, Dave Benson and Peter Symonds defined a new invariant γ_G(M) for a finite dimensional module M of a finite group G which attempts to quantify how close a module is to being projective. In this paper, we determine this invariant for permutation modules of the symmetric group corresponding to two-part partitions using tools from representation theory and combinatorics.

Appendix

Lemma A.1 (Chu-Vandermonde identity)

$$\sum\limits_{i} {r \choose i}{s \choose n-i}={r+s \choose n}$$

This can be generalized to the following:

Lemma A.2

$$\sum\limits_{\substack {\mu \models r }}{p \choose \mu_{1}}{p \choose \mu_{2}}{\cdots} {p \choose \mu_{d}}={dp \choose r}$$

If we fix the number of parts of the composition then using the inclusion - exclusion principle, we have:

Lemma A.3

$$\sum\limits_{\substack {\mu \models r \\ \text{into \textit{d} parts}}}{p \choose \mu_{1}}{p \choose \mu_{2}}{\cdots} {p \choose \mu_{d}} = \sum\limits_{i=0}^{d} (-1)^{i}{d \choose i} {(d-i)p \choose r}$$

A more general form of the above Lemma would be the following:

Lemma A.4

$$ \sum\limits_{\substack {\mu \models r \\ \text{into \textit{d}+1 parts}}}{p \choose \mu_{1}}{p \choose \mu_{2}}{\cdots} {p \choose \mu_{d}}{a_{0} \choose \mu_{d+1}} = \sum\limits_{i=0}^{d} (-1)^{i}{d \choose i} \Bigg({(d-i)p+a_{0} \choose r}- {(d-i)p \choose r}\Bigg) $$

Adding Lemma A.3 and Lemma A.4 we get:

Lemma A.5

$$\sum\limits_{\substack {\mu \models r \\ \text{into \textit{d} parts}}}{p \choose \mu_{1}}{p \choose \mu_{2}}{\cdots} {p \choose \mu_{d}}+\sum\limits_{\substack {\mu \models r \\ \text{into \textit{d}+1 parts}}}{p \choose \mu_{1}}{p \choose \mu_{2}}{\cdots} {p \choose \mu_{d}}{a_{0} \choose \mu_{d+1}}$$

$$ = \sum\limits_{i=0}^{d} (-1)^{i}{d \choose i} {(d-i)p+a_{0} \choose r} $$

Note that in the above Lemma, the compositions with any part > p do not contribute to the sum. So if we were to enforce each part of the composition to be less than p then we only need to avoid the partitions with some part equal to p. Using inclusion-exclusion principle, we have:

Lemma A.6

$$ \sum\limits_{\substack {\mu \models jp+b_{0} \\ \text{into \textit{d} parts}\\ \text{each} < \text{p}}} {p \choose \mu_{1}}{p \choose \mu_{2}}{\cdots} {p \choose \mu_{d}}+\sum\limits_{\substack {\mu \models jp+b_{0} \\ \text{into \textit{d}+1 parts}\\ \text{each} < \text{p}}} {p \choose \mu_{1}}{p \choose \mu_{2}}{\cdots} {p \choose \mu_{d}}{a_{0} \choose \mu_{d+1}} $$

$$= \sum\limits_{h=0}^{j} (-1)^{h}{d \choose h} \sum\limits_{i=0}^{d-h} (-1)^{i} {d-h \choose i} {(d-h-i)p+a_{0} \choose (j-h)p+b_{0}} $$

Lemma A.7

$$\sum\limits_{i=0}^{n} (-1)^{i} {n \choose i}{n-i \choose m}=\left\{\begin{array}{lr} 1, & m= n\\ 0, & \text{otherwise} \end{array} \right. $$

Proof

$$ \begin{array}{@{}rcl@{}} \sum\limits_{i=0}^{n} (-1)^{i} {n \choose i}{n-i \choose m}&=&\sum\limits_{i=0}^{n} (-1)^{i} {n \choose m}{n-m \choose i}\\ &=&{n \choose m}\sum\limits_{i=0}^{n} (-1)^{i} {n-m \choose i}\\ &=& \left\{\begin{array}{lr} 1, & m= n\\ 0, & \text{otherwise} \end{array} \right.\\ \end{array} $$

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Lemma A.8

$$(-1)^{m} \sum\limits_{i=m}^{k-1} (-1)^{i} {k-1 \choose i} {i \choose m} \sum\limits_{j=0}^{i-m} {i-m \choose j} {k-i \choose r-j} $$

$$=\left\{\begin{array}{lr} 1, & m=k-1, r=0,1\\ 0, & \text{otherwise} \end{array} \right. $$

Proof

$$(-1)^{m} \sum\limits_{i=m}^{k-1} (-1)^{i} {k-1 \choose i} {i \choose m} \sum\limits_{j=0}^{i-m} {i-m \choose j} {k-i \choose r-j}$$

$$ \begin{array}{@{}rcl@{}} &=& (-1)^{m} \sum\limits_{i=m}^{k-1} (-1)^{i} {k-1 \choose i} {i \choose m} {k-m \choose r} \text{ Using Lemma } A.1 \\ &=& (-1)^{m} {k-m \choose r} \sum\limits_{i=m}^{k-1} (-1)^{i} {k-1 \choose i} {i \choose m} \\ &=& (-1)^{m} {k-m \choose r} \sum\limits_{i=m}^{k-1} (-1)^{i} {k-1 \choose m} {k-1-m \choose i-m} \\ &=& (-1)^{m} {k-m \choose r} {k-1 \choose m} \sum\limits_{i=m}^{k-1} (-1)^{i} {k-1-m \choose i-m} \\ &=& (-1)^{m} {k-m \choose r} {k-1 \choose m} \sum\limits_{i=0}^{k-1-m} (-1)^{m+i} {k-1-m \choose i} \\ &=&\left\{\begin{array}{lr} 1, & m=k-1, r=0,1\\ 0, & \text{otherwise} \end{array} \right. \end{array} $$

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Theorem A.9

$$k+ (k-1)\sum\limits_{\substack {\mu \models p \\ \text{into 2 parts}}} {p \choose \mu_{1}} {a_{0} \choose \mu_{2}}+\sum\limits_{d=2}^{k-1} {k-1 \choose d}\Bigg(\sum\limits_{\substack {\mu \models p \\ \text{into \textit{d} parts}}}{p \choose \mu_{1}}{p \choose \mu_{2}}{\cdots} {p \choose \mu_{d}} +$$

$$\sum\limits_{\substack {\mu \models p \\ \text{into \textit{d}+1 parts}}}{p \choose \mu_{1}}{p \choose \mu_{2}}{\cdots} {p \choose \mu_{d}} {a_{0} \choose \mu_{d+1}}\Bigg) $$

$$=1+{n-p \choose p} $$

Proof

Simplifying the sum on the left hand side using Lemma A.1 for the first part and Lemma A.3 and Lemma A.4 for the second part, we have

$$k+ (k-1)\Big({p+a_{0} \choose p}-1 \Big ) +\sum\limits_{d=2}^{k-1} {k-1 \choose d}\Bigg(\sum\limits_{i=0}^{d-1} (-1)^{i}{d \choose i} {(d-i)p \choose p} +$$

$$ \sum\limits_{i=0}^{d-1} (-1)^{i}{d \choose i} \Bigg({(d-i)p+a_{0} \choose p}- {(d-i)p \choose p}\Bigg) \Bigg) $$

Now simplifying the first part and using Lemma A.5 for the second part we get

$$ \begin{array}{@{}rcl@{}} &=&1+ (k-1){p+a_{0} \choose p} +\sum\limits_{d=2}^{k-1} {k-1 \choose d} \sum\limits_{i=0}^{d-1} (-1)^{i}{d \choose i} {(d-i)p+a_{0} \choose p} \\ &=&1+ \sum\limits_{d=1}^{k-1} {k-1 \choose d} \sum\limits_{i=0}^{d-1} (-1)^{i}{d \choose i} {(d-i)p+a_{0} \choose p} \end{array} $$

(1)

The coefficient of

$${mp+a_{0} \choose p}$$

in the above expansion is

$$\sum\limits_{i=0}^{k-1} (-1)^{i}{k-1 \choose m+i}{m+i \choose i}$$

$$=\sum\limits_{i=0}^{k-1} (-1)^{i}{k-1 \choose i}{k-1-i \choose m}$$

which is equal to 1 when m = k − 1 and 0 otherwise by Lemma A.7. So (1) reduces to

$$1+ {(k-1)p+a_{0} \choose p}$$

which is

$$1+{n-p \choose p}$$

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Theorem A.10

$$ {a_{0} \choose r}+ \sum\limits_{d=1}^{k-1}{k-1 \choose d}\Bigg(\sum\limits_{\substack {\mu \models r \\ \text{into \textit{d} parts}}}{p \choose \mu_{1}}{p \choose \mu_{2}}{\cdots} {p \choose \mu_{d}} +$$

$$\sum\limits_{\substack {\mu \models r \\ \text{into \textit{d}+1 parts}}}{p \choose \mu_{1}}{p \choose \mu_{2}}{\cdots} {p \choose \mu_{d}} {a_{0} \choose \mu_{d+1}}\Bigg) $$

$$={n-p \choose r} $$

Proof

Simplifying the sum on the left hand side using Lemma A.3 and Lemma A.4, we have

$$ {a_{0} \choose r}+ \sum\limits_{d=1}^{k-1}{k-1 \choose d}\Bigg(\sum\limits_{i=0}^{d} (-1)^{i}{d \choose i} {(d-i)p \choose r}+$$

$$ \sum\limits_{i=0}^{d} (-1)^{i}{d \choose i} \Bigg({(d-i)p+a_{0} \choose r}- {(d-i)p \choose r}\Bigg) $$

$$ \begin{array}{@{}rcl@{}} &=&{a_{0} \choose r}+ \sum\limits_{d=1}^{k-1}{k-1 \choose d} \sum\limits_{i=0}^{d} (-1)^{i}{d \choose i}{(d-i)p+a_{0} \choose r} \\ &=&{a_{0} \choose r}+ \sum\limits_{d=1}^{k-1}{k-1 \choose d}\Bigg (\sum\limits_{i=0}^{d-1} (-1)^{i}{d \choose i}{(d-i)p+a_{0} \choose r}+ (-1)^{d}{a_{0} \choose r} \Bigg) \\ &=&{a_{0} \choose r} + \sum\limits_{d=1}^{k-1}{k - 1 \choose d} \sum\limits_{i=0}^{d-1} (-1)^{i}{d \choose i}{(d - i)p+a_{0} \choose r}+ \sum\limits_{d=1}^{k-1}{k-1 \choose d} (-1)^{d}{a_{0} \choose r} \end{array} $$

(2)

The coefficient of

$${mp+a_{0} \choose r}$$

in the expansion of the sum in the center is

$$\sum\limits_{i=0}^{k-1} (-1)^{i}{k-1 \choose m+i}{m+i \choose i}$$

$$=\sum\limits_{i=0}^{k-1} (-1)^{i}{k-1 \choose i}{k-1-i \choose m}$$

which is equal to 1 when m = k − 1 and 0 otherwise by Lemma A.7. So, (2) simplifies to

$$ \begin{array}{@{}rcl@{}} &=&{a_{0} \choose r}+{(k-1)p+a_{0} \choose r}-{a_{0} \choose r}\\ &=&{(k-1)p+a_{0} \choose r}\\ &=&{n-p \choose r}\\ \end{array} $$

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Theorem A.11

$$ {\left( \begin{array}{c}k \\ q \end{array}\right)}{\left( \begin{array}{c}a_{0} \\ b_{0}\end{array}\right)}+\sum\limits_{d=1}^{k-1} {\left( \begin{array}{c}k-1 \\ d\end{array}\right)} \left( \sum\limits_{j=0}^{d} {\left( \begin{array}{c}k-d \\ q-j\end{array}\right)} \left( \sum\limits_{\substack{\mu \models jp+b_{0}\\ \text{into \textit{d} parts}\\ \text{each} < \text{p}}} {\left( \begin{array}{c}p \\ \mu_{1}\end{array}\right)}{\left( \begin{array}{c}p \\ \mu_{2}\end{array}\right)}{\cdots} {\left( \begin{array}{c}p \\ \mu_{d}\end{array}\right)}+\right.\right. $$

$$ \left.\left.\sum\limits_{\substack{\mu \models jp+b_{0} \\ \text{into \textit{d}+1 parts}\\ \text{each} < \text{p}}} {\left( \begin{array}{c}p \\ \mu_{1}\end{array}\right)}{\left( \begin{array}{c}p \\ \mu_{2}\end{array}\right)}{\cdots} {\left( \begin{array}{c}p \\ \mu_{d}\end{array}\right)}{\left( \begin{array}{c}a_{0} \\ \mu_{d+1}\end{array}\right)} \right) \right) $$

$$ ={\left( \begin{array}{c}(k-1)p+a_{0} \\ qp+b_{0}\end{array}\right)} + \left( \begin{array}{c}(k-1)p+a_{0} \\ (q-1)p+b_{0}\end{array}\right) $$

Proof

Using Lemma A.6 we have,

$$\sum\limits_{d=0}^{k-1} {\left( \begin{array}{c}k-1 \\ d\end{array}\right)}\left( \sum\limits_{j=0}^{d} {k-d \choose q-j} \left( \sum\limits_{\substack {\mu \models jp+b_{0} \\ \text{into \textit{d} parts}\\ \text{each} < \text{p}}} {p \choose \mu_{1}}{p \choose \mu_{2}}{\cdots} {p \choose \mu_{d}}+\right.\right. $$

$$ \left.\left.\sum\limits_{\substack {\mu \models jp+b_{0} \\ \text{into \textit{d}+1 parts}\\ \text{each} < \text{p}}} {\left( \begin{array}{c} p \\ \mu_{1}\end{array}\right)}{\left( \begin{array}{c}p \\ \mu_{2}\end{array}\right)}{\cdots} {\left( \begin{array}{c}p \\ \mu_{d}\end{array}\right)}{\left( \begin{array}{c}a_{0} \\ \mu_{d+1}\end{array}\right)} \right)\right) $$

$$=\sum\limits_{d=0}^{k-1} {\left( \begin{array}{c}k-1 \\ d\end{array}\right)} \sum\limits_{j=0}^{d} {\left( \begin{array}{c}k-d \\ q-j\end{array}\right)} \sum\limits_{h=0}^{j} (-1)^{h}{\left( \begin{array}{c}d \\ h\end{array}\right)} \sum\limits_{i=0}^{d-h} (-1)^{i} {\left( \begin{array}{c}d-h \\ i\end{array}\right)} {\left( \begin{array}{c}(d-h-i)p+a_{0} \\ (j-h)p+b_{0}\end{array}\right)} $$

For any m and s, the coefficient of

$$ \left( \begin{array}{c}mp+a_{0} \\sp+b_{0}\end{array}\right) $$

is

$$ (-1)^{m} \sum\limits_{i=m}^{k-1} (-1)^{i} {\left( \begin{array}{c}k-1 \\ i\end{array}\right)} {\left( \begin{array}{c}i \\m\end{array}\right)} \sum\limits_{j=0}^{i-m} {\left( \begin{array}{c}i-m \\ j\end{array}\right)} {\left( \begin{array}{c}k-i \\ q-s-j\end{array}\right)} $$

Using Lemma A.8, we can simplify our left hand side expression to

$$ ={\left( \begin{array}{c}(k-1)p+a_{0} \\ qp+b_{0}\end{array}\right)}+{\left( \begin{array}{c}(k-1)p+a_{0} \\ (q-1)p+b_{0}\end{array}\right)} $$

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