A₂ edit

Consider the A2 root system, with positive roots ${\displaystyle \alpha _{1}}$ , ${\displaystyle \alpha _{2}}$ , and ${\displaystyle \alpha _{3}:=\alpha _{1}+\alpha _{2}}$ . If an element ${\displaystyle \mu }$  can be expressed as a non-negative integer linear combination of ${\displaystyle \alpha _{1}}$ , ${\displaystyle \alpha _{2}}$ , and ${\displaystyle \alpha _{3}}$ , then since ${\displaystyle \alpha _{3}=\alpha _{1}+\alpha _{2}}$ , it can also be expressed as a non-negative integer linear combination of the positive simple roots ${\displaystyle \alpha _{1}}$  and ${\displaystyle \alpha _{2}}$ :

${\displaystyle \mu =n_{1}\alpha _{1}+n_{2}\alpha _{2}}$ 

with ${\displaystyle n_{1}}$  and ${\displaystyle n_{2}}$  being non-negative integers. This expression gives one way to write ${\displaystyle \mu }$  as a non-negative integer combination of positive roots; other expressions can be obtained by replacing ${\displaystyle \alpha _{1}+\alpha _{2}}$  with ${\displaystyle \alpha _{3}}$  some number of times. We can do the replacement ${\displaystyle k}$  times, where ${\displaystyle 0\leq k\leq \mathrm {min} (n_{1},n_{2})}$ . Thus, if the Kostant partition function is denoted by ${\displaystyle p}$ , we obtain the formula

${\displaystyle p(n_{1}\alpha _{1}+n_{2}\alpha _{2})=1+\mathrm {min} (n_{1},n_{2})}$ .

This result is shown graphically in the image at right. If an element ${\displaystyle \mu }$  is not of the form ${\displaystyle \mu =n_{1}\alpha _{1}+n_{2}\alpha _{2}}$ , then ${\displaystyle p(\mu )=0}$ .

B₂ edit

The partition function for the other rank 2 root systems are more complicated but are known explicitly.^[1][2]

For B₂, the positive simple roots are ${\displaystyle \alpha _{1}=(1,0),\alpha _{2}=(0,1)}$ , and the positive roots are the simple roots together with ${\displaystyle \alpha _{3}=(1,1)}$  and ${\displaystyle \alpha _{4}=(2,1)}$ . The partition function can be viewed as a function of two non-negative integers ${\displaystyle n_{1}}$  and ${\displaystyle n_{2}}$ , which represent the element ${\displaystyle n_{1}\alpha _{1}+n_{2}\alpha _{2}}$ . Then the partition function ${\displaystyle P(n_{1},n_{2})}$  can be defined piecewise with the help of two auxiliary functions.

If ${\displaystyle n_{1}\leq n_{2}}$ , then ${\displaystyle P(n_{1},n_{2})=b(n_{1})}$ . If ${\displaystyle n_{2}\leq n_{1}\leq 2n_{2}}$ , then ${\displaystyle P(n_{1},n_{2})=q_{2}(n_{2})-b(2n_{2}-n_{1}-1)=b(n_{1})-q_{2}(n_{1}-n_{2}-1)}$ . If ${\displaystyle 2n_{2}\leq n_{1}}$ , then ${\displaystyle P(n_{1},n_{2})=q_{2}(n_{2})}$ . The auxiliary functions are defined for ${\displaystyle n\geq 1}$  and are given by ${\displaystyle q_{2}(n)={\frac {1}{2}}(n+1)(n+2)}$  and ${\displaystyle b(n)={\frac {1}{4}}(n+2)^{2}}$  for ${\displaystyle n}$  even, ${\displaystyle {\frac {1}{4}}(n+1)(n+3)}$  for ${\displaystyle n}$  odd.

G₂ edit

For G₂, the positive roots are ${\displaystyle (1,0),(0,1),(1,1),(2,1),(3,1)}$  and ${\displaystyle (3,2)}$ , with ${\displaystyle (1,0)}$  denoting the short simple root and ${\displaystyle (0,1)}$  denoting the long simple root.

The partition function is defined piecewise with the domain divided into five regions, with the help of two auxiliary functions.

Inverting the Weyl denominator edit

For each root ${\displaystyle \alpha }$  and each ${\displaystyle H\in {\mathfrak {h}}}$ , we can formally apply the formula for the sum of a geometric series to obtain

${\displaystyle {\frac {1}{1-e^{-\alpha (H)}}}=1+e^{-\alpha (H)}+e^{-2\alpha (H)}+\cdots }$ 

where we do not worry about convergence—that is, the equality is understood at the level of formal power series. Using Weyl's denominator formula

${\displaystyle {\sum _{w\in W}(-1)^{\ell (w)}e^{w\cdot \rho (H)}=e^{\rho (H)}\prod _{\alpha >0}(1-e^{-\alpha (H)})},}$ 

we obtain a formal expression for the reciprocal of the Weyl denominator:^[3]

${\displaystyle {\begin{aligned}{\frac {1}{\sum _{w\in W}(-1)^{\ell (w)}e^{w\cdot \rho (H)}}}&{}=e^{-\rho (H)}\prod _{\alpha >0}(1+e^{-\alpha (H)}+e^{-2\alpha (H)}+e^{-3\alpha (H)}+\cdots )\\&{}=e^{-\rho (H)}\sum _{\mu }p(\mu )e^{-\mu (H)}\end{aligned}}}$ 

Here, the first equality is by taking a product over the positive roots of the geometric series formula and the second equality is by counting all the ways a given exponential ${\displaystyle e^{\mu (H)}}$  can occur in the product. The function ${\displaystyle \ell (w)}$  is zero if the argument is a rotation and one if the argument is a reflection.

Rewriting the character formula edit

This argument shows that we can convert the Weyl character formula for the irreducible representation with highest weight ${\displaystyle \lambda }$ :

${\displaystyle \operatorname {ch} (V)={\sum _{w\in W}(-1)^{\ell (w)}e^{w\cdot (\lambda +\rho )(H)} \over \sum _{w\in W}(-1)^{\ell (w)}e^{w\cdot \rho (H)}}}$ 

from a quotient to a product:

${\displaystyle \operatorname {ch} (V)=\left(\sum _{w\in W}(-1)^{\ell (w)}e^{w\cdot (\lambda +\rho )(H)}\right)\left(e^{-\rho (H)}\sum _{\mu }p(\mu )e^{-\mu (H)}\right).}$ 

The multiplicity formula edit

Using the preceding rewriting of the character formula, it is relatively easy to write the character as a sum of exponentials. The coefficients of these exponentials are the multiplicities of the corresponding weights. We thus obtain a formula for the multiplicity of a given weight ${\displaystyle \mu }$  in the irreducible representation with highest weight ${\displaystyle \lambda }$ :^[4]

${\displaystyle \mathrm {mult} (\mu )=\sum _{w\in W}(-1)^{\ell (w)}p(w\cdot (\lambda +\rho )-(\mu +\rho ))}$ .

This result is the Kostant multiplicity formula.

The dominant term in this formula is the term ${\displaystyle w=1}$ ; the contribution of this term is ${\displaystyle p(\lambda -\mu )}$ , which is just the multiplicity of ${\displaystyle \mu }$  in the Verma module with highest weight ${\displaystyle \lambda }$ . If ${\displaystyle \lambda }$  is sufficiently far inside the fundamental Weyl chamber and ${\displaystyle \mu }$  is sufficiently close to ${\displaystyle \lambda }$ , it may happen that all other terms in the formula are zero. Specifically, unless ${\displaystyle w\cdot (\lambda +\rho )}$  is higher than ${\displaystyle \mu +\rho }$ , the value of the Kostant partition function on ${\displaystyle w\cdot (\lambda +\rho )-(\mu +\rho )}$  will be zero. Thus, although the sum is nominally over the whole Weyl group, in most cases, the number of nonzero terms is smaller than the order of the Weyl group.