I posted a question a short while ago on this but got no response. I have worked on this more and so now have a more specific question.

To start with we work with the $\mathbb{Q}$ version of Hamilton's quaternions and consider the maximal order $\mathfrak{O} = \mathbb{Z} + \mathbb{Z}i + \mathbb{Z}j + \mathbb{Z}\frac{1+i+j+k}{2}$.

Now consider the group:

$\Gamma_1 = \left\{A\in M_2(\mathfrak{O}^{\times})\,|\,A = \left( \begin{array}{ccc} a & 0 \\ 0 & b \end{array} \right) \text{ or } A = \left( \begin{array}{ccc} 0 & a \\ b & 0 \end{array} \right)\right\}$.

This is a subgroup of the Lie group Sp$(2)$ and is very simple to calculate.

I wish to use the Weyl character formula to evaluate $\sum_{\gamma\in\Gamma_1} \text{tr}(\rho_v(\gamma))$, where $\rho_v$ is the irreducible representation of Sp$(2)$ of highest weight $v(L_1+L_2)$ (using standard notation for $L_1,L_2$).

Now my plan (in order to be able to use the Weyl character formula to get actual traces and not just formal characters) is to shift focus from Sp$(2)$ into Sp$(4,\mathbb{C})\cap$ U$(4)$. There is an isomorphism between the two that behaves as follows on elements of $\Gamma_1$:

$\left(\begin{array}{ccc} a+bi+cj+dk & 0 \\ 0 & \alpha+\beta i+\gamma j + \delta k \end{array}\right)\longmapsto\left( \begin{array}{ccc} a+bi & 0 & c+di & 0\\ 0 & \alpha+\beta i & 0 & \gamma + \delta i \\ -c+di & 0 & a-bi & 0\\ 0 & -\gamma+\delta i & 0 & \alpha - \beta i \end{array} \right)$

$\left(\begin{array}{ccc} 0 & a+bi+cj+dk \\ \alpha+\beta i+\gamma j + \delta k & 0 \end{array}\right)\longmapsto\left( \begin{array}{ccc} 0 & a+bi & 0 & c+di \\ \alpha+\beta i & 0 & \gamma + \delta i & 0\\ 0 & -c+di & 0 & a-bi \\ -\gamma+\delta i & 0 & \alpha - \beta i & 0 \end{array} \right)$

My first question is whether the character of the rep $\rho_v$ of Sp$(2)$ is the same as the character of the similarly defined "$\rho_v$" on the Lie group Sp$(4,\mathbb{C})$ but restricted to Sp$(4,\mathbb{C}) \cap$ U$(4)$? The restriction is important here in order to get a rep on a compact Lie group (hence allowing me to guarantee conjugation into maximal torus to always be possible).

If so then I know that the maximal torus in Sp$(4,\mathbb{C}) \cap$ U$(4)$ consists of matrices of the form:

$\left( \begin{array}{ccc} x & 0 & 0 & 0\\ 0 & y & 0 & 0 \\ 0 & 0 & x^{-1} & 0\\ 0 & 0 & 0 & y^{-1} \end{array} \right)$

where $x,y$ are complex numbers of modulus $1$.

So, when I use the Weyl character formula for the Lie algebra $\mathfrak{sp}_4(\mathbb{C})$ I get the following large expression for the character:

$\frac{e^{(v+2)L_1 + (v+1)L_2} - e^{(v+2)L_1 - (v+1)L_2} - e^{-(v+2)L_1 + (v+1)L_2} + e^{-(v+2)L_1 - (v+1)L_2} - e^{(v+1)L_1 + (v+2)L_2} + e^{-(v+1)L_1 + (v+2)L_2} + e^{(v+1)L_1 - (v+2)L_2} - e^{-(v+1)L_1 - (v+2)L_2}}{e^{2L_1 + L_2}(1-e^{-2L_1})(1-e^{-2L_2})(1-e^{L_2-L_1})(1-e^{-L_1-L_2})}$

The rest of my question is as follows. First, is this correct? Also may I shift this into this formula for the character on the Lie group Sp$(4,\mathbb{C}) \cap$ U$(4)$ (where $T$ is an element of the maximal torus as above): $$\chi(T) = \frac{x^{2v+4}y^{2v+3}-x^{2v+4}y-y^{2v+3}+y-x^{2v+3}y^{2v+4}+xy^{2v+4}+x^{2v+3}-x}{x^vy^v(x^2-1)(y^2-1)(xy-1)(x-y)}$$

I got this by using the Lie algebra version above and rewriting multiplicatively, then rearranging. If I am correct so far then can this be simplified in any way?

  • $\begingroup$ Your Weyl character formula has a $2$ in a factor of the denominaotr that should be a $1$. Your formula for $\chi(T)$ appears weird to me; there shouldn't be any occurence of $v$ in the denominator, and the numerator does not have the required anti-symmetry. Can you explain how you obtained it? $\endgroup$ – Marc van Leeuwen Jul 11 '12 at 11:08
  • $\begingroup$ Well I am new to this so I have probably not used it correctly. Basically I rearranged it to get rid of all negative exponents. $\endgroup$ – fretty Jul 11 '12 at 11:22
  • $\begingroup$ I think you don't entire grasp the nature of the formula. Negative exponents are not a problem, the formula is symmtric w.r.t. $x\leftrightarrow x^{-1}$ and $y\leftrightarrow y^{-1}$ (and also $x\leftrightarrow y$). But you cannot expect to simplify the formula as long as $v$ has not been specialised. For every $v$ the division is exact as a Laurent polynomial, so you can "simplify", but the number of resulting terms grows with $v$. There is no "closed formula" in terms of $v$ that is simpler than the Weyl character formula itself. However combinatorial methods to find the quotient exist. $\endgroup$ – Marc van Leeuwen Jul 11 '12 at 11:49
  • $\begingroup$ I think with the version I write above there is a difference equation defining the exact division of the numerator by the denominator (ignoring the $x^v y^v$ bit). So I can generate these polynomials. Anyway, can I not just input the required values for $x,y$ into the "uncancelled" version to get the traces? $\endgroup$ – fretty Jul 11 '12 at 11:54
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    $\begingroup$ As I said at the other question, you can indeed substitute before doing the division, provided this does not happen to give you $\frac00$. $\endgroup$ – Marc van Leeuwen Jul 11 '12 at 12:10

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