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Let $v$ be a positive integer. I have a representation $\rho_v$ of $USp(4) = \{g\in M_2(\mathbb{H})\,|\,g^{T}\bar{g}\}$, where $\mathbb{H}$ is Hamiltons quaternions. The representation $\rho_v$ has young diagram consisting of two rows of length $v$.

I gather that by converting quaternions to matrices with complex entries that $\rho_v$ is really the irreducible representation of $Sp_4(\mathbb{C})$ that has highest weight $v(L_1 + L_2)$.

I now have to find $tr(\rho_v(\gamma))$ for certain matrices $\gamma$.

I understand that I have to use the Weyl character formula but how do I use this to find the trace of the rep acting on a single element? I have barely used the character formula so I apologise if this is simple.

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You are working in the compact group of unitary symplectic $4\times4$ matrices, in which every matrix is diagonalizable (because of the unitary condition); moreover there are only two independent eigenvalues, since for each eigenvalue the inverse is also an eigenvalue (because of the symplectic condition). Therefore each group element is conjugate to a diagonal matrix with diagonal entries $x,y,x^{-1},y^{-1}$ for some complex values $x,y$ of modulus $1$; it suffices to know the trace (which is invariant on conjugacy classes) on these matrices. The trace will be given by a Laurent polynomial in $x,y$ which in addition is symmetric in $x$ and $y$ and invariant under interchange of $x$ and $x^{-1}$.

The Weyl character formula gives you this Laurent polynomial as a quotient of two alternating polynomials (in the sense that they change sign under the symmetries indicated above). The result for the weight you indicated is not easy to express as a closed formula in $v$, but for instance for $v=3$ you get a Laurent polynomial with terms $x^3y^3+x^3y+x^2y^2+x^2+2xy+2$, to be completed with other terms like $x^{-3}y^3$ , $xy^3$ and $2x^{-1}y^{-1}$ to establish the required symmetry (there are $4+8+4+4+4+1=25$ terms in all; I'm too lazy to write them all down). That this is indeed the right formula can be checked by multiplying the $25$-term polynomial by the Weyl denominator $x^2y-xy^2+x^{-1}y^2-x^{-2}y+x^{-2}y^{-1}-x^{-1}y^{-2}+xy^{-2}-x^2y^{-1}$, which should give $$ x^5y^4-x^4y^5+x^{-4}y^5-x^{-5}y^4+x^{-5}y^{-4}-x^{-4}y^{-5}+x^4y^{-5}-x^5y^{-4}. $$ Thus that $25$-term polynomial gives you the trace of the action of any matrix in the group as function of its eigenvalues.

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Ok so to begin with you end up with 8 terms on both the numerator and the denominator of the WCF, but after cancellation you end up being left with a polynomial with 25 terms? Is the formula in my latest post correct then? (math.stackexchange.com/q/169084/25381) –  fretty Jul 11 '12 at 11:01
    
Is there even a need to cancel the denominator out or can I just sub in the values of $x$ and $y$ in the beginning? –  fretty Jul 11 '12 at 11:03
    
It is not really cancellation that is going on, it is polynomial division, which can create lots of terms (think of dividing $x^n-y^n$ by $x-y$). The division has to be exact (without remainder) by a subtle symmetry argument. But actually performing a division (not obvious with multiple indeterminates) is a rather difficult way to obtain the quotient –  Marc van Leeuwen Jul 11 '12 at 11:27
    
As for your question about "canceling" the denominator before substitution: this is not necessary for concrete values of $x,y$ for which the denominator is nonzero. But I cannot think of much use of actually performing this division for concrete numerical values of $x,y$ (unless you want to plot the character as a function); it is the formal quotient that seems to convey the information in a more useful form. –  Marc van Leeuwen Jul 11 '12 at 11:33
    
One more remark; the term "cancelling" is OK if you use it in a multiplicative sense. I was thinking of cancelling in an additive sense, which of course is nonsense here. –  Marc van Leeuwen Jul 11 '12 at 11:35

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