I have known of the following equation for characters of a $SO(2n)$ representation with highest weights $(h_1,...,h_n)$ and for $(t_1,t_2,..,t_n,t_1^{-1},t_2^{-1},..,t_n^{-1})$ being the eigenvalues of the matrix under consideration,

$$\chi (h_i,t_i) = \frac{det (sinh [ t_i(h_j + n-j)]) + det (cosh [ t_i(h_j + n-j)]) }{det (sinh [t_i(n-j)]) } $$

But when actually trying to put this into use I am faced with two confusions,

  • Given the eigenvalues how does one decide which eigenvalue is to be called $t_i$ and which one as $t_i^{-1}$?

  • I guess that (any?) elements in $SO(2n)$ can be given in terms of rotation angles in each plane. ($^{2n}C_2$ of them) right? So knowing the rotation angles how does one use the above equation for the character?

For a specific basic example consider this - Take the group element which rotates by an angle $\alpha$ in the $1-2$ plane. This as a $SO(2n)$ matrix is I guess a block diagonal matrix such that it has a $(2n-2)\times(2n-2)$ identity matrix in the lower block diagonal and the top $2\times 2$ diagonal block is $[\{cos \alpha, sin \alpha\},\{-sin \alpha, cos \alpha\}]$. Now from this way of picturing the rotation matrix how do I compute the character using the above formula?

  • $\begingroup$ The character formula you start with is apparently a version of Weyl's Character Formula, using physics notation. Keep in mind that such a quotient is usually not helpful computationally, so other methods like those of Freudenthal and Kostant come into play. Weyl's formula is mainly of theoretical value, with dimensions as the main practical byproduct after some manipulation. Compuation is nontrivial in your case, but perhaps more easily organized using the Lie algebra instead of the group. $\endgroup$ – Jim Humphreys Sep 2 '13 at 22:25
  • $\begingroup$ @JimHumphreys Can you kindly elaborate on what you said? If one knows the $t_i$ and the $h_i$ then this is an easy to use formula..right? But the problem is to be able to get the $t_i$ from the rotation picture. Can you kindly help with that? If for at least the simplest case of one single rotation as I have described... $\endgroup$ – user6818 Sep 2 '13 at 22:51
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    $\begingroup$ You should mention that this question is crossposted from math.SE. $\endgroup$ – Zev Chonoles Sep 2 '13 at 23:18
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    $\begingroup$ There is no standard way to say what is $t_i$ and what is $t_i^{-1}$ as switching which of a pair you call $t_i$ simply switches the representation to that of its dual. Also, the fact that you have the same number of weights $h_i$ as pairs $t_i$, it seems that you are dealing with reducible representations and your $t_i$ are implicitly assumed to be the highest weights of the various irreducible summands. $\endgroup$ – ARupinski Sep 2 '13 at 23:45
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    $\begingroup$ As far as which eigenvalue should be called $t_i$ and which should be $t_i^{-1}$, that doesn't matter too much as swapping the choice amounts to considering the dual of the representation you are looking at. But in the case of the $2n$-dimensional basic representation you mention directly, this representation is self-dual so you don't even have to worry about ambiguities coming from outer automorphisms, plugging in either $\lambda$ or its inverse as your highest weight will give the same character value due to the symmetry of this representation. $\endgroup$ – ARupinski Sep 2 '13 at 23:54

This is an attempt to clarify my comments; hopefully it will of use to the OP in determining how to use their formula. Like @JimHuphreys, I am not familiar with the particular version of the Weyl Character Formula (WCF) used by the OP (and in particular it seems a bit odd that it doesn't seem to involve the Weyl Group of $SO(2n)$ like the usual WCF does. If anyone else sees how the sinh/cosh take care of that, do enlighten me; perhaps it is related to the fact that we are working with $SO(2n)$ instead of its universal cover). Nevertheless, it seems most of your problem is stemming from an incorrect interpretation of the inputs as I shall try to outline below.

Anyhow, the quick background: $SO(2n)$ comes from the $D_n$ family of Lie groups; therefore it has rank $n$ and a highest weight of a representation is determined by a non-negative $n$-tuple of integers, say $(a_1,\ldots,a_n)$. The coordinates in this expression are the coordinates relative to a basis of fundamental weights for the group; these in turn are highest weights of the corresponding fundamental representations, so the weight can also be written as $\sum_{i=1}^n a_i\omega_i$ where the $\omega_i$ are these fundamental weights.

Now when I think of weights, I think of them in terms of logarithms of eigenvalues: when one applies the WCF one starts with a highest weight $\sum_{i=1}^n h_i\omega_i$ (where the $h_i$ should be the same as the ones in your formula) and applies each element of the Weyl group $W(G)$ to this weight according to a formula which looks a bit weird at first, but turns out to be fairly easy to program into a computer (at least when $W(G)$ is not too big). Each weight resulting from this action on the weight space is of the form $\sum_{i=1}^n b_i\omega_i$ where the $b_i$ are integral but no longer restricted to being all nonnegative. Each such weight then corresponds to a term of the form $\pm e^{b_1t_1+b_2t_2+\ldots+b_nt_n}$ which appears in the numerator of the WCF; the denominator of the formula has similar terms formed by applying the $W(G)$-action to the highest weight with all coordinates equal to 0.

To evaluate the character of the irreducible representation with highest weight $\sum_{i=1}^n h_i\omega_i$ at a particular element $g\in G$, put simply one must look at $g$ and determine what the eigenvalues of $g$ are in some fixed representation (usually a defining representation such as your action of $SO(2n)$ on $\mathbb{R}^{2n}$). The logarithms of these eigenvalues can then be expressed in terms involving the weights of the given representation and a certain set of coordinates; these coordinates are the $t_i$ alluded to above.

With that, lets address some of your specific questions in terms of the above background. Since your formula involves hyperbolic sines and cosines, your $t_i$ almost certainly should be the coordinates $t_i$ mentioned before and not the eigenvalues themselves.

Now to the specific element you mention, one is given the $2n$-dimensional representation for this element:

\[ g = \begin{pmatrix}\cos(\alpha)&\sin(\alpha)&0^T_{2n-2}\\-\sin(\alpha)&\cos(\alpha)&0^T_{2n-2}\\0_{2n-2}&0_{2n-2}&Id_{2n-2}\end{pmatrix} \]

This element is already in a form that gives the eigenvalues of $g$ within a certain representation; its eigenvalues in this representation are $\{\cos(\alpha)\pm i\sin(\alpha),1,\ldots,1\}$. These have logarithms $\{\pm i\alpha,0,\ldots,0\}$. Now one wants to write these logarithms in terms of the weights of this $2n$-dimensional representation being used to express $g$. The full set of weights can be obtained via the method of lowering weights starting at the highest weight, doing so one has the following set of weights for this particular representation:

\[ \pm(\omega_1), \pm(\omega_2-\omega_1), \ldots, \pm(\omega_{n-2}-\omega_{n-3}),\pm(\omega_{n-1}+\omega_{n}-\omega_{n-2}),\pm(\omega_{n-1}-\omega_{n}) \]

To determine our specific coordinates $t_j$ (which ought to be the $t_j$ you use in your formula instead of the eigenvalues) we have alot of choices to make. I think of $\omega_j$ as the coordinate vector with a 1 in its $j^{th}$ position and think of the logarithms of $g$ as the $n$-long vector $v(g) = (i\alpha,0\ldots,0)$ where in general we have arbitrarily chosen one logarithm from each $\pm$ pair (in this way, $v(g)$ now corresponds to the highest weight of this representation evaluated at $g$). My $s_j$ then correspond to the coordinates needed to express $v(g)$ in terms of a weight basis where one weight is chosen from each $\pm$ pair in the list above; i'll go ahead and choose all $+$ signs. So we have:

\[ v(g) = t_1(1,0,\ldots,0) + t_2(-1,1,0,\ldots,0)+\ldots+t_{n-2}(0,\ldots,0,-1,1,0,0)+t_{n-1}(0,\ldots,0,-1,1,1)+t_{n}(0,\ldots,0,1,-1) \]

So the "coordinates" $t_i$ one should use for this particular $g$ are simply $(t_1,\ldots,t_n) = (i\alpha,0,\ldots,0)$ and these coordinates are what should be used in your formula.

Note that we had alot of choices here, namely about which weight and which logarithm we should choose from each pair and even what order to choose the weights and logarithms in, but in the end each of these possible choices correspond to actions of different elements of $W(G)$ on the character. However, one useful fact about characters is that they are $W(G)$-invariant which is why the particular choices made will not affect your your calculation.

  • $\begingroup$ @ARupanski Thanks for your efforts! Let me study your reply in details and get back to you! It will take me some time! :) $\endgroup$ – user6818 Sep 7 '13 at 2:59
  • $\begingroup$ @ARupanski In the definition of the character formula, $t_i$ are the eigen values of the matrix in any given representation. Why/How are you interpreting that to be the coordinates of the logarithm of the eigenvalues in a basis of the weight? - I am not getting this - Are you saying that the formula I quoted is given in terms of the corresponding Lie algebra element of the group element whose caharacter one is after? Moreover $g$ is a $2n\times 2n$ matrix and hence the eigenvalue vector is $2n$ dimensional, but you have only $n$ weights - right? $\endgroup$ – user6818 Sep 16 '13 at 1:17
  • $\begingroup$ @ARupanski The weights of a representation of a Lie group can only span the dual of the Cartan of the Lie algebra. (...the weight vectors of a representation diagonalize the Cartan of the Lie algebra in the given representation...) $\endgroup$ – user6818 Sep 16 '13 at 1:17
  • $\begingroup$ @ARupanski The weights lie in the dual of the Lie algebra and they span only the dual of the Cartan of the Lie algebra. Why should the logarithm of the eigenvalues of a group element lie in the dual of the Lie algebra? $\endgroup$ – user6818 Sep 16 '13 at 1:19
  • $\begingroup$ Perhaps my explanation above was less clear than I intended; at any rate, in the above the weights behave as coefficients of the logarithms; once one knows the fundamental logarithms $t_i$ of an element $g$, then for any representation $\rho$ one can calculate the logarithms of all eigenvalues of $rho(g)$ by applying the various weights of $\rho$ to these fundamental logarithms. $\endgroup$ – ARupinski Sep 16 '13 at 22:35

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