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In a compact connected Lie group $G$, each element is conjugate to an element of a maximal torus $T$. For a classical group, one can pick a basis of the tautological representation such that $T$ is represented by diagonal matrices. The conjugacy classes in $G$ are in bijective correspondence with the orbits of the Weyl group $W$ on $T$. Thus, say, for $\operatorname{SU}(n)$ the conjugacy class of a unitary matrix $g$ is determined by its (unordered) eigenvalues, and the same holds for $\operatorname{SO}(2n+1)$.

For $\operatorname{SO}(2n)$, the torus $T$ can be represented by matrices $\operatorname{diag}(\gamma_1,\dotsc,\gamma_n,\gamma_n^{-1},\ldots,\gamma_1^{-1})$, and $W$ acts by permutations on $\gamma_1,\dotsc,\gamma_n$ and an even number of changes $\gamma_i\leftrightarrow\gamma_i^{-1}$ (contrary to $\operatorname{SO}(2n+1)$, where an arbitrary number of switches can be done, by virtue of an extra eigenvalue $1$). Thus the eigenvalues alone determine a pair of conjugacy classes which differ by one switch.

Q: How, given a matrix $g\in\operatorname{SO}(2n)$, to determine the corresponding element of the torus $T$?

It can be done by diagonalizing $g$ by a matrix $C$ and checking whether $\det(C)=1$ or $-1$, but I was hoping for a more efficient solution (calculating the eigenvalues of $g$ is faster then finding $C$). Are there any other invariants which can, given the eigenvalues $\gamma_1,\dotsc,\gamma_n,\gamma_n^{-1},\dotsc,\gamma_1^{-1}$ of $g$, tell whether $g$ is conjugate to $\operatorname{diag}(\gamma_1,\dotsc,\gamma_n,\gamma_n^{-1},\dotsc,\gamma_1^{-1})$ or to $\operatorname{diag}(\gamma_1^{-1},\gamma_2,\dotsc,\gamma_n,\gamma_n^{-1},\dotsc,\gamma_2^{-1},\gamma_1)$?

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  • $\begingroup$ Although, looking at a similar problem for the alternating groups, I am a bit pessimistic about this one. But who knows. $\endgroup$ Jul 31, 2022 at 14:13
  • $\begingroup$ Now I understand your question better, I think the answer to your last paragraph is "no". The information is embedded in the ordering of eigenvalues which comes from the $SO(2n)$ action and that is precisely the information that you are after. If you have just the unordered eigenvalues, the crucial information is not there. $\endgroup$ Aug 4, 2022 at 6:44
  • $\begingroup$ I could imagine that this information is encoded somehow in the $k\times k$ minors like the signature is encoded in Sylvester's criterion, but that doesn't seem more efficient than calculating eigenvectors. $\endgroup$ Aug 4, 2022 at 6:47

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One method, I don't know whether it is the most efficient, is to compute the following polynomial quantity $Q(g) = \mathrm{Pf}(g-g^t)$ for $g\in\mathrm{SO}(2n)$, where $g^t$ means the transpose of $g$ and $\mathrm{Pf}$, known as the Pfaffian, is a polynomial of degree $n$ on the skew-symmetric $2n$-by-$2n$ matrices that satisfies $\det(S) = \mathrm{Pf}(S)^2$ for skew-symmetric $2n$-by-$2n$ matrices.

This polynomial satisfies $Q(CgC^{-1}) = \det(C)\,Q(g)$ for $C$ in $\mathrm{O}(2n)$, so it takes different values on two elements of $\mathrm{SO}(2n)$ that are conjugate in $\mathrm{O}(2n)$ but not in $\mathrm{SO}(2n)$. (Also, if $Q(g)=0$, then the conjugacy class of $g$ in $\mathrm{SO}(2n)$ is the same as its conjugacy class in $\mathrm{O}(2n)$.)

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