Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Thank for your interest for my question and thank you very much if you can answer me.

In his book p.187, Daniel Bump says that a realization (a representation) of $SO(2n)$ is given by the unitary matrice $g$ which verify $gJ^t g=J$ i.e. $J\,\overline{g}J=g$.

I understand why $$\begin{eqnarray} \left ( \begin{array}{cccccc} e^{-i y}&&\newline &e^{-i b}&\newline &&e^{2 i y}\newline &&&e^{-2 i y}&&&&& \newline &&&&e^{ib}&&&& \newline &&&&&e^{iy}&&& \newline \end{array} \right ) \end{eqnarray}$$ with eigenvalues $1,e^{3 i y/2}, e^{-3 i y/2},1,e^{3 i y/2}, e^{-3 i y/2}$ is similar to a matrix belonging to $SO(2n)$ but I don't understand why $$\begin{eqnarray} \left ( \begin{array}{cccccc} &&&e^{-i y}&&\newline &&&&e^{-i b}&\newline &&&&&e^{2 i y}\newline e^{-2 i y}&&&&& \newline &e^{ib}&&&& \newline &&e^{iy}&&& \newline \end{array} \right ) \end{eqnarray}$$ is similar to a matrix from $SO(2n)$ because its eigenvalues are $1,e^{3 i y/2}, e^{-3 i y/2},-1,-e^{3 i y/2}, -e^{-3 i y/2}$.

share|cite|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.