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I want to show that SU(n) is a subgroup of SO(2n). I can show the first step, that is, for any $U=U_R + iU_I\in SU(n)$ with $U_R,U_I\in\mathbb{R}^{n\times n}$, we identify $U$ with the real $2n\times 2n$ matrix $ X = \begin{pmatrix} U_R & - U_I\\ U_I & U_R \end{pmatrix} $ and I can show that $X$ is orthogonal, i.e., $XX^T= Id_{2n\times 2n}$. But I do not know how to show that $\det X = 1$. Can someone show we how?

Thank you in advance!

Best,

Stefan

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    $\begingroup$ I think that we always have $\det X=|\det U|^2$. For all $U\in GL_n(\Bbb{C})$. $\endgroup$ Jul 18, 2022 at 10:10
  • $\begingroup$ See here. Does that answer your question? $\endgroup$ Jul 18, 2022 at 10:11
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    $\begingroup$ @JyrkiLahtonen more generally, if $L/K$ is a finite Galois extension, then by restriction of scalars any $U \in M_{n \times n}$ gives rise to a square matrix of dimension $n[L:K]$ with determinant $N_{L/K}(\det U)$ $\endgroup$ Jul 18, 2022 at 10:39

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Here is one nice approach. Denote $$ P = \pmatrix{I & -iI\\-iI & I}, $$ note that $P^{-1} = \frac 12 P^*$ (where $P^*$ denotes the conjugate-transpose of $P$). Now, note that $\det(X) = \det(P^{-1}XP)$, and verify (via block-matrix multiplication) that $$ P^{-1}XP = \frac 12 P^*XP = \pmatrix{U_R + iU_I & 0\\0 & U_R - iU_I}. $$ Thus, we have $$ \det(X) = \det(U) \det(\bar U) = \det(U)\overline{\det(U)} = |\det(U)|^2 = 1. $$ (In the above, $\bar \cdot$ denotes the complex conjugate, so $\bar U = U_R - iU_I$).

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