The Lie algebra associated to the group $SO(n)$ of real-valued special orthogonal matrices, is given by the set $\mathfrak{so}(n)$ of anti-symmetric real-valued matrices equipped with the commutator. Any $X\in\mathfrak{so}(n)$ has purely imaginary eigenvalues and is unitarily diagonalisable; thus, in an eigenbasis, we may write $X=\text{diag}(t_{1},\ldots,t_{n})$. It follows that $\exp(X)\in U(n)$, instead of $SO(n)$ as expected.

Question: What's going on here?

  • Guess: I'm working with some complex Lie subalgebra instead of a real Lie algebra, i.e. $\exp: \mathfrak{so}(n,\mathbb{C})\subset\mathfrak{u}(n) \to U(n)$ instead of $\exp: \mathfrak{so}(n,\mathbb{R}) \to SO(n)$. I've used $\mathfrak{u}(n)$ to denote the Lie algebra of skew-Hermitian $n\times n$ complex matrices. The real question now is what is meant by $\mathfrak{so}(n,\mathbb{C})$ -- these were meant to be real-antisymmetric matrices, but the 'realness' now seems to mean 'the subset of matrices in $\mathfrak{u}(n)$ such that there exists a choice of basis in which they are real'. Yes, I'm confused! Comments would be welcome.

Question: What's the inverse image (under $\exp:\mathfrak{u}(n)\to U(n)$) of the group $\Omega\leq SU(n)$ of special unitary matrices with real-valued entries? Note: I feel like I'd want to denote $\Omega$ by $SO(n,\mathbb{C})$, but understand that some people might just interpret this as $SU(n)$?

  • 4
    $\begingroup$ $SO(n) \subset U(n)$ indeed it is in $SU(n)$, so what is the problem? You can unitarily diagonalise $X$, sure but that isn't preserving $\mathfrak{so}(n)$ $\endgroup$
    – Callum
    Commented Oct 28, 2023 at 13:17
  • 3
    $\begingroup$ Note $\mathfrak{so}(n, \mathbb{C})$ is not contained in $\mathfrak{su}(n)$ as it comprises anti-symmetric complex matrices which only overlaps $\mathfrak{su}(n)$ in anti-symmetric real matrices (i.e. $\mathfrak{so}(n, \mathbb{R})$) $\endgroup$
    – Callum
    Commented Oct 28, 2023 at 13:19

1 Answer 1

  1. There is no contradiction here. As Callum says in the comments we have $SO(n) \subset U(n)$, so it is true that the exponential lands in $U(n)$, because it lands in $SO(n)$. The condition that $X \in \mathfrak{so}(n)$ is not just that it is unitarily diagonalizable with purely imaginary eigenvalues (that's the condition that $X \in \mathfrak{u}(n)$); in order for $X$ to be real-valued it also needs to have the property that its eigenvalues are closed under complex conjugation, and that if $v$ is an eigenvector with eigenvalue $\lambda$ then $\overline{v}$ is an eigenvector with eigenvalue $\overline{\lambda} = - \lambda$.

  2. $\mathfrak{so}(n, \mathbb{C})$ refers to complex skew-symmetric (not skew-adjoint / skew-Hermitian) matrices; this is the complexification of $\mathfrak{so}(n)$, and is not contained in $\mathfrak{u}(n)$ (which is skew-adjoint / skew-Hermitian matrices).

  3. A real unitary matrix is an orthogonal matrix, so $\Omega$ is just $SO(n)$ and its preimage under the exponential map is $\mathfrak{so}(n)$ (edit: together with some annoying extra stuff that I assume you don't care about, e.g. it also contains $2 \pi i I$ since this exponentiates to the identity. I'm assuming your real question is about the Lie algebra of $\Omega$.) $SO(n, \mathbb{C})$ is the group of complex matrices with determinant $1$ satisfying $M^T M = I$ which is a different group from either $SO(n)$ or $SU(n)$ (e.g. it is not compact).


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