# Lie algebra of real unitary matrices

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 $$SU(n)$$?

• $SO(n) \subset U(n)$ indeed it is in $SU(n)$, so what is the problem? You can unitarily diagonalise $X$, sure but that isn't preserving $\mathfrak{so}(n)$ Commented Oct 28, 2023 at 13:17
• 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. $\mathfrak{so}(n, \mathbb{R})$) Commented Oct 28, 2023 at 13:19

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).