# Counting unitary square roots

Let $$g$$ be an $$n \times n$$ unitary matrix. How many unitary matrices $$U$$ are there such that $$U^2=g$$ My first guess was that there are $$2^n$$ such unitary matrices since each of the $$n$$ eigenvalues has 2 square roots. However that already fails for the identity matrix $$I=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ since there are infinitely many unitary matrices that square to $$I$$. For example all unitary matrices of the form $$\begin{bmatrix} 0 & e^{i \theta} \\ e^{-i \theta} & 0 \end{bmatrix}$$ If there are infinitely many $$\sqrt{g}$$ then what can we say about the geometry of the real algebraic variety $$U^2=g$$? What is the dimension? Number of connected components? For connected components that aren't just points what kind of manifold are they?

If these questions are too difficult with a generic unitary $$g$$ what about just describing the structure of the variety $$U^2=I$$?

It is simpler to view $$g$$ as a unitary endomorphism of some hermitian space. Since $$g$$ is unitary, it is diagonalisable and the eigenvalues have modulus $$1$$. Eigenspaces of $$g$$ are stable under every endomorphism which commutes with $$g$$. Therefore, every endomorphism which commutes with $$g$$ (including all square roots of $$g$$) is completely determined by the endomorphisms it induces on the eigenspaces of $$g$$.
In particular, the choice of a square root of $$g$$ is given by the choice of a square root on each eigenspace of $$g$$, equivalently by the choice of a symmetry on each eigenspace of $$g$$ (which has to be multiplied by a fixed square root of the the corresponding eigenvalue). Calling $$m_1,\ldots,m_r$$ the multiplicities of the eigenvalues of $$g$$, we get a variety isometric to $$\Sigma_{m_1} \times \ldots \times \Sigma_{m_r}$$, where $$\Sigma_m$$ denotes the set of all $$\mathbb{C}$$-linear symmetries on $$\mathbb{C}^m$$.
The global number of connected components is $$(m_1+1)\cdots(m_r+1)$$. Indeed, each $$\Sigma_m$$ has $$m+1$$ connected components given by the dimension of the invariant subspace which may be any integer between $$0$$ and $$m$$, and $$\Sigma_m$$ can be viewed as a disjoint union of Grassmanian manifolds.
• Just checking for understanding: so for a $2 \times 2$ unitary with distinct eigenvalues there are exactly 4 unitary square roots. And for a $2 \times 2$ unitary with equal eigenvalues (i.e. a multiple of the identity) there are infinitely many square roots. Indeed the space of all unitary square roots of the $2 \times 2$ identity $I$ is a 2 dimensional variety of the form $V diag(1,-1) V^{-1}$ for $V \in U_2$ Commented Feb 15, 2023 at 0:22
• @IanGershonTeixeira There are two more square roots of $I_2$, namely $I_2$ and $-I_2$. Commented Feb 15, 2023 at 7:31