TITLE: Matrix equation $XAXBXC=I$
QUESTION [15 upvotes]: Let $A,B,C$ be unitary matrices. Does there always exist a unitary matrix $X$ such that $$(XA)(XB)(XC)=I,$$ where $I$ is the identity matrix? The quadratic equation $(XA)(XB)=I$ has the solution $A^*(AB^*)^{1/2}$, and I am hoping that the cubic and higher dimensional versions are always solvable.

REPLY [30 votes]: Here is an argument showing that the answer is 'yes'.  I'll let you check the details and that this result generalizes to all higher degrees.  
Consider the map $f_{ABC}:\mathrm{U}(n)\to\mathrm{U}(n)$ defined by
$$
f_{ABC}(X) = XAXBXC.
$$
Since the image of this map is compact, if this map were not onto, it would have to have topological degree equal to zero.
Next, since $\mathrm{U}(n)$ is connected, the map $f_{ABC}:\mathrm{U}(n)\to\mathrm{U}(n)$ defined by
$$
f_{ABC}(X) = XAXBXC
$$
is homotopic to the map $f_{III}$, and hence has the same degree as $f_{III}$. Thus, it suffices to show that the map $f_{III}$ has nonzero topological degree to show that $f_{ABC}$ is surjective.
I claim that the map $f_{III}$ has topological degree $3^n$.  
To prove this, it suffices to compute the local degrees around the pre-images of a regular value.  Let $Y = \mathrm{diag}(e^{i\theta_1},\ldots,e^{i\theta_n})$, where $0<\theta_1<\theta_2<\cdots<\theta_n<\pi$.  Then $Y$ has distinct eigenvalues.  Hence, any solution $X$ to $X^3 = Y$ has distinct eigenvalues and has the same eigendirections as $Y$. Thus, $X$, too, must be diagonal and must be of the form
$$
X = \mathrm{diag}(e^{i\tau_1},\ldots,e^{i\tau_n})
$$
where $3\tau_k \equiv \theta_k\ \mathrm{mod}\ 2\pi$ for $k=1,\ldots, n$.  Thus, there are $3^n$ solutions $X$ to $X^3=Y$.  
I claim that $Y$ is a regular value of the map $f_3:\mathrm{U}(n)\to \mathrm{U}(n)$ defined by $f_3(X)=X^3$ and that $f_3$ is orientation preserving at each solution $X$ of $X^3=Y$.  This follows from the following computation:  Consider the pullback under $f_3$ of the canonical left-invariant form $g^{-1}\mathrm{d} g$.
$$
f_3^*(g^{-1}\mathrm{d}g) = g^{-3}\mathrm{d}(g^3) 
= (I + \mathrm{Ad}(g^{-1})+\mathrm{Ad}(g^{-2}))(g^{-1}\mathrm{d}g)
$$
When one computes the determinant of $\bigl(I + \mathrm{Ad}(X^{-1})+\mathrm{Ad}(X^{-2})\bigr):{\frak{u}}(n)\to {\frak{u}}(n)$ for each solution $X$ of $X^3=Y$, one finds that, because $Y$ has distinct eigenvalues, this determinant is a positive number.  Thus, $Y$ is a regular value of $f_3$, and each of the $3^n$ solutions $X$ to $X^3=Y$ contributes a $+1$ to the topological degree in the usual degree formula.
(By a similar argument, the map $f_k:\mathrm{U}(n)\to\mathrm{U}(n)$ defined by $f(X) = X^k$ has topological degree $k^n$, which is nonzero, so it is necessarily surjective. This answers the higher degree cases as well.)