= Solution
Regard $\operatorname{Mat}_n(\mathbb C)$ as a real vector space of dimension $2n^2$, and let $\operatorname{Herm}_n(\mathbb C)$ be the real vector space of Hermitian matrices, of dimension $n^2$. The smooth Gram map
$$
F(A)=A^*A
$$
has $F^{-1}(I)=U(n)$ and differential $DF_A(K)=A^*K+K^*A$. At a unitary $A$, any Hermitian matrix $H$ is obtained by taking $K=AH/2$. Indeed $A^*K=H/2$ and $K^*A=H/2$. Hence $I$ is a <regular value>, and the <regular level set theorem> proves the <unitary group as a regular level set> result:
$$
\boxed{U(n)\text{ is a smooth manifold of real dimension }n^2.}
$$
Similarly let $G(A)=A^TA$ from $\operatorname{Mat}_n(\mathbb R)$ to the symmetric real matrices, whose dimension is $n(n+1)/2$. At an orthogonal $A$ its differential is $DG_A(K)=A^TK+K^TA$, again onto by choosing $K=AH/2$ for a symmetric $H$. Therefore the <orthogonal group as a regular level set> has
$$
\boxed{\dim_{\mathbb R}O(n)=n^2-\frac{n(n+1)}2=\frac{n(n-1)}2.}
$$
Matrix multiplication is smooth on both groups, and inversion is conjugate transpose or transpose, respectively, so these manifold structures also make them <Lie groups>. Both groups are closed and bounded in their finite-dimensional ambient matrix spaces, hence compact.
Realification identifies a unitary complex matrix with an orthogonal transformation of $\mathbb R^{2n}$. This is a smooth injective group homomorphism, with closed image, so $U(n)$ is the stated Lie subgroup of $O(2n)$. The assumed maximal-rank quotient map has fibers diffeomorphic to $U(n)$. Consequently the compact <homogeneous space>, also known as the <space of orthogonal complex structures>, has dimension
$$
d=\dim O(2n)-\dim U(n)=n(2n-1)-n^2=n(n-1).
$$
The original PDF's sphere is $S^{n^2}$, with a square on the exponent; its dimension exceeds $d$ by $n$.
For completeness, <maps to a sphere above the dimension of a compact smooth manifold are null-homotopic>. Let $X$ be a compact <smooth manifold> of dimension $d<q$, with $q\geq1$, and let $f:X\to S^q\subset\mathbb R^{q+1}$ be continuous. Choose a finite open cover on which $f$ is uniformly close to its value at a chosen point of that open set, and take a smooth <partition of unity> subordinate to the cover. The weighted sum of those chosen sphere values is a smooth map $g_0:X\to\mathbb R^{q+1}$ uniformly within $\epsilon<1$ of $f$. It never vanishes. Normalize it to $g=g_0/\|g_0\|$. Normalizing $(1-s)f+sg_0$ gives a <homotopy> from $f$ to $g$, since that vector remains within $\epsilon$ of the unit vector $f$.
Every value of $g$ is critical because its differential has rank at most $d<q$. The <Sard theorem> makes its image have measure zero in $S^q$, so it misses some point $p$. The punctured sphere $S^q\setminus\{p\}$ is homeomorphic to $\mathbb R^q$ and contractible. Thus $g$, and hence $f$, is null-homotopic. This reasoning does not require $X$ to be connected, and all constant maps are homotopic because $S^q$ is path connected.
Apply it with $X=O(2n)/U(n)$ and $q=n^2$, for positive integer $n$. We obtain
$$
\boxed{[O(2n)/U(n),\,S^{n^2}]=\{[\text{constant map}]\}.}
$$
The notation means <homotopy> classes of maps; no <homotopy> equivalence between the two spaces is asserted.
Back to article page
