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Regard ${\mathrm{Mat}}_n(\mathbb{C})$ as a real vector space of dimension $2n^2$, and let ${\mathrm{Herm}}_n(\mathbb{C})$ be the real vector space of Hermitian matrices, of dimension $n^2$. The smooth Gram maphas $F^{-1}(I)=U(n)$ and differential $D{F}_A(K)=A^{\ast }K+K^{\ast }A$. At a unitary $A$, any Hermitian matrix $H$ is obtained by taking $K=AH/2$. Indeed $A^{\ast }K=H/2$ and $K^{\ast }A=H/2$. Hence $I$ is a regular value, and the regular level set theorem proves the unitary group as a regular level set result:

Similarly let $G(A)=A^{T}A$ from ${\mathrm{Mat}}_n(\mathbb{R})$ to the symmetric real matrices, whose dimension is $n(n+1)/2$. At an orthogonal $A$ its differential is $D{G}_A(K)=A^{T}K+K^{T}A$, again onto by choosing $K=AH/2$ for a symmetric $H$. Therefore the orthogonal group as a regular level set hasMatrix 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 dimensionThe 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 $q\ge 1$, 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 $ϵ<1$ of $f$. It never vanishes. Normalize it to $g=g_0/\Vert g_0\Vert$. Normalizing $(1-s)f+s{g}_0$ gives a homotopy from $f$ to $g$, since that vector remains within $ϵ$ 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 obtainThe notation means homotopy classes of maps; no homotopy equivalence between the two spaces is asserted.