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The displayed torus-only Weyl integration formula needs $\phi$ to be a class function, meaning $\phi (xg{x}^{-1})=\phi (g)$. This hypothesis is implicit in the usual formula but is not printed here. We prove the general version for a continuous function, then specialize:Here $dg$ and $dt$ are probability Haar measures; $d\mu$ is the invariant probability measure on the homogeneous space $G/T$; and $W=N_G(T)/T$ is the finite Weyl group.

We use the following standard maximal-torus facts, as permitted: every element of a connected compact Lie group lies in some maximal torus; all maximal tori are conjugate; two elements of a fixed $T$ are conjugate in $G$ exactly when they are in the same $W$-orbit; and the complexified Lie algebra decomposes into ${\mathfrak{t}}_{\mathbb{C}}$ and one-dimensional root spaces. Opposite roots of a root system give real two-dimensional planes in ${\mathfrak{t}}^{\perp }$. At an element $t$ for which no root character has value one, the identity component of its centralizer is $T$.

Choose a positive root system ${\Phi }^{+}$. Regard each root $\alpha$ as a character $\alpha :T\to U(1)$ through the Adjoint representation. A convenient globally defined denominator isThe usual Weyl denominator additionally has a unit-modulus factor $e^{\rho }(t)$, with $\rho$ half the sum of positive roots. Where that half-weight is not a global torus character, it may be interpreted on a covering torus; the squared modulus is nevertheless unambiguous. In either convention,

Average an inner product on $\mathfrak{g}$ over the Adjoint representation to make it invariant. It defines a bi-invariant Riemannian metric on $G$ and the quotient metric on $G/T$. Consider the conjugation mapAt $(T,t)$, identify the source tangent directions with $X\in {\mathfrak{t}}^{\perp }$ and $Y\in \mathfrak{t}$, and translate the target tangent vector to the identity by $t^{-1}$. Differentiating givesThe two summands are orthogonal. On a real root plane, $\mathrm{Ad}(t^{-1})$ is a rotation with complex eigenvalues $\alpha (t{)}^{-1},\alpha (t)$. Thus the absolute real Jacobian determinant on this plane is $|1-\alpha (t){|}^2$. The torus direction contributes one. Therefore the conjugation Jacobian for a compact Lie group isThis is also the Jacobian for the normalized measures: the Riemannian submersion has fibers isometric to $T$, so $\mathrm{vol}(G)=\mathrm{vol}(G/T)\mathrm{vol}(T)$; the probability-volume normalization factors cancel.

Let $T_{\mathrm{reg}}$ be the elements for which $J_q\ne 0$. The restriction of $q$ is a local diffeomorphism onto the regular elements of $G$. It is $|W|$-to-one. Indeed, fixing $t_0\in T_{\mathrm{reg}}$, any pair with $xt{x}^{-1}=t_0$ has $xT{x}^{-1}=T$, because conjugation identifies the identity components of the two centralizers. Thus $x\in N_G(T)$, and there is exactly one pair $(xT,x^{-1}{t}_{0}x)$ for each coset in $N_G(T)/T$. This argument does not require the full centralizer of $t_0$ to be connected.

The nonregular locus has measure zero. In $T$ it is a finite union of sets $\alpha (t)=1$ of smaller dimension; their conjugation images are images of smaller-dimensional manifolds under $q$, and hence have zero volume in $G$. The change-of-variables formula for the regular covering therefore givesFor a class function, the inner integral is $\phi (t)$, proving the required formula with its correct hypothesis.

Without conjugation invariance, the torus-only expression can be false. For $G=SU(2)$ with diagonal $T$, take $\phi (g)=|g_{12}{|}^2$. It vanishes on $T$, whereas its group integral is $1/2$: the two squared coordinates of a Haar-distributed unit column have equal means and sum to one. Thus the conjugacy average cannot simply be dropped.

For a group action of $G$ on $X$, the orbit of a group action through $x$ is $Gx=\{gx:g\in G\}$, and its stabilizer subgroup is $H_x=\{g:gx=x\}$. A transitive group action has one orbit. A simply transitive group action has exactly one element carrying any given point to any other; equivalently it is transitive and every stabilizer subgroup is trivial.

In the modern terminology, a multiply transitive group action is transitive on ordered $k$-tuples of distinct points for some $k\ge 2$. In older geometric usage, the phrase can instead mean transitive but not simply transitive, so that there are several elements carrying one point to another. Under either interpretation the action is transitive, and that alone is what the following argument needs. These meanings should not be confused: a transitive action with nontrivial stabilizers need not be two-transitive.

Fix $x_0\in X$ and put $H=H_{x_0}$. DefineThis is well-defined because every element of $H$ fixes $x_0$. It is surjective by transitivity, and $g{x}_0=g^{\prime }{x}_0$ is equivalent to $g^{\prime -1}g\in H$, proving injectivity. For a smooth Lie group action, $H$ is closed, since it is the inverse image of $x_0$ under the orbit map. The orbit map has constant rank: translations in $G$ and the action diffeomorphisms of $X$ intertwine its differentials. Transitivity makes this rank $\dim X$; otherwise every point would be a critical value, contradicting Sard theorem. Its local submersion charts give smooth local sections and hence a smooth inverse for $F$. ThereforeChoosing $x_1=a{x}_0$ replaces $H$ by $aH{a}^{-1}$. Thus the stabilizer subgroup is determined by the base point, and intrinsically only up to conjugacy. A simply transitive group action gives $H=\{e\}$.

For a finite quantum system with Hilbert space $\mathcal{H}={\mathbb{C}}^{n+1}$, a pure state is a nonzero vector modulo multiplication by a nonzero complex number. Equivalently, it is a unit vector modulo its physically irrelevant phase. Consequently the space of pure states is the space of complex lines,The qualification “pure” is essential: arbitrary quantum states include mixed density matrices and do not form Complex projective space.

The unitary group $U(n+1)$ acts transitively on these lines: choose unit vectors in two given lines, complete each to an orthonormal basis, and map one basis to the other by a unitary transformation. The stabilizer of the line $\mathbb{C}{e}_0$ consists exactly of block matrices $\mathrm{diag}(z,V)$ with $z\in U(1)$ and $V\in U(n)$. Indeed a unitary transformation preserving the line also preserves its orthogonal complement. The homogeneous space description is thereforeIts real dimension is $(n+1{)}^2-1-n^2=2n$. This action is generally not two-transitive: the absolute inner product between unit representatives of two lines is invariant.

The Hopf fibration is the projection $\pi :S^{2n+1}\to {\mathbb{CP}}^n$, $z\mapsto [z]$. The right action $z\cdot \lambda =z\lambda$, $\lambda \in S^1$, is free, and its orbits are exactly the fibres. On the chart $U_i$ where $z_i\ne 0$, write $w_j=z_j/z_i$ for $j\ne i$. A smooth local section isEvery unit vector over this chart has the unique expression $z={\sigma }_i([z])\lambda$, where $\lambda =z_i/|z_i|$. These formulas provide local trivializations ${\pi }^{-1}(U_i)\cong U_i\times S^1$, proving that this is a principal bundle with fibre $S^1$. For $n\ge 1$ it has no global section: a global section would give $S^{2n+1}\cong {\mathbb{CP}}^n\times S^1$, contradicting their fundamental groups. For $n=0$ the base is a point and the bundle is trivial.

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.

A fiber bundle with base $B$, fiber $F$ and structure group $G$ acting on $F$ from the left consists of a projection $\pi :E\to B$ and local trivializations ${\varphi }_i:{\pi }^{-1}(U_i)\to U_i\times F$. On overlaps they satisfyThe smooth transition functions are valued in $G$. A principal bundle is the case $F=G$ with the left multiplication action on itself; its transition functions multiply the group coordinate from the left.

In each principal chart define $(x,h)\cdot a=(x,ha)$. This defines a global right action because $g_{ij}(x)(ha)=(g_{ij}(x)h)a$: coordinate changes commute with right multiplication. It is free, and acts transitively on each fiber. Thus the claimed right action follows from the left transition convention rather than being an additional choice.

If $s:B\to P$ is a global section of a fiber bundle, define $\Phi :B\times G\to P$ by $\Phi (x,a)=s(x)a$. Freeness and transitivity give exactly one $a$ for every point in the fiber, so $\Phi$ is bijective and equivariant. In a local trivialization where $s(x)$ has group coordinate $h_i(x)$, the inverse group coordinate is $h_i(x{)}^{-1}h$, proving smoothness of the inverse. Hence $\Phi$ is a principal-bundle trivialization. Conversely, the identity coordinate in a product trivialization gives $s(x)=\Phi (x,e)$. ThereforeThis is the principal bundle trivialization by a global section criterion. It is special to principal bundles; a section alone need not trivialize an arbitrary fiber bundle.

For the quotient examples, assume the Lie subgroups are closed when smooth homogeneous-space statements are intended. The map $G\to G/H_1$ is a principal bundle with structure group $H_1$ and right action $g\cdot h=gh$; $G/H_1$ is a homogeneous space for left multiplication by $G$. Similarly, $G\to H_2\backslash G$ has free transitive left $H_2$ action on its fibers. With the usual principal right-action convention one writes $g\cdot h=h^{-1}g$. The quotient $H_2\backslash G$ is a homogeneous space for right multiplication by $G$. These quotients are groups only when the respective subgroups are normal. Nonclosed immersed Lie subgroups can produce non-Hausdorff quotients, so an arbitrary Lie subgroup does not automatically give a smooth Hausdorff base.

Left $H_2$ and right $H_1$ multiplication commute, so taking their orbit equivalence relations in either order gives the same double coset setThis equality does not assert that the double quotient is a manifold or that either stage is a principal bundle. The $H_2\times H_1$ action $(a,b):g\mapsto ag{b}^{-1}$ can have stabilizers: they correspond to $H_2\cap g{H}_1{g}^{-1}$. If the induced action is free and proper, its quotient is a smooth principal-bundle quotient; without those hypotheses orbit dimensions can change. For example, with $G=SO(3)$ and both subgroups the rotations fixing the third axis, double cosets are parametrized by $e_3\cdot g{e}_3\in [-1,1]$. Interior and endpoint stabilizers differ, so this is not a uniform free principal-bundle construction.

For a principal bundle $P$ and a left $G$ action on $F$, the associated bundle iswith projection $[p,f]\mapsto \pi (p)$. A local principal section gives coordinates $[s_i(x),f]\leftrightarrow (x,f)$, and its changes give precisely the left action of the principal transition functions on $F$. A linear action on a vector space gives an associated vector bundle.

For a metric of signature $(r,s)$ on an $n$-manifold, an orthonormal frame at $x$ is an isometry $u:{\mathbb{R}}^{r,s}\to T_{x}M$. The orthonormal frame bundle $O(M,g)\to M$ has structure group $O(r,s)$, acting on the right by $u\cdot a=u\circ a$. Its associated bundle for the standard representation is the tangent bundle, sinceis well defined: $(ua)(v)=u(av)$. Every tangent vector has exactly one such associated equivalence class, and local frame coordinates show that this is a vector-bundle isomorphism. The full frame bundle works similarly with $GL(n,\mathbb{R})$.

Finally, choose a basis of the Lie algebra of a Lie group $G$. Its left translates are a global frame, yielding the explicit tangent bundle trivializationThis proves parallelization of a Lie group by left translations and triviality of the full frame bundle. Choosing an inner product at the identity and translating it gives a left-invariant Riemannian metric; an orthonormal basis translates to a global orthonormal frame. The global section criterion therefore trivializes the corresponding orthonormal frame bundle as well. For any other positive-definite metric on $G$, smoothly applying the Gram-Schmidt process to this global frame gives the same conclusion. Thus the tangent and frame bundles of every Lie group are trivial; the orthonormal statement is with respect to a chosen metric as specified above.

Because $F_1$ is projective twistor space, the printed column represents the line $p=[e_1]$, not a distinguished nonzero vector. Its stabilizer subgroup consists of matrices whose first column is a nonzero multiple of $e_1$:The block form is necessary to preserve the line, and conversely every such determinant-one matrix preserves it. By transitivity, the orbit map $g\mapsto g[e_1]$ is onto $F_1$. Its two values agree precisely when $g^{-1}{g}^{\prime }\in H_p$, equivalently when $g{H}_p=g^{\prime }{H}_p$. This gives the homogeneous space identificationThe dimensions also agree: ${\dim }_{\mathbb{C}}SL(4)=15$, while the free $B$ and $b$ parameters give ${\dim }_{\mathbb{C}}{H}_p=9+3=12$; the quotient has dimension three.

This coset space is not a quotient group, because $H_p$ is not a normal subgroup. For example $h=I+E_{12}\in H_p$. Let $g$ have upper-left block $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and identity on the other two coordinates. Then $g\in SL(4,\mathbb{C})$ but $gh{g}^{-1}=I-E_{21}$ sends $e_1$ to $e_1-e_2$, so it is not in $H_p$. Thus multiplying coset representatives does not define a well-defined group operation.

If instead one required the literal vector $e_1$ to be fixed, one would have $a=1$ and $B\in SL(3,\mathbb{C})$; its quotient would be ${\mathbb{C}}^4\setminus \{0\}$, not $F_1$. The projective reading is required by the printed target quotient and the flag manifold definition.

Because the metric coefficients are independent of $t$ and $y$,are Killing vector fields. The mapssatisfy ${\varphi }_s\circ {\varphi }_r={\varphi }_{s+r}$. Moreover, $dx$ is unchanged and both $e^{x}dy$ and $dt+\sqrt{2}{e}^{x}dy$ are invariant, so ${\varphi }_s^{\ast }g=g$. Differentiating at $s=0$ produces the third Killing fieldGiven two points, first use ${\varphi }_s$ to match their $x$ coordinates, then translations generated by ${\partial }_y$ and ${\partial }_t$ to match $y$ and $t$. The isometry group therefore acts transitively, so the spacetime is a homogeneous space.

The stabilizer of a line in $O(n)$ is $O(1)\times O(n-1)$, giving the homogeneous space ${\mathbb{RP}}^{n-1}$. Its classifying-space Serre fibration has rationally acyclic fiber for odd $n$, so its cohomological Serre spectral sequence has only its degree-zero row and gives the displayed restriction isomorphism. The factor $BO(1)={\mathbb{RP}}^{\infty }$ is itself rationally acyclic. For even $n$, the fiber has a top rational class, with determinant-sign monodromy, so the one-row argument fails. The top universal Pontryagin class $p_{n/2}$ is nonzero on $BO(n)$ but vanishes on restriction to $BO(n-1)$, witnessing genuine failure of stabilization.

For a centered static Skyrmion $U_0$, the effective rotation-isorotation group is $G=SO(3{)}_{\mathrm{space}}\times SO(3{)}_{\mathrm{iso}}$. Its stabilizer subgroup $H$ consists of pairs obeying $A{U}_0(R^{-1}x)A^{†}=U_0(x)$. Thus distinct rigid orientations form the homogeneous space $G/H$, by the orbit-stabilizer theorem. The unit hedgehog has diagonal $SO(3)$ stabilizer and three orientation coordinates; the toroidal two-Skyrmion has a one-dimensional continuous stabilizer and five; the tetrahedral and cubic cases have discrete stabilizers and six. Adding three translations gives zero-mode orbit dimensions 6, 8, 9, 9. These are symmetry-generated low-energy coordinates, not a claim that non-Bogomolny multi-Skyrmion solutions possess an exact flat moduli space of arbitrary separations.

On a Euclidean real space of dimension $2n$, an orthogonal complex structure is an orthogonal endomorphism $J$ with $J^2=-I$. Orthogonal changes of basis act transitively on these structures, and the stabilizer of the standard structure is $U(n)$. The compact homogeneous space has real dimension $n(n-1)$.

// Target: geometry-and-topology.bigb