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For an orthogonal group gauge transformation $R\in O(3)$, a scalar vector transforms as $\varphi \mapsto R\varphi$. The gauge connection belongs to the Adjoint representation, identified with axial vectors: the displayed cross-product identity makes its homogeneous transformation $A\mapsto (\det R)RA$. The matrix connection also has the usual inhomogeneous derivative term. Thus $D\varphi$ transforms as an ordinary vector, including under improper orthogonal transformations. Treating both fields as ordinary vectors would fail because a cross product changes sign under an orientation reversal. A nonzero vector vacuum has stabilizer subgroup $O(2)$; its connected part is $SO(2)$.

Without fixed gauge charges or interactions distinguishing particles from antiparticles, $N$ free Dirac fields give $2N$ left-handed Weyl fields after charge conjugation of the right-handed fields. Their massless kinetic term has unitary group $U(2N)$. Equal nonzero Dirac masses have equal positive singular values and can be written as $2N$ equal Majorana mass terms; their preserving group is the orthogonal group $O(2N)$. Explicitly, in a basis with symmetric mass matrix $m1$, invariance requires $V^{†}V=V^{T}V=1$, hence real $V$. Restricting to transformations preserving a chosen common fermion number gives the usual $U(N{)}_L\times U(N{)}_R$ at zero mass and $U(N)$ at nonzero mass.

Generate a real $n$-by-$n$ matrix $G$ with independent entries having the standard normal distribution. Take its QR decomposition with positive diagonal in the upper triangular matrix factor, $G=QR$. The orthogonal matrix $Q$ has normalized Haar measure on the full orthogonal group $O(n)$. If a numerical factorization returns $Q_0{R}_0$, set $S=\mathrm{diag}(\mathrm{sgn}{R}_{0,ii})$ and return $Q_{0}S$. The signs are nonzero almost surely: successively drawn Gaussian columns have probability zero of lying in the proper linear subspace spanned by earlier columns.

For correctness, every fixed orthogonal matrix $U$ preserves the standard Gaussian column density and column independence, so $UG$ and $G$ have the same probability distribution. The positive-diagonal QR decomposition is unique: two decompositions would give an orthogonal matrix that is also an upper triangular matrix with positive diagonal, hence the identity. Therefore $Q(UG)=UQ(G)$ and $Q(G)$ has a left-invariant probability law. Uniqueness of normalized Haar measure on the compact group $O(n)$ identifies that law. Each determinant sign has probability $1/2$ by reflection invariance.

The Lagrangian Grassmannian is the space of all Lagrangian subspaces of a symplectic vector space. The unitary group acts transitively on $\mathrm{LGr}({\mathbb{R}}^{2n})$, and the stabilizer of ${\mathbb{R}}^n$ is the orthogonal group $O(n)$, giving

Two canonically normalized complex scalar fields with potential $V=m^2{\Phi }^{†}\Phi +\lambda ({\Phi }^{†}\Phi {)}^2$ are four real fields with a radial potential. Consequently the isolated action is invariant under the orthogonal group $O(4)$, not only under complex-linear $U(2)$ transformations. Writing ${\varphi }_1=(q_1+i{q}_2)/\sqrt{2}$ and ${\varphi }_2=(q_3+i{q}_4)/\sqrt{2}$ makes the kinetic term ${\sum }_A(\partial q_A{)}^2/2$ and the potential a function of ${\sum }_A{q}_A^2$. The six real-plane rotations give currents $j_{AB}^{\mu }=q_B{\partial }^{\mu }{q}_A-q_A{\partial }^{\mu }{q}_B$. Gauge or Yukawa couplings can break this larger symmetry, so it should not be asserted for an arbitrary interacting doublet.

In odd dimension, the map $A\mapsto (\det A)A$ is a surjective group homomorphism from the orthogonal group to the special orthogonal group, with kernel of a group homomorphism $\{I,-I\}$. Scalar matrices commute, and $\det [(\det A)A]=(\det A{)}^{2m+2}=1$. Together with the determinant sign, it identifies $O(2m+1)$ with the direct product of groups $SO(2m+1)\times C_2$. In even dimension, scalar multiplication by $\det A$ does not repair the determinant, so this construction does not give the same splitting.

For particles in the vector representation of the orthogonal group, the two-body S-matrix is a linear combination of the identity $I$, permutation $P$ and trace contraction $K$: $S=S_{2}I+S_{3}P+S_{1}K$. Here $K(e_i\otimes e_j)={\delta }_{ij}{\sum }_a{e}_a\otimes e_a$, $P^2=I$, $PK=KP=K$ and $K^2=NK$. The orthogonal-invariant scattering channels diagonalize these three operators simultaneously.

The $O(N)$ model has an $N$-component real field and an orthogonal group symmetry. Its standard Landau-Ginzburg interaction is $g(\mathbf{\varphi }\cdot \mathbf{\varphi }{)}^2$; below the upper critical dimension its continuous transition is governed by an $O(N)$ Wilson-Fisher fixed point.

For a continuous real representation $\rho :G\to GL(V)$ of a compact group, average any positive-definite real inner product against normalized Haar measure:The result is positive definite and $G$-invariant. An orthonormal basis for it conjugates the representation into an orthogonal group.

A vector bundle trivialization is orthogonal when its coordinate frame is orthonormal for the fiber metric, so each fiber is identified isometrically with Euclidean space. Overlap changes then take values in the orthogonal group. Local orthogonal trivializations exist even when no global frame does.

An orthogonal representation is a real group representation preserving a positive-definite inner product, equivalently a group homomorphism into an orthogonal group. Every finite-dimensional continuous real representation of a compact group has this form after averaging an inner product over Haar measure, by orthogonalization of a compact-group representation.

An orthogonal structure is a reduction of the transition functions of a real rank-$r$ vector bundle to the orthogonal group. Its orthonormal frames define a fiber metric, independent of the chosen frame because orthogonal transitions preserve the inner product. Conversely a fiber metric gives smooth orthonormal frames by the Gram-Schmidt process, with orthogonal transition matrices. This equivalence does not require the vector bundle to be globally trivial.

Collect the two Dirac fields into $\Psi =({\psi }_1,{\psi }_2{)}^T$, with a common nonzero mass $m$. ThenAmong complex-linear rotations of the Dirac fields, the internal symmetry of a classical field theory is $\Psi \mapsto U\Psi$, $U\in U(2)$, with $\bar{\Psi }\mapsto \bar{\Psi }{U}^{†}$. Unitarity preserves the kinetic term, and the mass matrix commutes with every $U$. This unitary flavour symmetry of degenerate Dirac fields has group $(SU(2)\times U(1))/{\mathbb{Z}}_2$, where the quotient avoids counting the common central element twice.

Choose the four Hermitian matrices $T^0=1_2/2$, $T^a={\sigma }^a/2$ for $a=1,2,3$, with ${\sigma }^a$ the Pauli matrices. Promoting the four parameters to functions gives $\delta S=-\int d^{4}x({\partial }_{\mu }{\alpha }^A)\bar{\Psi }{\gamma }^{\mu }{T}^A\Psi$. The four Noether currents are thereforeTheir divergences vanish by the equal-mass Dirac equations. The singlet is half the total Dirac current; the other three currents rotate the two species.

Free-theory qualification. If “global internal symmetry” allows mixing the Dirac fields with charge-conjugate fields, the completely free theory written here has a larger enhanced internal symmetry of free Dirac fields. Write ${\psi }_i=({\lambda }_{2i-1}+i{\lambda }_{2i})/\sqrt{2}$ using four Majorana fermions. Up to a total derivative,The full orthogonal group $O(4)$ acts on the four real species; its connected part is $SO(4)$. Its six Noether currents can be chosen as $J_{AB}^{\mu }=i{\bar{\lambda }}_A{\gamma }^{\mu }{\lambda }_B$ for $A<B$, up to the common sign convention. The antisymmetry of this Majorana fermion vector bilinear makes these the six independent rotations. A concrete rotation outside the stated complex-linear $U(2)$ is ${\lambda }_2\mapsto \cos \theta {\lambda }_2+\sin \theta {\lambda }_3$, ${\lambda }_3\mapsto -\sin \theta {\lambda }_2+\cos \theta {\lambda }_3$: it preserves the displayed Lagrangian density while mixing ${\psi }_i$ with their charge-conjugate fields. The usual $U(2)$ answer specifies the subgroup preserving the chosen common fermion number.

For a real linear change of coordinates $x^{\prime }=Ax$, one has $d{x}^{\prime }=Adx$. The Euclidean metric therefore transforms toIt agrees with $d{x}^{T}dx$ for every displacement precisely when $A^{T}A=I$. For example, equality of the associated quadratic forms determines their symmetric coefficient matrices by polarization identity. The orthogonal group is defined by this condition, soTaking determinants gives $(\det A{)}^2=1$. Thus these matrices are invertible automatically; both signs of the determinant are allowed, since the metric alone does not specify an orientation.

Here the orthogonal group is the real group $O_n(\mathbb{R})$. Differentiating $g(t{)}^{T}g(t)=I$ at zero gives $X^T+X=0$. Conversely a real skew-symmetric matrix satisfies $(e^{tX}{)}^T{e}^{tX}=e^{-tX}{e}^{tX}=I$. ThereforeFor $n\ge 2$ this is not a complex Lie subalgebra of $M_n(\mathbb{C})$: multiplying a nonzero real member by $i$ leaves this real subspace. For $n=1$ the zero algebra is the trivial complex Lie algebra as well. The complex orthogonal group would be a different group from the real orthogonal group used here.

For a smooth manifold $M$ of dimension $n$, the tangent space $T_{p}M$ can be defined as the space of derivations at $p$: linear maps $v$ on germs of smooth real functions satisfying $v(fg)=f(p)v(g)+g(p)v(f)$. Equivalently, $v$ is the velocity of a smooth curve through $p$, with curves identified when their coordinate velocities agree. A chart $(x^1,\dots ,x^n)$ supplies the basis ${\partial }_i{|}_p$, so $v=v^i{\partial }_i{|}_p$.

The tangent bundle is the disjoint union $TM={\coprod }_{p\in M}{T}_{p}M$, with projection $\pi (v)=p$. Its smooth structure is defined by the local local trivializationsOn overlapping charts, the change of trivialization isThese are smooth changes of coordinates on a $2n$-dimensional smooth manifold, linear and invertible in each fibre. Thus $TM$ is a rank-$n$ vector bundle, and its projection is a submersion. Fibrewise addition, scalar multiplication and the zero section are smooth. A smooth section of a vector bundle $X:M\to TM$, satisfying $\pi \circ X={\mathrm{id}}_M$, is exactly a vector field; locally $X=X^i{\partial }_i$, with components transforming by the displayed Jacobian matrix. Its action on functions is a derivation, and the commutator of two such derivations is their Lie bracket.

The dual bundle of $TM$ is the cotangent bundle $T^{\ast }M$. Its smooth sections of a vector bundle are differential 1-forms, locally $\alpha ={\alpha }_{i}d{x}^i$. The coefficient transformation is the inverse transpose of that for tangent vectors, ensuring that $\alpha (X)$ is a well-defined smooth function. Tensoring these two bundles gives the tensor bundlesTheir sections of a vector bundle are tensor fields; their transition functions of a vector bundle are the corresponding tensor products of the tangent and cotangent transformations. For example, the bundle $\mathrm{End}(TM)=TM\otimes T^{\ast }M$ contains fields of linear endomorphisms, while $S^2{T}^{\ast }M$ contains symmetric bilinear fields. Taking alternating covariant tensors gives ${\Lambda }^k{T}^{\ast }M$, whose sections are differential forms. The wedge product, exterior derivative and pullback of a differential form make these bundles central to integration, Generalized Stokes theorem and de Rham cohomology.

A nowhere-vanishing section of ${\Lambda }^n{T}^{\ast }M$ chooses an orientation of a smooth manifold; such a section exists exactly when $M$ is orientable. Without an orientation of a smooth manifold, one can still integrate sections of the density bundle $|{\Lambda }^n{T}^{\ast }M|$, whose coordinate changes use the absolute Jacobian determinant. A Riemannian metric is a positive-definite section $g$ of $S^2{T}^{\ast }M$. It identifies $TM$ with $T^{\ast }M$ by the musical isomorphisms $X\mapsto g(X,\cdot )$ and gives the Riemannian volume densityOn an oriented smooth manifold, this density corresponds to a volume form.

Another organizing construction is the frame bundle $FM$. A frame over $p$ is a linear isomorphism $u:{\mathbb{R}}^n\to T_{p}M$, and the right action $u\cdot A=u\circ A$ makes $FM$ a principal bundle with general linear group $GL(n,\mathbb{R})$ as its structure group. Given a group representation $\rho :GL(n,\mathbb{R})\to GL(V)$, the associated bundle $FM{\times }_{\rho }V$ is obtained from pairs $(u,v)$ by the relation $(uA,v)\sim (u,\rho (A)v)$. The standard representation reconstructs $TM$; dual, tensor and exterior representations reconstruct the associated bundles described above. A Riemannian metric reduces the frame bundle to the orthonormal frame bundle, with orthogonal group $O(n)$ as structure group; an orientation of a smooth manifold reduces it further to the special orthogonal group $SO(n)$. A global section of $FM$ is a global frame, so it exists exactly when $M$ is a parallelizable manifold. Local triviality does not imply a global trivialization: the Hairy ball theorem prevents even one nowhere-zero vector field on $S^2$.

A connection on a vector bundle differentiates its sections in tangent directions and defines parallel transport. For $TM$, the Levi-Civita connection is distinguished by compatibility with a Riemannian metric and vanishing torsion tensor. Its coefficients are not tensor components, because a coordinate change introduces second derivatives; the difference of two connections is a section of $T^{\ast }M\otimes \mathrm{End}(TM)$. The Riemann curvature tensor is a genuine tensor measuring the failure of covariant derivatives to commute.

Finally, a smooth map $f:M\to N$ gives the pullback tangent bundle $f^{\ast }TN$. Its sections are vector fields along a map, including the velocity of a curve and Jacobi fields along a geodesic. For an embedded submanifold $i:M\hookrightarrow N$, the derivative identifies $TM$ with a subbundle of $i^{\ast }TN$, and the quotient is the normal bundle. With an ambient Riemannian metric, the normal bundle is identified with the orthogonal complement of $TM$ in $i^{\ast }TN$; the second fundamental form records the normal component of the ambient derivative of tangent fields. These constructions connect the bundle description to both intrinsic and extrinsic differential geometry.

For $X\in \mathfrak{g}=T_{e}G$, let $X^L$ be the left-invariant vector field with value $X$ at the identity. Its integral curve through the identity is the unique one-parameter subgroup ${\gamma }_X$ with ${\gamma }_X^{\prime }(0)=X$. It exists for all real time: local existence followed by translation and the identity ${\gamma }_X(t+s)={\gamma }_X(t){\gamma }_X(s)$ extends it indefinitely. The Exponential map of a Lie group isIt is smooth and $d_0\exp ={\mathrm{id}}_{\mathfrak{g}}$, because differentiating $t\mapsto \exp (tX)$ at zero gives $X$. The inverse function theorem therefore makes it a local diffeomorphism at zero.

If $G$ is abelian, the product $t\mapsto \exp (tX)\exp (tY)$ is a one-parameter subgroup with initial tangent $X+Y$. Uniqueness gives $\exp (t(X+Y))=\exp (tX)\exp (tY)$, henceConversely, if this additive homomorphism identity holds, all $\exp (tX)$ and $\exp (sY)$ commute. Differentiating their adjoint action shows $[X,Y]=0$. Alternatively, their commuting images contain an identity neighbourhood; that neighbourhood generates the identity component of a Lie group, so the identity component is abelian. Conversely, if the Lie algebra is abelian, its left-invariant vector fields commute, so their flows commute. Hence all one-parameter subgroups commute; their product with initial tangent $X+Y$ is the one-parameter subgroup for $X+Y$. The Lie exponential map is therefore additive, and its image generates an abelian identity component. Thus the stated equivalence holds for connected $G$. Without connectedness the exact assertion is that the Lie exponential map is a homomorphism precisely when $G^{\circ }$ is abelian. The converse implication to all of $G$ is false: in $O(2)$ the Lie algebra consists of $tJ$, whereRotations satisfy $R_{t+s}=R_t{R}_s$, but a reflection $S$ satisfies $S{R}_t{S}^{-1}=R_{-t}$, so the full orthogonal group is not abelian.

Now take a connected abelian Lie group of dimension $n$. Its Lie exponential map is a homomorphism whose image contains an identity neighbourhood, hence an open subgroup. A connected group has no proper open subgroup, so this map is onto. Its kernel $\Lambda$ is discrete, since it is injective on a neighbourhood of zero. The induced map is a bijective local diffeomorphism and therefore a Lie group isomorphismHere is the needed classification of $\Lambda$. Let $r$ be the dimension of its real span. Choose ${\lambda }_1,\dots ,{\lambda }_r\in \Lambda$ that form a real basis of that span. Their integer span ${\Lambda }_0$ has finite index in $\Lambda$: subtract integer multiples of the ${\lambda }_i$ to put each coset representative in a compact parallelepiped. Only finitely many points of $\Lambda$ lie there, since discreteness at zero gives a uniform positive separation between distinct points. Thus $\Lambda$ is finitely generated and torsion-free, and the structure theorem for finitely generated abelian groups makes it free abelian of rank $r$. Its integer basis is a real basis of the span. A linear change of coordinates consequently sends $\Lambda$ to ${\mathbb{Z}}^r\times \{0\}$ in ${\mathbb{R}}^n$.

This proves the classification of connected abelian Lie groups:The isomorphism depends on the choice of lattice basis and complementary vector space. Compactness is equivalent to $s=0$.

Work with canonically normalized real components of the scalar field; a complex representation can first be written in terms of its real and imaginary components. At a vacuum expectation value ${\phi }_0$, expansion of the scalar potential givesThe linear term vanishes by stationarity. Thus the scalar mass matrix is the Hessian matrix of the scalar potential. Invariance under each Lie algebra generator gives the identity ${\partial }_{r}V(\phi )(i{T}_a\phi {)}_r=0$. Differentiating this identity with respect to ${\phi }_s$ and evaluating at the stationary point removes the term containing ${\partial }_{r}V$, leavingThe kernel of the map $T\mapsto iT{\phi }_0$ is precisely the Lie algebra of the stabilizer subgroup H. Consequently its restriction to a complementary space of broken Lie algebra generators is injective: a linear combination that annihilates ${\phi }_0$ would belong both to the stabilizer algebra and its complement, and hence would vanish. There are therefore $\dim G-\dim H$ independent Goldstone directions in the scalar mass matrix, tangent to the vacuum manifold. There are $\dim G-\dim H$ symmetry-protected massless scalars.

This Goldstone theorem guarantees at least that many zero quadratic masses. The claim that these are all the massless scalars additionally requires nonzero curvature transverse to the vacuum orbit. For example, $V=\kappa ({\phi }^T\phi -v^2{)}^4$, with $\kappa >0$, has an ordinary spherical vacuum manifold but its entire Hessian matrix vanishes there. These accidental massless radial modes beyond Goldstone modes are not excluded by the hypotheses. For the indicated orthogonal group breaking, the protected number is

For the Higgs doublet, put $r=\sqrt{2{\varphi }^{†}\varphi }=v+H$ and $z=\sqrt{2}\varphi /r$ wherever $r\ne 0$. If $z=(z_1,z_2{)}^T$, thenThis explicitly exhibits the three angular coordinates as an SU(2) transformation of a real radial field. In particular, the polar coordinates for the electroweak Higgs doublet can locally be chosen in the formAt the vacuum the three variations generated by $-i{\sigma }_1$, $-i{\sigma }_2$, and $i(I-{\sigma }_3)$ acting on $(0,1{)}^T$ are independent over the reals. They span the three broken directions; the remaining generator $Q=({\sigma }_3+I)/2$ annihilates the vacuum. This proves that the exponential supplies local angular coordinates. An inverse local gauge transformation, applied also to the gauge fields, removes these three fields in unitary gauge. They supply the longitudinal polarizations of the massive W bosons and Z boson through the Higgs mechanism. The parameterization is local about a nonzero vacuum, rather than a coordinate system through every possible zero of the Higgs doublet.

Use the metric $(+---)$ and define the charged W bosons and neutral gauge fields bywhere $s_W=\sin {\theta }_W$, $c_W=\cos {\theta }_W$, and $\tan {\theta }_W=g^{\prime }/g$. Direct substitution into the gauge-covariant kinetic term givesand henceThe electroweak doublet gauge-boson mass matrix in the $(A^3,B)$ basis isIts normalized zero direction is $(s_W,c_W)$ and its massive direction is $(c_W,-s_W)$. Thus the scalar-vector terms in physical fields arewith the tree-level electroweak gauge-boson massesThese are also charge eigenstates. The unbroken U(1) gauge symmetry has generator $Q=T_3+Y$; on the Higgs doublet, $Y=I/2$. Since $[Q,T_{\pm }]=\pm T_{\pm }$, the combinations $W^{\pm }$ have charges $\pm 1$, while Z, the photon, and H are neutral. The photon disappears from the displayed scalar-vector terms because $g{A}^3-g^{\prime }B=\sqrt{g^2+g^{\prime 2}}Z$. The photon is massless and has no direct tree-level coupling to the remaining radial scalar in this sector. This does not remove its interactions with charged W bosons, or exclude a scalar-to-photon interaction generated by loops.

Finally, retain the printed normalization of the Higgs field potential:The kinetic term is $\frac{1}{2}(\partial H{)}^2$, so the Higgs radial mass for a half-normalized quartic potential is $\boxed{m_H^2=\lambda v^2}$. The extra factor of two familiar from a potential with prefactor $\lambda$, rather than $\lambda /2$, does not apply here.

An orthogonal structure on a real vector bundle $E$ of rank $r$ is a choice of local frames whose transition matrices lie in the orthogonal group $O(r)$. In such frames declareOrthogonal transitions preserve this expression, so it defines a smooth positive-definite fiber metric. Conversely, a fiber metric turns any local frame into a smooth orthonormal frame by the Gram-Schmidt process; the resulting transition matrices are orthogonal. The two constructions are inverse, giving the required equivalence.

An orthogonal local trivialization $E{|}_U\cong U\times {\mathbb{R}}^r$ identifies each fiber isometrically with standard Euclidean space. Equivalently, its coordinate frame satisfies $h(e_i,e_j)={\delta }_{ij}$. This is local and does not assert the existence of a global frame.

Every real vector bundle over a smooth manifold admits a fiber metric. Choose a trivializing open cover and local Euclidean metrics $h_i$. The partition of unity theorem for a Hausdorff second-countable smooth manifold supplies nonnegative smooth functions ${\rho }_i$, with locally finite supports contained in the respective cover members, such that ${\sum }_i{\rho }_i=1$. Setextending each weighted metric by zero off its chart. Local finiteness makes this smooth. For every nonzero fiber vector $v$, at least one positive weight gives $h(v,v)={\sum }_i{\rho }_i{h}_i(v,v)>0$. Thus $h$ is a fiber metric and provides an orthogonal structure.

One definition of an orthogonal connection is that its covariant derivative preserves the fiber metric:for all vector fields $X$ and smooth sections of a vector bundle $s,t$. Another is that its parallel transport along every smooth curve is an isometry between the endpoint fibers.

For the first implication, take parallel sections $s,t$ along a curve. Their covariant derivatives vanish, so the displayed identity makes $h(s,t)$ constant: parallel transport preserves a fibre metric. Conversely, if parallel transport preserves the inner product, choose parallel extensions of arbitrary initial fiber vectors along a curve with prescribed initial tangent $X$. Differentiating their constant inner product gives $({\nabla }_{X}h)(s,t)=0$ at the initial point. Since $X,s,t$ were arbitrary, the connection is metric-compatible.

The equivalent local description is that the connection matrix in an orthonormal frame is skew-symmetric. Writing $\nabla e_j={\sum }_i{A}^i{}_j{e}_i$ in an orthogonal trivialization, differentiate $h(e_i,e_j)={\delta }_{ij}$ to obtainConversely this matrix identity gives the metric-preservation formula by expanding arbitrary sections. It is the orthogonal group connection description of the same condition.

The orthogonal group acts transitively on unoriented lines in ${\mathbb{R}}^n$. The stabilizer of the line spanned by the first coordinate isThus $G/H={\mathbb{RP}}^{n-1}$; using only $O(n-1)$ would stabilize a vector and give a sphere instead. For the universal contractible free $G$-space $EG$, the associated bundle $EG/H\to EG/G$ gives the Serre fibration $G/H\to BH\to BG$. Its total space identifies up to homotopy with $BO(1)\times BO(n-1)$.

If $n$ is odd, $n-1$ is even and the rational cohomology of ${\mathbb{RP}}^{n-1}$ is just $\mathbb{Q}$ in degree zero. This follows from the integral projective cellular differentials, alternating zero and multiplication by two, with the even-dimensional top class killed rationally. The cohomological Serre spectral sequence therefore has only the rowwith all other rows zero. The local system in degree zero is trivial because the fiber is connected. There can be no differential or extension, and the multiplicative edge map $H^{\ast }(BO(n);\mathbb{Q})\to H^{\ast }(BH;\mathbb{Q})$ is an isomorphism of rings.

Also $BO(1)={\mathbb{RP}}^{\infty }$ has rational cohomology $\mathbb{Q}$ in degree zero only, by the same cellular calculation. The Künneth theorem identifies $H^{\ast }(BH;\mathbb{Q})$ with $H^{\ast }(BO(n-1);\mathbb{Q})$. The inclusion fixing the first coordinate factors through $H$, setting its $O(1)$ coordinate to one; hence the resulting isomorphism is the actual restriction along this inclusion:This includes $n=1$, with $BO(0)$ a point, and proves rational odd-rank stabilization of orthogonal classifying spaces.

For even $n$, the fiber is odd-dimensional real projective space and has a further rational class in degree $n-1$. The spectral sequence has a possible additional row, now with determinant-sign local coefficients: an orientation-reversing orthogonal matrix acts with degree $-1$ on the covering sphere and hence on the odd-dimensional projective space. Therefore the one-row collapse argument is unavailable; this top row must not be treated as automatically untwisted.

There is an actual failure of injectivity in every even rank. Write $n=2m$. The universal class $p_m\in H^{4m}(BO(2m);\mathbb{Q})$ is nonzero: pull the universal bundle back to the sum of the real underlying bundles of $m$ complex lines over $({\mathbb{CP}}^{\infty }{)}^m$. If their first Chern classes are $x_j$, the Pontryagin formula $p_m(E)=(-1{)}^m{c}_{2m}(E\otimes \mathbb{C})$ gives $p_m={\prod }_j{x}_j^2\ne 0$. On $BO(2m-1)$ the restricted bundle is ${\epsilon }^1\oplus {\gamma }_{2m-1}$; its top complex Chern class $c_{2m}$ vanishes by the rank bound and the trivial summand. Thus $p_m$ restricts to zero. The natural restriction is not injective for even $n$, as the extra fiber cohomology already warns.