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A vector bundle over $S^n$ can be constructed by gluing two trivial bundles over its hemispheres with a transition map $\gamma :S^{n-1}\to GL(k,\mathbb{R})$. Extend this map constantly in an equatorial collar to obtain local vector bundle trivializations. A transition map into the special orthogonal group gives compatible fiber orientations and a fiber metric. Homotopies of transition maps give isomorphic bundles.

A clutching function is the equatorial transition map used in the clutching construction. For an oriented real vector bundle one may choose a fiber metric and orthonormal trivializations, so its clutching function takes values in the special orthogonal group.

For real orthogonal matrices in odd dimension $d$, multiplying by their determinant defines a group homomorphism from $O(d)$ onto $SO(d)$. Scalar factors commute, so $\pi (AB)=\pi (A)\pi (B)$, while $\det \pi (A)=(\det A{)}^{d+1}=1$. The map fixes $SO(d)$ and has kernel $\{I,-I\}$. Since $-I$ is central and has determinant $-1$ in odd dimension, $O(d)\cong SO(d)\times C_2$. In even dimension the same expression does not map every orthogonal matrix into the special orthogonal group.

The Euclidean group $E(n)=O(n)⋉{\mathbb{R}}^n$ is the group of all isometries $x\mapsto Ax+b$ of Euclidean space, with $A$ an orthogonal matrix. Its orientation-preserving subgroup is $SE(n)=SO(n)⋉{\mathbb{R}}^n$, the semidirect product of the special orthogonal group with translations. The Euclidean motion of Euclidean three-space and the proper Euclidean motion of Euclidean three-space are the respective three-dimensional instances.

For $\mathrm{Spin}(2m)$, $m\ge 3$, the two complex half-spin irreducible representations have dimension $2^{m-1}$. Their weights are $\frac{1}{2}(\pm e_1\pm \cdots \pm e_m)$ with respectively even and odd numbers of minus signs, after a chirality convention is chosen; every weight multiplicity is one. A model is the even or odd part of the exterior algebra of ${\mathbb{C}}^m$, where an occupied mode reverses the corresponding sign. Each is a minuscule representation. These are representations of the spin group and its Lie algebra, but do not descend through the double covering to the special orthogonal group.

In a Hilbert-Schmidt orthonormal operator basis, the generator $-i[H,\cdot ]$ is represented on the traceless Hermitian subspace by a real skew-symmetric matrix. Its propagator is therefore in the special orthogonal group and preserves the purity of a density operator. For $N>2$, these rotations form the adjoint image of the accessible unitary group, generally a proper subgroup of the full Euclidean rotation group; preservation of the full spectrum of the density operator imposes further constraints.

On an oriented even-dimensional Riemannian manifold of strictly positive sectional curvature, a nonconstant closed geodesic has a length-decreasing smooth variation. Its parallel transport fixes the tangent and acts on the odd-dimensional normal space by a special orthogonal group element. An odd-dimensional special orthogonal transformation has a fixed vector, so there is a nonzero periodic parallel normal field $V$. The second variation of geodesic energy is then strictly negative. The Cauchy-Schwarz inequality converts lower energy into strictly lower length because the original geodesic has constant speed. For an embedded geodesic the small variation is a smooth isotopy; otherwise it is a deformation through immersed loops.

The action $(A,B):M\mapsto AM{B}^{-1}$ preserves the quadratic form $-\det M$ on real two-by-two matrices. Its kernel consists of the simultaneous scalar pairs $(I,I)$ and $(-I,-I)$. The differential $(a,b):M\mapsto aM-Mb$ is injective on the two traceless Lie algebras, and both source and target have dimension six. Its image is therefore open and, since the source is connected, equals the identity component of a Lie group $S{O}_0(2,2)$. This gives a two-sheeted covering map onto the identity component, not onto the disconnected full special orthogonal group $SO(2,2)$.

A Lie group is a group that is also a finite-dimensional smooth manifold, with multiplication $G\times G\to G$ and inversion $G\to G$ both smooth maps. Its tangent space at the identity element carries a natural Lie algebra structure. Smoothness lets us differentiate the group operation; being a group means that every transformation can be composed and undone. The dimension of a manifold counts independent local parameters, rather than the number of elements of the group. See MIT 18.745, Lecture 2, Definition 2.1 and Example 2.5.

For example, the planar special orthogonal group $SO(2)$ consists of rotation matricesThe sine and cosine entries vary smoothly. Four entries appear, but they are constrained by $R_{\theta }^T{R}_{\theta }=I$ and $\det R_{\theta }=1$: locally, one angle specifies the whole matrix. Multiplication adds angles, with angles differing by $2\pi$ representing the same rotation. Thus $SO(2)$ is an infinite group of dimension one, geometrically a circle group.

The positive-dimensional examples motivate the language of continuous symmetry, but the definition also admits finite groups with the discrete topology as zero-dimensional Lie groups. Conversely, being an infinite group alone says nothing about having a continuous angle parameter. The compatibility between smooth maps and the group operation is what makes differentiation useful.

All the special orthogonal groups are connected, and their second homotopy groups vanish. The rotation circle gives $SO(2)\cong S^1$. Conjugation by unit quaternions gives the twofold universal cover $S^3\to SO(3)$, so ${\pi }_{1}SO(3)=\mathbb{Z}/2$ and ${\pi }_{3}SO(3)=\mathbb{Z}$. The special orthogonal sphere fibration $SO(n-1)\to SO(n)\to S^{n-1}$ propagates the first two homotopy groups for $n\ge 4$. In dimension four its exact sequence has a zero map from the torsion group ${\pi }_4{S}^3=\mathbb{Z}/2$ to ${\pi }_{3}SO(3)=\mathbb{Z}$, giving ${\pi }_{3}SO(4)={\mathbb{Z}}^2$. For $SO(1)$ all positive groups vanish, and ${\pi }_{3}SO(2)=0$.

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.

Expand the matrix trace explicitly:Write ${\Phi }_{ij}=(x_{ij}+i{y}_{ij})/\sqrt{2}$ and collect the $2N^2$ real fields into a vector $X$. ThenBoth quadratic forms are preserved by every constant orthogonal matrix $R$, acting as $X\mapsto RX$. Thus the full linear internal symmetry contains $O(2N^2)$, with connected special orthogonal group $SO(2N^2)$ as requested. The enlargement beyond unitary transformations is possible because the real and imaginary components have identical kinetic and mass terms and no interactions distinguishing them.

For a Hermitian matrix $\Phi$, there are $N$ real diagonal entries and $N(N-1)/2$ independent complex off-diagonal entries, hence $N^2$ real fields. Choose an orthonormal basis $T_A$ of Hermitian matrices for the trace inner product, $\mathrm{tr}(T_A{T}_B)={\delta }_{AB}$, and put $\Phi ={\sum }_A{X}_A{T}_A/\sqrt{2}$. The same calculation giveswhere these are the connected rotation groups; their full orthogonal extensions are $O(2N^2)$ and $O(N^2)$. A purely quadratic theory with $m^2<0$ would have an unstable scalar potential, but this algebraic symmetry statement still holds.

Use the convention $A_{ij}=⟨{\alpha }_i,{\alpha }_j^{\vee }⟩$, so the rows of the supplied Cartan matrix are the simple roots in Dynkin label coordinates. Realize the B2 root system in an orthonormal plane byThus the first simple root is long and the second is short. The Dynkin diagram has two nodes joined by a double bond, with its arrow pointing to the short node:A Euclidean weight $w=x{e}_1+y{e}_2$ has Dynkin labelsIn particular the fundamental weights are ${\omega }_1=e_1$ and ${\omega }_2=(e_1+e_2)/2$. This long-first ordering is essential: sources that put the short simple root first interchange the two Dynkin labels.

The highest weight $(0,1)$ is the spinor representation. Its Euclidean weights are $(\pm \frac{1}{2},\pm \frac{1}{2})$, whose Dynkin labels areA lowering chain subtracts ${\alpha }_2=(-1,2)$, then ${\alpha }_1=(2,-2)$, then ${\alpha }_2$ again, passing through these four weights in the displayed order. All weight multiplicities are one.

The highest weight $(1,0)$ is the defining vector representation, with Euclidean weights $e_1,e_2,0,-e_2,-e_1$. ThusHere a lowering chain subtracts ${\alpha }_1,{\alpha }_2,{\alpha }_2,{\alpha }_1$. Each weight multiplicity is again one.

Finally $(0,2)$ is the highest root $e_1+e_2={\alpha }_1+2{\alpha }_2$, giving the Adjoint representation. Its nonzero weights are the eight roots $\pm e_1,\pm e_2,\pm (e_1-e_2),\pm (e_1+e_2)$. The Cartan subalgebra supplies zero with weight multiplicity two, soThe dimensions are $4,5,10$. Independently the long-first Weyl dimension formula iswhich gives those same values. These are Lie algebra representations; the four-dimensional spinor representation integrates to the Spin group $\mathrm{Spin}(5)$, and does not descend to the special orthogonal group $SO(5)$. A concrete obstruction is a $2\pi$ rotation: it is the identity in $SO(5)$ but acts as $-I_4$ on the spinor representation, as verified in solution.

Write $x=(x_1,x_2,x_3{)}^T$, $v=(x_4,x_5,x_6{)}^T$ and $z=x+iv$. Here products of differentials in a metric tensor are symmetric products, unlike the wedge products used later. ThenIndeed $d{z}_{k}d{\bar{z}}_k=d{x}_k^2+d{x}_{k+3}^2$. A transformation $z^{\prime }=Uz$ preserves this metric whenever $U^{†}U=I$, becauseThus the entire unitary group $U(3)$ preserves it, including its special unitary group $SU(3)$.

To obtain a real matrix embedding, write $U=A+iB$ with $A,B$ real. ThenThe unitarity equations imply $A^{T}A+B^{T}B=I$ and $A^{T}B=B^{T}A$, hence $\rho (U{)}^T\rho (U)=I$. Multiplication of the block matrices gives $\rho (UV)=\rho (U)\rho (V)$, and the blocks recover $U$, proving injectivity. FinallyFor this last identity, over the complex numbers the change of basis from $(x,v)$ to $(z,\bar{z})$ conjugates $\rho (U)$ to $\mathrm{diag}(U,\bar{U})$. This proves that the image lies in the special orthogonal group.

Let ${\rho }_3(P)=P\otimes P\otimes P$ act on the alternating coefficient tensor $\phi$, and setThe coefficient convention here is the one in the source's hint; using the usual covariant pullback gives the same stabilizer subgroup, since an invertible transformation preserves a differential form if and only if its inverse does. The identity matrix fixes $\phi$. If $P,Q$ fix it, then ${\rho }_3(PQ)\phi ={\rho }_3(P){\rho }_3(Q)\phi =\phi$. Multiplying ${\rho }_3(P)\phi =\phi$ by ${\rho }_3(P^{-1})$ shows that $P^{-1}$ fixes it as well. Thus $H$ is a subgroup of $SO(7)$. Its defining equations are polynomial, so it is closed and hence a Matrix Lie group.

For $P(t)=I+tT+O(t^2)\in H$, differentiating $P^{T}P=I$ gives $T^T=-T$. Differentiating the three factors in ${\rho }_3(P(t))\phi =\phi$ givesThe skew-symmetric matrix condition records the ambient special orthogonal group; it must be included when differentiating that group's definition.

These linear conditions are sufficient as well as necessary. On the tensor product, the infinitesimal action isIts three terms commute, so ${\rho }_3(e^{tT})=e^{tD(T)}$. If $D(T)\phi =0$, every positive power of $D(T)$ annihilates $\phi$, giving $e^{tD(T)}\phi =\phi$. Also $T\in \mathfrak{so}(7)$ implies $e^{tT}\in SO(7)$. Therefore $e^{tT}\in H$, proving the asserted Lie algebra description. For the given positive three-form this is the compact G2 Lie algebra ${\mathfrak{g}}_2$.

For a curve $O(t)=I+tX+O(t^2)$ in the special orthogonal group, differentiate $O(t{)}^{T}O(t)=I$ at the identity. This gives $X^T+X=0$. Conversely, the matrix exponential of any real skew-symmetric matrix is orthogonal and has determinant $\det e^{tX}=e^{t\mathrm{tr}X}=1$. HenceA basis is $L_{ij}=E_{ij}-E_{ji}$ for $1\le i<j\le N$, where $E_{ij}$ is a matrix unit. If $L_{ji}=-L_{ij}$ and $L_{ii}=0$, their Lie brackets areThere are $\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ independent off-diagonal entries. A maximal torus rotates independently in mutually orthogonal two-planes, generated by $L_{12},L_{34},\dots$. ThusThe maximality follows by simultaneously decomposing commuting real skew-symmetric matrices into rotation two-planes, with at most one remaining fixed axis.

The unbroken group is the stabilizer subgroup of the vacuum expectation value. For a nonzero vector vacuum expectation value, an orthogonal transformation puts $⟨\varphi ⟩=v{e}_N$. A special orthogonal transformation fixing it has the block form $\mathrm{diag}(A,1)$ with $A\in \mathrm{SO}(N-1)$. ConsequentlyThe zero vacuum expectation value leaves the entire special orthogonal group unbroken.

For a real symmetric tensor write its vacuum expectation value as a real symmetric matrix $Q$. The transformation is $Q\mapsto OQ{O}^T$. The spectral theorem for real symmetric matrices gives an orthonormal basis in whichBecause $OQ{O}^T=Q$ is equivalent to $OQ=QO$, an unbroken transformation preserves each eigenspace. The complete unbroken group is thereforeIts identity component is $H^0={\prod }_i\mathrm{SO}(m_i)$. This symmetric tensor vacuum stabilizer shows why the representation alone does not determine a unique breaking pattern:

For example a rank-one tensor $Q=v{e}_N{e}_N^T$ preserves $O(N-1)$ embedded by $A\mapsto \mathrm{diag}(A,\det A)$. Its continuous unbroken group is $\mathrm{SO}(N-1)$, but it also allows reversal of the distinguished axis. Restricting to a traceless symmetric tensor imposes ${\sum }_i{m}_i{q}_i=0$ and otherwise leaves the same classification.

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.

A maximal torus in a compact connected Lie group is a compact connected abelian Lie subgroup maximal among such subgroups. By the connected abelian classification it is isomorphic to a product of circles. Such subgroups exist: choose one of largest dimension among compact connected abelian subgroups; a proper inclusion of connected Lie subgroups would increase dimension. Write $\mathfrak{t}$ for its Lie algebra.

Here are the main steps of the conjugacy assertion, with the critical arguments supplied. Average an inner product on $\mathfrak{g}$ over the adjoint action to obtain an $\mathrm{Ad}(G)$-invariant positive definite inner product. It defines a bi-invariant Riemannian metric. Compactness makes this metric complete, and the Hopf-Rinow theorem joins the identity to any $g$ by a geodesic. For a bi-invariant metric the Levi-Civita connection on left-invariant fields is ${\nabla }_{X}Y=[X,Y]/2$, so the geodesic starting with $X$ is $\exp (tX)$. Consequently every $g$ has the form $\exp X$.

The Lie-algebra centralizer of $\mathfrak{t}$ is exactly $\mathfrak{t}$. If $X$ commutes with $\mathfrak{t}$, its one-parameter subgroup commutes with $T$. The closure of the subgroup generated by $T$ and $\exp (\mathbb{R}X)$ is compact, connected and abelian, so it is a torus; maximality forces it to equal $T$, giving $X\in \mathfrak{t}$. The torus action on ${\mathfrak{g}}_{\mathbb{C}}$ decomposes into finitely many character weight spaces. The zero-weight space is ${\mathfrak{t}}_{\mathbb{C}}$. Choose $H\in \mathfrak{t}$ outside the finitely many kernels of the nonzero weight differentials. Then the centralizer of $H$ is exactly $\mathfrak{t}$.

For an arbitrary $X\in \mathfrak{g}$, the continuous function $g\mapsto ⟨{\mathrm{Ad}}_{g}X,H⟩$ attains a maximum. Put $Y={\mathrm{Ad}}_{g_0}X$ at a maximizer. Differentiating along $\exp (sZ)g_0$ givesHence $[Y,H]=0$, so $Y\in \mathfrak{t}$. Exponentiating gives $g_0(\exp X)g_0^{-1}=\exp Y\in T$. This adjoint-orbit criterion for maximal-torus conjugacy proves

Restriction of representations is a ring homomorphism $\mathrm{Res}:R(G)\to R(T)$. If a virtual representation restricts to zero, its character vanishes on $T$. Since characters are class functions and every element is conjugate into $T$, it vanishes on all of $G$. The injectivity of the character map proved in Question 3 then givesThe Weyl group is $W=N_G(T)/T$. For $n\in N_G(T)$, the operator $\rho (n)$ intertwines the restriction of $\rho$ with its twist by $t\mapsto nt{n}^{-1}$. Therefore every restricted representation, and every virtual difference of them, is fixed by $W$:

For the special orthogonal group $G=SO(2m+1)$, takewhere $R_{\theta }$ is a real planar rotation. It is a maximal torus: its Lie-algebra centralizer consists of the independent infinitesimal rotations in those planes. Its character lattice is ${\mathbb{Z}}^m$, so, writing $z_i=e^{i{\theta }_i}$,Indeed commuting unitary torus matrices diagonalize simultaneously into one-dimensional characters, and the characters of the circle are exactly the integral powers.

The normalizer permutes the real rotation planes and can reverse each angle independently. Permutations of whole two-dimensional blocks have determinant one. A reflection in one rotation plane reverses its angle and has determinant minus one; combining it with reflection of the remaining fixed line gives an element of $SO(2m+1)$. Thus all independent inversions occur. Conversely a normalizer must permute the distinct nontrivial torus character pairs $\{\pm e_i\}$ and preserve the fixed line, so there are no additional actions. The kernel of this action is exactly $T$, giving

It follows that the restriction identifies $R(SO(2m+1))$ with a subring of the signed-permutation-invariant Laurent polynomials. In this case we can prove equality. Invariance under individual inversions first givesFor one variable, every invariant is an integral combination of $1$ and $z^k+z^{-k}$, and these satisfy the recurrence $s_{k+1}=x{s}_k-s_{k-1}$. Applying this one coordinate at a time proves the assertion over the other Laurent coefficient rings. Permutation invariance and the Fundamental theorem of symmetric polynomials then givewhere $e_k$ are the elementary symmetric polynomials.

Let $V$ be the complexification of the defining $(2m+1)$-dimensional real representation. Its torus weights are $1,z_i,z_i^{-1}$. The characters of its exterior powers obeyFor $k\le m$, choosing $j$ linear terms and $r$ quadratic terms shows that the coefficient $c_k={\chi }_{{\Lambda }^{k}V}$ isIts leading term is $e_k$ with coefficient one; all other terms use $e_j$ with $j<k$. By induction $e_k$ lies in the restriction image, because $c_k$ is the character of a genuine representation and all smaller $e_j$ already lie in the image. Hence every invariant polynomial lies there, proving the representation ring of an odd-dimensional special orthogonal group descriptionThe triangular change of generators is integral, so $[{\Lambda }^{1}V],\dots ,[{\Lambda }^{m}V]$ are algebraically independent polynomial generators. For $m=1$, $c_1=1+z+z^{-1}$ and this reduces to $R(SO(3))=\mathbb{Z}[V_3]$, with the three-dimensional defining representation as generator. These are genuine representations of the orthogonal group, not the half-integral spin weights of its simply connected double cover.

The complex Special orthogonal Lie algebra ${\mathfrak{so}}_5$ consists of endomorphisms preserving a nondegenerate symmetric bilinear form infinitesimally:In an orthonormal basis this is $\{X\in M_5(\mathbb{C}):X^T+X=0\}$, with the commutator bracket and dimension $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$. Its compact real form is obtained by taking real skew-symmetric matrices. For diagonal root calculations it is more convenient to use an isotropic basis $v_1,v_2,v_0,v_{-2},v_{-1}$ with $B(v_i,v_j)={\delta }_{i,-j}$, including $B(v_0,v_0)=1$.

A maximal torus of the complex special orthogonal group isIts Lie algebra is the Cartan subalgebraIn the compact real form the corresponding torus consists of independent rotations in two orthogonal coordinate planes, fixing the remaining direction. Define ${\epsilon }_i(H)=h_i$ for $i=1,2$, and ${\epsilon }_{-i}=-{\epsilon }_i$, ${\epsilon }_0=0$.

Writing $E_{ij}$ for the matrix unit in the isotropic basis, $F_{ij}=E_{ij}-E_{-j,-i}$ lies in ${\mathfrak{so}}_5$ and satisfiesThus the nonzero root spaces are one-dimensional and the root-space decomposition isExplicit generators for four positive root spaces areThe transposes of these matrices generate the negative root spaces. Each transpose still satisfies the bilinear-form condition and reverses the displayed diagonal weight. These eight root vectors and the two independent Cartan matrices exhaust the dimension ten; a matrix commuting with every $H$ is diagonal and the form condition makes it lie in $\mathfrak{h}$. This also proves that the torus has maximal dimension.