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Every two-qubit gate has the displayed form with local elements of the special unitary group and real interaction parameters. The three Pauli-product generators commute, so the central exponential is the product of three interactions. Local basis changes turn an Ising coupling into either of the other Pauli-product couplings. A fixed entangling controlled phase together with arbitrary local rotations can also synthesize each variable interaction, for example through two CNOT gates enclosing a variable $z$ rotation.

Identify the product with endomorphisms of the defining space. Special unitary group conjugation fixes scalar matrices and preserves the complementary traceless subspace. For $N\ge 2$, the latter is the complex Adjoint representation of a Lie algebra, which is irreducible because ${\mathfrak{sl}}_N$ is simple. With fundamental Dynkin index $1/2$, the trace-Casimir identity gives $C_2(\mathbf{N})=(N^2-1)/(2N)$; the tensor-product Casimir trace identity then gives $C_2(\mathbf{Adj})=N$.

An invariant tensor is fixed by every element of the represented group, so it spans or belongs to a trivial group representation. The special unitary group preserves its identity tensor and the upper and lower volume tensors. Such tensors make tensor contraction equivariant and permit invariant inner products and real structures.

Near the identity, the matrix logarithm of a unitary matrix is skew-Hermitian. Restrict the logarithm's operator norm below $\pi /n$; determinant one then forces its trace to be zero, since $\det U=\exp (\mathrm{tr}\log U)$ and no nonzero multiple of $2\pi i$ fits this bound. The matrix exponential gives the inverse chart on a ball in the displayed real Lie algebra, of dimension $n^2-1$. Left translation supplies compatible charts on the whole special unitary group.

A maximal torus is a connected compact abelian Lie subgroup maximal among such subgroups. Restricting a finite-dimensional unitary representation to it splits into one-dimensional weight characters. For special unitary group $\mathrm{SU}(2)$ it is the diagonal circle $\mathrm{diag}(z,z^{-1})$.

A Lie group is a group which is a finite-dimensional smooth Hausdorff second-countable manifold, with smooth multiplication and inversion. For the special unitary group, consider the real vector spaceThe diagonal entries of a skew-Hermitian matrix supply $n$ real parameters, and the entries above the diagonal supply $2\left(\genfrac{}{}{0ex}{}{n}{2}\right)$; the trace-zero condition removes one real parameter. Thus ${\dim }_{\mathbb{R}}\mathfrak{su}(n)=n^2-1$.

The matrix facts used to construct charts are these: the matrix exponential is analytic, its derivative at zero is the identity, and it has an analytic inverse matrix logarithm on a sufficiently small neighborhood of the identity. On those neighborhoods $\log (e^X)=X$, $e^{\log U}=U$, and $\det (e^X)=e^{\mathrm{tr}X}$. For a unitary matrix in that neighborhood, the spectral theorem writes $U=W\mathrm{diag}(e^{i{\theta }_j})W^{\ast }$ with principal angles close to zero, and $\log U=W\mathrm{diag}(i{\theta }_j)W^{\ast }$ is skew-Hermitian.

Choose a norm ball of radius $\epsilon <\pi /n$ small enough to lie in the exponential's inverse domain. If $U$ is unitary, $\det U=1$ and $\Vert \log U\Vert <\epsilon$ in operator norm, thenso its trace is zero. Conversely $X\in \mathfrak{su}(n)$ gives $(e^X{)}^{\ast }{e}^X=I$ and $\det (e^X)=1$. Thus the logarithm identifies an open neighborhood $V$ of $I$ in $SU(n)$, with its matrix subspace topology, with the open ball $B_{\epsilon }\subset \mathfrak{su}(n)$. Choosing a real linear identification $\mathfrak{su}(n)\cong {\mathbb{R}}^{n^2-1}$ turns this into a chart.

For every $A\in SU(n)$ take the translated chartTheir overlap maps are $X\mapsto \log (B^{-1}A{e}^X)$ wherever both charts are defined, so they are real analytic. They cover the group; the matrix topology is Hausdorff and second-countable. This constructs the real manifold directly, rather than merely counting equations. Matrix multiplication is polynomial in real/imaginary entries and inversion here is $U\mapsto U^{\ast }$; in the logarithm charts both are smooth. The logarithm charts for the special unitary group therefore proveFor $n=1$ it is the one-point group of dimension zero.

For $n=2$, let the first column be $(a,b)$, of unit norm. Every unit vector orthogonal to it is a scalar of modulus one times $(-\bar{b},\bar{a})$. The determinant of the resulting matrix is that scalar, so determinant one fixes it to $1$. Henceis bijective, with inverse given by taking the first column. The forward map is real polynomial in the four coordinates; composing with the local logarithm charts verifies smoothness into the constructed manifold. The inverse is a smooth matrix-coordinate projection and, in the usual embedded sphere charts, is smooth into $S^3$. Thus $\boxed{SU(2)\cong S^3\text{ by a diffeomorphism}}$, the explicit realization of SU(2) as the three-sphere.

Write $C_U=|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes U$ for the controlled unitary gate. For a normalized eigenvector $|u⟩$, the first Hadamard gate and quantum phase kickback giveThe final Hadamard gate therefore produces the unentangled outputThe auxiliary system is unchanged, while its eigenphase becomes accessible through interference of the control qubit. The special unitary group assumption is not needed for this calculation; unitarity and the normalized eigenvector suffice.

With determinant-one normalization, the necessary and sufficient condition for sphere rotations as special-unitary Möbius transformations isEquivalently, the matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ belongs to the special unitary group $SU(2)$. The two determinant-one representatives $M$ and $-M$ induce the same Möbius transformation, and both satisfy this condition.

Use north-pole stereographic projection in the direction from the plane to the sphere:Its chord-distance identity is $|\phi (z)-\phi (w)|=2\chi (z,w)$, where $\chi$ is the chordal metric. Direct subtraction, using $ad-bc=1$, givesThis is first calculated away from poles and then extended continuously to the Riemann sphere. A sphere rotation preserves chord distances, so for distinct finite nonpole pointsUsing three distinct points shows all these positive ratios equal one. Hence $H(z)=1+|z{|}^2$ everywhere. Comparing its coefficients givesThus $M$ is a unitary matrix. Comparing $M^{-1}=M^{\ast }$ with $M^{-1}=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$ proves the boxed relations.

Conversely, these relations give $H(z)=1+|z{|}^2$, so the induced sphere map $T=\phi g{\phi }^{-1}$ preserves all Euclidean chord distances. Such a bijection of the unit sphere is the restriction of an orthogonal matrix: let $v_j=T(e_j)$; preservation of distances makes these an orthonormal basis, and $T(p)\cdot v_j=p\cdot e_j$ forces $T(p)={\sum }_j{p}_j{v}_j$. Finally the Möbius transformation is holomorphic with positive real Jacobian at every ordinary coordinate point. Conjugating by stereographic projection preserves orientation, so this orthogonal matrix has determinant $+1$. It is therefore a genuine rotation in three dimensions, proving sufficiency rather than just spherical isometry.

For the circle assertion, a nondegenerate generalized circle has an equationFor $A\ne 0$ this is an ordinary circle, and for $A=0$ it is a line with infinity included. On the sphere write coordinates $(p_1,p_2,p_3)$, with $z=(p_1+i{p}_2)/(1-p_3)$ and $|z{|}^2=(1+p_3)/(1-p_3)$. Multiplication by $1-p_3$ gives the planeIf $n=(2\mathrm{Re}B,-2\mathrm{Im}B,A-C)$ and $h=-(A+C)$, then $|n{|}^2-h^2=4(|B{|}^2-AC)>0$. Thus the plane intersects the unit sphere in a proper circle. It contains the north pole exactly when $A=0$, accounting for the point at infinity on a line.

Conversely, a circle on the sphere is the intersection with a plane $n\cdot p=h$, $|h|<|n|$. Set $A=(n_3-h)/2$, $C=(-n_3-h)/2$ and $B=(n_1-i{n}_2)/2$. The preceding equation is recovered and has $|B{|}^2-AC=(|n{|}^2-h^2)/4>0$. Its inverse stereographic image is an ordinary circle or a line with infinity attached. This proves both directions of the circle-plane relation under stereographic projection.

The two unheaded requests are addressed here; the individual group-cover calculations are in the scoped parts below. A quaternion $q=a+bi+cj+dk$ has norm $|q{|}^2=a^2+b^2+c^2+d^2$. Thus the unit quaternions form exactly the sphere $S^3\subset {\mathbb{R}}^4$, with quaternion multiplication giving its smooth group structure. The explicit identification with the special unitary group $SU(2)$ isThe matrices representing $i,j,k$ satisfy the quaternion multiplication relations. The displayed matrix obeys $U^{†}U=|q{|}^{2}I$ and $\det U=|q{|}^2$; conversely every $SU(2)$ matrix has that form with $|q|=1$. Hence $S^3$ is the manifold of unit quaternions, and its group is $SU(2)$.

For the exterior-square double cover of SO(3,3), fix a basis $e_1,\dots ,e_4$ and volume element $v=e_1\wedge e_2\wedge e_3\wedge e_4$. Define the symmetric bilinear form $Q$ on the six-dimensional exterior square byIt is symmetric because two-forms commute under the wedge product. PutThen $Q(a_i,a_j)=Q(b_i,b_j)=0$ and $Q(a_i,b_j)={\delta }_{ij}$. The Gram matrix is $\left(\begin{array}{cc}0& I_3\\ I_3& 0\end{array}\right)$: the vectors $(a_i+b_i)/\sqrt{2}$ have norm $+1$, while $(a_i-b_i)/\sqrt{2}$ have norm $-1$. Therefore the quadratic form $\omega \mapsto \omega \wedge \omega$ has signature $(3,3)$.

For $A\in SL(4,\mathbb{R})$,Thus $\rho (A)={\Lambda }^{2}A$ preserves $Q$. The special linear group is connected: polar decomposition of an invertible real matrix writes $A=OP$ with $O\in SO(4)$ and $P$ positive definite of determinant one, and both factors can be joined to the identity. The image consequently lies in $S{O}_0(3,3)$.

If ${\Lambda }^{2}A=I$, then $Au\wedge Aw=u\wedge w$ for every pair, so $A$ preserves every two-plane. A line is an intersection of two such planes, hence every line is preserved. A linear map preserving every line is scalar: apply it to basis vectors and then to their pairwise sums. Thus $A=\lambda I$, and ${\Lambda }^{2}A=I$ forces ${\lambda }^2=1$. The kernel is exactly $\{\pm I\}$. It is discrete, so the derivative of $\rho$ is injective. Both Lie algebras have dimension $15$, making the image open; an open subgroup of a connected group is the whole group. HenceThe identity-component restriction is essential. The same principle of an explicit form-preserving representation, kernel calculation and dimension argument supplies the five remaining covers.

The special unitary group is $\mathrm{SU}(2)=\{g\in M_2(\mathbb{C}):g^{\ast }g=I,\det g=1\}$. The special orthogonal group is $\mathrm{SO}(3)=\{R\in M_3(\mathbb{R}):R^{T}R=I,\det R=1\}$. Conjugation $A\mapsto gA{g}^{-1}$ preserves trace and skew-Hermiticity: for unitary $g$, $(gA{g}^{-1}{)}^{\ast }=g{A}^{\ast }{g}^{-1}=-gA{g}^{-1}$. It therefore defines the representation on the real space $V$ described in the following parts. The bar over the transpose in the PDF is essential; the condition is $A+A^{\ast }=0$, not ordinary antisymmetry over the complex numbers.

An orthonormal basis for the pairing proved below isPut $J_j=\sqrt{2}{E}_j$, so $J_j^2=-I$. For any unit real vector $v$, $J(v)={\sum }_j{v}_j{J}_j$ also squares to $-I$. Thus $g=\cos (\theta /2)I+\sin (\theta /2)J(v)$ belongs to $\mathrm{SU}(2)$. Conjugation fixes $J(v)$ and rotates its perpendicular plane by angle $\theta$ up to the orientation convention of this basis, as follows either by multiplying the displayed matrices or from their anticommutator relation $J(v)J(w)+J(w)J(v)=-2(v\cdot w)I$. Since every three-dimensional proper rotation has an axis and an angle, this proves surjectivity onto $\mathrm{SO}(3)$.

If an element is in the kernel, it commutes with each $E_j$. Commutation with $E_3$ forces it to be diagonal; commutation with $E_1$ forces its two diagonal entries to agree. Thus it is a scalar matrix, and determinant one and unitarity give exactly $\{I,-I\}$. Therefore $\boxed{\mathrm{SO}(3)\cong \mathrm{SU}(2)/\{\pm I\}}$.

The finite-dimensional complex irreducible representations of $\mathrm{SU}(2)$ are ${\mathrm{Sym}}^m({\mathbb{C}}^2)$, $m\ge 0$, of dimension $m+1$. The central element $-I$ acts by $(-1{)}^m$, so precisely the even ones descend. The irreducibles of $\mathrm{SO}(3)$ are $\boxed{{\mathrm{Sym}}^{2\ell }({\mathbb{C}}^2),\ell =0,1,2,\dots }$, viewed as representations of the quotient, with dimensions $2\ell +1$.

Use the full definition of the special unitary group: $A^{†}A=I$ and $\det A=1$. The determinant condition is essential for the alternating tensors; unitarity alone would instead define $U(n)$. Assume $n\ge 2$ when discussing proper invariant subspaces.

Let $V={\mathbb{C}}^n$ be the fundamental representation. A $(j,k)$ tensor belongs to $V^{\otimes j}\otimes (V^{\ast }{)}^{\otimes k}$. Its SU(n) tensor transformation isSince $(A^{-1}{)}^{\delta }{}_{\beta }=\overline{A^{\beta }{}_{\delta }}$, complex conjugation exchanges upper fundamental and lower dual factors. Reordering the two index groups shows that $\overline{T}$ is a $(k,j)$ tensor.

The invariant tensors follow directly. The Kronecker delta transforms to $A^{\alpha }{}_{\gamma }(A^{-1}{)}^{\gamma }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. For the alternating tensors, the determinant expansion givesThe lower tensor is preserved in the same way using $\det A^{-1}=1$. Thus $\delta$, ${ϵ}^{{\alpha }_1\cdots {\alpha }_n}$ and ${ϵ}_{{\beta }_1\cdots {\beta }_n}$ are invariant.

When $j\ge 2$, the permutation of two upper slots commutes with the tensor product representation. Its $+1$ and $-1$ eigenspaces are nonzero proper invariant subspaces. The same argument uses two lower slots when $k\ge 2$. The remaining rank-two case $(j,k)=(1,1)$ splits into scalar multiples of $\delta$ and traceless second-rank tensors, since tensor contraction is equivariant. This covers every case except the fundamental, dual fundamental and scalar cases specified in the question. Those three are irreducible: the special unitary group acts transitively on unit vectors in its fundamental space, so a nonzero invariant subspace must be the full space, and the dual has the same property.

A symmetric tensor of upper rank $j$ has one component for each multiplicity vector $(r_1,\dots ,r_n)$ with nonnegative entries and $\sum r_i=j$. Equivalently it is a homogeneous degree-$j$ polynomial in $n$ variables. Counting these monomials givesFor a totally antisymmetric tensor, any repeated index gives zero and the remaining components are indexed by increasing $j$-element subsets. HenceThese are the dimensions of the symmetric power and exterior power, respectively.

For SU(4), the two-index antisymmetric representation has complex dimension six. To establish a six-dimensional real representation, one must exhibit a real structure; simply counting six complex components would give twelve real components. Normalize ${ϵ}^{1234}=1$ and define the antilinear mapThe invariant Hermitian form identifies the conjugate indices with lower indices, and the invariant volume tensor then returns an upper pair. Thus $\mathcal{J}$ commutes with SU(4). The identitygives ${\mathcal{J}}^2=1$. Its fixed space hasThree arbitrary complex entries specify all the others, giving exactly six real degrees of freedom. The real structure of the SU4 exterior square therefore provesThe ordinary Hermitian norm restricts to an invariant positive real inner product on this fixed space.

Irreducible representations classify states because an exact symmetry commutes with the Hamiltonian, preserving energy eigenspaces. Irreducible multiplets are the smallest sets closed under all symmetry transformations; their labels and Casimir operators provide quantum numbers. Approximate flavour symmetry gives approximate multiplets and mass relations, with splittings caused by its breaking. Spin and internal quantum numbers are distinct labels.

For the light-quark flavour symmetry in Quantum chromodynamics, the quark flavours form $3$. Their Triple tensor decomposition for the defining sl3 representation isThis algebraic decomposition alone does not determine the allowed ground-state baryons. The three-quark colour singlet is antisymmetric. The no-orbital-excitation ground state has a symmetric spatial wavefunction, so the Pauli exclusion principle requires the combined spin-flavour wavefunction to be symmetric. The three spin-one-half factors haveHere the four-dimensional spin representation has spin $3/2$, while each doublet has spin $1/2$. Symmetric decuplet flavour pairs with symmetric spin $3/2$. The two flavour-octet copies and two spin-doublet copies carry the two-dimensional standard permutation multiplicity space; its tensor square contains one symmetric singlet. Their invariant pairing therefore supplies one octet with spin $1/2$, not two independent ground-state octets. Antisymmetric flavour would need a completely antisymmetric three-quark spin state, which does not exist.

Equivalently the symmetric spin-flavour SU6 representation hasThus the Pauli constraint on three-quark flavour multiplets gives the concise spectrum classificationThese are the baryon octet and baryon decuplet. The three-quark flavour singlet requires a different spatial permutation symmetry and belongs to excited states, outside the symmetric ground-state approximation used here. If low-lying spatially excited states are also included while their orbital labels are merely suppressed, singlets can occur; the full flavour tensor cube has types $1,8,10$. The absence of the singlet from the unexcited ground multiplet is a permutation-symmetry constraint, not a prohibition by $SU(3)$ alone. Light-quark mass differences make the flavour symmetry approximate rather than exact.

For a Lie bracket over $\mathbb{R}$ or $\mathbb{C}$, antisymmetry of a Lie bracket means $[x,y]=-[y,x]$. In particular $[x,x]=0$ in these characteristic-zero fields. The Jacobi identity isIt expresses compatibility of the bracket with its own adjoint action. Bilinearity must hold over the chosen base field, and the bracket must take its values in the same vector space.

For the matrix commutator, bilinearity and antisymmetry follow directly from distributivity. Associativity of matrix multiplication givesadding the two cyclic permutations cancels every monomial. Thus the Jacobi identity holds in the entire matrix algebra. For a specified linear subspace, the only additional bracket condition is closure: $AB-BA$ must belong to the subspace whenever $A,B$ do. It is unnecessary to require closure under the separate products $AB$ and $BA$.

To determine the special unitary Lie algebra, let $g(t)$ be a differentiable curve in the special unitary group with $g(0)=I$ and $A=g^{\prime }(0)$. Differentiating $g(t{)}^{†}g(t)=I$ gives $A^{†}+A=0$. Differentiating $\det g(t)=1$ at the identity gives $\mathrm{tr}A=0$. Conversely a traceless skew-Hermitian matrix has $e^{tA}$ unitary and $\det e^{tA}=e^{t\mathrm{tr}A}=1$, so it really is a tangent vector. ThereforeThis is a real Lie algebra of complex matrices: multiplication by $i$ generally leaves this real subspace. Its complexification is ${\mathfrak{sl}}_n(\mathbb{C})$, not the compact algebra itself. For $A,B\in \mathfrak{su}(n)$,Thus closure holds, and the already verified commutator identities establish all the Lie algebra axioms.

The cross-product Lie algebra on ${\mathbb{R}}^3$ is bilinear, antisymmetric and closed because the cross product has those properties. Its Jacobi identity follows from the vector triple-product identity:In the cyclic sum, the coefficients cancel by symmetry of the scalar product.

For the explicit relation to the SU(2) Lie algebra, take the three Pauli matrices and defineThese matrices are traceless and Skew-Hermitian, and the three images of the standard basis form a real basis of $\mathfrak{su}(2)$. Using the Pauli matrix commutator identity,Hence $F$ is a real Lie algebra isomorphism. The factor and sign $-i/2$ are essential for preserving the unscaled cross-product bracket. At the group level there is an Adjoint double cover from SU(2) to SO(3); the isomorphism of their tangent algebras does not identify the two global groups.

Use Hermitian generators $T^a$ of the special unitary group, with $[T^a,T^b]=i{f}^{ab}{}_c{T}^c$. An adjoint field is the Lie-algebra-valued matrix $\psi ={\psi }^a{T}^a$, transforming as $\psi \mapsto U\psi U^{-1}$. The adjoint covariant derivative isIn components, $(D_{\mu }\psi {)}^a={\partial }_{\mu }{\psi }^a+g{f}^{abc}{A}_{\mu }^b{\psi }^c$. With $A_{\mu }\mapsto U{A}_{\mu }{U}^{-1}-(i/g)({\partial }_{\mu }U)U^{-1}$, direct substitution gives $D_{\mu }\psi \mapsto U(D_{\mu }\psi )U^{-1}$. This is covariance in the Adjoint representation of a Lie group. In the following BRST symmetry formulas, absorb the coupling into the connection, so $D_{\mu }={\partial }_{\mu }-i[A_{\mu },\cdot ]$.

The character of a representation is ${\chi }_V(g)=\mathrm{tr}{\rho }_V(g)$. Average any positive inner product over the compact group using Haar measure to obtain a unitary representation. On the maximal torus $\mathrm{diag}(z,z^{-1})$, the commuting unitary operators admit a simultaneous eigenbasis. Each eigenline carries a circle character of a representation $z^n$ by part (a). Hencewith finitely many nonzero coefficients. Every element of special unitary group is conjugate to an element of this torus, so this describes the full character of a representation via its torus restriction. Here the coefficient notation includes zero multiplicities.

In a manifold chart, write a smooth differential form as $\eta ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$, with $I$ increasing. Define its exterior derivative byTo check that this definition is global, change coordinates to $y$. The chain rule expresses $d{x}^i$ as ${\sum }_j({\partial }_j{x}^i)d{y}^j$. Differentiating it gives zero because the second partial derivatives of $x^i$ are symmetric and $d{y}^k\wedge d{y}^j$ is antisymmetric. The coordinate definition satisfies the graded Leibniz rule, so differentiating the transformed expression for $\eta$ leaves exactly ${\sum }_{I}d{a}_I\wedge d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. This agrees with the original formula and proves independence of the manifold chart. The same cancellation of mixed derivatives gives $d^2=0$.

The de Rham cohomology is thereforethe vector space of closed differential forms modulo exact differential forms.

Here is a Green-theorem proof of exactness on a disc. Identify the disc with a round disc centred at zero, and let $\eta =Pdx+Qdy$ be a closed differential form. Set $f(x)={\int }_{[0,x]}\eta$. For $x$ and $x+h$ in the disc, Green theorem on the oriented triangle with vertices $0,x,x+h$ gives zero boundary integral, since ${\partial }_{x}Q-{\partial }_{y}P=0$. HenceTaking $h=\epsilon v$ and dividing by $\epsilon$ yields $d{f}_x(v)={\eta }_x(v)$. The radial integral defining $f$ is smooth, so $\eta =df$ andDegenerate triangles follow by continuity. The same proof works on any star-shaped plane domain, including the whole plane.

Integration requires an orientation: take $M$ to be an oriented compact $n$-dimensional smooth manifold without boundary. The Generalized Stokes theorem says ${\int }_{M}d\beta =0$ for every smooth $(n-1)$-form $\beta$. Every $n$-form is closed by dimension, and changing it by an exact differential form leaves its integral unchanged. Thusis a well-defined linear map $H_{\mathrm{dR}}^n(M;\mathbb{R})\to \mathbb{R}$. Compactness alone does not supply the orientation omitted from the printed integration statement.

For an $n$-dimensional Lie group, choose a nonzero $\lambda \in {\Lambda }^n{T}_e^{\ast }G$ and define the invariant volume form on a Lie group byEach $d{L}_{g^{-1}}$ is an isomorphism, so $\omega$ is nowhere zero and defines an orientation. The composition identity $L_{(hg{)}^{-1}}\circ L_h=L_{g^{-1}}$ gives $L_h^{\ast }\omega =\omega$. Moreover, every left-invariant differential form of degree $n$ is determined by its value at $e$, so the space of such forms is one-dimensional.

Under the supplied invariant-representative assumption, all top-degree de Rham cohomology classes are multiples of $[\omega ]$. For compact $G$, orient it so $\omega$ is positive; then ${\int }_G\omega >0$, whereas the Generalized Stokes theorem makes every exact top form have integral zero. Therefore $[\omega ]\ne 0$ andunder that assumption. For actual compact Lie groups, this argument applies when $G$ is connected. The supplied assumption fails for disconnected groups: on $S^1\times \mathbb{Z}/2$, the two components have independent degree-one classes, while globally left-invariant top forms form only a one-dimensional space. In general the top-degree de Rham cohomology of a compact Lie group is ${\mathbb{R}}^c$, where $c$ is its number of connected components.

For $S^2$, use $U=S^2\setminus \{N\}$ and $V=S^2\setminus \{S\}$. Stereographic projection identifies each with the plane. The disc argument gives $\eta {|}_U=d{f}_U$ and $\eta {|}_V=d{f}_V$ for any closed differential one-form $\eta$. The overlap $U\cap V$ is connected, so $d(f_U-f_V)=0$ makes the difference a constant. Adjust one potential by that constant; they glue to a global smooth function $f$ with $df=\eta$. ThusFinally, $S^3$ is a compact connected Lie group: the identificationidentifies it smoothly with the special unitary group $SU(2)$. Matrix multiplication and inversion supply its smooth group operations. The preceding invariant-volume argument therefore gives

Use the Riemann curvature tensor convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. A Jacobi field is a smooth vector field $J$ along $\gamma$ satisfyingwhere $D_t$ is the covariant derivative along a curve for the Levi-Civita connection. Differentiating a geodesic variation gives this equation, and conversely every Jacobi field arises from a geodesic variation by varying its initial point and velocity.

Choose a parallel frame $E_1(t),\dots ,E_n(t)$ along $\gamma$ and write $J={\sum }_a{y}_a{E}_a$. The Jacobi equation becomes the linear systemThe existence and uniqueness theorem for linear ordinary differential equations says that each pair $(y(0),y^{\prime }(0))\in {\mathbb{R}}^n\oplus {\mathbb{R}}^n$ determines exactly one solution on $[0,1]$. Addition and scalar multiplication preserve the equation. Thus evaluation of initial position and covariant velocity is a linear isomorphismand the dimension is

For the convexity assertion, use the convex-normal meaning of a geodesically convex open set: its points are joined by a unique geodesic within the set, and the joining geodesic depends smoothly on its endpoints. Equivalently, the appropriate star-shaped restriction of ${\exp }_p$ is a diffeomorphism onto the set. The convex-normal-neighbourhood theorem supplies such sets around every point. Mere existence of some minimizing geodesic, without this uniqueness and normality, would not imply the assertion.

Let $p=\gamma (0)$ and $v=\dot{\gamma }(0)$. For a Jacobi field with $J(0)=0$ and $D_{t}J(0)=w$, differentiating ${\exp }_p(t(v+sw))$ gives the standard differential-of-the-exponential map identityIn a convex normal neighbourhood, $(d{\exp }_p{)}_v$ is invertible. Therefore a Jacobi field vanishing at both endpoints has $w=0$ and hence is identically zero. The difference of two fields with the same endpoint values consequently vanishes. More precisely, the mapis an isomorphism: it is injective and both spaces have dimension $2n$. The constant-geodesic case follows directly from $D_t^{2}J=0$. This uses the convex-normal-neighbourhood theorem, the differential-of-the-exponential map identity, and linear ordinary differential equation uniqueness.

For the special unitary group example, use Jacobi fields from conjugation at a central endpoint. WriteThe endpoints of $\gamma (t)=\exp (tA)$ are $I$ and $zI$. The latter is central. For every $X\in \mathfrak{su}(3)$ considerConjugation is an isometry for a bi-invariant Riemannian metric, and geodesics of a bi-invariant metric are one-parameter subgroups. Thus $F_X$ is a geodesic variation. Its Jacobi field iswhich vanishes at $t=0$ and $t=1$ because both endpoints are central. Its initial covariant derivative is $[X,A]$.

The linear map $X\mapsto J_X$ has kernel equal to the centralizer of an element of a Lie algebra of $A$. Indeed, $J_X\equiv 0$ implies $[X,A]=0$ on differentiating at $t=0$, and that commutation conversely implies $J_X\equiv 0$. The repeated first two eigenvalues giveAs ${\dim }_{\mathbb{R}}\mathfrak{su}(3)=8$, rank-nullity theorem givesExplicit independent generators are the fields belonging to $E_{13}-E_{31}$, $i(E_{13}+E_{31})$, $E_{23}-E_{32}$ and $i(E_{23}+E_{32})$. Their initial derivatives $[X,A]$ are independent because the eigenvalue differences in the $13$ and $23$ entries are $2\pi i$. This also demonstrates directly why the endpoint-value conclusion fails along this geodesic.

Orient both spheres in the standard way and normalize their area forms to total area $4\pi$. The degree of a map between oriented manifolds can be obtained in two ways. For a regular value $y$, use the degree as a sum of local degrees:Each inverse image is isolated by the inverse function theorem, and compactness makes the set finite. The determinant is computed in consistently oriented local coordinates. A second method is spherical degree by area pullback:where the last formula represents $g$ as a unit vector in ${\mathbb{R}}^3$. The pullback of a differential form already contains the signed Jacobian determinant; no extra $\sin \theta$ is to be inserted in that last coordinate expression.

To relate the methods, replace $\omega$ by a smooth top-degree differential form with the same total integral supported in a small neighbourhood of a regular value. Two such top-degree forms with equal integral differ by an exact differential form on $S^2$, by its top-degree de Rham cohomology. Their pullbacks therefore have the same integral by Stokes theorem. Over the chosen neighbourhood, $g$ splits into local inverse branches; the change of variables formula makes the contribution of each branch its orientation sign times $4\pi$. Their sum is precisely the first formula. Thus the area integral is an integer and agrees with the signed inverse-image count.

For a nonconstant rational map, first use common-factor reduction of a rational map so $p$ and $q$ are coprime. Write $k=\max (\mathrm{deg}p,\mathrm{deg}q)$ for these reduced polynomials. A generic finite target value $w$ has inverse images at the roots of $p-wq$: avoiding exceptional values makes its degree $k$ and its roots simple. The fundamental theorem of algebra supplies $k$ roots. A holomorphic map has positive real Jacobian determinant $|R^{\prime }{|}^2$ at a regular point, so every local sign is $+1$. HenceThe source leaves coprimality implicit. In an unreduced representation the answer is $\max (\mathrm{deg}p,\mathrm{deg}q)-\mathrm{deg}\gcd (p,q)$, including degree zero for a constant reduced map. For example $(z^2-1)/(z-1)$ extends to $z+1$ and has degree one, although the unreduced maximum degree is two. Exceptional inverse images at infinity or multiple roots do not change the degree of a rational map of the Riemann sphere.

For the rational map approximation for Skyrmions, use stereographic projection $z=\tan (\theta /2)e^{i\phi }$ and the unit target vectorCombine this rational map with a radial profile to form a special unitary group field:where $\mathbf{\sigma }$ are the Pauli matrices. The endpoint values make $U(0)=-1$ independent of angle and $U(\infty )=1$. Appropriate radial behaviour gives an admissible finite-energy field configuration. With $L_i=U^{†}{\partial }_{i}U$, choose the topological baryon number in the Skyrme model conventionSeparating the radial and angular factors givesThus the degree of a rational map of the Riemann sphere supplies the Skyrmion charge.

In conventional dimensionless massless Skyrme model units, its static energy reduces towith the angular Jacobian of a rational mapThe Cauchy-Schwarz inequality gives $\mathcal{I}\ge k^2$. These formulas follow from the radial strain $|f^{\prime }|$ and the two equal angular strains $\sin f\sqrt{J_R}/r$: the quadratic energy sums their squares and the quartic Skyrme term sums their pairwise products of squares. Minimize the angular integral in the rational map approximation over degree-$k$ maps, then minimize the remaining radial energy with the stated endpoints. This replaces a three-dimensional field minimization by finitely many map coefficients and an ordinary differential equation for $f$.