Special unitary group ($SU(n)$)

The special unitary group is $SU(n)=\{U\in U(n):\det U=1\}$.

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.

The four-by-four unitary matrices of determinant one form this special unitary group. Its two-index antisymmetric representation has complex dimension six and the real structure of the SU4 exterior square realizes it on a six-dimensional real space.

For $n\ge 2$, the special unitary group acts transitively on the complex unit sphere. Extend any unit vector to an orthonormal basis, then adjust the phase of a remaining column to give determinant one. The stabilizer subgroup of the first vector consists of special unitary transformations of its orthogonal complement. Thus the sphere is a homogeneous quotient, not just a set with a norm-preserving action.

The SU(3) group consists of unitary complex three-by-three matrices of determinant one. Its defining action on ${\mathbb{C}}^3$ preserves the Hermitian norm. It contains block-diagonal copies of SU(2) and the real SO(3) group. Its Lie algebra consists of traceless anti-Hermitian matrices.

The central element $\omega I$ of SU(3), with $\omega =e^{2\pi i/3}$, acts on an irreducible representation of Dynkin labels $(a,b)$ as ${\omega }^{a+2b}$. This residue is its triality. In a tensor realization the representation has $a+2b$ fundamental boxes, and a complete column of three is a singlet; removing such columns preserves the residue. Tensor products add trialities and complex conjugation reverses them.

A mixed tensor has $k$ defining indices and $l$ dual indices. The SU(3) action applies $U$ to every upper index and $U^{-1}$ to every lower index. Young symmetrizers select permutation types, while invariant Kronecker deltas take traces and invariant alternating tensors convert an antisymmetric pair of defining indices into a dual index, or conversely. Repeated fundamental and dual tensor rules decompose it with multiplicities: $(p,q)\otimes (1,0)=(p+1,q)\oplus (p-1,q+1)\oplus (p,q-1)$ and the conjugate rule, omitting negative labels.

For $T^{abc}$, split its first pair into symmetric and antisymmetric parts. The fully symmetric part $S^{abc}$ is a symmetric cubic representation of SU(3). Write $M^{abc}=T^{(ab)c}-S^{abc}$ and $O^a{}_d={ϵ}_{dbc}{M}^{abc}$. Then $O$ is traceless and $M^{abc}=({ϵ}^{bcd}{O}^a{}_d+{ϵ}^{acd}{O}^b{}_d)/3$. The antisymmetric part is $T^{[ab]c}={ϵ}^{abd}{N}^c{}_d$, with $N^c{}_d={ϵ}_{dab}{T}^{abc}/2$. Splitting $N$ into its traceless part and trace gives the second adjoint representation of SU(3) and the singlet. These maps use invariant tensors, so they commute with the group action and explicitly realize the four irreducible summands.

Symmetrize the $p$ upper and $q$ lower indices separately, and remove every upper-lower trace. The resulting representation has highest weight $p{\omega }_1+q{\omega }_2$. Its dimension isThe subtraction is valid because contraction is onto: in the polynomial realization its adjoint, for the monomial-factorial inner product, is multiplication by the nonzero polynomial ${\sum }_i{x}_i{y}_i$, an injective map. The resulting kernel contains the highest-weight constituent and has precisely its Weyl dimension formula dimension, hence is irreducible. This highest-weight constituent is not always the largest-dimensional constituent of the full mixed tensor.

The degree-three monomials $z_1^a{z}_2^b{z}_3^c$, with $a+b+c=3$, form a basis of this ten-dimensional highest-weight representation. Its highest monomial is $z_1^3$, and $E_{21}=z_2{\partial }_{z_1}$ and $E_{32}=z_3{\partial }_{z_2}$ generate every basis monomial. Its Dynkin labels are $(3,0)$, and the weight of a monomial is $(a-b,b-c)$. In coordinates $(h_1,(h_1+2h_2)/\sqrt{3})$, the ten weights form four horizontal rows of lengths $4,3,2,1$, all with weight multiplicity one. The baryon decuplet realizes this representation of approximate flavor symmetry.

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For $d(p,q)=(p+1)(q+1)(p+q+2)/2$, the exact largest constituent dimension of a mixed SU(3) tensor representation isEvery constituent highest weight lies below $k{\omega }_1+l{\omega }_2$, so its labels have the form $(p,q)=(k-2a+b,l+a-2b)$ for nonnegative integers $a,b$. If both are positive, reducing each by their minimum increases both $p,q$ and strictly increases the dimension. The resulting boundary labels occur: antisymmetrize disjoint upper pairs into lower indices, or lower pairs into upper indices, then take the symmetric traceless component. This proves the maximum formula. For $(k,l)=(5,0)$, the fully symmetric dimension is 21 but $(3,1)$ has dimension 24. For $k=l$, the fully symmetric traceless dimension $(k+1{)}^3$ is the maximum.

The SU(2) group is the group of two-by-two unitary matrices of determinant one. It is a compact three-dimensional Lie group whose underlying manifold is a three-sphere.

An SU(2) matrix is a two-by-two complex unitary matrix with determinant one. It has the form $\left(\begin{array}{cc}a& b\\ -\bar{b}& \bar{a}\end{array}\right)$ with $|a{|}^2+|b{|}^2=1$. Its action on the Riemann sphere is a rotational Möbius transformation; opposite matrices induce the same sphere rotation.

Every element of $SU(2)$ has the unique formThus $SU(2)$ is diffeomorphic to the unit three-sphere in ${\mathbb{C}}^2$ and is a parallelizable manifold.