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A bivector is an element of the second exterior power ${\Lambda }^{2}V$. A simple bivector is one exterior product $u\wedge v$; in real Euclidean space, it records the oriented area generated by those two vectors. A general bivector is a sum of such elements and need not describe just one plane. Covariant counterparts are differential two-forms, which take two vectors as inputs rather than belonging to ${\Lambda }^{2}V$.

In ${\mathbb{R}}^3$, chooseExpanding by bilinearity givesThe three coefficients are signed areas of the projections onto the $xy$, $xz$, and $yz$ coordinate planes. They are three components of one bivector, not the three coordinates of either input vector.

With the usual dot product and a fixed orientation of a vector space, the cross product represents this area element by the normal vector $(6,-3,1)$. The order of the components matters: $e_1\wedge e_2$ corresponds to $e_3$, while $e_1\wedge e_3$ corresponds to $-e_2$. The Gram determinant checks the full area:Here the norm on bivectors is the one induced by the Euclidean inner product. The exterior product exists without that metric and in arbitrary dimension; the familiar cross product description relies on the additional three-dimensional geometry.

For eigenvalues ${\lambda }_1,\dots ,{\lambda }_d$ of a finite-order linear action, its symmetric power characters satisfy ${\sum }_{n\ge 0}{\chi }_{S^{n}V}{t}^n={\prod }_j(1-{\lambda }_{j}t{)}^{-1}$. The denominator is the alternating generating polynomial of its exterior power characters.

A connection on a vector bundle induces a connection on its top exterior power by differentiating each factor in turn. Its local connection one-form is $\mathrm{Tr}A$ and its curvature form of a connection is $\mathrm{Tr}{\Theta }_D$.

The top exterior power of a rank-$r$ vector bundle is its determinant line bundle. A short exact sequence of vector bundles gives a canonical isomorphism of the middle determinant with the tensor product of the other two. Locally, a subbundle frame followed by lifts of a quotient frame defines it; changing lifts only adds off-diagonal blocks. This is the determinant step in adjunction for a smooth submanifold.

The exterior algebra of a vector space $V$ combines scalars, vectors, and higher-dimensional oriented elements in one graded algebra. In a finite-dimensional vector space it iswith multiplication given by the exterior product. Degree zero is the scalar field, degree one is $V$, and degree two is the space of bivectors. Multiplication adds degrees: combining two vectors produces a bivector, rather than another ordinary vector.^[ref]

The essential rule is $v\wedge v=0$. Repeating the same direction cannot produce a nonzero oriented area. Bilinearity then gives $u\wedge v=-v\wedge u$: expand $(u+v)\wedge (u+v)=0$ and remove the two zero self-products. The alternating rule remains part of the definition in characteristic two, where sign reversal alone would not force a self-product to vanish.

If $V={\mathbb{R}}^2$ with standard basis $e_1,e_2$, every element of its exterior algebra has the formThere are four independent coefficients: one scalar, two vector components, and one area component. In ${\mathbb{R}}^3$, the respective degree dimensions are $1,3,3,1$, giving eight components in total. More generally, the exterior power basis gives total dimension ${\sum }_{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)=2^n$.

This language is useful when direction and orientation matter together. Determinants measure the action on top-degree exterior powers; differential forms use exterior powers of a dual space to describe quantities integrated along paths, surfaces, and higher-dimensional regions. The Grassmann algebra viewpoint also makes the sign changes in anticommuting generators systematic. A single multiplication rule connects geometry and algebra.

The $k$th exterior power ${\Lambda }^{k}V$ is the quotient vector space of the tensor power $V^{\otimes k}$ that makes every elementary tensor with two equal factors zero. More precisely, take the quotient vector space by the linear span of all such tensors. Its elementary products $v_1\wedge \cdots \wedge v_k$ turn alternating multilinear maps into linear maps: every alternating $k$-linear map on $V$ factors uniquely through ${\Lambda }^{k}V$.^[ref]

If $e_1,\dots ,e_n$ is an ordered basis, the increasing productsform a basis of ${\Lambda }^{k}V$. There is one for each $k$-element choice of basis directions, soRepeated directions vanish; products with a different order can be sorted using their sign. Coordinate determinants distinguish the remaining increasing products, establishing linear independence.^[ref]

For $V={\mathbb{R}}^3$, a basis of ${\Lambda }^{2}V$ is $e_1\wedge e_2,e_1\wedge e_3,e_2\wedge e_3$. This explains the three bivector components in the worked example. The top exterior power ${\Lambda }^{3}V$ has one basis element, $e_1\wedge e_2\wedge e_3$, while ${\Lambda }^{4}V=0$: four independent directions cannot fit into that vector space. By convention ${\Lambda }^{0}V$ is the scalar field, even when $V$ has dimension zero.

A Lie algebra representation on $V$ induces one on each exterior power by acting on every factor and summing. The tensor product of Lie algebra representations preserves the defining alternating relations, so this descends to the quotient in every characteristic of a field. In characteristic two define the exterior algebra by the relations $v\wedge v=0$, rather than by dividing a tensor antisymmetrizer by $2$.

For a rank-$k$ smooth vector bundle, take the exterior power ${\Lambda }^r{E}_p$ in each fibre. A local frame gives a trivialization with fibre ${\Lambda }^r{\mathbb{R}}^k$, and a transition matrix $A$ acts by ${\Lambda }^{r}A$. Functoriality preserves the cocycle identities, producing a smooth vector bundle of rank $\left(\genfrac{}{}{0ex}{}{k}{r}\right)$, or the zero bundle when $r>k$. Its top exterior power is the determinant line bundle.

Take $U$ to be the kernel of a $t$-row Vandermonde matrix on distinct coordinates. For each $(r+t)$-set $T$, $U$ intersects its coordinate subspace in dimension $r$. Wedge a basis of that intersection: its coefficient on every $r$-subset of $T$ is nonzero, because projection onto that subset is an isomorphism by independence of the complementary $t$ columns. Modulo the exterior power of $U$, this gives a dependence involving every clique edge with nonzero coefficient. Strong saturation expresses every missing-edge vector in the span of present ones. Since all coordinate-edge vectors span the quotient, edge count is at least its dimension.

On the torus ${\mathbb{R}}^d/{\mathbb{Z}}^d$, an integer matrix $A$ defines an endomorphism. The coordinate circles give a basis of the first homology group, and the induced map on that group is $A$. The Künneth theorem identifies higher integral homology with the exterior powers of ${\mathbb{Z}}^d$, so the degree-$q$ map is ${\bigwedge }^{q}A$. In top degree it is multiplication by $\det A$. The map is a homeomorphism precisely when $A$ is a unimodular matrix; a merely nonsingular integer matrix need not have an integer inverse.

Assume $0\le k\le n$. Let $V=L_{n-1}$ be the $n$-dimensional irreducible sl2 Lie algebra representation. Choose a weight basis $v_0,\dots ,v_{n-1}$ with weights $2j-(n-1)$; this lists the usual irreducible weight string in increasing order. Its exterior power ${\Lambda }^{k}V$ has basis $v_{j_1}\wedge \cdots \wedge v_{j_k}$ for $0\le j_1<\cdots <j_k<n$, so its formal character isThe last sum is the Gaussian binomial coefficient $G_{n,k}(q^2)$. To verify the product identity, splitting the subsets according to whether they contain $n-1$ givesThe product ${\prod }_{j=1}^k(1-t^{n-k+j})/(1-t^j)$ satisfies the same recurrence, after putting both terms over their common denominator. Thus induction identifies it with the subset sum, proving it is a polynomial with nonnegative integer coefficients.

The quantum integer convention in the PDF hasMultiplying these factors in the factorial quotient givesThis establishes the required representation-theoretic interpretation.

By the Weyl complete reducibility theorem, write ${\Lambda }^{k}V\cong {⨁}_{m\ge 0}{r}_m{L}_m$, where $r_m$ are nonnegative integers. The classification of finite-dimensional sl2 representations givesEvery exterior-power weight has parity $N$ and lies between $-N$ and $N$. Therefore only $m\le N$ with $m\equiv N(\mathrm{mod}2)$ occur. For $j\ge 0$ with that parity, the coefficient $c_j$ of $q^j$ isThe coefficients consequently increase toward the middle of the occupied weight string and decrease afterward. If $G_{n,k}(t)={\sum }_{s=0}^N{a}_s{t}^s$, then $a_s=c_{-N+2s}$. HenceThis is the unimodality of Gaussian binomial coefficients, proved through parity unimodality of an sl2 character. It includes $k=0,n$, when the character is one.

There is a normalization qualification in the printed wording. The symmetric quantum-binomial quotient is a Laurent polynomial in $q$, with exponents spaced by two; the unimodality just proved is for that occupied parity, or equivalently for the ordinary polynomial $G_{n,k}(t)$ after setting $t=q^2$ and removing the monomial shift. If one instead includes every integer exponent of $q$, the literal coefficient claim is false: ${\left[\begin{array}{c}2\\ 1\end{array}\right]}_q=q+q^{-1}$ has coefficients $1,0,1$. The representation-theoretic conclusion is unimodality on the parity string, and ordinary unimodality of the Gaussian polynomial.

Let $H$ be the $r$-uniform hypergraph and use strong saturation of a uniform hypergraph in its immediate-completion sense: each missing edge $e$ lies in an $(r+t)$-set $T$ whose other $r$-subsets are already edges of $H$. The argument allows existing hypergraph cliques. Take $n\ge r+t$, the usual parameter range; if $r\le n<r+t$, strong saturation forces every possible edge to be present, and the inequality is immediate whenever its binomial terms are defined. The case $t=0$ has the trivial lower bound zero.

For $t\ge 1$, choose distinct real numbers $z_1,\dots ,z_n$ and form the $t\times n$ Vandermonde matrix $M$ whose $i$th column is $(1,z_i,\dots ,z_i^{t-1}{)}^T$. Every $t$ columns are linearly independent by the Vandermonde determinant. Set $U=\ker M\subseteq {\mathbb{R}}^n$, so $\dim U=n-t$, and work in the quotient vector spaceHere the inclusion of ${\bigwedge }^{r}U$ is injective: extend a basis of $U$ to a basis of ${\mathbb{R}}^n$ and use the resulting wedge bases of the exterior powers. Write $w_e$ for the image in $W$ of the coordinate wedge indexed by an $r$-set $e$. All these images span $W$.

For any $(r+t)$-set $T$, let $V_T$ be its coordinate subspace. The restriction of $M$ has rank $t$, so $U\cap V_T$ has dimension $r$. Wedge a basis of this intersection to obtain a nonzero elementEvery coefficient $c_e$ is nonzero. Indeed, projection of $U\cap V_T$ onto the $e$ coordinates is an isomorphism: a vector in its kernel is supported on $T\setminus e$, whose $t$ columns of $M$ are independent. The coefficient is the determinant of that projection in the chosen bases. Passing to $W$ therefore gives a linear dependence

If $e$ is missing from $H$, take its strong-saturation witness $T$. The dependence expresses $w_e$ in the span of the $w_f$ for the other edges $f\subseteq T$, all of which are present. Thus the vectors indexed by the present edges span all of $W$. The exterior-power quotient bound for strong hypergraph saturation follows:The bound is sharp: include precisely the $r$-sets meeting one fixed $t$-set $S$. A missing edge $e$ is disjoint from $S$, and every other $r$-subset of $e\cup S$ meets $S$, so adding $e$ immediately completes a hypergraph clique of order $r+t$. The number of included edges is exactly the displayed bound.

When the three light-quark masses are equal, the mass term is proportional to the identity in flavour space, so rotating $q=(u,d,s)$ by SU(3) leaves it unchanged. The gluon coupling is also flavour blind. Nearly equal masses therefore give an approximate flavour symmetry, distinct from the exact colour gauge symmetry. The quark transforms as $3$ and the antiquark as $\overline{3}$.

For mesons, a quark-antiquark tensor is a $3\times 3$ matrix. Its trace and traceless parts are invariant subspaces, givingThe trace is the eta singlet state; the traceless part is the meson octet. For baryons, first separate a pair of quarks into symmetric and antisymmetric tensors:The antisymmetric pair transforms as $\overline{3}$ by contraction with the invariant epsilon tensor. In $6\otimes 3$, complete symmetrization gives ${\mathrm{Sym}}^{3}3$ of dimension $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$; the remaining mixed-symmetry subspace has dimension eight. In $\overline{3}\otimes 3$, trace and traceless parts give a singlet and another octet. Thus the tensor cube of the defining SU(3) representation isThe two octets carry the two mixed permutation-symmetry components. Their weight diagrams are identical; the extra copy is a multiplicity, not a new arrangement of weights.

The printed charge relation fixes the half-hypercharge convention, rather than the usual Flavor hypercharge normalization:The quark weights $(I_3,Y)$ are $u:(1/2,1/6)$, $d:(-1/2,1/6)$, $s:(0,-1/3)$; adding three weights gives the baryon weight diagrams in the half-hypercharge convention. The baryon octet has $(p,n)$ at $Y=1/2$, the isotriplet $({\Sigma }^{+},{\Sigma }^0,{\Sigma }^{-})$ and the isosinglet ${\Lambda }^0$ at $Y=0$, and $({\Xi }^0,{\Xi }^{-})$ at $Y=-1/2$. The baryon decuplet has rows $\Delta$ at $Y=1/2$, ${\Sigma }^{\ast }$ at zero, ${\Xi }^{\ast }$ at $-1/2$, and ${\Omega }^{-}$ at $-1$. The three-quark flavour singlet is the antisymmetric $uds$ state at $(0,0)$.

The fully symmetric flavour representation is precisely the baryon decuplet, $10={\mathrm{Sym}}^{3}3$. For a spatial ground state, the orbital wavefunction is symmetric. Spin $3/2$ is the symmetric part of three spin-$1/2$ quarks, so a spin-$3/2$ decuplet would have a completely symmetric space-spin-flavour wavefunction. In particular the $uuu$ state with all spins aligned has no antisymmetric factor. This conflicts with the Pauli exclusion principle for the three fermions.

An additional colour degree of freedom resolves the conflict. With three colour labels the normalized colour state isIt is completely antisymmetric and an SU(3) colour singlet. Multiplying it by the symmetric remaining wavefunction gives the required fermionic antisymmetry. Three orthogonal labels are the minimum needed for a nonzero third exterior power: ${\bigwedge }^3{\mathbb{C}}^m=0$ for $m<3$. Thus the decuplet suggests at least three colour states, and three supply the minimal colour-singlet construction used in the quark model. This argument motivates three colours rather than deriving their exact number without further physical input.

The Pauli constraint on three-quark flavour multiplets also explains the absence of a flavour-singlet baryon in the lowest spatial multiplet. After extracting the antisymmetric colour factor, spin-flavour must be symmetric. Singlet flavour is completely antisymmetric, so its spin factor would also have to be completely antisymmetric. Butso three spin-$1/2$ quarks have no such spin state. By contrast, mixed flavour and mixed spin can combine symmetrically to produce the ground-state baryon octet with spin $1/2$. The two abstract flavour-octet copies do not imply two independent ground-state octets after the full Pauli constraint on three-quark flavour multiplets is applied. Excited orbital wavefunctions can permit three-quark flavour singlet; the exclusion is for the spatially symmetric low-energy ground multiplet, not for every baryon at every energy.

A flavour-singlet meson is allowed: a quark and an antiquark are not three identical fermions, and the invariant flavour combination $(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$ can combine with an appropriate spin and colour singlet. The missing three-spin antisymmetric state therefore causes no meson exclusion.

Finally, ${\pi }^{+}p$ has baryon number one, strangeness zero and $I_3=3/2$, so it is a pure $I=3/2$ channel. These quantum numbers match the $uuu$ Delta baryon ${\Delta }^{++}$, the low-energy resonance with spin $3/2$. For $K^{+}p$, the kaon $u\bar{s}$ contributes strangeness plus one, giving total $B=1$, $S=+1$. No ordinary $qqq$ baryon can have positive strangeness, since every $s$ quark contributes minus one. Hence that channel has no matching state in the ordinary low-energy three-quark multiplets. A resonance would require exotic quark content, for example $uuud\bar{s}$, and is not mandated by this spectrum. This is strangeness selection in pion-proton and kaon-proton resonances.

Put $V={\mathbb{C}}^n$. We use complete reducibility of compact-group representations, and the highest-weight classification: the multiplicity of $V_{\lambda }$ in a completely reducible representation is the dimension of its weight-$\lambda$ subspace killed by all positive root vectors. Each irreducible summand contributes just its one-dimensional highest-weight line to that subspace.

Symmetrizing and antisymmetrizing the three tensor factors give invariant subspaces ${\mathrm{Sym}}^{3}V$ and ${\Lambda }^{3}V$ of $V^{\otimes 3}$. The vectors $e_1^{\otimes 3}$ and $e_1\wedge e_2\wedge e_3$ are highest-weight vectors of weights $(3,0,\dots ,0)$ and $(1,1,1,0,\dots ,0)$ respectively. The unitary version of the Weyl dimension formula isIt givesThese are exactly the dimensions of the third symmetric power and exterior power, so each is irreducible and occurs once in those subspaces.

The weight $(2,1,0,\dots ,0)$ in $V^{\otimes 3}$ has basisThe only positive root vector that acts nontrivially on this weight space is $E_{12}$. Acting on the tensor product as the sum of its actions on the three factors, it sends each of $a,b,c$ to $e_1^{\otimes 3}$. Consequently the highest-weight vectors in this space are preciselyThis is a two-dimensional space, so $V_{(2,1)}$ occurs exactly twice. Its dimension isThe dimensions of the summands already found add toThus no other irreducible summand can remain. We have proved the full decompositionThere are four summands counted with multiplicity and three distinct isomorphism types.

For $n=2$, the exterior cube vanishes. The third symmetric power still has highest weight $(3,0)$ and dimension $4$, while the same two-dimensional highest-vector calculation gives two copies of the representation of weight $(2,1)$ and dimension $2$. It is $\det \otimes V$, since a determinant twist shifts the standard highest weight $(1,0)$ to $(2,1)$. Hence

The loop group here is $LU(1)=C^{\infty }(S^1,U(1))$, with pointwise multiplication. Although it is abelian, its nontrivial positive-energy quantum representations are projective: implementers can commute only up to a phase. A positive energy representation of the loop group of U(1) has rotations $R_t$ whose self-adjoint generator is bounded below and which implement $g(z)\mapsto g(e^{it}z)$ in the projective unitary group. The fermionic construction gives both that phase and an explicit bosonic description.

Choose the one-particle Hilbert space $H=L^2(S^1)$ with modes $e_r(z)=z^r$, $r\in \mathbb{Z}$, and projection $P$ onto $r\ge 0$. Use the polarized fermionic Fock spacewhere the exterior powers are Hilbert-completed and the tensor product is graded. Its vacuum fills all modes $r<0$ and leaves $r\ge 0$ empty; excitations are particles in the empty modes and holes in the filled modes. The fermionic annihilation operators $c_r$ and fermionic creation operators $c_r^{\ast }$ satisfy the canonical anticommutation relationsFinite particle-hole configurations form an orthonormal basis. As in the finite-dimensional exterior algebra construction, their creation and annihilation operators act irreducibly: an operator in the commutant sends the vacuum to a scalar vacuum, since the joint kernel of its vacuum annihilators is one-dimensional, and then acts by that scalar on the dense configuration basis.

Segal's quantization criterion for fermions says that a unitary one-particle operator $u$ has a unitary implementer $\Gamma (u)$ on this Fock space, satisfying $\Gamma (u)c^{\ast }(f)\Gamma (u{)}^{\ast }=c^{\ast }(uf)$, exactly whenEquivalently $P-uP{u}^{\ast }$ is Hilbert-Schmidt. Implementers are unique up to phase, hence form a projective unitary representation of the restricted unitary group. This is the fermionic, polarized criterion; it is not a condition that the entire operator $u-I$ be Hilbert-Schmidt. The criterion and the corresponding continuity topology are stated in Theorem III.2.1.1 of the fermionic construction of loop-group representations. For the compact principal-angle part of a pair of polarizations, the threshold has a concrete interpretation. Its two-dimensional blocks are rotations through angles ${\theta }_j$; finitely many unmatched modes can be handled separately. Each angle ${\theta }_j$ rotates the vacuum into $\cos {\theta }_j$ times its original component plus $\sin {\theta }_j$ times a particle-hole pair. After discarding finitely many exceptional blocks, the successive tensor products are Cauchy precisely when ${\prod }_j\cos {\theta }_j$ converges to a nonzero value, equivalently ${\sum }_j{\sin }^2{\theta }_j<\infty$. In this block model the sum is exactly half the squared Hilbert-Schmidt norm of the projection difference; finitely many unmatched modes account for a change of charge. Infinitely many unmatched modes already violate the criterion. This explains why infinitely many arbitrarily small mixings can fail to be implementable.

For $g(z)={\sum }_n{g}_n{z}^n\in LU(1)$, multiplication $M_g$ is a unitary one-particle operator. Its matrix entry from $e_s$ to $e_r$ is $g_{r-s}$. ThereforeFor a fixed difference $n$, exactly $|n|$ pairs cross the polarization boundary. Smoothness makes the sum finite, so Segal's quantization criterion for fermions constructs a continuous projective unitary representation of every smooth loop. Smooth convergence gives operator-norm convergence of $M_g$ and Hilbert-Schmidt convergence of its commutator, which is sufficient for the strong projective continuity of the implementers.

Normal ordering fixes the charge and rotation energy. On the dense finite-configuration domain putThe sums are finite on each configuration. A particle at $r\ge 0$ adds charge one and energy $r$; a hole at $r<0$ adds charge minus one and energy $-r$. Consequently $L_0$ is a nonnegative self-adjoint diagonal operator with integer spectrum and finite-dimensional eigenspaces. The unitaries $R_t=e^{it{L}_0}$ are strongly continuous, have period $2\pi$, and implement $e_r\mapsto e^{itr}{e}_r$. Conjugating multiplication by this one-particle rotation gives $M_{g(e^{it}\cdot )}$, so their Fock implementers rotate the loop representation projectively. This is an actual positive-energy circle action.

The scalar anomaly is visible in the Heisenberg current algebra of a fermion on the circle. Define on finite-energy vectorsFor fixed $n$ the sum has only finitely many nonzero actions on any configuration. Direct use of the canonical anticommutation relations givesOn summing, the two noncentral sums cancel. If $m=-n$ the central sum is ${\sum }_r(1_{r<0}-1_{r+n<0})=n$: for $n>0$ the terms are precisely $r=-n,\dots ,-1$. ThusIn particular $J_n\Omega =0$ for $n>0$. The coefficient one in the central term is the level of this basic representation.

Let $h(\theta )={\sum }_n{h}_n{e}^{in\theta }$ be smooth and real. The normal-ordered infinitesimal implementer of multiplication by $h$ isThe sign of the current index matters: multiplication by $z^n$ raises the one-particle mode by $n$, so it corresponds to $J_{-n}$. It follows either directly from the bilinear commutator or from the fermionic currents on the circle that $[A(h),c_r^{\ast }]={\sum }_n{h}_n{c}_{r+n}^{\ast }$. For trigonometric $h$ these are oscillator fields on the finite-energy domain; their self-adjoint closures exponentiate to the implementers of $e^{ih}$. For general smooth $h$, their extensions are obtained by the oscillator-field construction below: ${\sum }_{n>0}n|h_n{|}^2<\infty$ guarantees the square-summable field coefficients, and Fourier truncation converges in the corresponding Weyl relations implementers. Thus no unjustified infinite operator product is required.

SetIt is real and antisymmetric for real $h,k$. The Heisenberg current algebra of a fermion on the circle relations give $[A(h),A(k)]=i\omega (h,k)I$. The Weyl relations, or the finite-mode Baker-Campbell-Hausdorff computation followed by the square-summable limit, yieldThis is an explicit two-cocycle on the identity component of the loop group. A different phase choice changes it by a group coboundary, but its antisymmetric commutator remains $e^{-i\omega (h,k)}$, and hence cannot be removed.

The winding components also matter. Every smooth loop is $z^w{e}^{ih}$ with $w\in \mathbb{Z}$, real periodic $h$, and mean $h_0$ determined modulo $2\pi$. Let $S$ implement multiplication by $z$. It moves the filled boundary one place upwards, so it raises charge by one; phases of charge ground states can be chosen so that $S{\Omega }_q={\Omega }_{q+1}$. Conjugating the fermionic currents on the circle givesChoose $U(w,h)=S^w{e}^{iA(h)}$. Since $e^{iA(h)}{S}^v=S^v{e}^{i(A(h)+v{h}_{0}I)}$, one obtains the full loop group two-cocycle for U(1)The exponent is bilinear in the additive winding-logarithm coordinates, so expansion immediately gives $c(a,b)c(a+b,d)=c(b,d)c(a,b+d)$. Replacing $h$ by $h+2\pi \ell$ changes $A(h)$ by $2\pi \ell Q$, whose exponential is identity since charges are integers; the scalar cocycle also stays unchanged since $v\in \mathbb{Z}$. Thus the expression is well-defined on loops, not just on chosen logarithms. It records both the derivative anomaly and the commutator of winding with constant loops.

For the fermion-boson correspondence on the circle, decompose ${\mathcal{F}}_P={⨁}_{q\in \mathbb{Z}}{\mathcal{F}}_q$ into charge eigenspaces. The charge-$q$ ground state ${\Omega }_q$ fills all integer modes $r<q$. Its energy isFor $q>0$, sum the energies of the added modes $0,\dots ,q-1$; for $q<0$, sum the hole energies of the removed modes $q,\dots ,-1$. Every other charge-$q$ configuration is obtained by moving finitely many occupied levels upwards, so this is the lowest energy. Since $J_n$ preserves charge and lowers energy by $n>0$, $J_n{\Omega }_q=0$ for $n>0$.

Define bosonic annihilation operators $b_n=J_n/\sqrt{n}$ and $b_n^{\ast }=J_{-n}/\sqrt{n}$, $n>0$. They obey $[b_n,b_m^{\ast }]={\delta }_{nm}I$. Their occupation vectorsare orthonormal by commuting positive fermionic currents on the circle through negative fermionic currents on the circle. Their energies are $q(q-1)/2+{\sum }_{n>0}n{m}_n$. It remains to prove that they span, rather than assume the boson correspondence. Count configurations by formal charge $u$ and energy $t$. Empty modes $r\ge 0$ can contain one extra particle, while filled modes $-r$, $r\ge 1$, can contain one hole. Therefore the fermionic character isThe charge form of the Jacobi triple product proved in Question 4 identifies this withThe reciprocal product counts bosonic occupations, or partitions of an integer. For fixed charge and energy, the finite-dimensional fermion space has exactly as many basis configurations as the independent fermionic current on the circle monomials already constructed. Hence those monomials exhaust every such space, proving their dense completeness. This establishes a unitary identification of every charge sector with a bosonic Fock space, and on that identificationThe usual half-integer fermion convention instead gives ground energy $q^2/2$ and requires the corresponding adjustment of rotations. Here integer modes were deliberately chosen to obtain a genuine circle action with integral nonnegative energies. The two ground states of energy zero, at charges zero and one, are therefore consistent, not a negative-energy defect.

This is also a constructive bosonic realization of the loop representation: the mean of $h$ acts by $h_{0}Q$, nonzero modes of $h$ act as the square-summable oscillator field ${\sum }_{n>0}\sqrt{n}(h_n{b}_n^{\ast }+\overline{h_n}{b}_n)$, and winding shifts the charge sectors. Taking $k$ tensor copies adds the fermionic current on the circle central terms and gives level $k\in \mathbb{N}$. Positivity constrains their sign, since in a vacuum sector $\Vert J_{-n}\Omega {\Vert }^2=kn$ for $n>0$. The level-one construction, its explicit loop cocycle, and its complete bosonic occupation basis show how an abelian classical loop group acquires noncommuting quantum currents while retaining positive rotation energy.

Apply the lowering operator $F$ to the highest-weight vector from part (ii). Since $F{w}_{03}=4w_{13}+w_{04}$ and $F{w}_{12}=2w_{13}$, we obtainThis is nonzero because $w_{04},w_{13}$ are distinct elements of the exterior power basis. Both have weight zero. As a consistency check, the raising operator sends this vector to $2v_{+}$, as required by the sl2 highest-weight lowering formula.

Take a basis $x,y$ of the defining representation of the sl2 Lie algebra, with $Hx=x$, $Hy=-y$, $Ey=x$, $Ex=0$, $Fx=y$ and $Fy=0$. A monomial basis of the symmetric power ${\mathrm{Sym}}^{4}V$ isFor the exterior power $U={\bigwedge }^2{\mathrm{Sym}}^{4}V$, put $w_{ij}=v_i\wedge v_j$ for $i<j$. These ten elements form a basis, and their weights of a representation add:Thus the complete weight-space decomposition, with explicit bases, isThe weight diagram therefore has multiplicities $1,1,2,2,2,1,1$ at weights $6,4,2,0,-2,-4,-6$, respectively.

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.