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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.

Let $H$ be the tautological bundle on ${\mathbb{CP}}^1\cong S^2$. One sign convention for the Bott element iswhere $[1]$ is the rank-one trivial vector bundle. It has rank zero, and its Chern character has degree-two term $c_1(H)$, a generator up to sign. Choosing the dual line bundle reverses that sign.

The $n$-fold reduced external product in topological K-theory gives a generator ofThis product uses the smash product of different sphere factors. It is distinct from taking exterior powers of a vector bundle over one fixed base. Multiplication by the Bott element realizes Bott periodicity.^[ref]

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 top exterior power of an $n$-dimensional group representation is one-dimensional, and $g$ acts on it by $\det R(g)$. It is trivial for the defining representation of $SU(n)$. This is why a full $n$-box column in an $SU(n)$ Young diagram can be removed without changing the representation.

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 exterior cube of a vector space $V$ over a field is its third exterior power ${\Lambda }^{3}V$. It is generated by products $u\wedge v\wedge w$, multilinear in the three vectors and zero whenever two inputs coincide. Every alternating trilinear map out of $V^3$ factors uniquely through a linear map from ${\Lambda }^{3}V$.

If $e_1,\dots ,e_n$ is a basis, the products $e_i\wedge e_j\wedge e_k$ for $i<j<k$ form a basis, so $\dim {\Lambda }^{3}V=\left(\genfrac{}{}{0ex}{}{n}{3}\right)$. In particular it vanishes for $n<3$ and is one-dimensional for $n=3$.

In an oriented three-dimensional coordinate space,Thus the exterior product packages oriented volume; swapping two vectors reverses its sign. The alternating condition, including vanishing repeated inputs, remains the definition even in characteristic two, where minus and plus signs agree.

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.

Multilinear algebra studies multilinear maps, which are linear in each input separately while the other inputs stay fixed. Bilinear forms, determinants, and tensor operations are basic examples.

For instance, $B(x,y)=x^{T}Ay$ is bilinear, but $x\mapsto B(x,x)$ is a quadratic form rather than a linear map. The tensor product replaces multilinear dependence by linear dependence on a larger space: each bilinear map $V\times W\to U$ factors uniquely through a linear map $V\otimes W\to U$. Symmetric powers and exterior powers then impose symmetry or alternation, organizing polynomial and oriented-volume calculations in a coordinate-independent way.

For the special unitary group $SU(N)$, denote a one-column Young tableau with $k$ empty boxes by ${\mathcal{C}}_k$. Explicitly the columns needed here areA column ${\mathcal{C}}_k$ is the exterior power ${\Lambda }^{k}V$, with highest weight ${\omega }_k$ and dimension $\left(\genfrac{}{}{0ex}{}{N}{k}\right)$. Contraction with the invariant volume tensor identifies $V^{\ast }\cong {\Lambda }^{N-1}V$ and ${\Lambda }^2{V}^{\ast }\cong {\Lambda }^{N-2}V$. This gives all the requested Dynkin labels, Young tableaux, dimensions and tensors:Here $1\le I,J\le 6$ and $1\le i,j\le 5$. Lower indices denote the dual representation, equivalent to the conjugate representation for these compact groups. The SU(6) $15$ in this question is an exterior square; the SU(5) $15$ in solution is a symmetric square. The ambient group is essential to distinguish them.

For each complex coordinate, exchanging the factors of a wedge product of differential forms reverses its sign, givingConsequently the Kähler form is $\omega =d{x}_{14}+d{x}_{25}+d{x}_{36}$, and $\omega \wedge d{x}_7=d{x}_{147}+d{x}_{257}+d{x}_{367}$. Expanding the holomorphic volume form givesThe sign of $d{x}_{426}$ changes on exchanging $4,2$, while reordering $4,5,3$ to $3,4,5$ requires two exchanges. This verifies every sign inUnder $z\mapsto Uz$, unitarity preserves $\omega$:The determinant is the factor by which a transformation acts on a top exterior power, so ${\Omega }^{\prime }=(\det U)\Omega =\Omega$ for $U\in SU(3)$. Extend the real embedding by fixing $y$. It preserves $\omega$, $dy$ and $\Omega$, and therefore $\varphi$. ThusThe induced seven-dimensional representation is exactly the restriction already computed in c. In the conventional complex notation its branching rule isOver the real numbers, the six-dimensional summand is the underlying real space of the complex fundamental representation, and the last real line is the fixed $y$ axis.

Let index $0$ denote the first coordinate, which the embedded $SU(5)$ fixes, and let $a,b=1,\dots ,5$ denote the remaining coordinates. The fundamental representation and its conjugate representation immediately giveFor a symmetric tensor, $S^{00}$ is a trivial representation, $S^{0a}=S^{a0}$ is a $5$, and $S^{ab}$ is the fifteen-dimensional symmetric square of a group representation of $5$. For an antisymmetric tensor, $A^{00}=0$, $A^{0a}=-A^{a0}$ is a $5$, and $A^{ab}$ is the ten-dimensional exterior power ${\Lambda }^{2}5$. HenceHere the left-hand $15$ is the antisymmetric tensor of $SU(6)$, while the right-hand $15$ in the first equation is the symmetric tensor of $SU(5)$; equal dimensions alone do not identify representations of different groups.

Finally decompose the traceless block matrixThe traceless part of $D$ is the Adjoint representation $24$, the column $c$ is $5$, the row $b$ is $\bar{5}$, and the one remaining diagonal trivial representation is generated by $\mathrm{diag}(-5,1,1,1,1,1)$. ThusAll dimensions match: $6=1+5$, $21=15+5+1$, $15=10+5$ and $35=24+5+5+1$.

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.