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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 a connection on a vector bundle written $D=d+A\wedge$ in a local frame, its curvature form of a connection is represented by $\Theta =dA+A\wedge A$. The products combine matrix multiplication with the exterior product. The formula follows by squaring $D$ and using the graded Leibniz rule.

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 product, also called the wedge product, combines oriented elements through the multiplicationFor two vectors, it is bilinear and alternating:These rules imply $v\wedge u=-u\wedge v$. They distinguish the exterior product from both the symmetric dot product and the unconstrained tensor product.^[ref]

In an oriented orthonormal plane, write $u=(a,b)$ and $v=(c,d)$. Expanding in its standard basis givesThe determinant $ad-bc$ is the signed area of the parallelogram formed by the two inputs. Its absolute value is the ordinary area. Reversing the input order reverses the sign; parallel inputs produce zero.

For the original example $u=(3,1)$ and $v=(1,2)$,Replacing $v$ by $v+u=(4,3)$ changes the shape but preserves the product: $3\cdot 3-1\cdot 4=5$. Both parallelograms have the same oriented area element. If $v=2u$, the exterior product is zero, since $u\wedge (2u)=2(u\wedge u)=0$.

The same rule adds directions in higher dimensions. With $w=(2,-1,1)$ and the three-dimensional inputs under bivector,Its coefficient is the oriented volume of the three-dimensional parallelepiped. Repeating or linearly depending on earlier inputs collapses the volume, giving zero.

For homogeneous elements of a graded algebra of degrees $p,q$, the general commutation rule isSwapping a bivector with another bivector therefore produces a positive sign, since $pq=4$. It is the number of individual vector swaps that controls the sign, not a rule that every pair of objects must anticommute.

For the special unitary group $SU(N)$, put ${\epsilon }_i=e_i-N^{-1}{\sum }_j{e}_j$. Its defining representation has weight system $\{{\epsilon }_i\}$, and its exterior square has weight system $\{{\epsilon }_i+{\epsilon }_j:i<j\}$; each has multiplicity one. Their dual representations negate the weights of a representation. The Dynkin label coordinates are $({\epsilon }_i{)}_r={\delta }_{ir}-{\delta }_{i,r+1}$. These follow by evaluating diagonal traceless generators on coordinate vectors and their exterior products.

Under the standard embedding $SU(N-1)\subset SU(N)$, the last defining weight of a representation restricts to zero and the others become the defining weights of a representation of $SU(N-1)$. Splitting pairs according to whether they contain the last index proves the branching of a representation $V_N\downarrow =V_{N-1}\oplus 1$ and ${\Lambda }^2{V}_N\downarrow ={\Lambda }^2{V}_{N-1}\oplus V_{N-1}$, including multiplicities.

For the polynomial ring $A=k[X_1,\dots ,X_r]$ in characteristic zero, antisymmetrization identifies $H{H}^{\ast }(A,A)$ with ${\bigwedge }_A^{\ast }{\mathrm{Der}}_k(A)$. The map on a wedge of $p$ derivations is $\frac{1}{p!}{\sum }_{\sigma \in S_p}\mathrm{sgn}(\sigma ){\prod }_j{D}_{\sigma (j)}(a_j)$. The Hochschild cup product becomes the exterior product, and the left Gerstenhaber bracket becomes the left Schouten-Nijenhuis bracket. This assertion requires a smoothness hypothesis when generalized beyond polynomial rings.

An element of a single graded component $A_n$ of a graded algebra is homogeneous of degree $n$. The degree sign rules for an exterior product or a graded Lie bracket apply first to homogeneous elements, then extend by linearity.

On an oriented Riemannian manifold, the metric volume form is the volume form taking value one on positively oriented orthonormal frames. In an oriented manifold chart it has the displayed expression. Under an orientation-preserving coordinate change with Jacobian $A$, $\det g$ acquires the factor $(\det A{)}^2$, precisely matching the transformation of the top exterior product. Thus the expressions agree on chart overlaps. Orientation is needed to make a globally positive form, though the volume density exists without it.

Work in three-dimensional projective space, with four homogeneous coordinates indexed $0,1,2,3$. Let $A$ and $B$ be independent representatives of two points. The Plücker coordinates of their line areThis antisymmetric matrix has rank two and is defined only up to a nonzero scalar. Replacing the pair by another basis of the same two-dimensional subspace multiplies $L$ by the determinant of that basis change. Its six independent entries satisfy $L^{01}{L}^{23}-L^{02}{L}^{13}+L^{03}{L}^{12}=0$, since the exterior product $(A\wedge B)\wedge (A\wedge B)$ vanishes.

To describe the same line as an intersection of planes, choose a volume orientation with ${\epsilon }_{0123}=1$ and define the dual Plücker coordinates byThe independent components pair asConversely, $L^{ij}=\frac{1}{2}{\epsilon }^{ijkl}{L}_{kl}$ with the matching volume convention. The lower-index form annihilates both defining points: $L_{ij}{A}^j=L_{ij}{B}^j=0$. Its columns span the pencil of planes containing their line. This dualization uses the volume form, not a Euclidean metric for raising and lowering indices. Two-index point and plane representations are dual in projective dimension three; in higher dimensions the corresponding dual exterior form has a different degree.

For a general plane covector $F_i$, contract the upper-index form:This is a linear combination of $A,B$, so it is on the line, and $F_i{P}^i=0$ by antisymmetry, so it is on the plane. Provided the resulting vector is nonzero, its homogeneous coordinates give the unique intersection, including intersections at infinity. If $F(A)=F(B)=0$, the whole line lies in the plane and the zero result correctly indicates that there is no unique intersection. These are the line-plane incidence in Plücker coordinates operations used in the subparts.

A connection on a vector bundle $E\to M$ is a linear map $D:\Gamma (E)\to {\Omega }^1(M;E)$ with $D(fs)=df\otimes s+fDs$. Write a local frame as a row $e=(e_1,\dots ,e_r)$, and a section as $s=eu$ with a coefficient column $u$. If the connection one-form is the matrix $A={\sum }_i{A}_{i}d{x}^i$, thenThis fixes the matrix convention for all subsequent signs.

For an $E$-valued differential form of degree $k$, represented locally by a column $\alpha$ of ordinary forms, its covariant exterior derivative isThe defining graded Leibniz rule with a scalar form $\omega$ of degree $j$ isIt follows by moving the one-form $A$ past $\omega$, which introduces $(-1{)}^j$. It uniquely extends the connection on sections.

The induced endomorphism bundle connection is characterized on an endomorphism section $T$ by $(D^{\mathrm{End}}T)s=D(Ts)-T(Ds)$. For a matrix-valued form $\beta$ of degree $k$, the extension isIndeed it is exactly the formula that makesThe endomorphism-valued exterior product uses matrix composition as well as the ordinary exterior product, so factors cannot in general be interchanged.

Define the curvature form of a connection by $D^2$ acting on sections, or equivalently byIt is tensorial in both vector arguments and in $s$: the derivatives of multiplying functions cancel by the connection rule. For example $D^2(fs)=D(dfs+fDs)=f{D}^{2}s$, since the two $df\wedge Ds$ terms cancel and $d^{2}f=0$. Thus curvature is a two-form with values in the endomorphism bundle. Locally, applying the graded rule givessoFor a change of frame of a vector-bundle connection $e^{\prime }=eg$, the coefficients satisfy $u^{\prime }=g^{-1}u$, and differentiation givesSquaring yields $F^{\prime }=g^{-1}Fg$, precisely the transformation rule of an endomorphism-valued two-form. This independently proves that the local expressions define one global curvature form of a connection.

For the Bianchi identity, use the endomorphism formula with degree two:Every term cancels with its indicated sign. Thereforewhere $d_A$ here means the induced derivative on the endomorphism bundle.

For a line bundle, the endomorphism bundle is canonically the trivial scalar bundle: every fibre endomorphism is a scalar times the identity. Scalar multiplication commutes, so $A\wedge A=0$ and the endomorphism covariant derivative on scalar-valued forms reduces to $d$. The Bianchi identity consequently says $dF=0$.

If $D^{\prime }$ is another connection, their difference is $C^{\infty }$-linear in sections because the Leibniz terms cancel. Hence $D^{\prime }-D=\eta$ is a global endomorphism-valued one-form; for a line bundle it is an ordinary global scalar one-form. Locally $A^{\prime }=A+\eta$, soThus the two closed curvatures differ by an exact form, proving the connection-independent curvature class of a line bundle:No global trivialization of the line bundle itself was assumed; it is its endomorphism bundle that has the canonical scalar identification.

Since the Riemannian orthonormal frame is positively oriented, its dual coframe satisfies ${\omega }_g={\omega }_1\wedge \cdots \wedge {\omega }_n$. The form $\nu$ is the interior product ${\iota }_X{\omega }_g$. The contraction rule for an exterior product givesOne can also obtain the sign by moving $E_i$ from the first argument into slot $i$ of the alternating form, using $i-1$ transpositions. Linearity in $X={\sum }_i{f}_i{E}_i$ proves the interior contraction of the Riemannian volume form formula:The exponents $i-1$ and $i+1$ have the same parity.

The exterior algebra of a vector space $V$ is ${\Lambda }^{\ast }{V}^{\ast }={⨁}_{r=0}^{\dim V}{\Lambda }^r{V}^{\ast }$, whose degree-$r$ elements are alternating $r$-linear forms. Its exterior product satisfies $\alpha \wedge \beta =(-1{)}^{rs}\beta \wedge \alpha$ for degrees $r,s$. For example, one-forms anticommute, so $d{x}^i\wedge d{x}^i=0$. A differential form of degree $r$ on a smooth manifold is a smooth section of ${\Lambda }^r{T}^{\ast }M$; in coordinates it is an alternating sum of products of $r$ coordinate one-forms. Its exterior derivative increases degree by one, satisfies $d^2=0$, and obeysThe coordinate definition and pullback of a differential form are compatible, making differential forms suitable for coordinate-independent integration.

An orientation and a nondegenerate metric tensor define the Hodge star operator by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\mathrm{vol}}_g$ for equal-degree forms. It maps degree $r$ to degree $d-r$. If the metric has $s$ negative directions, its square is ${\ast }^2=(-1{)}^{r(d-r)+s}$. Thus Hodge duality turns antisymmetric tensors into complementary-degree tensors, using the metric and orientation rather than merely the underlying vector space. The Generalized Stokes theorem says ${\int }_{M}d\eta ={\int }_{\partial M}\eta$, with the induced boundary orientation, for appropriate smooth forms and compactness or support assumptions. It generalizes the fundamental theorem of calculus, circulation version of Stokes theorem, and divergence theorem.

For the field calculation, choose signature $(-,+,+)$ and a fixed orientation on Minkowski spacetime. Write $a=\delta A$, so $\delta F=da$. Symmetry of the metric pairing and the Hodge star operator gives $\delta (F\wedge \ast F)=2da\wedge \ast F$. Applying the exterior derivative product rule,Since $F\wedge a=a\wedge F$, variation of the action isFix the boundary data or use compactly supported variations. Arbitrary interior one-forms $a$ then imply the bulk field equationWith $B=\ast F$, the Lorentzian Hodge star operator squares to $-1$ on both one- and two-forms, hence $F=-\ast B$. Thus the Maxwell–Chern–Simons one-form equation becomesThese two displayed equations are equivalent. Also $dF=0$ gives $d\ast B=0$. For $k\ne 0$ the latter follows from applying $d$ to the field equation; for $k=0$ it must be retained as the identity inherited from $F=dA$. Using the opposite metric-sign convention changes the corresponding ${\ast }^2$ sign; the signature-independent equation is $d\ast F+kF=0$.

Under the Abelian gauge transformation $A\mapsto A+d\Lambda$, the curvature $F=dA$ and the one-form $B$ are unchanged because $d^2\Lambda =0$. The field equation is therefore exactly gauge-invariant. The kinetic term is unchanged, while the other term changes byThe last equality uses $dF=0$ and the Generalized Stokes theorem. Thus the action density changes by a total derivative and the integrated action can change on a boundary. The source's statement that the action is not invariant needs this qualification: on a boundaryless spacetime, or when the boundary integral vanishes, the integrated action is invariant too. No such boundary restriction is needed for invariance of the bulk field equation.

First use a genuine simple bivector of the specified chirality, with spinor part ${\varphi }^{AB}={\alpha }^A{\alpha }^B$. It is a exterior product: choose independent primed spinors ${\rho }^{A^{\prime }},{\sigma }^{A^{\prime }}$ and take $u^{A{A}^{\prime }}={\alpha }^A{\rho }^{A^{\prime }}$, $v^{B{B}^{\prime }}={\alpha }^B{\sigma }^{B^{\prime }}$. Their antisymmetric product is proportional to $w^{A{A}^{\prime }B{B}^{\prime }}={\alpha }^A{\alpha }^B{ϵ}^{A^{\prime }{B}^{\prime }}$.

For any antisymmetric $w^{ab}$, the Riemann curvature tensor acts on both tensor indices. In the printed curvature convention its complete contraction isbecause the Ricci tensor is symmetric. Substituting the one-chirality bivector and the chiral spinor curvature operator decomposition, the primed-operator term vanishes on contraction with the symmetric ${\alpha }^A{\alpha }^B$. The other term is $2{\Delta }_{AB}({\alpha }^A{\alpha }^B)$. Hence ${\Delta }_{AB}({\alpha }^A{\alpha }^B)=0$.

Repeat with $\alpha +\beta$ and subtract the identities for $\alpha$ and $\beta$. Linearity leaves twice the mixed symmetric product, giving the contracted chiral curvature identityThe polarization works for spinor fields as well as pointwise spinors: the chiral spinor curvature operator is algebraic by the preceding part.

The displayed source expression ${\alpha }_A{\beta }_B{ϵ}_{A^{\prime }{B}^{\prime }}$ needs this clarification. For independent $\alpha ,\beta$ it is not antisymmetric in the spacetime indices. Its antisymmetric part is ${\alpha }_{(A}{\beta }_{B)}{ϵ}_{A^{\prime }{B}^{\prime }}$, which is generally not simple. The proof above uses a genuinely simple bivector first and then polarization, so it establishes the requested conclusion without relying on the incorrect simplicity assertion.

An $r$-uniform hypergraph is strongly saturated with respect to a complete uniform hypergraph of order $r+t$ if adding any missing edge creates a new complete uniform hypergraph on $r+t$ vertices containing that edge. Equivalently, for every missing $r$-set $e$, there is an $(r+t)$-set $T\supseteq e$ for which all other $r$-subsets are edges. Existing complete subhypergraphs are allowed in this strong-saturation definition.

We prove the bound by an exterior-power quotient bound for strong hypergraph saturation. Assume $r,t\ge 1$ and $n\ge r+t$. Let $V={\mathbb{R}}^n$ with basis $e_1,\dots ,e_n$. Choose distinct real numbers $a_1,\dots ,a_n$ and let $U$ be the kernel of the $t\times n$ matrix whose column $i$ isEvery $t$ columns are linearly independent by the Vandermonde determinant, so $\dim U=n-t$. Form the quotient vector spaceFor each $r$-set $e=\{i_1<\cdots <i_r\}$, let $v_e$ be the image in $W$ of the exterior product $e_{i_1}\wedge \cdots \wedge e_{i_r}$. These images span $W$.

For an $(r+t)$-set $T$, let $V_T$ be its coordinate subspace. The matrix restricted to $T$ has rank $t$, so $U\cap V_T$ has dimension $r$. The wedge of a basis of this intersection is a nonzero vectorEvery coefficient $c_e$ is nonzero. Indeed, projection from $U\cap V_T$ onto the $r$ coordinates in $e$ is an isomorphism: a vector in its kernel would be supported on $T\setminus e$, where the $t$ columns are independent, and hence would be zero. That projection's determinant is precisely the corresponding wedge coefficient. Thus, in the quotient vector space,For a missing edge, choose its strong-saturation witness $T$. All other vectors in this relation come from present edges, so the missing edge's vector is in their span. Consequently the present-edge vectors already span all of $W$. Their number is at least its dimension:This is sharp. Take all $r$-sets that meet a fixed $t$-set $D$. A missing edge $e$ is disjoint from $D$, and on $e\cup D$ every other $r$-set meets $D$; adding $e$ completes the required clique. The number of edges is exactly the boxed difference. If $r\le n<r+t$, no missing edge could have a witness, so a strongly saturated hypergraph must be complete; the substantive construction above is for the usual range $n\ge r+t$.

For the set-pair problem, write $m=|I|$ and choose $x_i\in R_i\cap S_i$. If $i\ne j$, the cross-disjointness implies that $x_i$ is in neither $R_j$ nor $S_j$. In particular, the witnesses are distinct. Let $D=\{x_i:i\in I\}$; each $R_i$ and $S_i$ contains exactly its own witness from $D$.

Choose two different indices $i,j$. The sets $R_i\setminus \{x_i\}$ and $S_j\setminus \{x_j\}$ are disjoint, lie outside $D$, and have sizes $r-1$ and $s-1$. HenceFor equality, partition the ground set into disjoint sets $P,Q,D$ with sizes $r-1,s-1,n-r-s+2$, respectively, and enumerate $D$ as $x_1,\dots ,x_m$. SetThen $R_i\cap S_j$ is $\{x_i\}$ for $i=j$ and empty otherwise. This realizes equality whenever the bound permits $m\ge 2$. The argument is a diagonal-intersection set-pair bound, with no additional assumptions on intersections among the $R_i$ or among the $S_i$.

Fix the input-index conventionExtend the connection on a vector bundle by the graded Leibniz rule, $D(\alpha \otimes s)=d\alpha \otimes s+(-1{)}^{\mathrm{deg}\alpha }\alpha \wedge Ds$. Applying it twice givessoThis is the Cartan curvature equation with input indices. In the more usual coefficient-column convention $A={\theta }^t$, and $(A\wedge A{)}^t=-\theta \wedge \theta$ because the entries have degree one. Thus $F=dA+A\wedge A={\Theta }^t$; the sign difference is entirely conventional.

The Chern connection of a Hermitian vector bundle with holomorphic structure is characterized by metric compatibility and $D^{0,1}={\bar{\partial }}_E$. To construct it, use a holomorphic local frame and let $h_{ij}=⟨e_i,e_j⟩$, with the Hermitian inner product linear in the first variable. The second defining property makes $\theta$ type $(1,0)$. The first givesand therefore the only possible matrix isConversely, this formula has type $(1,0)$ and satisfies both type components of the metric identity, because $h={\overline{h}}^t$. Under a holomorphic frame change $e^{\prime }=ge$, $h^{\prime }=gh{\overline{g}}^t$ and direct differentiation givesThis is precisely the transformation obtained by differentiating $e^{\prime }=ge$. The locally constructed connections glue, proving existence, and the forced formula proves uniqueness of the Chern connection.

The derivative of $h^{-1}$ gives $\partial \theta =\theta \wedge \theta$. Consequently the Chern curvature isThe second assertion is ${\bar{\partial }}^2\theta =0$ in a holomorphic local frame; holomorphic transition functions make it intrinsic. Thus the curvature Dolbeault class is $\Psi =[\Theta ]\in H^1(M,{\Omega }_M^1\otimes \mathrm{End}E)$ by the Dolbeault theorem. Here ${\Omega }^1(\mathrm{End}E)$ denotes this tensor product. We use the identification with bundle-valued Dolbeault forms in which the antiholomorphic exterior factor is placed first, so $\bar{\partial }\theta$ has the stated positive descent sign.

For the stated transitions $e^{(\beta )}=g_{\alpha \beta }{e}^{(\alpha )}$, differentiating the frame relation yieldsTransport the alpha matrix to the beta frame before subtracting it. The difference is the holomorphic sectionOn a triple intersection, $g_{\alpha \gamma }=g_{\beta \gamma }{g}_{\alpha \beta }$, soThis is exactly the Čech cocycle condition after transporting all representatives to the gamma frame.

To identify its class, let $D_{\alpha }$ be the local holomorphic connection making the alpha frame parallel. Then $t_{\alpha }=D-D_{\alpha }$ is a genuine local section of $A^{1,0}(\mathrm{End}E)$, represented by ${\theta }_{\alpha }$, with $\bar{\partial }{t}_{\alpha }=\Theta$. Its Čech coboundary, using $t_{\beta }-t_{\alpha }$, is the intrinsic section represented by ${\sigma }_{\alpha \beta }$ in the beta frame. This local-primitive construction is the Čech–Dolbeault identification, and proves that $({\sigma }_{\alpha \beta })$ maps to $\Psi$. For an arbitrary trivializing cover use its natural map to sheaf cohomology; a Stein refinement gives the usual Čech computation. The resulting object is the Čech cocycle for the curvature Dolbeault class.

If the metric changes, the difference $B=D^{\prime }-D$ is a global smooth $(1,0)$ form with values in $\mathrm{End}E$. The local curvature formulas give ${\Theta }^{\prime }-\Theta =\bar{\partial }B$, so $\Psi$ is metric independent. If holomorphic frames change by $e_{\alpha }^{\prime }=u_{\alpha }{e}_{\alpha }$, the corresponding local tensors $t_{\alpha }^{\prime }-t_{\alpha }$, expressed in the old frame, are $B_{\alpha }=u_{\alpha }^{-1}\partial u_{\alpha }$. They are holomorphic, and their differences change the cocycle by $\delta B$. Hence $\Psi$ depends only on the holomorphic bundle, not on its metric or trivializations.

Apply exterior product and endomorphism composition to curvature to defineThese are closed for the Dolbeault operator by the product rule. Their independence can also be verified without commuting matrices. If ${\Theta }_1-{\Theta }_0=\bar{\partial }B$, thenThis follows by the noncommutative telescoping identity and the even total degree of curvature. The matrix products are interpreted in the fixed frame convention; powers of the degree-two curvature transpose to the corresponding usual curvature powers. The trace contraction is invariant under frame changes, so its class is well defined. This proves metric independence of curvature Dolbeault powers; for $k>\dim M$ the forms and groups in question vanish.

Finally take a smooth unitary frame. Metric compatibility gives ${\theta }^{†}=-\theta$, and the Cartan curvature equation with input indices gives ${\Theta }^{†}=-\Theta$. ThusThe second expression is therefore a real differential form. It is also $d$-closed: differentiating Cartan's equation gives $d\Theta =\theta \wedge \Theta -\Theta \wedge \theta$, and cyclicity of trace cancels the resulting commutator in $d\mathrm{tr}({\Theta }^k)$. On a compact Kähler manifold, the Hodge decomposition theorem for compact Kähler manifolds identifies $H^k(M,{\Omega }_M^k)$ with the $(k,k)$ component of complex de Rham cohomology. HenceIt is $k!$ times the degree-$2k$ component of the Chern character. Reality does not assert integrality of these individual power-sum classes.

Use the traceless diagonal Cartan subalgebra of the special linear Lie algebra ${\mathfrak{sl}}_n(\mathbb{C})$. Write ${\epsilon }_i(\mathrm{diag}(h_1,\dots ,h_n))=h_i$, so ${\sum }_i{\epsilon }_i=0$. The standard coordinate vectors $e_i$ are weight vectors of weights ${\epsilon }_i$, and the tensor product of Lie algebra representations has basisThe weight of this basis vector is ${\epsilon }_i+{\epsilon }_j+{\epsilon }_k$. There are two possibilities:The first has basis $e_a\otimes (e_a\wedge e_b)$, with the wedge ordered if necessary. In the second, the tensor's first factor can be any one of the three indicated coordinate vectors. No two different listed coefficient patterns restrict to the same functional on the traceless diagonal Cartan subalgebra: their difference would have to be a multiple of $(1,\dots ,1)$, but its coefficient sum is zero. All unlisted weight spaces are zero. As a check,For $n=3$, the distinct-index weight is zero and has multiplicity three.

To determine the irreducible summands, take the upper triangular raising operators. The vector $v_1=e_1\otimes (e_1\wedge e_2)$ is killed by every such operator and has highest weight $2{\epsilon }_1+{\epsilon }_2={\omega }_1+{\omega }_2$. The exterior product mapis a Lie algebra representation homomorphism. It has an equivariant right inverseIndeed $A\iota =I$. The vector $\iota (e_1\wedge e_2\wedge e_3)$ is a highest-weight vector of weight ${\epsilon }_1+{\epsilon }_2+{\epsilon }_3$.

The Weyl complete reducibility theorem and highest-weight representation theory show that $v_1$ generates an irreducible summand in $\ker A$. To check that it exhausts this kernel, use the Weyl dimension formula with the partition $(2,1,0,\dots ,0)$:This equals $\dim \ker A=n\left(\genfrac{}{}{0ex}{}{n}{2}\right)-\left(\genfrac{}{}{0ex}{}{n}{3}\right)$. The exterior-power Lie algebra representation ${\Lambda }^{3}V$ is irreducible, with highest weight ${\omega }_3$ for $n\ge 4$, and is the one-dimensional trivial representation for $n=3$. Therefore the tensor product of the standard representation with its exterior square givesEach summand occurs once; their dimensions are $n(n^2-1)/3$ and $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$.

Let $V={\mathbb{C}}^4$ with its nondegenerate alternating bilinear form. Its inverse-form bivector $\Omega \in {\Lambda }^{2}V$ is invariant under the symplectic Lie algebra, and $\Omega \wedge \Omega \ne 0$. Choose $\mathrm{vol}=\Omega \wedge \Omega /2$. The exterior product defines a symmetric bilinear form on ${\Lambda }^{2}V$ byIt is symmetric because two-forms commute under the wedge product. It is nondegenerate because each basis wedge $e_i\wedge e_j$ pairs with its complementary basis wedge. Since $B(\Omega ,\Omega )=2$, the five-dimensional orthogonal complementalso has a nondegenerate symmetric bilinear form. This is the primitive exterior square, equivalently the kernel of symplectic contraction of an exterior square.

Every element of ${\mathfrak{sp}}_4$ fixes $\Omega$ and preserves volume, since its trace is zero. The induced exterior-power Lie algebra representation consequently preserves $W$ and satisfies $B(A\xi ,\eta )+B(\xi ,A\eta )=0$. Thus it gives a Lie algebra homomorphismWe check injectivity explicitly. If $A$ acts as zero on $W$, it also kills $\Omega$, and therefore acts as zero on all of ${\Lambda }^{2}V=W\oplus \mathbb{C}\Omega$. For three distinct indices $i,j,k$, the coefficient of $e_k\wedge e_j$ in $A(e_i\wedge e_j)$ is $A_{ki}$, so all off-diagonal entries of $A$ vanish. Its action on $e_i\wedge e_j$ is then $(A_{ii}+A_{jj})e_i\wedge e_j$. All these sums vanish, which in characteristic zero forces every diagonal entry to vanish. Hence $A=0$.

Finally,The injective Lie algebra homomorphism is therefore surjective. Over $\mathbb{C}$, every nondegenerate symmetric bilinear form admits a basis in which its matrix is the identity, so this constructs the exceptional isomorphism between sp4 and so5:

A connection on a vector bundle is a complex-linear map from smooth sections to vector-bundle-valued one-forms satisfyingChoose a locally finite trivializing cover, its componentwise flat connections on a vector bundle $D_i$, and a subordinate smooth partition of unity ${\rho }_i$. The formula $D={\sum }_i{\rho }_i{D}_i$ is globally meaningful: each weighted term extends by zero outside its chart. Its Leibniz rule follows from ${\sum }_i{\rho }_i=1$, proving every smooth complex vector bundle admits a connection.

Extend the connection on a vector bundle to bundle-valued differential forms byThe curvature form of a connection is ${\Theta }_D=D^2$. Applying this twice to $fs$ shows that the $df$ terms cancel, so $D^2$ is tensorial and defines an $\mathrm{End}(E)$-valued two-form. With column coordinates for sections and connection one-form $A$, the local expression is $D=d+A\wedge$. Direct expansion on an arbitrary bundle-valued form givesThe terms containing a derivative of the argument cancel by the graded Leibniz rule. Thus Cartan curvature matrix equation isHere multiplication includes matrix multiplication and the exterior product of form entries. For a frame change $e^{\prime }=eg$, the connection one-form and curvature form of a connection transform asThe trace is consequently frame independent. Alsothe diagonal terms vanish and the off-diagonal terms cancel in pairs. Locally $\mathrm{Tr}\Theta =d\mathrm{Tr}A$, and hence $d\mathrm{Tr}\Theta =0$. The local primitives need not agree, but the two-form does. We obtain a global closed two-form $\mathrm{Tr}{\Theta }_D$.