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If ${\nabla }_X{e}_i={\sum }_j{\omega }_{ij}(X)e_j$ and ${\varphi }_i$ is the dual coframe, the first index of $\omega$ is the input frame vector. Expand the connection derivative of $Y={\sum }_i{\varphi }_i(Y)e_i$ and antisymmetrize in $X,Y$. The formula for the exterior derivative of a one-form evaluated on vector fields gives ${\tau }_j=d{\varphi }_j-{\sum }_i{\varphi }_i\wedge {\omega }_{ij}$. This is the first structure equation with the indicated index convention. Reversing the wedge order changes the sign.

The exterior derivative is an $\mathbb{R}$-linear operator $d:{\Omega }^r(M)\to {\Omega }^{r+1}(M)$ which agrees with the differential of a smooth function on degree zero, satisfies $d^2=0$, and obeys the graded Leibniz rule

For the uniqueness of the exterior derivative from its axioms, first note that these axioms force locality. If $\alpha$ vanishes near $p$, choose a smooth bump function $\chi$ equal to one near $p$ and supported where $\alpha =0$. Then $0=d(\chi \alpha )=d\chi \wedge \alpha +\chi d\alpha$ gives $d\alpha (p)=0$. Thus $d\alpha (p)$ depends only on the form near $p$.

In a coordinate chart write $\alpha ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}$. Since $d(d{x}^i)=d^2{x}^i=0$, the axioms forceThis also proves existence. Define $d$ by that formula in each chart. The product rule for partial derivatives gives the graded Leibniz rule, and symmetry of mixed partial derivatives against antisymmetry of the wedge product of differential forms gives $d^2=0$. On overlaps it still agrees with the differential of every function, by the chain rule, so the preceding uniqueness argument in the second coordinate system makes the formulas agree. They therefore glue to a global exterior derivative.

For a differential one-form $\omega ={\sum }_i{a}_{i}d{x}^i$, the exterior derivative of a one-form evaluated on vector fields follows directly:In $X(\omega (Y))-Y(\omega (X))$, the additional terms are ${\sum }_i{a}_i[X(Y^i)-Y(X^i)]=\omega ([X,Y])$, where the bracket is the Lie bracket of vector fields. Hence

The de Rham cohomology iswith zero denominator in degree zero. The Poincaré lemma states that a closed differential form of positive degree on a star-shaped open subset of Euclidean space is an exact differential form; consequently every closed positive-degree form is locally exact on a smooth manifold.

For the suspension isomorphism for top de Rham cohomology of spheres, put $U=S^n\setminus \{\text{south pole}\}$ and $V=S^n\setminus \{\text{north pole}\}$. Both are diffeomorphic to ${\mathbb{R}}^n$, while $W=U\cap V$ deformation retracts onto the equatorial $S^{n-1}$. The Mayer--Vietoris sequence for de Rham cohomology comes from the short exact sequence of cochain complexesThe final map is surjective: a partition of unity ${\chi }_U+{\chi }_V=1$ lets a form $\eta$ on $W$ be lifted by extending ${\chi }_V\eta$ to $U$ and $-{\chi }_U\eta$ to $V$ by zero. The Poincaré lemma makes $H^{n-1}(U),H^{n-1}(V),H^n(U),H^n(V)$ vanish when $n>1$. Exactness givesIf $j:S^{n-1}\hookrightarrow W$ is the equatorial inclusion, homotopy invariance of de Rham cohomology makes $j^{\ast }$ an isomorphism. Thereforeis injective, indeed an isomorphism. Concretely, choose local primitives $d{\alpha }_U=\omega {|}_U$, $d{\alpha }_V=\omega {|}_V$, and send $[\omega ]$ to $[j^{\ast }({\alpha }_U-{\alpha }_V)]$. Their difference is closed; changing either primitive changes it by an exact form, so this agrees with the cohomological construction.

The given equation is a Maurer-Cartan equation, with the sign appropriate to the chosen invariant coframe. Write $\lambda =({\lambda }_{ij})$ as an antisymmetric matrix of one-forms, so $d\lambda =\lambda \wedge \lambda$. Using the graded product rule for the exterior derivative,The two sums coincide after exchanging their dummy indices $k,\ell$. This is also associativity of matrix-valued wedge products of differential forms.

To make the relation to the Jacobi identity explicit, let ${\theta }^a$ be any independent coframe and write $d{\theta }^a=-\frac{1}{2}{c}^a{}_{bc}{\theta }^b\wedge {\theta }^c$. Its dual invariant vector fields satisfy $[E_b,E_c]=c^a{}_{bc}{E}_a$, by the exterior derivative of a one-form evaluated on vector fields. Evaluating $d^2{\theta }^a=0$ on three basis fields gives the cyclic coefficient identityThese are exactly the components of $[[E_b,E_c],E_d]+[[E_c,E_d],E_b]+[[E_d,E_b],E_c]=0$. Thus the structure constants obey the Jacobi identity.

For the real four-dimensional case, abbreviate $a={\lambda }_{12}$, $b={\lambda }_{23}$, $c={\lambda }_{31}$, $d={\lambda }_{43}$, $e={\lambda }_{41}$, $f={\lambda }_{42}$; here $d$ without an argument is a one-form, whereas $d(\cdot )$ denotes the exterior derivative. Directly expanding the given equations givesDefine ${\alpha }_1^{\pm }=a\pm d$, ${\alpha }_2^{\pm }=b\pm e$, and ${\alpha }_3^{\pm }=c\pm f$. For example,The minus combination and the other two cyclic combinations giveThe six forms are independent, since the original pairs are their half-sums and half-differences. Their dual vector fields $E_i^{+},E_i^{-}$ consequently obeyEach triple is a three-dimensional Lie algebra ideal with the SO(3) Lie algebra brackets; the two ideals commute and span the original six-dimensional algebra. This proves the real four-dimensional rotation algebra splittingThis is a Lie-algebra statement. Globally $SO(4)$ is a central quotient of $SU(2)\times SU(2)$; a direct product of the corresponding $SO(3)$ groups is not asserted.

Let $e$ be the identity. For a left-invariant differential form $\omega$ and a left-invariant vector field $X$, one hasThe value is therefore independent of $g$, even if the Lie group is disconnected. The same calculation applies to another left-invariant field $Y$.

The standard exterior derivative of a one-form evaluated on vector fields isBoth differentiated functions are constant in the present situation. ThusThis identity is a direct consequence of left invariance and the stated general formula for the exterior derivative.

Write the differential one-form as $\omega ={\sum }_i{\omega }_{i}d{x}^i$, and the vector fields as $X=X^i{\partial }_i$, $Y=Y^i{\partial }_i$. The coordinate definition of the exterior derivative givesExpand the right side of the invariant formula:The last line is $\omega ([X,Y])$, by the coordinate formula for the Lie bracket of vector fields. Subtract it and interchange the two indices in the second sum. The remaining terms are exactly $d\omega (X,Y)$. ThereforeThis is the exterior derivative of a one-form evaluated on vector fields; it is valid in any frame, not just a commuting coordinate frame.

A connection on a vector bundle $E\to M$ is an $\mathbb{R}$-linear mapsatisfying $\nabla (fs)=df\otimes s+f\nabla s$. Equivalently its directional version is real-bilinear and satisfiesFor a complex bundle it is complex-linear in the section variable and has the same Leibniz property.

Every smooth vector bundle over an ordinary Hausdorff, second-countable smooth manifold has such a connection. Choose trivializing open sets $U_a$, local frames $e_{a,i}$ and a subordinate locally finite partition of unity ${\rho }_a$. In each frame define the flat local connectionThen set $\nabla s={\sum }_a{\rho }_a{\nabla }^{(a)}(s{|}_{U_a})$, extending each weighted term by zero outside $U_a$. The supports lie inside the corresponding trivializing sets, making the extensions smooth, and local finiteness makes the sum smooth. Since ${\sum }_a{\rho }_a=1$, its Leibniz rule isThis is the construction of a vector bundle connection by a partition of unity.

The difference $A={\nabla }^{\prime }-\nabla$ of two connections is $C^{\infty }(M)$-linear in both its vector-field argument and its section argument. The derivative terms cancel, so $A_{X}s$ depends only on $X_p,s_p$ at each point. Smooth local coefficients therefore identify it withConversely, any such $A$ added to a connection preserves its axioms. Thus connections form an affine space modeled on ${\Omega }^1(M;\mathrm{End}E)$. Choosing a base connection ${\nabla }^0$ gives the noncanonical bijection $A\mapsto {\nabla }^0+A$; there is no distinguished zero connection and hence no canonical vector-space identification. This is the affine space of vector-bundle connections.

For a connection on $TM$, the torsion tensor is the alternating operation $\tau (X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]$. The connection and bracket Leibniz rules giveAlternation then gives $\tau (X,fY)=f\tau (X,Y)$ as well. Hence it is tensorial, of type one contravariant and two covariant indices:It is equivalently a $TM$-valued two-form, a torsion form.

For the local frame $e_i$ with dual coframe ${\varphi }_j$, defineThe connection is $C^{\infty }$-linear in $X$, so these are smooth differential one-forms. Expansion in the frame gives ${\nabla }_X{e}_i={\sum }_j{\omega }_{ij}(X)e_j$, and uniqueness follows by applying each ${\varphi }_j$. Here the first index is the input frame vector and the second the output coefficient; this fixes the connection one-form convention.

Expand $Y={\sum }_i{\varphi }_i(Y)e_i$ and apply the connection Leibniz rule:Subtract the corresponding expression with $X,Y$ exchanged, and then subtract ${\varphi }_j([X,Y])$. By the exterior derivative of a one-form evaluated on vector fields,Since all pairs of tangent vectors can be tested, equality of the two-forms follows:This is the Cartan first structure equation with input-first indices. Its plus sign is correct with the stated index order and wedge order; the common output-first convention writes the same identity with a minus sign and reversed wedge factors.

In coordinates, expanding $d\alpha$ and the Lie bracket of vector fields showsThe terms differentiating components of $X,Y$ cancel against the bracket, leaving $({\partial }_i{\alpha }_j-{\partial }_j{\alpha }_i)X^i{Y}^j$. This is the exterior derivative of a one-form evaluated on vector fields.

A connection on a vector bundle is a linear map $\nabla :\Gamma (E)\to {\Omega }^1(E)$ satisfying $\nabla (fs)=df\otimes s+f\nabla s$; evaluation on $X$ defines ${\nabla }_{X}s$. Its curvature form of a connection is $d_A^2$, locally $F_A=dA+A\wedge A$. Direct expansion, using the displayed exterior-derivative identity, givesThe derivative terms on a scalar multiplying $s$ cancel, and the expression is also linear over smooth functions in $X,Y$. Thus it is an alternating tensor with values in $\mathrm{End}(E)$, namely an element of ${\Omega }^2(\mathrm{End}(E))$.

The dual connection and tensor product connection induce the endomorphism bundle connection on $E^{\ast }\otimes E$:Expanding twice cancels the cross terms and givesHere $\varphi ,s$ are local smooth sections; derivatives are not defined for isolated fiber elements without extensions. Subtract the corresponding ${\tilde{\nabla }}_{[X,Y]}$ identity to get the curvature of an endomorphism bundle connection:This vanishes for all $\varphi$ exactly when each $F_A(X,Y)$ is central in the full matrix algebra, hence scalar. For rank $r>0$, $\omega =r^{-1}\mathrm{tr}{F}_A$ is a smooth two-form andConversely scalar curvature commutes with every endomorphism. This is the scalar-curvature criterion for a flat endomorphism connection; it permits a nonflat connection on $E$ itself.