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Let $V$ be a real vector space. A $k$-tensor on $V$ is a multilinear function $$T:V^k\to ℝ.$$ The vector space of $k$-tensors on $V$ is denoted ${𝒯}^k\left(V\right)$.

If $S\in$${𝒯}^k\left(V\right)$ and $T\in {𝒯}^l\left(V\right)$, define the tensor product $S\otimes T\in {𝒯}^{k+l}\left(V\right)$ by $$(S\otimes T)(v_1,\dots ,v_{k+l})=S(v_1,\dots ,v_k)T(v_{k+1},\dots ,v_{k+l}).$$

A $k$-tensor $\omega$ on $V$ is alternating if interchanging any two arguments changes its sign. The vector space of alternating $k$-tensors on $V$ is denoted ${\mathrm{\Lambda }}^k\left(V\right)$.

If $T\in$${𝒯}^k\left(V\right)$, define $$\mathrm{Alt}\left(T\right)(v_1,\dots ,v_k)=\frac{1}{k!}\sum _{\sigma \in S_k}\mathrm{sgn}\left(\sigma \right)T(v_{\sigma \left(1\right)},\dots ,v_{\sigma \left(k\right)}).$$

If $\omega \in$${\mathrm{\Lambda }}^k\left(V\right)$ and $\eta \in {\mathrm{\Lambda }}^l\left(V\right)$, define the wedge product using the alternating operator and tensor product: $$\omega \wedge \eta =\frac{(k+l)!}{k!l!}\mathrm{Alt}(\omega \otimes \eta ).$$

An orientation of an $n$-dimensional real vector space $V$ is one of the two classes of ordered bases, where two ordered bases are equivalent if the change-of-basis determinant between them is positive.

If $V$ is an oriented $n$-dimensional inner product space, its volume element is the unique alternating tensor $\omega \in {\mathrm{\Lambda }}^n\left(V\right)$ such that $$\omega (v_1,\dots ,v_n)=1$$ for every positively oriented orthonormal basis $v_1,\dots ,v_n$.

A vector field on an open set $A\subseteq {ℝ}^n$ is a function $F$ assigning to each $x\in A$ a vector $F\left(x\right)\in {ℝ}_x^n$.

A $k$-form on an open set $A\subseteq {ℝ}^n$ assigns to each $x\in A$ an alternating $k$-tensor on the tangent space ${ℝ}_x^n$. In coordinates, a $k$-form can be written $$\omega =\sum _{i_1<\cdots <i_k}{\omega }_{i_1,\dots ,i_k}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_k}.$$

If $f:A\to B$ is differentiable and $\omega$ is a $k$-form on $B$, then the pullback $f^{\ast }\omega$ is the $k$-form on $A$ defined by $$\left(f^{\ast }\omega \right)\left(x\right)(v_1,\dots ,v_k)=\omega \left(f\left(x\right)\right)(Df\left(x\right)v_1,\dots ,Df\left(x\right)v_k).$$

If $$\omega =\sum _{i_1<\cdots <i_k}{\omega }_{i_1,\dots ,i_k}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_k},$$ then its exterior derivative is $$d\omega =\sum _{i_1<\cdots <i_k}\sum _{j=1}^n{D}_j{\omega }_{i_1,\dots ,i_k}d{x}^j\wedge d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_k}.$$

A form $\omega$ is closed if its exterior derivative satisfies $d\omega =0$. It is exact if there is a form $\eta$ such that $$\omega =d\eta .$$ Every exact form is closed.

A singular $k$-cube in a set $A\subseteq {ℝ}^n$ is a differentiable map $$c:{\left[\mathrm{0,1}\right]}^k\to A.$$

A singular $k$-chain in $A$ is a finite formal integer combination $$c=\sum _{i=1}^r{a}_i{c}_i$$ of singular $k$-cubes $c_i$ in $A$.

For a singular $k$-cube $c$, let $c_{(i,\alpha )}$ denote the $(k-1)$-cube obtained by setting the $i$-th coordinate equal to $\alpha \in \left\{\mathrm{0,1}\right\}$. The boundary of $c$ is $$\partial c=\sum _{i=1}^k\sum _{\alpha =0}^1{(-1)}^{i+\alpha }{c}_{(i,\alpha )}.$$ This extends linearly to chains.

If $\omega$ is a $k$-form on $A$ and $c$ is a singular $k$-cube in $A$, define the integral using the pullback $${\int }_c\omega ={\int }_{{\left[\mathrm{0,1}\right]}^k}{c}^{\ast }\omega .$$ For a chain $c=\sum a_i{c}_i$, define $${\int }_c\omega =\sum a_i{\int }_{c_i}\omega .$$