I read, again in Spivak's Calculus on Manifolds, that the integral of 1-form over a 1-cube is equivalent to a line integral. And indeed, if I consider the 1-form w = Pdx + Qdy on(adsbygoogle = window.adsbygoogle || []).push({}); R², and c a given 1-cube inR², I find that

[tex]\int_c\omega = \int_0^1 F(c(t))\cdot c'(t)dt[/tex]

where F=(P,Q), which is the integral of the vector field F along the curve c.

Next, I read that the integral of a 2-form over a 2-cube is equivalent to a surface integral. Let [itex]\omega = Pdx\wedge dy + Qdx\wedge dz + Rdy\wedge dz[/itex] be a 2-form onR³ and c be a 2-cube inR³. I am guessing that [itex]\int_c\omega[/itex] corresponds to the flux integral of the vector field F = (P,Q,R) through c([0,1]²). Recall that this integral is given by

[tex]\iint_{[0,1]^2}F(c(s,t))\cdot \left(\frac{\partial c}{\partial s}\times \frac{\partial c}{\partial t}\right) dsdt=\iint_{[0,1]^2}P(c(s,t))\left| \begin{array}{cc}

\frac{\partial c^2}{\partial s} & \frac{\partial c^3}{\partial s} \\

\frac{\partial c^2}{\partial t} & \frac{\partial c^3}{\partial t}

\end{array} \right| - Q(c(s,t))\left| \begin{array}{cc}

\frac{\partial c^1}{\partial s} & \frac{\partial c^3}{\partial s} \\

\frac{\partial c^1}{\partial t} & \frac{\partial c^3}{\partial t}

\end{array} \right| + R(c(s,t))\left| \begin{array}{cc}

\frac{\partial c^1}{\partial s} & \frac{\partial c^2}{\partial s} \\

\frac{\partial c^1}{\partial t} & \frac{\partial c^2}{\partial t}

\end{array} \right| \ dsdt[/tex]

Alright, so I get something close to this, but not quite, so I'm wondering where I'm mistaken... I say,

[tex]\int_c\omega = \int_{[0,1]^2}(P\ \circ \ c) c^*(dx\wedge dy)+(Q\ \circ \ c) c^*(dx\wedge dz)+(R\ \circ \ c) c^*(dy\wedge dz)[/tex]

Now, [itex]c^*(dx^i\wedge dx^j))[/itex] is a 2-form on [0,1]² so it is of the form [itex]f_{ij} \ ds\wedge dt[/itex] for some 0-form f_ij on [0,1]². In order to find f_ij, let p belong to [0,1]² and [itex]v_{p}, \ w_{p} \in \mathbb{R}^2_{p}[/itex].

Using the definitions, I find that

[tex]c^*(dx^i\wedge dx^j)(v_p,w_p)=dx^i\wedge dx^j(c_*(v_p),c_*(w_p))=dx^i\otimes dx^j(c_*(v_p),c_*(w_p))-dx^i\otimes dx^j(c_*(w_p),c_*(v_p))[/tex]

where [itex]c_*(v_p)=(Dc(p)(v))_{c(p)}[/itex], and therefor

[tex]dx^i(c_*(v_p))=\frac{\partial c^i}{\partial s}(p)v^1+\frac{\partial c^i}{\partial t}(p)v^2=\left(\frac{\partial c^i}{\partial s}(p)ds+\frac{\partial c^i}{\partial t}(p)dt\right)(v_p)[/tex]

So I conclude that

[tex]c^*(dx^i\wedge dx^j)(v_p,w_p)=\left(\frac{\partial c^i}{\partial s}(p)dx+\frac{\partial c^i}{\partial t}(p)dt\right)\wedge \left(\frac{\partial c^j}{\partial s}(p)dx+\frac{\partial c^j}{\partial t}(p)dt\right)=\left(\frac{\partial c^i}{\partial s}(p)\frac{\partial c^j}{\partial t}(p)-\frac{\partial c^i}{\partial t}(p)\frac{\partial c^j}{\partial s}(p)\right) ds\wedge dt(v_p,w_p)[/tex]

and hence

[tex]f_{ij}=\frac{\partial c^i}{\partial s}\frac{\partial c^j}{\partial t}-\frac{\partial c^i}{\partial t}\frac{\partial c^j}{\partial s}=\left| \begin{array}{cc}

\frac{\partial c^i}{\partial s} & \frac{\partial c^j}{\partial s} \\

\frac{\partial c^i}{\partial t} & \frac{\partial c^j}{\partial t}

\end{array} \right|[/tex]

Finally,

[tex]\int_c\omega = \iint_{[0,1]^2}P(c(s,t))\left| \begin{array}{cc}

\frac{\partial c^1}{\partial s} & \frac{\partial c^2}{\partial s} \\

\frac{\partial c^1}{\partial t} & \frac{\partial c^2}{\partial t}

\end{array} \right| + Q(c(s,t))\left| \begin{array}{cc}

\frac{\partial c^1}{\partial s} & \frac{\partial c^3}{\partial s} \\

\frac{\partial c^1}{\partial t} & \frac{\partial c^3}{\partial t}

\end{array} \right| + R(c(s,t))\left| \begin{array}{cc}

\frac{\partial c^2}{\partial s} & \frac{\partial c^3}{\partial s} \\

\frac{\partial c^2}{\partial t} & \frac{\partial c^3}{\partial t}

\end{array} \right| \ dsdt[/tex]

So, 3 things differ: the factors of P and R and the sign of Q.

Thanks!

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# Integrating a 2-form on a 2-cube

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