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I am working out of Munkres Analysis on Manifolds and I see that he claims for $\eta = f dx_1 \wedge \ldots \wedge dx_k$.

$$ \int_A\eta = \int_{x \in A} \eta(x)\big((x;a_1) ,\ldots, (x;a_k) \big)$$

when $a_1,\ldots,a_k$ is a right handed orthonormal basis for $\mathbb{R}^k$.

I believe I understand the definition of each of these objects independently except for the right hand side of the equals. I am not sure how exactly the integral over a k-form can be the same as an integral over a k-tensor, though it seems clear that it has to do with the description of the basis.

I would like a hint as to how to show this or understand these types of integrals better.

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    $\begingroup$ The rightmost integral is an ordinary Riemann integral. You are just integrating a function: $x \in A \mapsto \eta(x)(a_1(x)\ldots a_k(x))\in \mathbb{R}.$ $\endgroup$
    – Giuseppe Negro
    Dec 10, 2012 at 22:47
  • $\begingroup$ Ok, then on the left hand side of the equals, I can use $\int_A \eta = \int_A f$ to compute and they will both be an integral over $k$ variables essentially. What problems with the equality would arise then in the case where the basis was either not right-handed or orthonormal? $\endgroup$
    – ackshooairy
    Dec 10, 2012 at 23:11
  • $\begingroup$ If the basis is not right-handed but left-handed, then the integral will turn out to be opposite in sign. If the basis is not orthonormal the integral will turn out to differ by the determinant of one of the matrices of change of basis (which one exactly I cannot tell you at the moment). $\endgroup$
    – Giuseppe Negro
    Dec 10, 2012 at 23:15

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The integral of a $n$-form $\eta$ on a orientable differentiable manifold of dimension $n$ (in the present case, an open submanifold of $\mathbb{R}^n$) is defined in such a way that it agrees with its expressions in coordinates. Precisely, if a coordinate system $(U, \phi=(x_1\ldots x_n))$ on $M$ is given, then on $U$ the $n$-form has a unique expression as $$\tag{1}\eta= f dx_1\wedge \ldots \wedge dx_n.$$ Then we define $$\tag{2}\int_U \eta = \int_{\phi(U)} f(t_1\ldots t_n)\, dt_1\ldots dt_n,$$ where the rightmost is the ordinary Riemann integral of multivariable calculus. Turns out that the way $\eta$ changes if we pass to another coordinate system (with the same orientation) is precisely the change of variable formula of multiple integrals, so that (2) is well defined (modulo some technicalities regarding the convergence of the integral, partitions of unity and the role of orientation).

Now, to answer your question. If $\eta$ is a $n$-form on an open subset $A$ of $\mathbb{R}^n$, and if $a_1\ldots a_n$ is an orthonormal basis of $\mathbb{R}^n$, then the assignment $$f(t_1\ldots t_n)=\eta(t_1\ldots t_n)(a_1\ldots a_n)$$ yields exactly the function $f$ we had in equation (1). So, by definition, $$\int_A \eta = \int_A f(t_1\ldots t_n)\, dt_1\ldots dt_n,$$ which is the formula the book is giving.

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  • $\begingroup$ To make sure I understand, you're saying that we can define $f(t_1,\ldots,t_n) = \eta(x)(a_1,\ldots,a_n)$ and because of this, then by the way you define the integral in (2), you can say that this function f is equal to the function in (1)? That is where I am caught up. $\endgroup$
    – ackshooairy
    Dec 10, 2012 at 23:28
  • $\begingroup$ Not really. I edited a small typo that I'm afraid was confusing. I'm saying that the assignment $f(t_1\ldots t_n)=\eta(t_1\ldots t_n)(a_1\ldots a_n)$ gives you exactly the expression of $\eta$ with respect to the coordinate system $$t_1 a_1 +\ldots + t_n a_n \mapsto (t_1 \ldots t_n).$$ $\endgroup$
    – Giuseppe Negro
    Dec 10, 2012 at 23:34
  • $\begingroup$ By definition, the integral of $\eta$ is exactly the Riemann integral of one of its coordinate expressions. So we get the formula in the book. $\endgroup$
    – Giuseppe Negro
    Dec 10, 2012 at 23:35
  • $\begingroup$ I am still having difficulty with the assignment and how I can conclude that it is $\eta$. Are you saying that these are simply two different coordinate expressions for $\eta$ and therefore each should yield the same result? $\endgroup$
    – ackshooairy
    Dec 10, 2012 at 23:48
  • $\begingroup$ Yes: precisely I am saying that $$\eta=f dt_1\wedge \ldots \wedge dt_n,$$ that is, $$\eta(t)(v_1\ldots v_n)=f(t)(dt_1\wedge\ldots \wedge dt_n)(v_1\ldots v_n)$$ for all points $t=(t_1\ldots t_n) \in \mathbb{R}^n$ and $n$-uples of vectors $v_1\ldots v_n$ in $\mathbb{R}^n$. $\endgroup$
    – Giuseppe Negro
    Dec 11, 2012 at 0:17

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