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So, given the $2$-form $$\omega=x_1\text{d}x_{2,3}+x_2\text{d}x_{1,3}+x_3\text{d}x_{1,2}$$ and the simplex $$\sigma=[e_1+2e_3,2e_2,e_2-e_1],$$ I defined the function $$g:\mathbb{R}^2\to\mathbb{R}^3$$ $$(x_1,x_2) \mapsto (\frac{3x_1-x_2}{2}+1,x_2,x_1-x_2+2),$$ and then did the following calculation, using the Change of Variables Theorem and Stokes' Theorem:

$$\int_{\sigma}\omega = \int_{g^{-1}(\sigma)}g^*\omega =$$ $$\int_{g^{-1}(\sigma)}(\frac{3x_1-x_2}{2})\text{d}x_{2,1} + x_2[(\frac{3\text{d}x_1}{2}-\frac{\text{d}x_2}{2})\wedge(\text{d}x_1-\text{d}x_2)]+\frac{3(x_1-x_2+2)}{2}\text{d}x_{1,2}=$$ $$\int_{g^{-1}(\sigma)}(-\frac{x_1}{2}+2x_2+2)\text{d}x_{1,2} = \int_{\partial g^{-1}(\sigma)}(x_2^2\text{d}x_1+(-\frac{x_1^2}{4}+2x_1)\text{d}x_2).$$

Now, first question is: Is everything I did here correct?

Second question: Is there a better choice of function $g$ to do the pullback (or, more generally, is something better to do here, instead of pulling it back)? How do I decide such a thing?

Third question: I could have calculated the double integral, following the third equal sign, instead of using Stokes' Theorem; which is better here, and how do I decide something like that?

Fourth question: If I'm going to use Stokes' Theorem here, the best next step is to divide the domain of integration into three (one variable) integrals and parametrize each, doing the pullback thing again, correct?

Arnaud Mortier
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    How did you choose this map $g$? If I was to make a choice I would try and make it so that the vertices of $\sigma$ correspond to $(0,0)$, $(0,1)$, $(1,0)$ just to have a nice picture in mind. – Arnaud Mortier Feb 02 '18 at 19:22
  • For no real good reason I decided I wanted to have $g(-1,1)=(-1,1,0)$, $g(0,2)=(0,2,0)$ and $g(0,0)=(1,0,2)$. –  Feb 02 '18 at 19:25
  • (Those are $e_2-e_1$, $2e_2$ and $e_1+2e_3$.) –  Feb 02 '18 at 19:31

2 Answers2

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Question 1: That's $\dfrac{dx_2}{2}$ in the second term (see edit). And in the second line after everything simplifies out I find $-2x_2+3$ in the brackets.

Question 2: It is easier imho to use a map $g$ so that the simplex gets pulled back onto the standard simplex in $\mathbb{R}^2$. That's because this puts essentially all the messy parts of the problem together in one object.

Question 3: It is a matter of choice. But note that if your simplex is nice enough (see Question 2) then its parametrisation is straightforward (no more error-provoking stuff).

Question 4: Yes. Except if you have a nice simplex, there is really only one pull-back where you need to think, on the slanted segment $x+y=1$.

Arnaud Mortier
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Following Arnaud Mortier's suggestions, I'm going to leave a (hopefully) correct answer here:

Define $f: \mathbb{R}^2\to\mathbb{R}^3$ by $$f(x_1,x_2)=(x_1+2x_2-1,x_1-x_2+1,2x_2).$$ Then $$f^*(\omega)=4x_1-4x_2\text{d}x_{1,2},$$ and $$\int_{\sigma}\omega=\int_{f^{-1}(\sigma)}f^*(\omega)=\int_{[0,1]}\int_{[0,1-x_1]}(4x_1-4x_2)\text{d}x_2\text{d}x_1=\cdots=0.$$