I'm following Do Carmo's book Differential forms and Applications and I'm working on the following:
Given the $2$-form \begin{align} \omega_{p} = e^{2x-y^{2}}dx \wedge dy - 3x^2z dx \wedge dz + (z^2x-7y)dy\wedge dz \end{align} with $p=(x,y,z)$, and a function $f\colon \mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$ defined by \begin{align*} f(x,y,z) = (xe^{y},z^{2}y-2x,y^{2}e^{z} - 3x), \end{align*} I have to find and simplify $(f^{\ast}\omega)_{p}$.
Here is my attempt: first, by definition we have that \begin{align*} (f^{\ast}\omega)_{p}(v,u) = \omega_{f(p)}(df_{p}(v),df_{p}(u)), \;\; \forall v,u\in \mathbb{R}^{3}_{p}, \end{align*} where $df_{p}\colon \mathbb{R}^{3}_{p}\rightarrow \mathbb{R}^{3}_{f(p)}$ is the differential of $f$ in $p$. Of course we can express $f$ as $f=(f_1,f_2,f_3)$. Now, if I denote $\beta_{1} = f_1(x,y,z), \beta_{2}=f_{2}(x,y,z), \beta_{3}=f_{3}(x,y,z)$ and replace:
\begin{align} \omega_{f(p)} = e^{2\beta_{1}-\beta_{2}^{2}}(df_{1} \wedge df_{2}) - 3\beta_{1}^2\beta_{3} (df_{1} \wedge df_{3}) + (\beta_{3}^2\beta_{1}-7\beta_{2})(df_{2}\wedge df_{3}). \end{align}
So now I have to find each $df_{i}$, and this is a straightforward computation: \begin{align} df_{1} &= e^{y}dx+ xe^{y}dy\\ df_{2} &= -2dx+z^{2}dy + 2zydz\\ df_{3} &= -3dx+2ye^{z}dy +y^2e^{z}dz \end{align}
Here is my problem, I'm not sure how to compute $df_{1} \wedge df_{2}$, $df_{1} \wedge df_{3}$ and $df_{2}\wedge df_{3}$. In Do Carmo's book he has the following definition:
Let \begin{align} \omega = \sum a_{I}dx_{I}, && I=(i_{1},\dots,i_{k}), && i_{1}<\cdots<i_{k}\\ \varphi = \sum b_{J}dx_{J}, && J=(j_{1},\dots,j_{s}), && j_{1}<\cdots<j_{s} \end{align} By definition \begin{align} \omega \wedge \varphi = \sum_{IJ} a_{I}b_{J} (dx_{I}\wedge dx_{J}) \end{align}
But the indexes confuse me and I don't know how to make these calculations. I would appreciate your help. If I am not being clear on something or you need more details please let me know.