Inequality: norm of differential forms and mass of singular simplex Let M be a Riemannian manifold of dimension $n$ with a differentiable $p$-form $\omega\in \Omega^p(M)$. Let $\sigma\colon \Delta^p\to M$ be a differentiable singular $p$-simplex. We may assume that $\sigma$ is lipschitz if needed.
I'm interested in the following inequality from $[1]$:
$$|\omega(\sigma)| \leq \|\omega\|_\infty \cdot  \text{mass}(\sigma)$$
though I'm not 100% sure I know what all the ingredients in the inequality are, as my diffentialgeometryskills are not the best. (Also, there could be more assumptions missing, but I don't think so.)
So I tried to make sense of the objects appearing. First the LHS:
Having a $p$-form on $M$, I can pull that back by $\sigma$ to obtain a $p$-form on $\Delta^p$. Which I then may integrate. That would be
$$\omega (\sigma) = \int_{\Delta^p} \sigma^*(\omega)$$
The mass of a simplex is$^{[2]}$ the integral of the Jacobian over $\Delta^p$.
$$\text{mass}(\sigma) = \int_{\Delta^p} |(J\sigma)| \text{vol}_{\Delta^p}$$
Apparently, the Jacobian $J\sigma$ at $x$ is the (absolute value of? the) determinant of the transformation matrix of the map $T_x\sigma\colon T_x \Delta^p \to T_{\sigma(x)}M$, where I restrict the codomain to the image. I have to do this since the dimension won't match up.
Last but not least, the norm of $\omega$. With the Hodge-star-operator, we may turn $p$ into an $n-p$-form $\star \omega$. Wedging both, we obtain an $n$-form, which is a multiple of the volumeform on $M$. This multiple is the norm.
$$\|\omega\|^2 = \star(\omega \wedge \star \omega)~~~\text{ or }~~~ \omega\wedge \star \omega = \|\omega\|^2 \text{vol}_M$$
Now $\|\omega\| \colon M\to \mathbb R$, so we may take the supremum $\|\omega\|_\infty = |\|\omega\||_\infty$ over $M$.
In the explicit case, this is bounded by a constant.
So how do these things match up? How to get the above inequality (I guess it shouldn't be that hard, ass ECHLPT don't say anything about it).
[1] The inequality is from Epstein-Cannon-Holt-Levy-Paterson-Thurston's Word Processing in Groups, p.239.
[2]  according to Word Processing in Groups, p. 221
Edit (solution if $n=p$):
If $n=p$ holds, this is not hard. As $\omega$ is now a $n$-form, we can write $w=f \cdot \text{vol}_M$, where $f\colon M\to \mathbb R$. Even more, we have $\|\omega\|=f$.
$$\omega(\sigma) = \int_{\Delta^p} \sigma^*(f\cdot \text{vol}_M) = \int_{\Delta^p} \sigma^*(f) \cdot \sigma^*(\text{vol}_M) = \int_{\Delta^p} \underbrace{f\circ \sigma}_{\leq |f|_\infty} \cdot \sigma^*(\text{vol}_M) \leq |f|_\infty \int_{\Delta^p} \sigma^*(\text{vol}_M) = |f|_\infty \int_{\Delta^p} |J\sigma| \text{vol}_M = \|\omega\|_\infty \cdot \text{mass}(\sigma)$$
However, the transformation formula (first to second line) cannot be easily applied for $p\neq n$ and the characterization of $\omega$ as a multiple of the volume form won't work either.
 A: I think i got it (with the help of a friends, thank btw).
Let $x\in \Delta^p$. We show that $\sigma^\ast(\omega)\leq \|\omega\|_\infty |J\sigma| \text{vol}_{\Delta^p}$.
Let's assume that $\sigma_\ast \colon T_x\Delta^p \to T_{\sigma(x)}M$ is injective, otherwise the Jacobian $|J\sigma|$ is zero, as well as $\sigma^\ast(\omega)$. Let $(c_1,\dots,c_p)$ be an orthonormal basis of $T_x \Delta^p$ and $(b_1,\dots,b_n)$ be an orthonormal basis of $T_{\sigma(x)}M$ such that $\langle \sigma_\ast(\{c_1,\dots,c_p\})\rangle = \langle b_1,\dots,b_p\rangle$. We can do this by applying Gram-Schmidt to the image of the first basis. In the following, I denote by $(b^1,\dots,b^n)$ and $(v^1,\dots,v^p)$ the dual basis.
Therefore, we can decompose the images of the $v_i$:
$$\sigma_\ast(v_i)=\sum_{j=1}^p a_{ij} b_j$$
In particular, $|J\sigma|=\det(a_{ij})_{ij}$.
Furthermore, we have
$$\sigma^\ast(b^j)(v_i) = b^j(\sigma_\ast(v_i))=\langle \sigma_\ast(v_i),b_j\rangle = \begin{cases}a_{ij}&\text{if } j\leq p\\0 &\text{if } j>p\end{cases}$$
Therefore, $\sigma^\ast(b^j)=\sum\limits_{i=1}^p a_{ij} v^i$ for $j\leq p$ and $=0$ else.
Now, $\sigma^* (b^1\wedge \dots \wedge b^p) = \sigma^*(b^1)\wedge\dots\wedge\sigma^\ast(b^p) = \bigwedge\limits_{i=1}^p \sum\limits_{j=1}^p a_{ij}v^i = |J\sigma| v^1\wedge\dots\wedge v^p = |J\sigma|\text{vol}_{\Delta^p}$.
Let $\omega=\sum\limits_{\substack{|I|=p\\I\subset\{1,\dots,n\}}}\omega_I \bigwedge\limits_{i\in I}b^i$. Then
$$\sigma^\ast(\omega) = \sum\limits_{\substack{|I|=p\\I\subset\{1,\dots,n\}}}\sigma^\ast(\omega_I) \bigwedge\limits_{i\in I}\sigma^\ast(b^i) = \sigma^\ast(\omega_{1,\dots,p})\sigma^\ast(b^1\wedge \dots \wedge b^p) \leq \|\omega\|_\infty |J\sigma| \text{vol}_{\Delta^p}$$
Integrating the inequality yields the result. I guess there are some absolute values missing at many points, but I don't really care.
What do you think? Is this the way to prove it? Probably there is a way to do it coordinate free.
