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Suppose $U$ is an open subset of $\mathbb R^n$, and $\omega:U\to\operatorname{Alt}^k(\mathbb R^n)$ is a smooth differential $k$-form. By $\operatorname{Alt}^k(\mathbb R^n)$ I mean the set of alternating $k$-linear forms, and by "smooth" I mean that $\omega$ is infinitely differentiable (since $\operatorname{Alt}^k(\mathbb R^n)$ is a real normed space, we can make sense of $D\omega$ and $D^2\omega$, etc., where $D$ denotes the Fréchet derivative).

We can write $\omega$ in coordinates as $\sum_{I}\alpha_Idx^I$, where each $\alpha_I:U\to\mathbb R$ is a function. However, how do we conclude that each $\alpha_I$ is smooth from the fact that $\omega$ is smooth?

Joe
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    You have a fixed basis for $\text{Alt}^k(\Bbb R^n)$. So you're just asking why an $\Bbb R^N$-valued function is smooth in one sense if and only if it's smooth in the other, and this reduces to the same question for $\Bbb R$-valued functions. – Ted Shifrin Jan 12 '24 at 19:27
  • @TedShifrin: Thanks for your response. Could you elaborate on how the fact that $\operatorname{Alt}^k(\mathbb R^n)$ having a fixed basis reduces the question to one about $\mathbb R^N$-valued functions? Do you mean that we should use the fact that $\operatorname{Alt}^k(\mathbb R^n)$ isomorphic to $\mathbb R^N$ for some natural number $N$? – Joe Jan 12 '24 at 22:08
  • Yes, and the basis of $dx^I$ ($I$ increasing) gives the isomorphism, of course. – Ted Shifrin Jan 12 '24 at 22:21

1 Answers1

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Fréchet-differentiability of $\omega:U\to \bigwedge^k (\mathbb R^n)^*$ (at $x\in U$) means that a linear operator $D_x:\mathbb R^n \to \bigwedge^k (\mathbb R^n)^*$ exists so that $$\omega(x+\Delta)=\omega(x)+D_x(\Delta)+o(\Delta).$$

We'll show that the components of $D_x$ in the $\{dx^I\}_I$ basis are the derivatives of $\omega_I$, thus rendering $\omega_I$ differentiable, too.

The equation above is in $\bigwedge^k (\mathbb R^n)^*$, so we can expand all terms in the $dx^I$ basis (including $o(\Delta)$): \begin{align} \omega(x+\Delta)&=\omega(x)+D_x(\Delta)+o(\Delta)\\ &= \omega_I(x)dx^I + \left(D_x(\Delta)\right)_Idx^I + o(\Delta)\\ &= \left(\omega_I(x) + \left(D_x(\Delta)\right)_I + o(\Delta)\right)dx^I.\tag{1} \end{align} (omitting $\sum_I$ signs). Expand the left-hand side too $$\omega(x+\Delta)=\omega_I(x+\Delta)dx^I\tag{2}$$ and equate the components in front of $dx^I$ in (1) and (2):

$$\omega_I(x+\Delta)=\omega_I(x) + \left(D_x(\Delta)\right)_I + o(\Delta).$$

This is precisely the definition of differentiability of $\omega_I$ with a derivative operator $(D_x)_I$.
I hope you can do higher derivatives on your own from here on.

If you're wary about abuse of notation, keep in mind that in the vector equalities $o(\Delta)$ denotes some $\bigwedge^k(\mathbb R^n)^*$-valued function, while in the scalar equalities (from (1) onward) it is an $\mathbb R$-valued one, in both cases with $\lim_{\Delta\to 0}\frac{o(\Delta)}{\|\Delta\|}=0$.

Al.G.
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