Recover the differential form by its pullback

I have a smooth manifold $M$ in $\mathbb{R}^n$ given by $M = \{ x \in \mathbb{R}^n \mid g(x) = 0 \}$. Its atlas consists of a single chart $(M,\varphi)$, where $$\begin{array}{rcl} \varphi(x_1,...,x_n) & = & (x_2,...,x_n), \\ \varphi^{-1}(y_1,...,y_{n-1}) & = & (x_{1}(y),y_1,...,y_{n-1}) \end{array}$$ where $g(x_{1}(y),y_1,...,y_{n-1}) = 0$ (such $x_{1}(y)$ is unique) and $\varphi(M) = \mathbb{R}^{n-1}$.

A differential form $\omega$ on $\mathbb{R}^{n-1}$ is given by $$\omega = \frac{a(x_{1}(y),y_1,...,y_{n-1})}{\partial_{1} g(x_{1}(y),y_1,...,y_{n-1})} dy_{1} \wedge ... \wedge dy_{n-1}.$$

Is it possible to find a form $\alpha$ on $M$ such that $\omega = (\varphi^{-1})^{*} \alpha$?

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What do you mean by "a form $\alpha$ on $M$"? By specifying $\omega$ on $\mathbb{R}^{n-1}$, you've specified a differential form on $M$. Do you mean that you are looking for a differential form on the ambient space $\mathbb{R}^n$ such that $\omega$ on $M$ is its pullback under the embedding map? –  Willie Wong Feb 3 '12 at 12:20
$\int\limits_{M} \alpha = \int\limits_{\mathbb{R}^{n-1}} \omega$, so the question is to specify $\alpha$ by $\omega = (\varphi^{-1})^{*}\alpha$ –  Nimza Feb 3 '12 at 12:29
Let me put the question to you in a different way: if $\alpha$ solves $\omega = (\varphi^{-1})^*\alpha$, then so would $\alpha + dg\wedge\beta$ for any $n-2$ form $\beta$ defined on $\mathbb{R}^n$. My question is: are you asking about the "intrinsic" problem, where $M$ is treated as a manifold in itself, and for which I cannot understand why you want to distinguish between $\alpha$ and $\omega$, or are you asking about the "extrinsic" problem, in which you take $M$ to be a particular submanifold, and "$\alpha$ on $M$" means you are thinking about the restriction of a form $\alpha$ defined on –  Willie Wong Feb 3 '12 at 13:25
$\mathbb{R}^n$ to $M$. In the latter case you don't have uniqueness, but existence is easy, as long as $\varphi^{-1}$ is a smooth map. –  Willie Wong Feb 3 '12 at 13:27
The problem is extrinsic, so $\alpha$ is not unique, but do such $\alpha$'s have the same restriction to $M$? –  Nimza Feb 3 '12 at 13:54

Assuming the map $\varphi^{-1}:\mathbb{R}^{n-1}\to\mathbb{R}^n$ is smooth (which, if you are defining $\varphi$ the way I think you are, should be automatic). Then by definition $\mathrm{d}\varphi^{-1}: T_x\mathbb{R}^{n-1} \to T_pM$ where $\varphi(p) = x$. To find an $\alpha$ defined along $M$ such that it pullsback via $\varphi^{-1}$ to $\omega$, it suffices to specify, since $\alpha(p) \in \wedge^{n-1} T_p^*\mathbb{R}^n$, how it acts on a basis of $\wedge^{n-1} T_p\mathbb{R}^n$.

For the following we´ll also need to assume that $\iota := \mathrm{d}\varphi^{-1}$ is injective. Let $e_i$ be the standard basis vector fields on $\mathbb{R}^{n-1}$. Then by construction $f_i = \iota\circ e_i$ are linearly independent vector fields along $M$ that span $TM$. Smoothly complete this with a smooth vector field $f_0$ along $M$ that is transversal to $TM$, then $\{f_0, f_1,\ldots, f_{n-1}$ define a basis for $T\mathbb{R}^{n}$, which induces a basis for $\wedge^{n-1}T\mathbb{R}^n$.

Now define $\alpha$ such that $\alpha(f_i\wedge f_j \wedge \cdots \wedge f_k) = \omega(e_i\wedge e_j\cdots\wedge e_k)$ if none of the indices are 0. And define $\alpha(f_0\wedge\cdots) = 0$. It is easy to check that by definition $\omega = (\varphi^{-1})^* \alpha$.

Notice that the $\alpha$ depends on how you ¨complete the basis¨, that is, which vector field you choose as $f_0$. But the restriction of all the different $\alpha$s to $TM$ are the same, since $f_1, \ldots, f_{n-1}$ span $TM$. Notice that if $\alpha$ and $\alpha´$ are two such forms, we have that $(\alpha - \alpha´)(X_i\wedge\cdots\wedge X_j) = 0$ whenever all $X_i\in TM$. By dimension counting and a bit of linear algebra, you´d see that $\wedge^{n-1}T\mathbb{R}^n$ is $n$ dimensional, and so the space of $\beta$s with the property that it vanish on $(X_i \wedge \cdots \wedge X_j)$ for $X_i\in TM$ is $n-1$ dimensional. Since the image of $dg \wedge : \wedge^{n-2}T\mathbb{R}^n \to \wedge^{n-1}T\mathbb{R}^n$ is also $n-1$ dimensional, you have that necessarily $\alpha - \alpha´ = dg \wedge \gamma$ for some $n-2$ form $\gamma$.

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Thank You, one of such forms in my case is $\frac{ a(x_1,...,x_n)}{\partial_{1} g(x_1,...,x_n)} dx_2 \wedge ... \wedge dx_{n}$, isn't it? –  Nimza Feb 3 '12 at 15:25
Indeed, that is one of the solutions. It is interesting to think about why the form, which is in fact the pullback of $\omega$ under the canonical projection $(x_1,x_2,\ldots,x_n) \mapsto (x_2,\ldots,x_n)$, is a solution to your problem. –  Willie Wong Feb 3 '12 at 16:00
Dear Willie, $M$ is just the graph of the function $x_1:\mathbb R^{n-1}\to \mathbb R$ (in Nimza's notation), right? –  Georges Elencwajg Feb 3 '12 at 17:25
Yes, it is the graph of $x_1(\cdot)$. –  Nimza Feb 3 '12 at 18:16
Let $\alpha = \sum\limits_{k=1}^{n} (-1)^{k-1} \frac{a(x) \partial_{k}}{ |\nabla g(x)|^2 } dx_1 \wedge ... \wedge \overline{dx_k} \wedge ... \wedge dx_n$ (so $dg \wedge \alpha = dx$). I want to test if $\int\limits_{M} \alpha = \int\limits_{\mathbb{R}^n} \omega$, but $\alpha$ doesn't solve my initial problem. So is there another way to test it? –  Nimza Feb 3 '12 at 18:38