Exercise about a 1-form on a manifold

Question

I'm struggling a little bit thinking about this apparenlty innocent exercise. I'll provide an incomplete solution.

Exercise

Let $M$ be an $n$-dim manifold and let $S\subset M$ be an embedded $(n-1)$-dim submanifold. For every $P\in M$ let's identify the tangent space $T_PS$ with its image in $T_PM$ through the differential of the inclusion of $S$ in $M$.

$(i)$ Let $\alpha\in A^1(M)$ be a $1$-form on $M$ such that $T_PS\subseteq \ker(\alpha_P)$, for every $P\in S$. Prove that $(\alpha\wedge d\alpha)_P=0$, for every $P\in S$.

Solution (i)

Now, if $\alpha$ is a $1$-form on $M$ such that $T_PS\subseteq \ker(\alpha_P)$ for every $P\in S$, then, for every $P\in S$ there exist a local chart $\phi=(x^1,\ldots,x^n)$ in $P$ such that, with respect to that coordinates,

$$\alpha=\alpha_n(x)dx^n,$$

i.e. all the coefficients of $\alpha$ are zero except for the $n$-th (but I'm not at all sure about it!). So we get

$$d\alpha=\sum_{i=1}^{n-1}\frac{\partial\alpha_n(x)}{\partial x^i}dx^i\wedge dx^n$$ and

$$(\alpha\wedge d\alpha)=\alpha_n(x)dx^n\wedge\sum_{i=1}^{n-1}\frac{\partial\alpha_n(x)}{\partial x^i}dx^i\wedge dx^n=\sum_{i=1}^{n-1}\alpha_n(x)\frac{\partial\alpha_n(x)}{\partial x^i}dx^n\wedge dx^i\wedge dx^n=0$$

$(ii)$ Let's now suppose that $M$ is oriented and let $\omega\in A^n(M)$ be a volume form associated to the fixed orientation. Prove that for every $(n-1)$-form $\phi\in A^{n-1}(M)$ there exists a unique vector field $X^\phi\in\mathcal{T}(M)$ such that

$$\phi_P(v_1,\ldots,v_{n-1})=\omega_P(v_1,\ldots,v_{n-1},X_P^\phi)$$

for every $P\in M$ and every $v_1,\ldots,v_{n-1}\in T_PM$. Besides, prove that $\Theta\colon A^{n-1}(M)\to\mathcal{T}(M)$ defined by $\Theta(\phi)=X^\phi$ is an isomorphism.

For this second problem I really don't know how to begin.

Answer 1

(1) I am still patching together a solution but thanks to @Ted Shifrin, I can clarify your first remark about the expression $\alpha$. Let $\iota: S \to M$ be the inclusion map then $\iota^*: T_pM \to T_pS$. Now take $(U,\phi) = (U, x^1,...,x^n)$ to be a chart on $M$ then $\omega \in \Omega^1(M) \Rightarrow \omega_p = \sum_{i=1}^n a^i(p)\ dx^i$. Consider;

$$\iota^*\omega = \sum_1^n (a^i \circ \iota) \ d(x^i \circ \iota) = \sum_1^{n-1} (a^i \circ \iota) \ d(x^i \circ \iota)$$

i.e if you want a 1-form on $M$ that has your property, you need to dispose of $dx^j$ for $1 \leq j \leq n-1$. Hence you have $\alpha = f dx^n$ where $f$ is non-vanishing.

Answer 2

Your approach for (i) only works when $\alpha$ has it's kernel in a foliation defined by some function. However, the situation you are in is more delicate and we have to do some things pointwise. For instance, we could have $\alpha = x \alpha_0$ where $\alpha_0 = dz-ydx$ is the standard contact structure in $\mathbb{R}^3$. This certainly cannot be locally expressed as $fdx$ near the $x = 0$ plane.

First, using the inclusion $i : S \subset M$, we have that $i^*\alpha = 0$ and hence $i^*d\alpha = d i^*\alpha = 0$. This means that for any vectors $u,v \in T_p M$, tangent to $S$ at $p \in S$, we have $\alpha(u) = 0 = d\alpha(u,v)$.

If we contract with two tangent vectors $u,v$ to $S$ at $p \in S$, we thus have \begin{align*} i_u i_v(\alpha \wedge d\alpha)|_p &= i_u(i_v\alpha|_p \wedge d\alpha|_p + (-1)^2\alpha|_p \wedge i_vd\alpha|_p)\\ &= i_u(\alpha|_p \wedge i_vd\alpha|_p)\\ &= i_u\alpha|_p \wedge i_vd\alpha|_p + (-1)^1\alpha|_p \wedge i_ui_vd\alpha|_p\\ &= 0. \end{align*} Using basic linear algebra, it suffices to check the claim with two kinds of triples of vectors $(n,u,v) \in (T_p M)^3$. The first case is when each vector is tangent to $S$ and the second is when $n$ is transverse to $S$ at $p$ while the rest are tangent. Our above calculation shows that the result will always be zero. Thus, $\alpha \wedge d\alpha|_S = 0$.

Concerning (ii), this is can be analysed at a point. That is, in a vector space $V$ of dimension $n$ with volume form $\mu$, one can show that the $(n-1)$ forms are of dimension $n$. After this, just note that the map (a contraction) has trivial kernel because $\mu$ is a volume form and apply rank-nullity. That your map sends smooth things to smooth things just follows from looking at everything in a chart.