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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.

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  • $\begingroup$ Yes, for the first part, you're choosing local coordinates on $M$ in which $S$ is given by $x^n=0$. For the second part, maybe you should think about the inverse map $\Theta^{-1}$ (remembering that $\Lambda^{n-1} (T^*_PM)$ and $T_PM$ are each $n$-dimensional). $\endgroup$ – Ted Shifrin Sep 19 '16 at 22:32
  • $\begingroup$ @TedShifrin: Shouldn't there also be a comment about the fact that $\alpha_p = \sum_{i=1}^{n-1} a_n(p) \ (dx^i)_p$? (i.e a sum, not just $a_n dx^n$). $\endgroup$ – Faraad Armwood Sep 19 '16 at 22:37
  • $\begingroup$ @Faraad: No, since $S$ is locally given by $x^n=0$, $dx^n$ is the $1$-form that annihilates its tangent spaces. So $\alpha = f dx^n$ for some nowhere-zero smooth function $f$. $\endgroup$ – Ted Shifrin Sep 19 '16 at 22:38
  • $\begingroup$ @TedShifrin: Hmm, I need to read some more then because if anything the sum should be up to $n-1$ since $(x^1,...,x^{n-1},0)$ locally defines $S$. $\endgroup$ – Faraad Armwood Sep 19 '16 at 22:40
  • $\begingroup$ Nope, @Faraad. Think about $1$-forms $\omega$ on $\Bbb R^2$ whose restriction to the $x$-axis is $0$. (Officially, pull back by the inclusion map.) Do they look like $f\,dx$ or $f\,dy$? $\endgroup$ – Ted Shifrin Sep 19 '16 at 22:42
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(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.

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