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Here is an extract of my Symplectic Geometry course notes.

Remark 2.57. Let $M$ be a smooth manifold, $Q\subseteq M$ a compact submanifold and $\omega_0,\omega_1\in\Omega^2(M)$ symplectic forms such that for all $t\in[0,1]$ and $q\in Q$, $\omega_t:=(1-t)\omega_0+t\omega_1$ is non degenerate in $q$ (hence on $T_qM$). Then there exist an open neighbourhood $U$ of $Q$ such that for all $t\in[0,1]$, $\omega_t$ is symplectic on $U$.
If this were false infact, there would be a sequence $(t_j)_{j\in\mathbb{N}}$ in $[0,1]$ and one $(m_j)_{j\in\mathbb{N}}$ in $U_{1/j}$ such that $\omega_{t_j}$ is degenerate in $m_j$, but $[0,1]$ and $Q$ are compacts, so we can extract a subsequence $j_k$ with $k\in\mathbb{N}$ such that the two sequences converge.

$$t_{j_k}\xrightarrow[k\to\infty]{}t_\infty\qquad\qquad m_{j_j}\xrightarrow[k\to\infty]{}m_\infty$$

Now, since $m_{j,k}\in U_{1/j_k}$, $m_\infty\in U_{j_k}$ for all $k$ (otherwise we can find an open neighborhood of $m_infty$ disjoint with all the $m_{j_k}$ for all $k$ large enough). Thus $m_\infty\in\bigcap_kU_{1/j_k}=Q$. Now in every local chart centred in $m_\infty$, the matrix representing $\omega_{t_\infty}$ in $m_\infty$ is the limit of the matrices representing $\omega_{j_k}$ in $m_{j_k}$ which are all degenerate, so they have determinant 0. Since the determinant is continuous, also the determinant of the matrix representing $\omega_\infty$ in $m_\infty$ must be 0 and thus $\omega_\infty$ is degenerate in $m_\infty$, which is against our hypotheses for $m_\infty\in Q$.

Note: $U_\delta$ is a tubular neighborhood of $Q$ of radius $\delta$, that is, we have any Riemann structure on $M$, nd we consider $N_\delta$, the subset of the normal bundle of $Q$ consisting of only vector of norm at most $\delta$, and for $\delta$ small enough the exponential map will map $N_\delta$ diffeomorphically onto an open neighborhood of $Q$, which is what we call $U_\delta$. This $U_\delta$, whenever defined, can be deformation retracted onto $Q$.

Now, apart from certain English errors (which are transcribed verbatim from those notes), and some typos (also verbatim, AFAIK), my problem in this proof is that, though $Q$ is indeed compact, $m_j$ is not a sequence in $Q$. What "one $(m_j)_{j\in\mathbb{N}}$ in $U_{1/j}$ means is that $m_j\in U_{1/j}$ for all $j$. So what has $Q$ got to do with the sequence, except for the fact that the $U_{1/j}$ are defined starting from it, and that any limit point of $(m_j)$ will have to be in $Q$ because the $U_{1/j}$ decrease towards it and so any $m_k$ is in any $U_{1/j}$ with $j\leq k$, meaning a limit point must be in all of them, hence in $Q=\bigcap_kU_{1/k}$? Or in other words:

How does compactness of $Q$ give me a converging subsequence of $(m_j)$ when $\{m_j\}_j\not\subseteq Q$?

Edit: one idea

Naturally $\{m_j\}_j\subseteq U_1$, so if one could prove that $U_1$ is (relatively) compact, then we would be done. Actually, if any of those tubular neighborhoods are relatively compact, a "tail" of the sequence (i.e. $\{m_j\}_{j\geq k$ for some $k\in\mathbb{N}$ – what's that in English? In Italian it's "coda", hence "tail") would be in a compact set and hence have a converging subsequence, and again, we'd be done. But how can I prove that they are compact? Could it be that $N_\delta$ is compact? I mean, $Q$ is, and we are limiting the norm, so perhaps $N_\delta$ is compact, and hence $U_\delta$ is compact since it's diffeomorphic (hence homeomorphic) to $N_\delta$. Is that so though?

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  • $\begingroup$ The standard tubular neighborhood theorem gives you a compact neighborhood of any smoothly embedded compact submanifold. If $U_{1/k}$ is the closed disk normal bundle over $Q$ with radius 1/k (wrt some Riemannian metric), this is automatically compact. $\endgroup$ Aug 28, 2017 at 22:15
  • $\begingroup$ @PVAL-inactive "this is automatically compact": why? The "standard tubular neighborhood theorem", I am afraid, is not something I am familiar with. We have only constructed tubular neighborhoods as described in the question. We have never proved they were compact. If the standard t.n. theorem uses the construction, we have only partially proved it. I believe my answer proves compactness of the $N_\delta$'s which are diffeomorphic to the $U_\delta$'s involved in this question, thus solving the problem. $\endgroup$
    – MickG
    Aug 28, 2017 at 22:28

1 Answer 1

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$N_\delta$ is indeed compact.

To see this, for every $q\in Q$, consider a chart of $N$ (the whole normal bundle) centered at $(q,0)$. Let $\phi_q$ be the chart map $\phi_q:U_q\to A_q$, where $U_q\subseteq N$ is the neighborhood of $(q,0)$ where $\phi_q$ is defined, and $A_q\subseteq\mathbb{R}^k$, where $k:=\dim Q$, is the open image of $\phi_q$. For any $(q',0)\in U_q$, and any $v\in(T_{q'}Q)^\perp$, we will have $(q',v)\in U_q$ as well, because the topology in the normal bundle is a subspace topology from $TM$ and the charts there have this property. Now let's get back to $N_\delta$: we restrict $\phi_q$ to $N_\delta\cap U_q$. The image of $\phi_q$ will be of the form $A'_q\times\mathbb{R}^{d-k}$, with $d:=\dim M$, and the image of that restriction will be $A'_q\times B_{\mathbb{R}^{d-k}}(0,\delta)$. Let $\phi_q(q,0)=(a_q,0)$. Inside $A'_q\ni a_q$, we can surely find a compact set (e.g. a closed ball) containing $a_q$. So let us consider e.g. $O_q:=B_{\mathbb{R}^k}(a_q,\delta_q)\times B_{\mathbb{R}^{d-k}}(0,\delta)$, where $\delta_q$ is such that that ball is included in $A'_q$. The family $\{\phi_q^{-1}(\mathring O_q)\}$ is an open cover for the set $\{(q,0):q\in Q\}\subseteq N_\delta$, which is compact since $Q$ is, so we can find a finite subcover $\{\phi_{q_j}^{-1}(\mathring O_{q_j})\}_1^n$. That is in fact a cover of $N_\delta$ with finitely many sets, and their closures are $\phi_{q_j}^{-1}(O_{q_j})$, that is preimages via homeomorphisms of compact sets, so we have just covered $N_\delta$ with finitely many compact sets. But a finite union of compact sets is itself compact, and hence $N_\delta$ is compact.

So there you go. Problem solved.

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