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I'm studying the proof of the Darboux-Weinstein theorem, but I'm very confused about a step.

Let $M$ be a smooth manifold, $Q\subseteq M$ be a compact submanifold, and $\omega_0, \omega_1 \in \Omega^2(M)$ be two symplectic forms on $M$ such that $\omega_0|_Q = \omega_1|_Q$, in the sense that for all $q \in Q$, the bilinear maps $(\omega_0)_q,(\omega_1)_q\colon T_qM \times T_qM \to \Bbb R$ are equal. For all $t \in [0,1]$, define $\omega_t \doteq \omega_0 + t(\omega_1-\omega_0)$.

I want to construct a neighbourhood $N_0$ of $Q$ such that all the $\omega_t$ are symplectic along $N_0$. Note that $\omega_t$ is non-degenerate along $Q$ because of the restriction hypothesis. I know we must use compactness of both $Q$ and $[0,1]$, but I'm messing up the order.

Attempt: given $t \in [0,1]$, for all $q \in Q$ there is an open set $U_{t,q}$ around $q$ such that $(\omega_t)_p$ is non-degenerate for all $p \in U_{t,q}$. Then $\{U_{t,q}\}_{q \in Q}$ is an open cover for $Q$ and we get $q_1,\ldots,q_k \in Q$ with $$Q \subseteq U_{t,q_1}\cup \cdots \cup U_{t,q_k},$$for all $t \in [0,1]$. I'd like to take the intersection in the right side, but the result will remain open only if we consider a finite amount of $t$'s.

Fixed one of these points $q_i$, for all $t \in [0,1]$ we get an open interval $I_{t,q_i}$ around $t$ such that $(\omega_s)_{q_i}$ is non-degenerate for all $s \in I_{t,q_i}$. The $\{I_{t,q_i}\}$ is an open cover of $[0,1]$ and we get $t_{1,i}, \ldots, t_{r_i,i} \in [0,1]$ such that $$[0,1] \subseteq I_{t_{1,i},q_i}\cup \cdots \cup I_{t_{r_i,i},q_i}$$for all $1 \leq i \leq k$. Intersecting we get $$[0,1]\subseteq \bigcap_{i=1}^n(I_{t_{1,i},q_i}\cup \cdots \cup I_{t_{r_i,i},q_i}).$$Now, for good or worse we have a finite quantity of $t$'s. Intuition says that putting $N_0$ as the intersection of $U_{t,q_1}\cup \cdots \cup U_{t,q_k}$ with $t$ running over these values should work.

But this does not seem uniform. I can only control non-degeneracy using continuity on each variable separately. If I actually can control both at the same time, I need to see a proof.

I am not interested in other references or constructions of tubular neighborhoods in general. I want to fix this argument. Thanks.

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2 Answers 2

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Let $(q,t)\in Q\times[0,1].$ You know that $\omega_t$ is non-degenerate at $q$. Hence there is an open rectangle $U_{q,t}\times V_{q,t}\subset M\times[0,1]$, containg $(q,t)$, such that $\omega_s$ is non-degenerate at $p$ for $(p,s)\in U_{q,t}\times V_{q,t}.$ Due to compactness of $[0,1]$, a finite number of the $V_{q,t}$'s cover $[0,1]$. Call them $V_{q,t_1},\ldots,V_{q,t_n}$. Write $$U_q:=U_{q,t_1}\cap\ldots\cap U_{q,t_n}.$$ Then $U_q$ is an open neighborhood of $q$ in $M$, such that $\omega_t$ is non-degenerate at $p$ for every $(p,t)\in U_q\times[0,1]$. Finally, take $$N_0:=\bigcup_{q\in Q}U_q.$$ I think $Q$ doesn't have to be compact.

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  • $\begingroup$ I'm sorry, I don't see why we can get a rectangle with this property, and that's exactly the problem. It comes back to what I wrote before: "I can only control non-degeneracy using continuity on each variable separately". I'd think that $(\omega_s)_p$ is non-degenerate only if $p = q$ or $s = t$, a priori. $\endgroup$
    – Ivo Terek
    Apr 4, 2018 at 17:46
  • $\begingroup$ @IvoTerek Being non-degenerate is an open condition, and you have a smooth (continuous, in particular) family of bilinear forms indexed by $M\times[0,1]$. In other words, the set of points $(p,t)\in M\times[0,1]$ such that $\omega_t$ is non-degenerate at $p$ is open in $M\times[0,1]$. The existence of the open rectangles is just the usual product topology. $\endgroup$ Apr 4, 2018 at 18:52
  • $\begingroup$ @IvoTerek If you like, you could express $\omega_0$ and $\omega_1$ locally by means of coordinates on some $U\subset M$ and then write the expression for $\omega_t$ at the point $p\in U$. This is an optional way to convince oneself that the family of bilinear forms in question is indeed smooth on $M\times[0,1]$. $\endgroup$ Apr 4, 2018 at 18:57
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    $\begingroup$ @AmitaiYuval You might want to add your comments to the answer. As is, your answer is essentially a proof of the tube lemma which does not address OP's concerns of continuity of the "full" map $M \times [0,1] \to TM^* \otimes TM^*$. That said, I agree that $Q$ doesn't have to be compact. It also does not have to be compact for what follows in the proof (e.g., a neighbourhood such that the flow of a vector field given by Moser's trick exists up to time $1$). Maybe compactness is being assumed to simplify the tubular neighbourhood theorem, or it is a "safety" assumption by the author. $\endgroup$
    – Aloizio Macedo
    Apr 5, 2018 at 1:57
  • $\begingroup$ I sent the professor an email and he said that in trying to simplify the proof, he might have complicated it, so we will review it in the start of the next lecture (which is a few hours from now). He also thinks that $Q$ does not need to be compact. I'll get back to you guys soon. $\endgroup$
    – Ivo Terek
    Apr 5, 2018 at 12:48
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As Amitai says in the comments, the map $$f:M \times [0,1] \to TM^* \otimes TM^*$$ $$(p,t) \to (p,t\omega_{0,p}+t(\omega_{1,p}-\omega_{0,p}))$$ is continuous. This is immediate from the local representation: $$\pi^{-1}(U) \stackrel{\Phi}{\simeq} U \times L(\mathbb{R}^n,\mathbb{R}^n;\mathbb{R})$$ $$(p,\omega) \to (p,\Phi_{2}(\omega)), $$ since we then have that the map is locally \begin{align*} f|_{U \times [-1,1]}:(p,t) &\to (p,\Phi_{2}(\omega_{0,p}+t(\omega_{1,p}-\omega_{0,p})) \\ &\to (p,\Phi_{2}(\omega_{0,p})+t\Phi_2(\omega_{1,p})-t\Phi_2(\omega_{0,p})), \end{align*} which is obviously continuous. So the result that around every $q \in Q$ there exists a neighbourhood $U_q$ such that $\omega_t$ is non-degenerate for every $t \in [0,1]$ follows from the tube lemma and the fact that non-degeneracy is an open condition.

Since the issue of whether $Q$ being compact is relevant or not arose, I must say that I don't see why compactness of $Q$ should play a role other than in the tubular neighbourhood theorem (and I think that it is not necessary in this case either). Note also that the neighbourhood constructed by Amitai is not necessarily a tubular neighbourhood. Let me then elaborate a bit on the proof you are probably following:

It amounts to

  • Showing that there is a neighbourhood $N_0$ such that $\omega_t$ is non-degenerate for all $t$.
  • Showing that there is a neighbourhood $N_1$ such that the flow of the vector field from a potential Moser's trick exists up to time $1$.
  • Showing that it is indeed possible to use Moser's trick by showing that $\frac{d}{dt}\omega_t$ is exact on a neighbourhood $N_3$.
  • Getting a neighbourhood which satisfies all that.

Bullet point $1$ does not need compactness, as we have seen.

Bullet point $2$ does not need compactness as well. What is behind it is the fact that we can look at the issue locally, and once we do so, we can remember that the point up to where the flow of a vector field is guaranteed to be defined by the existence and uniqueness theorem depends essentially in an inverse proportional way to the $\sup$ norm of the vector field (a vector field which will be $0$ on $Q$).

Bullet point $3$ (and $4$) is where a tubular neighbourhood appears: instead of taking an arbitrary neighbourhood $N_3$, we consider a tubular neighbourhood $N_3$ which is already inside $N_0 \cap N_1$ (this might seem circular based on how I made the constructions chronologically, but the circularity is easily circumvented by going to bullet point $3$, then $2$, then returning to $3$ with a smaller open set).

If $Q$ is compact, the way to obtain $N_3$ is simple: we have a neighbourhood $N_0 \cap N_1$ of $Q$, thus (fixing a metric on $M$), there exists a small enough $\epsilon>0$ such that the $\epsilon$-neighbourhood of $Q$ is inside $N_0 \cap N_1$. Taking a perhaps smaller $\epsilon$, we arrange a tubular neighbourhood.

If $Q$ is not compact, such a tubular neighbourhood is not necessarily of the form of an $\epsilon$-neighbourhood, and you need a bit more of work to show that there exists a tubular neighbourhood inside $N_0 \cap N_1$ (to be honest, I don't recall the details in this case).

The relevant fact of being a tubular neighbourhood is that the inclusion $i: Q \to N_3$ is an homotopy equivalence. Once we know that, it follows immediately from the homotopy invariance of de Rham cohomology that $\frac{d}{dt}\omega_t=\omega_1-\omega_0=d\sigma$ on $N_3$, since $\iota^*(\omega_1-\omega_0)$ is zero by assumption (hence, $\omega_1-\omega_0$ is exact). Relevant to say: the argument given for this part tends to be rather convoluted in some standard texts, for no apparent good reason.

So, unless there is some subtlety in the tubular neighbourhood theorem for non-compact submanifolds which I'm overlooking (or I overlooked some other reason which I can't point), I don't see why $Q$ being compact should be necessary.

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    $\begingroup$ @IvoTerek $\Phi_2$ is $\pi_2 \circ \Phi$, where $\Phi$ is the local homeomorphism on the definition of a vector bundle. The reason for $f$ being continuous is the same reason for why $g: \mathbb{R}^n \times \mathbb{R} \to \mathbb{R}^m$ given by $g(x,t)=g_1(x)+tg_2(x)-tg_3(x)$ is continuous if $g_1,g_2,g_3: \mathbb{R}^n \to \mathbb{R}^m$ are continuous functions (quite literally). $\endgroup$
    – Aloizio Macedo
    Apr 6, 2018 at 2:33
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    $\begingroup$ @IvoTerek Not very relevant for this case, but very important by itself: we can't necessarily arrange for the closure of a tubular neighbourhood to be diffeomorphic to $Q \times [-\epsilon,\epsilon]$. The normal bundle may be twisted. $\endgroup$
    – Aloizio Macedo
    Apr 6, 2018 at 2:38
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    $\begingroup$ @IvoTerek The version of Moser's trick you have assumes that $M$ is compact for the sake of not bothering about the fact that the flow may not be defined for all time. If you want to have the result in OP for non-compact $M$, I don't see how to evade the fact that you must take a small neighbourhood of $Q$ such that the flow is defined up to $1$ (which is really not that much a hassle anyway). If you can live with having the result in OP for compact $M$, then there is nothing to fear, and you can apply your version of Moser's trick directly. $\endgroup$
    – Aloizio Macedo
    Apr 6, 2018 at 2:42
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    $\begingroup$ @IvoTerek You don't need the relatively compact tubular neighbourhood. The vector field you obtain is defined on $M$, which is assumed compact, so its flow exists up to any time you want. $\endgroup$
    – Aloizio Macedo
    Apr 6, 2018 at 2:47
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    $\begingroup$ The point is that a "small enough" neighbourhood of the zero section in $NQ$ is not necesarrily something like $Q \times (-\epsilon,\epsilon)$. Firstly, $Q$ need not be a codimension $1$ manifold. But even if it were, consider the "standard" $S^1$ inside the Moebius band. You only get something like what you suggest if the normal bundle is trivial. $\endgroup$
    – Aloizio Macedo
    Apr 6, 2018 at 2:53

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