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This is about proving connectedness of level sets of moment map of a $\mathbb T^n$ $\implies$ convexity of image of moment map for $\mathbb T^{n+1}$ action. I am following Ana Cannas's wonderful lecture notes but I got stuck at a point in the proof where it says for any two points $p_0,p_1 \in M$ there is a sequence of points $q_n \rightarrow p_0,l_n\rightarrow p_1$ such that $\mu(p_1) - \mu(p_0)$ is in $ker A^t$ for some integer $(n+1) \times (n)$ matrix $A$ with rank n.

I realised that it'd be enough to get a sequence such that $\mu(p_1) - \mu(p_0)$ has all rational coefficients is enough, but I am confused about how to do this.

I looked at this part in Dusa Mcduff's book and it doesn't elaborate on this topic either, so I have a feeling it might be really trivial but I guess I am missing some trick/ easy observation.

Soham
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1 Answers1

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There may be a simpler way of doing this, but here is an approach:

It suffices to consider the case when the moment map is effective (see part 4 HW21).

To show that any point $q$ of $\mu(M)$ can be approximated by points with rational coordinates, if suffices to prove that $\mu(M)$ has non-empty interior. Indeed, if $q_0$ is an interior point of $\mu(M)$, then $\mu(M)$ contains an open ball $B\subseteq \mu(M)$, and hence (by "rational slope segment convexity" proved before) $\mu(M)$ contains all "rational slope" segments from $q$ to points in $B$, and hence the closure of the set of all such segments. This closure is the solid cone connecting $q$ to the closure of $B$, and it contains points with rational coordinates arbitrarily close to $q$.

To show that $q_0$ exists we just need to find $x\in M$ with $d\mu_i(x)$ linearly independent. Let's call such points $x$ "good", and all others "bad". A "bad" point $y$ is one where exists $X$ such that $<\mu, X>$ is critical at $y$, and so $y$ is a fixed point of some 1 paramter subgroup of the acting torus, so is fixed point of the subtorus that is the closure of this subgroup. This subtorus contains a subcircle generated by rational $X^r$ meaning that $y$ is critical for $<\mu, X^r>$. That means that the set of bad points is the union of critical submanifolds of function $<\mu, X^r>$ as $X^r$ varies over rational vectors. For each individual $X^r$ the critical set is a symplectic submanifold of $M$ (HW21, part 2), and if the action is effective it is proper submanifold, so of codimension at least 2. Since a countable union of codimension 2 submanifolds can not cover $M$ we know that a good point exists and we are done.

Max
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  • I'm sorry but I can't follow last part of your argument.

    Why does action being effective $\implies$ the critical submfd are proper and of codim 2? Do you really need codim 2? I guess codim 1 would suffice as the measure of codim 1 mfd is 0 thus countable union of codim 1 can't cover M, but in this argument I need some orientation on M, how to bypass that?

    – Soham Apr 16 '19 at 15:52
  • Okay, I figured out the first part of my question. As each critical set is the fixed point set of a subcircle of the torus, as action is effective, it can't be a codim0 submfd thus it has to be codim 1, but it is symplectic thus codim can't be 1.

    I am still unsure about the covering contradiction though.

    – Soham Apr 16 '19 at 16:24
  • A symplectic $M$ is orientable but that is beside the point. A countable union of codimension at least 1 submanifold can not cover any neighborhood of any $M$, orientable or not -- their image in any local chart will be of measure 0. – Max Apr 16 '19 at 17:40
  • Oh yes, that was stupid of me. Now another issue I have is that when you're saying using HW21 Part 4, you're using a different definition of effectiveness. The moment map is effective if $d \mu_1,d\mu_2,\dots d\mu_n$ are linearly independent but how does that imply the group action is effective? The example I have in mind is that if $S^1$ acts on $S^2$ by rotation by twice the normal speed, then the action is not effective but the moment map is still so. – Soham Apr 16 '19 at 18:08
  • My intention was to use the "moment map is effective" in both cases. I'll edit the answer. – Max Apr 16 '19 at 18:40
  • One can see that the case of effective moment map is sufficient directly, but also an effective group action will have an effective moment map (your example disproves the converse implication). – Max Apr 16 '19 at 18:54