Here's the question I actually want to answer: Say $W$ is smooth, compact with boundary $M_0\sqcup M_1$ and $f\colon W\rightarrow[0,1]$ is a submersion with $f(M_0)=0$ and $f(M_1)=1$. Is $f^{-1}(0)=M_0$?
Benjamini-Schramm limits have some kind of "time-reversal" symmetry always, which is known as unimodularity in literature. The big open problem is if every unimodular graph a Benjamini-Schramm limit, which would imply every group is sofic.
(I'm working with a very different kind of limit which is much closer to an inverse limit of discrete graphs to get some zero dimensional metric graph in the end)
@tigre Just consider them Set-valued, since after all when I talk about a module I'm asking that there be a set-map R x M -> M with certain properties.
But yes it is @Thorgott. Take $f^{-1}(0) = S$. If $S$ has an interior component then the differential maps $df: NS \to \Bbb R$ isomorphically on each fiber. In particular $f$ decreases on "one side" of $N_x S$ and increases on the other side. If $f(x) = 0$ this prohibits $x$ being an interior point.
In general if $f: M \to N$ is a submersion of manifolds with boundary this argument shows that $f^{-1}(\partial N) \subset \partial M$.
Ah wait, it's much easier to see contrapositively. If $p$ is interior in $M$, every tangent vector at $p$ can be represented by a curve with domain $(-\epsilon,\epsilon)$, but then, since $f$ is a submersion, every tangent vector at $f(p)$ can be represented by a curve with domain $(-\epsilon,\epsilon)$ and so the image is an interior point.
If we let P(n) be the set of all positive definite, symmetric nxn matrices and T(n) be all the upper-triangular matrices with positive diagonal, then is there an obvious way that the map $A \mapsto A^\top BA$ is an isomorphism (even bijection) from $T(n)\to P(n)$, where $B\in P(n)$ is fixed? I was trying to fiddle around with the spectral decomposition which it allegedly relates to but I am not sure how.
@anakhro This is correct. You have an action of the group T(n) on the space of positive definite symmetric matrices. To see that it acts transitively on some element B it suffices to show it acts transitively on I, aka, that every pos def symmetric guy is of the form A^T A for A upper triangular with positive diagonal.
You know that as soon as you know (a) that every pos def symmetric guy is A^T A for some A and (b) the QR decomposition.
To see that it acts without stabilizer on one element it now suffices to show it acts without stabilizer on the identity matrix, aka, that if A^T A = I for A upper triangular with positive diagonal entries, then A = I. This seems clear from an inductive argument.
Actually I guess induction is unnecessary ...
Seems pretty easy to construct A by hand given the information that A^T A = B
In particular if you have a group G acting on a set X so that the action is transitive and G_x = {1} for some element x, then in fact G acts freely on X; the map g mapsto gx is a bijection for any x in X
(If the action-on-x map gives a bijection G -> X for one x, it does for all x)
Yes
It's probably not a standard phrase, I hope it's clear why those are meant to mean the same thing.
Your "mild amusement" in observing folks coming on the chat and asking questions that require just a modicum of their own effort to answer is leaking out @copper.hat....😝
ah....lol.....well I have peeped those the last few weeks whenever you've hyperlinked a vent of frustration here......🤣..............but folks have been getting real lippy with you when you are providing help. Which to me is amazing...considering you are doing this out of your own good will...
young whoopersnappers haven't fathomed a time when you didn't have the privilege of being able to search on the internet for instant help.......
