whereas for pi/3 i immediately think of a semicircle divided into 3 wedges
Where this comes into physics is stuff like $\omega$ vs. $\nu$
($\nu$ is another case: I think it's better practice to use $\nu$ than $f$ for frequency, but in relating to $\omega$ I always think of $\omega=2\pi f$. but that can be chalked up to intro-level classes always using $f$)
I always forget the Lie derivative. If $T$ is a tensor field in general say, $\mathcal{L}_X T$ is $\partial_t \rho_t^* T$ or $\partial_t (\rho_t)_* T$?
god i have started writing like a physicist
2
i mean do you pull $T$ back by flowing for a little while or do you flow $T$ for a little while?
At least on exam attendance it looks good. Of the people enrolled in the course, only 4 did not qualify for the exam, and all those 4 had completely abandoned their studies. And of the people qualified for the exam, only 4 did not show up
My favourite experience in grading was the statement "$G$ is isomorphic, but $H$ is not" to question whether $G$ and $H$ are isomorphic. I also liked the student who spend over four pages computing the determinant of a 5x5 matrix repeatedly using Laplace expansion on rows that had no zeroes at all, although it was a god damn lower block matrix made of 1x1,2x2 and 2x2 blocks
@MatheinBoulomenos One of the exam problems here has them show that a set of congruences has precisely 2 solutions in the interval from $0$ to $118$. One student claimed to have checked them all
Tobias: In the US, if you fail the exam typically you fail the course. I have been generous on a few occasions and broken rules to allow people a last chance to pass if they needed it to graduate (open-book retake of the final but you need to score 80 or some such) — not once has a student ever passed.
@Mathein so I graded for discrete in the fall and one of the problems was that if you have an $n$-letter alphabet, find the probability that a word of length $k$ has no repeated letters
For one thing, Tobias, an exam can only (fairly) test a certain tiny amount of the core material. I generally want to use exams to determine competence (and maybe a little bit to delineate Bs and As), but the real challenges come on homeworks.
If $X$ is a completely normal topological space, then for every pair of separated sets $A,B$ (i.e., $\overline{A} \cap B = \emptyset = A \cap \overline{B}$), there exist disjoint open sets $U$ and $V$ such that $A \subseteq U$ and $B \subseteq V$ [hint: consider $X-(\overline{A} \cap \overline...
@Balarka: What do you think about this? The OP put the same wrong tags back after I removed them ... I actually have never thought about homotopy type of 3-element topological spaces. Have you?
Hint: This is $[0,1]$ quotiented by the equivalence relation whose equivalence classes are $\{0\}$, $(0,1)$, and $\{1\}$. — Akiva Weinberger57 secs ago
I think I worked with a scheme that has the topology from the question before. It's what you get when glue two copies of the Sirpinski space (e.g. the Spec of a DVR) together along their generic point. I used that as a counterexample to something, but I can't remember for what
Yeah, this reminds me of the exam question I tried to write for my take-home topology final a few years ago -- a leaf space of a foliation where, I thought, there would be a dense point.
I think I gave you the question. I was doing things like $[-1,1]\times\Bbb R$ with the leaves the two vertical lines at the end and the vertical translates of $y=\tan(\pi x/2)$.
@AkivaWeinberger Like, triangulate S^3 and S^2 reasonably so that the Hopf map becomes simplicial. Then write it down at the level of corresponding finite topological spaces given by squishing the interior of the simplices.
It'd be a map between finite spaces, which would be kinda cool
"When translational symmetry is broken in the ground state, the homotopy theory of defects of ordered media has to be supplemented with integrability conditions, coming from the theory of foliations. These show how some homotopy classes split into several distinct defects, while other homotopy classes do not occur physically. This framework can also be used in order to discuss defects of gauge fields, where in a first approximation classifying spaces play the role of the manifolds of internal states. "
I think it'd be kinda trolly and fun. You have this map {null, a, b, c, {a, b}, blah blah } --> {more bullshit} given by f(a) = x y z, this is the Hopf map
Totally symbolic description
@Semiclassical I don't really know the math or physics behind this but the familiar names from topology makes me happy
(If you have a simplicial attachment of 4-disk to S^2 then extend the triangulation of S^3 under which you have the Hopf map as simplicial by coning off to a triangulation of D^4 then attaching that bikh)
So I looked up "completely normal" on Google hoping to find a definition, but I forgot to include the word "mathematics" in the search and I got this
> Comedy. Awkward romantic Greg falls in love from afar when he crosses paths with the beautiful but reclusive Gwen on the New York subway, little does he know she is suffering from multiple personality disorder. Touching hilarity ensues.
So normal is "every two disjoint closed sets of X have disjoint open neighborhoods" and completely normal is "every subspace of X with subspace topology is a normal space"
If we had an example of a Tychonoff which is not normal, then this would also give rise to a a normal, not completely normal space, because for Tychonoff spaces, the map into its Stone-Cech compactification is an embedding, and the Stone-Cech compactification is compact Hausdorff, so it's normal
I can't think of a Tychonoff not-normal space though
I see that there are various definitions for Radon measure and they are NOT equivalent, but they are equivalent on locally compact Hausdorff spaces. I think this is the reason why Radon measure has several different definitions, but I'm curious what is the standard one. On locally compact Hausdor...
@quallenjäger the answer says that it's possible to extend this to non-Hausdorff spaces, but that nobody does it, because it's not useful and not well-behaved
But technically speaking, the definition doesn't really rely on the hausdorff property?
Radon measures are only defined for Hausdorff spaces. This is because of both the local finiteness condition as well as the inner regularity condition. Just to make sure we are on the same page:
If $X$ is a completely normal topological space, then for every pair of separated sets $A,B$ (i.e., $\overline{A} \cap B = \emptyset = A \cap \overline{B}$), there exist disjoint open sets $U$ and $V$ such that $A \subseteq U$ and $B \subseteq V$ [hint: consider $X-(\overline{A} \cap \overline...
