@Claudio Something that you will eventually come to realize is that there is a lot of jargon in mathematics. You don't have to be particularly smart or clever to use it and understand it. You just have to be immersed in it long enough for it to stick.
I failed out of college in my first two years, dropped out three years later, and eventually finished a bachelors degree 10 years after I started. After that, I did okay, but I was not a model student.
I have a basic doubt. If $f=g$ a.e. and $h$ is some function, under what conditions is it true that $f\circ h=g\circ h$ a.e.? I found this, and it makes me worried.
Hmm ok. Anyways, it's gotten way too late, my time has come since I have lectures at 8 am tomorrow. Bye everyone and thanks for the inetersting discussions :p
@Claudio Connectedness is a fairly broad topic in topology, with lots of different notions that all try to capture some aspect of what it means for a space to be connected. But "simply connected" is one that has always felt like a bit of an odd duck to me.
@psie i must be missing something. Folland defines the product measure as the outer measure with respect to the measurable rectangles, and in my world, so far, an outer measure is complete.
@ILikeMathematics I can't paste the javascript into the address bar. I need to have it as a link on the shortcut bar. I don't know if Chrome has that. There is a suggestion on the installation page for Android and Chrome.
@copper.hat well, we start with a premeasure $\pi$ on the algebra of finite disjoint unions of measurable rectangles. This induces an outer measure $\mu^\ast$ and then we take the restriction of this outer measure (note restriction) of this outer measure to the $\sigma$-algebra generated by algebra of finite disjoint union of rectangles. We could also get a complete measure by restricting to the $\mu^\ast$ measurable sets, if we would like to have a complete measure.
@psie i suspect that what i am missing is that the $\sigma$ algebra on $(0,\infty) \times S^{n-1}$ is incomplete, so any concerns would be about measurability.
let me mull this a bit more and consult my expert when he is awake
today I've come to appreciate that the join operation behaves much better on augmented simplicial sets than plain simplicial sets, for it actually supplies them with a biclosed monoidal structure
could it be that there is a causal relationship? certainly. but note that nobody in the study actually developed blood clots, and if cnn's reading is to be believed the estimated clotting risk doubled
“Consumers should interpret the results of this pilot with extreme caution. The limited number of participants, a total of 10, were given an excessive amount of erythritol, nearly quadruple the maximum amount approved in any single beverage in the United States,” the council’s president, Carla Saunders, said in an email.
If I were you I wouldn't worry too much, as long as you use it sparingly. The fact that it is on the market already means that it has undergone testing and is generally considered safe in the recommended doses.
On a completely unrelated subject, do you suppose that there is a Cameo-like service where one could pay to have Shohreh Aghdashloo read the phonebook to you?
@BenSteffan To be fair, quadruple isn't that much---people often have four beverages.
The Ihara zeta function (for a finite connected undirected graph) is invariant under graph automorphisms, because the adjacency structure is preserved and hence the primitive cycles are preserved. Is there anything more to be said about this?
i don't know. you were talking about finishing a bag of somthing, sort of like it was cocaine, so i made a joke based on how you can make crack out of powder cocaine
Consider a co-dimension $1$ surface of revolution $L^{n-1}$ and an embedding $e:L^{n-1} \hookrightarrow X^n$ for $X^n=[0,1]^n$ with points $p,q$ elements of $\partial X^n$ where $\partial X^n=X^n-(0,1)^n$ for $\mathrm {sup}~ \mathrm{dist}(p,q)=\sqrt{n}$. Assume $L^{n-1}$ has constant positive sectional curvature and maximal volume.
Next replicate this symmetrical manifold, $L^{n-1}$, $2^{n-2}$ additional times. In each replication, the cone points on $L^{n-1}$ align with unique pairs of vertices on the boundary of the $n$-cube. Take the union of all these surfaces - there $2^{n-1}$ total. …
For $n=3$ we have $\chi(\mathcal I^{n-1})=0$. I don't know how go further
@XanderHenderson I’m not sure exactly where yet, but I’m hoping to study somewhere with strong programs in pure mathematics. My country’s college education system, especially in math, isn’t very strong, so I really want to study abroad—maybe in the US, Europe, or Canada, or anywhere with a good education in pure math. I know it would be difficult to achieve, but it’s my dream.
pie as others have said it depends on the program. generally speaking the kind of program that would do that would not be competitive to get into, and also probably expect you to pay for it (whereas a more selective program might provide funding to a more limited number of students).
the master's in math in the US is kind of a weird degree, many graduate programs basically don't offer one, or only provide it on the way to getting a phd. there is often a specific focus to a masters program, e.g. it might be designed with a K-12 education focus or something. if there's a research focus it would most likely be something you worked out with a specific advisor and not something the department was just set up to provide people.
@pie for programs that aren't funded, i would expect almost anywhere else to be cheaper, the cost of education in the US is kind of out of control. haha
@pie As @leslietownes, there aren't a lot of places in the US where one would apply to a masters program in mathematics. Typically, you apply to a PhD program, and end up with a masters as a consolation prize if you drop out after finishing your coursework.
Though there are pure math masters programs without that set of strings (I, myself, completed one before applying to phd programs).
(Not sure if that program still exists as it did when I did it, since the department now has a phd program.)
yeah, in california, a few CSUs offer them, or used to. i don't know of a UC that does, except maybe rolled into something else (e.g. a bachelors program with an education focus)
@leslietownes Yeah, the primary distinction between the CSUs and UCs is that the CSUs are supposed to offer phds (except, maybe, in a few limited cases).
funding is the tricky part. the last time i looked into this it was like, any number of places were happy to have you, but you'd probably need to pay tuition and things. which seems not great to me.
Officially, the UCs don't admit people to masters degree programs in math, but it is pretty simply to apply to the phd program and masters out after two years.
@leslietownes In the US, most graduate programs come with funding in the form of teaching assistantships. That doesn't cover everything, but it helps. And if they want you, as a foreign student, they'll typically find a way to make sure that they can have you.
most [non exploitative] graduate programs [in math] :)
math is better about this than a lot of disciplines.
but the kind of department that will admit you without regard for whether you might be able to TA "right out of the box," for example, will probably expect you to pay
i once had a friend who discovered this guy she was dating was a serial liar because he said he had a masters degree from a department that didn't offer one (or have a phd program that might, on the off chance, provide one)
But I was focused on math programs---they can generally afford to have a lot of TAs, as they have a lot of service classes to teach. English departments are also often good at this.
@leslietownes My father dumped on ed degrees in general (particularly doctorates in "educational leadership"). But he didn't have an ed degree. A phd and an jd, but no edd.
xander in my dads case it was just 'a masters you can do on the GI bill,' utterly no intention behind it. which feels better than getting one, like, on purpose.
it was pretty common for public school teachers in my district to get them because some salary formula meant if you had one you could be paid more. which is also a defensible use of the degree. i think most of the 'scholarship' that goes on in ed schools is horseshit
like the scholarship in law schools, i might add. just a kind of bubble, formed by people needing to go to graduate school in something, and professors at such schools needing something to do
i forget if it's my grandfather or grandfather's brother who had a phd in education. my grandfather was a high school principal, which feels like, yeah why not.
@leslietownes I am more dismissive, I think, of ed programs. The real work of understanding how to teach people is in psychology and sociology and related fields. But ed programs have to be able to claim that they are doing good research, so they end up being somewhat anti-intellectual.
at some point far enough back it was probably just a penmanship test or something, no theory
something i have detected in a number of ed papers i have read is an unusual hostility to the notion of subject matter expertise. it's like everything needs to be a blank slate so that a field of Education, divorced from anything more specific that specific people might actually need to know, can say something about it
What I am trying to convey to my audience is, a monodromy representation, associating each homotopy class of loops at a certain basepoint $p_1$ to a holonomy transformation
pi_1(X,p) has fairly self explanatory meaning to me as a fundamental group, the rest of that stuff is meaningless to me
this doesn't strike me as a situation where you would want to rely on notation to do any explaining for you
even in the fairly standard setting of pi_1 meaning what it usually does there is static in the background of what kind of space are you working in, such that the choice of base point might or might not matter. prose is going to have to address that
@ModularMindset No, but you are asking the wrong question. You should be using notation in context to explain some idea. If you are asking "is my notation clear?", then I assume that you are inventing some new notation. That's fine, as long as you include some explanation of what that notation is supposed to mean.
Not being in your area of mathematics, my assumption would be that $\operatorname{Mon}$ is some kind of function (or possibly functor), that it acts on the fundamental group ($\pi _1$) of some topological space that I know nothing about, and lands in some place called $\operatorname{Hol}(\Gamma_{\mathcal{I}})$, which is a completely mystery to me.
But, again, I am only reading a bit of notation in isolation. Presumably, everything there will have been defined elsewhere, previously.
@Jakobian maybe that wasn't the statement of completeness. But if $\mu$ is complete, i.e. it is a measure on a complete $\sigma$-algebra, then I'd guess so is $\nu(E)=\int_E f\,d\mu$, since they are measures on the same space. Right?
what is a complete sigma-algebra? sorry if i am butting into some longer conversation here, feel free to ignore this. the notion of completeness i am most familiar with assumes not just a sigma algebra but a measure and it is the measure or measure space or whatever that is said to be complete, not the sigma algebra.
@psie Potentially, $\nu$ can have more measure zero sets than $\mu$, so you need to prove that either this doesn't happen, or that subsets of these additional sets are still measurable
@XanderHenderson yes, it is about measures, but we can only speak of a complete measure if its domain includes all subsets of null sets. So the domain comes first (the complete $\sigma$-algebra), then the measure.
@XanderHenderson From an outsider perspective(e.g. not a math student), I kinda feel the same about it. I have a question, which probably will sound stupid to you: is it possible that simple-connectedness gets much harder to prove as the dimension of the space gets larger (I mean $R^2 \to R^3 \to ...$), than connectedness
@psie then it definitely doesn't look like $\nu$ is complete. For example, if $f\equiv 0$ then for $\nu$ to be complete we'd have to include the whole power-set, but $\sigma$-algebra of $\mu$ is usually not defined on the whole power-set
@Jakobian maybe that wasn't the statement of completeness. But if $\mu$ is complete, i.e. it is a measure on a complete $\sigma$-algebra, then I'd guess so is $\nu(E)=\int_E f\,d\mu$, since they are measures on the same space. Right?
@SoumikMukherjee I should mention that physicists don't have topology courses hahahah, it'd be pointless to even try to understand what you linked as of now
@Claudio what I linked is something that kinda answers your question. Fundamental group is something that tells wheather a (path connected) space is simply connected or not.
@Claudio maybe I am wrong about this, but I think what you really want is de Rham cohomology, and in particular, conditions for every $1$-form on an open connected set to be exact
@Jakobian This is an interesting question and the way it's written makes it understandable even by me, the answer, on the other hand, is the exact opposite :p
that being said, I've seen these names pop up in mathematical physics (but only by user like Urs schreiber, who's pretty much a mathematician)
I do not understand what he does to begin with :p, I didn't think you guys knew who he was. Why do you say so? I'm kinda interested now
@Thorgott wait, maybe it's better not to investigate any further, forget what I said, let us act as though I didn't mention his name in the first place :p
