Watching Elon Musk seamlessly moving from "the free market solves everything" to "trade unions sow social division among classes" and "we need a watchdog to discredit lying journalists" is like watching capitalism decay into fascism in real time on twitter.
I just think the idea of a private news-reviewing agency is good, but Musk's intention behind them trying to silence news organizations - which provide data that may or may not be systematically categorized as false by counterdata - is similar to the anti-media movement that's cropping up in the US which is dangerous. Being skeptic about data is good, silencing it isn't.
(That he's trying to silence a section of the media is evident from his tweets)
@Zee I think I'd feel what Mike said. Kind of sad, a little pointless, but I'll pick up other things to do - like listening music and watching movies :P
look, watching a movie is more fun than math, but i truly dont think its possible for a person born as a mathematcian to be truly happy doing anything else, movies are just a distraction
He basically says that you sorta find two different kinds of people. Some who are principally preoccupied with logic, every step is slow but bulletproof, the other type of people who are bold and intuitive but precarious
The former are analysts and the latter geometers (even if some of the former work in geometry and vice versa)
Your not fooling me @BalarkaSen , its just a defense mechanism since you may actually be forced to abandon math by your circumstances but i truly believe, it is the most important thing in your life
And he was like yeah I don't think that's a question of just being trained one way or another, some folk just are how they are. That's the context of the quote "Mathematicians are born, not made"
@Zee Yeah it started as a defense mechanism 'cuz I did badly at the multiple-choice section of the admission I'm hopeful about. I was sad a lot. But I asked people around and they said I'll make the cut-off just fine. So I think it's much more than simply a defense mechanism at this point, because I genuinely think broader life-goals are harmful to my sanity as a human being.
I have given a lot of thought to it over the last month
I don't value my goals of being a mathematician over my general well-being.
@XanderHenderson there's probably a case that among careers that have less of an entry barrier, that STEM (really I think people look at engineers and CS folk) is generically more likely to be lucrative
The whole thing that goes on in India is a race towards the engineering and IT placements because of the whole socio-economic elitism of applied STEM fields
also, what is the spread? someone with a teaching credential likely won't ever make as much as the average IT jock, but teachers typically have very stable employment, stable hours, predictable downtime, and a much narrower range of salaries.
so they might not make as much on the top end, but they are also less likely to make very little on the low end
The entertainment (and artistic) market is a prime example of where free market economy fails. Only an exponentially handful of actors are exponentially successful, say. Similarly, an exponentially handful of pop records get exponentially streamed.
@Zee He made his money as a hedge fund manager, no? The fact that a math degree helped him out there does not indicate that "hedge fund manager" is a STEM career; only that it is a career in which a mathematics background can be helpful.
well Xander, you are right in the general case, but in this specific sense, this guy used fellow mathematicians, using mathematical methods to beat the market
Most hedge fund managers have no more background in mathematics than the average Joe, and most people with degrees in STEM fields will have very little opportunity to make it rich as a hedge fund manager.
this guy was able to cash in big using his math background now, that means you can sometimes use that background, nonetheless, i do think , there are much easier ways to make money than learning math
if I take $ax + by = c$ as my straight line and map $(x,y) \mapsto \left( \frac{x}{x^2 + y^2}, \frac{y}{x^2 + y^2}\right)$ I get a circle not centred at the origin don't I?
it's interesting but I've an exam in it in about 5 days and I've spent all of the time I could've spent learning it doing algebra and number theory for my dissertation
In fact, I think $\langle x,y\rangle$ is necessarily isomorphic to $S_3$, with $x\mapsto(12)$ and $y\mapsto(123)$ defining the isomorphism, so the above example is essentially unique.
Odd-ball weirdness: two copies of RP^3, delete 3-balls from each. Glue them by (1) the identity homeomorphism $S^2 \to S^2$ along the boundary of the deleted balls or (2) by the antipodal homeomorphism $S^2 \to S^2$ along the boundary of the deleted balls. The results are $\Bbb{RP}^3 \# \Bbb{RP}^3$ and $\Bbb{RP}^3 \# \overline{\Bbb{RP}^3}$ respectively
The first is orientable (as RP^3 is orientable), the second isn't.
I guess it's not that hard to see. Take two RP^1's in RP^3. After the second operation, they give a loop given by an arc on both copies of the connected sum so that they are flip-glued along the $S^2$ it's connected summed along
It's also a spherically symmetric way to represent it
Both connect-summed manifolds just look like $S^2\times I$ (a spherical annulus) to me where each of the two components of the boundary is identified to itself by the antipode map
Wait, I'm confused. $\Bbb{RP}^3$ is homeomorphic to the unit tangent bundle of $S^2$. Fill in the circle fibers by disks, that gives a 4-manifold with boundary $M$ such that $\partial M = \Bbb{RP}^3$. Consider $M \times I$ and delete a neighborhood of $\{x_0\} \times I$ from it for some $x_0 \in M$. The boundary of that manifold is $\Bbb{RP}^3 \# \overline{\Bbb{RP}^3}$. $M$ is orientable as fuck, how can boundary of an orientable manifold be non-orientable. That would be garbage.
It shouldn't be, because some other computation says it's nonorientable.
Meh, I'm sleepy. RP^3 x I, delete an nbhd of {x0} x I from it. That bounds RP^3 # overline{RP^3}, is what I meant. But that manifold is orientable as hell.
Because RP^3 is orientable.
I don't get it.
This is like a general argument. If $M$ is orientable, take $M \times [0, 1] \setminus N_{\epsilon}(p \times [0, 1])$. The boundary of that orientable manifold-with-boundary is $M \# \overline{M}$, so it has to be orientable. ???
The correct bit of my calculation says RP^3 # overline{RP^3} is a circle bundle over RP^2. I was mis-identifying which circle bundle it is (a certain nonorientable manifold is also a nontrivial circle bundle over RP^2)
The connected sum of orientable manifolds is indeed orientable. It's really not so hard to prove it yourself, so you should try! — Mike MillerNov 6 '14 at 2:52
I was rather confused because it seemed plausible that if you sum a manifold and it's orientation reversed copy togather you'd get something not orientable.
@Akiva Is RP^3 # RP^3 actually homeomorphic to RP^3 # overline{RP^3}?
@AkivaWeinberger The proof exploits that $H^*(\Bbb{CP}^2) \cong \Bbb Z[x]/(x^3 = 0)$, I think. Any homeomorphism gives a ring automorphism of this ring.
If it reversed orientation it'd send $x^2$ to $-x^2$.
But that's impossible: Send $x$ to $nx$, then $x^2$ goes to $n^2x^2$, and $n^2 > 0$.