porridgemathematics

Right

But that’s just economics

So obviously if you are studying something that is very hot, you will find innately talented people fail at a higher rate than in niche sub fields (again putting aside how you can’t really compare innate talent across fields like this) despite working just as hard as people with less innate talent

For fields that are more niche it might be slightly easier in a sense, but obviously you still need to have done good work and your supervisor needs to be well connected in that niche subfield

And some fields are just less hot than others which means the process of getting a postdoc or PhD in one or the other may look different as well

Also a lot of this is also pretty field specific, that is, innate ability in some subfield of analysis looks different to innate ability in algebraic geometry

It’s probably equally insulting to highly innately talented people with less than they deserve to show for it at the career level to say that

But I’m not sure it’s helpful to say innate ability is what makes or breaks deals which amount to career progression in life

But yeah, I also don’t think “innate ability” is what makes people get postdocs as Leslie suggests, it’s a lot of luck and networking too, of course you need to have innate ability too

That’s basically what I was saying

My sentiments are more or less echoed by what leslie just wrote

To be clear, I’m not at all implying that someone who doesn’t “make it” in regards to getting a PhD position or postdoc didn’t work hard enough

Not implying it’s easier in the sense that you can just do nothing and reach it (for most people)

Here by easier I mean in terms of requiring innate ability

And getting to the “door” of math is much easier than you might think. You don’t need to be a genius.

And it’s this latter thing that’s actually important.

But beyond that, you can’t measure someone’s general ability in getting beyond the door with much success

So I agree with the sentiment that likely to do well at math you need some kind of innate ability to get to the “door” of math

That is to say, there comes a point where if you reach it, either through your “innate ability” or a mixture of that and something else, the next limiting factor depends disproportionally less on “innate ability”

The bottleneck for doing math at some point shifts from innate ability to persistence

But they are far from being guarantees of being able to do research

so it’s not as meaningful as you might think to speak of someone’s innate ability in it broadly, of course there are still statistical tools you can use to measure these things which have good chances of being accurate

That is to say, math is so vast that there is not really a good way of measuring one’s propensity to ever be able to contribute significantly to it in some meaningful respect

@pie math is sufficiently vast that innate ability however you measure it will not necessarily pick up on a lack of innate ability with grasping certain niche/understudied ideas or techniques

I guess studio ghibli movies are in a way similar to some Disney ones

In terms of anime movies I kinda only know about the studio ghibli stuff and evangelion, but both are obviously very different to any Disney movie

Also that’s a tv show not a movie

I mustn’t be familiar with the ones that were actually Disney inspired, since they’re pretty old

@Thorgott oh for some reason I was thinking of dragon ball as a possible kind of Asian Disney inspired cartoon, but I guess it wasn’t really since it came out in the 80s

They’re sometimes way too on the nose with their injection of morals into the story though

@Thorgott oh that’s interesting. I didn’t know toei was trying to emulate disneys success . It makes sense in retrospect that anime was sort of supposed to be like the asian answer to Disney cartoons and they even sort of fit that theme now that I think about it

Seems pretty cut and dry to me

@skullpatrol don’t we know? He had dreams in which his family goddess namagiri planted equations on his tongue in ink

here is another question about polynomials math.stackexchange.com/questions/5045670/…

oh and I should also add, that $B$ can be taken to be an open set, but Im not sure how important that actually is

Initially I thought the answer is "yes, obviously", but writing down a proof I realized I couldn't write a very short proof, which makes me doubt whether what I wrote down is even correct

sorry I keep asking questions about polynomials :/

and $\epsilon > 0 $ is some fixed number

I should also clarify, I mean to say, I am fixing some bounded subset of the plane, $B$, and I want to know if it is reasonable to say that the sums of these indicator functions converge a.e. to the corresponding sum for $Q$ on $B$

so the sums are over the critical points counted with multiplicity

Here $Z(P_n')$ is the set of critical points of $P_n$ , as a multiset (counted with its corresponding multiplicity as a zero of $P_n'$) and $Z(Q')$ defined the same way but for $Q'$.

Suppose $P_n $ is a sequence of polynomials of degrees between $2 $ and $m$, satisfying $P_n'(0) = 1$, and converging locally uniformly in $\mathbb{C}$ to a polynomial $Q$ also of degree between $2$ and $m$ and satisfying $Q'(0) = 1$. Should it be true that $\sum_{b \in Z(P_n')} \chi_{\Delta(b ; \epsilon)}(z) \rightarrow \sum_{b \in Z(Q')} \chi_{\Delta(b ; \epsilon)}(z)$ pointwise almost everywhere on all bounded subsets of the plane?

But what you were asking was whether any smooth structure “comes from” a complex one, so I think the answer is still yes? Although which one it “comes from” isn’t unique

Because then as you say the latter would always be a point

Uh sorry, I misspoke, this shows that every smooth structure does “come from” a complex structure, in the sense that there is a compatible complex structure , it does not imply the moduli spaces in terms of diffeomorphism coincide with the moduli spaces in terms of biholomorphism

So if you have a smooth map $f$ between orientable surfaces, then by choosing any metric on the base, and the induced metric on the target, and choosing the atlases on base and target compatible with the smooth structures (maximal smooth atlases) which makes $f$ a conformal map, one finds $f$ is a biholomorphism with respect to certain biholomorphic subatlases of the base and target maximal atlases

So I have shown that if you give me a smooth atlas $\mathcal{A}$, I can find you a smooth atlas $\mathcal{B}$ compatible with $\mathcal{A}$, whose transition maps are biholomorphisms.

*post-composed

Of course the moduli space does equal a point for the sphere

It implies that the moduli space of an orientable surface in terms of diffeomorphism coincides with the moduli space of an orientable surface in terms of biholomorphism