@Ultradark 11 is prime. The primes less than or equal to 11 are 2,3,5,7, and 11. 2+3+5+7+11 = 28 = 2*2*7 which is neither prime nor semiprime

@Semiclassical Meant to say that that integral yields primes or semiprimes when $k$ is prime or semiprime

big if true

why

and I gave this problem to my dad and he solved it

it's probably false

$k>11$

@LeakyNun I don't know zeta2. But the problem amounts to showing that $\lim_{x \to 0^+}\sum_{n=1}^\infty \frac{\sqrt{x}}{1+n^2x}=\frac{\pi}{2}$? How do we do this?

some Fourier analysis might help

@LeakyNun I have not studied Fourier analysis

geometric series

that is very sad

@LeakyNun Ok. But how can I do this without using that?

no idea

@user330477: Have you tried estimating the series with a (perhaps improper) integral?

@TedShifrin Improper integrals have not been taught.

Do you not know the integral test for series?

@TedShifrin I think that $\frac{\sqrt{x}}{1+n^2x} \to 0$ uniformly, so we can exchange the limit and the sum. As a result, we would end up getting 0. Where am I going wrong?

@TedShifrin I know the integral test for series.

Saying a sequence of functions converges uniformly is different from saying the series of functions converges uniformly.

Think about setting $n^2x = u^2$ and doing an integral.

@TedShifrin What would then be the limits of the integral. Because n varies from 1 to \infty, so u varies from x to \infty, right?

But $x\to 0$, so think about the integral from $0$ to $\infty$.

Oh, and $u^2$ starts at $x$, so $u$ starts at $\sqrt x$.

I hate turning $n$ into a continuous variable. But, no, you have a mistake.

@TedShifrin What is the mistake?

Remember, $u=n\sqrt x$, formally.

@TedShifrin Yes. But how does this lead to a mistake?

You shouldn't have a $\sqrt x$ in the denominator, should you?

So now you see where the $\pi/2$ estimate is coming from.

if $H$ is a char. subgrp. of $G$ then we have a short exact sequence $1 \to \operatorname{Aut}(G,H) \to \operatorname{Aut}(G) \to \operatorname{Aut}(H)$

@TedShifrin Yes. Thank you so much.

what's an example where the map Aut(G) -> Aut(H) is not surjective?

Well, you probably have a bit more work to do, but now you're in the right frame of mind, @user330477.

This is like asking for a subvariety that is not projectively normal, @Leaky.

no idea what that means

Good. :)

This is what you do to people all the time.

well I'm all ears if you want to teach me that :P

is it related to a variety being normal, i.e. around each point the ring is integrally closed?

A subvariety of $\Bbb P^n$ is projectively normal if every section of $\mathscr O(k)$ is the restriction of a section of $\mathscr O_{\Bbb P^n}(k)$.

O(k) being the tautological bundle?

I mean, the k iterate tensor of the tautological bundle

Dual thereof. Tautological bundle is $\mathscr O(-1)$.

On $\Bbb P^n$ sections of $\mathscr O(k)$ are in bijection with homogeneous polynomials of degree $k$.

interesting

Presumably you want $H$ sitting inside $G$ suitably twistedly. I'm sure I "know" an example, and so do you, but I don't know one offhand.

why are they in bijection?

what do you mean that I "know" an example?

Well, the presheaf is defined precisely that way.

if I knew I wouldn't be asking...?

I'm guessing we have all encountered such creatures.

I put "know" in quotes to signify that we aren't aware of the answer.

Pay attention!

so we're thinking of Frobenius-like groups?

or anti-Frobenius-like groups?

I dunno what those are.

is G char in the tautological semidirect product G x| Aut(G)?

presumably no

I have no idea. Summon @Mathein.

@TedShifrin I don't really have that much work to do. The integral test for convergence gives us that $\frac{\pi}{2}-\arctan{\sqrt{x}} \leq \sum_{n=1}^\infty \frac{\sqrt{x}}{1+n^2x} \leq \frac{\sqrt{x}}{1+x}+\frac{\pi}{2}-\arctan{\sqrt{x}}$ and so as $x \to 0^+$, we have $\sum_{n=1}^\infty \frac{\sqrt{x}}{1+n^2x} \to \frac{\pi}{2}$.

ah any Sylow p-subgroups are char

I might want to think about those groups

D8 char S4

Aut(D8) = D8, Aut(S4) = S4

no surjection from S4 to D8

qed

no

that was utter nonsense

how about V4 char A4, Aut(V4)=S3, Aut(A4)=S4, no surjection from S4 to S3

hii chat!! any SE community where one can discuss about MATLAB?

so 1 -> Aut(A4,V4) -> Aut(A4) -> Aut(V4) becomes 1 -> ? -> S4 -> S3... the image must be A3 but I don't think D8 in S4 is normal?

this doesn't make any sense

|S4|=24=1+6+8+6+3

can't form 6 or 8

oh wait what am I doing

ok S4 -> S3 is surjective with kernel 1+3 = V4...??

ok so that example didn't work

@user330477 Oh right ... There was really no question about dealing with uniform convergence.

@TedShifrin do you know Frobenius theorem?

@Leaky: I think of $S_4$ as the symmetries of the cube, and, no, the symmetries of the square is not a normal subgroup. There are $3$ conjugate subgroups.

the one saying that |{x:x^n=e}| is divisible by n

@TedShifrin surprisingly V4 is normal in S4... what?

Hmm ... I'm not sure I know that.

Yes, the Klein subgroup is very different.

where does it live in your picture?

It comes from $180^\circ$ rotations about segments joining midpoints of opposite edges.

aha

nice

group theory: a geometric approach

You should work out $A_4$ and $A_5$ and $S_4$ as symmetries of regular polyhedra :P

LOL, it's all in my algebra book (stolen from Artin).

let's classify finite subgroups of SU(2,C)

@TedShifrin do you know about Sylow 2-subgroups of Sn?

I have once, but I don't now. Leaving in moments.

see you

Okey dokey.

I think I've found the Sylow p-subgroup of S_n

write n in base p, so n = sum a_i p^i

then it is the product of the Sylow p-subgroups of the S_(p^i)

and there it is the i-fold Wreath product of C_p

for the char. subgrp, it will never work with Dn or Cn or Sn or An

but those are the only groups I know, lol

(what a lie, two messages above I mentioned wreath product)

Guys, how do you from equation 2 to equation 3 in this answer? Maybe I'm retarded but I can't seem to factor this bad boy... math.stackexchange.com/a/2265930/12631

Hey, if P(x) has degree n, what is an upper bound on the number of solutions to P(x)^2=1?

@Anubhab $P(x)^2 - 1$ is a polynomial as well

hello all , $$x + x^2 + x^3 + ....... x^{100} = 0$$

can we prove that the above ^^^ equation have only 2 real roots ( I can only think of 2 i.e. 0 or -1) , but can we prove it ??

@AdvilSell multiply by $x-1$ to get $x^{101}-x = 0$

@LeakyNun Okay

yeah I got it , thanks :D

gap> for n in [1..30] do for G in AllSmallGroups(n) do for u in NormalSubgroups(G) do if IsCharacteristicSubgroup(G,u) then if Size(AutomorphismGroup(G)) mod Size(AutomorphismGroup(u)) <> 0 then Print(StructureDescription(G),";  ",StructureDescription(u),"\n"); fi; fi; od; od; od;
C4 x C2;  C2 x C2
(C4 x C2) : C2;  C2 x C2 x C2
(C4 x C2) : C2;  C2 x C2
C4 : C4;  C2 x C2
C8 x C2;  C2 x C2
C8 : C2;  C2 x C2
QD16;  Q8
C4 x C2 x C2;  C2 x C2 x C2
C2 x D8;  C2 x C2
C3 x S3;  C3 x C3
C12 x C2;  C6 x C2

@TedShifrin ^

hey @MatheinBoulomenos

@LeakyNun hi

find H char G such that the map Aut(G) -> Aut(H) is not surjective

@MatheinBoulomenos also, suppose $P$ is an $R$-module and $M, N \le P$ are submodules such that $M_\mathfrak m = N_\mathfrak m \le P_\mathfrak m$ for each maximal $\mathfrak m \triangleleft R$. Show that $M = N \le P$.

what is $(R_\mathfrak m)_\mathfrak n$?

if $R=\Bbb Z$ then it seems to be $\Bbb Q$

@LeakyNun Hmm, are there groups all of whose automorphisms are inner, but with non-trivial center?

@TobiasKildetoft that looks like another GAP question

@LeakyNun That would provide an example of such a characteristic subgroup

oh I already found such an example through GAP

@TobiasKildetoft $\Bbb Z/2\Bbb Z$

@MatheinBoulomenos Ahh, and with non-trivial automorphism group I suppose

I can't find any example of order <= 100 other than... C2

gap> for n in [1..100] do for g in AllSmallGroups(n) do if Size(Centre(g)) > 1 then if Size(InnerAutomorphismsAutomorphismGroup(AutomorphismGroup(g))) = Size(AutomorphismGroup(g)) then Print(StructureDescription(g),"\n"); fi; fi; od; od;
C2
gap>

is $(R_\mathfrak m)_\mathfrak n = R_{\mathfrak m \cap \mathfrak n}$?

how are you localizing at something which is not necessarily a prime ideal?

no idea

then what on earth should the left hand side be?

If $S$ and $T$ are submonoids of $(R,\cdot)$, then $(S \cdot T)^{-1} R = S^{-1}(T^{-1} R)$

ok so it's $((R-\mathfrak m)(R-\mathfrak n))^{-1} R$

but what on earth is $(R-\mathfrak m)(R-\mathfrak n)$?

$(R-\mathfrak m)(R-\mathfrak n) = R^2 - \mathfrak m R - \mathfrak n R + \mathfrak m \mathfrak n$

@LeakyNun lol

corollary: Hodge conjecture

I have a coursework on Numerical Analysis due tomorrow but I'm addicted to commutative algebra plz halp

okay I found a group of order 8315553613086720000 with the required properties

groupprops.subwiki.org/wiki/Conway_group:Co0

oh and I found no other groups of order <= 200 with such properties

@MatheinBoulomenos how do you know it satisfies those properties?

just going by what I read, I haven't checked myself

what did you read?

@MatheinBoulomenos From your link I read that Co1 is the inner automorphism group Co0, but I can't see anything about outer automorphism group

@TobiasKildetoft but if there's a theorem "Out(G) = 1 and Z(G) > 1 implies G = C2" then you would have known it...

I read that the center is C2 and that the outer automorphism group is Co1 is trivial and the morphism Out(Co0)->Out(Co1) is injective, I think

where did you read that?

(mathematics has devolved into he said she said)

not a good source, but alas: en.wikipedia.org/wiki/Outer_automorphism_group#In_finite_groups

Co1 is Co0/Z(Co0)

Co1 itself is centerless

your link only says Out(Co1) = 1

Hey there, is there someone here who's really sharp at math and who might have 15 minutes to spare ?

@MatheinBoulomenos my prof suggests to look at GxC2 where G is sporadic simple with Out(G) = 1

arxiv.org/pdf/1106.3760.pdf

so e.g. G=M11

(for some odd reason 11 < 16 but |M16| <<< |M11|)

Any pi day puzzles

[Random]

On gambler's fallacy and the nature of probability:

Most events are independent events, thus say there is 50% probability that a Russian roulette will land on a red space (ignoring the green space for now for simplicity)

Gambler's fallacy says that if you keep on landing on a sequence of blacks, then the probability of landing on red will increase in order to balance out

The reason why this is a fallacy has to do with the independence of the events:

Because each event is independent, the probability of landing at e.g. 50 regardless of what happens before does not change. This also means that you can land on 50 again

If Gambler's fallacy is to be true, then it means the events are dependent in the following way:

Sorry typo, I mean Roulette, not Russian roulette

Let $x_i$ be a random variable on what number will be landed on the Roulette after the ith game. Then Gambler's fallacy is true if for all $i,j$:

$$\text{Pr} (x_i|x_j) < \text{Pr} (x_i)$$

This means, the probability of landing in the same place again will diminish over time, and hence by the conservation of probability $\int \text{Pr}(x) dx = 1$ the complementary outcomes has to increase

The most extreme of these cases coincides with the typical ball without replacement problem, the case where $\text{Pr}(x_i|x_j) = 0$ thus an outcome will not be landed twice

Now as for the nature of probability. What probability told us is given any n trials, how likely is for some event x to occur (Or for the Bayesian interpretation, first make a guess on how much you put your belief that x happens, and then refine your guess iteratively as more evidence became available).

By itself, $\text{Pr}(x)$ does not give you any information on whether the event will happen on the next trial, or the next next trial and so on. This is how it captures one aspect of unpredictability

If you want to say predict how likely that $x$ happens after say, n trials, then what you need is really all the conditional probabilities:

$$\text{Pr}(x|y_1,...,y_n)$$

If that list that follows after the "given" is ordered, then it tell us something about the system. A computation of all of these thus gives you some idea on the ordering that a sequence of events happens

Therefore, if a process is deterministic, it basically means something like this:

$$\text{Pr}(x_0) = \delta_{0i}, \text{Pr}(x_1|x_i) = \delta_{0i}, \text{Pr} (x_2|x_i,x_j) = \delta_{0i}\delta_{1j}, ...$$

Put it in another way, let $\mathbf{x}$ be an ordered sequence of events. Let $x_0 \in \mathbf{x}$. Then for every $x_0$, known as the initial condition, there is a unique $\mathbf{x}$ such that $\text{Pr}(\mathbf{x})=1$ and all the rest are zero

Might return on studying stochastic process in more detail when I need to generate some probabilistic dynamics for reasons

hey i have a question

imgur.com/9a71j2y

look at example 10

they say that this function is only discontinuous at the point x = 1

but what about the point x = 1+dx ?

isnt that technically discontinuous too?

Happy $(-\frac12)!^2$ day!

@MartianCactus How big is dx?

infinitely small

its basically positive limit of 1

so 1+

Technically, one needs a topology to talk about continuity, thus when something is discontinuous at a point, it usually means the neighbourhood of the point has a different value than the point itself

Thus arbitrarily close to the point, the value is always different from that of the point itself. If I recall stuff from nonstandard analysis properly, it means there are points yinfinitesimally close to some point x such that |f(x)-f(y)| is not infinitesimal

so in a sense, you can say "f(x+dx)" is technically discontinuous, though one cannot really wrote x+dx as an argument rigourously

@MartianCactus That's not a real number so it doesn't count

(The cool thing of phrasing every analysis results in terms of general topology is that one no longer need to worry about freak things like "f(x+dx)")

because generally topology does not really care what the underlying number system is

I'm trying to decide whether the line with two origins over $k$ algebraically closed is a regular scheme or not, but I'm confused by its structure sheaf and the stalks of the latter

I still have not yet get to schemes and sheaves yet, is Balarka still around these days?

@AlessandroCodenotti it is certainly regular

being regular is a local property and locally it looks like the affine line

That's my intuition too, but I'd like to understand what its structure sheaf looks like more precisely

this is not just intution, f you want to make that precise: if $X$ is a topological space and $\mathcal F$ is a sheaf on $X$ and $U \subset X$ is an open subset and $x \in U$, then $\mathcal F_x={\mathcal F_{\mid U}}_x$

Let $X$ be the line with two origins, obtained by glueing together two affine lines $Y_1$ and $Y_2$. A section of $\mathcal O_X(U)$ looks like a section of $\mathcal O_{Y_1}(U\cap Y_1)$ and a section of $\mathcal O_{Y_2}(U\cap Y_2)$ agreeing on $Y_1\cap Y_2\cap U$ (there's a fibered product of sets hidden in here that is too painful to type out in latex)

@MatheinBoulomenos Oh, yeah, of course, if you restrict to an open set you have a cofinal family in the directed system defining the stalks

right

That makes a lot of sense, thanks

this means any property of schemes which is defined in terms of local rings can be checked "locally", i.e. by showing that any point has an open neiborhood for which the property holds

Nice, in particular the same applies when deciding whether a scheme is normal, which is pretty nice

Do you have time (and patience :P) to help me with another AG exercise?

I can try

So I'm looking at $\pi:X=\mathrm{Spec}\Bbb Z[i]\to\mathrm{Spec}\Bbb Z$ induced by $\Bbb Z\to\Bbb Z[i]$ and I'm asked to determine $\pi_\ast\mathcal O_X$ and whether it is locally free

So I want to describe $\pi_\ast\mathcal O_X$ on the standard basis for the Zariski topology on $\mathrm{Spec}\Bbb Z$

I'm pretty sure that's not just localy free, but is even globally isomorphic to ${\mathcal{O}_{\mathrm{Spec}(\Bbb Z)}}^2$

Happy $(-\frac12)!^2$ day!

I don't see why is that, I'm trying to understand explicitely what is going on at the moment

If I look at $\pi_\ast\mathcal O_X(D(p))$ for an odd prime $p$ for example

You're probably wondering what 14/3 has to do with pi

(kidding)

14/3 is actually a good approximation for the first Feigenbaum constant

so maybe on this day we should be celebrating chaos

if $f:R \to S$ is a ring morphism and $\varphi:S \to R$ is the associated morphism of schemes and $M$ is a $S$-module, then we have $\varphi_*(\widetilde{M})=\widetilde{f_*M}$

(or causing?)

$f_\ast M$ being $M$ as an $R$-module where $rm$ is defined as $f(r)m$?

right

restriction of scalars along $f$

so in this case $R=\Bbb Z$ and $S=\Bbb Z[i]$, but $S$ as a $\Bbb Z$-module is just $\Bbb Z^2$

Aha that makes sense

If $\sigma = (i_1,...,i_r)$, then is it true that $\sigma^{m}(i_{k}) = i_{k + m \mod r}$? The few experiments I ran seemed to confirm it.

But I'm not seeing this by looking explicitely at the sheaf. $\pi_\ast\mathcal O_X(D(p))$ is by definition $\mathcal O_X(\pi^{-1}(D(p)))$, and $\pi^{-1}(D(p))$ should be $\mathrm{Spec}\Bbb Z[i]$ minus $p$ if $p$ is $3$ mod $4$ or minus two points if $p$ is $1$ mod $4$, right?

yeah, but you don't really need to worry about that

$\pi^{-1}(D(p))$ is always again a basic open set, now matter who many elements it happens to contain

Right

$\pi^{-1}(D(p))=D(p)$ in $\mathrm{Spec}\Bbb{Z}[i]$

So what I get is $\pi_\ast\mathcal O_X(D(p))=\Bbb Z[i]_p$

that's a free $\Bbb Z[1/p]$-module of rank two

@MatheinBoulomenos Aha, that's what I was missing

So yeah it is globally a free sheaf of rank $2$

If $M$ is a free $R$-module, then $S^{-1}M$ is a free $S^{-1}R$-module (with the "same" basis)

Makes sense

I'm confused at what does it mean geometrically for this sheaf to be free of rank $2$ though

Also since $A=\Bbb Z[i]\simeq\Bbb Z[x]/(x^2+1)$ the module of Kähler differentials $\Omega_{X/\mathrm{Spec}\Bbb Z}$ should just be the $A$-module generated by $dx$ with relation $2xdx$

I mean geometrically it's like a rank 2 vector bundle. Which makes sense since $\Bbb Q(i)$ has dimension $2$ over $\Bbb Q$ (that's the stalk at the generic point)

@AlessandroCodenotti right

more generally if $A=R[x]/(f)$, then $\Omega_{A/R}=R[x]/(f,df)$

Yeah, this can be generalized to whenever $A$ is a f.g. $R$-algebra

Anyway I'm leaving now to think about another exercise which is really an algebra exercise, thanks a lot for your help! Working out an explicit example was rather iilluminating!

That is the curve r=θ, what are the limits if we are finding the area? I thought it had to be split up to avoid double counting but I was told you can just integrate from 0 to pi/2. Who is correct?

@AfronPie: Based on your picture, you need $\theta$ to go from $2\pi$ to $5\pi/2$.

@santimirandarp Good luck telling us what $0.999\dots 8$ means.

Ok @TedShifrin I think I get it

Hi @Ted

@TedShifrin happy pi day

Hi, demonic @Alessandro.

Thanks, @Leaky.

So you have lots of examples answering that automorphism question, @Leaky. Do you have a conceptual understanding of what to look for?

no idea :P

I'm discovering I don't know commutative algebra

@AlessandroCodenotti what happened

I'm thinking about supports of modules (actually supports of quasicoherent sheaves on affine schemes, but there's no difference)

Quick sanity checks: If I have $A$ as an $A$-module then $\mathrm{Supp}(A)=\mathrm{Spec}(A)$. If I have $A/I$ as an $A$-module then $\mathrm{Supp}(A/I)=\mathrm{Spec}(A/I)$

So, do you understand how it can happen that you end up with somebody supported at a single point?

@AlessandroCodenotti yes and yes

@TedShifrin No I don't think so

lol one thing i learnt in grad school is that i like algebra less than i thought

I mean I can cook up some examples using the $A/I$ fact from above and some maximal $I$

@loch :c

I'm thinking about the cokernel of the map that's multiplication by $x$, say.

@loch: So have you learned what you like more?

@TedShifrin im fine with algebraic geometry!

Well, that's still lots of algebra unless you do it as complex geometry :P

but put in some rep theory and my soul starts to die

Algebraic geometry already has a lot of algebra in it for my taste

wow, even I like a bit of representation theory now and then.

I was probably exaggerating :p

you think?

@TedShifrin Wait why do you get a single point? I'm thinking of multiplication by $6$ in $\Bbb Z$, the cokernel should be supported in $2$ points as a $\Bbb Z$-modules if I'm not doing stupid mistakes

Because away from $x=0$ multiplication by $x$ is invertible (locally).

I am working over a field, of course.

I'm thinking the affine line.

Ohhh ok I see now

Even though I'm not algebraic, @loch, I love how the geometry of a symmetric space can be read off all the Lie algebraic data and basic representations.

Or how you give the tangent bundle of $G/H$ by the adjoint representation, etc.

That is also an example of $A/I$ with $I$ maximal, but I like the intuition of multiplication by $x$ being locally invertible

Glad to help, @Alessandro :)

@TedShifrin oh nice

I never thought of that :P

Is there any standard notation to write that f is a function to B, that is, f: dom(f) -> B but dom(f) also has some a set A as a subset. the "f:A -> B" does not work since that means that A = dom(f) and I want to have A subset dom(f)

of which, @Leaky?

@TedShifrin multiplication by x

ah ... it is standard in Riemann surface theory and with divisors in general.

@famesyasd: Give the domain a name, regardless.

More sanity checks: if $M_1$ and $M_2$ are two $A$-modules then $\mathrm{Supp}(M_1\oplus M_2)$ is the union of the supports

@TedShifrin yeah ok i was exaggerating then - probably just PTSD from a noncommutative algebra class

LOL, OK :P

@TedShifrin Divisors are going to be in algebraic geometry II next term I think

@TedShifrin what do you mean? did not get what you said

They're super important, @Alessandro.

I don't know if I'm going to take that course though

@AlessandroCodenotti yes

in general for $0 \to M' \to M \to M'' \to 0$, Supp(M) = Supp(M') U Supp(M'')

@famesyasd: I'm just saying you should define the domain of $f$. But then you need to restrict that domain to $A$. It's no longer the same function $f$. You have to write $f|_A$ or some similar notation.

The professor mentioned effective Cartier divisors extremely briefly when talking about invertible sheaves but they should be explained in detail in AG II

that's not very effective then

groans

that's effectively explaining nothing

Yup, invertible sheaves are in bijective correspondence with Cartier divisors.

Effective is a special case ...

@TedShifrin yes but I do not want to take restrictions I just want to highlight that the very same function "f" also has "A" as a subset of its domain. like when you write that "f" is a surjection above the arrow or replace staright arrow with some fancy arrow to denote that "f" is a surjection from A to B.

There is no convenient notation, @famesyasd.

Hmm I'm unsure about the support of $A_f$ as an $A$-module (this is the last one, I promise)

Promises, promises, promises ...

@AlessandroCodenotti support doesn't need to be closed

No quite the opposite, I think it should be $D(f)$

yes

Makes sense, thanks

no problem

Does this fit with my example earlier, @Alessandro?

Isn't looking at the cokernel of multiplication by $x$ more like looking at $k=k[x]/(x)$ as a $k[x]$-module, whose support is then $\mathrm{Spec}(k)$ so a single point?

That sounds right.

But I agree that it makes sense that localizing at $x$ gives something supported everywhere except $x=0$

So my cokernel example can be rephrased in terms of this, I think.

Did I tell you I'm running in the Jerusalem Marathon tomorrow

(5k)

'Cause that's a thing that's happening

Better rest up, DogAteMy.

I don't know what picture you have of me in your head, but I am not the most athletic person

so I'm a bit scared

I remember you had operations on your leg(s) or back?

Did you train for the marathon?

@TedShifrin Yes, on both legs (not at the same time)

and also not a surgery on my back but I had a back brace for scoliosis for a while

@Dair Not as much as I should've

I can't actually do five minutes of sustained running

OK, so my memory is basically still functioning :)

Stay safe man.

Yeah, maybe you're nuts.

so it'll be an alternating runing/walking affair

Even just walking that distance can be brutal...

@AkivaWeinberger 5km, not a full marathon

Like an eighth-marathon

You @ed yourself lol.

Yeah, to get the arrow that points to a previous message

Basically the radius of Jerusalem

nvm lol. i'm an idiot.

Jerusalem isn't a perfect circle but you can think of it as a 5km-radius circle and you won't be too far off

Anyway I'm not happy about admitting this but you were right when you said I should see more explicit examples together with the fancy theory :P @Ted

Admitting Ted is right is always a painful process, @Alessandro :P

That's not the placement I expected for the "always" in that sentence

examples are great!

LOL

I'm glad you think I'm a total and utter egotist, @Alessandro :P

(I was joking of course)

Maybe :P

So now that I know 3 basic facts about the support of modules it's time to start playing the big kids games and start computing supports of sheaves of differentials (which is really the same thing, but anyway)

Yes, just a special case ...