Mathematics - 2017-08-08 (page 3 of 3)

Lucas Henrique

12:00

so I have a parametric line and a point. I want to have the span of them; is there a general way? my line does not pass through origin

Tobias Kildetoft

@VincenzoOliva I don't think it got read by many people anyway. It had like 10 comments of which something like 8 were really just hidden links meant to improve search scores for various companies.

user8469759

hi guys, out of curiosity

say I have a sequence in a normed vector space which is cauchy

say $x_n$

Secret

math.blogoverflow.com/tag/induction

Lucas Henrique

since the vector of the point is linearly independent to the lin, I thought of using the span of the point and the directive vector of the line. But seems weird

user8469759

why can I draw a subsequence $y_n$ such that $ d(y_{n+1},y_n) < 2^{-n}$?

I'm sure I can get there using the definition of cauchy sequence

given $\forall epsilon \exists N$ s.t. $\forall m,n \geq N d(x_n,x_m) < \epsilon$

Tobias Kildetoft

12:07

@Secret Ahh, neat. I didn't know they had archived the posts

user8469759

I suppose I can do this construction by induction

for $\epsilon = 1$ there's an $N$ such that for all $m,n \geq N$ I have $d(x_m,x_n) < 1$

Tobias Kildetoft

@user8469759 Yeah, just pick the epsilon suitably small and pick something like $y_n = x_N$

user8469759

ok but I don't know why

I struggle with rigorous details

Simply Beautiful Art

morning

user8469759

can you please help me to work out all the construction if there's any?

Tobias Kildetoft

12:09

@user8469759 Pick for each $n$ an $N_n$ (the subscript is just so we can see it depends on $n$ such that this is your $N$ for $\varepsilon = 2^{-n}$.

Vincenzo Oliva

@Secret Nice!

Secret

12:37

> Proof: Let $B\subseteq X$ be a subset and assume that $B$ does not have a smallest element. Let $A=X∖B$. We need to show that $A=X$ and thus by assumption it is enough to show that if $x\in X$ and $y \in A$ for all $y<x$ then $x\in A$. But if $y\in A$ for all $y<x$ then $x$ cannot be in $B$, since it would then be the smallest element in $B$ (since all strictly smaller elements are not in $B$), and thus $x\in A$ as we needed.

But what if $A=X∖B$ is done in the middle and thus $B$ is a disjoint union of two subets, one where $y \in A, z \in B, y < z$ and $y \in A, w \in B, w < y$, then even for all $y < x$, $x$ can still be somewhere in the portion of $B$ that is smaller than all elements of $A$, or it is sufficient that we only need to pick one such $A$ and $B$ such that the required result is shown?

Secret

12:52

To be checked: $\{0,1\}\times \Bbb{N}$ under the lexicographic order is order isomorphic to $\omega^2$

Secret

13:05

Proof: Let $f$ be a map such that $(0,0)\mapsto 0$ and the recursive definition $(0,n)\mapsto n,\forall n \in \Bbb{N}$. In addition $(1,0)\mapsto \omega$ and recursively defined, $(n,0)\mapsto \omega n, n\in \Bbb{N}$. We can now check that $\forall n,n',m,m' \in \Bbb{N},\omega m + n < \omega m' + n'$ if $m < m'$ or $(m = m'$ and $n < n'$. We also see the only elements without predecessors are $0,\omega n$, corresponds to $(n,0), n\in \Bbb{N}$

Semiclassical

hi chat

Simply Beautiful Art

heyo

Secret

(almost forgot) therefore the ordering is preserved by $f$ and is bijective, hence $f$ is an order isomorphism between $\{0,1\}\times \Bbb{N}$ and $\omega^2$

By generalisation, $\{0,1,...,n\} \times \Bbb{N}$ will be order isomorpic to $\omega^n$ and $\Bbb{N}\times \Bbb{N}$ to $\omega^{\omega}$

o wait, maybe not...

$$\{n,\omega n, \omega^{n+1}, \omega^{\omega}(n+1), \omega^{\omega+n}, \omega^{\omega n}, \omega^{\omega^n}, \omega^{\omega^{\omega}n},\omega^{\omega^{\omega +n}},\omega^{\omega^{\omega n}},\omega^{\omega^{\omega^n}},...\}$$

$$\{0,1,2,3,4,5,6,7,8,9,10,...\}$$

Therefore $\Bbb{N}^2$ is order isomorphic to $\epsilon_0$

Danu

13:48

Hey @TedShifrin I'm having trouble with something which should be very basic... Say I have an ample line bundle $L$ on an $n$-dimensional complex manifold $M$, and $c_1(M)=(n+1)c_1(L)$ (as far as I'm concerned you may even assume $K_M=L^{-(n+1)}$). Now I want to find that $H^k(M,L)=0$ for every $k>0$. The obvious thing to use is Kodaira vanishing, right: $H^k(M,K_X\otimes L)=0$ for every $k>0$. But I cannot seem to relate this to $H^k(M,L)$, just (by Serre duality) to $H^k(M,L^{-1})$!

In particular, I cannot seem to make sense out of the following:

Waiting

@robjohn hey. How is it going?

Danu

Everything is easy/clear until "[...] and $K_M-L=-(n+2)L$ is negative, so $H^k(M,L)=0$"?!?!?!?! @TedShifrin

I must be missing something simple.

Semiclassical

"so =0 if k>0 by Kodaira vanishing."

makes me wish I understood what Kodaira was

Danu

Kodaira was a famous complex geometer; he proved a famous "vanishing theorem" for sheaf cohomology groups; the above is one special case.

Secret

Now, I finally see how ordinal tetration really works, and how the exponential tower of ordinal really grows:

$$\{+,*,*,*, ...,*,*,*, ...\}$$
$$\{0,n; \omega, (\omega)^n; \omega^{\omega}, \left(\omega^{\omega}\right)^n; \omega^{\omega^{\omega}}, \left(\omega^{\omega^{\omega}}\right)^n, ..., \epsilon_0,\epsilon_0^n; \epsilon_0^{\omega},...\}$$

That means $"{}^{\omega+1}\omega" = \epsilon_0^2, "{}^{\omega 2}\omega" = \epsilon_0^{\omega}$

$"{}^{\omega (n+1)} \omega" = \epsilon_0^{\omega n}, "{}^{\omega^2} \omega" = \epsilon_0^{\omega^2}, "{}^{\omega^n} \omega" = \epsilon_0^{\omega^n}, "{}^{{}^{2}\omega} \omega" = \epsilon_0^{\omega^{\omega}}, "{}^{{}^{n}\omega} \omega" = \epsilon_0^{{}^{n}\omega},"{}^{{}^{\omega}\omega} \omega" = \epsilon_0^{\epsilon_0},"\omega [5] n" = {}^{n}\epsilon_0,,"\omega [5] \omega" = \epsilon_1$

robjohn

14:12

@Waiting pretty good. How about you?

Waiting

@robjohn Glad to hear that. I'm writing up the solution to a problem from my project. It's almost done.

@robjohn these days are hot here, I barely can work on it.

Secret

$"\omega [n+4]\omega = \epsilon_n"$
And finally:
$"\omega [\omega] \omega = \epsilon_{\omega}"$

$\omega [\alpha] \omega = \epsilon_{\alpha}"$

So, the arithmetic of ordinal exponentiation. (In particular the result $(\omega^{\omega})^{\omega^{\omega}} = \omega^{\omega\omega^{\omega}} = \omega^{\omega^{\omega}}$) convert all hyperoperations in terms of exponential towers of the epsilon numbers

robjohn

@Waiting It's been in the 100s here and when its down in the 80s, the humidity is around 50%. All in all, it's pretty miserable.

Secret

and from this we can finally answer the question. The smallest fixed point of arbitrary ordinal hyperoperation: $\omega[\alpha]\omega=\alpha$ when $\alpha = \zeta_0$

Waiting

@robjohn I see. Something unpleasant happens with the weather all over. I never met such hot days here, hard to bear.

Balarka Sen

14:20

@Semiclassical The only thing by Kodaira that I know is his embedding theorem. That gives a nice condition for when a complex projective manifold can be embedded in $\Bbb {CP}^n$ for some $n$ (this is a sort-of-Whitney in the complex category).

Actually I think there are several equivalent conditions for when it's possible.

Danu would know about those

Danu

The embedding theorem uses, and is harder than, the vanishing theorem. It tells you that if you have a positive line bundle (i.e. a holomorphic line bundle whose first Chern class can be represented by a real (1,1)-form $\alpha$ such that $-i\alpha(v,\bar v)>0$ for every tangent vector $v$), then $X$ embeds into a projective space.

Balarka Sen

Ah

Danu

The condition on the form just means that it is given by $\alpha(-,-)=b(J-,-)$, where $J$ is the complex structure and $b$ is a positive definite, symmetric bilinear form

(think: fundamental form)

Balarka Sen

I suppose it's not obvious why this condition implies the line bundle has as many section as we want to embed it in CP^n?

Or maybe one would look at some iterated tensor product of that line bundle and sections of that chap.

(I know embarrassingly little of this)

Danu

@BalarkaSen Positive implying ample is what you need to prove, yeah

Balarka Sen

14:33

Gotcha. The definition of positivity I saw used the Chern connection but I don't really get that condition either.

Tobias Kildetoft

14:57

Hmm, I think I recall Kodaira vanishing being a special case (the characteristic zero one to be precise) of Kempfs vanishing theorem. I just can never remember how to formulate that in terms of the geometric objects

Semiclassical

Kodaira vanishing for non-experts: What happens to how confident you are about a subject as soon as Kodaira is mentioned. :>

Balarka Sen

@Tobias yeah it probably has a sheaf cohomological equivalent

Tobias Kildetoft

@BalarkaSen I only know it in terms of induced representations

Balarka Sen

oh

hides back into his mousehole

Tobias Kildetoft

There is a connection between the induced functors of induction from $B$ to $G$ and cohomology of certain associated line bundles on $G/B$.

Ahh, no, Kodaira implies Kempf in characteristic $0$

Secret

15:11

actually nvm, what I just said about tetration makes no sense

Tobias Kildetoft

@Secret You seemed to be claiming that $\{0,1\}\times\mathbb{N}$ was order isomorphic to $\omega^2$ when really it is $\omega + \omega$.

Fawad

Hi chat

Secret

uh, I am thinking about enumerating the elements of the set like so:
Each subsequent row is larger than the previous:

(0,0) < (0,1) < (0,2) < ... < (0,n) ...
(1,0) < (1,1) < (1,2) < ... < (1,n) ...
...
(n,0) < (n,1) < (n,2) < ... < (n,n) ...
...

Is this how the lexicographical ordering works in $\{0,1\} \times \Bbb{N}$ ?

Tobias Kildetoft

@Secret Yes

Wait, no, because that is not $\{0,1\}\times \mathbb{N}$, it is $\mathbb{N}\times\mathbb{N}$.

which is indeed $\omega^2$

you only get the first two rows for $\{0,1\}\times\mathbb{N}$.

Secret

Ah I see

@TobiasKildetoft Btw, I also have question about your induction proof in your blog post previously. Basically the question boils down to whether it is legal to take $B$ as a set made of the disjoint union of two sets $B1$ and $B2$ such that $"B1 < A < B2"$ (i.e. all elements in each set has the relation as shown), because if I did that, then my $x$ can avoid becoming the smallest element of $B$ since it could well be located in $B1$?

Tobias Kildetoft

15:21

@Secret I am not sure which part you are referring to

Secret

> Proof: Let $B\subseteq X$ be a subset and assume that $B$ does not have a smallest element. Let $A=X∖B$. We need to show that $A=X$ and thus by assumption it is enough to show that if $x\in X$ and $y \in A$ for all $y<x$ then $x\in A$. But if $y\in A$ for all $y<x$ then $x$ cannot be in $B$, since it would then be the smallest element in $B$ (since all strictly smaller elements are not in $B$), and thus $x\in A$ as we needed.

(above inline is excerpt from your blog post)

Tobias Kildetoft

@Secret I am not sure what the sets $B1$ and $B2$ would have to do with this.

Secret

So if I pick $A$ such that it is "sandwiched between" $B$, then I cannot necessary conclude that x cannot be in B because otherwise it will be the smallest element of B

Like this

Because the proof only said $B \subseteq A$, it does not said whether you can allow $B$ to be "disconnected"

Tobias Kildetoft

@Secret I don't see why being connected is relevant.

Secret

If $B$ is a set shown in the bottom rectangle, then when $x \in X, y \in A$ for all $y < x$, $x$ can still be in $B$ (by being present at the lower portion of the set $B$) and thus will not be the smallest element of $B$

Tobias Kildetoft

15:30

@Secret $x$ is picked arbitrarily

Ohh wait, it is not

I don't see where you want to pick $x$ in that picture

Secret

The ordering of the set is smallest to largest from left to right, I can pick $x$ in the left portion of $B$ since all $y \in A$ only need to satisfy $y < x$

Tobias Kildetoft

@Secret But if $x$ is in the left most part that not all $y\in A$ satisfy that

user193319

Suppose $G$ is a nonabelian group of order $6$ in which every element has order $3$. Is this a contradiction; does it, perhaps, imply that $G$ is abelian? Obviously it implies that every cyclic subgroup is normal, but I can't draw any further implications. I am trying to obtain a contradiction.

Secret

Ah right, I see

Tobias Kildetoft

@user193319 Note that any two subgroups of order $3$ will either be identical or intersect in only the identity element

(actually, it is also a standard exercise to show that a group of even order has an element of order $2$).

SteamyRoot

15:36

The identity element does not have order 3 :P

user193319

@TobiasKildetoft Ah. Obviously if they all only intersect at the identity, then this would contradict the order of $G$.

@SteamyRoot You are right: I should have said all cyclic groups with nonidentity generator.

Astyx

It's order divides 3 though @SteamyRoot

SteamyRoot

That's a rather vacuous statement, @Astyx, its order divides any positive integer

user193319

@TobiasKildetoft And if only two groups overlap, then we get a similar contradiction. But if three cyclic groups overlap, this is consistent with $G$'s order, right? I don't see how to get a general contradiction.

Any thoughts?

Astyx

@SteamyRoot Well if it divides three and it's not three, then it's ....

SteamyRoot

15:54

@Astyx I don't see what you're getting at. All I did was point out that the original statement was flawed, because every group has an element that does not have order 3.

@user193319 I don't see what you're getting at here, and what you mean with "overlap"

Tobias mentioned that subgroups of order 3 either intersect in the identity, or are identical.

So, if $a$ has order $3$, then $a$ belongs to the subgroup $\{1,a,a^2\}$. If $b$ is another element in the group, not $1$, $a$ or $a^2$, then $b$ belongs to the subgroup $\{1,b,b^2\}$.

user193319

@SteamyRoot Sorry. I should have used "coincide" or "equal." I think I figured it out: as I already said, if none of the cyclic groups coincide, then $G$ will be forced to have 10 elements, contradicting the fact that it has 6. If only two coincide, we get a similar contradiction. Now, if 3 or more of the cyclic groups coincide, then at least one of the 5 elements has to be an identity element, which contradicts the fact that none of them are.

Hence, if $G$ is a nonabelian group of order $6$, there must be at least one element of order $2$.

SteamyRoot

I think you're making this way harder than it should be, but fair enough.

Also, a more general statement is given by Cauchy's theorem: if $F$ is a finite group and a prime $p$ divides the order of $F$, then $F$ contains an element of order $p$.

user193319

Oh shoot...but how do I know that that element of order $2$ forms a nonnormal subgroup...That's ultimately what I am trying to prove. Here is the original problem I am working with: Prove that every nonabelian group of order 6 has a nonnormal subgroup of order 2.

Ted Shifrin

@Danu: One way Kodaira vanishing is stated is that $H^q(M,L)=0$ for $q<n$ when $L$ is negative.

(I mean, there's usually $H^q(M,\Omega^p(L))$ with $p+q<n$.)

Danu

Let's keep $L$ to be positive. Then by Serre duality that's just saying $H^{n-k}(M,L^{-1})=H^k(M,K_X\otimes L)=0$ for $k>0$

Are you saying I should take $\mathcal L=K_X\otimes L^{-1}$ as my negative line bundle, and write $H^{n-k}(M,\mathcal L)=0$ for $k>0$? And then what? How do I get back to $H^k(M,L)$?

Oh, huh, I think I see now

So I say $H^{n-k}(M,\mathcal L)=0=H^k(M,K_M\otimes \mathcal L^{-1})=H^k(M,L)$

(I'm being sloppy about taking duals of the $H^k$ but whatever)

Tobias Kildetoft

16:15

@user193319 If the subgroup of order $2$ is normal, show that the element of order $2$ commutes with everything

S.C.

16:32

Hi all. Any ideas with this simple problem:
I have a set S = {2,3,...,2n}. How many 3 elements subsets A of S i can have of the form A={i,j,k} where difference between each element in A is >=2?

Balarka Sen

Can you not count 3-elements subsets of S such that at least one difference is <= 2? Does that get complicated quickly?

I guess.

GFauxPas

hey guys, I'm trying to write an algorithm and I'm not sure how to go about it, I don't think it's a stackoverflow question because I don't need the answer in any particular language, I'm more looking for pseudocode

does that belong here?

S.C.

@BalarkaSen i am not too sure. Is a direct counting difficult?

Balarka Sen

I mean if you count the complement, then the subsets are of the form {i, i+1, j}.

Those are easy to count, right?

S.C.

@BalarkaSen yeah. They are \binom{2n-1}{2} right? So is the answer \binom{2n-1}{3}-\binom{2n-1}{2} something like that?

Balarka Sen

16:43

Yeah

Well, binom{2n, 3} minus binom{2n-1, 2}

Uhh no what you wrote is correct

S.C.

How come binom{2n,3}?? Note S={2,3,...,2n}

Balarka Sen

Yeah S doesn't have 1

just noticed

S.C.

@BalarkaSen This was simple. Thanks for the kind help bro :)

Balarka Sen

No problem.

GFauxPas

I'm trying to populate a matrix with 1's and 0's, and I want to impose a constraint that puts a 1 only in a column with no 1's above it

so that I ensure the column sum is at most 1

how do I go about that?

Danu

16:50

@TedShifrin Is there anything more to "irreducible (analytic) variety" than essentially "connected manifold"? If so, they how do I understand why (apparently) if $M=V_1+V_2$ is reducible then $\int_M \omega=\int_{V_1}\omega+\int_{V_2}\omega$ for a top degree form?

Buca Hajdini

@GFauxPas You pick a position in the column put a 1 in there and fill the rest with zeros. Shouldn't be hard, actually it's so easy that I don't think you were asking that, or am I wrong?

GFauxPas

the row sums have to be 6

forgot to say its random

Buca Hajdini

How big are the rows?

Tobias Kildetoft

@Danu That sounds odd, since being irreducible is much stronger than being connected (there being no requirement that the two spaces be disjoint in the definition of irreducible).

GFauxPas

let's say 24

Danu

16:57

@TobiasKildetoft Right. All the more reason for me to be confused :D

Do you know more?

Tobias Kildetoft

@Danu I don't know anything about the analytic version

S.C.

@BalarkaSen @ Looks like we are over counting. There is a problem with counting {i,i+1,j}. We can't have i=2n right?

Ted Shifrin

17:15

@Danu: A variety needn't be a manifold, first of all, as it may be singular. But if it's reducible, the two components meet in something of codimension $\ge 1$ and hence the integral isn't affected by the overlap.

GFauxPas

hm, looks like I want something called the "partition" of an integer

or a magic square?

is there a such thing as a magic rectangle?

Danu

@TedShifrin I know, that's why I said "essentially"...

Hmm, okay. That sounds plausible...

Buca Hajdini

@GFauxPas Let (i,j) be the position in the matrix, where i represents the ith column an j represents the jth row. The value in that position is represented by V(i,j).
1. Start from the first column, that is; i = 0,
2. Generate a random number for j,
3. Check if the row of j isn't equal to 6, if it isn't set P(i, j) := 1, if it is move back to step 2,
4. Set other values of that column to 0,
5. Set i := i+1, and repeat the steps 2-5 until you're done.

A note: The row's length must be multiple of 6 .

Ted Shifrin

The integral claim is obvious. You define integrals by integrating over the smooth locus, anyhow.

Danu

@TedShifrin (I don't know the basic theory when things are not smooth :P)

Ted Shifrin

17:22

There's no big deal. You can read about it in G/H.

Danu

Yeah, I will start reading G/H as soon as I finish my thesis ;)

Algebraic geometry is #1 on the priority list

(besides my actual thesis I guess...)

Rajib Bahar

can someone point me to a good article that lists every math involved in the field of Data Science? I have been refreshing undergrad stat lessons lately... it's been a decade and my brain have degraded so much :P

Secret

[Random]

Experiment: Left handed ordinals
\begin{align}
\text{Right handed} &&& \text{Left handed}\\
\alpha^{{}^{n}\alpha} = {}^{1+n}\alpha &&& \alpha^{{}^{n}\alpha} = {}^{1+n}\alpha\\
\omega+\omega\omega = \omega(1+\omega) = \omega^2 &&& \omega\omega+\omega = \omega(\omega+1) = \omega^2\\
\omega\omega^{\omega} = \omega^{1+\omega} = \omega^{\omega} &&& \omega^{\omega}\omega = \omega^{\omega+1} = \omega^{\omega}\\
\omega^{\epsilon_0} = {}^{1+\omega}\omega = \epsilon_0 &&& \epsilon_0^{\omega} = {}^{\omega+1}\omega =\epsilon_0\\

Details on implementation later. Expect a couple of ordinal identities will be borked when flipping from right handed to left handed (We can also see how in the usual ordinals, tetration is forced to be left tetration as the exponential tower can only grow on the right. Also see math.stackexchange.com/questions/2313576/… for another implementation of left tetration)

GFauxPas

thanks a bunch @BucaHajdini :)

Ted Shifrin

Hi Demonark. Aren't you in class?

Secret

17:32

actually, the two epsilons will be different $\epsilon_0 = \omega^{\omega^{⋰}}$ while $\epsilon_0' = {}_{{}_{⋰}\omega}\omega$ and thus $\epsilon_0' \geq \epsilon_0$

Semiclassical

oh hey

skullpatrol

heya

Ted Shifrin

hi @Semiclassic and @skull

skullpatrol

Hello @TedShifrin

Started preparing for the AoPS?

Secret

Likewise, for right handed (usual) ordinals, the tetration relation is defined as ${}^{n}\alpha=\alpha^{\alpha^{⋰^{\alpha}}}$ while for left handed ordinals, it is defined as ${}^{n}\alpha=\alpha^{⋰^{\alpha^{\alpha}}}$. Order of operations is left associative for the right handed ordinals and right associative for the left handed ordinals

For finite height, the exponential towers have identical appearance, but the value is different (or perhaps may be the same due to the properties of infinite ordinals). Therefore even the base case $\epsilon_0$ there will be two versions, one for the left and one for the right, contrary to how it is treated previously

Ted Shifrin

17:37

No, @skull. I don't figure I need to prepare that much, and I don't have the book yet :) I'll get it when I get back from my drive to and from San Francisco.

Nice name, @MichelAngello. That's actually my brother-in-law's real name.

hi anon :)

Eric Silva

hi chat

Ted Shifrin

Hi Eric

skullpatrol

hi

Eric Silva

People keep interrupting my lunch

Ted Shifrin

Don't blame me!

Eric Silva

17:45

It's very disconcerting

Ted Shifrin

Get over it.

Eric Silva

I just wanna have a peaceful day :(

Ted Shifrin

Naïve boy :P

Balarka Sen

Eric is getting rekt by Ted

Eric Silva

Every where I turn is unforgiving

skullpatrol

17:49

I forgive you :P

Eric Silva

Thank you skull

Ted Shifrin

I certainly wouldn't call that "rekt."

Secret

typo:
\begin{align} \text{Right handed} &&& \text{Left handed}\\ \alpha^{{}^{n}\alpha} = {}^{n+1}\alpha &&& \alpha^{{}^{n}\alpha} = {}^{1+n}\alpha\\ \omega+\omega\omega = \omega(1+\omega) = \omega^2 &&& \omega\omega+\omega = (\omega+1)\omega = {}_{2}\omega\\ \omega\omega^{\omega} = \omega^{1+\omega} = \omega^{\omega} &&& {}_{\omega}\omega\omega = {}_{\omega+1}\omega = {}_{\omega}\omega\\ \omega^{\epsilon_0} = {}^{1+\omega}\omega = \epsilon_0 &&& \epsilon_0'^{\omega} = {}_{\omega+1}\omega =\epsilon_0'\\ \therefore \epsilon_0^{\omega} = {}^{\omega+1}\omega > \epsilon_0 &&& \there…

skullpatrol

Now that^ is #rekt!

Danu

Rick and Morty - Riggity Riggity Wrecked Son!

►

skullpatrol

17:53

\o @Danu

Secret

hate those timeout edits. Need to type agai

Danu

hi

Secret

typo ver 2:
\begin{align} \text{Right handed} &&& \text{Left handed}\\ m^{{}^{n}m} = {}^{n+1}m,({}^{n}\alpha)^{\alpha} = {}^{n+1}\alpha &&& \alpha^{{}^{n}\alpha} = {}^{1+n}\alpha\\ \omega+\omega\omega = \omega(1+\omega) = \omega^2 &&& \omega\omega+\omega = (\omega+1)\omega = {}_{2}\omega\\ \omega\omega^{\omega} = \omega^{1+\omega} = \omega^{\omega} &&& {}_{\omega}\omega\omega = {}_{\omega+1}\omega = {}_{\omega}\omega\\ \omega^{\epsilon_0} = {}^{1+\omega}\omega = \epsilon_0 &&& \epsilon_0'^{\omega} = {}_{\omega+1}\omega =\epsilon_0'\\ \therefore \epsilon_0^{\omega} = {}^{\omega+1…

Ted Shifrin

puts Secret on ignore

Daminark

Hey @Ted, we don't have class today per se (also I just woke up)

There are some office hours but I haven't worked on the homework problems enough on my own yet so I'm not going

Eric Silva

17:58

isn;'t someone giving a lecture ont he zeta function or something

Daminark

True, but it's the second in a sequence, the first of which happened while I was out

Danu

@TedShifrin So if I have a complex manifold... Is it automatically an irreducible variety?

anon

I guess the question is if a reducible variety can be a complex manifold

Danu

Yeah

Balarka Sen

What if it's disconnected

Danu

18:11

Let's assume connected

Ted Shifrin

@Danu: Is it projective?

Danu

Not a priori, no

Ted Shifrin

Then I don't know what it means to be a variety.

Danu

Right

Ted Shifrin

You need to live somewhere to be cut out by analytic or polynomial functions.

Danu

18:12

I guess the entire space is trivially cut out by the empty collection of functions

Balarka Sen

Canonical example of a reducible variety is $xy = 0$

in $\Bbb A^2$

that's not a manifold

Ted Shifrin

Right, a connected, reducible variety is always singular.

Danu

OK, coolbeans

Astyx

Hi chat

Ted Shifrin

Salut, @Astyx.

Astyx

18:17

Quoi de neuf ?

Le film était bien en fin de compte ?

Leaky Nun

19:25

What's your favourite method to find $\displaystyle \int \sec x \ \mathrm dx$?

Danu

Mathematica

4

Balarka Sen

lol

mine is to multiply top and bottom by sec(x) + tan(x)

I hope

Lucas Henrique

hey.

SteamyRoot

19:40

@LeakyNun Any FLOSS maths software (i.e. not Mathematica)

Akiva Weinberger

20:17

In the same vein as @Danu, tables

@SteamyRoot First Last Out Something Something?

Semiclassical

F L open source software?

Lucas Henrique

21:10

hey @Ted

Akiva Weinberger

21:44

chat.stackexchange.com/rooms/63/mathematics

Perturbative

Hi chat!

Daminark

O hai

Astyx

haio

Perturbative

ohi @Dami

I have like 3 different books on diff geom, and all of them seem to use different notation for the same concepts.

^My feelings towards those textbook authors

SteamyRoot

@AkivaWeinberger Free/Libre Open Source Software :P

Akiva Weinberger

21:57

Why is it randomly French

Daminark

Rip @Perturbative

Akiva Weinberger

I think the motto of the ISO is "Great things happen when the world agrees"

or something

SteamyRoot

Well, some people preferred "libre" instead of "free" because it's less likely to be confused with "you don't have to pay for it"

Akiva Weinberger

Wait what does 'free' mean here

SteamyRoot

Free software

Free software, freedom-respecting software, or software libre is computer software distributed under terms that allow the software users to run the software for any purpose as well as to study, change, and distribute the software and any adapted versions. Free software is a matter of liberty, not price: users, individually or collectively, are free to do what they want with it, including the freedom to redistribute the software free of charge, or to sell it, or charge for related services such as support or warranty for profit. The right to study and modify software entails availability of the...

Perturbative

22:00

On another note, another book which didn't want to make use of any topology whatsoever decided to call an open set a 'domain' (presumably just for luls)

Daminark

Loring Tu?

SteamyRoot

@Perturbative at least each book was consistent with its own notation, I hope?

Perturbative

@Daminark Nah Tu's book called open sets, open sets I think, this was another book I picked up from my uni's library

@SteamyRoot Yeah they were, its just that converting notation back and forth is a pain :(

Semiclassical

@SteamyRoot also, better acronym

Danu

@Perturbative I always thought "domain" means connected, open set

SteamyRoot

22:05

@Danu A nonempty open, connected set is sometimes called a region :^)

At least, that's terminology often used in complex analysis

Perturbative

@Danu I've never heard that definition before, but regardless the first mention of open set (that I've seen) in this book was on pg 196

@Daminark How's things going in your diff geo course?

Daminark

It's going aight, we've done some stuff on parametrizations and the first fundamental form

I'm actually gonna work some on the pset right now

:P

Benjamin

Are there any standard programs or standards for modelling logic?

Perturbative

Have fun @Dami :p

LucasTizma

Is there a function or method such that the rounding method changes along a range? E.g., from 0.0 (floor) ... 0.5 (round) ... 1.0 (ceil)

I don't know if that makes any sense or not. Hahah.

A more evenly distributed "weighting" of which side to round to.

Benjamin

22:13

@LucasTizma It could be done as a simple piecewise function. You would need to define your boundaries though.

LucasTizma

Ya, I had thought of that. I was hoping for something more continuous rather than picking discrete intervals to switch methods.

Benjamin

@LucasTizma Rounding is by its nature not continuous. Depending on how you define boundaries you could make it nearly continuous for certain levels of precision.

LucasTizma

I wonder if I could add a bias value or something to the input before rounding.

Perturbative

I've gotta go study for a test now, cheers everyone!

Ted Shifrin

Hi/bye Perturbative

LucasTizma

22:15

Yeah, I suppose that's what I meant. More-or-less continuous for reasonable precision.

Maybe something like -value...0.0...+value can be added along the range to coerce input one way or the other.

Benjamin

@LucasTizma The easiest way to do this would be to drop a level of precision from input to output.

LucasTizma

Or perhaps I'm just overcomplicating things.

What do you mean?

Benjamin

@LucasTizma Input a number with precision to the tenth decimal, output with precision to the 8th decimal.

LucasTizma

Ah I see.

Benjamin

@LucasTizma Depending on what impacts you want to have, you could use a bias value.

LucasTizma

22:18

I'm displaying an approximate "time remaining" in minutes, disregarding the seconds. The problem is that I don't want it to do something like 0 min for the last 30 seconds.

Benjamin

@LucasTizma How precisely are you willing to display the minutes.

LucasTizma

I guess realistically, I should only care when the minutes themselves change rather than rounding... So at 4:01 I would still show 5 min and at 0:01 I would still be showing 1 min.

So it's more of an upper bound.

Benjamin

@LucasTizma That should just be doable with a ceiling function.

LucasTizma

Ya, sounds like it now that I think about it.

But, assuming a less trivial case, I imagine I could bias the rounding along the range if I really wanted more human-like rounding behavior.

I'll just use a ceiling function!

Danu

@TedShifrin I have the following SES of sheaves

Semiclassical

22:30

I momentarily read that as SOS

Ted Shifrin

It probably is SOS.

Danu

$0 \to \mathcal O_X \to \mathcal O_X\otimes L \to \mathcal O_Y \otimes L\to 0$, where $Y\subset X$ is a divisor and the first map is induced by multiplication by a section of the line bundle corresponding to the divisor

No, that's probably nto quite what I have

Ted Shifrin

That isn't right unless the $L$ is very much related to the divisor $Y$.

Danu

yeah

it is

I wanted to omit unnecessary context

Ted Shifrin

So you should start with $0\to \mathscr I_Y \to \mathscr O_X \to \mathscr O_Y\to 0$ and tensor with $L$.

Danu

22:33

So $Y$ is exactly defined by $Z(s)\cap X$ where $s$ is the section of $L$

Ted Shifrin

OK.

I don't get the $\cap X$, though, @Danu. We're living on $X$.

Danu

So here $\mathcal O_Y$ should a priori at least be a sheaf on $X$. Huybrechts told me to think of it as just equivalence classes of sections of $\mathcal O_X$ up to holo sections that vanish along $Y$

Ted Shifrin

Well, no, it's supported on $Y$. It's a sheaf on $X$, if you insist.

Danu

What do you mean by supported?

Ted Shifrin

Your sentence you just typed is what my $\mathscr I_Y$ tells you. That's the ideal sheaf of $Y$ ... things vanishing on $Y$.

Danu

22:35

You mean it only has nonzero sections along $Y$?

Ted Shifrin

The stalks of the sheaf off of $Y$ are all $0$.

Danu

Right, ideal sheaf

@TedShifrin yeah, right

OK. So then you can sort of mentally think of it as just the sheaf on $Y$

Ted Shifrin

Yuppers.

Danu

OK, so I understood what's going on.

Neat

I suddenly need this stuff to understand some proof, that's why I've been asking all these questions today :P

Ted Shifrin

So the punchline is that $\mathscr I_Y \cong \mathscr O_X(L^*)$, so tensoring with $L$ gives what you had.

Danu

22:36

Right

Ted Shifrin

I guess I shouldn't use divisor notation with a bundle.

Danu

It's just a line bundle, so it's OK

I have a slightly pedantic question

So I have this line bundle $L$ with $\dim H^0=n+1$, and some basis $\{s_j\}$ of $H^0$. Now I have (by induction) an exact sequence $0\to \operatorname{span}(s_1,\dots,s_{k-1})\to H^0(X,L) \to H^0(X_{n-(k-1)},L)$. I now have some $X_{n-k}$ inside $X_{n-k+1}$ and I have also proven that the kernel of the restriction $H^0(X_{n-(k+1)},L)\to H^0(X_{n-k},L)$ is spanned by $s_k$. This makes it intuitively clear that I get the analogous sequence for $k$ (rather than $k-1$)

But I don't see how to precisely formalize it

Ted Shifrin

I don't know what your $X_j$ are. You can have zillions of sections when $\dim X = 1$.

Danu

Could I write something like this: $0\to \operatorname{span}(s_1,\dots s_k)\to H^0(X,L) \to H^0(X_{n-(k-1)},L)/\operatorname{span}(s_k)$ is clearly exact, and the latter is precisely $H^0(X_{n-k},L)$?

Akiva Weinberger

Whoa Mauritania changed its flag three days ago

(I feel like there might be an overlap between people who like flags and people who like math)

Danu

22:52

@TedShifrin Ehm, things whose dimension goes down by one, as k goes up (so n-k is the dimension of $X_{n-k}$)

Ted Shifrin

DogAteMy, my interest is flagging.

Danu

I love flag manifolds...

Ted Shifrin

I know, @Danu, so I'm questioning the whole thing.

Danu

@TedShifrin ?

Ted Shifrin

If I start on a curve, I can't go down very far.

Are you assuming $\dim X\ge n+1$?

I mean I can have yuge $n$.

Danu

22:53

$\dim X=n$, sorry for not clarifying

So $X=X_n$ basically in more harmonious notation

Ted Shifrin

Are you assuming these sections form a regular sequence ... so that the divisors intersect transversely? Otherwise you're in trouble.

Danu

Yeah, everything nice

I thought I can take the basis $\{s_j\}$ to be like that, no? I just need them to be a basis of $H^0$, isn't that an open condition?

Ted Shifrin

A priori, your linear system could have base points so that there's bad intersection of these things.

Danu

I mean... I don't think I need everything to be non-singular at this point

But I honestly don't know exactly what I'm doing. I'm just reading Kobayashi & Ochiai's paper and making sense out of it

Ted Shifrin

I'm going to have to abandon you for a while. I'm having computer issues — due mostly to my own insistence on restoring my iPad. Everything's messed up.

Bye.

Danu

22:56

Oh, okay.

Cya, and good luck!

Never restore anything is the takeaway message!

Akiva Weinberger

23:08

Like monarchies

mdave16

23:50

idk monarchies are okay

they generate more money than they consume, much like a good zoo exhibit

ah, i'm an hour late

@eyeballfrog, so you wondered what was wrong with the fundamental theorem of arithmetic (for an ultrafinitist)

eyeballfrog

yeah

mdave16

so FTA states that any integer can be factorised uniquely (up to order) into primes

eyeballfrog

right

mdave16

but how do you know primes exist? since there are only a finite amount of primes

eyeballfrog

hmm

it seems I can prove that there exist primes

that there are an infinite number is a known proof, but I'm sure ultrafinitism as some problem with it

mdave16

23:58

so you can have numbers which exist, (because they have enough "information density" to be written down and thought about), but the primes which divide them, are too large and cannot be written down as nicely, so do not exist in the ultrafinitist world

yeah

so take, (i think Euler's?) proof

eyeballfrog

but said primes are smaller than this number which exists

if it's OK for me to write down this number that exists, surely it's OK to write down smaller ones

mdave16

finite primes $p_1, \dots, p_n$, then $p_1\cdotsp_n + 1$ is a new number, must have a prime divide it, and so a new prime $p_{n+1}$ exists

not exactly

Transcript for

$Mathematics$ Mathematics