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00:01
Right. So in $H_0(S^1)$, where I'm considering the simplicial homology of 3 1-simplicies, I just set the boundaries of the 1-simplices to 0.
yup.
Here everything's abelian, so it's a bit simpler.
So assuming I named the vertices of the "triangle" $u$, $v$, and $w$, that means $u = v = w.$
Yup. So you get a single generator for $H_0$.
But what can I conclude from this?
Ohhhh.
For a connected simplicial complex, $H_0 \cong \Bbb Z$.
00:04
I feel enlightened.
well, turn off the light :)
Do computers use simplicial homology? Or cellular?
when do computers do homology?
On point clouds?
perhaps ... :)
00:09
Why is simplicial homology taught if cellular is typically easier?
you have to learn various theories to understand things deeply
you do simplicial and singular first because they're fundamental, and cellular is more intricate
Do you mind walking me through the simplicial homology of the torus @TedShifrin?
First you need to have a triangulation.
Right.
Can I just use the square divided into two triangles?
The one that identifies opposite edges.
That's not a legal triangulation, but it's a so-called delta-complex.
00:13
Why isn't it legal?
what are the requirements for a triangulation?
Oh. Is it because the intersection of two simplices must itself be a simplex?
or empty, yup.
Your intersections are very baddddddd.
Oh I see, because all the simplices will have one vertex which is impossible.
No, you overlap on way too much.
Like on three simplices?
So to get a legitimate triangulation you need lots more pieces.
00:17
captain obvious here: How many pieces would you need to get a legit triangulation?
I always assigned this in my diff geo course, but I don't know the minimal number. Dividing into 18 triangles will do.
:/
that's pretty disgusting
I know there's a paper somewhere with the optimal.
Ah, 14.
00:20
"For the particular case of a simplicial complex structure for a torus, David Eppstein is right: the minimal triangulation has 7 vertices, 21 edges, and 14 triangles."
ew
But how do you form such a triangulation?
You take your square and you draw lots of little squares to make sure overlaps are legal.
Then you turn the squares into triangles.
But then you realize you can save a few if you really care.
This is why ultimately people don't use simplicial to compute.
I think the first time I saw this question I thought they meant a triangulation in $\Bbb R^3$.
Singular is important if you're going to tie things in to manifolds and differential forms.
I still don't know whats the number of simplices you need for that.
00:23
Well, @PVAL, you can draw it on the torus in $\Bbb R^3$. I've had plenty of students do that.
I don't think there's any difference.
Unless you're requiring convexity or something.
Okay @TedShifrin, let's just work with the 18 triangles.
Did you draw the picture?
Yes. I have a 3 x 3 square of squares that are divided into triangles.
And you agree that the 3x3 arrangement makes everything cool?
OK, so write down your chain complex $C_2\to C_1\to C_0$. Each one is $\Bbb Z^m$ for some $m$. (Ugh.)
Yes I checked how each face would glue.
00:27
We obviously don't want to label everything. It's going to be ridiculous.
$C_n$ for $n > 2$ will be trivial because there are no n-simplices.
@ted I want the simplices to be linear. Maybe its the same but its not obvious to me that when you linearize you'll get the right thing.
LOL, yeah, I'm not worried about those.
@PVAL: What do you mean by linear? You're going to take a PL torus?
I see. So maybe my 18 triangles won't all work.
Like I want to build a polyhedron out of regular simplices that is homeomorphic to the torus.
or thats probably impossible
so throw out regular
Throw out regular. That was stooopid anyhow.
But my 18 probably won't all work. You might need a few more.
@gian: So we have $\Bbb Z^{18} \to \Bbb Z^? \to \Bbb Z^?$.
You can see this is not going to be fun. What's the number of vertices?
00:31
16?
Nope. Don't forget identifications.
9?
Yup. Now count edges (tee hee).
21?
I don't think so.
00:37
27?
I don't think @Ted counted.
27 is correct.
All I'm doing is identifying opposite edges on the perimeter of the triangulation.
Oh yay :D
I actually just did to be sure, @PVAL sticks out 10 tongues
9 vertices, 27 edges, 18 faces... :)
00:38
Well, it does check, but @gian doesn't know why/how.
What checks?
Something called Euler characteristic.
9+18=27 is not a coincidence.
Obviously we do not want to write out matrices for the boundary maps $\Bbb Z^{18}\to\Bbb Z^{27}$ and $\Bbb Z^{27}\to\Bbb Z^9$.
But in principle ...
You should still be able to convince yourself that $H_2 \cong \Bbb Z$ and $H_0 \cong \Bbb Z$.
How would I handle im$ \partial_1$? $\partial(C_1 \cong \Bbb{Z}^{27})$ looks scary.
00:45
This is like what we discussed for the single simplex.
You can join the bottom left-hand corner to every other vertex with edges.
For each $1$-simplex, it's giving you $v-w$.
But would that then mean that im $\partial_1$ is generated by 27 boundaries?
Yeah, but they all involve the same 9 vertices.
Oh that's right.
But if you pick $v_0$ and any other $v_i$, you can get there by following $1$-simplices.
So they're equivalent mod the image.
Hello
00:49
So there are really 9 distinct boundaries. So there's a single generator?
No, not 9 distinct boundaries.
I don't know what you're thinking there.
But, yes, there's a single generator for $H_0$.
I had a quick quesiton about extensions. If $S$ is a collection of elements in the closure of $F$, can a general element $\alpha\in F(S)$ be represented as an $F-$linear combination of elements in $S$?
the elements of S are algebraic?
so let $\alpha$ in $S$
we have $P(\alpha)=0$ for some irreducible polynomial with coefficients in $F$
now let $a_0$ be the constant coefficient of $P$.
Irreducible implies that $a_0 \ne 0$, but even if it wasn't you could divide by $x$ a bunch of times
00:54
So is ker $\partial_1 \cong \Bbb{Z}^{27}$?
No, that's not right, @gian. That would say everything mapped to $0$.
What's $\ker \partial_0/\text{im}\,\partial_1$?
Now just by high school algebra we have $-(a_0)^{-1}(P(\alpha)-a_0) = 1$
I concur
and now mulitplying both sides of this equation by $alpha^{-1}$ gives $\alpha^{-1}$ as an element of $F[S]$.
I know that ker $\partial_0 \cong \Bbb{Z}^9$.
00:58
@gian: Do you know nullity-rank in linear algebra or the fundamental homomorphism theorem in group theory? Surely you do, or you shouldn't be playing with homology.
So what's the dimension of $\text{im}\,\partial_1$?
@Prototank You should be able to figure out why $\alpha^{-1}(P(\alpha)-a_0)$ is still a polynomial in $F[\alpha]$
throws @PVAL a backslash
whew :)
I JUST SEE THE CODE
Ohhhh.
don't you be bitchin' at me, boy!
01:01
So im $\partial_1 \cong C_1/$ ker $\partial_1$.
Right.
So think dimensions (if this were linear algebra).
@PVAL on your phone?
G'morning, @MikeM.
Also hey guys
@PVAL-inactive, I see why it is still a polynomial in $F[\alpha]$. Since we removed the constant, we were able to decrement powers of $\alpha^n$.
01:03
Bye, Demonark.
@Prototank Yeah that's right.
Last night dream: (forgot except there's an n-cube lattice where a graph consists of two connected components attached at a single vertex is traced and the total number of some unspecified type of component is given by the expression: $$\sum_{r=1}^{n}\binom{n}{r}(n-r)+1$$. In particular, I asked the tutor about the infinite dimensional case $n\to \infty$
I guess to start a proper proof you should take an element $\alpha \in F[S]$ and use the fact that any such alpha is algebraic.
So then the dim $C_1$ - dim ker $\partial_1$ = dim im $\partial 1$.
So for any $\alpha\in F(S)$, we need only consider $\alpha^{-1}\in F(S)$, run the argument you just made to get that $\alpha$ is an $F-$linear combination.
01:05
Wait let me double check.
if you swap $r\mapsto n-r$, then the sum becomes $\sum_{r=1}^{n-1} \binom{n}{r} r+1$ @secret
@PVAL Should I try to come to the UT thing?
Yeah I think that's the dimension analog.
@Mike Sure.
And that's not too hard to do if you note that $\sum_{r=0}^n \binom{n}{r}x^r=(1+x)^n$ and differentiate
01:06
except that dimension is usually reserved for vector spaces and here we have free $\Bbb Z$-modules, @gian, but don't sweat it.
It might be a livable temperature here by november.
if there isn't a hurricane
@TedShifrin I have reservations about your reservation.
I appreciate your time
I use the word rank, not the word dimension, @PVAL :P
01:08
But why are we finding the dimension of im $\partial_1$ in this way? Can't I just consider the number of generators of im $\partial_1$?
That's not so easy to see, @gian.
I mean, you'd need independent generators.
Oh gotcha.
So how is dim ker $\partial_1$ easier?
We can see $H_0 \cong \Bbb Z$, so ...
If I have a collection of proper submodules of some module $M$, and my book tells me their sum is also a submodule of $M$, then surely they don't mean direct sum right? Since say $\Bbb R\oplus \Bbb R\not\subset \Bbb R$. Is this normal terminology, and is it equivalent to just taking a union of their basis (say they are free modules)
No, they don't mean direct sum.
01:12
im $\partial_1 \cong \Bbb{Z}^8$
Sum means $\{v+w: v\in V, w\in W\}$.
Right, @gian, so that tells you that $\ker \partial_1\cong \Bbb Z^?$.
heya SBA.
@TedShifrin Thanks, this makes me happy.
Does anyone know of some math scholarships that don't involve competitions? I'm... interested. :-)
01:13
Just $\Bbb{Z}$
@PVAL-inactive I guess this is just $F[S]\subset F(S)$ is obvious, and then we show that any element in $F[S]$ is invertible by your argument, which makes $F[S]$ a field. The proof is done by minimality?
Glad to make you happy, @user462339.
No, no, that's not right, @gian.
@Prototank Yeah, that sounds fine to me.
If you like, F[S] is the minimal ring containing F and S
so clearly the minimal field contains the minimal ring
as there are more assumptions.
Hm? What am I doing wrong? Don't the dimensions of the image and kernel add up to the dimension of the domain?
But the domain was $\Bbb Z^{27}$?
01:17
Ohhh silly me...
I was looking at $C_0$
@Semiclassical $$\sum_{r=0}^{n}\binom{n}{r}rx^{r-1}=n(1+x)^{n-1}$$. Then set x=1 in the GF to get the required sum $$\sum_{r=1}^{n-1}\binom{n}{r}r+1=(n-1)2^{n-1}$$ which diverges as $n\to \infty$ as suspected
So ker $\partial_1 \cong \Bbb{Z}^{19}$
You missing an exponent, @Secret?
OK, @gian. Now what about $H_2$?
Hey guys
So much math going on in this room. What gives?
01:19
Hush, @PVAL. :)
In Did's answer here: math.stackexchange.com/questions/2436627/… I'm trying to show that $G^{\prime} = -g$ like he said, but I don't know how to use the fundamental theorem on an improper integral.
I'm doing algebraic topology, @ALannister, so I'm not gonna pay attention.
@TedShifrin lucky
Anybody else here want to do something mundane?
Like explain how to do the derivative of an improper integral using fundamental theorem?
It's no different from a proper integral.
Write $\int_x^\infty = \int_x^b + \int_b^\infty$.
Well im $\partial_3$ is trivial.
01:21
Sure, @gian.
How do we get an element of $\ker\partial_2$?
Draw some pictures in your picture.
@Ted and that's supposed to give me -g?
hmmm...
I'm not looking at your question, @ALannister, but remember that if we take $G(x)=\int_x^b g(t)\,dt$, then $G'(x)$ is what?
@Ted Dont think that's relevant
$g(b)$ isn't it?
Er, I mean $g(x)$
No, negative.
Think about it geometrically.
01:23
Oooh, because you've got to turn the beat around.
He want's to differentiate something under the integral sign where the bounds are constant.
As $x$ increases, the area decreases.
aaah
ok
That's just putting the partial under there.
@PVAL
all right, i'm gonna go Vicki Sue Robinson my way through the rest...Thanks @Ted
01:24
LOL, sure, @ALannister.
@gian: So how do you combine $2$-simplices and get something with no boundary?
0
Q: $f ^{\prime \prime}(x) = f(x) f^{\prime \prime} (x-1) $

mickI know the equation $ f^{\prime} (x) = f(x) f^{\prime} (x-1) $ is solved by $f(x) = C$ or by tetration ( $ f(x+1) = \exp(f(x)) $). So I wonder What are the solutions to $$f ^{\prime \prime}(x) = f(x) f^{\prime \prime} (x-1) ?$$

@mick: UGGGHHHHHHHHHHHH.
That's all I have to say.
Well you just need to ensure that you have 2-simplices sharing common edges so that orientations cancel, right?
Absitively right, @gian.
All of them have to cancel.
So that means that you need all the scalars from the ground ring to be equal.
01:30
Yikes. Huh?
Heyo, someone finally answered my ordinal collapsing function question.
Oh, I see.
So what do you think $H_2$ must be?
Well I guess there's only one generator so $H_2 \cong \Bbb{Z}$
Well done, @gian.
Now you can put the pieces together and "solve" for $H_1$.
H_1 seems a little more difficult to me.
01:32
We don't need to do it. We have nullity-rank to put the pieces together. (I'm not doing the Hopf lemma, or whatever it's called.)
@gian is about to figure out one of the basic tools in homological algebra, but don't tell him.
Well, good night all.
night, SBA
Oh I got it!
$H_1 \cong \Bbb{Z}^2$
You're great, @gian :)
And you can see it geometrically :)
You should be able to see two independent $1$-cycles in the torus.
Right, just like the path homotopy equivalence classes in its fundamental group.
And then I'm assuming the 2-hole is the one it encloses.
01:40
The fundamental group is a non-abelian mess.
of the torus?
Yeah but isn't $H_1$ just the abelianization of the fundamental group?
Oops. I wasn't paying attention.
Well, @gian, but that wouldn't help.
I know but is that why the two are similar?
In this case, $\pi_1$ is abelian, so $H_1=\pi_1$.
Ted was being a dope.
And it's time for Ted to go cook dinner.
01:42
It's not hard to show [A, B \times C] \cong [A, B] \times [A, C] where A, B and C are pointed spaces.
What does it mean the logarithmic derivative of $R_{x}(t) = \frac{G(x+t)}{G(t)}$? The regular derivative of $R_{x}(t)$ is $\frac{ G(t)\cdot G^{\prime}(x+t) - G(x+t)G^{\prime}(t)} {[G(t)]^{2}}$ And since $G(x) = -g(x)$, this derivative comes out to $\frac{g(t)G(x+t) - g(x+t)G(t)}{[G(t)]^{2}}$
@Gian: You did great with the homology. I hope you sit down and think about what you learned.
and [A,B] denotes the homotopy classes of pointed maps [A, B]
Okay, thank you for your time @TedShifrin.
@ALannister: Logarithmic derivative means take log (ln) and then take derivative.
$\log(f/g) = \log f - \log g$.
01:45
Hmm...I'm not sure how doing this gets you that...
let me try it
Actually, I just did. And the denominator did not go away for me like it did for him
@ALannister: You should have taught logarithmic differentiation in calc 1 ... it's a good technique.
Calc 2. I know how to do it. I've just never done it with integrals before.
And at the end, you have to re-exponentiate
It's also not giving me his result.
No, you don't have to re-exponentiate. I don't know what you're solving for.
01:46
The result he has is essentially just the numerator of the regular derivative.
Which doesn't make sense.
Oh, blah. Give me the link, and then I'll go cook dinner.
1
Q: Behavior of Gamma Distribution over time

ALannisterI need to prove that the age behavior of the Gamma Distribution with probability density function $$ f(x)=\frac{\lambda^{\alpha}x^{\alpha-1}e^{-\lambda x}}{\Gamma(\alpha)}$$ For $x\geq 0$, $\lambda, \alpha >0$; I.e., the conditional pprobability $P(X>x+t|X>t)$, increases in $t$ whenever $\alpha >...

Oh Ted!! You're the greatest!! <3
it's the thing he calls $h(t,x)$.
I don't know why he said logarithmic derivative — you're right — he's doing the quotient rule. But the denominator is positive, so he's looking just at the sign of the numerator of the quotient rule.
Oooh, okay.
I knew I wasn't going nuts.
It's what you get when you do logarithmic derivative and take a common denominator. Same thing.
Nope, you're not walnuts, yet. Maybe hazelnuts.
01:50
I hate hazelnut
too sweet.
I love hazelnut. So hush.
Fine, be bitter and have ... hmm ... walnuts.
Toasted hazelnuts and chocolate ... yummmm ...
goes to cook dinner :D
Here we just put P cans on everything.
In Georgia, too. But now I'm a Californian. I still like pecans, too.
Bon appetit! Thank you!!
And Josh just won Big Brother!!! YAAAAY!
or pickahns
I forget how its pronounced here.
01:54
LOL, @ALannister ... keep me posted :P
Big Brother? Really? ... Oy. Well, I suppose anything is better than our president.
Can anyone help with this maximal ideal question?
0
Q: If $A$ is a semilocal ring and $f : A \twoheadrightarrow B$, then $f(\text{rad}(A)) = \text{rad}(B)$?

EyesOnBudI know that if $f$ is surjective then $f(\mathfrak{a})$ is an ideal for every ideal $\mathfrak{a}$ in $A$. I want to first prove that if $\mathfrak{m}$ is maximal in $A$ then $f(\mathfrak{m}) = \mathfrak{m'}$ is maximal in $B$. $A/\mathfrak{m}$ is a field by maximality of $\mathfrak{m}$ and $f$...

Bye for now, all.
@EyesOnBud A/m is a field so any surjective ring map with A/m as the source is automatically an isomorphism so B/m' is a field.
Yeah, I know that
What about my last question at the bottom
wait...
If $f : A \twoheadrightarrow B$
and $A/(\mathfrak{a} \cap \mathfrak{b}) \approx B / (f(\mathfrak{a}) \cap f(\mathfrak{b}))$ then is it true
that $f(\mathfrak{a} \cap \mathfrak{b}) = f(\mathfrak{a}) \cap f(\mathfrak{b})$?
I know $\subset$ is already true since it's true for any maps
@PVAL-inactive
02:05
Lemme see I'm very rusty on comm. algebra.
thx
I think this might come in handy:
Third isom. theorem for rings
let $f(x)$ in $ f(\mathfrak{a}) \cap f(\mathfrak{b})$ but not in $f(\mathfrak{a} \cap \mathfrak{b})$ . I think you can say $x$ is in the kernel of the map A to B.
or it differs from something in $\mathfrak{a} \cap \mathfrak{b}$ by an element of the kernel
but since you know what the kernel is contained in $\mathfrak{a} \cap \mathfrak{b}$ you are done.
so you should have that equality.
Mind making an answer to be more precise?
When you say map $A$ to $B$ do you mean original $f$?
yeah
I think I'm tired and confused and you should just ignore me
If $S$ is a collection of separable elements over $F$, not necessarily finite, is $F(S)$ separable over $F$? I know that it is true if $S$ is finite.
02:16
and wait for someone who knows this stuff better.
@Prototank any element of $F(S)$ is contained in a field $F(S')$ where $S'$ is finite.
This is probably easier to prove if you think about $F[S]$ and $F[S']$.
This is what I feared
It just comes down writing what an arbitrary element in $F[S]$ looks like, which shouldn't be too hard.
I worry because this implies, as far as I can see, something that isn't true: every normal extension $E/F$ is galois
Let $K$ be the separable closure of $F$ in $E$. So we have $E/K/F$. By some lemma, $E/K$ is separable. Since $K$ is just $F$ adjoin some separable elements, $K/F$ is separable also. By some tower theorem, $E/F$ is separable.
holy crap forgive me
I found the mistake
:( gosh it sucks when you spend so much time on something... trudging through a mistake
02:49
Morning
Hey @Daminark
@Prototank One standard example of a non-separable extension is F_p(x^(1/p)) over F_p(x). In this case the separable closure is just the base field, so your "some lemma" isn't true.
[Harmonic reciprocal series]
I wish there's a more obvious geometric meaning of harmonic numbers, because they do have nice properties
\begin{align}
\sum_{m=1}^n\frac{H_{m}}{m} = \frac{1}{2}(H_nH_n+H_{n,2})\\
\sum_{m=1}^n\frac{H_{m-1}}{m} = \frac{1}{2}(H_nH_n-H_{n,2})
\end{align}
Yo @Faust
03:27
Let k,n be two positve integers and let d be their greatest common divisor.
in the group $\mathbb {Z}_n $ prove that $ < [k]_n> = < [d]_n> $
for some reason im having alot of trouble showing this
03:52
If $F$ is characteristic zero, can't we say that the only elements which are purely inseparable over $F$ are elements in $F$?
@Prototank usually a purely inseparable extension requires positive characteristic
04:11
I think there should be a more general notion of p.i. so that it can be hypothetical at least, but not interesting
in the case of p=0
05:09
Back.
How are you able to turn $\int_{z}^{\infty}$ into $z\int_{1}^{\infty}$?
@Faust Bezout
@PVAL-inactive's avatar keeps reminding me of Secret
05:25
Hi @EricSilva
05:38
[Random]
NBG
$A \in V$
$V \in A$ is nonsense
e.g. $A \in S \in \cdots \in V$
$Rp(A,a) := \forall x (x \in A \leftrightarrow x \in a)$
$Rp(V) := udf \because V \not \in V$
Let $(W,\in)$ be a well order
$P := \{\forall x \exists ! y P(x,y)=T\}$
$F_P(X)=Y$ iff $P(X,Y)$
$F_P[A] :=\{\forall y \in B, \exists x \in A : F_P(x)=y\}$
(and someone please use 2nd order logic so I don't need to write that lengthy axiom of replacement statement!)
Construction 1:
05:56
Hi
Why am inot getting a matrix after I do this
\[ $$A =

\begin{bmatrix}

0.975 & 0.0125 & 0 \\

0.0125 & 0.975 & 0.0125\\

0 & 0.0125 & 0.975\\
\end{bmatrix} \] $$
Let $0 \in On$. Then pick function $0 \in x$ where $x \in On$ already constructed:
$0 \in 1 \in 2 \cdots$

to be continued...

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