Mathematics - 2024-03-14

leslie townes

00:00

yet another person turned down the path of fascism by (math) chat rooms on the internet

sigh

Jakobian

reminds me of our current government

Jakobian

01:15

I think this sort of makes sense, and does make me less anxious

EE18

03:18

Any hints here? Induction obviously suggested but the inductive step is weird because it's between two different rings...$R[X_1,...,X_m][X]$ and $R[X_1,...,X_m,X_{m+1}][X]$. I guess I need to show some isomorphism between $R[X_1,...,X_m][X]$ and some subring of the larger ring in the inductive step?

leslie townes

uh, have the s_k been defined elsewhere? or if not, what are the s_k required to be?

[that formula could easily be a definition of s_k, and in some texts might actually be "the" definition of s_k, hence my questions]

[if you get to choose what the definition is, try that one. there's nothing to prove if that is the definition]

EE18

oops i should have

included that

leslie townes

07:05

EE18: it is regrettable that they have notationally suppressed the dependence of "s_k" on m, and that is way more "..." than i would hope to see in a definition that a textbook is asking me to prove anything about, but OK

if you write something like s_k(m) for their "s_k for the variables X_1, ..., X_m" and induct on m, it likely reduces everything to proving that your "definition" satisfies s_k(m+1) = s_{k-1}(m) + x_{m+1} s_{k-1}(m) for all 1 <= k <= m

which is either a separate induction on k, or (more likely given the amount of ... appearing in the definitions themselves) something you can wave the magic of "..." at yourself and just see that it's true

should be = s_k(m) + x_{m+1} ... on the rhs of that equation up there

EE18: i would see the point of exercises like this to be 'getting comfortable working with sigma notation,' and not really anything specific to these polynomials, or to proof writing generally

Pizza

09:10

Equivalently, $f_k$ *uniformly converges in* $I$ towards $f$ if, for every $\epsilon > 0$, there exists $\nu_\epsilon \in \Bbb N$ such that
$$sup\{|f_k(x) - f(x)| : x \in I\} < \epsilon, \quad \forall k > \nu_\epsilon$$ yet, equivalently, $f_k$ *converges uniformly* in $I$ towards $f$ if the limit relation is satisfied ( of numerical succession)
$$\lim_{k\to\infty} sup \{|f_k(x) - f(x)| : x \in I \} = 0.$$

Can someone explain this to me, pls

robjohn

09:34

what part do you not understand? It is just a rewriting of the limit.

Pizza

09:52

I'm trying to understand for a moment on my own

I'm reading some stuff from the internet

Ryder Rude

10:25

what is the geometric interpretation of the non-free generators of $H_r(K)$?

what kind of holes are these and how are they differentiated from the usual holes that free generators represent?

one potato two potato

yeeeeeees

@BalarkaSen That's for Borromean rings and I checked the polyhedral convergence using figure eight knot complement.

It would be useful if it has $\Bbb D^2$ for Fuchsian too

Ryder Rude

10:50

in the proof of Euler Poincare theorem, it says $\sum _{r=0}^n (-1)^r (dim(C_r(K))+dim(B_{r-1}(K)))=\sum _{r=0}^n ((-1)^r dim(C_r(K)) - (-1)^r dim(B_r(K)))$

Pizza

cdn.discordapp.com/attachments/702893241691013182/…;

Among all these differences $f_n(x) - f(x)$ who is the largest?

Ryder Rude

i dont understand how $dim(B_{r-1}(K))=-dim(B_r (K))$ in the above step

Pizza

(the link is a figure)

Ryder Rude

this makes no sense as the dimension of a vector space is always positive

but this step is necessary to get the Betti numbers on the right

what is happening here

Balarka Sen

12:36

@onepotatotwopotato Great!

I should eventually learn how to read the "cusp shapes".

Thorgott

@RyderRude that's not the implication

Balarka Sen

@Jakobian Epicurus was also a big hater

"Epicurus postulated the least conceivable length called elahiston, declared that geometry is based on falsehoods, and banned it from the curriculum at his school, he similarly rejected Eudoxian mathematical astronomy because it was based on geometry."

Ryder Rude

@Thorgott how else do u get that equality

Thorgott

many applications of the rank-nullity formula

the two sums are equal, but nobody is claiming the individual terms are each equal

Ryder Rude

maybe it means $\sum (-1)^r dim(B_{r-1}(K))=- \sum (-1)^r dim(B_r (K))$

ooh makes sense

$\sum _ {r=0}^{n} (-1)^{r+1} dim(B_r(K))= \sum _{r=-1}^{n-1} (-1)^r dim(B_{r-1}(K))$

and $dim B_{-1}(K)=0$

they defined this to be zero

@Thorgott thanks

Thorgott

12:49

ah yes, this step is just the index shift

the rank-nullity step comes before/after

Ryder Rude

but after this step, they just get the Betti numbers on the right, which is the final result

how are the holes represented by non-free generators of $H_r(K)$ different from the holes represent by free generators?

Balarka Sen

@RyderRude The "hole" analogy is bad.

Do you know an example where $H_r(K)$ has torsion?

Thorgott

i don't find it that meaningful to think about "holes" at this point

Ryder Rude

ive been thinking of these as multi-dimensional holes. is this not accurate "The generators are not boundaries themselves and they dont have a boundary?"

@BalarkaSen yeah. the projective plane and the Klein bottle

Balarka Sen

@RyderRude this is accurate

@RyderRude What does the generator for $H_1$ of the projective plane look like?

Do you know how to visualize the projective plane?

Ryder Rude

13:02

yeah. it's the hemisphere with the opposite points on equator identified

Balarka Sen

But what does it look like

Ryder Rude

idk. maybe the Klein bottle would be better to consider. i know its look

Balarka Sen

You can't make the identifications if I give you a physical sphere (stretchable). Or can you?

No, the projective plane is better.

Ryder Rude

oh

lemme check images

Balarka Sen

Just think about it

The fact that you cannot seem to explain torsion in homology should be an indication that the hole point of view is moot, and you should rethink what homology class mean. With the simplest spaces in mind.

The point is to learn to see the spaces you work with, not do silly algebra with them

Ryder Rude

13:05

yeah

Balarka Sen

The algebra becomes obvious if you can see the spaces well enough (a point made clear by Poincare when we invented his invariants)

Ryder Rude

can the free generators always be thought of as multi dimensional holes?

ive been having trouble visualising becuz it's so alien

Balarka Sen

First step: Learn to see RP^2.

Ryder Rude

okay

shud i be able to visualise in higher dimensions

Balarka Sen

First learn to see dimension 2. That's complicated enough

RP^2 is 2-dimensional

Ryder Rude

13:09

yes. is the embedding possible in 3d tho

Balarka Sen

"Seeing" doesn't have to be "embedding". Seeing doesn't have a formal meaning. And that's fine.

Jakobian

young man, you can't visualize algebraic topology. You only get used to it

Ryder Rude

thats re-assuring becuz i suck at visualising

Jakobian

Its a parody of quote by von Neumann

Ryder Rude

what would u say is the general interpretation of generators of homology groups of things homeomorphic to simplical complexes

Balarka Sen

13:13

@RyderRude Prove that RP^2 is D^2 with antipodal points glued along the boundary circle. Then decide if it helps you see it better.

Don't think too much about general questions before learning enough examples you're comfortable with.

Once you learn enough examples you'll find an interpretation of your own

Ryder Rude

okay

@BalarkaSen i tried this a while ago but the whole boundary became a point

Balarka Sen

Try easier. What is S^1 with antipodal points glued? Can you visualize it?

Ryder Rude

no, it all becomes a point. im doing it wrong

sorry it becomes a semicircle

Balarka Sen

More correct, but still wrong.

Ryder Rude

with the two ends identified

Balarka Sen

13:17

Which is?

Ryder Rude

it's a circle

Balarka Sen

What is S^1 x [0, $\varepsilon$] with antipodal points in S^1 x 0 glued?

Ryder Rude

was the previous answer correct

Balarka Sen

Yes. I decided to ask you a harder question, but easier than RP^2.

(How did you hit upon the semicircle idea?)

Ryder Rude

ive seen it before with the spehere in which antipodals are identified. we first make it a hemisphere

Balarka Sen

13:19

What does "making it a hemisphere" mean?

Ryder Rude

it's like compressing it to make the two hemispheres meet so it becomes a disx

disx

disc

Balarka Sen

Yes, great

Ryder Rude

with the boundary identified

Balarka Sen

You're doing "identification in stages".

Ryder Rude

yeah

but i cant carry out the boundary identification visually

so we have a disc so far

and..

Balarka Sen

13:21

@BalarkaSen Think about this one first.

Ryder Rude

i guess we make it a semi-disc

Thorgott

the issue with the "holes" idea is that all the typical invariants (homology, homotopy, bordism) "count holes" in one way or another, yet are radically different

Balarka Sen

You can do the boundary identifications, you just demonstrated S^1 with antipodal points identified is S^1.

You just can't do it with the presence of the disk.

So I'm suggesting, do it in the presence of a little annular neighborhood of the boundary. Forget the whole disk for now.

Ryder Rude

@BalarkaSen so this is a cylinder and i hav to identify the boundary of the circle at the bottom?

Balarka Sen

Yep

Ryder Rude

13:23

okay so we still get the same cylinder

Balarka Sen

(FWIW, here is my internal image for why S^1 with antipodal points identified is S^1. First pick a pair of antipodal points x, -x. Identify them. This gives "a figure eight", two circles attached along a point. The other antipodal identifications now means that you have to "flip-glue" the pair of circles. And that gives you... a circle)

@RyderRude Wrong.

Ryder Rude

i first compressed the base to get a cylinder with a line base

Balarka Sen

That's not antipodal identification.

Ryder Rude

ooh and then we have to join the ends of the line

@BalarkaSen oh

Balarka Sen

You're identifying $(x, y)$ with $(x, -y)$. You have to identify $(x, y)$ with $(-x, -y)$, along the circle.

Ryder Rude

13:25

sorry

Balarka Sen

Be back in a bit.

Ryder Rude

ok so we first get a cylinder with 8 as the base

and then we flip over the two eggs of 8 and glue

i can visualise the process

this one seems embeddable in 3d

Thorgott

this part is embeddable in 3d

it's subtle, though. the resulting space is not again a cylinder. what is it?

Ryder Rude

is this a cylinder with the bottom base sealed?

PM 2Ring

13:40

Jos Leys has some nice topology anims. josleys.com/galleries.php?catid=13 Eg, here's a cross-cap josleys.com/show_gallery.php?galid=373

Thorgott

not quite

Ryder Rude

@PM2Ring thanks!

Thorgott

I'd cut the cylinder open at $\{x-,x\}\times[0,\varepsilon]$ to get two disjoint cylinder $S^1_+\times[0,\varepsilon]$ and $S^1_-\times[0,\varepsilon]$ ($S^1_{\pm}$ are semi-circles). The identifications on this space are $(z,0)\sim(-z,0)$ for all $z$ and $(x,t)\sim(-x,t)$ for all $t$. The latter first and then the former gives us the space we're meant to describe, but we may also do it in other order. This yields a cylinder $S^1_+\times[-\varepsilon,\varepsilon]$ with some extra identifiactions

(you can theoretically carry out the entire process with an actual piece of paper in 3D)

PM 2Ring

Greg Egan has an applet illustrating SO(3).

> Any rotation can be specified by a vector pointing along the axis of rotation, with a length equal to the amount of rotation; using this correspondence, each cube here has been rotated by its own position vector, relative to the central cube [...] the true topology identifies opposite points on the surface, which represent rotations of 180° around opposite axes.

gregegan.net/APPLETS/01/01.html

^ That may help you to get a feel for identifying antipodal points on the sphere.

Ryder Rude

14:03

thanks everyone. i will also try to do it with paper

Balarka Sen

14:39

@PM2Ring Greg Egan is fantastic

I love his writing

@RyderRude You forgot to think about what happens to the rest of the annulus when you flip over the two circles in 8

(In fact, if you try to work it out, you'll find your process will never work)

PM 2Ring

14:57

@BalarkaSen Not only is he a top sci-fi author his technical writing is excellent too.

Here's some hyperbolic tiling stuff he's currently playing with: mathstodon.xyz/@gregeganSF/112059794401525844

PM 2Ring

15:31

Ooh. An inside-out Mandelbrot set. From mathstodon.xyz/@[email protected]/112000083660466499

I'm almost certain that I tried doing that a couple of decades ago, but that image doesn't look familiar.

Xander Henderson

@PM2Ring yozh.org/2011/08/22/mmm047

Koro

@Thorgott: Suppose that f:M1--->M2 is a diffeomorphism of smooth manifolds Mi. Can it be said that Df: M1--> tangent bundle of M1 is a smooth map?

Here is how I tried to solve it:
Let Y be a pushforward of X by f. For every p in M1, we have: $Y(f(p))= Dfp(X(p))\implies Y\circ f(p)= Df_p\circ X(p)= Df\circ id{M1}\circ X (p)\implies Y= Df\circ id{M_1}\circ X\circ f^{-1}$
If every map on RHS is smooth, then clearly $Y$ is smooth as well.

I'm confused about smoothness of Df.

If Df is not smooth, say, then how do I even prove the pushforward to be smooth? What is the technique?

EE18

@leslietownes Thanks very much for all your helpful comments Leslie (apologies, am just seeing them this morning). I guess the tricky bit for me is that the $s_k(m)$ for different $m$ (and likewise for things like $(X-X_j)$, which is really shorthand for $(1_{R[X_1,...,X_m]}X - (X_1^0...X_j^1...X_m^0)X^0)$) are different objects entirely for different $m$, and I'm therefore struggling to see how to induct. Hopefully that makes sense, I'm admittedly not sure if it does

Koro

15:47

My guess is that it is usually assumed that Df is smooth.

Balarka Sen

@Koro Check locally using smoothness of $f$.

$Df$ is an expression involving $f$ and the first partial derivatives of $f$.

Koro

But Df need not even be continuous, right?

f is not given to be smooth.

Balarka Sen

It's a diffeomorphism. Diffeomorphisms are smooth, or at least $C^1$, by definition.

So $Df$ is at least $C^0$.

You can assume everything is $C^\infty$, I think that's what they mean by a diffeomorphism. Most people do.

Koro

@BalarkaSen this

Balarka Sen

But regardless, $Df$ has to be continuous.

Koro

15:53

I think that's the underlying assumption there.

Balarka Sen

No matter what definition of diffeomorphism you choose.

Koro

@BalarkaSen Is this some convention? Because diffeomorphism to me is a map between manifolds which is differentiable with a differentiable inverse.

nonetheless, I'm happy assuming everything to be C^oo. $\ddot\smile$

Balarka Sen

@Koro Where did you get this definition?

What you said is a horrible thing to do, because differentiable maps do not satisfy inverse function theorem.

Thorgott

any degree of differentiability less than $\infty$ is a crime

also $Df$ is a map from $TM_1$ to $TM_2$

Balarka Sen

@EE18 $R[X_1, \cdots, X_m][X]$ is isomorphic to $R[X_1, \cdots, X_m, X]$... are we going to distinguish the identity element of two rings $A, B$ if $A$ is a subring of $B$, now?

In fact, $R[X, \cdots, X_m][X]$ and $R[X_1, \cdots, X_m, X]$ are so isomorphic that you might as well say they are equal

If I were keeping track of all the canonical isomorphisms in math I have implicitly been using throughout my entire mathematical career (which has barely even started) I'd be dead by now

"Ah yes what I mean when I say $2/2 = 1$ is actually under the canonical isomorphism between $\Bbb Z$ and the embedded copy of it inside $\Bbb Q$, the equivalence class of the order pair $(2, 2)$ is mapped to $1$"

Get rid of the set theory book.

Koro

16:05

@BalarkaSen My definition is wrong. I imagined it somehow. Diffeomorphism in my case is defined for C^oo smooth maps only.

Balarka Sen

That's all settled then.

The problem is smootheomorphism is a silly name

Koro

yeah, thanks a lot.

But my solution is wrong.

It needs more work as Thorgott said.

@Thorgott yes, I made a mistake there.

Koro

16:20

$Y(q=f(p))= Df_p(X(p))=\sum_j Df_p (X(p))(y_j)e_j,$ where $e_j= \partial/{\partial y_j}|_q$ so $Y(q)= \sum_j X(f^{-1}(q))(y_j\circ f) e_j$ and the coefficient maps $q\mapsto X(f^{-1}(q))(y_j\circ f)$ are smooth.

So $Y$ is smooth at $q$.

$q\in M_2$ is arbitrary so Y is smooth on $M_2$.

I should consider charts too.

Jakobian

@Thorgott What are examples of properties of rings, such that if $R$ has it then $R[x_1, ..., x_n]$ has it as well, but there exists infinite set $S$ such that $R[S]$ doesn't have it?

sort of, finite dimensional

Koro

The representation of Y(q) as combination of e_j is local that is valid in a chart of q.

Thorgott

@BalarkaSen ok but what if you need MUTUALLY ALIEN COPIES

@Jakobian Noetherian

Jakobian

UFD?

Thorgott

I think that holds in the infinite case, too

Balarka Sen

16:32

@Thorgott cries in hexagon

Thorgott

cause any one polynomial only involves finitely many variable and can thus be factored into primes

of course, finite-dimensionality itself is also such a property

I think normality is another property that gets inherited in both the finite and infinite case

EE18

@BalarkaSen One has domain $\Bbb N^{m+1}$ and the other has domain $\Bbb N$ right?

Balarka Sen

what is domain of a ring

Koro

is my solution correct? Or there are any objections?

Thorgott

in essence, for the property to be inherited under adjoining finitely many free variables and not under adjoining infinitely many is to ask for it to not be stable under very nice colimits

Jakobian

16:39

@Thorgott yeah

Balarka Sen

another property which all R[x1, ..., xn] have but R[S], |S| = infty doesn't have is that the the former are obtained by adjoining finitely many indeterminates to R and the latter isnt

:P

EE18

They are both rings given the operations we define on them but underneath they are sets of functions right? I'm really not trying to get too down in the weeds on this, I'm just trying to understand. I want to be able to consider $(X- X_1)...(X-X_{m})(X-X_{m+1})$ and to use the induction hypothesis on the first part ($(X- X_1)...(X-X_{m})$) of this, but the induction hypothesis only applies to $R[X_1,....,X_m][X]$ and not to $R[X_1,....,X_{m+1}][X]$ as is being considered here

Thorgott

if R=S[x_n, n in N], this isnt true

EE18

So obviously I've got to carve out some subring of $R[X_1,....,X_m,X_{m+1}][X]$ which is ring isomorphic to $R[X_1,....,X_m][X]$ so that I can use the induction hypothesis

Thorgott

adjoining finitely or countably infinitely many variables to it yields isomorphic rings

Balarka Sen

16:43

@Thorgott what im saying is distinct from saying R[x1, ..., xn] cannot be obtained, upto ring isomorphism, from R by adjoining finitely many variables

PM 2Ring

@XanderHenderson Lovely!

Jakobian

Is there a condition that makes $R[x_1, ..., x_n]$ not isomorphic to each other for all $n$? Fields I suppose? Krull dimension?

Balarka Sen

@EE18 Just prove $R[X][Y]$ is isomorphic to $R[X, Y]$ for gods sake, then you can treat all of those rings as sitting inside some big ring and bob's your uncle

these rings are not sets of functions. functions on what? R is some arbitrary abstract ring

EE18

$R$ isn't but $R[X]$ is right?

Balarka Sen

$x^p - x \in \Bbb F_p[x]$ is zero when evaluated on all elements of $\Bbb F_p$. It's not the zero polynomial

EE18

16:45

$R[[X]]$ is, i should say

Balarka Sen

There's no need to bring the Laurent polynomial ring

Jakobian

so this should hold for any ring with finite Krull dimension

Balarka Sen

It's not correct to think of these polynomials as functions over anything

EE18

@BalarkaSen I see that this is Exercise 16 for me in my book :)

Jakobian

is there any good family of rings of finite Krull dimension? Or characterization of rings of finite Krull dimension?

Balarka Sen

16:47

even if you think of them as functions over something, a function of two variables naturally curries into a function of one variable when you suppress one of the variables, so R[x][y] = R[x, y] should still make sense

$(x, y) \mapsto f(x, y)$ same thing as $x \mapsto (y \mapsto f(x, y))$

Jakobian

I think this should be true for arbitrary Noetherian ring too

EE18

Totally agreed. I am just gonna sit with this for a little bit I guess because sure I can prove $R[X_1,...,X_m][X]$ is isomorphic to $R[X_1,...,X_m,X_{m+1}]$ but what I need is to show that it's isomorphic to a particular subring of $R[X_1,...,X_m,X_{m+1}][X]$ and its what that subring ought to be that I'm struggling to convince myself of

Balarka Sen

no, you don't need to do that. R[x1, ..., xm][x] is isomorphic to R[x1, ..., xm, x]. And for any m, R[x1, ..., xm, x] is a subring of R[x1, ..., xm, xm+1, x]

Jakobian

yeah you can prove that if $R$ is Noetherian then none of the $R[x_1, ..., x_n]$ are isomorphic

If $R$ is Noetherian and there existed isomorphism $R\cong R[x_1, ..., x_n]$ then composing with $R[x_1, ..., x_n]\to R$ the obtained map $R\to R$ is surjective, hence it must be bijective, but this is impossible since the map $R[x_1, ..., x_n]\to R$ is not bijective

EE18

OK it's that latter statement which I don't follow, and please take it easy on me, but what do we mean its a subring? The contents of the former set are different (functions on $\Bbb N^{m+1}$) than the contents of the latter ($\Bbb N^{m+2}$). My suspicion has been that perhaps by subring you mean the subring of $R[x1, ..., xm, x_{m+1}, x]$ which has nonzero output values only when $\alpha \in N^{m+2}$ has its $m+1$th coordinate equal to 0

I think that's vaguely right and appreciate you helping me sort this out

Jakobian

16:56

now if $R[x_1, ..., x_m]$ and $R[x_1, ..., x_n]$ are isomorphic the same argument shows that if $m < n$ we obtain a contradiction (since $R[x_1, ..., x_n]$ is Noetherian for all $n$)

EE18

its this notion of "embedding" (if that's the right word) some structure which is isomorphic to some substructure of a larger structure which is different

that's what's proved elusive to me

Balarka Sen

what is the definition of R[x] for you?

EE18

For example, how we "consider" $R$ as a subring of $R[X]$, namely the constant formal power series

$R$ isn't the same thing as those constant formal power series, but it is isomorphic thereto and so we identify them

So here, I am trying to make precise/convince myself of what it means to "take" $R[X_1,...,X_m]$ as a subring of $R[X_1,...,X_m,X_{m+1}]$, inter alia

Jakobian

3

A: If $R$ is Noetherian, show $R\ncong R[x]$ as rings

Suppose there is an isomorphism $R \cong R[x]$. Compose this with the quotient map to get $R \cong R[x] \to R$, which is surjective. By part (a) we know that this composition must be bijective. However, we know that $R[x] \to R$ has a kernel generated by $x \neq 0$, which is a contradiction. I b...

EE18

$R[X]$ is the subset of $R[[X]]$ where only finitely many coefficients are nonzero ($R[[X]]$ is the set of maps $R^{\Bbb N}$ endowed with the usual operations)

Balarka Sen

16:59

My definition of R[x] is set of elements of the form a0 + a1x + ... + an x^n where ai are in R, x is a formal variable, and stuff in R and x commute. period.

then R[x] is canonically a subset of R[x, y]

and the ring structure restricts, subring thereof

its messed up to think of a polynomial as a map from R to N with weird properties

EE18

From $N$ to $R$, but fair enough

I don't know, I suspect this is the way that such things are formalized but I am far from expert. My gripe with your definition would be that I have no idea what a formal variable is :)

Balarka Sen

what is 1?

EE18

The map where $0 \mapsto 1_R$ and all other $n \in \Bbb N$ map to $0_R$

Balarka Sen

no no i mean the integer 1

what is the integer 1?

we dont know what it is but we sure can do arithmetic with it. peano axiom starts by declaring a variable called 1

no one defines what 1 is. do you also have no idea what 1 is?

its a formal symbol

EE18

Sure that's fair and well taken, but in the same vein one could say that we define function as a rule with a domain and codomain and that's the end of it. One certainly can, and there may or may not be problems. At least in my context though I am trying to work with what my book has presented me thus far

I had assumed that this definition of $R[[X]]$ etc was standard but perhaps not

Balarka Sen

17:04

first of all, no one in their right mind defines R[x] by defining R[[x]] first

thats what is really nonstandard

i recommend reading a standard textbook in algebra, like Mike Artin's, to get a less myopic perspective on ring theory

or mathematics in general

EE18

One day I will definitely be picking up an algebra book

I look forward to it :)

Thorgott

17:19

@Jakobian I'm not sure if finite Krull dimension implies that the $R[x_1,\dotsc,x_n]$ are pairwise non-isomorphic. Noetherian does imply it, as you've noted. (Also, these are orthogonal hypotheses.)

@BalarkaSen cries in Spec(R[X])

Xander Henderson

@Thorgott Which dimension is Krull from?

EE18

having thought about it, at least in the context of my book I think I need to consider the following: $R[X_1,...,X_m][X]$ and $R[X_1,...,X_m,0][X]$ are ring isomorphic, where $R[X_1,...,X_m,0]$ denotes the subring of $R[X_1,...,X_m,X_{m+1}]$ wherein all elements (again, elements of $R[X_1,...,X_m,0]$ are functions which map to $0_R$ if the $m+1$th component of the element $\alpha \in N^{m+1}$ is nonzero.

Intuitively, the only nonzero terms in $p \in R[X_1,...,X_m,0][X], p = \sum_\alpha X^\alpha$ are those terms wherein $X_{m+1}^j$ has $j = 0$.}

Then I need to use that $S$ a subring of $R$ implies $S[X]$ a subring of $R[X]$ to use that $R[X_1,...,X_m,0][X]$ a subring of $R[X_1,...,X_m,X_{m+1}][X]$

And from there I should be able to use the isomorphism and the inductive hypothesis. Does that make sense as a sketch of a strategy at least?

Oops look like one of my comments above got gut off but hopefully makes sense still

Thorgott

@EE18 if you can identify $R$ with a subring of $R[X]$ (the polynomials where all monomials involving $X$ have coefficient $0$), then you can identify $R[X]$ with a subring of $R[X][Y]$ (the polynomials in $Y$ with coefficients in $R[X]$, where any monomial involving $Y$ has coefficient $0$). then you identify $R[X][Y]=R[X,Y]$ as indicated, this makes $R[X]$ the subring of polynomials in $X,Y$, where all monomials involving $Y$ have coefficient $0$.

@XanderHenderson oh lol, that's an awesome poster

EE18

That poster is terrifying

Cheers, thanks Thorgott and balarka. Time to work on some less mathy stuff but hopefully this will all percolate while I'm doing that...

Jakobian

@Thorgott If $dim(R) = n$, then $n+1\leq dim(R[x])\leq 2n+1$

Thorgott

17:29

I highly advise you adopt the following perspective: any polynomial in $R[X_1,\dotsc,X_n]$ can be written as an $R$-linear combination of the monomials $X^{\alpha}=X_1^{\alpha_1}\cdot\dotsc\cdot X_n^{\alpha_n}$, where $\alpha=(\alpha_1,\dotsc,\alpha_n)\in\mathbb{N}^n$ is a multi-index, in a unique way. This is equivalent to thinking about "finitely supported functions", but much more algebraically useful phrasing.

EE18

Is there even anything to prove re uniqueness? If even one monomial term differs then the function differs? But that’s well taken

Jakobian

but I was thinking that finite Krull dimension implies Noetherian so good you mentioned that

Thorgott

For example, I can tell you without reference to functions and whilst being perfectly rigorous that the canonical isomorphism $R[X_1][X_2]\cong R[X_1,X_2]$ is given by mapping $\sum_{n\in\mathbb{N}}(\sum_{m\in\mathbb{N}}a_{m,n}X_1^m)X_2^n\mapsto\sum_{\alpha=(m,n)\in\mathbb{N}^2}a_{\alpha}X^{\alpha}$

This is the convenient way in which to algebraically think about polynomial rings.

In fact, you can ponder the following: If $R,S$ are commutative rings, then any ring homomorphism $\varphi\colon R[X_1,\dotsc,X_n]\rightarrow S$ determines a ring homomorphism $\varphi\vert_R\colon R\rightarrow S$ by restricting to the subring $R$ and $n$ elements $\varphi(X_1),\dotsc,\varphi(X_n)\in S$.
I claim that, conversely, given a ring homomorphism $\psi\colon R\rightarrow S$ and elements $x_1,\dotsc,x_n\in S$, there exists a unique ring homomorphism $\varphi\colon R[X_1,\dotsc,X_n]\rightarrow S$ s.t. $\varphi\vert_R=\psi$ and $\varphi(X_i)=x_i$ for $i=1,\dotsc,n$.

Taking this perspective, I can specify a unique homomorphism $R[X_1]\rightarrow R[X_1,X_2]$ by demanding it restricts to the canonical embedding $R\subseteq R[X_1,X_2]$ and maps $X_1\mapsto X_1$. Then, I can specify a unique homomorphism $R[X_1][X_2]\rightarrow R[X_1,X_2]$ by demanding it restricts to the homomorphism just constructed and maps $X_2\mapsto X_2$.

In the converse direction, I can specify a unique homomorphism $R[X_1,X_2]\rightarrow R[X_1][X_2]$ by demanding it restricts to the canonical embedding $R\subseteq R[X_1]\subseteq R[X_1][X_2]$ and maps $X_1\mapsto X_1,X_2\mapsto X_2$.

These will be precisely the inverse isomorphisms I gave above.

The bottom line is that both objects are "free" $R$-algebras on $X_1,X_2$ (these elements exist and satisfy no additional relations), in some sense.

@Jakobian yeah, but that doesn't force the dimensions to be pairwise distinct

Jakobian

@Thorgott why not?

Thorgott

oh wait

yeah I did a silly

Balarka Sen

18:07

Trying to think through a simple point. Suppose $S = M \times \Bbb C^2 \to M$ is the trivial bundle over a closed oriented 3-manifold. Recall $TM \cong M \times \Bbb R^3$ is the trivial bundle. Let $e_1, e_2, e_3$ be a basis of orthonormal frames with respect to a metric on $M$.

I have a representation $\rho : TM \to \mathrm{End}(S)$ given by sending $\rho(e_i) = \sigma_i$, the Pauli matrices (recall these are matrices such that $\sigma_i^2 = -1$ for all $1 \leq i \leq 3$, $\sigma_i \sigma_j + \sigma_j \sigma_i = 0$ for $1 \leq i \neq j \leq 3$ and $\sigma_i^\star + \sigma_i = 0$, $1 \leq i \leq 3$).

I want to cook up a connection $\nabla$ on $S$ such that for all vector fields $X, Y$ on $M$ and a section $s$ of $S$, the condition $\nabla_X \rho(Y)s = \rho(\nabla_X Y)s + \rho(Y) (\nabla_X s)$ holds.

$\nabla_X Y$ being the Levi-Civita connection on $M$.

Is there an obvious thing that works

I don't want to solve a system of equations, but I do want a more or less explicit answer.

Suppose $Y$ is unit. Replace $s$ by $\rho(Y)s$. I get $-\nabla_X s = \rho(\nabla_X Y) \rho(Y) s + \rho(Y) \nabla_X (\rho(Y)s)$.

Huh. For all $Y$...

Ah this doesn't help.

Balarka Sen

18:43

OK, neat approach: write $\nabla_X s = D_X s + A(X) s$ where $D_X$ is directional derivative and $A(X) \in End(S)$.

Then the equation boils down to $[A(X), \rho(Y)] = \rho(\nabla_X Y)$.

$\nabla_{e_i} e_j = \Gamma_{ij}^k e_k$. Then $\rho(\nabla_{e_i} e_j) = \Gamma_{ij}^k \sigma_k$ is a matrix. I need to solve for $[A(e_i), \sigma_j] = \Gamma_{ij}^k \sigma_k$, $1 \leq i, j, k \leq 3$.

Write $A(e_i) = \alpha_{il} \sigma_l$. Then we need to solve for $\alpha_{il} [\sigma_l, \sigma_j] = \Gamma^k_{ij} \sigma_k$. This is easy enough, because $[\sigma_l, \sigma_j] = \sigma_k$ upto scale.

user 85795