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Bob
12:01 AM
am I correct that $|Z_4 \times $Z_4| = 16?
 
Zee
Is the times a Cartesian product ?
 
That's not quite how rings come up in algebraic geometry. I don't know much about the stuff but say, one theorem in ring theory is that all ideals in $k[x_1,\ldots,x_n]$ are finitely generated where $k$ is a field
 
Bob
yes
 
Zee
Then yes
 
Why is this important? Well, many people care about solving polynomial equations simultaneously, that's for sure. So let's say you're trying to solve infinitely many polynomials $f_1(x) = f_2(x) = \ldots = 0$
And they're all in $n$ variables
 
Zee
12:05 AM
I don’t hate rings... am just tired
 
Well it turns out that there's some $N$ such that if you solve $f_{1,\ldots, N}$, you solve everything, and that's by the above theorem
Fair
What are you doing with rings at the moment?
 
Zee
Actully nothing
I have very little knowledge about Algebra
I don’t know if that’s becouse I hate it or I hate it couse I have very little knowledge of it
 
Or I guess what I'm wondering is, what interaction did you have with rings in particular that pissed you off?
Either way is kind of possible, I know some folk IRL who were just predisposed to hate algebra from the start and for that reason mostly avoided it, other people end up growing into it quite a lot with time
 
Zee
Algebra on the whole I don’t like couse it’s very non visual , but groups I’ll accept since they are simple
 
Hey @Antonios!
 
12:08 AM
hi @Daminark
@Zee don't get me wrong I haven't read all the conversation but groups are far from simple :P
 
@Zee you'd be surprised, things can be quite visual depending on how you approach it
 
Zee
Am sure they get very complicated but atleast they have one operation
Yes @Daminark that’s why I have an interest in geometric group theory
Unfortunately, there isn’t geometric ring theory
Maybe that’s just AG lol
 
That's basically it, it's one of those things where you can put some geometry on the rings themselves by putting a topology called the "Zariski topology" on their prime ideals
 
Zee
The prime ideals form the open sets ?
 
Not quite, what happens is you call Spec(R) the set of prime ideals
 
Zee
12:12 AM
Of a single ring ?
 
And the topology on Spec(R) is that a closed set is one of the form $V(I) = \{\mathcal{p}\in \text{Spec}(R) : I\subset \mathcal{p}\}$
Yeah
Here $I$ is an ideal
 
Zee
Ehh I can’t compile Latex
 
Oh, it's just the set of prime ideals containing $I$
 
Zee
So a closed set is the set of all prime ideals containing a single ideal I ?
 
Yup
People who know AG can tell you more about what you do with it and why you care
 
12:15 AM
I sometimes wonder, is there a nice way to visualise ideals? To me, they still look like black holes
 
But in commutative we talked about this briefly and apparently these lead to a lot of interesting geometry
 
Zee
no , you have done a good job @Daminark
 
@Daminark its actually nice, because in the case of $\mathbf{C}[x]$, for example, the prime ideals are exactly of the form $0\cup \{(x-a):a\in \mathbf{C}\}$.
 
in that when you multiplied any ring element by something, you never escaped from some subset
 
Give a mini-lecture on the Nullstellensatz, @Antonios.
 
12:16 AM
This because Nullstellensatz?
 
So, if you study $\text{spec}(\mathbf{C}[x])$ you recover $\mathbf{C}$ with its usual zariski topology as an affine space, along with a "generic point"
I haven't the time :P in the middle of writing a talk
but it turns out that affine spaces i.e. C^n etc can be viewed naturally as spectra of polynomial rings, i.e. C[x_1,...,x_n] and so on
but I guess this is the reason why "schemes" (ring specs glued together) are a good way to go about AG
the situation is a little less pretty when you don't have an algebraically closed field
but you can still do some interesting things
 
Zee
All my AG profs do AG over the complex numbers , what’s the point of a scheme if people rarely use it’s full generality
 
You can have interesting scheme structures even over $\Bbb C$ ... we don't all have to worship at the temple of characteristic p.
 
Zee
But introducing all that extra machinery must come at a cost, is it worth the cost ?
I suppose I broke a social norm by asking this , oh well , c ya
 
I started writing an answer but decided it would take too long :P
 
12:25 AM
how does one visualise a polynomial ring?
 
@Zee I don't think it's necessarily that rare, I have professors who care about characteristic $p$ as well. I got the vibe that it tends to be more of use of number theorists though
 
But there are powerful techniques in AG that allow you to deduce stuff in char 0 by proving it in char p. This technique has been used to great success.
 
Oh that's nifty
I should probably check that subject out at some point, it does seem quite fun
 
This could be of interest @Daminark
In mathematics, Helmut Hasse's local–global principle, also known as the Hasse principle, is the idea that one can find an integer solution to an equation by using the Chinese remainder theorem to piece together solutions modulo powers of each different prime number. This is handled by examining the equation in the completions of the rational numbers: the real numbers and the p-adic numbers. A more formal version of the Hasse principle states that certain types of equations have a rational solution if and only if they have a solution in the real numbers and in the p-adic numbers for each prime...
 
Nice
 
12:32 AM
@ted oh, I finished that note I mentioned
 
Good for you!
 
@TedShifrin hi
 
hi Leaky
 
The link is here, if you’re curious
 
I've been writing notes for Hatcher Ch.2
(writing manually)
Hatcher doesn't like to number things
my Thm 68 is his Eg 2.18
 
12:35 AM
You're way overnumbering.
 
he doesn't number his definitions
 
LaTeX has made people do that, and it's absurd.
I don't either.
 
and there is quite a lot of definitions
 
I number sparingly.
That's what an index is for.
 
I over-number :P
 
12:36 AM
Yeah, you're obnoxious.
 
@TedShifrin fair enough
 
@TedShifrin Hey Ted :D
@LeakyNun leaky :D
 
Hi Kasmir
 
@KasmirKhaan hi
 
Ted I got a question ._.
 
12:37 AM
An argument I’ve seen for numbering most equations in a paper is that you really don’t know what equations people will want to cite
 
homology is the most beautiful thinig I've ever seen @TedShifrin
 
what is this notion about degenerate is about ?
 
after linear algebra (sorry :P)
 
So it’s a favor to the reader in that scenario
 
@LeakyNun you using different account ?
 
12:38 AM
I will disagree, Semiclassic.
 
Ted answer me :D
 
@KasmirKhaan what?
 
I like cohomology even more sometimes, @Leaky.
 
degenerate what does it mean ?
 
Too vague, Kasmir. Context?
 
12:38 AM
@TedShifrin I haven't seen cohomology yet
 
hmm well when defining the dot product
 
I can see the logic for it but eh. Numbering everything just gets distracting
 
there is something called positive definite
but the author said , we dont allwasy need that
 
Oh hey
 
it is enough to work with non degeneracy
like in R , AA >= 0
 
12:39 AM
Nondegenerate means that $\langle v,v \rangle \ne 0$ whenever $v\ne 0$.
 
but if we pick A = (1,i ) it is negative
 
There are no null vectors.
Heya, DogAteMy.
 
please keep explaing that Ted :D
 
@KasmirKhaan Like so many problems in math, we define it away
 
12:40 AM
I mean the use of that word
 
how many years, in your opinions, do we need to wait until mathematics is mature enough to determine whether $e+\pi$ is rational?
(open question to everyone)
 
The matrix representing the inner product will no longer be positive definite, but it will have all nonzero eigenvalues.
 
because degenerate triangle is soemthing i heard of and understand
 
Different, Kasmir.
 
Why do they use the same word?
 
12:41 AM
@KasmirKhaan note that for complex vectors $v$ and $w$, we define $\langle v,w \rangle$ to be $v^*w$ instead of $v^Tw$
 
aha okay well that clears up one comfusion
 
degenerate is one of those words like normal, regular, and a few others that gets tossed around with high frequency
 
depends whether we're doing hermitian inner product or orthogonal inner product, @Leaky.
 
aha thanks yall
 
I feel like the first line of Harry Potter:
> Mr. and Mrs. Dursley, of number four, Privet Drive, were proud to say that they were perfectly normal, thank you very much.
is about topology
 
12:42 AM
so i dont need to worry about the word?
just accept the defintion of it ?
witch makes me wonder btw
 
so abnormal psychology is about topology, DogAteMy?
 
is there some vector A such that A dot X = 0 for all X
without A being 0 ?
 
@KasmirKhaan mathematics is the art of giving the same name to different concepts
 
That's an equivalent definition of nondegeneracy, @Kasmir.
 
would this case be a non-non degeneate?
 
12:43 AM
> Mr. and Mrs. Dursley, of number four, Privet Drive, were proud to say that every two disjoint closed sets $E$ and $F$ can be precisely separated by a continuous function $f$ from $X$ to the real line $\Bbb R$, thank you very much.
3
 
that's degenerate, Kasmir :)
 
yeah but
does there exist such thing ?
 
Of course.
 
such vector being dotted with all vectors and being zero ?
without that vector being 0 ?
example ? :D
@Antonios-AlexandrosRobotis Long time no see!
 
Define $\langle e_i,e_i\rangle = 1$ if $i=1$ and $0$ if $i=2$ on $\Bbb R^2$.
 
12:44 AM
:P I've been busy
still am, I should say
 
Antonios doesn't like us anymore.
 
ted's onto me!
 
<<< smarter than he looks
 
I reckon that's true for a lot of math profs :P
 
hmm Ted that was a made up example :D
 
12:45 AM
All examples are made up.
 
I wanted something that is not contructed to surve us
hmm
 
The symmetric matrix has to have 0 as an eigenvalue. What do you expect me to do?
 
nah nah am not complaining about that
 
If $A$ is a linear map from $\Bbb R^n$ to $\Bbb R^m$, then define $\langle x,y\rangle^A$ to be $\langle Ax,Ay\rangle$. Then $\langle\cdot,\cdot\rangle^A$ is nondegenerate iff $A$ is injective.
(I think)
 
I just wondered if it is something that is there
i mean like the difference between the function x^2
and defining a function pieacewise
 
12:47 AM
DogAteMy. Your definition is wrong.
You want $\langle x,y\rangle = x^\top Ay$.
 
I just made up the definition
 
I think what you’re getting at is that the example sounds artificial
 
For $A$ symmetric.
 
exactly Semi !
 
I just made it up, I wasn't trying to build on something existing
 
12:48 AM
Well, tough. Kasmir. I quit.
 
sorry Ted :D
and thanks for the help !
 
@TedShifrin I believe he means a coordinate-free definition
 
DogAteMy: Yours will always be positive-semidefinite. The matrix will be $A^\top A$.
 
So I guess you’d want an example of something that you’d want to be an inner product but isn’t
 
I don't have a coordinate free definition.
 
12:48 AM
okay so the main point is now, should I worry about this " degeneracy thing " ?
you said no null vector Ted
and i forgot to ask you about that
 
I can define in terms of a decomposition $\Bbb R^2 = V\oplus W$.
A null vector is a nonzero vector whose inner product with itself is $0$ (length $0$).
 
The point is that you can prove more stuff if your thing is not degenerate
and we like proving stuff, it gives us that high
 
A=diag(1,-1) is the natural example for me
 
That's nondegenerate!!
 
null for me seems like the zero vector
 
12:50 AM
Be careful.
 
@TedShifrin right, I don't think any map $\Bbb R^2 \approx \Bbb R \oplus \Bbb R$ is computable
 
It's not positive-definite, but it's nondegenerate.
 
Yeah. Not sure what I was thinking
 
but $\Bbb R^2$ isn't a very constructive object to begin with
 
12:51 AM
ted so (1,i) is a nullvector?
 
A={{1,-1},{-1,1}} should work but eh
 
Sure, Semiclassic. After change of basis, it's what I gave.
 
@Semiclassical well we all know that its rank is 1
 
Yeah.
 
can you handsome folks
 
12:53 AM
@KasmirKhaan we're working in $\Bbb R^n$
 
With the real inner product, @Kasmir.
 
Any outer product $vv^\top$ will do, naturally
 
give me something simple so i see the difference betwen degenerate and non degenerate
 
If you're doing complex vector spaces, usually there's a hermitian inner product.
 
@TedShifrin I could not find an example thats y !
also i know for a fact in field of reals, there is no such thing
 
12:54 AM
@Kasmir: Consider $\langle x,x\rangle = x_1^2$, $=x_1^2-x_2^2$.
 
A non-degenerate bilinear form?
 
4 mins ago, by Ted Shifrin
A null vector is a nonzero vector whose inner product with itself is $0$ (length $0$).
 
The first is degenerate. The second is nondegenerate, but indefinite.
 
@KasmirKhaan inner product, not dot product
 
@Leaky: same thing.
 
12:54 AM
(so whether a vector is null depends on your choice of inner product)
 
@TedShifrin okay I see :D
@Daminark dami yes bilinear form
I want to understand why they include that as definition
 
Wow, DogAteMy got 7 stars for being an eigenpainintheass :P
 
i mean i dont think they call that property a name if it was not important
 
Why they include is as the definition of an inner product?
Oh oh
 
12:56 AM
no as defnition
 
no, you need positive-definiteness for inner product.
 
Well, I've heard that some folk relax to non-degeneracy
 
Yall got me more comfused then i first came here :D
thanks for that -.-
 
There's a moral to that story, Kasmir. Stay away.
 
haha
 
12:57 AM
one way to see what’s going on is to pick $A={{1,a},{a,1}}$ and consider what $v^T A v$ would be as a function of $v=(x,y)^\top$
 
well i was gonna add
exept for Ted's contribution
but you was faster
 
Demonark: I don't think that's legitimately called an inner product. (Like the Lorentz metric is indefinite, but not an inner product.)
 
In that example it comes out as $x^2+2axy+y^2$. If $a=0$, then the level sets of that function are circles
 
@Semiclassic: I think diagonalizing makes things far clearer.
 
So if $FA(X)$ is the free abelian group generated by $X$, then any set theoretic function $X \to FA(Y)$ extends to a group homomorphism $FA(X) \to FA(Y)$?
 
12:59 AM
Probably.
 
You know what would be a paradox that might cause a huge inblance in our universe? if our Ted ! went to do talk in " Ted's talks " I don't think anyone would be ready enough for the outcome :D
 
Not a single person would watch, Kasmir.
 
Where I’m headed is that my matrix A only has a zero eigenvector when $a=\pm 1$
 
WHY TED!
I WOULD!
 
eigenvalue @Semiclassic
Eigenvectors can never be 0.
 
1:00 AM
and I know for a fact that youll get most views :D
that anyone has on that talk thing :D
 
I appreciate the enthusiasm, Kasmir, but you far overestimate interest in mathematics.
 
Eh. I mean it in the sense of “an eigenvector corresponding to a zero eigenvalue”..,which is pretty terrible now that I say it
 
grrrrrr
 
I would slap you for that, @Semiclassic. Seriously.
 
1:01 AM
Anyway thanks yall ! ill be back later :D
 
Anyways. When $a=\pm 1$ the matrix is singular
But then the function becomes $x^2\pm 2xy+y^2=(x\pm y)^2$
 
Back to my change of basis comment. shrug
 
Actually I am wondering what are some examples of bilinear forms "in nature" aside from inner products?
 
Intersection forms in topology.
 
In which case the level sets go from being conic sections to lines
 
1:04 AM
And special relativity, of course.
 
One inner product I’ve run into a bunch lately is $\langle A,B\rangle = \mathrm{tr}(A^T B)$
 
The Killing form on a Lie algebra is what that is.
 
(Frobenius inner product)
Same idea, anyways
 
It's nondegenerate if and only if you have a semisimple Lie algebra.
 
Hmm. All I want A and B to be in my case are matrices
 
1:08 AM
Yeah, well most Lie algebras are matrix algebras.
OK, I'm off to cook dinner. Bye, all.
 
See you, and thanks!
@Leaky yeah that sounds right
 
If I have short exact sequences $0 \to A' \to A \to A'' \to 0$ and $0 \to B' \to B \to B'' \to 0$, and maps $f':A' \to B'$ and $f:A \to B$ such that the obvious square commutes, do I have a map $f'':A'' \to B''$?
 
I think so. Say you have $a''$ in $A''$. It's the image of some $a$ in $A$, so you can take it as the image of $f(a)$ in the map $B\to B''$. So now we need well-definedness.
 
So let $a'' = g(a) \in A''$, you want to define $f''(g(a)) = g''(f(a))$, but this doesn't have anything to do with $f'$ unless that has to do with well-definedness
 
Sorry, I'm back
Right so say we have $a_1$ and $a_2$ in $A$ with the same image in $A''$, we need that $f(a_1)$ and $f(a_2)$ have the same image in $B''$
In other words, if $a:=a_1-a_2$ is in the kernel of $A\to A''$, then $f(a)$ is in the kernel of $B\to B''$.
But since it's in the kernel of $A\to A''$, it's the image of some $a'$ in $A'$.
 
1:24 AM
does it have a name?
 
And thus since stuff commutes, $f(a)$ is the image of $f'$ of the thing in $A'$ so it's in the kernel of the other thing so in the end the thing we wanted to be zero is zero
Probably. I don't know a whole lot of homological algebra though @LeakyNun
 
1:45 AM
@Semiclassical today in analysis the professor, after explaining that analytic functions can't be so, asked us if we could think of any $C^{\infty}$ functions with compact support
and I knew of one becuase of a conversation we had, and he afterwords said he was surprised
you take a triangle and convolve it with itself infinitely
presumablyyou can define such a function $\mathbb R^n \to \mathbb R$ by having a pyramid with $n+1$ sides, right?
 
Nice
Yeah
 
the one he used in class, though he said mine would be fine, was
$\exp\left({ \dfrac 1 {1 + {\Vert x \Vert}^2}}\right)\cdot \Large{\chi_{\{\Vert x \Vert \le 1\}}}$
 
Yeah, that’s the standard one
 
how do I get the $^2$ to just the $\Vert x \Vert$
 
You have it outside one too many braces.
 
1:54 AM
Phew, 8 seconds left
 
There you go
 
thanks
 
ikr
It’s great when you sneak that edit in just in time
 
did we find a closed form for ours?
even just for $\mathbb R^1 \to \mathbb R$
 
Don’t remember tbh
 
1:55 AM
that is the largest chi i've ever seen
 
I made it Large because I couldn't see the indicator set very well otherwise
 
...don't do that
squint if you have to
 
well I had to think fast because I only had 8 seconds!
 
I know you probably have heard of the game before but I hadn't really known what it was about and I want it now
Someone have 15 dollars I can borrow indefinitely?
 
I’ve heard of it yeah
 
1:57 AM
@Semiclassical Today I asked a seminar speaker if his work had applications in string theory (similar stuff does)
 
That’s an interesting definition of borrow
 
he asked me "what's that"
 
...haaaah
 
I'll give it back after the never
 
and then after some silence "oh, that physics problem?"
 
1:58 AM
Well
 
The thing what you're made of maybe
 
You can tell where his priorities are
 
I'm made of sugar and spice and possibly strings?
 
not knowing some string theory is just nuts
 
I know that it involves strings, so I know that much
High dimensional vibratey strings
 
1:59 AM
Not knowing string theory is a thing is a bit weird yeah
 
That's about it for me
Not knowing of it, at least
 
My advisor then asked me "you thought that guy knew anything about string theory?"
TIL "knowing string theory" is something one should be able to tell from someone's demeanor
 
“Considering the usual applications of this stuff, yes”
 
Oh, no, it's not a usual application of this stuff.
 
What sort of stuff was it?
Ahh.
 
2:00 AM
Gauss curvature flow
 
The other curvature flows (Ricci, mean curvature) show up in renormalization group flow in string theory
 
Then yeah, I’m a lot less surprised
 
hi
 
Ugh, I just got a YouTube advert for Scientology
do not want
 
2:03 AM
lol
 
this book writes $R_\nu$ for $\Bbb R^n$
very strange
 
(More specifically, it was a link to DavidMiscavige.org... which is somehow even worse in my book than a generic Scientology ad)
 
@Semiclassical idk, I'm often disappointed in mathematican's lack of interest in physics (and vice versa)
 
I hear you on the vice versa
 
@Semiclassical I'm giving a presentation on linear stability for stars in my fluids 2 class
I imagine when I write out what the Riemann tensor is, people will just stop paying attention
 
...ow
 
2:50 AM
I think a lot of people in math have very vague ideas, like they might know of string theory without knowing string theory?
Though I'm not sure how many of my math profs have serious knowledge of physics. Definitely at least a few but not that many
 

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