« first day (4523 days earlier)      last day (794 days later) » 

D Left Adjoint to U
D Left Adjoint to U
03:57
en.wikipedia.org/wiki/File:Bertrand.jpg Bertrand throwing up math gang signs
He was very ahead of is time
Semiclassical
Semiclassical
04:17
the part of grading which drives me up a wall (aside from sheer volume) : deciding on a rubric
sometimes it works out easily
sometimes...not
D Left Adjoint to U
D Left Adjoint to U
05:01
@Semiclassical I love graded rings
"modules" I mean
lol
I see you're applying them to rubrik's cube
:D
@Semiclassical you could always pay me to automate your grading process
employing lots of AI to recognize handwriting
@ParamanandSingh hi, I left adjoint to you.
Unknown x
Unknown x
0
Q: Could you suggest a textbook for learning basic properties of spectrum of closed operators?

Unknown xI wish to learn the very basic properties of the spectrum of Closed operators. I may use these properties in the research in the Fluid dynamics/Differential equations. My attempts:- I searched in google. Search was disappointing. I asked in Chat. I couldn't find the book. I searched in the follo...

functional-analysis reference-request hilbert-spaces book-recommendation closed-map
D Left Adjoint to U
D Left Adjoint to U
@Unknownx what was disappointing about your google search?
@Unknownx I can teach you
first you tell me (quiz) what a closed operator is
I will use the Socratic method
Rubrik is this: 1,2,3 puff, pass.
5 puffs, A+
I can preach to you some math, you can teach me, my guru
@Unknownx how would you show that $\sum_{d \mid p_n\#}(-1)^{\Omega(d)}\sum_{r^2 = 1 \pmod d} \lfloor\dfrac{x - r}{d} \rfloor$ does not vanish eventually?
Could you somehow use differential equations?
Because derivatives of polynomials show up in Hensel's lemma
maybe the connection is stronger than that
Unknown x
Unknown x
@DLeftAdjointtoU A closed operator is an operator $T$ such that if $\{x_n\} ⊂ \text{Dom}(T)$ converges to $x ∈ X$ and $\{Tx_n\}$ converges to $y ∈ X$, then $x ∈ \text{Dom}(T) $and $Tx = y .
D Left Adjoint to U
D Left Adjoint to U
here though, $p_n\# = p_1 p_2 \cdots p_n$ (primorial)
Oh cool. Your turn
What is Dom($A$)?
Unknown x
Unknown x
domain of A
sorry
T
D Left Adjoint to U
D Left Adjoint to U
05:11
Oh, just say $\text{dom}(A)$ please
Unknown x
Unknown x
dom(T)
D Left Adjoint to U
D Left Adjoint to U
Rewrite, ples
I have to see it completely to come up with the next part of the lesson plan
What you're missing, etc
Or could use guidance on
What space is this?
$\text{dom}T \subset ?$
Unknown x
Unknown x
yes
D Left Adjoint to U
D Left Adjoint to U
What is the ambient space?
Unknown x
Unknown x
A closed operator is an operator $T$ such that if $\{x_n\} \subset \text{Dom}(T)$ converges to $x ∈ X$ and $\{Tx_n\}$ converges to $y ∈ X$, then $x ∈ \text{Dom}(T) $and $Tx = y .
how to render latex?
D Left Adjoint to U
D Left Adjoint to U
05:13
Okay, but what are we working in?
Vector space?
Real or complex?
You have to declare these things
Unknown x
Unknown x
user image
@DLeftAdjointtoU Hilbert space
@DLeftAdjointtoU complex
D Left Adjoint to U
D Left Adjoint to U
Okay, define that: it's a complex inner product space, such that?
Just give me a sentence or two
to check your understanding
Unknown x
Unknown x
@DLeftAdjointtoU Yes. It is a complex complete inner product(Hilbert space).
D Left Adjoint to U
D Left Adjoint to U
Complete with respect to the metric induced by the norm $|\cdot|$, which is?
Unknown x
Unknown x
I am interested in Hilbert space.
D Left Adjoint to U
D Left Adjoint to U
05:16
Okay now that we have that covered for the most part
What is the motivation behind finding closed operator spectra?
QM?
Unknown x
Unknown x
$d(x,y)=sqrt{\langle x-y,x-y\rangle}$
D Left Adjoint to U
D Left Adjoint to U
What applications does it have? Because if you have no goal or motivation, it could even be because they're beautiful,
then it's harder to motivate to keep studying
I see, perfect
Unknown x
Unknown x
@DLeftAdjointtoU yes. I wish to do a research in Mathematical Physics. My professor assign me to learn this topic.
D Left Adjoint to U
D Left Adjoint to U
Okay, they didn't mention why they were cool?
Unknown x
Unknown x
@DLeftAdjointtoU Most probably, I will move to Quantum mechanics
D Left Adjoint to U
D Left Adjoint to U
05:18
Cool
Okay, next
is define spectra
for such $T$
You must know this definition by heart, but it might take a few weeks
Unknown x
Unknown x
@DLeftAdjointtoU $\text{Spec}(T)=\{\lambda\in \mathbb C: T-\lambda I \text{ is not invertible}\}$
@DLeftAdjointtoU :(
D Left Adjoint to U
D Left Adjoint to U
Okay, now doesn't that immediately imply $\text{Spec}(T) = \{ \lambda \in \Bbb{C}: \det(T - \lambda I) = 0\}$ at least for finite dimensional $\text{dom}(T)$?
Unknown x
Unknown x
I know only basic properties and definitions in Hilbert space.
Paramanand Singh
Paramanand Singh
@DLeftAdjointtoU hello! I will need the original matrix as well and not just the adjoint to do anything useful (I may be wrong here as I am no linear algebra expert). Nice username!!
D Left Adjoint to U
D Left Adjoint to U
I will pass you meatrices and adjoints
:D
@Unknownx I found what you need to study first
Fredholm determinant
In mathematics, the Fredholm determinant is a complex-valued function which generalizes the determinant of a finite dimensional linear operator. It is defined for bounded operators on a Hilbert space which differ from the identity operator by a trace-class operator. The function is named after the mathematician Erik Ivar Fredholm. Fredholm determinants have had many applications in mathematical physics, the most celebrated example being Gábor Szegő's limit formula, proved in response to a question raised by Lars Onsager and C. N. Yang on the spontaneous magnetization of the Ising model. =...
Because invertibility probably has a condition stated in terms of that determinant
So you take the definition of spectra, and break it down into equivalent statements
Unknown x
Unknown x
05:25
@DLeftAdjointtoU Thank you. Let me read it.
D Left Adjoint to U
D Left Adjoint to U
I take that back, I'm not seeing any immediate $\det T = 0$ thing related to that
Unknown x
Unknown x
@DLeftAdjointtoU Hai, I didn't do the exterior products. I have only basic knowledge. Is there anything like weak convergence related theorem. For compact operator in a hilbert space, Compact operator maps a weakly convergent sequence to strongly convergent sequence, 0 is always in the spectrum of compact operator.
I need such a basic results.
D Left Adjoint to U
D Left Adjoint to U
So you need a book on functional analysis?
Try Rudin: 59clc.files.wordpress.com/2012/08/…
Unknown x
Unknown x
Yes. Which covers basic properties of Spectrum of closed operators.
D Left Adjoint to U
D Left Adjoint to U
Start reading on page 252
Read 5 pages
I will review it and we'll meet back here in 1 hour
If there are exercises
skip them and then we'll go over them together
You will read through this more thoroughly at a later date
Unknown x
Unknown x
05:33
okay
Thank you very much:)
D Left Adjoint to U
D Left Adjoint to U
@Unknownx which is more general: Hilbert or Banach space?
(pop quiz)
Unknown x
Unknown x
@DLeftAdjointtoU I didn't get what a Banach Algebra is?
@DLeftAdjointtoU Banach is general
D Left Adjoint to U
D Left Adjoint to U
Banach space
In mathematics, more specifically in functional analysis, a Banach space (pronounced [ˈbanax]) is a complete normed vector space. Thus, a Banach space is a vector space with a metric that allows the computation of vector length and distance between vectors and is complete in the sense that a Cauchy sequence of vectors always converges to a well-defined limit that is within the space. Banach spaces are named after the Polish mathematician Stefan Banach, who introduced this concept and studied it systematically in 1920–1922 along with Hans Hahn and Eduard Helly.Maurice René Fréchet was the first...
Unknown x
Unknown x
I don't want to learn more general stuff. I don't wish to learn abstract items.
D Left Adjoint to U
D Left Adjoint to U
Click that to see the deifniion
It's not super abstract
Unknown x
Unknown x
05:39
I wish to learn the essence for application.
D Left Adjoint to U
D Left Adjoint to U
It's only one-two axioms less
Unknown x
Unknown x
@DLeftAdjointtoU Okay
Koro
Koro
Banach = complete normed space.
D Left Adjoint to U
D Left Adjoint to U
Sometimes generalizing a slight bit makes it faster / easier
Unknown x
Unknown x
@DLeftAdjointtoU Banach space I know. Algebra???
@Koro I know this defintion
D Left Adjoint to U
D Left Adjoint to U
05:40
Banach space
In mathematics, more specifically in functional analysis, a Banach space (pronounced [ˈbanax]) is a complete normed vector space. Thus, a Banach space is a vector space with a metric that allows the computation of vector length and distance between vectors and is complete in the sense that a Cauchy sequence of vectors always converges to a well-defined limit that is within the space. Banach spaces are named after the Polish mathematician Stefan Banach, who introduced this concept and studied it systematically in 1920–1922 along with Hans Hahn and Eduard Helly.Maurice René Fréchet was the first...
Click that
it goes to Algebra
Koro
Koro
So what is your doubt?
D Left Adjoint to U
D Left Adjoint to U
essentially, there is defined/definable a ring-like multiplication
that distributes
And that is also a topological monoid $(X, \cdot)$ i.e.
Unknown x
Unknown x
@DLeftAdjointtoU okay. Thanks
I will ping you, when I complete the reading.
is that okay?
D Left Adjoint to U
D Left Adjoint to U
That would make $(B, +, -, \cdot, ...)$ into a topological ring or algebra
That just means all arithmetic operations are continuous
Unknown x
Unknown x
okay
Koro
Koro
05:42
Oh I see. You want to know Banach Algebra. I don't yet know that.
D Left Adjoint to U
D Left Adjoint to U
with respect to the norm-induced metric space
Unknown x
Unknown x
okay
D Left Adjoint to U
D Left Adjoint to U
@Unknownx yes, let's chat in a bit
Unknown x
Unknown x
@DLeftAdjointtoU okay
D Left Adjoint to U
D Left Adjoint to U
Let's just do one page though
The bottom of pg 252
Unknown x
Unknown x
05:44
okay
D Left Adjoint to U
D Left Adjoint to U
+ pg 253
Also, try to understand the Proof somewhat
I don't know if you can do this without writing stuff down. It helps me to write it as I'm reading
Unknown x
Unknown x
@DLeftAdjointtoU okay.
@DLeftAdjointtoU compliment means the set compliment. right?
D Left Adjoint to U
D Left Adjoint to U
Yes
Unknown x
Unknown x
user image
@DLeftAdjointtoU Hello, I am able to read and understand the logic. I mean I am able to understand the proof. But I am not able to understand the application :9
:(
D Left Adjoint to U
D Left Adjoint to U
$e = I$ here
Unknown x
Unknown x
05:56
@DLeftAdjointtoU yes
D Left Adjoint to U
D Left Adjoint to U
I would say your first goal should be to understand Theorem 10.7
If it refers to something else, we'll start there instead, and so on
but it looks like it's not too terrible of a proof
First step, is prove to me that $s_n = I + x + x^2 + \dots + x^n$ is Cauchy
By using $|x^n| \leq |x|^n$.
That's by $|ab|\leq |a||b|$ in a Banach algebra, basically $|\cdot|$ is continuous, so this is an equivalent condition to that
Unknown x
Unknown x
@DLeftAdjointtoU yes. I understood the theorem
Lemon
Lemon
Guys if $g(x,y,z) = (u,v)$ and I want to find $\partial_{x} (f \circ g)$, it should be $\partial_x (f \circ g) = \partial_u f \partial_x u + \partial_v f \partial_x v$ right?
Semiclassical
Semiclassical
@Lemon what is $f$ here? $\mathbb{R}^2\to \mathbb{R}$?
Lemon
Lemon
@Semiclassical Yes.
Semiclassical
Semiclassical
06:09
seems right, yeah
Unknown x
Unknown x
@DLeftAdjointtoU $||x^{-1}h||<(1/2) \implies e-(-x^{-1}h)$ is invertible.
Right?
x is invertible implies $x+h$ is invertible. product of two invertible elemnts are invertible.
Unknown x
Unknown x
07:04
@DLeftAdjointtoU But, results are abstract type. I think it has no use in reallity. :(
one potato two potato
one potato two potato
07:21
Why don't I have a hat?
Lemon
Lemon
07:42
If $f$ and $g$ are differentiable maps such that $| f(x) - g(y) | < \epsilon$ for any $\epsilon$, does this also imply there is a $\delta>0$ such that $|x - y| < \delta$
If the answer is no, does the result change if $g$ is now the identity map?
one potato two potato
one potato two potato
08:30
Now I'm wearing a sun!
 
4 hours later…
Lucas Henrique
Lucas Henrique
12:34
hey there chat
 
1 hour later…
one potato two potato
one potato two potato
13:38
The covariant derivative on a tensor field behaves similarly to the Lie derivative on a tensor field.
 
2 hours later…
ILikeMathematics
ILikeMathematics
15:28
$f(x) = 2 - |x|$.

We can define $f$ with cases in three different ways:

$f(x) = \begin{cases} 2 - x& x \geq 0\\
2 + x & x < 0\end{cases} = \begin{cases} 2 - x& x > 0\\
2 + x & x \leq 0\end{cases} = \begin{cases} 2 - x& x > 0\\
2 + x & x < 0 \\2 & x = 0\end{cases}$.

Using only these cases, differentiating the piecewise function $f$ gives us:

$f'(x) = \begin{cases} -1& x \geq 0\\
1 & x < 0\end{cases} = \begin{cases} -1& x > 0\\
1 & x \leq 0\end{cases} = \begin{cases} -1& x > 0\\
Koro
Koro
If f is given as you have written it, one would assume its domain to be all of R in which case, as per the definition of the derivative, one would check if the limit $\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}$ exists. This certainly exists if the left hand and right derivatives at 0 exist.
If however the domain of f is given an [0, 1], say then the derivative at 0 would stand for right hand derivative at that point so you -1= f'(0), etc.
ILikeMathematics
ILikeMathematics
"one would check if the limit $\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}$ exists" why would you check if that exists?

All of the representations are valid, if you only took one representation using cases, for example $f(x) = \begin{cases} 2 - x& x \geq 0\\
2 + x & x < 0\end{cases}$, then $f'(x) = \begin{cases} -1& x \geq 0\\
1 & x < 0\end{cases}$.
This would give you the result $f'(0) = -1$ which is wrong, but you wouldn't have any way of knowing, all the operations you did are valid, you are allowed to piecewise differentiate
Since one would arrive at this wrong solution like this, where is the step you wouldn't be allowed to take or would have to be cautious about?
Koro
Koro
16:21
@ILikeMathematics Because that's the definition of derivative at a point (0 in this case). No you can't conclude f'(0)=-1. Why? Because it is not known if $f$ is differentiable at $0$.
ILikeMathematics
ILikeMathematics
16:36
@Koro But then you also couldn't right away conclude that $f'(2) = -1$ because it is not known if $f$ is differentiable at $2$ yet
nickbros123
nickbros123
How to formally prove that if f:I-R has intermediate value property then f(I) is an interval
shintuku
shintuku
16:58
@nickbros123 first find the definition of an interval of R
nickbros123
nickbros123
17:08
In Bartle and sherbert the definition is If a,b belong to R, then the open interval with a and b as endpoints is (a,b)={x: a<x<b}. This suffices right?
shintuku
shintuku
{x in R: a<x<b}
nickbros123
nickbros123
Yes
shintuku
shintuku
if f:I -> R, what is the definition of f(I)
nickbros123
nickbros123
Set f(I)={f(x) x in I}
shintuku
shintuku
given the intermediate value theorem, what is another definition for f(I) if f is continuous?
leslie townes
leslie townes
17:20
mm, you will need the other possibilities for 'interval' too. this is kind of a weird exercise, in that the 'right' definition of interval for this purpose is basically the intermediate value property.
Koro
Koro
@ILikeMathematics Yes, one can't conclude that right away. But it is an exercise to prove that polynomials are differentiable everywhere. Knowing this one concludes f'(2)=-1. But note that it is different from making conclusion at x=0.
Leslie: I distinguish between rays and intervals. To me, an interval is with real values as end points and ray is what one would call an unbounded interval.
leslie townes
leslie townes
sure. but, i wasn't talking about 'interval' in any ideal sense, but rather what 'interval' needs to be for that particular exercise to be true. :D
i don't like reverse-engineering definitions from assuming that an exercise is properly posed, but, it seems like that is called for here.
nickbros123
nickbros123
@shintuku another definition? Range?
My teacher basically treated IVP without teaching definitions of intervals, and he just gave this theorem, and swiftly moved on
shintuku
shintuku
the intermediate value theorem implies that there is another way to write the set f(I) if f is continuous
leslie townes
leslie townes
side note, this seems to come up an awful lot in chat. there's 'interval' in the sense of an enumerated list of types of sets that one calls an interval (whether each endpoint is included, whether each endpoint can be +infty or -infty). then there's 'connected subset of R' or 'convex subset of R' or any number of order theoretic characterizations, which all amount to the same thing - the same enumerated list - but are often easier to work with and prove things about.
and for some reason, apparently, textbooks don't want to discuss this.
i note that nobody's asked what I is. if I = [0,1] or any closed interval, then you can definitely simplify the exercise if f is assumed continuous, because then only one 'kind' of interval is possible.
shintuku
shintuku
17:31
@nickbros123 so, we know f(I) = {f(x): x in I}. but this is very general, this doesn't care about whether the function is continuous or not. the intermediate value theorem implies we can write this differently, in the case that f is continuous
leslie townes
leslie townes
but if f is assumed only to have the IVP, more kinds of intervals are possible.
Koro
Koro
I think Munkres' discusses this while discussing order topologies: rays form subbasis for the order topology on a simply ordered set X.
ILikeMathematics
ILikeMathematics
@Koro Thank you.
Koro
Koro
@ILikeMathematics you're welcome. Glad that you found that helpful. :)
nickbros123
nickbros123
@shintuku there's another definition of f(I) for continuous f? Should I use the the definition of IVP itself?
shintuku
shintuku
17:36
there's a stricter possible meaning for f(I) in the case where IVT holds
@nickbros123 can you write a set-theoretical statement for what the intermediate value theorem states?
Koro
Koro
((I should have said open rays form subbasis for the order topology on a simply ordered set.)
nickbros123
nickbros123
The ivp says if a,b in I, and some r lies between f(a) and f(b) , then there is x in I so that r=f(x). So I can rewrite this like r belongs to f(I) ?
Ok sry that's wrong
I would have to define a new interval for f(a) to f(b) right
shintuku
shintuku
if IVT holds for f, then what is the set-theoretical statement for "f(I) takes any value between f(a) and f(b)"?
Koro
Koro
Does it make sense to teach rings and modules in semester 1 and group theory in semester 2?
I don't see any justification of doing so.
nickbros123
nickbros123
@shintuku I'm confused with regards to f(I), what we are concerned about is only the set of (a,b) for which we are saying some x exists in this set so that r=f(x) for some r belongs to f(I) ?
shintuku
shintuku
17:54
for any r belonging to f(I)
nickbros123
nickbros123
Yes, sorry, for any r in f(I), but the set I is (a,b) right?
shintuku
shintuku
yes
nickbros123
nickbros123
Then is it right to say f(I) is a subset of ((f(a),f(b))
leslie townes
leslie townes
sorry, what is f(a) if the domain of f does not include a?
more generally, even if f is continuous, and [c,d] is a subset of the domain of f, it is not generally true that {f(x): x in [c,d]} = [f(c),f(d)]
for example because a continuous function can assume the same value in two places without being constant in between those two places
nickbros123
nickbros123
Oh, you're right, i got the definition wrong
The problem vanishes on closed sets right? Cos you can find atleast one r between f(a) and f(b) (closed set)
shintuku
shintuku
18:05
heh, my bad the mentioned interval is closed, to make sense of f(a) and f(b)
nickbros123
nickbros123
So on the closed set I=[a,b] the function is defined. For all r belonging to [f(a),f(b)] there exist x belongs to I such that f(x)=r
I can say this?^
leslie townes
leslie townes
yes although i don't know that it is particularly helpful to single out [f(a),f(b)] when the exercise is asking about f(I)
also some minor annoyances around how you interpret that notation if f(a) happens to be greater than or equal to f(b)
nickbros123
nickbros123
Since there exist x in I for every r in [f(a),f(b)] so that f(x)=r, can I say f(I) supre of [f(a),f(b)] ?
Superset*
leslie townes
leslie townes
18:24
sure
Ted Shifrin
Ted Shifrin
18:36
@Koro Yes. I wrote a text doing rings/fields first, group theory next.
There are sound pedagogical reasons, but I would not do that for graduate students.
Koro
Koro
18:52
@TedShifrin But why is that so?
Why I find it difficult to digest is because if groups are taught first, then 1) the students will find it 'easy' as a group has only 1 binary operation. 2)Definition of ring will then become shorter (as one is now licensed to use the word 'abelian'.
Feynman_00
Feynman_00
@onepotatotwopotato Not quite. They coincide for scalar functions but $C^{\infty}$-linearity $\nabla_{f_1X_1+f_2X_2}Y=f_1\nabla_{X_1}Y+f_2\nabla_{X_2}Y$ is an important difference
Koro
Koro
3) Isomorphism theorems follow from the similar theorems from gt.
but as you said; and as per my next sem's syllabus, there must have been some reason behind it.
Ted Shifrin
Ted Shifrin
Rings — particularly commutative rings — generalize one’s experiences with integers and polynomials. Quotients are totally intuitive. Normal subgroups and quotient groups require far more sophistication.
Koro
Koro
But then quotient rings should not be covered in such a course? For, they will also require at least the same sophistication that normal subgroups would do. Isn't that right?
Ted Shifrin
Ted Shifrin
No, it’s totally wrong.
leslie townes
leslie townes
19:04
the 'only one operation, therefore easier' is a thing, but maybe more with students who have some degree of familiarity or comfort with working with systems of axioms already. which definitely isn't everybody. (and the theory of the integers, including ring stuff, is definitely a big part of group theory, e.g. for properties of cyclic and abelian groups.) two operations can be more intuitive than one.
Ted Shifrin
Ted Shifrin
Working with modular arithmetic is easy. Understanding $F[x]/(f(x))$ as $F(\alpha)$ is intuitive. Fill in the necessary things.
Normality of a subgroup is either total symbol-pushing or else conceptually challenging.
leslie townes
leslie townes
more structure often makes things easier. fields even easier than rings, for example.
Koro
Koro
Oh I see. So it is because it probably is more intuitive. I was thinking based on no. of operations.
leslie townes
leslie townes
finite groups are really weird. (finite rings are too)
Ted Shifrin
Ted Shifrin
I’ve taught it to undergrads both ways, and I stand by my pedagogical decision to write the book the way maybe only 30% want to teach it. Sadly, many algebra courses are just formal symbol-pushing.
Thorgott
Thorgott
19:13
finite groups are cool
moreso than infinite ones
Ted Shifrin
Ted Shifrin
Yuck.
Lie groups rule.
Thorgott
Thorgott
Lie groups suck because they're related to Lie algebras
I'd love to like them, but guilt by association is too strong
Koro
Koro
ssi=iff :)
Feynman_00
Feynman_00
@Thorgott Can you expand more about your hate towards Lie algebras? :P
Ted Shifrin
Ted Shifrin
You’re just throwing your highest weight around.
Feynman_00
Feynman_00
19:27
I have a problem with Lie groups too, though. Most physics books are ambiguous when dealing with Lie groups
Thorgott
Thorgott
root systems are by far the ugliest piece of algebra I've had to lay my eyes upon and they will never recover
the fields of Lie algebras and representation theory are the most unintuitive and least accessible branches of algebra and I will die on that hill
Feynman_00
Feynman_00
That's not comforting to read for someone studying representation theory right now...
Yai0Phah
Yai0Phah
@leslietownes Maybe I am mistaken, but why do fields carry more structures than rings?
Feynman_00
Feynman_00
Although I'm not sure I need to delve too much into the topic
leslie townes
leslie townes
yai i did not mean it in any formal sense. i just mean, if you add more to the pile of things you assume about something ("structure"), sometimes things get easier and not harder.
i'm not one of those people who says structure (or the perennial favorites natural or canonical) with any specific meaning behind them.
Thorgott
Thorgott
19:35
this is actually a fairly subtle issue, I'd say
Yai0Phah
Yai0Phah
The category of fields is a full subcategory of that of rings.
Thorgott
Thorgott
I don't think there's universal agreement on how to formalize what a field really is, as opposed to a ring.
leslie townes
leslie townes
part of koro's intuition was, the minute you bring in more axioms, things automatically get more complicated. all i meant was, not necessarily. add "G is finite and order 3" to group theory and it gets a lot simpler.
Feynman_00
Feynman_00
By the way this is one the ambiguous things I find written about Lie groups
Thorgott
Thorgott
it's an infamous weirdness that the theory of fields is not an algebraic theory
an interesting example, I think, is trying to internalize the notion of a field in the category of topological spaces
since we agree on what a topological field is, and it's not just a topological ring that is a field
"inversion" here is not really an axiomatic condition on the algebraic structure, but is expressed through a structural morphism $K\setminus\{0\}\rightarrow K\setminus\{0\}$
so that's a case in which I would argue fields do have more structure than rings
leslie townes
leslie townes
19:40
would any of this have started if i had used a word like 'rigidity,' or 'information,' instead of 'structure'? i need to know so that i don't trip over this next time.
Yai0Phah
Yai0Phah
The category of topological spaces is very bad. Even if you replace it by a very good alternative such as a topos, there does not seem to be a good definition of internal fields.
Ted Shifrin
Ted Shifrin
You always mess up, leslie, regardless.
leslie townes
leslie townes
that does seem to be the rule.
Thorgott
Thorgott
well, the definition given in the category of topological spaces is very much sensible
there's no debate on what a topological field is
it just isn't clear how well the pattern generalizes
Yai0Phah
Yai0Phah
I mean, it does not have a very rich theory of internal language, in comparison to topoi.
It is not locally presentable.
Thorgott
Thorgott
19:46
sure
 
1 hour later…
Jakobian
Jakobian
21:10
@Thorgott in the universal algebra sense?
Thorgott
Thorgott
in the sense of Lawvere theory
but that may or may not be related, I'm not knowledgeable enough to tell you
leslie townes
leslie townes
21:37
it ends up, as it must, at lawvere theory
CroCo
CroCo
22:06
hi everyone, could you please take a look at my proof and correct it if it needs
user image
I'm not very strong at proofs but this is my attempt.
leslie townes
leslie townes
22:19
so far so good. i might include more detail about why the result is in R^n. (you're implicitly using the fact that you don't leave R in each component, i.e., that the set R from which each component is taken is also convex.) not the most interesting detail in the world, but a detail nonetheless.
CroCo
CroCo
@leslietownes thanks for the feedback. I think each element z_i is in R so z should be in R^n. That is why I stopped
leslie townes
leslie townes
yeah, i would just add in that observation, each one of these components z_i is in R, so z is too. it almost feels like it goes without saying, but it wouldn't go without saying if, for example, R^n were replaced with a subset of R^n with a more complicated definition.
Ted Shifrin
Ted Shifrin
Why is it true in one dimension? Somewhere we need to prove something.
leslie townes
leslie townes
as a side note, there's definitely a type of definitions-checking exercise that is pedagogically maybe suboptimal for not giving the person enough to do. i might put this exercise in that category.
unless this class is littered with 'but wait, is R closed under addition' type checking, this kind of thing sort of comes out of left field.
Ted Shifrin
Ted Shifrin
I don't like the exercise, personally.
CroCo
CroCo
22:31
user image
Ted Shifrin
Ted Shifrin
But other than saying "it's true because it's true," there has to be something to do.
I don't see that writing it out in coordinates does anything.
leslie townes
leslie townes
as you say, in R^1 there's definitely something to prove, but, i'm guessing this is arising in the kind of class where R having addition and scalar multiplication operations is assumed and not proved.
Ted Shifrin
Ted Shifrin
We know that $\Bbb R^n$ is a vector space, so is closed under scalar multiplication and addition.
Where is this exercise coming from?
CroCo
CroCo
@TedShifrin I'm reading elementary linear programming by Bernard Kolman and Robert E. Beck
2nd
@TedShifrin so my proof is not convincing? :<
Ted Shifrin
Ted Shifrin
I don't see why looking at components is necessary. I would just say that $\Bbb R^n$, being a vector space, is closed under scalar multiplication and addition. Therefore, $\lambda x, (1-\lambda)y\in\Bbb R^n$ and so their sum is ...
leslie townes
leslie townes
22:38
this reminds me a bit of the kind of physics class where the instructor spends 30 min on some more complicated than necessary proof of the schwarz inequality in R^n before, like, doing all kinds of unjustified stuff with unbounded operators
e.g., weird place for formalism, or at least this kind of formalism
Ted Shifrin
Ted Shifrin
Yeah, it would be a better exercise to prove that the unit ball is convex and that a punctured ball is not.
CroCo
CroCo
@TedShifrin your approach makes sense but my understanding for proofs is the fact that we need to show all steps or may be I don't know which crucial steps are necessary to be shown.
leslie townes
leslie townes
the details that are necessary to be shown may depend on formalism that the book maybe hasn't even bothered to set up.
Ted Shifrin
Ted Shifrin
You have no more steps than I do. Why is it any more obvious in $\Bbb R$ than in $\Bbb R^n$? As leslie and I said, this is a stupid exercise.
leslie townes
leslie townes
e.g. are the usual properties of the vector space operations on R^n known, or are they defined in terms of the R^1 operations (whose properties are assumed known), or what.
Ted Shifrin
Ted Shifrin
22:43
I guess they want to start to develop the notion of convexity for piecewise-linear subsets of $\Bbb R^n$.
CroCo
CroCo
The authors didn't provide the answer so I'm guessing my solution is stupid. :)
why the exercise is not good?
Ted Shifrin
Ted Shifrin
Students have trouble proving totally obvious things. What do we assume is known and what is not? That's what leslie's been saying.
Your solution is no more stupid than mine. I think you should do a different exercise.
CroCo
CroCo
Thank you so much both of you for being informative and helpful. I clearly see your points now.

« first day (4523 days earlier)      last day (794 days later) »