Mathematics - 2020-04-27 (page 1 of 2)

Knight

02:54

Hi Ted

What does this statement really mean: $$ \sqrt{\textrm{any real-valued function}} \geq 0$$

Ted Shifrin

03:08

By definition, $\sqrt x$ is the unique nonnegative number whose square is $x$.

CaptainAmerica16

03:32

Is anyone here?

I'm trying to figure this out. I know what I need to show, I'm just having trouble figuring out how to show it

I think I need to show that the composition $(g \circ f)(x)$ is invective and surjective

So maybe I can start with let $x_1, x_2 \in A$ and assume $f(x_1)=f(x_2)$

for the injective part

Knight

05:36

@CaptainAmerica16 Since, $$\left ( g \circ f \right) ^{-1} = f^{-1} \circ g^{-1} $$ and $f^{-1}$ and $g^{-1}$ exist such that $$f^{-1} \circ g^{-1} : C \rightarrow A $$ hence, the inverse of $g \circ f$ exist, therefore, it is bijective

Captain Bohemian

Is it fine to say a function is "monotonously increasing"? I think one should say "monotonically increasing", right?

Knight

@CaptainBohemian According to the dictionary, both “monotonously” and “monotonically” both are correct adjectives.

Captain Bohemian

@Knight but I remember my MSc advisor told me that in this circumstance when we want to describe a function, using monotonous is wrong and we must use monotonic. He corrected once when I used monotonous in this kind of circumstance.

Knight

05:53

@CaptainBohemian How about asking it on English Langauge and Usage Stackexchange ?

Captain Bohemian

but now I see an instructor uses "monotonically increasing" in his slide.

@Knight but I doubt people in English Language and Usage know that because I think only mathematicians know that since this is a math jargon.

Knight

@CaptainBohemian Well I suggested that because I thought you were writing some kind of theses for publication and you didn’t want any mistake, so people over there quite used to publication and hence would have helped you with that.

Ted Shifrin

@CaptainBohemian monotonous describes a tone of voice; monotonic describes a function.

Captain Bohemian

06:09

@TedShifrin so my MSc advisor, who is a senior professor in physics, is correct while the instructor, who is a postdoctor in physics, is wrong?

CaptainAmerica16

@Knight Dang, I didn't want the answer

I probably should have clarified.

Thanks though

@TedShifrin You here?

Ted Shifrin

Monotonic is the only math option!

Captain Bohemian

@TedShifrin so I should correct that instructor.

Ted Shifrin

Yup.

Knight

06:28

@CaptainAmerica16 Sorry :( . By the way, How you doing Capitan?

Ted you are awake so late in the night?

CaptainAmerica16

@Knight It's cool. I'm doing pretty good. Getting ready to wrap up the school year

Knight

@CaptainAmerica16 Okay, which course will you opt in Univeristy ? I mean which major in Bachelors ?

CaptainAmerica16

@Knight I plan on majoring in Math

Knight

@CaptainAmerica16 Wow! That’s great

Which university?

CaptainAmerica16

Yeah, I'm pretty excited

PennState

Knight

06:32

Pennsylvania Stae Univeristy ?

CaptainAmerica16

yep

CaptainAmerica16

06:44

This is what I get for working so late at night, @Knight. I just realized what I was doing wrong for a different approach I was working on for the problem.

$(g \circ f)(x_1) = (g \circ f)(x_2)$. By definition of function composition, we have $g(f(x_1)) = g(f(x_2))$. Since $g$ is injective, we have $f(x_1)=f(x_2)$. Since $f$ is injective, we have $x_1=x_2$. I was overthinking again.

I'll write this up properly and do the surjective part tomorrow

Rudi_Birnbaum

07:16

Hi all!

I have posted that one here math.stackexchange.com/questions/3644438/… , maybe some one of you has an idea?

Captain Bohemian

09:02

can we type math symbols in Microsoft Teams chat?

skullpatrol

math.ucla.edu/~robjohn/math/mathjax.html

that^ is what works here

Captain Bohemian

but is MathJax installed in Microsoft Teams chat?

skullpatrol

don't know, sorry

vesii

12:32

Is it right to say that $C=\{p(x)\in \mathbb{R}_3[x]|p'(1)=p(0)\}$ is subspace of $\mathbb{R}_3[x]$ because:
1. $p(x)=0\in C$ because $p'(1)=0=p(0)$.
2. $(p+q)'(1)=p'(1)+q'(1)=p(0)+q(0)=(p+q)(0)$.
3. $(\lambda p)'(1)=\lambda p'(1)=\lambda p(0)=(\lambda p)(0)$
Just making sure I'm proved it right.

Simply Beautiful Art

sounds right

vesii

thanks :)

courge9

12:46

If f is a Schwartz function on R, is it then clear that the integral of f(x)/x over all of R converges?

Thorgott

I don't think it's clear for I think it's not true

courge9

I also have that impression, but I'm struggling to find a counterexample

Thorgott

$x\mapsto e^{-x^2}$

courge9

ah, you're right, thanks

but what about P.V.?

Does the integral exist in a principal value sense?

Thorgott

in this case yes

but I think generally probably not

let's see if we can modify this

courge9

12:54

as P.V.[1/x] is a distribution on Schwartz space this should work, right?

Is there some converse condition? P.V.\int f(x)/x exists only if f is Schwartz?

Thorgott

wait, do you want to take the principal value over $\mathbb{R}\setminus(-\varepsilon,\varepsilon)$ as $\varepsilon\searrow0$ or over $[-R,R]$ as $R\rightarrow\infty$?

courge9

the first one

Thorgott

yeah, that always exists

courge9

you made me curious... what about the second limit option?

Thorgott

$e^{-x^2}$ still works as counter-example then, no? it blows up in any neighborhood of the origin

courge9

13:03

right

Thorgott

as for the converse condition, I don't think something like that works: take an even $\mathcal{L}^1$-function $f$, then $\int_{\mathbb{R}\setminus(-\varepsilon,\varepsilon)}\frac{f(x)}{x}\mathrm{d}x=0$ for all $\varepsilon>0$, but such a function can be very badly not Schwartz (it can be everywhere discontinuous, say)

courge9

correct

I should have seen that

thanks for your help and time

Thorgott

no problem

Mike Miller

@BalarkaSen Some good ol commutative algebra for you mathoverflow.net/questions/358665/…

Leaky Nun

@MikeMiller oh no

@MikeMiller OP has had too much Jacob Lurie to read

Pongal

13:48

Hello everyone. I am not sure if its against the rules, but I have a small problem to solve. I have an excel sheet with a formula (exponential smoothing). I am looking for a universal formula that when i enter the current value and expected value, I want to know how many days it will take to reach the expected value. I can upload the excel sheet somewhere online for you to see.

Could any one help me with that ?

ibb.co/G5sYZ9b

So, in the above example, if I give current value as 85 and expected value as 90, I am looking for the result 7

Could some one prepare a generic formula for that pls

Edward Evans

14:20

Splitting of a short exact sequence in an abelian category is equivalent to the middle module being the direct sum of the outer modules right?

idk what to call the "exact sequence-ends"

factors?

lol "modules"

geocalc33

can you do 4mod(rational)

topologicalorientablesurface

14:42

if A is connected and I add limit points to A

is the new set connected?

Mike Miller

@EdwardEvans That's false in abelian groups

Edward Evans

@MikeMiller but true in modulesÜ

?

altho

feynhat

@EdwardEvans Yes.

Edward Evans

Okay :) thanks both, I know that it's false in the category of groups because that'S not an abelian category

feynhat

What.

Mike Miller

14:54

@feynhat No, this is subtle.

@EdwardEvans Abelian groups are Z-modules. There is probably a counterexample for the vast majority of rings R (maybe all R that aren't fields?)

@feynhat The correct statement is that a short exact sequence splits iff it is isomorphic to the standard SES $0 \to A \to A \oplus B \to B \to 0$, where the first map is inclusion and the second map is projection. This is stronger than an isomorphism of the middle term to $A \oplus B$.

Edward Evans

yeah right

Mike Miller

The simplest example I know in Abelian groups is $\bigoplus_{k \geq 1} (\Bbb Z/2^k)[t]$ and $B = \bigolus_{k \geq 1} (\Bbb Z/2)[t]$.

feynhat

I see.

Mike Miller

The polynomial generator $t$ is not really relevant to the additive structure, it's just an easy way for me to say "take infinitely many copies"

Balarka Sen

@MikeMiller Yikes

Mike Miller

14:59

I think it might be true if you assume everything in sight is Noetherian?

Edward Evans

I have $0 \to \Bbb Z \to A \to \Bbb Z \oplus \Bbb Z/7 \to 0$ and I'm trying to find a $\Bbb Z$-module such that the sequence is exact but doesn't split

just on a pset

Balarka Sen

Jazz up $7\Bbb Z \to \Bbb Z \to \Bbb Z/7\Bbb Z$

Tobias Kildetoft

@BalarkaSen And now Balarka will provide the precise definition of what it means to Jazz up an exact sequence, directly from nCat

Edward Evans

lol

Balarka Sen

Lmfao

@MikeMiller I am not sure what you are giving an example of. An abelian group $G$ isomorphic to $A \oplus B$ such that there is an exact sequence $0 \to A \to G \to B \to 0$ which doesn't split?

Alessandro Codenotti

15:03

lol

Looks like I joined at the right moment

Edward Evans

Hey @Alessandro

Mike Miller

@BalarkaSen Yes

Edward Evans

I have a candidate for $A$ but it seems a bit contrived, since the map out of $A$ needs to be the zero map for exactness

as in, out of my $A$

so to speak

Mike Miller

There can be no $A$ so that the map out of it is zero in that sequence

Because then Z -> Z/7 would be injective...

Edward Evans

alright then I gotta go back and do something else

Tobias Kildetoft

15:05

@EdwardEvans Exactness just means that the maps are the inclusion of a kernel and the projection onto the cokernel

Edward Evans

Okay thanks all :P

Mike Miller

My sequence is just the direct sum of every Z/2^k -> Z/2^{k+1} -> Z/2. You take infinitely many copies of this so that the middle term is indeed isomorphic to the sum of the outside two

Balarka Sen

If I remember, there are two exact sequences $1 \to \Bbb Z_p \times \Bbb Z_p \to H_3(\Bbb Z_p) \to \Bbb Z_p \to 1$ one of which splits and one of which doesn't, where $H_3(\Bbb Z_p)$ is the Heisenberg group. If this is right, you should get a $\Bbb Z$-Mod example by taking the group algebras over $\Bbb Z$

Nah

Tobias Kildetoft

@BalarkaSen But the Heisenberg group is not abelian

Balarka Sen

Yeah I was thinking maybe you could take the group algebra over $\Bbb Z$ and get an exact sequence

That's not correct

geocalc33

15:11

can you not just take the mapping space of the sequence, project it onto the cokernal and then split via the kernal

Tobias Kildetoft

So the sequence is also only split on one side, not both sides

Balarka Sen

Funny, there should be natural examples of this phenomenon.

@MikeMiller Seems like it. Do you see a proof

geocalc33

I have a prophecy. Ted will enter the chat in 20 seconds.

darn

is ln(1-x)e^{1/x} a sort of convolution?

without the integral

can an almost uncrackable security system be built from some sort of dscrete convolution?

feynhat

15:34

@topologicalorientablesurface Yes.

If $A$ is connected, then anything that lies between $A$ and $\bar A$ is also connected.

SimoBartz

Can you help me in understanding something about representation theory?

its now something that can be written in a single question post

it's not*

Tobias Kildetoft

@SimoBartz What sort of representation theory?

SimoBartz

group representation

I'm very new to the topic

Tobias Kildetoft

Try writing the question then

Don't worry about introducing the terminology

SimoBartz

if I consider the three-dimensional representation of the group SU(2), the 3x3 matrices that I find are themself a group?

Tobias Kildetoft

15:41

Yes

SimoBartz

is this group SU(2) again?

Tobias Kildetoft

Depends on which representation you take

Mike Miller

I assume he means 3x3 real matrices

SimoBartz

in general is just in homomorphism with SU(2)?

Mike Miller

So it's clear the desired representation is the standard one on imaginary quaternions, I think

The image of that is SO(3)

The first faithful representation of SU(2) is in real dimension 4

SimoBartz

15:43

what do you mean with faithful?

Mike Miller

No elements act as the identity

You can find that in your 3dim representation, -I in SU(2) acts as the identity

SimoBartz

but a group need the identity element to be group

Tobias Kildetoft

@SimoBartz He meant apart from the identity

SimoBartz

Ah ok, so in general the group of the operator that I find when I look at a representation are not the original group

Tobias Kildetoft

In general, you get a quotient

SimoBartz

15:51

Can I represent SO(3) in two dimensions?

Tobias Kildetoft

depends on what properties you want from the representation

SimoBartz

Ah wow, now I got the point, not all the property are conserved in the representation

I didn't realize it thanks

which property I can't preserve if I represent SO(3) in two dimensions?

because I think this is the reason why we need SU(2)

Leaky Nun

@BalarkaSen one day left

SimoBartz

16:07

@TobiasKildetoft ?

Balarka Sen

16:21

@leakynun of what

Leaky Nun

@BalarkaSen our correspondence game

SimoBartz

are the rotations SO(3) a representation of SU(2)

?

Balarka Sen

o shit i forgot about that completely

dont think ill be able to play today tho lol

Leaky Nun

rip

Mike Miller

16:50

@SimoBartz Out of curiosity, are you a physicist? It's hard to answer some of your questions because they're not phrased the way a mathematician might.

SimoBartz

yes I am, I'm sorry for the confusion

I'm trying to understand the idea of this theory in order to understand physics

but I think you can help me

Because I think there is some connection between this two groups

I've seen SO(3) is a double cover of SU(2)

Mike Miller

Yes, that's true. The standard 3-dimensional representation gives a surjective homomorphism $\rho: SU(2) \to SO(3)$, and the kernel is $\{\pm I\}$, the group of order 2.

That's why it's called a double cover; as a map of manifolds $\rho$ is a covering map with fibers of cardinality 2

SimoBartz

but are there some representations of SO(3) that are the same as the representation of SU(2)?

Balarka Sen

17:08

@SimoBartz You mean it the other way around, SU(2) is a double cover of SO(3), as Mike explained

The physicist's way of seeing this is the Dirac's plate trick (not quite this, but a related picture), in which to get the plate to do at least one full rotation and bring it back to the usual state you need to swirl it around in your hand two full rotations. Hence a 2-1 map.

Of course, what is SU(2) and what is SO(3) is up in the air in my simplistic description

SimoBartz

mm I mean, the representation of SO(3) in 2 dimension is the same as the representation of SU(2)?

Balarka Sen

I don't know representation theory, so someone else might be better equipped to answer this question. What is "the representation in 2 dimensions"?

A 2-D representation of SO(3), which is a group homomorphism SO(3) -> GL2(R), automatically gives you a group homomorphism SU(2) -> GL2(R) by composing with the double cover $\rho$ that Mike described.

SimoBartz

is the map T acting of the group G such that T(g1)T(g2)=g1g2

Balarka Sen

So given any 2D representation of SO(3) you get a 2D representation of SU(2) with the same image, if that's what you meant.

SimoBartz

what's the image of a representation?

Balarka Sen

17:14

To a mathematician, a representation of a group $G$ on a vector space $V$ is an action of $G$ on $V$ by linear transformatons. This is equivalent to a group homomorphism $G \to \mathrm{GL}(V)$. I refer to the image of this group homomorphism

SimoBartz

Ah ok, does it means that the matrices that acts

sorry

Farhad Rouhbakhsh

Can someone answer this please?

0

Q: A subtle problem regarding tangent line when gradient is non-zero

In the section of "Applications of Partial Derivates" in Adams' calculus book, 7th edition, page 756, there is a part which talks about "Lagrange Multipliers". It says: "Suppose that $f$ and $g$ have continuous first partial derivatives near the point $P0=(x0,y0)$ on the curve $C$ with equation ...

Its hours since I've asked this question and there is no answer

Balarka Sen

That's right, for you, a representation of a group $G$ on a real vector space $\Bbb R^n$ will mean associating to each element of $G$ a nonsingular $n\times n$-matrix. The nonsingular $n \times n$-matrices form the group $\text{GL}_n(\Bbb R)$, so you get a map $G \to \text{GL}_n(\Bbb R)$, which is the homomorphism given by the representation that makes it precise what this "association" is.

Mike Miller

@SimoBartz A representation of dimension $n$ is a homomorphism $f: SU(2) \to GL_n(\Bbb R)$ (sometimes people will use complex coefficients instead). Any such map so that $f(-I) = I$ factors through $\rho$ --- meaning there is $f': SO(3) \to GL_n$ so that $f = \rho f'$. (This is a statement about what quotient groups are, SU(2) and SO(3) and GL_n are not particularly relevant.)

Farhad Rouhbakhsh

0

Q: A subtle problem regarding tangent line when gradient is non-zero

In the section of "Applications of Partial Derivates" in Adams' calculus book, 7th edition, page 756, there is a part which talks about "Lagrange Multipliers". It says: "Suppose that $f$ and $g$ have continuous first partial derivatives near the point $P0=(x0,y0)$ on the curve $C$ with equation ...

multivariable-calculus lagrange-multiplier

SimoBartz

17:19

does it means that the matrices that I find with SO(3) and SU(2) on $C^2$ are the same?

Mike Miller

There is a single non-trivial representation $f_n: SU(2) \to GL_n(\Bbb C)$ up to conjugacy / isomorphism of representations for each $n>1$. When $n$ is odd, this factors through $SO(3)$, because $f_n(-I) = I$. (This comes from knowing what the maps $f_n$ are.)

@SimoBartz what matrices do you find for SO(3) acting on C^2

the answer is no, but i want to know exactly what you mean

Balarka Sen

I think he's trying to say the image of the rep homs using our correspondences between reps of SO(3) and reps of SU(2) are the same, @MikeMiller

Mike Miller

I can't parse your sentence

The standard representation of SU(2) on C^2 doesn't factor through SO(3), so it doesn't make sense

SimoBartz

when I represent SU(2) on $C^2$ the image of the homomorphism is the same as if I represent SO(3) on $C^2$?

Balarka Sen

Whenever it does factor, @MikeMiller.

Ah ok he's trying to do what you suggest, so yeah, that doesn't make sense

SimoBartz

17:23

why it doesn't make sense?

Balarka Sen

@SimoBartz Not all representations of SU(2) give rise to representations of SO(3): Mike wrote down the condition for when this happens (if the matrices representing -I and I are the same). The other direction is true.

SimoBartz

I don't think I'm understandign

Ok let's try in this way (I'm sorry I'm so confuse about this topic) can I find SO(3) from SU(2) some how?

Farhad Rouhbakhsh

any idea on my question guys?

0

Q: A subtle problem regarding tangent line when gradient is zero

In the section of "Applications of Partial Derivates" in Adams' calculus book, 7th edition, page 756, there is a part which talks about "Lagrange Multipliers". It says: "Suppose that $f$ and $g$ have continuous first partial derivatives near the point $P0=(x0,y0)$ on the curve $C$ with equation ...

multivariable-calculus lagrange-multiplier

Mike Miller

@SimoBartz There is a subgroup $\{\pm I\} \subset SU(2)$. The quotient $SU(2)/\pm I$ is isomorphic to the group $SO(3)$. This means there is some explicit formula for a map $f: SU(2) \to SO(3)$ so that $f(I) = f(-I) = I$, and so that $f$ is surjective, and so that the $f^{-1}(I) = \{\pm I\}$.

I don't know that I can recommend anything more than trying to understand and use the definitions.

SimoBartz

thank you very much

Mike Miller

17:31

If you're just looking for the formula for $f$ it's on the bottom of the fourth page here.

Oops, fourth page. On that fourth page $U(x,y)$ is a 2x2 unitary matrix ($x$ and $y$ are vectors in $\Bbb C^2$) and $R(U)$ is a 3x3 orthogonal matrix.

Leaky Nun

@BalarkaSen Fabiano playing white wins Armageddon against Hikaru playing black

Ted Shifrin

@FarhadRouhbakhsh it's sufficient (but not always necessary) to guarantee that the level curve of $g$ is smooth at the point. Simple example: $g(x,y)=xy$ with $P=(0,0)$.

Farhad Rouhbakhsh

@TedShifrin Why is it sufficient? Actually in this question by "smoothness" I mean differentiability

@TedShifrin I cant figure out why the fact that gradient is non-zero at a point is sufficient for differentiability in this question

Tobias Kildetoft

17:50

Hmm, the preparation materials for the certification I am about to take start off with being mainly a sales pitch. Including the classic "you can do all these thing without even coding", which is pretty much the opposite of what makes it interesting for me (fortunately, actually doing anything meaningful without code in it is not anywhere as easy as they try to make it seem).

Ted Shifrin

@FarhadRouhbakhsh it's called the Implicit Function Theorem.

Balarka Sen

Oh what? Bela Bollobas was a student of Frank Adams

That doesn't make any sense

Tobias Kildetoft

(if it were that easy to do a ton of stuff without coding, I would risk no longer getting paid so well :) )

Farhad Rouhbakhsh

@TedShifrin From what I read about "Implicit Function Theorem" in my notes, it says that certain equations of some dependent and independent variables can be solved.

It doesnt talk about differentiability

Ted Shifrin

Then your notes are no good.

You can find several of my YouTube videos on this. (Link in my profile.)

Farhad Rouhbakhsh

18:02

@TedShifrin My notes are from Adams Calculus

Sha Vuklia

what does it mean for an object in a category to be "fully characterised (by ...)"? would it mean that for each morphism from and to our object, we know how it looks like?

this is the context by the way

It seems to me that we have only characterised the morphisms to the limit of $X$, but not from the limit of $X$

and I feel like Yoneda doesn't give us much more..?

Tobias Kildetoft

@ShaVuklia It just means that any other object satisfying the given will be isomorphic to that limit

Sha Vuklia

but the theorem already says that, so why do they say that Yoneda characterises the limit?

Leaky Nun

oh wow @ShaVuklia is studying category theory now

HokieFan7

Hey guys quick question

Leaky Nun

18:08

@ShaVuklia beucase we're taking limit over Set, which is "unique"

the above theorem characterizes Hom(-,L), so by Yoneda characterizes L

Thorgott

doesn't the theorem characterize the functors, which characterize the limit by Yoneda or sth

Leaky Nun

s n i p e d

HokieFan7

I have u(x,y) = (x^2)/2 - (y^2)/2 + 2xy - y, how do i write that in terms of z if z = x + iy

Thorgott

I admit my defeat

Farhad Rouhbakhsh

@TedShifrin In Adams' Calculus we read:"The Implicit Function Theorem guarantees that systems of equations can be solved for certain variables as functions of other variables under certain circumstances, and it provides formulas for the partial derivatives of the solution functions."

Ted Shifrin

18:09

That's not the statement of a theorem.

Whatever happened to the @Sha I used to like so much?!!! :)

Sha Vuklia

@TedShifrin waaat?:0

Astyx

Does studying Category theory make her unworthy ?

Ted Shifrin

NO comment.

Sha Vuklia

I am unaware of the math camps here it seems, but that is okay

Ted Shifrin

@HokieFan7 What have you tried?

Astyx

18:11

(also hi)

Ted Shifrin

(also hi @Astyx)

Farhad Rouhbakhsh

@TedShifrin Its not the actual theorem, its just an explanation of what it is breefly. I can send the photo of the actual "STATEMENT" of the theorem which comes next if you like

Ted Shifrin

@FarhadRouhbakhsh: You don't need to send me anything. You need to read it and understand it! Or look at other resources.

Farhad Rouhbakhsh

@TedShifrin and believe me there is no talk about gradients there

HokieFan7

Mostly just guess and check but I can't get anything to work, I

I'm not sure the method behind it

Ted Shifrin

18:13

@Farhad: there is talk about one partial derivative being nonzero.

Farhad Rouhbakhsh

@TedShifrin perhaps you misunderstood gradient with Jacobian?

Ted Shifrin

@Farhad: There's no difference in this situation.

@Hokie: It's going to be a mess, because you need real parts and imaginary parts, so you're going to have to use both $z$ and $\bar z$. What is the context of this question?

Farhad Rouhbakhsh

@TedShifrin It talks about " systems of equations can be solved for certain variables as functions of other variables under certain circumstances". Man Im talking about smoothness and differentiability. Who talks about solving systems of equations??

0

Q: question on the relation between smoothness and gradient

Suppose that $g(x,y)$ has continuous first partial derivatives near the point $P0=(x0,y0)$ on the curve $C$ with equation $g(x,y)=0$. What does the fact that $∇g(P0)≠0$ tells us about the smoothness and the existence of tangent line of $C$ on $P0?$ Why? What happens to the tangent line and smoo...

multivariable-calculus

HokieFan7

I had that f(z) = u(x,y) + iv(x,y) and z = x+iy. I was given v(x,y) and I need to find u(x,y) which I already did. Then I just need to write f(z) in terms of z which is what I'm stuck on

Farhad Rouhbakhsh

@TedShifrin I would appreciate if you look at the link I sent right now

Thorgott

18:17

what is the precise definition of "the curve $C$ with equation $g(x,y)=0$"

Ted Shifrin

@Farhad: I'm not going to keep arguing with you. I explained exactly this point in lectures 39 and 40 [3500] and then the general theorem in lecture 20 [3510] here.

@Hokie: So you started with a harmonic function, found its harmonic conjugate. So you didn't give us $f$. You just gave us $u$.

Tobias Kildetoft

@Thorgott $R[x,y]/\langle g\rangle$ where $R$ is whatever ring you want it over.

Sorry, Spec of that of course

HokieFan7

I ended up getting the answer, thank you though Ted

Ted Shifrin

LOL, glad to be of no use :P

Balarka Sen

@TobiasKildetoft lol

Thorgott

18:23

my temporary goal for the commutative algebra lecture I'm listening to this semester will be to understand this joke

Farhad Rouhbakhsh

@Thorgott I dont have any formula for that. It is just what is written. The surface on which g(x,y) is 0

Tobias Kildetoft

@BalarkaSen I suppose in the context that it really should have been MaxSpec

Farhad Rouhbakhsh

@Ted Shifrin Never keep arguing with anybody. Thank you for your response

Tobias Kildetoft

@Thorgott It was not entirely a joke

Thorgott

I mean, more precisely what is $g$ in this context and what's a curve. If say, $g$ is the constant zero function, I don't think anyone would call the set of $x,y$ such that $g(x,y)=0$ a curve.

@TobiasKildetoft that's even more imposing then

Farhad Rouhbakhsh

18:27

@Thorgott Actually my detailed question is in this link math.stackexchange.com/questions/3646300/…

@Thorgott See if you can help me to understand this with chatting not by videos

Thorgott

My point is that this question is not detailed enough for me to know what precisely is being asked

loch

take g to be nonconstant then :)

if you find out that the speaker wants the curve to be 'irreducible', then you want g to be irreducible as a polynomial
if you find out that the speaker wants the curve to be 'smooth', then you want the derivatives to not vanish simultaneously on the curve

Sha Vuklia

@LeakyNun ah, I think I see. By Yoneda, it's not possible to have to nonisomorphic $L$ and $L'$ such that $\hom(-,L)$ and $\hom(-,L')$ are isomorphic. But by the statement of the theorem, each such $\hom(-,L)$ is isomorphic to a "fixed" functor, so these $\hom(-,L)$'s will be isomorphic.

Farhad Rouhbakhsh

@Thorgott which part is not clear?

Thorgott

@loch what if $g$ isn't a polynomial

(algebraists hate this trick)

Balarka Sen

18:30

rekt

loch

what's a non-polynomial???

Farhad Rouhbakhsh

@Thorgott "Together (i) and (ii) imply that C is smooth enough to have a tangent line at P0." I have problems understanding this

Tobias Kildetoft

@loch These people are clearly mad :)

Alessandro Codenotti

@ShaVuklia Also note that you get for free from there that you can pull out limits from the right half of an Hom set

Balarka Sen

@Farhad The answer to your question is implicit function theorem. If the book you're reading doesn't mention this it's time to get a different book.

Thorgott

18:32

@FarhadRouhbakhsh What "the curve $C$ with equation $g(x,y)=0$" means. If you want to talk about curves defined by an equation in a reasonably general analytical context, then you will most likely have to think of it in terms of the implicit function theorem, I think that's what Ted was getting at.

Balarka Sen

When $\nabla g = 0$ there is no guarantee that $g$ is even a curve, like Thorgott said: take $g$ to be the constant function

loch

oh
i see

for some reason i saw spec above and thought this is a AG discussion

Balarka Sen

Without $\nabla g \neq 0$ you have no guarantee of smoothness and tangent lines

@loch we only talk about AG when we mock AG

3

Tobias Kildetoft

@loch That was my fault I suppose

loch

shouldve known

Tobias Kildetoft

18:34

Someone asked what a curve was and I gave an "answer"

Leaky Nun

@ShaVuklia right

Farhad Rouhbakhsh

@BalarkaSen What happens to differentiability when ∇g(P0)=0? Can you explain this?

Balarka Sen

@Farhad Differentiability of what? Tangent line to the "curve given by $g = 0$" is not guaranteed when $\nabla g = 0$. Consider $g(x, y) = x^3 - y^2$.

Take $P_0 = (0, 0)$

Farhad Rouhbakhsh

and why ∇g(P0)≠0 guarantees smoothness?

Balarka Sen

Implicit function theorem, like two people already told you

Look up the statement in a book which explains the theorem properly.

Farhad Rouhbakhsh

18:38

perhaps a quick favor explaining it breefly?

*briefly

or in form of an answer on the link I posted :D?

Balarka Sen

Intuitively, it's what you think it is. If $g : \Bbb R^2 \to \Bbb R$ is a $C^1$-function, let $C = \{(x, y) \in \Bbb R^2 : g(x, y) = c\}$ be a level set and suppose $\nabla g(P) \neq 0$ for all $P \in C$. Then $C \subset \Bbb R^2$ is a "curve with well-defined tangent line at every point of $C$"

Leaky Nun

@BalarkaSen it's trippy how Hikaru refers to the chat almost as a person

Balarka Sen

"you guys"

Leaky Nun

"it's not alphabetical, chat, because"

Balarka Sen

Lmao

Leaky Nun

18:41

he says "chat" like every sentence

as if referring to "chat"

"chat" is an entity with its own mind

Thorgott

it's a hivemind

Leaky Nun

"one second chat; it's not loading; okay there we go chat. here are the actual standings. i think you guys can still see it right. here are the actual standings chat."

I mean, if you're hikaru, you're probably talking to air, so maybe you need to imagine chat as a person whom you're talking to

Balarka Sen

he's so high energy i love it

keeps repeating the same thing 5 times with interjections in between

Farhad Rouhbakhsh

@Balarka Sen Thank you. But this is somehow advanced for me. Im an student of Calculus II and I dont have any familiarity with C1 functions. Isnt there any easier explanations of these? I say it because when I was talking to my teacher assistant about this problem, he told me the actual definition of "smoothness" is something related to C infinity classes. (Something I've never heard of before) But then he insisted here we mean just "differentiability"

Balarka Sen

C^1 just means continuously differentiable

I am sure the implicit function theorem is written down in standard multivariable calculus texts. Try Spivak

Leaky Nun

18:47

29 mins ago, by Ted Shifrin

@Farhad: I'm not going to keep arguing with you. I explained exactly this point in lectures 39 and 40 [3500] and then the general theorem in lecture 20 [3510] here.

Balarka Sen

Or watch that video ^ yeah

Leaky Nun

@BalarkaSen find the winning move for Fabiano

Balarka Sen

Bf3 seems like it should work

Leaky Nun

black pawns go down

Balarka Sen

Oh

Farhad Rouhbakhsh

18:50

@BalarkaSen I will try Spivak. Thanks. But when ∇g(P0)=0 can there be some cases which it is differentiable nevertheless?

Balarka Sen

There can be yeah but those are fluke

Leaky Nun

lol

Farhad Rouhbakhsh

for example x^2 + y^2 at point(0,0)?

the gradient is 2xi + 2yj and is 0 at (0,0)

Leaky Nun

x^2+y^2=0 is a single point at the origin

so in particular not a curve

Farhad Rouhbakhsh

Do you have any example of a curve at which the gradient is 0 and it is differentiable?

those fluke curves

Thorgott

18:55

your example does that

it just doesn't define a curve

Farhad Rouhbakhsh

I just want to make sure ∇g(P0)≠0 is sufficient not necessary. A curve example would do it :D

Leaky Nun

@FarhadRouhbakhsh $(x^2+y^2-1)^2 = 0$

Balarka Sen

@LeakyNun It should be Re1 but don't see what happens after Kf7 instead of Kf8 (which is an obvious lose)

Leaky Nun

@BalarkaSen what would you do if Kf8?

Balarka Sen

Re8

Leaky Nun

18:59

this also works if Kf7

very beautiful geometry pattern

the rook entrapped by two rooks

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