Mathematics - 2013-10-18 (page 1 of 4)

Karl Kronenfeld

00:00

derp

@anon But still, we don't really need $x^M\in\mathfrak p[x]$, do we?

Oh, derp again.

anon

how do we arrange for the integrality condition?

Karl Kronenfeld

lol I am of no help.

@anon I have a feeling it is really some determinant trick.

Note that multiplication by $x$ induces a map $\mathfrak p\to\mathfrak p$ which can be made into a matrix.

anon

00:15

indeed

I was unable to make headway in that direction

mmm, is End_Z(Z^n) integrally closed?

Karl Kronenfeld

What about the $\mathbb Z$-module $\mathbb Z+\mathfrak p$?

anon

ooh, that is f.g. and contains 1! so we can just work in (Z+p)[x]. that is very clever.

care to post that as an answer to the question on main?

wait, but we need (Z+p)[x] f.g. somehow

Karl Kronenfeld

Yes, I was noticing that.

Wow, wtf could we be missing.

anon

found the answer in an Algnumthry book

we do use a determinant trick

Alec Teal

00:32

Evening all

Kat

Hi, I need help on an algebra problem

0

Q: Elementary Algebra Inequality question

$$ \frac{1}{x-1} < -\frac{1}{x+2} $$ (see this page in wolframalpha) Ok, so I think the main problem is that I don't really know how to do these questions. What I tried to do was move $-1/(x + 2)$ to the LHS and then tried to get a common denominator. I ended up with $$ \frac{(x + 2) + (x -1)...

algebra-precalculus

I don't know what I'm doing wrong

Enjoys Math

If you could somehow solve for $x$ as if $\lt$ mean equation, then you could solve for $x$ in some equations. However this isn't $=$ and the difference is that when you multiply by a negative or you invert, both sides of $\lt$ you have to change signs.

@Kat

Kat

I don't get what that means

Enjoys Math

So pretend like you're solving an algebra equation and each time you need to multiply by possibly a negative value, then handle two cases

Karl Kronenfeld

@anon Is it specific to a particular ideal $\mathfrak p$ or something?

Enjoys Math

00:37

$\frac{1}{x-1} \lt -\frac{1}{x+2} \iff (\text{case 1}; x - 1 \gt 0) \dots$

Kat

So then when I move the -1/x+2 to the LHS I have two cases?

anon

@KarlKronenfeld nah

Enjoys Math

Do this: first find all $x \in \Bbb{R}$ such that $x \gt 0$ and your inequality holds. Then do the same for all $x \lt 0$.

*I mean $x - 1$

Kat

I can't see MathJax

Enjoys Math

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

go there to render mathjax

Kat

00:41

I still can't see it

Enjoys Math

Did you make the bookmark and click in while in this tab?

Kat

yeah

oh wait, it works now

Enjoys Math

\o/

Kat

thanks

ok so i can't multiply (x +2 ) because it was negative?

Enjoys Math

it's weird; why don't they put it in the site <script> area

Imagine this.

If you broke your problem into cases

and took the union of all the solutions at the end then that's valid right?

Kat

00:43

yeah

Enjoys Math

Well, break into cases so that you can somehow get those expressions out of the denominator

Kat

i don't know how

i don't remember learning any of this

Enjoys Math

if $x - 1 > 0$, then $x \gt 1$, so $x+2 \gt 0$ automatically see?

Kat

right

so I'm seing if x -1 is > 0 or if x -2 is > 0 ?

Enjoys Math

So in this first case we're working, multiply both sides by $(x-1)$ and both by $(x+2)$

Kat

00:45

yeah

well I moved -1 / x + 2 to the right side

Enjoys Math

what's do you have so far?

Kat

(1 / x- 1) + ( 1/ x +2) < 0

Karl Kronenfeld

@anon $\DeclareMathOperator{\end}{End}$So what we are dealing with is the usual determinant trick and the ring homomorphism $\mathbb Z[x]\to\end(\mathfrak a)$. I think what I failed to do is establish that $x$ can be determined by this homomorphism and the endomorphism $x$ induces on $\mathfrak a$.

Enjoys Math

$ x - 1 \gt 0 \implies \left ( \frac{1}{x-1} \lt -\frac{1}{x+2} \iff (x + 2 ) \lt -(x-1) \right )$ after multiplying both sides by said above

Pedro Tamaroff

@KarlKronenfeld Dude.

Kat

00:50

ok, but i still don't know how to solve this

Pedro Tamaroff

\newcommand{}{}

@Charlie I am not.

Enjoys Math

@Kat Do you see now?

Kat

no...

Enjoys Math

well do you see how I got the rightmost expression in my last post?

Kat

I really don't get it

Enjoys Math

00:52

The property I used was this...

Kat

i don't get the <-> part

Enjoys Math

That means the operation is reversible, certainly you can divide again by those linear expressions and get back to where you were, correct?

Kat

ok

Enjoys Math

You keep it iff in case you need to make use of that later

but I don't think we'll use the backwards implication arrow for this one

Pedro Tamaroff

@anon KCd is Keith Conrad? (i.e. the user here?)

Kat

00:53

ok, but then how do i solve it and find x?

Enjoys Math

Well, continuing from above you have $(x + 2) \lt - (x-1)$. So solve a linear equation for $x$

Karl Kronenfeld

27

A: newcommand vs. DeclareMathOperator

\DeclareMathOperator is designed to create commands that should typeset operator names such as sin and lim. Some of these are already defined in base TeX or LaTeX so one writes 2\sin\theta instead of 2sin\theta giving correct spacing and font. If you need an operator of this type that i...

Kat

x + 2 +x - 1 < 0

so then 2x + 1 < 0

and x < -1/2

but x != 1 or -2

so then what is the solution?

Enjoys Math

@Kat, the properties of inequalities we are using are if $a \lt b$ for two real numbers $a,b$, then $c a \lt c b$ for all $c \gt 0$. Even more generally, $a \lt b \implies a + c \lt b + c$ for ANY real number $c$. So we can still use those if our expressions involve variables but evaluate to real numbers, right? That's what we're doing

Kat

ok so the 2 < 1?

Mikael Öhman

00:57

I'm reading about greens functions and fixed point theorems (in regards to existence and uniqueness of solutions). Both texts from my course literature only ever considers homogeneous boundary values. Are nonzero b.c.s really so problematic?

Enjoys Math

I get $x + 2 \lt 1 - x \iff 2x + 1 \lt 0 \iff x \lt -1/2$

Pedro Tamaroff

@KarlKronenfeld Oh, don't get LaTeX righteous on me now!

Enjoys Math

In other words, for $x - 1 \gt 0$, there is no solution to the inequality since we have a contradiction, but Idk for sure, someone?

Kat

I don't know. This is from a high school pre-calc textbook

it shouldn't be this complicated :(

Enjoys Math

@Kat get it yet?

Mikael Öhman

01:10

@Kat, At the point where you have the single fraction, for a/b < 0, then a < 0 & b > 0, or a > 0 & b < 0. Two cases to check.

Twink

01:23

hi

Enjoys Math

hello

Twink

could someone tell me how to graph this polar function? $r(\theta)=1+\frac{\theta}{\sin(\theta)}$

but don't tell me what it is

I wanna know some technique to graph it without seeing the graph

Enjoys Math

blind fold yourself?

Twink

yes

Enjoys Math

what inverval does $\theta$ run on?

Twink

01:26

they don't say it

I guess I just have to deduce it

but it's not gonna be defined on $k \pi$ for $k \in \Bbb Z$

I think it's enough to draw it on $[0, 2\pi]\backslash \{0, \pi, 2\pi \}$

Mikael Öhman

Compute the value at a few points.. connect the dots?

Twink

ok I was just trying to find a more descriptive method

because I don't know what happens on $0$ for example

I think I'll get two U's as the graph

U U

Kevin Driscoll

This function sounds like its not single-valued if you let the angle continue to increase

Twink

well then maybe it's convenient to consider only $[0, 2\pi]$

I'm computing some values

Mikael Öhman

After having seen the plot, I actually got some ideas on what you could have done to determine it, but its a lot easier after having seen it.

Twink

01:35

don't tell me

Alec Teal

You should know it's a line... try and define a polar line, you'll see this is a horizontal one then

Wait, that's theta/sin theta

Twink

¬¬

thanks for telling me it's a line

Kevin Driscoll

its not a line

Twink

¬¬ !!!!

Alec Teal

No that's one over sin theta

Twink

01:36

don't tell me what it is please

I need some technique

Kevin Driscoll

it's not a horse

Twink

to graph it before seeing the graph

Alec Teal

Think of the asymptotes. There will be a range of theta where r gets really big.

There'll be another range where it's basically a circle (linear in polar terms)

Mikael Öhman

Actually, i dont have any really good ideas. I'm sticking to my "compute a bunch of values" approach. Is this a "sketch the function" type exercise?

Twink

yes I know there must be asymptores in $0$, $\pi$ and $2\pi$

Kevin Driscoll

01:38

I agree with Alec, look at the limiting behavior

what happens when $\theta$ is near 0? what about when its near $\pi$?

Alec Teal

If theta is a bit bigger than zero, it's basically 2. start there.

sin(theta)'s gradient is less than 1, but theta's gradient (wrt theta) is 1, so theta/sin(theta) is getting bigger as theta increases for as long as sin is getting bigger (0-pi/2)

Then sin stays positive but is getting smaller, as it's bounded by 1, theta/sin(theta) starts getting huge

Then it gets undefined at pi, obviously, then the same thing happens in reverse! (think of the interval (-pi,pi] I noticed you were using [0,2pi] this is wrong.

Now @KevinDriscoll I need some analysis help. I have a question that makes no sense.

Pedro Tamaroff

One doesn't have the chance to use the word Eulerian every day.

Kevin Driscoll

@AlecTeal Analysis!?!?!? @PEDRO!!!!!!!!!!!!!!!!!!!!!!!!!!!

Pedro Tamaroff

01:53

@KevinDriscoll WHHHhhhhAAAAAAAAAAT?

Kevin Driscoll

@Pedro @Alecteal needs help with analysis....... I didn't know what else to do........

Pedro Tamaroff

@KevinDriscoll "I panicked."

Alec Teal

@PedroTamaroff $f:[a,b]\rightarrow\mathbb{R}$, it is a squelch function if for each partition $a=p_0<p_1<...<p_{k-1}<p_k=b$ $f$ is constant on $(p_{j-1},p_j)$

Pedro Tamaroff

@AlecTeal Hit me.

Alec Teal

Show that such a function must have $f|(a,b)$ constant. What is the maximum cardinality of the set of values f can take.

This is apparently a common error students make when defining a step function.

Twink

01:55

one

Alec Teal

@PedroTamaroff what does f|(a,b) actually mean? "over", if it is true then obviously 1, but I cannot show that f|(a,b) must be constant.

Thanks @Davith for that.

Pedro Tamaroff

@AlecTeal The restriction of $f$ to $(a,b)$.

Twink

suppose it's not constant

and you can find a partition

Pedro Tamaroff

@AlecTeal squelch?

Twink

that has not those propierties

Alec Teal

01:56

@PedroTamaroff I searched for it, nothing. I think they made it up.

Pedro Tamaroff

@AlecTeal Take the partition $P=\{a,b\}$?

Daniel Fischer

@Twink Or three, $f$ is defined on $[a,b]$, and we know nothing about the values at the end points.

Alec Teal

@PedroTamaroff I don't follow.

Is it saying "there is a function, it is "squelch" if you can take any partitions you like and do this" rather than "there is a function, over these partitions"

Pedro Tamaroff

@AlecTeal A partition of $[a,b]$ is a finite set of points of $[a,b]$, one them which is $a$, another which is $b$.

Thus $P=\{a,b\}$ is a valid partition of $[a,b]$.

Fernando Martin

What's up?

Alec Teal

02:00

So it is a squelch because the partitions are ot defined with the function, because we can take any partition we like and this "squelch" function is over it, I could choose any open region I like?

@PedroTamaroff what's the formal definition of a step function (in contrast to this)

Pedro Tamaroff

@AlecTeal Well, you can say a function is a step function if it is a finite sum of characteristic functions of intervals.

Daniel Fischer

@AlecTeal "There is a partition such that..."

Alec Teal

I said "a function is a step function if and only if on an interval I it can be devided into a finite number of sub intervals ($I_1,I_2,...,I_n$) such that $f(x)=c_i$ for all $x$ interior to $I_i$, where $c_i$ is constant for i=1,2,3,...,n

@PedroTamaroff @DanielFischer

Daniel Fischer

@AlecTeal Yes, that's one way to describe them.

Alec Teal

@DanielFischer and this differs from the "squelch" definition how..... In the sense that I define the intervals bsed on the function? I'm not sure (I get brief flashes, but I am not confident) how they differ.

Daniel Fischer

02:05

@AlecTeal You: $\exists \text{partition}$, squelch: $\forall \text{partition}$.

Twink

guys do you kow a technique to graph the polar function $f(\theta)=1+\frac{\theta}{\sin \theta}$?

Pedro Tamaroff

@FernandoMartin Yo.

Twink

without seeing the graph?

Alec Teal

@Davith I just went through that.

Pedro Tamaroff

@Twink ...?

Alec Teal

02:05

@DanielFischer oh okay! thanks, each=all.

Twink

to plot it without seeing the graph

without cheating

Alec Teal

Again;

If theta is a bit bigger than zero, it's basically 2. start there.
sin(theta)'s gradient is less than 1, but theta's gradient (wrt theta) is 1, so theta/sin(theta) is getting bigger as theta increases for as long as sin is getting bigger (0-pi/2)
Then sin stays positive but is getting smaller, as it's bounded by 1, theta/sin(theta) starts getting huge
Then it gets undefined at pi, obviously, then the same thing happens in reverse! (think of the interval (-pi,pi] I noticed you were using [0,2pi] this is wrong.

It is not quite symmetrical about the x axis remember.

Twink

yes Alec but I'm asking if maybe there's a technique that we don't know

yes it is symmetrical

Fernando Martin

hey @Pedro

Twink

because $f(\theta)=f(-\theta)$

Alec Teal

02:07

Not true.

Wait, true.

But seriously sketch the top half and then reflect it.

Ive just told you what to do

Twink

well I'd like to have an easier technique to apply it in general

maybe there's a way to tansform it onto rectangular coordinates

Alec Teal

There is but it'd be tedious and not good because of that +1

Kevin Driscoll

@Twink As far as a I know there is no general method for graphing functions by hand and getting in right. Its an art.

Alec Teal

x^2+y^2+arctan(y/x)/(y/(x^2+y^2)^0.5) isn't easier

Wait that's not quite correct (bad night :P), the thing after the x^2+y^2 is though, and that alone should scare ya off

@PedroTamaroff @KevinDriscoll math.stackexchange.com/questions/529900/… have a gander at that, 10h in now.

Pedro Tamaroff

@AlecTeal Observe that if $s(x)$ is constant over $(a_i,b_i)$ then $s(x/k)$ is constant $(ka_i,kb_i)$ and conversely.

That is, you have a bijection between partitions $I\mapsto kI$.

Alec Teal

02:27

So @PedroTamaroff I ought not be using linear algebra

Pedro Tamaroff

@AlecTeal I don't see where I used linear algebra.

Harshdeep

Heyy everyone, I stumbled upon a thought regarding calculating discrete logarithm problem and now I am super confused...

Alec Teal

Exactly @PedroTamaroff

Pedro Tamaroff

@AlecTeal Where did you use linear algebra?

Alec Teal

I wanted a linear map $L:[ka,kb]\rightarrow[a,b]$

Pedro Tamaroff

02:29

@AlecTeal Just because you used the word "linear" doesn't mean you're using "linear algebra" ^_^

Alec Teal

Even

@PedroTamaroff changed it

then $a(L):[ka,kb]\rightarrow\mathbb{R}$

$a(L)=b$

I wanted to use the fact that a linear map like L is a bijection between the sets [ka,kb] and [a,b]

Harshdeep

Can someone explain whether all numbers can be summed u to be in Zp* or not.. The exact query is what if we chose a random number, r and raised it to say a. Can we calculate what that number was after taking mod p of answer.

r^a (mod p) = x. Can x be used to tell what a was....r is random number not a number from Zp*

Pedro Tamaroff

@AlecTeal You're thinking too hard.

Alec Teal

@PedroTamaroff it's really obvious, f(x/7) stretches f out 7 times.

Proving it though...

Anyway, @PedroTamaroff will you put that as an answer please (just C&P)

Pedro Tamaroff

@AlecTeal As a sidenote, using $a,b$ both for functions and real numbers is really untidy.

I'd rather use $f,g$ for functions, like always.

Alec Teal

02:34

@PedroTamaroff I just noticed that.... I don't know the letters I meant to use

One is a .... u with a | through it, the other is a u with a | where the top right of the u joins the top of the |

Mikael Öhman

Hmm.. sanity check: $\max_{x=[0,1]} |\arctan(u(x)) - \arctan(v(x))| \leq \max_{x=[0,1]} |u(x) - v(x)|$ is this valid?

Pedro Tamaroff

@MikaelÖhman Mean value theorem.

Alec Teal

@DanielFischer because my step function definition was a mouthful, using all and exists, how would you define one?

Pedro Tamaroff

$(\tan^{-1}x)'=\dfrac{1}{1+x^2}\leqslant 1$ everywhere.

anon

yes, KCd is keith conrad

Pedro Tamaroff

02:43

@anon Is he boss in his field?

Alec Teal

@PedroTamaroff if I have an interval I = [a,b] is there a notation for the open interval (a,b). Or do I just say "internal to"?

Last question I promise!

anon

I believe he is known best for his expository notes. I do not know about his research.

Pedro Tamaroff

@AlecTeal The notation $(a,b)$ is ubiquitous.

@anon Ah.

Twink

@anon how do you know it's him?

Alec Teal

@PedroTamaroff I never defined the interval in terms of a and b, i have a bunch of closed ones, $I_i$ I want to denote the open "version" of them.I'm guessing no notation, thanks.

anon

02:45

@Twink check his MO account associated to the MSE account

(for example)

Pedro Tamaroff

@AlecTeal $(a,b)=\{x\in{\bf R}:a<x<b\}$.

@anon Hey, I tell you about Halmos' discussion of the tensor product?

I guess you can enlighten me further.

Mikael Öhman

@Pedro thanks

Pedro Tamaroff

03:01

@TedShifrin Hey!

Fernando Martin

@PedroTamaroff: Today I mentioned what we were discussing yesterday about Set^op

It turns out that it has a more or less understandable interpretation

You can think of it as a subcategory of Rel

an arrow $f^{op}$ is just the relation formed by the pairs $(f(x), x)$

Pedro Tamaroff

@FernandoMartin Ah. Now that is not bad.

=)

Fernando Martin

Yes, but it's not a bad idea to get used to formally dualizing stuff either

Pedro Tamaroff

@FernandoMartin So if $f\subseteq A\times B$ then $f^{\rm op}\subseteq B\times A$ and $(a,b)\in f\iff (b,a)\in f^{\rm op}$.

@FernandoMartin Time to time.

Fernando Martin

Exactly

Pedro Tamaroff

03:10

@FernandoMartin Today I had the chance to talk about Eulerian numbers. They count stuff. But they also appear naturally when using the $x\dfrac{d}{dx}$ operator.

Fernando Martin

That's impressive

Where did you learn that?

Pedro Tamaroff

@FernandoMartin I read.

Fernando Martin

I know!

I mean, which book did you read that from?

Pedro Tamaroff

I honestly cannot remember.

Ted Shifrin

@Pedro: Hey (later)

Pedro Tamaroff

03:12

I think I found them in Polya & Szego again, @Fer.

@TedShifrin You're leaving? =/

Ted Shifrin

I did leave, I returned, and I'll soon leave again!

skullpatrol

We all have to leave sometime :-)

Ted Shifrin

Indeed @skull

Fernando Martin

@PedroTamaroff: I don't get your second-to-last line

Why is it that $D^k \frac{1}{1-x}$ is that sum?

Pedro Tamaroff

@FernandoMartin $$\sum\limits_{k = 0}^\infty {{x^k}} \mathop \to \limits^d \sum\limits_{k = 1}^\infty {k{x^{k - 1}}} \mathop \to \limits^x \sum\limits_{k = 1}^\infty {k{x^k}} \mathop \to \limits^d \sum\limits_{k = 1}^\infty {{k^2}{x^{k - 1}}} \mathop \to \limits^x \sum\limits_{k = 1}^\infty {{k^2}{x^k}} $$

Fernando Martin

03:18

Duh

You're absolutely right

Pedro Tamaroff

@TedShifrin Do you have a geometric interpretation of the tensor product?

@FernandoMartin You didn't even try! =D

Fernando Martin

I never thought of replacing $\frac{1}{1-x}$ with its series

Pedro Tamaroff

@FernandoMartin Ah.

Note that "occur" means something different that "se me ocurrió". =P

Fernando Martin

Haha, you're right

How should I say it?

I never thought of?

It doesn't have the same ring to it...

Pedro Tamaroff

@FernandoMartin I was going to go with that.

@FernandoMartin True.

Fernando Martin

03:21

@PedroTamaroff: remember the smooth homotopies problem?

Pedro Tamaroff

@FernandoMartin Ya.

Ted Shifrin

All over @Pedro, but it's too late to talk about vector bundles :)

Fernando Martin

It was slightly ill-posed

Pedro Tamaroff

@TedShifrin But I just started drinking my coffee!

@FernandoMartin ORLY?

Fernando Martin

Yeah, I'm not even sure the relation is transitive! To prove that, one would like to compose homotopies

Ted Shifrin

03:22

LOL ... But I'm teaching in 9 hours ... And I got up at 6 am

Fernando Martin

But we can't guarantee that the composition will be smooth in general...

Pedro Tamaroff

@TedShifrin Ah. What are you teaching?

Ted Shifrin

Compose or concatenate? @ Fernando

Fernando Martin

Concatenate would be the correct word, though my topology teacher calls it "vertical composition"

Ted Shifrin

Symmetric spaces and allied Lie algebra stuff

Fernando Martin

03:23

It isn't composition in the usual sense, that's for sure

Pedro Tamaroff

@FernandoMartin Who's teaching?

Fernando Martin

Minian is teaching topology, Cukierman is teaching geometry

Ted Shifrin

You can do it smoothly if you use bump functions

Fernando Martin

Both are good, but Minian is amazingly good

skullpatrol

What makes him so "amazing"?

Fernando Martin

03:26

He's a great teacher and he really knows how to motivate topics

Ted Shifrin

@Fernando: Maybe I'm missing something, but it's sorta standard to patch smooth homotopies with a bump function.

Fernando Martin

@Ted: I've never really studied bump functions so I can't know for sure

skullpatrol

The motivation behind topics is indeed something that is missing in most textbooks.

Ted Shifrin

Essential tool for differential topology, diff geometry, and analysis.

Fernando Martin

@TedShifrin: The problem also included some other restrictions on the homotopy

Ted Shifrin

03:30

Well, send it to me sometime ... :)

Fernando Martin

Haha, let's see if I can write it down without forgetting any hypotheses

Take the set of injective, smooth, regular curves in $\mathbb{R}^3$

Define to curves $\alpha$ and $\beta$ to be equivalent if there's a smooth homotopy $H$ between them, such that $H_t$ is an injective, smooth, regular curve for each $t$

Ted Shifrin

Isotopy, yes

Fernando Martin

Is there just one equivalence class?

oh, we consider our curves to have domain $[0,1]$

Ted Shifrin

My gluing comment works fine.

Closed curves, I assume?

Fernando Martin

That's great! Then I have a proof I guess

No, we consider them to be injective at endpoints too

Pedro Tamaroff

03:34

@TedShifrin Me wonders.

Ted Shifrin

Oh, weird.

Fernando Martin

@TedShifrin: If we allow closed curves then we have all sorts of knots there, I think

Ted Shifrin

Then can't we use straight-line homotopy?

Fernando Martin

We want $H_t$ to be injective

so with most curves the straight-line homotopy won't work

Ted Shifrin

Not most, but occasional, yes.

Wonders what, @Pedro?

Pedro Tamaroff

03:37

Suppose that $w$ is a bilinear form on $U\oplus V$.

Then for each $y_0\in V$ we obtain an element of $U^\ast$ by $\psi(u)=w(u,y_0)$.

Halmos asks: is it true that if $w$ is nondegenerate, then every linear functional in $U$ can be obtained in this way?

Ted Shifrin

Doesn't this depend on dimensions of $U$ and $V$?

Pedro Tamaroff

@TedShifrin My thought was the same.

Ted Shifrin

Unless you prove nondegeneracy implies such.

Fernando Martin

@Pedro: Are $U$ and $V$ finitely generated?

Pedro Tamaroff

@FernandoMartin I would think so.

Ted Shifrin

03:44

Isn't the book titled Finite Dimensional ...?

Fernando Martin

Isn't the map from $U$ to $U^*$ defined by $u\mapsto w(u,y_0)$ injective by non-degeneracy?

Ted Shifrin

Huh?

Pedro Tamaroff

@FernandoMartin The map is $y_0\to w(u,y_0)$.

Ted Shifrin

Yup.

Pedro Tamaroff

So $V\to U^\ast$.

Fernando Martin

03:47

Duh, you're right

Karl Kronenfeld

Well, the map $V\to U^\ast$ is injective, and similarly you can construct an injection $U\to V^\ast$.

Thus, the dimensions are equal.

Pedro Tamaroff

Wonders if the @anon is around.

Ted Shifrin

Why do you need @anon? :)

Pedro Tamaroff

@TedShifrin I like his explanations.

@KarlKronenfeld Cool.

Ted Shifrin

Pshaw. You can do this yourself! Check out my comment on math.stackexchange.com/questions/530513/…

Pedro Tamaroff

03:55

@TedShifrin No, I wanted to talk about tensors.

Not about this.

Ted Shifrin

Ah, ok :) we'll talk about Möbius strips re tensors this weekend :)

anon

@PedroTamaroff you rang?

Pedro Tamaroff

@anon Yeah.

anon

anything particular you want to know, or do you want an intuitive idea of what a tensor is?

Pedro Tamaroff

1 hour ago, by Pedro Tamaroff

@anon Hey, I tell you about Halmos' discussion of the tensor product?

(can)

anon

03:58

you mentioned that Halmos mentioned that his treatment is slightly unorthodox, but I don't recall anything else

Pedro Tamaroff

@anon Well, first he says the following.

Ted Shifrin

Night, all!

Pedro Tamaroff

(I will not quote, just talk.)

anon

night

Pedro Tamaroff

Consider $\Bbb C[s]$ and $\Bbb C[t]$. He says we would like to call $\Bbb C[s,t]=\Bbb C[s]\otimes \Bbb C[t]$, for example.

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$Mathematics$ Mathematics