Almost as good as the many $1=2$ proofs.

Is here someone who feels confident about tensor products?

"just ask, don't ask to ask"

(which is to say: if you have a question about tensor products, just ask it. if someone here has expertise and feels like answering it, they will)

@Semiclassical okey let me write it. I only don't wanted to spend too much time if everyone say "no I have no idea"

Our resident algebra experts are not around.

They're scarcer these days.

@Wave even if no one here knows now, someone may come along who knows. You can always post a permalink to the question here later.

what is a permalink?

not all the time, but once in a while

if you place your cursor just to the left of where your chat text starts, you'll see a dropdown icon you can click

if you right-click that and copy the link address, then you can paste that

e.g.

Our European algebraists have mostly disappeared.

that'll link directly to that part of the transcript

i sorta wish the dropdown icon was visible without having to put your cursor in the right spot, but i guess it'd be too visually noisy

click to the left of the comment

yeah, that

right-clicking has the advantage that you can copy the link address directyl

it's not the most intutive thing but once you get used to it it's fine

I always used to do it the draconian way. Getting a new page from which I copied the link. Stooopid me.

Hi, how can I show the straight lines through of $a,b,c$ and $d$ in $\mathbb{C}$ they're orthogonal if $arg\left( \frac{a-b}{c-d}\right)=\pm \frac{\pi}{2}$?

Why are you having difficulties?

Sorry I wrote it like this so it's easier to read

I know the hypothesis implies $\frac{a-b}{c-d}$ is is purely imaginary, then $\frac{a-b}{c-d}=ik$ with $k\in \mathbb{Z}$

Not $\Bbb Z$, just $\Bbb R$.

Ah in the last line I wanted to take $f$ outside of the sum.. I messed this up

@Event: So what is the issue?

@Wave You're writing things that don't make sense.

Rational multiples of elements of $A$ are not in $A$.

In particular, you cannot apply $f$ to them.

so $a-b=ik(c-d)$ and then $|a-b|=|ik|(c-d)$, but I don't know how to continue. Can I show the ortiogonality of the line straight @TedShifrin?

If $v=iw$, then the line spanned by $v$ is orthogonal to the line spanned by $w$. Remember what multiplication by $i$ means geometrically.

Hi

The complex multiplication means geometrically add angles and multiply lenght, so in this case we have a add angles $\pi/2$, then the lines $v=span(a,b)$ and $w=span(c,d)$ are orthogonal. Is it correct @TedShifrin?

Right. But writing $\text{span}(a,b)$ is wrong. The direction vector of the line is $\pm(a-b)$. I was writing $v$ and $w$ for complex numbers.

@TedShifrin sorry where do you mean?

ah I see

For example, in the final term of your displayed equation.

okey thanks!

Thank you @TedShifrin! It's more clear the geometric interpretation of the problem. I was thinking only the algebraic part :-(

so when I arrived to $a-b=ik(c-d)$ I didn't see the geometric part about the multiplication by $i$ factor here.

You're welcome, @Event.

Yes, one of the beautiful and powerful things about complex numbers is the mixture of algebra and geometry.

By the way, I had a problem with my previous MSE account related to email. I am the person who once asked here how to open a thermos flask with water. I don't know if you remember professor @TedShifrin?

By the way, I had a problem with my previous MSE account related to email. I am the person who once asked here how to open a thermos flask with water. I don't know if you remember professor @TedShifrin?

LOL, no, I don't remember. I might not have been here for that.

Well, maybe @copper.hat remembers me or @robjohn for the integrals as the user with a double life in AoPs. I don't frequent the site much anymore, because I'm working.

Also thank you because thanks to your vector calculus lectures which are very good @TedShifrin

But I need to practice more, of course

What was your previous name?

Alex

I remember

Now I'm event horizont haha

@Koro hi :3

koro found me in AoPS

Ah, I remember vaguely. I taught for AoPS for a few years, but I never hung out in their chats or whatever.

Hi @EventHorizon

Well, I was mentioning AoPs because that's where the issue of writing styles came up. I used to have more time to solve problems, but nowadays work keeps me a bit away from the forums.

Is AoPS good?

I am planning to buy a math olympiad course there

it is when ted's teaching it.

for everyone else, caveat emptor.

Ha ha ... I never taught their competition-style courses.

i remember the thermos incident

That is their name to fame.

Oh Are you guys mentors in AOPS?

@leslietownes Ohh

yes @Koro also the one who had difficulty in opening a thermos with water haha

@Belucat I think there are some problems in AoPs like the review. You post answers, but you don't necessarily get feedback. The texts are very good, the calculus one is brilliant.

Ohh books?

Do they ship books to India?

@EventHorizon

@leslietownes :3 hi Well, it's good to be back here. It's a nice community.

@EventHorizon I totally disagree with you. Their calculus text is sub-par.

Their strengths are in the lower-level stuff and the competition stuff. The precalculus and calculus were very disappointing to me.

Well, I won't compare it to a classic Spivack calculus, but it has a lot of problems. That's usually a good thing for me. But I'm not an expert.

Spivak was great

But not many problems ig

And few concepts I couldn't understand!

@"Not many problems"? WHAT?

@Belucat @Belucat I don't know. Maybe?

Sorry to interrupt your discussion. If I have a ring $R$ and a maximal ideal $m$ Then $R/m$ is a field. If in addition N is an $R$ module and I look at the tensor product (R/m)\otimes_R (mN) then we know from class that this is zero but without explanation. But is it zero because mN is an R/m-module? or why is this zero?

@Event It doesn't pretend to have any theory at all in it, but I think the better standard college calculus texts are better.

Yeah the limit problems given there were easy for me and I was actually looking for some harder problems

I approach the problem with RHL and LHL

That's usually a waste of time.

If you were working from an early edition of Spivak, there weren't as many problems. The last two editions added zillions of problems.

Ohh

I didn't know that

@Wave Try some simple examples. Always.

A classic in multivariable calculus is Stewart, I think.

@Belu I know that because (a) I taught the course 15 times and, more significantly, (b) I am the one who wrote hundreds of new problems for the later editions.

We're not talking multivariable. And Stewart is popular but not great.

Oh Great @TedShifrin

AoPS calculus is single variable. They ran out of energy and there aren't even interesting problems.

You are very much talented!

Nyc to meet you

Not necessarily talented. Just a mathematician/teacher for 50 years.

@TedShifrin Yes, well. It makes sense to me. So we'll stick with Spivak.

😊😊

Which edition u meant @TedShifrin

Spivak calculus

Spivak is a great book for the best students who want to write and wrestle with hard theory; it's not a book for everyone.

has lot of problems

Third and fourth ...

Thank you @TedShifrin

Actually, we added a lot to the second, but even more later in the book in the third and fourth.

@TedShifrin so I wanted to take $R=\Bbb{Z}$, $m=3\Bbb{Z}$, $N=\Bbb{Z} then I need to compute $\Bbb{Z}/3\Bbb{Z}\otimes 3\Bbb{Z}$ right?

But we have never computed such things. So the problem is we had this theory one lecture without seen it befor and the prof said it's easy theory we don't need to make examples ect

Hi everyone, is the following red circled matrix positive definite? I know it is symmetric but any special properties other than being symmetric.

K is positive definite.

a matrix of the form MM^T will be positive. it might not be positive definite.

i see a rank assumption below that line, that might imply it. is J square?

Generally speaking it is not square.

ok. so J is mxn. JJ^T is going to be mxm. is m <= n? in that case, J having "full rank" would be enough, because JJ^T will have rank m in that case.

What will happen to the positiveness property if the matrix loses its maximal rank?

n >= m for redundant robots.

OK, so if J has full rank, then JJ^T will be positive definite, at least for "redundant robots," whatever those are.

croco: it will stay positive in the sense of being a "positive" matrix (e.g. its quadratic form is nonnegative), but it will also have one or more 0 eigenvalues.

redundant robots have more joints than it needs to achieve a particular task. For example, in 3D we need 6 degree of freedom for rotation and translation. If the robot has more than 6 joints, then it is redundant.

redundant, but hopefully not surplus to requirements and back on the dole.

lol. it seems I can only prove it is stable but not asymptotically stable.

uh oh. i hope we programmed asimov's laws of robotics into it.

Safety is a higher priority in our lab. Endless instructions :<

I need permission to enter the lab and always our hands on the emergency button.

@leslietownes can I draw same conclusion about pseudoinverse of J?

@leslietownes thermos incident?

@CroCo an extra degree of freedom can prevent gimbal binding.

@Wave Yes. Use the properties of tensor product to move the $3$ from the right factor to the left.

@TedShifrin then I get $\Bbb{Z}/\Bbb{Z}\otimes \Bbb{Z}$ which is isomorphic to $1\otimes \Bbb{Z}$ right?

I must show that $F$ closed implies $F=\overline{F}$. Since I already know that it is always true that $F$ closed implies $F \subseteq \overline{F}$, it remains to show that $\overline{F} \subseteq F$. In order to do this, I noticed that $\overline{F} \subseteq F \iff F^c \subseteq (\overline{F})^c$. In a previous lemma, I proved that for any set $A$ it is $(\overline{A})^c=\text{int}(A^c)$, hence $F^c \subseteq (\overline{F})^c \iff F^c \subseteq \text{int}(F^c)$.

However by hypothesis $F$ is closed, hence $F^c$ is open and so $F^c=\text{int}(F^c)$. So the inclusion $F^c \subseteq \text{int}(F^c)$ is true, because set equivalence means both set inclusions true. Hence, the inclusion $(\overline{F})^c \subseteq F$ is true as well because equivalent to the inclusion $F^c \subseteq \text{int}(F^c)$. Is this proof correct?

Nope. You can’t cancel symbols without thinking. What is $3$ times any element of $\Bbb Z/3\Bbb Z$?

@TedShifrin 9.

:P

Close!

Oh, I forgot the brackets! $[9]$.

Now it’s right.

So I mean the elements of $\Bbb{Z}/3\Bbb{Z}$ are $[0],[1],[2]$. And then when I multiply them with $3$ I get $[0],[3],[6]$ or am I wrong?@TedShifrin

But what is the equivalence class of $6$?

its again $[0]$

There ya go.

Ah so I always get $[0]$

So $3[x] = [0]$ for any $[x]\in \mathbb{Z}/3\mathbb{Z}$, no?

And therefore my tensor product is zero

Remember, you are working with equivalence classes here---don't get hung up on representatives of those classes.

sorry what do you mean by that

You first wrote that multiplying by $3$ gives $[0], [3], [6]$. These are all the same object, but you are representing them in different ways.

don't assume that just because you have two different representatives, that you must have two different calsses

The fact that you are representing them in different ways appears to be a point of confusion. Focus on the classes, not the specific representatives of those classes.

Exactly.

Ah sure yes I understand. No I only wanted to be clear in orther that I'm wrong so that you see where the mistake is. I knwo that they are the same class

Now prove the general result.

@TedShifrin So I mean I take the $m$ from the left to the right using the properties of tensor right? so I get $m(R/m)\otimes N$. But then an equivalence class in $R/m$ is of the form $r+m$ for some $r\in R$ then $m(r+m)=mr+m$ but $mr=0$ in $R/m$ hence we get $0$ on the right hand side of the tensor product

But $m$ is an ideal, not necessarily principal. So be precise and correct.

but taking it on the other side is correct right?

But sorry where did I used that it's principal?

@TedShifrin do you mean when I multiply m with m?

@robjohn :-)

Unless you’re just pushing symbols around without understanding, you need to pick an arbitrary element of the tensor product and show that it is $0$.

@TedShifrin yes you are right I'm a bit lost since I have never computed with tensor product neither in the exercise class nor in the lecture. So I'm really sorry for the stupid questions. Do I need to start again from the beginning?

Do I need to work with the universal property or with the properties

@copper.hat Yes? Was that in reply to something?

@Wave Properties. The example we did shows you how to proceed. Just pick a general element of the tensor product.

@TedShifrin If I pick a general element of the tensor product then it is of the form $(r+m)\otimes mn$ for some $r\in R$ and $n\in N$ is this right?

Then I would use the property that I can write this as $r\otimes mn+m\otimes mn$

oh sorry I should take another letter instead of $m$ since this denotes the ideal. let me take x

so I have $r\otimes xn+x\otimes xn$ @TedShifrin

An element of $R/m$ is an equivalence class $[r]=r+m$ (sorry yes I'm lost with notation...)$ And an element of $mN$ is $mn$?

@TedShifrin am I again totally wrong I think

$m$ is an ideal, not an element. I’m fine with the $[r]$.

But $mN$ is really the multiplication of an ideal with an $R$-module right?

I think you're trying to prove something false @Wave, generally $R/m\otimes_RmN$ is nonzero.

If you want to transfer scalars across $\otimes$ to get $0$, then you are not computing in $R/m\otimes_RmN$ but rather $R/m\otimes N$.

In general, you have $R/I\otimes_RM\cong M/IM$ for ideals $I$ and $R$-modules $M$, and this allows you to figure out what $R/m\otimes_RmN$ is. (E.g., consider the case $N=R$).

@KarlKroningfeld hmm strange but in the notes they also said it is zero

If $N$ is finitely generated and $R$ is Noetherian, then the tensor product is $0$ if and only if $mN=0$ already, by Nakayama's lemma.

@KarlKroningfeld sorry I need to take a look at Nakayama again I don't know this by heart

Hmm sorry I don't have such a nakayama lemma. I only have the one saying that if IM=M then the annihilator...

You have to localize at $m$ first to apply Nakayama's lemma.

I'm just saying, what's in your notes is false.

@KarlKroningfeld okey now I'm absolutely confused. I thought about understanding something on tensors where no one before told me it is wrong and then this... sad

Hi,

I am reading about probability mass functions and then I stumble upon this footnote:

What is “quotienting”?

at this level of generality, 'regarding distinct things as the same' might be the best way of looking at it

you could be much more precise in this setting but it's not gonna help or really give the full idea of the notion

@Wave Math is hard lol. I jumped on it mainly because I remembered making a similar mistake before.

@KarlKroningfeld okey so I need to talk to the assistant because then a whole proof he gave us in the exercise class does not work

@leslietownes was this meant for me?

@Karl: I’m totally confused, then, by the example Wave and I did at the beginning.

My algebra is rusty, but not that rusty!

schn: yes

$\mathbb Z/3\mathbb Z\otimes 3\mathbb Z$ is cyclic of order $3$.

Because $3\mathbb Z\cong \mathbb Z$ as $\mathbb Z$-modules.

How so?

But every element I take is 0.

I'm guessing you were computing in $\mathbb Z/3\mathbb Z\otimes \mathbb Z$ and not $\mathbb Z/3\mathbb Z\otimes 3\mathbb Z$

Everything on the right of the tensor has to stay a multiple of $3$ if you're computing in $\mathbb Z/3\mathbb Z\otimes 3\mathbb Z$.

Yes, I see your point. I’m thinking inside the first tensor product. Very subtle. I knew this 50 years ago.

Sorry could maybe someone explain me what was the problem of our example?

Read what Karl just wrote.

But can't I take the 3 on the other side using the properties of tensoring

Read the second sentence. You have to stay in $3\Bbb Z$, not $\Bbb Z$.

ooookey I don't know why we need this because I mean $3$ is an element of the ring and we had the property that $rm\otimes n=m\otimes rn$ where $r$ is in the ring and $m,n$ in some R-modules $M,N$ but I need to ask the assistant

this would be a huuuuuuuge mistake because everything is wrong then

But here $N$ is multiples of $3$, so you can’t strip off the $3$.

but with elements I could do it

When you factor out $3$ from $3\cdot 1$, then $1$ is no longer in your module $N$.

but there is the property that $r(m\otimes n)=rm\otimes n=m\otimes rn$ right

Sure..

okey thank you very much for your time. I really think I need to talk to the assistant.

Here $m$ is in $\Bbb Z/3$ and $n\in 3\Bbb Z$.

Yes I see now$ $3$ is not an own R-module

Yeah, talk to the assistant. I’m sorry I made it worse.

No no problem I mean I still got some Ideas how to work with it even if the basic idea fails there are some right steps if we would not have overseen the fact karl mentionned

For me it still was a useful nice discussion

in some points;)

So see you!

I’ll stick to differential geometry now :)

Thanks @Karl

@copper.hat Ah, yes. I'd forgotten about that.