Mathematics - 2013-10-19 (page 1 of 3)

I havn't really done measure theory so I am not sure how they are concluding their proof. Does anyone have a proof where it doesn't use measure theory

Ted Shifrin

You need to understand Fubini's Theorem. But the first bit after your line is analogous to saying $\sum c_ke_k=0$ with $e_k$ a basis implies all $c_k=0$.

So @Pedro, how do you understand tensor products?

Pedro Tamaroff

@TedShifrin I don't really know.

Halmos conviced me that $\Bbb C[t]\otimes \Bbb C[s]=C[s,t]$ and that $\Bbb C^n\otimes \Bbb C^m=\Bbb C^{mn}$, say.

Ted Shifrin

12:43 AM

Do you understand $V\otimes_{\Bbb R}\Bbb C$ when $V$ is a real vector space?

Pedro Tamaroff

@TedShifrin Isn't that what Halmos calls the complexification of $V$?

Ted Shifrin

Yes.

Pedro Tamaroff

Oh, I read about it briefly.

Ted Shifrin

It's called extension of scalars in commutative geometry.

Pedro Tamaroff

@TedShifrin Why do you write $\otimes_{\Bbb R}$ instead of $\otimes$ alone?

masfenix

12:46 AM

@TedShifrin I understand fubini so I understand the change of the order of integrand. Thanks for pointing out that $c_k = 0$ in your statement. That was key.

Ted Shifrin

Because we're tensoring $\Bbb R$ modules.

Pedro Tamaroff

@TedShifrin So basically we put complex numbers in front of the elements of $V$ and multiply this coordinate to obtain a $\Bbb C$-vector space.

Ted Shifrin

Here's another example, @Pedro. Take the vector space of differential $k$-forms and tensor with a vector space $E$.

Pedro Tamaroff

As in $w(z\otimes v)=wz\otimes v$.

Ted Shifrin

Yes, @Pedro.

This can be interpreted as $E$-valued $k$-forms.

Now here comes the fun.

Pedro Tamaroff

12:50 AM

@TedShifrin How?

Ted Shifrin

Well, choose a basis for $E$, say $e_i$. Then you have an element $\sum\omega_i\otimes e_i$

Pedro Tamaroff

@TedShifrin Aha.

And?

Ted Shifrin

So when you feed it $k$ vectors (as always with $k$-forms) it spits out a vector in $E$ — hence a form with values in $E$

Now here's some geometry/topology.

Pedro Tamaroff

@TedShifrin The form is in $\Bbb R^n$?

Ted Shifrin

Sure, or, more generally, a manifold.

Pedro Tamaroff

12:54 AM

OK.

90intuition

Trying to understand this question: How many elements does the ring $ℤ[X]/(X^2-3,2X-4)$ have?

Pedro Tamaroff

@90intuition Could you link instead of that?

90intuition

This is an answer:

In the quotient ring, we have $X^3 = 3$ (I willl continue to write $X$ to denote the image of $X$ in the quotient)). Thus $(X-2)(X+2) = X^2 - 4 = - 1,$ and so in the quotient we have $X-2$ is a unit. Thus the equation $2(X-2) = 0$ simplifies to $2 = 0$. Thus the quotient is equal to $(\mathbb Z/2\mathbb Z)[X]/(X^2 -3)$.

Now $\mathbb Z/2\mathbb Z$ is a field of two elements, and in particular in this field $(a+b)^2 = a^2 + b^2$. Using this, we see that $X^2 - 3 = X^2 + 1 = (X+1)^2$. If we make a change of variables $T = X+ 1$, then we can write the quotient as $(\mathbb Z/2\mathbb Z)[T]/(T^2).$

Pedro Tamaroff

@TedShifrin So if I have a form $\omega $ and a vector $v=\sum \alpha_ie_i$ I can write $\omega\otimes v=\sum\alpha_i( \omega\otimes e_i)$

Karl Kronenfeld

@90intuition I read it, what do/don't you understand?

Ted Shifrin

12:56 AM

$\alpha_i$ inside the sum?

Pedro Tamaroff

@TedShifrin The coordinates of $v$.

Ted Shifrin

Right ... Put them inside the sum.

Pedro Tamaroff

@TedShifrin Ah, sorry.

90intuition

@Karl I understand most steps. I understand that $X^2=3$ and $2X-4=0$ imply that $2=0$. Therefore all the coefficients are in $ℤ/2ℤ$. But I dont see why this implies that $ℤ[X]/(X^2-3,2X-4)=ℤ/2ℤ[X]/(X^2-3)$. Why can you leave out $(2X-4)$ ?

Pedro Tamaroff

@TedShifrin And $\omega$ is supposed to be $\omega:\Bbb R^n\to \Lambda^k(\Bbb R^n)$, right?

Ted Shifrin

12:59 AM

So, the idea so far is that we can take something like a vector and give it "twisted" coefficients, @Pedro. ... Yes.

Pedro Tamaroff

And when I feed that a vector I get an alternating $k$-form times basis vectors?

Ted Shifrin

No, you evaluate at a point, then feed $k$ vectors.

Pedro Tamaroff

@TedShifrin Yes, that is what I meant, feed a vector $\Bbb R^n$, to get a $k$-form.

Then I can go around an evaluate.

Ted Shifrin

Thinking of the first vector as a point, yes.

Pedro Tamaroff

@TedShifrin Well, yes.

Still, it is weird. =O

Karl Kronenfeld

1:01 AM

@90intuition The expression "all of the coefficients are in $\mathbb Z/2\mathbb Z$" is nice and intuitive, but it isn't the formal picture I have in mind. You have a map $$\mathbb Z[X]\to\mathbb Z[X]/(X^2-3)$$ and $(X^2-3,2X-4)$ maps to $(2)$. Then pullback the ideal $(2)$ to $(2,X^2-3)$.

Ted Shifrin

Ok, moving to more geometry. Möbius strip.

Pedro Tamaroff

@TedShifrin Buckles up.

Ted Shifrin

Think of this as an open interval moving around a circle. If it didn't twist, you'd have a cylinder.

Pedro Tamaroff

@TedShifrin Aha.

Perpendicular, yes?

90intuition

@KarlKronenfeld pullback ?

Karl Kronenfeld

1:04 AM

@90intuition Inverse image.

"Pullback" is probably technically incorrect anyway.

90intuition

But why are we looking at this map: $\mathbb Z[X]\to\mathbb Z[X]/(X^2-3)$ ?

Ted Shifrin

Sure. Think of the open interval as $\Bbb R$. Cylinder is direct product $S^1\times \Bbb R$, Möbius twists once as we go around.

Pedro Tamaroff

@TedShifrin Aha.

@TedShifrin So at every points of $S^1$ we're rotating $I$.

Karl Kronenfeld

@90intuition Because establishing $(X^3-3,2X-4)=(2,X^2-3)$ in $\mathbb Z[X]$ is the same as establishing $(2X-4)=(2)$ in $\mathbb Z[X]/(X^2-3)=\mathbb Z[\sqrt 3]$.

Ted Shifrin

This is called a line bundle (one-dim vector spaces param by the circle) on the circle

Yes @Pedro.

Pedro Tamaroff

1:07 AM

@TedShifrin Wait, but how is this a one dimensional vector spaces?

Ted Shifrin

I said to think of the interval as $\Bbb R$

Karl Kronenfeld

@90intuition Really, this is just how I look at it. It's easier for me to work with $\sqrt 3$ instead of $X$.

Pedro Tamaroff

@TedShifrin Oh, you mean each lines is a one dim vector space?

Ted Shifrin

Yup.

90intuition

But establishing that $(2X-4)=(2)$ in $ℤ[\sqrt{3}]$ sounds strange :P

Or should I say that $(2 \sqrt{3} -4 ) = (2)$ ?

Kevin Driscoll

1:09 AM

@Pedro @Ted Good evening

Ted Shifrin

Exercise: Topologically, this is $S^1\times\Bbb R$ mod $(x,t)\sim (-x,-t)$

Hi, @Kevin... pedro is learning line bundles :)

Karl Kronenfeld

@90intuition When you reinterpret $X\mapsto\sqrt3$, you change notation $(2X-4)\mapsto(2\sqrt3-4)$ simultaneously.

That's hard to do in writing though.

Kevin Driscoll

@Ted Oh dear.....

I'm learning about black holes and the firewall problem

90intuition

What you do mean with hard to do in writing ?

Ted Shifrin

Better you than I.

Karl Kronenfeld

1:11 AM

@90intuition I can't represent that change immediately in writing, but I am still imagining it.

Kevin Driscoll

@Ted I've decided that these people know a lot of things

@Ted but almost none of them are about quantum gravity or black holes

Ted Shifrin

Which people?

@Pedro: Have you keeled over! :)

Pedro Tamaroff

@TedShifrin Sorry, I'm back.

Fernando Martin

Hey hey hey

Kevin Driscoll

@Ted The physicists who do research in this area

Ted Shifrin

1:14 AM

You can't walk out on me, @Pedro :D

Pedro Tamaroff

@TedShifrin I'm drawing now. =)

Ted Shifrin

Tensor will come back soon @Pedro

90intuition

But why is $(2 \sqrt{3} -4 ) = (2)$ in $ℤ[\sqrt{3}]$ ?

Pedro Tamaroff

@TedShifrin I'm trying to figure that out.

@Fer Heya.

Fernando Martin

What's up @Pedro?

Karl Kronenfeld

1:16 AM

@90intuition First $(2\sqrt3-4)\subseteq(2)$.

Pedro Tamaroff

@FernandoMartin I'm trying to learn about tensors.

Karl Kronenfeld

@90intuition Now, prove $\sqrt3-2$ is invertible, so that the inclusion is actually equality.

Ted Shifrin

Hi @Fernando.

Fernando Martin

Hi @Ted

90intuition

If $x \in (2 \sqrt{3}-4)$, then $x = (a+b\sqrt{3})(2 \sqrt{3}-4) = ...$ lots of terms :p

Pedro Tamaroff

1:19 AM

@TedShifrin I cannot see how the Möbius strip is the cylinder quotient that.

Karl Kronenfeld

@90intuition Just factorize $2\sqrt3-4=2(\sqrt3-2)$.

Pedro Tamaroff

You identify antipodal points?

Karl Kronenfeld

@90intuition I was really hinting at how you would discover $\sqrt3-2$, which you want to show is invertible.

Ted Shifrin

So, here's where we're headed, @Pedro. Think of our Möbius strip as a line $L_x$ for each $x\in S^1$. Yes.

Pedro Tamaroff

@TedShifrin Yes, that I got.

Ted Shifrin

1:21 AM

We're going to instead do $L_x\otimes L_x$. Then what do we get for our space?

90intuition

Why does the invertability of $\sqrt{3}-2$ imply the equality ?

Ted Shifrin

@Pedro: Do you know how to make a Möbius strip out of a rectangle?

Pedro Tamaroff

@TedShifrin What does $L_x\otimes L_x$ mean?

@TedShifrin Yes.

Fernando Martin

@Pedro: I learned a little about tensors in Algebra II, but only from a purely algebraic perspective. I'm going to read this through to see if I can gain some geometric perspective :)

Pedro Tamaroff

You take two sides, draw opposing arrows and connect appropriately.

Ted Shifrin

1:22 AM

You flip the interval upside down and glue. That's what that equiv reln does.

Pedro Tamaroff

@TedShifrin But you already have the cylinder. Shouldn't it be $(s,t)\sim (s,-t)$?

Ted Shifrin

I hate teaching remotely! :)

Karl Kronenfeld

@90intuition Let $x$ be the inverse of $\sqrt3-2$; then, $2=2(\sqrt3-2)x$ implies $2\in(2\sqrt3-4)$ and moreover $(2)\subseteq(2\sqrt3-4)$.

Fernando Martin

Oh, forms.

crawls back to where he came from

Ted Shifrin

Try $[0,1]\times \Bbb R$ with $(0,t)\sim (1,-t)$.

90intuition

1:24 AM

Okay I see.

It has inverse $-\sqrt{3}-2$

Ted Shifrin

Then convince yourself what I gave you is homeo to that, @Pedro.

Pedro Tamaroff

@TedShifrin Yes, that is OK.

@TedShifrin I don't understand what you use all $\Bbb R$.

Why not $[0,1]\times [0,1]$?

Ted Shifrin

I'm using $\Bbb R$ because I want a vector space for each $x$.

90intuition

@KarlKronenfeld So we have the equality. And therefore $(2X-4)=(2)$ in $\mathbb Z[X]/(X^2-3)=\mathbb Z[\sqrt 3]$. And therefore $(X^3-3,2X-4)=(2,X^2-3)$ in $ℤ[X]$. Wohooo

Karl Kronenfeld

:)

Pedro Tamaroff

1:28 AM

@TedShifrin OK, but then the thing is a little bit nastier. It crosses itself and goes off to infinity up and down, like a double cone, for example.

Right?

90intuition

Thanks for helping me out :) @KarlKronenfeld

Ted Shifrin

No ... Do it abstractly, not in $\Bbb R^3$

Pedro Tamaroff

@TedShifrin What do you mean?

Karl Kronenfeld

@90intuition You're welcome. :) It might be interesting to see how Matt E responds to the comment on his answer with a similar question.

Ted Shifrin

Think of intervals in your visualization, but for vector space structure the interval is $\Bbb R$.

Pedro Tamaroff

1:30 AM

@TedShifrin OK.

90intuition

@KarlKronenfeld Yes, indeed. Actually a classmate of me asked that question.

Pedro Tamaroff

Now you said $S^1\times \bf R$ with $(s,t)\sim (-s,-t)$.

Ted Shifrin

I done did.

Note: what is $S^1/x\sim -x$?

90intuition

Do you also understand why he doesn't write $(ℤ/2ℤ)[T-1]/(T^2)$ but $(ℤ/2ℤ)[T]/(T^2)$ after doing that substitution ?

Pedro Tamaroff

@TedShifrin Come again?

Ted Shifrin

1:33 AM

If you identify antipodal points on $S^1$ what do you get?

Karl Kronenfeld

@90intuition You map $F_2[T]\to F_2[X]/(X^2-3)$ by sending $T$ to $X+1$. Then $T^2\mapsto (X+1)^2=X^2-3$.

Pedro Tamaroff

Doesn't $(s,t)\sim (-s,-t)$ collapse the spheres into lines?

Ted Shifrin

No.

Pedro Tamaroff

Hm.

Ted Shifrin

No. Think upper semicircle and lower ...

Pedro Tamaroff

1:35 AM

@TedShifrin Well, isn't that a line?

Ted Shifrin

Endpoints?

Pedro Tamaroff

@TedShifrin Ah, it has none. So it is another circle.

Ted Shifrin

Right!

90intuition

@KarlKronenfeld And then show that this a surjective ringhomomorphism with kernel $T^2$ ?

Pedro Tamaroff

@TedShifrin You kinda twisted the circle into itself.

Like turn, get a figure eight and fold?

Ted Shifrin

1:37 AM

Right.

Karl Kronenfeld

@90intuition Yes. I actually meant to write $F_2[T]\to F_2[X]$.

Pedro Tamaroff

@TedShifrin Cool.

90intuition

But he is saying: $(\mathbb Z/2\mathbb Z)[X]/(X^2 -3)=(\mathbb Z/2\mathbb Z)[X]/(X+1)^2 = (\mathbb Z/2\mathbb Z)[T]/(T)^2$

Pedro Tamaroff

@TedShifrin I need to see how this is a Möbius strip.

90intuition

with as argument for the last step to use the change of variable $T=X+1$

that sounds a lot quicker, but I don't really see it

Alec Teal

1:39 AM

@TedShifrin Do you recommend anything for the analysis of integration? Note wise?

Ted Shifrin

Draw your interval centered along the circle, @Pedro.

90intuition

Why you can change the variable, and you don't have to write: $(ℤ/2ℤ)[T-1]/(T)^2$

Karl Kronenfeld

@90intuition Because it really is a map like I gave above in conjunction with the lattice theorem for rings.

Ted Shifrin

@alec: I'm not acquainted with on-line texts. What level?

Pedro Tamaroff

@TedShifrin OK.

90intuition

1:41 AM

I don't know the lattice theorem for rings.

Alec Teal

@TedShifrin any, I'm gathering resources. For practice. If I find one useful line it would have been worth it.

I'm searching for "real analysis integration pdf"

But you might know something good :)

Karl Kronenfeld

@90intuition You actually use it all the time. It says that a surjective homomorphism $\varphi:R\to S$ of rings induces a one-to-one order-preserving correspondence of ideals by $J\subseteq S\mapsto \varphi^{-1}(J)\subseteq R$.

Ted Shifrin

You mean Lebesgue, @Alec?

Pedro Tamaroff

@TedShifrin Err, cannot see it.

90intuition

@KarlKronenfeld I don't see how I use that all the time :P ?

Alec Teal

1:44 AM

@TedShifrin maybe, Riemann seems more applicable.

Pedro Tamaroff

@TedShifrin I don't see how all the new circles get together.

Ted Shifrin

When you get back around your circle, the interval is glued to itself upside-down, @Pedro.

90intuition

@KarlKronenfeld It is surjective ? And injective for the ideals ? I don't really get that statement I think.

Karl Kronenfeld

@90intuition It is surjective to the set of all ideals of $R$ containing the kernel of $\varphi$.

Ted Shifrin

Ah, @Pedro, only the one for $t=0$ glues back up!

Karl Kronenfeld

1:46 AM

@90intuition It also injective yes.

Ted Shifrin

I don't understand where you are and what you want to apply to, @Alec.

Karl Kronenfeld

But it (and the isomorphism theorems) is the reason why you can say that $\varphi:F_2[T]\to F_2[X]:T\mapsto X+1$ determines an isomorphism $\psi:F_2[T]/(T)^2\to F_2[X]/(X^2-3)$.

Alec Teal

@TedShifrin I was just looking for lecture notes really. I've got plenty of search results, don't worry about it, thanks.

90intuition

@KarlKronenfeld Do you mean $\varphi$ is an injective map ?

Karl Kronenfeld

@90intuition No.

Pedro Tamaroff

1:47 AM

@TedShifrin Oh, I was reading it all wrong, but I cannot imagine the quotient. Give me a second.

90intuition

It is injective for the ideals ?

Karl Kronenfeld

@90intuition Yes.

90intuition

Ah okay.

@KarlKronenfeld Aaaaaaah. That's why !

Pedro Tamaroff

@TedShifrin I cannot see how the quotient will not be cylindrical.

Ted Shifrin

Because $(1,1)\sim (-1,-1)$, so when you identify the endpoints of the upper semicircle, the interval glues to itself upside-down.

90intuition

1:55 AM

But wait a second, I'm not sure I really understand. So you $\varphi$ is surjective and injective for the the ideals. So an isomorphic for those ideals ? Something like that ? But why do you get the isomorphism : $\psi:F_2[T]/(T)^2\to F_2[X]/(X^2-3)$ ?

Karl Kronenfeld

@90intuition The kernel of $F_2[T]\xrightarrow{\varphi}F_2[X]\to F_2[X]/(X^2-3)$ is exactly $(T)^2$ by the lattice theorem.

90intuition

Aaah cool. Thanks for explaining that:) I'm going to bed now.

Ted Shifrin

Night @90.

Karl Kronenfeld

@90intuition Goodnight.

Pedro Tamaroff

@TedShifrin OK, I think I kinda see it. But I find this strange.

AGH.

Ted Shifrin

2:04 AM

If I understand you, yes.

nsanger

If we have a function $f$ that has only removable discontinuities, and the function $g$ defined as $g(a) = \lim_{y \to a}f(y)$, the problem asks to prove that $g$ is everywhere continuous. The solution to this problem states that the epsilon-delta definition tells us that $|y-a|<\delta \implies |f(y)-g(a)|<\varepsilon$, and this second inequality expands to $g(a) - \varepsilon < f(y) < g(a) + \varepsilon$.

They then say that if $|x-a| < \delta$ we have $g(a) - \varepsilon \leq \lim_{y\to x}f(y) \leq g(a) + \varepsilon$, which implies $g$ is continuous. Can anyone explain to me why this substitution from $f(y)$ to the limit in the last inequality works? This result makes some intuitive sense ($f(y)$ is bounded, and therefore since the limit exists it will be bounded as well for certain values), but I can't see how to make a rigorous argument out of it.

Pedro Tamaroff

@nsanger If $f(y)<C$ then $\lim_{y\to x}f(y)\leqslant C$.

Ted Shifrin

Go back to $[0,1]\times\Bbb R/(0,t)\sim (1,-t)$.

Pedro Tamaroff

@TedShifrin OK, let's try to move on.

Ted Shifrin

Well, what I wanted you to do was to replace $L_x$ with $L_x\otimes L_x$.

Pedro Tamaroff

2:07 AM

@TedShifrin But what on Earth is $L_x\otimes L_x$?

nsanger

@Pedro That doesn't seem right. What about a function $f$ that is equal to $x$ for $x \neq 10$, and then is equal to $0$ at $x = 10$. $lim_{x \to 10}f(x) = 10$, but $f(10) = 0$, so your inequality doesn't hold there, right?

Ted Shifrin

It's still a $1$-dimensional vector space, right?

Pedro Tamaroff

@TedShifrin Well, $1\times 1=1$.

I can give you that.

Ted Shifrin

Thank you :)

Pedro Tamaroff

@nsanger How does it not hold?

More precisely, suppose that $f(y)\leqslant C$ for $y$ in a nbhd of $x$. Then $\lim_{y\to x}f(y)\leqslant C$.

Ted Shifrin

2:11 AM

Imagine $L_x$ twisting by $e^{\pi ix}$ as $x$ goes from $0$ to $1$.

nsanger

@Pedro Ah, I see. I wasn

Pedro Tamaroff

This is analogous to $a_k\leqslant C$ then $\lim_{k\to\infty}a_k\leqslant C$.

nsanger

@Pedro I wasn't sure what $x$ represented, I guess.

Pedro Tamaroff

@TedShifrin Why should it twist?

nsanger

@Pedro So this is what I was talking about when I meant I understood the intuition behind it. I just need to make it into a formal argument.

Pedro Tamaroff

2:12 AM

I see $L_x\otimes L_x$ as pure symbolism for the time being.

Ted Shifrin

Because it does as you go around the circle. And when you're done it's turned through $\pi$ when you glue.

Pedro Tamaroff

$L_x$ alone?

Ted Shifrin

Well, $L_x\otimes L_x$ must twist twice as fast, because the twist factor squares! :)

Yes to your last.

Pedro Tamaroff

@TedShifrin Well, as I said I just see $L_x\otimes L_x$ as symbols. I don't see why they should represent something.

Ted Shifrin

Well, remember all these one- dim vector spaces are gluing together to give us our topological space.

Pedro Tamaroff

2:16 AM

@TedShifrin I am compelled to say you get a Möbius strip with two twists.

But I'm just nodding here.

Ted Shifrin

Compelled, but right!

You asked how I think about tensor products. One way I do is in terms of twisting vector spaces like this.

I'll be glad to teach you a rigorous course off-line sometime :)

nsanger

@Pedro So how can you make your statement earlier into a formal proof?

Ted Shifrin

We need the notion of local trivialization and transition functions to rigorize. I'm sure Mariano can explain better than I :)

Pedro Tamaroff

@TedShifrin I see. I just cannot yet figure out what $V\otimes W$ should stand for given $V,W$.

Ted Shifrin

Suppose $V$ is $e^{i\theta}$ times $\Bbb R$ and $W$ is $e^{i\phi}$ times $\Bbb R$.

Pedro Tamaroff

2:23 AM

@TedShifrin The angles are fixed?

Ted Shifrin

Tensor allows us to pull the multipliers out and multiply the real numbers.

Yes.

Pedro Tamaroff

@TedShifrin So you get two lines.

Ted Shifrin

No, we get one line.

Tensor, not sum.

Pedro Tamaroff

I mean $e^{i\theta}\Bbb R$ is a line in $\Bbb C$.

Ted Shifrin

Oh, sorry.

Pedro Tamaroff

2:26 AM

@TedShifrin So you mean $e^{i\theta}\Bbb R\otimes e^{i\phi}\Bbb R=e^{i(\theta+\psi)}\Bbb R$?

Ted Shifrin

Yup.

Pedro Tamaroff

@TedShifrin How can you see that?

anon

I find it restrictive to think in terms of modulus 1 complex numbers alone. My mental visual is that of a hyperbola (since the pairs $(a,b)$ s.t. $a\otimes b$ is a given pure vector form a hyperbola), but complicated by the fact that the scalar ring over which we are tensoring may be varied on demand.

Ted Shifrin

@anon: I'm trying to explain tensoring as twisting, in the context of line bundles, but using the rotation to represent the twisting.

anon

oh

Ted Shifrin

2:30 AM

Avoiding coherent sheaves :) just real line bundles on the circle.

Pedro Tamaroff

@TedShifrin So...?

Ted Shifrin

@Pedro: I've tried to be heuristic here. I'm suggesting thinking of these lines as sitting in a fixed copy of $\Bbb C$ that rotates through angle $\pi$ as we go once around the circle. It turns twice as fast when we tensor square, three times when we cube, etc.

Pedro Tamaroff

@TedShifrin I was talking about $e^{i\theta}\Bbb R\otimes e^{i\phi}\Bbb R=e^{i(\theta+\psi)}\Bbb R$.

Ted Shifrin

Yes, I know. I guess I have problems making that formal in an algebraic way. i may have to teach you trivializations and transition functions to do so. Let me ponder to see if I can give a cheap rigorous out.

Pedro Tamaroff

@TedShifrin Wait wait.

My point is that what I wrote is a guess.

And well, the $=$ should be $\simeq$.

My point is how to see it, from the definition of what $\otimes$ is.

Ted Shifrin

2:40 AM

Yes, I understand. What we do is tensor the isomorphisms that make the space look like $U\times \Bbb R$ on two pieces of the circle.

I was trying to bypass that with heuristics ... Never a good idea with you :)

Pedro Tamaroff

I give up for the day. I need to wake up early tomorrow.

This will just upset me further.

Ted Shifrin

Yeah, I was asking for time out, too. :) tennis lessons?

Pedro Tamaroff

@TedShifrin Yes.

Ted Shifrin

Please don't be upset. It's my failure to communicate appropriately.

Pedro Tamaroff

@TedShifrin Well, I am not angry.

I'm just unsettled because I don't understand $\otimes$.

Ted Shifrin

2:43 AM

Well, it takes years ...

Good night :)

Pedro Tamaroff

Bye.

Thanks.

Don Larynx

Is $(0, 1), (0.5, 1.5)$ an open cover of $(0.2, 1.2)$?

Fernando Martin

2:59 AM

@DonLarynx: why not?

Don Larynx

just making sure

J L

6:24 AM

hey guys

is there a way to confirm whether one answer is equivalent to another (say integral answers) automatically like say on wolfram alpha or something? i keep answering these trig integrals and always get dif answers than in solutions manual but dont wanna have to ask whether each one of these is right or wrong on this site... is there a way to confirm two answers are equal or not?

anon

7:48 AM

@JL use trig identities and algebra manipulations

Chris's sis

10:52 AM

Greetings amazing people!

Two things are hard to bear for any human being: 1) to be deprived of love and 2) to live a life without meeting the beauty of math.

skullpatrol

11:34 AM

Greetings

:-)

Mathematical beauty

Mathematical beauty describes the notion that some mathematicians may derive aesthetic pleasure from their work, and from mathematics in general. They express this pleasure by describing mathematics (or, at least, some aspect of mathematics) as beautiful. Sometimes mathematicians describe mathematics as an art form or, at a minimum, as a creative activity. Comparisons are often made with music and poetry. Bertrand Russell expressed his sense of mathematical beauty in these words: Mathematics, rightly viewed, possesses not only truth, but supreme beauty — a beauty cold and austere, like...

Charlie

12:39 PM

Hi @skull there?

skullpatrol

@Charlie hi there :-)

@Charlie how are you?

Charlie

@skullpatrol :D how are you?

@skullpatrol good

skullpatrol

@Charlie Fine thanks for asking.

Charlie

@skullpatrol :) it's a pleasant morning here

skullpatrol

@Charlie sunny?

Charlie

12:46 PM

@skullpatrol yipyipyip not very hot

skullpatrol

@Charlie :D

Charlie

@skullpatrol :D

skullpatrol

@Charlie Sooo quite in here...almost 8 hours?

Charlie

@skullpatrol almost 8.hours of.what?

Kevin Driscoll

Silence

skullpatrol

12:50 PM

3 hours later…1 hour later…3 hours later…1 hour later…

Charlie

Sound of silence

Kevin Driscoll

What sound does 1 hand clapping make!?

Clap clap

cl

Hehe

12:53 PM

:D

Charlie

@skullpatrol I'm only.listening to birds singing and the clock ticking

skullpatrol

@KevinDriscoll "cl" or "ap"

Kevin Driscoll

clever

Charlie

@KevinDriscoll wassup, Kevin?

Kevin Driscoll

@Charlie Went to bed quite early last night because I wasn't feeling well, so I'm up early today. Feeling just fine now. Gonna finish grading these tests so I can drop them off at my advisor's office later

also watching videos about the black hole information problem and the firewall effect in the background

its a pretty crazy time in quantum gravity research right now

skullpatrol

1:00 PM

is that Hawking's field?

Charlie

@KevinDriscoll sounds exciting

Kevin Driscoll

@skullpatrol Yes and then Firewall problem is intimately related to Hawking's prediction that black holes evaporate through radiation

skullpatrol

@KevinDriscoll would you like to meet him one day?

I know I would.

Kevin Driscoll

Hawking? It could be interesting, but he's not a particular hero of mine because I can't say that I understand anything that he's done

skullpatrol

yes Hawking

Charlie

1:04 PM

My heroes are all dead

I can be your hero

superman dat ho

@skullpatrol :D

@Charlie :D

@skullpatrol did you here that herr a group of of people.invaded a laboratory and freed 200 dogs?

skullpatrol

1:12 PM

@Charlie no, I didn't hear about it; where?

Charlie

@skullpatrol here , in my city

skullpatrol

@Charlie icic

@Charlie :?

Charlie

thejournal.ie/brazil-dogs-free-laboratory-1136817-Oct2013

@skullpatrol :/

Enjoys Math

1:45 PM

You should catch a beagle and become a dog owner

skullpatrol

@EnjoysMath Why did you call her a "ho"?

Enjoys Math

who?

skullpatrol

@EnjoysMath You called Charlie a "ho"

Enjoys Math

no, I just said "superman dat ho"

you were talking about heroes, thought it was relevant

Kevin Driscoll

Souljah Boi is ALWAYS relevant

Turn Mah Swag On!!!

N3buchadnezzar

1:50 PM

erh, no.

skullpatrol

@EnjoysMath So now you're saying I'm "Superman that hero"?

Enjoys Math

I'm just saying superman dat ho, bra!

skullpatrol

@EnjoysMath Explain what you mean?

Enjoys Math

lol

it's a song lyric, it has no meaning

skullpatrol

@EnjoysMath Stop trying to change the subject, and give me a straight-up answer.

3 mins ago, by skullpatrol

@EnjoysMath Explain what you mean?

robjohn

2:10 PM

@skullpatrol: I hope I wasn't too harsh in this comment

skullpatrol

@robjohn sounds good to me pal

robjohn

@skullpatrol Well, I will await the response.

I hate saying that my answer is better than that one. Of course, he did hint at plagiarism in his comment...

I don't think anyone would look at our answers and think that one was plagiarized from the other.

skullpatrol

I think it was Einstein who said "genius is knowing how to hide your sources"

robjohn

@skullpatrol Heh. However, Giraffe's answer was definitely not one of my sources.

At first, I didn't even realize they were using the same idea of splitting the sum, but even when that seemed the case, it took me a while to fully understand his answer.

In any case, I have to go to the park. BBL

skullpatrol

later

user43418

2:36 PM

Can someone help me with this: math.stackexchange.com/questions/531963/…

I have a question regarding one of the hints..

Anyone ?

Don Larynx

3:25 PM

Is it necessary to do all the problems per section

like in Dummit he has like 30 questions on Dihedral groups, should I do them all

Don Larynx

3:37 PM

"Rarely if ever expressible as a ratio of integers. See below for guidelines." What does that mean?

skullpatrol

irrational

Chris's sis

3:55 PM

@robjohn have you seen this one? $$\sum_{n=1}^{\infty} \log\left(1+\frac{1}{n }\right) \log\left(1+\frac{1}{2n }\right) \log\left(1+\frac{1}{2n+1 }\right)$$ It's divine.

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$Mathematics$ Mathematics