Mathematics - 2017-03-11 (page 5 of 5)

Krijn

23:01

How many years in high school do you have left

Balarka Sen

@MikeMiller If you have two 3-planes in R^6 intersecting, what do you do to "tube" them up? Ie homologue a bit so that it becomes an embedded manifold

what's the resulting thing you get in RP^6, if you want that to be precise? RP^3 disjoint union RP^3 becomes what?

@Krijn 1

Benjamin

@BalarkaSen This is how I wish more people thought.

Krijn

@BalarkaSen Can you not already get some kind of degree in mathematics?

Benjamin

@Krijn As I always like to explain to people, "It would make too much sense to allow the gifted to be accelerated, therefore it may not be permitted."

Balarka Sen

I'm happy enough with my free time to be carrying around the woes of grad school

just ask Mike

Krijn

23:09

Also, just read this quote by Deligne "I was not afraid to ask completely stupid questions"

10

Benjamin

@Krijn One who is afraid to ask stupid questions is afraid to answer stupid questions and by extension the big questions.

Meow Mix

hi

Balarka Sen

@Krijn eg

Meow Mix

someone give me motivation

Krijn

@MeowMix You get one cookie if you finish every exercise in Hartshorne

I'll even double it if you manage to do that in 2017.

Meow Mix

23:16

no, this is Ted's textbook we're talking about

Balarka Sen

@MeowMix Is this linear algebra or multivariable calculus?

Meow Mix

linear algebra

in "Multivariable Mathematics"

Balarka Sen

which chapter

Meow Mix

I just don't have the motivation to LaTeX my answers, so they're left in the limbo of Zach's mind

Balarka Sen

linear algebra is ok but I really prefer calculus

I never latexed any of my stuff

Meow Mix

23:18

@BalarkaSen Well, in the words of Ted, "Strong foundations... [other stuff]..." Paraphrased, learn your shit Zach.

MickLH

Shia LaBeouf "Just Do It" Motivational Speech (Original Video)

►

Meow Mix

Oh that spongebob video might be motivational

Balarka Sen

I can try to trick Ted into convincing him to get you into the calculus, @Meow. Anyhow, which chapter?

Meow Mix

Chapter 1, just looking over some stuff. Particularly the cool geometry stuff in section 1, because that's the only thing I find a bit interesting.

Balarka Sen

Oh. Well, you're into cool stuff.

I personally skipped to Chapter 2

Meow Mix

23:20

And maybe a little miscellaneous exercises here and there, and then I'll continue to chapter 4 or 5 or whatever the next linear algebra chapter is

@BalarkaSen I have to read Spivak's before I do multivariable

My calculus is weak :[

(by that, I mean I was taught without rigor)

Balarka Sen

A'right then

Meow Mix

Which I bought... I mean, obtained

creeps into shadows

Balarka Sen

i obtained a lot of books

@MeowMix I can tell you a little something. You know what dot product is, right?

Meow Mix

Yeah... lol

Balarka Sen

What - to you - dot product actually tells? Why's it a big deal?

Meow Mix

23:27

Well, first you can find the angle between two vectors

Also you know that if it's 0 then the vectors are perpendicular

Balarka Sen

The second is just a special case of the first (90 degrees = perp), but yes.

Measuring angles between two vectors is indeed fundamental.

The other fundamental thing about it is that it in particular gives you a notion of length. If $v$ is a vector $\|v\|^2 = v \cdot v$. Agree?

Meow Mix

umm

ah, yeah

Balarka Sen

Right. So this is weaker, of course, also known as "norm" but it's what a dot product gives anyway.

Meow Mix

is there anything else?

Balarka Sen

You know about abstract vector spaces a little bit, right?

Meow Mix

23:30

I guess

Balarka Sen

Not much other than what I said and what you said, @Zach :)

Meow Mix

I've heard of stuff like

Inner product spaces

Don't know what they really are though.

Balarka Sen

Exactly!

The point is an abstract vector space doesn't have a notion of a dot product in built.

Inner product is exactly that: a dot product on abstract vector spaces.

Meow Mix

Is it a field with a module and some "inner product" from two elements of the module to a member of the field?

At least that's how it works in $\Bbb R^n$ I think

I don't remember, but I think they called it a bi- something mapping

eh, I'm just blabbering what I think.

Balarka Sen

The inner product must also satisfy $(x + y) \cdot z = x \cdot z + y \cdot z$, $(cx) \cdot y = c (x \cdot y)$, $x \cdot y = y \cdot x$ and $x \cdot x \geq 0$ with equality iff $x = 0$.

Just like dot product.

Meow Mix

23:34

Let me look it up.

Balarka Sen

But yes.

Billinear.

Meow Mix

Yep

I guess "linear" makes sense here :P

Hi @Mike and @arctictern

Mike Miller

hi

Alessandro Codenotti

If you know what a bilinear form is then an inner product is a symmetric positive definite bilinear form

Meow Mix

@AlessandroCodenotti Remember, I'm stupid. So I don't. But I guess Balarka just told me:P

Does positive definite imply the last condition Balarka mentioned?

Balarka Sen

23:36

Indeed.

Meow Mix

And symmetric the one before it? :P

Balarka Sen

Obviously.

@MeowMix Look at chapter 4.3 section 3.1. in Ted's book if you want to learn a bit about abstract vector spaces.

Meow Mix

Are there any other examples of inner products on $\Bbb R^n$?

Balarka Sen

Anyway, I was trying to lead down a more concrete direction. The point is dot product is not god given on a vector space; it's by fate that we have a natural one in R^n.

@MeowMix Yes, there are quite some. I don't know an easy one to write out off the top of my head.

arctic tern

$\langle x,y\rangle=x^TAy$ for any symmetric invertible matrix $A$

Balarka Sen

23:40

If you want an example where the last condition (positive definiteness) fails look up Lorentzian/Minkowski inner product.

arctic tern

also hi

Balarka Sen

Ah yeah, sure

Meow Mix

Ah, alright. Thanks for the interesting talk, I'm off to eat dinner. Bye math.SE :)

Balarka Sen

(In fact, those are all the inner products there are.)

@arctictern I guess you want positive eigenvalues though.

Otherwise you just get a symmetric billinear form.

arctic tern

true

Semiclassical

23:44

[insert discussion of the Minkowski metric special relativity here. I can't be arsed to actually do it.]

Akiva Weinberger

Hi all!

Balarka Sen

@MeowMix Take it as an exercise to prove that the examples arctictern wrote down are indeed inner products and are all of them. Derive a 1-1 correspondence between nonsingular symmetric positive definite (ones which have positive eigenvalues) matrices and inner products on R^n.

Hi @Akiva.

Akiva Weinberger

@BalarkaSen I thought about your Zariski proof of the Cayley–Hamilton theorem

I think the main idea behind it is this:

Balarka Sen

Cool, what do you have?

Akiva Weinberger

If $f$ and $g$ are multivariate polynomials, and $fg=0$ (the zero polynomial), then either $f=0$ and $g=0$. Which sounds obvious enough, but it has an interesting consequence

which is that "$fg=0$" is the same as "for all $x$, either $f(x)=0$ or $g(x)=0$". And "$f=0$ or $g=0$" is the same as, "(for all $x$, $f(x)=0$) or (for all $x$, $g(x)=0$)".

(That actually looks a lot more confusing than I meant it to be)

but as a consequence, it means that, if you ever need to prove that $f$ is the zero polynomial, you just need to prove that $f(x)=0$ outside the zero set of some other polynomial $g$.

(Assuming $g$ isn't the zero polynomial, because then its zero set would be the entire space)

Balarka Sen

23:57

Exactly.

This is a correct 1-line summary of what it means to extend the 0-function on an open set in Zariski topology to the whole space by continuity.

Akiva Weinberger

So, to apply it to this problem, all you need to do is find a $g$ such that $g(x)\ne0$ implies $x$ is diagonalizable.

And that's readily provided by the discriminant.

Balarka Sen

yup

Alessandro Codenotti

How weird can the zero set of a polynomial in 2 variables be in $\Bbb R^2$?

Balarka Sen

It's a curve, unless it's a point

Akiva Weinberger

@BalarkaSen One problem — this doesn't work for all rings. In $\Bbb Z_2$, for example, if $f(x)=x$ and $g(x)=x+1$, then $\forall x,f(x)g(x)=0$ is true, but $(\forall x,f(x)=0)\lor(\forall x,g(x)=0)$ is false.

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