Mathematics

Miguelgondu

12:16 AM

But the sum of the maxes $\geq$ the max of the sums, amirite?

robjohn

@MikeMiller I only got 40... I spent too long on a problem, evaluating a sum.

Mike Miller

oof :(

morphic

This text says "Let $X_i$ be spaces and let $\pi_i : \Pi_i X_i \to X_i$ be the projection." What is the general definition of a "projection?"

Mike Miller

the projection $X \times Y \to X$ is the map $(x,y) \mapsto x$

generalize

Miguelgondu

$\pi_i(a,\dots,d,x_i,e,\dots,f) = x_i$, if I'm not mistaken.

where $x_i \in X_i$.

morphic

12:25 AM

Oh, thank you

Mike Miller

if you're just talking categorically, though, the product $\prod_i X_i$ has a "projection map" $\pi_j: \prod_i X_i \to X_j$ just as part of the data of the product

in an arbitrary category there's not much reason to believe it relates the notion of projection of a cartesian product of sets

Anthony

@MikeMiller I'm reading through Lang's Algebra book, and I came across the below- it's really obvious, but I'm blanking on what a good way to show it might be.

It's talking about a commutative monoid.

I mean I guess since it says almost all pairs, I could just make one product by indexing over the pairs, and use the rule for commutativity with just one variable... I just dunno. And does this fail if it's not the identity for almost all pairs?

I guess it could.

Mike Miller

It says "almost all pairs" so that that product is well-defined; otherwise, how do you multiply infinitely many elements? When you expand it ends up just being a product of finitely many terms, like you said, and then it's just commutativity in one variable.

Anthony

You multiply infinitely many elements... from left to right. :P

And yeah, gracias.

This is gonna be a loooooong semester.

Ted Shifrin

12:40 AM

Who's teaching the course, @Anthony? Oh, and I just emailed you.

Anthony

I'll go check- and I'm taking algebra with Ken Ribet!

Ted Shifrin

Oh, wow. He's an old friend (even on Facebook). He's a fantastic mathematician and teacher.

I wanted to see him when I'm up there, but he'll be out of town.

Anthony

Oh! I see. I'm looking forward to being less horrible at algebra by the end of the semester.

I'll reply in a bit to your email. I should be free anytime after 12, though!

Ted Shifrin

Awesome. Looking forward to it. Maybe I can help you be slightly less horrible.

Anthony

:D

Anthony

1:05 AM

@TedShifrin Are you still there?

Ted Shifrin

Where?

Anthony

hohoho

In Lang, in his discussion of monoids, he talks about the homeomorphism classes of compact surfaces

Mike Miller

i remember this. it was very silly.

Ted Shifrin

@morphic: I will complain about that notation. I would insist on writing $$\pi_i\colon \prod_j X_j \to X_i.$$

Anthony

He says that the 2-sphere is the unit- why is this the case? My understanding of the monoidal operation is that you just stick one shape onto another?

Mike Miller

1:07 AM

define connected sum for me

Ted Shifrin

So sticking a sphere onto another surface gives you the same surface back, just with a bulge. But homeomorphic.

Sorry, @Mike. I forgot I'd been fired.

Mike Miller

meh

Anthony

@MikeMiller You identify the inside of one circle on one surface with the inside of another circle on another surface, right?

Mike Miller

i don't like that. let's be rigorous.

Anthony

I mean I can retype what he wrote lol

Ted Shifrin

1:09 AM

Rigorous? This is an algebra course :D

Anthony

I haven't worked with connected sums before, but I guess that doesn't matter.

Mike Miller

the connected sum of $\Sigma_1$ and $\Sigma_2$ is given by an embedding of a ball $i_i: D^n \to \Sigma_i$, deleting the interior of $i_i(D^n)$, and then gluing the resulting manifolds together by the induced identification of the boundary spheres $i_i(S^n)$

Anthony

Oh, man you beat me.

Mike Miller

connected sum with a sphere is trivial because the complement of an open disc in a sphere is... a disc again

but ted's right, this is an algebra course, who even cares

Anthony

:P

Ted Shifrin

1:12 AM

Lang is a really tough book unless you know a lot of it already.

But I know Ken has personal fondnesses for it.

Anthony

OH

I get it

I'm dumb

Mike Miller

you're not dumb

connected sums are hard

Anthony

I mean, they aren't hard to visualize right?

Ted Shifrin

@Anthony: If you aspire to go to grad school in math, you need an attitude adjustment.

Mike Miller

i also implicitly assumed connected sum was well-defined. ;)

Anthony

1:14 AM

@TedShifrin I've actually been thinking about this a bit.

Ted Shifrin

I don't think I've ever worried about well-definedness, @MikeM

We can talk about it if you want, @Anthony.

Mike Miller

@Ted: Since you only work with smooth manifolds, that's ok, because it's not hard to prove

modulo the standard concerns that your manifolds are oriented, or non-orientable, or one supports an orientation-reversing diffeomorphism.

Anthony

@TedShifrin I'll think about what questions I should ask you. I'm sure there's a lot.

Anyway, I think I should go outside for a bit.

Talk to ya'll later

Ted Shifrin

Bubye.

Mike Miller

i don't think well-definedness of connected sums of topological 4-manifolds was known before the 80s.

Ted Shifrin

1:17 AM

Now you have me confused about the orientation-reversing diffeo.

columbus8myhw

How many people here are bilingual, out of curiosity?

(אני יודע עברית)

Ted Shifrin

What's your definition of bilingual, @columbus? Equally good in both? Or good in both?

columbus8myhw

Good in both.

Ted Shifrin

Then I'll raise my hand.

columbus8myhw

What language?

Ted Shifrin

1:18 AM

I'd almost go for trilingual, but the third is weaker.

columbus8myhw

Same

Ted Shifrin

English, French, German ...

columbus8myhw

(I have one year of Spanish classes behind me)

Ted Shifrin

One year of Spanish doesn't count :P Make it 4 years.

Mike Miller

@Ted: The issue is you need to be able to identify your embedded discs; if you reverse the orientation of one of the discs, you can change the topological type of the resulting manifold. This is why $\Bbb{CP}^2 \# \Bbb{CP}^2$ is not homeomorphic to $\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$. If one of your manifolds is non-orientable, or supports an orientation-reversing diffeomorphism, it doesn't matter

columbus8myhw

1:19 AM

@TedShifrin They'll be over before I know it. I hope.

Mike Miller

Because you can reverse the orientation of that first disc by moves that don't change the topological type of the connected sum.

Ted Shifrin

Oh, I see what you meant, @MikeM.

LOL, @columbus.

I did 3 years of college German, but it's still behind my French and English.

columbus8myhw

Why do so many mathematicians know French?

Ted Shifrin

Typically, 4 years of high school is about 3 semesters of college

columbus8myhw

(Or is that not accurate?)

Mike Miller

1:21 AM

Because there's so much good French mathematics.

Not to say I know French. I can read French mathematics. I still need subtitles if I'm watching a French movie.

columbus8myhw

@MikeMiller In a conversation about languages, you might want to watch your English, sir.

:P

Oh wait

Mike Miller

No.

Ted Shifrin

To get a Ph.D. mathematicians are supposed to have a reading knowledge in at least one language (French, German, Russian are the usual ones). In my day, it was two of those. But now a lot of schools use computer programming skills instead of one.

columbus8myhw

Shoot

I'm sorry

I misread

Your English is fine

Mike Miller

That's unlikely. Maybe it was in those two lines, though.

Ted Shifrin

1:22 AM

@MikeM and I have argued about English syntax before. He really doesn't believe in it.

Mike Miller

That is incorrect.

Ted Shifrin

LOL

columbus8myhw

It is something up with which I shall not put, as Churchill said that one time.

Hey, you guys both live in CA (in cities with Spanish names), I'd expect you guys to know Spanish.

Ted Shifrin

I never learned Spanish, but it's on my list of things to do the next few years.

Mike Miller

You expect too much from me.

columbus8myhw

1:24 AM

I mean, instead of French.

Ted Shifrin

I lived in god-forsaken Georgia for 34 years, @columbus.

columbus8myhw

(Your profile has French on it, so I assumed you know French)

Ted Shifrin

French was chosen for academic reasons and because I loved the literature.

columbus8myhw

@TedShifrin Ah.

Ted Shifrin

Spanish will now be practical.

columbus8myhw

1:25 AM

I live in NYC. There are Spanish ads on the subway.

(Which helps with the Spanish practice)

I know Hebrew, though, since I'm Jewish. (I wouldn't say I'm fluent, but I think I have a decent amount.)

Ted Shifrin

well, @Columbus, I'm Jewish, and I know some Yiddish through German, but I know 0 Hebrew.

columbus8myhw

It's not a strict implication.

Ted Shifrin

LOL

We've had issues with implications, haven't we? :D

columbus8myhw

I think your joke just flew over my head there

Ted Shifrin

I was referring to that logic question we both were answering, somewhat confusedly, the other night.

columbus8myhw

1:31 AM

Ahh

Right

You at least know some Hebrew, though:

$\aleph$ and $\beth$.

And quite a bit of Greek, for the same reason, of course

Ted Shifrin

I used to know some of the alphabet from playing dreidl, but I've forgotten.

columbus8myhw

נ ג ה ש

From right-to-left, "nun gimmel heh shin"

Ted Shifrin

That was a long time ago :)

OK, time for dinner and packing. You all misbehave without me.

columbus8myhw

Packing?

Ted Shifrin

2-week trip

columbus8myhw

1:33 AM

Nice, have fun!

Tanx.

$\tan x$

no, not that.

Goodbye!

Enjoy yourself @TedShifrin

You deserve it

Mike Miller

1:39 AM

Up for debate.

morphic

lol

user147690

1:52 AM

Can I have a hint or some guidance in general for showing that if $|G|=112$ and $H=A_7 \cap G$ then $H\trianglelefteq G$?

user147690

Actually I think it isn't simple oops.

Mike Miller

How do you intersect $A_7$ with $G$?

dREaM

Intersection (road)

An intersection is the junction at grade (that is to say, on the same level) of two or more roads either meeting or crossing. An intersection may be three-way (a T junction or Y junction – the latter also known as a fork), four-way (a crossroads), or have five or more arms. Busy intersections are often controlled by traffic lights and/or a roundabout. This article primarily reflects practice in jurisdictions where vehicles are driven on the right. If not otherwise specified, "right" and "left" can be reversed to reflect jurisdictions where vehicles are driven on the left. == Types of inte...

Let me google that for you

morphic

lol

user1667423

2:12 AM

Does anyone know what the metric tensor is for an n-sphere in ambient n+1-Cartesian coordinates?

dREaM

Does anyone here know how UVA works?

I am having problem submitting stuff in java.

user147690

@MikeMiller Well $|G|=2^4 * 7$ so we have 7 sylow 2-groups, and it gives us an embedding of $G\hookrightarrow S_7$

user147690

2:27 AM

But I did that with $G$ is simple, so I got confused :\

Mike Miller

@AlexClark: Your goal was to prove that $G$ is not simple, yes?

user147690

Yep

user147690

It asked me to(treat $G$ as simple in the first question), and then it says that we can conclude that there is no simple group of order $112$(in the second question)

Mike Miller

@AlexClark: OK, you start by embedding $G \hookrightarrow S_7$. (This is what I was bugging you about before - we need an embedding into some group that $A_7$ is a subgroup of, or else intersection makes no sense.)

user147690

Ok sure

Mike Miller

2:31 AM

Now that you have this, there are two options. Either $G$ is a subgroup of $A_7$ or, well, it's not.

If it's not, then $G \cap A_7$ is a subgroup of index 2 in $G$. (Why?) And then you use the fact that subgroups of index two are automatically normal.

dREaM

@MikeMiller how can I prove this?

user147690

@dREaM Are you simulating me or? :P

dREaM

lol, no

Transcript for

$Mathematics$ Mathematics