I got side tracked revieiwing linear algebra

Lots to review/relearn :P

Yeah! But I love math so it's all fun

Im more strategic than I used to be

now i read quickly

and i get to the problems

and start trying to do them

and learn by doing and reread

yeah, just doing exercises doesn't suffice ...

What do you mean?

I mean I do both

but before I read too much

and thought that was enough

Yes, I know, you do try to digest the text/proofs, too, but some of my students skip that

I used to

But then I spent more time talking to the profs

and I realized, boy, this guy is smart, I probably shouldn't skip. why would they write something that isn't useful.

Just because we want to feel important?

I mean, maybe the textbooks you read when you are young set a bad rep. yes, some of the books i read when i was little were terrible

so you build up this impression like they are going to be bad

"little"? you mean two years ago?

uh yeah something like that. although i'm still pretty little. usually the short guy.

Lol same thing with French high school textbooks :(

middle school was horrible

We runts are smarter to make up for it, @Stan :)

LOL i notice that. it seems to work too. I have a doctor friend of mine and apparently he finds me intimidating. and he is this huge hunk buff guy

he's like a doctor/physical therapist

@MikeMiller: Can two spaces be homotopy equivalent but not basepoint-preserving homotopy equivalent?

Also, feel free to tell me to shut up.

Great question, @MichaelA.

Thanks @TedShifrin.

Do you know a famous example of a contractible space that's not contractible rel $x_0$ for certain $x_0$?

@MichaelAlbanese: Wasn't that yesterday's question? In any case, yes: put the basepoint in different components. I guess Ted just gave the non-stupid answer, though.

Comb space?

Hello @Hippalectryon

So there's your "yes".

I think that's the first time @MikeM has called me "non-stupid."

@evinda Hello.

Just the answer, @Ted

LOL

@Hippalectryon How are you?

@MikeMiller: It may have been but it was mixed up with category theory.

Thanks @TedShifrin.

@TedShifrin Okay, sent! :)

@evinda Fine; it's midnight here so I'll be off soon.

Alright my mathematicians friends, I must skedaddle. Hasta luego amigos!

@Hippalectryon A ok... Is it 12o'clock there?

See ya, @Stan. I'll look at your homework after dinner.

It's not that you should shut up, @MichaelAlbanese, just that I'm the wrong person to ask.

@evinda yep

@Hippalectryon Aha

There's a couple things I know something about, but homotopy theory isn't one of them.

Fair enough.

I guess Ted is, though.

Ted is what, though?

Hello @Ramanewbie

The right person to ask about homotopy theory.

Hi @evinda

NOOOOO ...

@TedShifrin Guten Abend

It's way past @Ramanewb's bedtime.

guten Abend, @evinda.

@Ramanewbie How was your day?

@ted no, Im in Ireland so its one hour earlier than in France here

oh, right :P

@evinda good and yours

@Ramanewbie I didn't do something special. Did you go out with your host family?

You just successfully answered a homotopy theory question, @TedShifrin, so yes.

The first week of algebraic topology is not homotopy theory, @MikeM.

I'm going to stick to undergraduate stuff, I think :)

You don't get to eat any of them, @Ramanewb?

Why torture the fish, then?

@Ramanewbie :/ I don't know...

I guess my days of fishing (younger than you) I didn't have to worry about that.

@Ramanewbie What did you have for lunch then? Again potatoes?

LOL again potatoes :) I'm about to finish my potato salad I made for salade niçoise :)

@evinda chicken from a fast food

Ugh ...

@ted potatoe is small in rather small amount...

@TedShifrin Nice... :)

Even in France one eats potatoes :)

good jeu de mots, @evinda :)

ok ... dinner time. Bubye, all. Schlaf gut, @evinda. Dors bien, @Ramanewb.

See you @ted

@TedShifrin Guten Appetit und gute Nacht!!! @TedShifrin

@TedShifrin: Do you know if there is a classification of compact almost complex surfaces? By an almost complex surface, I mean a four-dimensional manifold with an almost complex structure.

The integrable case is the Kodaira-Enriques classification.

I would bet you can construct any fp group as the fundamental group of an almost complex 4-manifold.

Is that true of symplectic manifolds?

If you restrict to simply connected a 4-fold has an almost complex structure iff $b_2^+$ is odd.

Yes, it is.

Oh, good point.

But I suspect it should be more elementary for AC manifolds.

I think so too.

Hi @Aleksandar

I trying to become a better mathematician by mastering the art of proof. I've taken lots of plug-n-chug courses. Is their any basic theorem that I could prove?

Anyone here?

@Aleksandar get a book with ("Real Analysis" or "abstract algebra") and "introduction to" in the title

Okay. I've tooken Calc I,II, III and complex analysis and have some knowledge of real analysis.

@MichaelA: Let $M$ be a 4-manifold with fundamental group $G$. If $w_2$ has an integral lift (that is, $\beta(w_2) = 0$), you should be able to connect sum with an appropriate simply connected 4-manifold to obtain an almost-complexable 4-fold with given fundamental group.

So the question is how to build $M$ so that $\beta(w_2) = 0$.

Easy question: the common zero set of an infinite number of continuous functions mapping to R is closed, right?

The book I'm reading puts the qualifier "finite" on that statement (not "infinite") so I'm wondering if I'm missing something.

This is clearly true because the intersection of any collection of closed sets is closed, yes?

Yes, that's correct. Perhaps they like the finiteness constraint because they can just pass to $(f_1 \cdots f_n)^{-1}(0)$ instead of thinking about intersections of closed sets or whatever.

Thank you.