I just wrote an answer about Floer homology. It was great, nobody ever asks about that. But the OP didn't seem to read it.

@MikeMiller It does seem great.

I'll try to return to it in a few years (??)

I meant it was great to see a question about Floer homology. :p

I know.

reminds me of

G'night @MikeM, heya @Danu

Hi

morning

Belated hello @Jasper

@MikeMiller Weirdly realistic animation

It's an edit of an actual commercial

It's hilariously/disturbingly strange

I know that.

I know.

It's in the book.

You know, I used to have a Corolla

I had two of 'em ages ago.

^lol

You broke the spell

Who was spelling?

You should watch the video to understand

Or not.

:p

I think it's worth it :) Anyways, how are you doing?

I'm doing ok, thanks. You?

I've decided to drop a course.

Hi all.

Which one is the winner?

hi Skull

Supersymmetry

So now you're destined to be asymmetric, @Danu.

He decided it's all garbage :)

Just like us.

That course was, indeed, garbage.

but isn't symmetry suppose to be a large part of you're speciality? @Danu

Indeed, @Ted.

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Sure, but this was the useless kind of SUSY course.

I see.

it was all too symmetric

@MikeMiller omg there's a 10 hour version of that commercial?!

Guys, I need a hint: How do I prove that the trace of a nilpotent matrix is zero?

What are the eigenvalues of a nilpotent matrix?

characteristic of this sort of exercise is a desire to think about polynomials

cute pun, @MikeM, but not even necessary :D

yeah yeah

I guess they should be $0$?

just putting off thinking about spectra

@MikeMiller That book looks pretty nice.

Well, when you're done guessing, DogAteMy, let me know :)

@Danu It does, but I've never read it.

Seems like the right kind of thing for me.

And the trace should be the sum of the eigenvalues, but I thought that's only for when it's diagonalizable

(if it's not too hard :( )

take a look!

Nope, always true, DogAteMy. Prove it.

@AndrewThompson Do you know something about Tate cohomology of finite groups?

If so, can you explain it to me?

Does the characteristic polynomial have to be $\lambda^n$ because anything else of degree $n$ has other roots, giving more eigenvalues? And if there's another eigenvalue, then multiplying the matrix by the corresponding eigenvector $k$ times just multiplies the eigenvector by $\lambda_0^k$, but it should be $0$ since $A^k=0$.

OK, a bit contorted, but right.

So, the trace is the coefficient of $\lambda^{n-1}$, which is $0$.

so does that prove that the trace is always the sum of the eigenvalues, btw?

Wait, trace is $-[\lambda^{n-1}]p(\lambda)$, not $[\lambda^{n-1}]p(\lambda)$. Doesn't change anything. (Where $[-]$ is the coefficient.)

But, yeah. 'Cause they're both $-[\lambda^{n-1} ]p(\lambda)$. @TedShifrin

Depends on your definition of $p(t)$ whether there's a minus sign or not.

note ofc that you're counting eigenvalues by algebraic multiplicity / dim of generalized eigenspace, or else it's not correct

$\det(\lambda I-A)$ @TedShifrin

OK, I usually use $A-tI$ for computational reasons. Students will mess up all those negatives.

That would make it $-(-1)^n$ times the coefficient, I think

I'm not going to try to sort it out.

Calling them (repeatedly) at risk is probably not appropriate for most of them, although I may have once used that term.

Yeah, the year is over. The seniors have graduation on Saturday.

Nice.

Kudos for your efforts :)

In the end, I may have made a little bit of difference to a couple of the students. Perhaps helping them realize that more challenges would be in store in college.

@MikeMiller Maybe but after seeing various questions from people reading Scorpan I'm convinced its not a good source to learn anything concrete.

heya @PVAL

@Ted Hiya

I think it's meant as an overview with slight details.

wouldn't want to be overburdened with heavy details

I think it's hard to write down too many details if you want to survey all of the 4-manifolds literature even upt o 1990.

I mean I think thats what Gompf-Stipsicz is supposed to be.

But that book does things.

I don't know what Scorpan does.

Does it cover Freedman's work though? I don't remember any details about that. It's all the smooth stuff.

Of course GS is a great source.

It states the theorems

I'm sure Scorpan doesn't do anything more.

Bob said that when he told Kirby he wanted to do 4-manifolds, Kirby handed him a 5 page packet and said "this is everything currently known about 4-manifolds".

I'm glad we got a little better since then.

What is Bob's last name?

Gompf

Thanks.

Okay, the thing hes having trouble with is explained explicitly in Freedman-Quinn

I guess I should try to understand that stuff someday.

The first two chapters and the last two chapters are pretty nice.

I've already wasted too much of my life on Ch.3 of that book.

Or

I might be getting the numbers mixed up.

Sure.

But you mostly think about smooth things anyway right?

The surface intersection theory is all quite cute. They prove Wall's theorem and other nice smooth results before doing any decomposition theory.

Well the first two chapters in that book are really about smooth things.

Oh, that's good. I've wanted a source for that lately.

Oh you probably mean a different Wall's theorem.

I think its proven in Gompf-Stipsicz too

I guess you mean the h-cobordism result.

I was thinking of the stabilization theorem.

That two h-cobordant manifolds are diffeomorphic after connect summing with S^2 \times S^2 a bunch of times.

Thats what I call Wall's theorem.

I'm certain he has others.

That is both about h-cobordism and stabilization

That's the one I was thinking about I guess. There's a theorem that manifolds with the same intersection form are smoothly h-cobordant. I thought that was also due to Wall.

@MikeMiller no thats to due to Milnor and someone else

ok

Apparently its due to Milnor and Whitehead though the sources only say up to homotopy equiv.

So maybe Wall showed that.

@robjohn nice to see you approaching the 200K club :)