Hey Ted, say I have $x_1,\dots,x_n$ independent in $V^{2n}$ (a vector space). Then does $x_1,J x_1,\dots,x_n,J x_n$ determine $J$ ($J$ is an almost complex structure)?

For the second one : Add the top to the bottom. Remove half of the bottom from the top. Remove twice the top from the bottom. Add $1/4$ of the bottom to the top @Ted

doesn't really matter. I'm stuck at understanding ordering, and inducing an ordering.

If someone has time for that, I'd be glad.

I'll check that in a moment, @Astyx. Granting your first sentence, then the path is easy!!

isn't this correspondence the "projection" $G\to G/N$? (I say "projection" because the domain is all wrong but you get what I mean)

@AlessandroCodenotti Yes

I'm asking because I want to understand why the set of complex structures is $SO(2n)/U(n)$. I see that $SO(2n)$ acts transitively on oriented, orthonormal bases and that $B\cdot (x_1,Jx_1,\dots,x_n, Jx_n)$ is of the same form if and only if $B\in U(n)$.

ok, I see

@Danu: You need to know that $x_k,Jx_k$ determine independent $2$-dimensional subspaces, and that $J^2 = -I$.

I guess the latter implies the former. But just independent $x_k$ won't do it.

@TedShifrin That latter thing is included in the definition.

No, because you asked if the vectors determined $J$?

Now do I know if $Jx_j$ is orthogonal to $x_k$?

@TedShifrin Given that $J$ is an almost complex structure

See my second message for motivation

@TobiasKildetoft more unrelated. Basically the difference between an ideal and a principal ideal is that the conditions on an ideal is restricted to the subring (we pick a \in I). But principal ideal is just picked from a ring.

Oh ...

@Hawk No. A principal ideal is a specific type of ideal (one generated by a single element)

@Danu SO(2n)/U(n) is the set of complex structures on R^(2n) that are compatible with the inner product right?

So what precisely is the question, @Danu? Two oriented orthonormal bases obviously are equivalent almost complex if and only if if they differ by $U(n)$.

Right @tern

then that's orbit-stabilizer right?

@Danu: We're building in that $J$ is an isometry.

OH, and hi @ tern

@TedShifrin I'm just trying to see what you say is obvious

hello

I think it's just what I said earlier but I don't know the definitions

Your original question was ill-posed.

Never mind it.

I meant my second message, not my first.

@Displayname this only works for sets where division by two is always defined. But 0 being positive or negative is still a problem under $\mathbb{Z}$ (if you even see it as a problem)

So we start by defining $J$ by $Jx_{2k-1}=x_{2k}$ and $Jx_{2k}=-x_{2k-1}$.

We check that this is an almost complex structure and an isometry.

We then observe that we set $e_k=(1+J)x_{2k-1}$, this is a unitary basis, and that we get an equivalent complex structure by doing any unitary transformation to it.

So I guess the question is why any complex structure that's an isometry arises in this manner. But isn't that immediate?

being an isometry and complex structure on $\Bbb R^{2n}$ is equivalent to being in ${\rm SO}(2n)\cap{\frak so}(2n)$. these are right-angle rotations in $n$ orthogonal oriented plane (with orientations adding up to that of the whole space). ${\rm SO}(2n)$ acts on this space transitively by conjugation (the effect of conjugation by $R$ simply applies $R$ to each plane of rotation), and the stabilizer is $U(n)$ essentially by definition.

not sure what the original question was but hey

Why $SO(2n)/U(n)$ is the space of complex structures compatible with the metric, yeah.

@TobiasKildetoft oh okay now getting back to the original question. they basically showed that the ideal R/M{a + M} cannot exist because its preimage is an ideal containing a maximal ideal. So this shows R/M contains only 2 trivial ideals? ${M}$ and $R/M$?

@Hawk Right, or rather that it must equal one of those two

@TobiasKildetoft But not every ideal in a ring is principal. How does this eliminate the possibility of a non-principal ideal in between the 2 trivial ideals?

@Hawk Right, but if a ring has non-trivial proper ideals then it also has principal ones

@TobiasKildetoft that is supposed to be a very simple statement...but I am having trouble comprehending it.

@Hawk If it has a non-trivial proper ideal, consider any non-zero element in that ideal

I am not sure what you mean by the first part

hello, could someone help me with determining which of those vectors v1(x)=x^3+2x^2+x-1 v2(x)=2x^3+4x^2-x-2 v3(x)=x v4(x)=x^3+2x^2+x+1 are linearly independent?

@user379685 That does not really make sense. Are you looking for the largest subset of them which is linearly independent?

i think so

By the way @Tobias I still don't understand how you can define a topology on $M_n(\Bbb Z/2\Bbb Z)$ such that any matrix is a limit of a sequence of $GL_n(\Bbb Z/2\Bbb Z)$ and $\det$

hi chat

@TobiasKildetoft if I transform your statement: "a ring has non-trivial proper ideals, then it has principal ones" to its contrapostive, it makes sense to me. Can we prove contrapostive directly? The original is just $a \in I$, then $ra \in I$, so the ideal generated by $a$, $r(a) \subset I$

@Astyx I am not sure if you can do that

hi @semi

Then I didn't understand what you meant yesterday

@Astyx Ahh, right. I never really think of being dense in terms of sequences since that tends to be impractical in the Zariski topology

@Tobias Then how can density be defined ?

@Astyx As intersecting any non-empty open subset non-trivially

And as it turns out, the space we are considering is irreducible, so all non-empty open subsets have this property

(actually, we may need to move up an an algebraic closure of the field we were working over, but it obviously suffices to prove the statement there)

I'm gonna pretend I know what that means :p

@Astyx The whole argument does take a bit of algebraic geometry to get going (since we also need morphisms to be uniquely determined by their value on dense subsets and this usually requires the space to be Hausdorff which this is not, so we need a substitute)

Yeah, I don't have that context yet .. Thanks for the answer, as always

but is my approach wrong or not?

I mean you can do that, what do you get?

that they are independent

@Hawk typo with bmatrix ?

Maybe...

hello, what'd be the best way to solve this inequality: $(1/2)^x < (1/3)^2x$

It's just that it's not computing on my screen

Should that be $(1/3)^{2x}$?

yes, sorry

@Dartek12 Simple. Take the "x" root of both sides:

XD

Take the log

:-)

@Simply :-)

Actually, inverse both sides

One useful property of the log: If $x<y$ for real $x,y$ then $\log x <\log y$.

$$2^x>9^x$$

So taking logs of both sides can't change the inequality.

which happens only for $x<0$

Or notice that your innequality is $(1/2)^x \lt (1/9)^x$, ie $(9/2)^x \lt 1$. This function is strictly increasing and its value at $x=0$ is 1.

Then what Simply said

$\begin{bmatrix} -1 &-2 &0 &1 \\ 1&-1 &1 &1 \\ 2&4 &0 &2 \\ 1& 2& 0& 1 \end{bmatrix}$

@Hawk

ok, thank you

Hm...

did you guys know MSE gets about 571 questions per day?

Yep.

@Hawk you are right that it is not invertible, but why?

it's sad, since SO gets over 7k questions per day

and I want to answer more questions XD

Well I feel some questions on MSE are already too much

Too much homework or lazy questions without any interest

huh

the det

@Astyx Nah, I just get kinda bored

is 0

True

I like the good questions

@Hawk Computing the det is overkill

@Hawk yes. So what does that mean for the vectors?

@Astyx You meant "debt", as in US debt?

:-)

$$\huge\overbrace{\left(\ddot{\stackrel{\quad>}{\smile}}\right)}_{\begin{align}---‌​\quad\ \end{align}}$$

Remember that a necessary and sufficient condition for being invertible is full rank

So see if you can show that the rows/columns are linearly dependent

There's a particular instance which stands out to me

Heya @Daminark

Hi @TedShifrin

Hi @TedShifrin

Bye guys, gotta do life

Hi, @Simply. Are you someone with a different previous name?

Bye @Simply

@TedShifrin Yeah. Was Simple Art. That one guy

Ah.

I just like this username better is all

No big deal.

I'm going, bye everyone

Enjoy your day

See you @Astyx

Hey @Ted, now I'm looking at the diffeomorphism $\Bbb P(TS^6)\to \Bbb P(T^*S^6)$. With some help from Balarka, I was convinced that this should be orientation-reversing---at least the map $TS^6\to T^*S^6$ should be orientation-reversing, should it not (because $V\to V^*$ is complex anti-linear)?

Um ...

Where did complex come from?

almost complex structure on $S^6$

But the dual bundle has nothing to do with that.

Well, $TS^6$ is a complex bundle because of it.

Are you saying that $T^*S^6$ is not its dual as a complex vector bundle?

I am not caught up to you.

I don't see why it should be complex dual at all.

I just assumed it would be

So I see that we use a Riemannian metric to give a vector bundle isomorphism.

So how do I think about this?

I have never thought about it.

So what happens if we look at $S^2$. The holomorphic tangent and cotangent bundles are clearly not isomorphic bundles. But in the smooth world we have an isomorphism.

We should probably just be thinking about the regular (co)tangent bundles---no complexifications.

Right.

Since I guess that's what I'm doing in the $S^6$ case.

So we're looking at a rank 6 real bundle.

Which might or might not be a rank 3 complex bundle.

yeah

So let's go back to $S^2$.

I guess I don't see anything orientation-reversing.

I don't even know how to think about whether or not it could be orientation-preserving or reversing.

So what I decided to do is to pretend like it's a product of manifolds and try to understand the fiber and the base separately.

I guess you guys were trying to explain why the Euler class became the negative in terms of orientation things.

A (smooth) bundle isomorphism to me looks like it should be orietnation preserving onthe base since it preserves base points

It's the intersection number with the zero-section that changes.

@TedShifrin Right, the Euler class switches sign. But the Euler number can't...

Yes, right. We're just looking at fibers.

What? What do you mean by Euler number? The underlying base manifold doesn't change.

I meant the Euler characteristic.

Multiple choice questions are fun XD

Euler characteristic is just for the tangent bundle. Forget about it.

So in the situation I'm thinking about BOTH spaces are diffeomorphic to a third thing which itself is the tangent bundle of another manifold. That's why I said Euler characteristic, and why I know it cannot change.

Are they isomorphic bundles, though? This sounds like that subtlety we talked about months ago.

Diffeomorphic as manifolds doesn't mean isomorphic as bundles.

Probably not iso as bundles, I guess.

So characteristic classes can be different.

They are, in fact, different (Chern numbers).

So I don't see anything going wrong with orientations of the fibers. But intersections with the zero section change sign.

I'm confused.

That's because local index does change sign.

Basically because of the tensor transformation rules.

Why do intersections with the zero section change sign

and what do you mean by local index

Index of a zero of a section (like index of a vector field) is the local intersection number with the zero section.

So why do they change sign?

In fact any complex rank 3 bundle over $S^6$ is self-dual, because $\pi_5 U(3) = \Bbb Z/2$.

Aha.

I was trying to avoid the whole complex thing at the moment, but that's interesting.

Are you back stateside, @MikeM/

In Chicago. Flight has been badly delayed.

:-(

In any case, I think you're right that we shouldn't be thinking about the complex dual @Ted

D'you enjoy the UK at least, Mike?

Oh ... A physics prof friend of mine at UGA was badly delayed for a day at Chicago a few weeks ago.

@Simple Haha, can I adopt you just because of your username?

@SimplyBeautifulArt sure

:)

:D Ok

The workshop was good, yes.

@Danu: Spiritually, what I'm talking about is that a vector field with one singularity on $S^2$ has index $2$, whereas a one-form will have index $2$. In the holomorphic category, we see a pole of order $2$ because setting $w=1/z$, $dw=-1/z^2\,dz$, and so $dz$ has a pole of order $2$ at infinity.

@MikeM: Any news on your laptop?

Oh, that's been recovered. But I can't pick it up til Tues.

Recovered back in LA?

yeah

Oh, cool. :)

@Semiclassical I'm curious whether my answer to math.stackexchange.com/questions/2104820/… is at all helpful/interesting/etc

I was afraid you'd "forgotten" it at an intermediary airport.

@Astyx inflational new users

whats the difference between an anti derivative and an indefinite integral ?

By definition they're the same, @WDUK.

@MikeMiller Did you live without computer? :)

He still had phone @Danu

oh - people seem to use the terms differently

I don't like referring to $\int_a^x f(t)\,dt$ as an indefinite integral, @WDUK. That is a definite integral.

Typically, $\int f(x)\,dx$ is an indefinite integral.

@TedShifrin Okay. So... can we conclude anything about the orientation?

Dualizinf a real rank k oriented vector bunxle changes the sign of the orientation of the fiber by (-1)^n, changing the orientation of the base of the manifold does just that.

@Danu: I guess I don't see canonical orientations in the first place.

Slow down, @MikeM.

No.

You mean $(-1)^k$?

Or you meant $(-1)^n$ where $n$ is the dimension of the base?

And I'm confuzled. I'm feeling very dumb.

k

sorry

@TedShifrin I've felt like this all day.

@TedShifrin hi Ted i need some help with flux integral and divergence theorem, am doing a review of the course

Also, that makes me sad because in my case everything is even-dimensional so I don't get any signs. But I need a sign

BTW, @Danu, since you know Huybrechts's book so well, you should answer this. I haven't yet found where that gets said in the book.

dualizing a rank k complex goes by the same formula - that's why e(T*P1) is different than e(TP1)

@TedShifrin if your not busy may i ask you couple questions ?

I am busy, @Kasmir.

okay thanks anyway :)

@WDUK We can private chat about calculus: chat.stackexchange.com/rooms/51337/simply-beautiful-arts-room

but T(-P1) is T*P1, but when you compute the oriented intersection number the two orientations cancel out

leaving you the Euler characteristic

cant right now maybe later :)

Ok :-)

how does one calculate cos(x)=0? Using $$cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$. What would be the things one have to know?

@Socrates Easy

@TedShifrin The guy didn't read the book carefully (?)

I've taught all this stuff, but my brain is dead. Where does the $(-1)^k$ come from?

$e^0=?$

@Semiclassical 1

$$0=\frac{e^{ix}+e^{-ix}}2\implies e^{ix}=-e^{-ix}$$

Right.

@Semiclassical I pretend I didn't read that

Lol

@MikeMiller So now what about the projectivizations and the induced map $\Bbb P(E)\to \Bbb P(E^*)$? Any idea? Just the same as for $E$?

And don't forget $-1=e^{(2n+1)\pi i}$

Dropping all the arbitrary stuff, we are left with

$$x=-x+(2n+1)\pi$$

But I supposed I assumed that one step

certainly the 2 is irrelevant for this, so we search x, such that $e^{ix}=-e^{-ix}$

You can multiply both sides by $e^{ix}$ to get something convenient.

True, that is good

Lol, this is how we solve things like $\cos(x)=0$.

0 as the right exponent

so $e^{2ix}=-1$

@Socrates Square both sides

and drop the cases when $e^u=1$.

Hm...

Now here I have a problem, because I don't even understand $e^{\pi i}=-1$

Huh?

@Socrates Add one to both sides

$$e^{\pi i}+1=0$$

Which is Euler's identity (I think)