Okay. Here's another way that seems to avoid induction...not sure if it's any better. Given $A \in SO(n)$, the columns of $A$ form an orthonormal basis. These basis should be able to be continuously rotated to the standard basis vectors which represents the identity...

This should form a path from $A$ to $I$...but I'm not sure how to make it more precise...

The only thing that worries me is that such reasoning would seem to prove that $O(n)$ is path connected, but it isn't...so I must be missing something.

Boring linear algebra problem: Suppose I have two unit vectors $u,v$ and their corresponding projection operators $P_u:=u u^\top$, $P_v:=vv^\top$. If $u,v$ are orthogonal, then the projector onto the span of $u,v$ is just $P_u+P_v$.

Is there an obvious modification to make when $u,v$ fail to be orthogonal? (i'm assuming $u\neq \lambda v$)

I suppose the smart thing to do is take $v$ and project it onto the orthogonal complement of $u$, i.e. $v':=(I-uu^\top)v$

Then $\|v'\|^2 = v^\top (I-uu^\top)v = 1-(u^\top v)^2$, so the projector along $v'$ would be $$P_{v'} := \frac{v'(v')^\top}{\|v'\|^2} = \frac{v(I-P_u)v^\top}{1-(u^\top v)^2}$$

so therefore the projector onto the span of $u,v$ should be $P_u+P_{v'}=uu^\top+\dfrac{v(I-P_u)v^\top}{1-(u^\top v)^2}=uu^\top+\dfrac{vv^\top -(u^\top v)^2}{1-(u^\top v)^2}$

@user193319 look up Givens rotations

@LeakyNun What do I do with them? Is $SO(n)$ homeomorphic to upper hemisphere of $S^n$?

no

Hmm...shoot.

@user193319 as a corollary of the spectral theorem, you can show that for any $A \in SO(n)$ there is a basis such that $A$ is block diagonal with matrices of the form $\begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{pmatrix}$, you can continuously deform such a matrix via the path to the identity matrix given by $t \mapsto \begin{pmatrix} \cos(t\alpha) & -\sin(t\alpha) \\ \sin(t\alpha) & \cos(t\alpha)\end{pmatrix}$

you can do this for each block diagonal matrix (each with a potentially different $\alpha$) and then compose with the needed base change and this will give you a path from $A$ to the identity matrix

there's no induction in this of course, but I don't see how to do it inductively

@MatheinBoulomenos perhaps a fibration $SO(n) \to SO(n+1) \to S^n$ is inductive

Where do we use, in this proof, that $f(a,b)$ is not onto?

the induced LES ends with $\pi_1(S^n) \to \pi_0(SO(n)) \to \pi_0(SO(n+1)) \to \pi_0(S^n)$

for $n>1$ this becomes $0 \to \pi_0(SO(n)) \to \pi_0(SO(n+1)) \to 0$

so $SO(n)$ has as many path-components as $SO(n+1)$

I like this

there you go @user193319

are the rationals a small set w.rt. the reals

@MatheinBoulomenos do you know if the curves $y^2=x^3-x$ and $y^2=x^3-x+\varepsilon$ are isomorphic as curves over $\Bbb C[\varepsilon]/(\varepsilon^2)$?

@LeakyNun lol i think my friend solved it

how?

I think my j-invariant thing didn't work

@LeakyNun I don't

@loch what is this infinitesimal deformation thing that doesn't seem to correspond to any real (i.e. complex) numbers and live in a world of its own

@MatheinBoulomenos Shouldn't it be a direct sum of 2x2 rotations possibly together with number of $1$'s? How do you prove the corollary? It is a quick proof? It's not clear to me how to prove it.

Given $A \in SO(n)$, $A$ is normal and therefore unitarily diagonalizable. I.e., there exists a unitary $V$ and a diagonal matrix $D$ such that $A = V^* D V$. But why is $D$ block diagonal?

@user193319 yeah there is also a number of $1$s

Wait, can't there also be some $-1$'s? Doesn't show that the map you defined might not be continuous?

there can be $-1$, but only an even number of them

and if we have two $-1$, there is no problem, note that $\begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{pmatrix}=-\mathrm{Id}$ for $\alpha=\pi$

Oh, yes. I see.

this is why this works for $SO(n)$ but not for $O(n)$

@MatheinBoulomenos Okay. So $A = Q^{-1}DQ$, where $D$ is the block diagonal of the form described above. I know the columns of $Q$ are the eigenvectors of $A$; but why does this imply that $Q \in SO(n)$?

@user193319 I didn't state the corollary in full strength, you can even find an orthogonal basis such that the matrix has the desired form

Oh, so $Q$ will be in $SO(n)$?

yes

or at least in $O(n)$

For the corollary I don't remember the full proof, but the idea was that you diagonalize over $\Bbb C$ and since the matrix is real, eigenvalues and eigenvectors come in conjugate pairs and then you combine them like $v + \overline{v}$ and $v - \overline{v}$, then rescale by $\sqrt{2}$ to keep unit length and this will give you your real basis

here conjugation in $\Bbb C^n$ is done componentwise

Oh,...if it's in $O(n)$, that might be problematic. You've shown there is a path from $D$ to $I$ in $SO(n)$; but when you conjugate this path by $Q$, it might land outside of $SO(n)$, right?

no, conjugation doesn't change the determinant

Ah, crap!

I'll get all these details in my head eventually. Thanks for the help.

no problem

Evening @Mathein

Evening @ÍgjøgnumMeg

hyvaeae yoetaeae @ÍgjøgnumMeg

:P

hahaha nice @Leaky

@ÍgjøgnumMeg I proved that SO(n) is path-connected inductively

χαῖρε @Leaky

servus @MatheinBoulomenos

lmao

wait salve is also right?

salve is a Latin greeting

also used in Italian still

servus?

servus is a Bavarian greeting

wait what I thought servus was Latin lmao

ok til

servus is Latin, meaning slave

when I heard it in Bayern ich dankete warum sprichst du Latin

daaachte

but I don't think it's a greeting in Latin though

I think the etymology is that it's short for "servus sum" "I'm your servant" in the sense of "at your service"

or something like that

yeah I heard that too

ciao has the same etymology lol

ciao as in time to ciao down on this delicious food that I have, as your slave, prepared for you

lol

sup

hi @Ryan

I was searching for this exact thing and was shocked that it exists

@MatheinBoulomenos have you done any graduate courses?

@LeakyNun do master courses count?

I guess

then yes

how many hours per week do they cost you?

depends on the course and how diligent I am, ANT1 was pretty easy imo. I'd guess something between 6 and 7 hours

I have no idea how much each course would cost me

@MatheinBoulomenos how do I derive such a number?

not sure I was just guessing based on my experience

I may be wrong with the guess

my prof is saying that ag1 will take up half my time

probably grad courses at MIT are more intense than masters courses here

after I correctly answered what the Nullstellensatz is

another student couldn't define primary ideals so he said ag1 will take him more than half his time

do you know about Fourier transform of sheaves?

Fourier Mukai or something

yeah

not really

@MatheinBoulomenos for every $A$ there is $S_\bullet$ such that $\operatorname{Proj}(S_\bullet) = \operatorname{Spec}(A)$

does $A[x]$ work?

idk about Proj anymore it doesn't have any UMP

@AkivaWeinberger you might be interested

can't you just take $S_{\bullet}=A$ with the trivial grading (everything has degree $0$)?

idk

@LeakyNun I don't know the original

I think Proj(S_0) is always the empty set

Sounds nice though

@AkivaWeinberger it's like super famous how can you not know

What makes it negative?

when you flip everything, major becomes minor and all that stuff

Ohhhh

Looked up the original

Yeahh I know that one

just not by name

Didn't make the connection

nice

@LeakyNun right

ok so we definitely have a map Proj(A[X]) -> Spec(A)

Yeah so the inverted version sounds really nice

cool

A[X] doesn't work, consider $A=\Bbb C$

what's Proj(C[X])?

isn't that just Spec(C)

well, it's $\Bbb{P}^1_{\Bbb C}$

no that's Proj(C[X,Y])

ah

ok so every element of degree > 1 in A[X] is just a x^n

so prime ideals must have something of degree <= 1

and since it can't be the irrelevant ideal it must have something from A...

what happened to homogeneous

do we get $\mathfrak p A[X]$ and $\mathfrak p + (X)$ being different prime ideals :o

oh the latter is invalid as it contains the irrelevant ideal

what if I have $\mathfrak p \le \mathfrak q$ and then consider $\mathfrak p + \mathfrak q (X)$

is that a prime ideal

right, the ideal must generated be generated by elements of degree 0. Because all homogenous elements are of the form ax^n, and it doesn't contain x

so it's the extension of a prime ideal in $A$

it seems to be a prime ideal

so there's a bijection between Spec(A) and Proj(A[x]) by extension and retraction

but my example contradicts you

consider $A=\Bbb Z$, $\mathfrak{p}=0$, $\mathfrak{q}=(q)$ for $q$ prime, $(qX)$ isn't prime

hmm

more generally if $q \in \mathfrak{q} \setminus \mathfrak{p}$, then $qX \in \mathfrak{p}+ \mathfrak{q}(X)$, but $q,X \notin \mathfrak{p} + \mathfrak{q}(X)$

how do you prove that it's a bijection?

If $\mathfrak{p}$ is a homogenous prime ideal not containing $x$, then it must be generated by elements in $A$, since all homogenous elements are of the form $ax^n$

so if we intersect with $A$ and extend again, we get the same ideal

hmm

and if we take an ideal $I$ of $A$, extend to $A[x]$ and intersect with $A$, then clearly we get the same ideal back, since the extended ideal just consists of all polynomials with coefficients in $I$

@MatheinBoulomenos so Proj(A[X]) = Spec(A) !

yes, you were right

@AkivaWeinberger ^^^

@MatheinBoulomenos no it doesn't help me understand Proj lol

@LeakyNun you can think of proj as a quotient. A grading on $A$ is equivalent to a $\Bbb{G}_m$ action of $\mathrm{Spec}(A)$ and Proj is the quotient by that action

or at least a quotient of some subset of $\mathrm{Spec}(A)$

G_m?

the multiplicative group

base?

$\mathrm{Spec}(\Bbb Z[x,x^{-1}])$

:o

why?

So given a grading $A=\bigoplus_{d \in \Bbb Z}A_d$, we can define a homomorphism $A \to A[x,x^{-1}]$ given by sending $a \in A_d$ to $ax^d$

and conversely if we have a homomorphism $f:A \to A[x,x^{-1}]$, we get a grading on $A$ given by $A_d=f^{-1}(A \cdot x^d)$

hmm

and what's with the $x^{-1}$?

we want $\Bbb Z$-gradings for this I think

oh

I thought you do $\Bbb N$-gradings for Proj

now $A[x,x^{-1}]=A \otimes_{\Bbb Z} \Bbb Z[x,x^{-1}]$, so taking $\mathrm{Spec}$, we get a map $\mathrm{Spec}(A) \times \Bbb{G}_m \to \mathrm{Spec}(A)$

well, any $\Bbb N$-grading is also a $\Bbb Z$-grading

@LeakyNun Sick beats bro