Mathematics

12:02 AM

wow it's so quiet here

12:29 AM

@TedShifrin I've got a small question.

So if I have a (Weil) divisor $D$ and its associated Cartier divisor is given by $f_i\in \mathcal K^*_X(U_i)$ then $\mathcal O(D)$ is determined by the cocycle $\{U_i,f_i f_j^{-1}\}$, right (not $f_j f_i^{-1}$!)

This is fine, but it runs me into a small problem in a proof later on---consider the canonical bundle of the blow-up of $0\in\Bbb C^{n+1}$

Simple Art

Hi all

Danu

I want to determine $K_{\hat X}$ ($K$ is canonical bundle, $\hat X$ the blowup). Consider the maps $\varphi_i( (x,z)\in \mathcal O(-1)=\hat X)\mapsto (x_0/x_i,\dots,x_n/x_i,z_i)$ which are isomorphisms from the open sets $V_i=\{x_i\neq 0\}$ to $\Bbb C^{n+1}$.

Oh, never mind! I know what happened---but it seems like Huybrechts makes a rather "big" mistake in his proof.

He determines the cocycle of $\det \mathcal T_{\hat X}$ but pretends like it's that of $K_{\hat X}$

Then he computes the cocycle of $\mathcal O(E)$ where $E$ is the exceptional divisor

Of course it doesn't match, but he pretends that it does by saying that for a divisor, $\mathcal O(D)$ is given by the cocycle $\{U_i,f_jf_i^{-1}\}$

This way, him not dualizing the $\det \mathcal T_{\hat X}$ and getting the line bundle of a divisor wrong cancels out, so he gets the inverse answers for both

They match, and he pretends like it's okay.

I ran into trouble because I calculated $\mathcal O(D)$'s transition functions correctly, then got the inverse of the ones that were supposed to be $K_{\hat X}$'s...but they're really $\det\mathcal T_{\hat X}$'s.

Hi @MikeMiller

Nice monologue, huh?

OneRaynyDay

Hey guys - how's everyone doing?

I'm starting a numerical linear algebra class, and they told me to review some terms I've never heard of. After learning up about them, I'm curious to their applications

Danu

I'm doing good, I just finished that proof ^^^

OneRaynyDay

So I was wondering if I could have a conversation about them right now :) Just to get a little exposure on the topic

@Danu Haha yep, can see that

Are you doing real analysis?

Danu

12:42 AM

Heh, no

I'm doing complex geometry.

OneRaynyDay

ahh~ Well, I'm not even close to that level of mathematics - I'm a filthy CS major ^_^

Simple Art

Hi @Fargle

I wanna learn complex analysis

and analytic number theory

they sound like fun subjects to learn

Danu

@OneRaynyDay So what terms?

OneRaynyDay

But anyways, so I learned some interesting matrices, but not sure why they're important

@Danu Just a glossary of them: hermitian matrix, unitary matrix, LU decomposition, Hadamard product and Kronecker product

Danu

@OneRaynyDay I know a few of those.

OneRaynyDay

12:45 AM

I've learned QR decomposition, but never LU. I understand how each of these are created from a matrix along with their definitions, but I don't really understand

Danu

Hermitian matrix is one that's invariant under transposing + complex conjugating

OneRaynyDay

their usefulness in the actual world

right :) Unitary is when you take the hermitian transpose of A, call it A' and $A*A' = I_n$

Danu

It's the complex analog of symmetric---lots of important theorems that hold for symmetric matrixes in the "real" context hold for Hermitian ones in the complex context.

So things like spectral theorem

OneRaynyDay

I see. But when do we actually use complex matrices in the real world? (Other than fourier transforms)

Danu

Decompositions, I don't know and don't really care.

OneRaynyDay

12:46 AM

No idea what that is, but I do see that on my syllabus for later on

Danu

@OneRaynyDay All of quantum mechanics

OneRaynyDay

Ah I see. That's a field I'm completely blind about as well haha

Danu

@OneRaynyDay Spectral theorem tells you that you have a basis of eigenvectors

So that you can diagonalize

OneRaynyDay

oh - I've indirectly learned it then

Danu

Also about complex matrices---I recall some nice application

You can model some things like resonances in terms of complex eigenvalues of matrices

OneRaynyDay

12:47 AM

So basically every square matrix of nxn has n eigenvalues, and those eigenvalues correspond to subspaces in which the eigenvectors exist in?

Danu

I had a prof. for linear algebra who told us this story of how he worked on trying to fix this train that kept on breaking

OneRaynyDay

resonance? like the physics term?

Simple Art

XD quantum mechanics fun stuff?

all so interesting

wish I understood more :(

Danu

It turned out there was a small resonance when riding the tracks, that manifested itself as a tiny nonzero complex eigenvalue of a matrix in his mathematical model

OneRaynyDay

ahhh interesting

Danu

12:49 AM

@OneRaynyDay Now, you have to be careful. Real matrix, or complex matrix?

OneRaynyDay

Oh, I suppose for complex.

Danu

Do you remember why?

OneRaynyDay

My previous professor only talked about matrices with real eigenvalues.

Because the characteristic polynomial equation

may have imaginary roots?

Which are in turn the eigenvalues of the matrix

Danu

So why does it work over complex numbers?

OneRaynyDay

Not 100% sure of my answer, feel free to correct me :)

Danu

12:50 AM

It's the fundamental theorem of algebra---which tells you that any polynomial over $\Bbb C$ has a root.

OneRaynyDay

It works over complex numbers because there exists n solutions for a polynomial of ^n when complex numbers are involved?

Danu

Of course, once you found one root you factor it out and are again left with a (lower order) polynomial, so inductively you find $n$ roots for an order $n$ polynomial

Exactly

Simple Art

$\ddot\smile$

OneRaynyDay

Oh I see. gotcha :) proof by induction~~ yay

Simple Art

$$\ddot{\ddot\smile}$$

4-eyed

OneRaynyDay

12:51 AM

Awesome, I feel like I'm getting excited about linear algebra already thanks to you

Danu

Easy way to remember that this doesn't work over the reals is to take a nontrivial rotation in $\Bbb R^2$ (can you find the exception?).

It has no eigenvectors! This is obvious, because it rotates everything.

Simple Art

? Are we solving polynomials/systems of equations with matrices in linear algebra?

Danu

@SimpleArt Finding the eigenvalues of a matrix is solving a polynomial equation.

OneRaynyDay

Oh right - the matrix $\begin{bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix}$ ?

Simple Art

Oh, huh, that's interesting.

Danu

12:52 AM

@OneRaynyDay That'd be a 90 degree rotation.

Simple Art

Hm, maybe I'll go learn linear algebra

OneRaynyDay

Right - and it would have no eigenvectors since the charpoly(A) is $x^2+1$

Simple Art

cya guys

OneRaynyDay

cya art!

It's not?

Simple Art

$\ddot{\ddot\smile}$

OneRaynyDay

12:54 AM

$A^2$ gives the matrix $\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$

Danu

It's a clockwise rotation

I was thinking anti-clockwise

OneRaynyDay

ah yeah

oh god what is going on with my LaTeX

there we go

Danu

Anyways, I find it much clearer to think geometrically about that example

Now, do you know a nontrivial rotation in $\Bbb R^2$ that nevertheless has eigenvectors?

OneRaynyDay

yep!

the $A^2$ matrix I said above?

Danu

Which is...?

(in terms of rotations)

OneRaynyDay

12:56 AM

Simply because given any vector $v$ multiplied by that vector $A$ gives the vector $-v$

It's 180 degrees :)

Danu

As you can tell, I like to think visually

Right

OneRaynyDay

Yeah - no I totally get it. I learned linear algebra by drawing diagrams

Danu

So only $\operatorname{id}$ and $-\operatorname{id}$

lol... used to having my custom commands :(

OneRaynyDay

\id?

Danu

identity

OneRaynyDay

12:58 AM

ahh gotcha :)

Hmm yup - so the later topics talk about linear regression which is fairly simple, and PCA

Danu

brr, doesn't look like my cup of tea

OneRaynyDay

With a slight gloss over online, it seems like PCA chooses a couple eigenvectors from the linear transformation and uses the ones that preserve the original vector the most

Danu

Principal component analysis

Principal component analysis (PCA) is a statistical procedure that uses an orthogonal transformation to convert a set of observations of possibly correlated variables into a set of values of linearly uncorrelated variables called principal components. The number of principal components is less than or equal to the number of original variables. This transformation is defined in such a way that the first principal component has the largest possible variance (that is, accounts for as much of the variability in the data as possible), and each succeeding component in turn has the highest variance possible...

OneRaynyDay

ah - no problem :)

Danu

Lots of info

OneRaynyDay

12:59 AM

Righty - I find wikipedia, however, to be extremely hard to read(maybe it's just me)

but not as bad as research papers

Ted Shifrin

@Danu: your transition functions are, I think, upside down. Remember that you have poles along $D$.

Danu

@TedShifrin No, wait what?

The equations I'm using to define the exceptional divisor have zeros, not poles.

Perhaps it's best to look at the full proof together

Mike Miller

morning

OneRaynyDay

Anyways, thanks @Danu! You have a nice day :)

I'm off, got some filthy CS assignments to do.

Danu

It's night (3 AM) here :P

OneRaynyDay

1:10 AM

Oh.. 6PM here.

Danu

@TedShifrin So what I said about the proof: (i) he calculates the cocycle of $\det \mathcal T_{\hat X}$, not of $K_{\hat X}$. (ii) His rule for getting the cocycle from a divisor is upside down---this is confirmed by the corollary he refers to. Also, once you flipped the cocycle to get that of $K_{\hat X}$, one had better find an excuse to flip the other one too, right? :) Otherwise they won't match.

Do you still think I'm getting it the wrong way around?

Please let me know what you think @Ted---I have to get to sleep now.

I'm playing tennis in 5.5 hours...

sigh

Transcript for

$Mathematics$ Mathematics