It is amusing that inverse computations are pretty much never done in applications. Even in CG when one needs explicit inverses usually the matrices are given as explicit compositions of things you already know the inverses of.

It was very satisfying for me, though, to work out exactly how Mathematica draws pictures (including the PointOfView command) and to include that as a long exercise.

Most common application of by hand GJ inversion of matrices is job interviews.

To my knowledge.

GJ?

Oh Gauss-Jordan

Yeah, if you do elementary matrices in the right way, then inverse will be simple, but you have to be careful for things to commute (and I'd have to look at the book to remember)!

Of course classical adjoint is the way everyone should compute inverses.

Using adj(A)/det(A)? That takes so long beyond 3x3!

Gaussian elimination is the easiest way to compute inverses aren't they

Balarka clearly missed my trenchant sarcasm.

Oh

lol

I love PQ, LU and QR as well, they're nice

I also like generalized inverses

What is PQ?

Useful for solving underdetermined systems

@TedShifrin Rank factorization

Writing mxn as mxr (full rank) times rxn (full rank)

QR is one way to do it

Oh, I guess I have not seen this. It's sorta like the rank theorem, I guess.

Compose immersion and submersion.

Yeah.

There's a very natural way to obtain PQ from Gauss-Jordan

It's magic

I think you go to row reduced echelon form, look at the pivot entries. The pivot rows in the original matrix is P, the pivot columns in the row reduced form is Q

Linear algebra is the art of finding unused two-letter pairs for factorization theorems.

Something like this

Indeed, @Fargle.

"Now you've learned your ABCs ...."

I have done so much linear algebra throughout my undergrad

It's all coming back

Most people in the US don't do nearly enough.

It's linear algebra all the way down.

And Ted's right, but don't tell him I said that.

hi @TedShifrin

What didn't you say, @Fargle?

hi Karim.

Here is an insane fact. Let $A, B$ be $n \times n$ matrices over any commutative ring $R$. Then $x I_n - A$ and $x I_n - B$ are equivalent over $R[x]$ iff $A$ and $B$ are similar over $R$

I think the last one I took was in middle school or high school, but I have no idea.

@Balarka: It seems to me that came up somewhere in Artin's treatment of linear algebra/modules.

That is possible

Mensa isn't an acronym

I made the kids show that if A is an invertible linear map then it's square and its inverse is linear (as the two most notably proofy problems this week). Other more standard stuff about understanding what "this particular system has a unique solution" says about rank, etc.

a good linear algebra book *

@LeakyNun maybe he was excited about his test results

that is great @MikeMiller

@MikeMiller can you list some if it is ok with you.

Actually, most everything should be learned eclectically.

No, @Karim, I don't really know. The biggest problem with complex geometry comes at the beginning with complexifications. Does $d\bar z(v)$ actually equal $\overline{dz(v)}$ or does it equal the anti-holomorphic part of $v$? So much confusion.

@TedShifrin oh yeah... I should do that. easy enough if $b$ is a unit...

Nothing should be learnt

The only one I consistently recommend is Treil. I don't memorize these lists of books.

Sounds good @MikeMiller I will try it. Thanks.

I find that the more I understand something the less I like any one exposition of it

There are lots of things I've learned by teaching and then having my own favorite way to do it. Whence writing too many books.

@MikeMiller The converse is not true

Proof: I hate this book I am reading right now

Even though I do not know probability

Yeah have few issues about complexification. I was reading this book where he defined conjugation as operator $C : E := E \oplus E \rightarrow E \oplus E$ though he doesn't explicitly mention that actually the space $\bar{E}$ is $(C(E),-J)$

For instance, there are no good smooth manifolds books. There are some pretty good differential topology books.

amazon.ca/Holomorphic-Functions-Complex-Manifolds/dp/0387953957

It is this book it is good but I have some quibbles

Yeah agree things should be learned eclectically especially Algebraic Geometry.

@MikeMiller I still know only a handful of 3-manifolds

It's insane that I can confidently say I know smooth manifolds but not know any examples of smooth manifolds

You can learn all the seifert fiber spaces and still not understand 3-manifolds unless you get hyperbolic geometry. (I do not yet understand 3-manifolds.)

I never have attempted.

The worst thing is I will never learn 3-manifolds because all of 3-manifolds has been "done"

No one in their right mind will offer a course about a dead subject

I've taught mostly (95%?) courses about dead subjects.

It isn't dead if it's a staple tool :)

let E be the tautological line bundle over CP^n. this comes naturally with a bundle metric. let D(E) be the corresponding disk bundle and S(E) be the corresponding sphere bundle. is it true that D(E)/S(E) is CP^{n+1}? I only need a y/n.

What is $p$, @BigSocks?

I understand the classification of surfaces pretty well, but not the symmetries of hyperbolic surfaces.

wait no that's wrong hold on

I don't understand anything living

@Thorgott Yes, tell me why.

Hyperbolic geometry is too hard

Is tautological bundle what I mean by tautological line bundle or its dual?

$\mathscr O(-1)$ or $\mathscr O(+1)$?

Topologically doesn't matter, but you'd like to think about O(1)

If you understand why the explanation will be very short

So it's the dual of mine.

Right

I want to think about the normal bundle of $\Bbb P^n$ in $\Bbb P^{n+1}$.

I don't know what that notation means, but I mean the bundle where the fiber over a point of CP^n is the 1-dimensional subspace of C^{n+1} represented by that point

That's O(-1), but it's better to think about the dual like Ted described.

Thorgott means the one for which the answer to his question is yes. :D

Oh, @Thor is using the "right" tautological. So I want to dualize to get it right for complex geometry.

@Balarka: You like to say topologically it doesn't matter, but Chern classes change sign, even topologically.

I think Thor needs the dual.

O(-1) is isomorphic to O(1) as real bundles, is what I mean.

And, yes, this is always confusing, especially when topologists mess up. I had the fight with Rob Kirby back around 1978 when he was learning complex geometry :P

Ok @TedShifrin so it's usually that $a \vert n$ and $b = n/a$ that I can imagine getting the same ideal from zero divisors.

So you basically always multiply by $-1$ to get the other

maybe there is another way, but it's not coming to mind

Thorgott is famous for taking the definition which makes it hardest to see why something is true

@BigSocks: Well, in $\Bbb Z/n$, we don't necessarily have $\bar a = -\bar b$, do we?

what is $\overline{a}$ here?

The representative of integer $a$ in $\Bbb Z/n$. You must distinguish between integers and equivalence classes.

Use brackets if you prefer.

sure, ok

no we don't

i didn't choose this definition man

Yeah I got confused because I often multiply by a unit that squares to $-1$

I guess to $[-1]$

You're right, @BigSocks, in that the way it's going to arise is by taking the negative. I didn't understand your $[b]=[n/a]$, though.

@TedShifrin Be careful. They are different as bundles over fixed base, which means when you do not allow yourself to use symmetries of the base. And when you allow only holomorphic symmetry then they are also distinct.

But complex conjugation takes the one to the other.

Ted was making the distinct argument, I was arguing against. I clarified what I meant :)

Yes, which accounts for the sign change on Chern class.

As a complex geometer, I just refuse to muddle myself the way topologists do.

This is always quite irritating. For instance, it is often said that there are Z many oriented plane bundles over a surface. Not even up to oriented diffeomorphism! There are N many. But yes as oriented diffeomorphisms covering the identity.

Hmm but in $\Bbb Z / 12 \Bbb Z$, $(2) = (10)$, and $[2] = [5]*[10]$ $\wedge [2]*[5] = [10]$, and $[5]^2 = [25] = [1]$, so that's kind of odd that it's a class that isn't $[-1]$ that squares to $[1]$. hmm

Z doesn't make sense in real world

Hmm, I don't feel muddled. I am just giving a complete discussion.

and I meant $[b] = [n - a]$ right

Well, topologists still want Chern class to classify topological isomorphism :)

We have had this conversation before

No, I feel muddled, @MikeM.

I shouldn't have said "topological" yeah

I meant "real"

Yes, I agree. I think it is just important to point out the difference between isomorphism covering the identity and isomorphism covering isomorphism.

The latter is harder but sometimes what one is really asking for

But it was funny having these discussions with Rob Kirby many years ago. Of course, he was shortly thereafter an expert. :P

Two step problem

I don't even contemplate isomorphisms covering anything other than the identity.

Rob Kirby has a strange essay about sexism in mathematics

Fun fact

Aha

I don't want to leave that there without being 100% sure.

Yeah lol good idea.

But what you said makes sense, given essay

Yeah, lots of male chauvinism in mathematics, I'm afraid. And it propagates in many ways.

The UGA math department had a huge scandal a few years ago not unrelated.

That is not good mathematicians should be rational.

I think I know the one you speak of

I think it may have been mentioned here, yes.

Anyhow, I didn't mean to waylay Thor's discussion with Mike.

I would be impressed if one could give a coherent, universally applicable, and consistent definition of rational. I think this is a famous open problem.

There are Tromp-supporting mathematicians. Ergo, @Karim must be wrong.

hides

Hilbert's 10th problem over $\Bbb Q$ yeah

Mathematics is not epitome of rationality. It's only a game. Gamers are full of shit, even the self-respecting ones: Kasparov famously thinks women are biologically bad at chess (which is nonsense lol).

I need to write some lectures anyway.

Anyhow, @BigSocks, we still want a proof. I suggest you lift up to $\Bbb Z$ and then try to deduce what happens downstairs.

ok ok

The proof should generalize to a quotient of any PID.

That's provably false, is the problem.

schizophrenics are really good at making connections

Back to work, @Big.

scampers off

@MikeMiller I think perfect rationality is inertia. Action is irrational. Any other definition will be self-contradictory or inconsistent.

are there any alggeo'ers here that now why for an affine variety X and $U=X\cap D(g)$ we have the iso $O_X(X)[g^{-1}]=O_X(U)$?

I understand that we have a surjective morphism $O_X(X)[g^{-1}]\to O_X(U)$

since $g$ is invertible, and I have a result which basically says that this map is surjective

(basically what I asked some hours ago, but I realised that I might like to think of it this way)

@BalarkaSen Still love to think about the time he got rekt by Judit Polgar.

yeah lol

@Sha Can't you give the inverse mapping?

@ShaVuklia It will be hard for me to write it correctly, but this is like saying $\Bbb A^1 \setminus \{(0, 0)\}$ (which is an affine open in $X = \Bbb A^1$) is isomorphic to $xy = 1$, whose coordinate ring is $k[x]$ with $x$ inverted.

well @TedShifrin in $\Bbb Z$ you don't have zero divisors so there's not a lot to check. Also the only units are $-1$ and $1$ so the only way $(a) = (b)$ is to have $a = b$ or $a = -b$

@TedShifrin I wouldn't know how to extend a function uniquely

so nopes

Well I mean, I do know this:

If $X=Z(f_1,\dots,f_r)$ then $X\cap D(g)=Z(f_1,\dots,f_r,x_{n+1}g-1)$

I think that's what Balarka is using?

That's right.

OK, @BigSocks. Now I start with $[b]=[k][a]$, where $[k]\ne [0]$.

You have to be a little careful, because $\pi^{-1}([b]) \ne (b)$, right?

The coordinate ring of $Z(f_1, \cdots, f_r, x_{n+1} g - 1)$ is precisely $k[x_1, \cdots, x_n]/(f_1, \cdots, f_r)$, with $g$ inverted, yes?

ohh

OOOOOOO

Why am I like this

ok thanks

Balarka has learned the standard trickery.

This is a famous trick and it has a name.

The irony is

Oh, Sha was saying that the other day.

2nd time in 2 days it comes up

I actually mentioned that trick not too long ango

Lol ya

I never knew the name!

It takes time to internalize, I have forgotten it a few times before too

Sha = burned.

Lol, Ted

I said that only with affection :P

(I know xD)

More things in math should be called tricks. Let's promote intellectual honesty

AGREE

AGREE SO MUCH

Munkres taught me 50 years ago or so that a method is any trick that you use three times.

Damn, I'm old.

Product rule? No. Leibniz trick.

loool

I'm sure that he wasn't the first to pass that along.

Quotient rule?

Upside-down-Leibniz?

What's the chain rule?

Leibniz trick, plus the exponent trick, plus the composition trick.

LOL ... well, this could go on ad infinitum.

Moser's trick is still a trick to me, even if I have seen it used more than 3 times.

Moser's trick is insane.

Yes, it is perennially called that.

@TedShifrin Reminds me of infinite descent---which I propose calling the Fermat trick.

To solve a differential equation you do not integrate, you differentiate

Right @Fargle

Nuts idea. Also used by Gromov later

renaming "inverse function theorem" to "inversion trick"

and implicit function theorem to the implicit trick

ohhh, snake lemma = snake trick

thinks too much of a good thing may be toxic

Banach fixed point theorem does not deserve to be called a theorem

@TedShifrin you started it this time

ackchyually, it's the salamander trick

It's like a directed graph

Follow the arrow

To the proof

@BalarkaSen iteration trick

Banach maze

Labyrinth is a favourite synonym of maze

Makes it sound more lofty.

@anakhro Fraid so.

NotBurnside's lemma is a pretty good name.

lol

quotient trick instead of isomorphism theorem

What's the one theorem about algebras that is like the PCBASDFHEJ theorem?

Groups should be defined as a presentation

No abstract nonsense

@Thorgott awhile back on wikipedia there was like an edit war on the Isomorphism Theorems page where someone kept on trying to name the page as Noether's Isomorphism Theorems.

Or something like that.

Led me to being very confused as at the time I was taking a ring theory course and I couldn't locate the page on google to double check which theorem was which number.

@BalarkaSen why does it not deserve to be called a theorem?

Are they due to Noether? I assumed they'd be older.

@copper.hat the ease with which one can prove it with.

Though it's no easier than Bolzano-Weierstrass.

It's a directed graph

i.e. "cage the animal theorem"

@BalarkaSen surely the establishment of a useful fact deserves such a label?

everything is obvious afterwards.

@BalarkaSen ok, define $\operatorname{Aut}(\mathbb{R})$ via presentation

I would agree to call it Banach's directed graph

@Thorgott LMAO

Triangle inequality was called "the lazy ass" theorem by a professor who went on to explain that a donkey will take the shortest path.

Triangle inequality is not a theorem, it's a hypothesis

Triangle hypothesis is the hypothesis that space is not curved

does the implicit function theorem fall under this too simple guise?

you mean the "pretend analysis is linear algebra trick"?

@Thorgott I said groups, not abominations

weak response for 5 minutes worth of time

the geometry of $\langle r,s\mid r^n,s^2,(sr)^2\rangle$ is, of course, self-evident

Groups, like (wo)men, should be defined by their actions.

I don't remember who said that.

so free groups are lumberjacks?

loopy day at mse?

GROUPOID INTENSIFIES

Algebra is the offer made by the devil to the mathematician...All you need to do, is give me your soul: give up geometry --Michael Atiyah

lmao

@BalarkaSen Fortunately we have RB to take that same childish step repeatedly for 50 years

Lol oh no

And when we want to do something useful with it, we have

Do not ask whether a statement is true until you know what it means. -- Errett Bishop

This is so condescending

How so?

I'd have to effectively stop asking questions by that logic.

I don't know what anything means lol

The amount of students who come to office hours without knowing what the question means/is asking is basically testament to the quote.

I can only speak for myself

Balarka "Eiron" Sen.

@BalarkaSen For this I do need to show that $(f_1,\dots,f_r,x_{n+1}g-1)$ is radical. It's clear that $x_{n+1}g-1$ is irreducible, so for $X=\mathbb A^n$, I'd be done.

Or does taking the radical not mess with the localisation argument?

You do not mean $X = \Bbb A^n$. You mean $X \subset \Bbb A^n$ is a reduced affine subscheme in our argument, yes, I agree.

There must be a fix that I haven't thought about

No, I mean, if $X=\mathbb A^n$, then we just have $I(X)=(0)$, and since $x_{n+1}g-1$ is prime, we know that $k[x_1,\dots,x_{n+1}]/(x_{n+1}-g)$ is the coordinate ring

(Also, I'm not familiar with schemes yet)

@Thorgott I believe so. Probably some version of each of them was known before her, but the results as being general and central, I believe, is due to her.

Wait

Maybe I'm confusing one thing

I don't understand how you're writing things, I'm sorry. Why doesn't this work, simply? Let $X \subset \Bbb A^n$ be an affine algebraic set, then $k[X] = k[x_1, \cdots, x_n]/I$ where $I$ is the defining ideal of $X$.

Oh right, it's also radical

The defining ideal of $X \cap D(g)$ is simply the one generated by $I$ and $g$, yes?

sry I thought that it might have no been radical

@ShaVuklia No, I don't need radicality, is my point, I believe.

so by the Nullstellensatz we might get something bigger

interesting

$I$ does not have to be radical.

@BalarkaSen Ye, I believe so (that's what I meant when I said I might be confusing sth)

OK, cool :)

@BalarkaSen Hm, the statement in my book goes that it does take the radical

So if we have $X$ an affine variety, and we define by $I(X)$ its corresponding ideal (as by the Nullstellensatz)

Then $O_X(X)\cong k[x_1,\dots,x_n]/I(X)$

My issue now is that $I(X\cap D(g))=\sqrt{(f_1,\dots,f_r,x_{n+1}g-1)}$

We do know that $\sqrt{(f_1,\dots,f_r)}=(f_1,\dots,f_r)$, and $x_{n+1}g-1$ is irreducible

So I'm guessing $(f_1,\dots,f_r,x_{n+1}g-1)$ might still be radical

I guess... I should try proving it directly

So your varieties are reduced, by definition?

Well in this case I'm just working with an (embedded) affine variety

So just a closed subset of $\mathbb A^n$

Got it, so reduced guys :)

Ok @TedShifrin so doing some stuff with $\Bbb Z_{16}$ I saw that $(2) = (14) = (2)(7)$ so $(2)((1) - (7)) = 0$ and $(1) - (7) = 0$ since they are the same thing. so I think that probably if $a$ and $b$ are divisible by different primes you'll be able to factor some out, so at the most we will be looking at $a$ and $b$ powers of the same prime

Recall $D(g) = \{g \neq 0\} \subset \Bbb A^n$. There is an isomorphism of algebraic sets $D(g) \to \{x_{n+1} g - 1\} \subset \Bbb A^{n+1}$, given simply by $(x_1, \cdots, x_n) \mapsto (x_1, \cdots, x_n, 1/g(x_1, \cdots, x_n))$. Under this isomorphism the closed subset $X \cap D(g)$ is sent isomorphically to some closed subset. If $X \subset \Bbb A^n$ is cut out by $f_1, \cdots, f_r$, then the image of $ X \cap D(g)$ under this map is cut out by $f_1, \cdots, f_r, x_{n+1} g - 1$.

but idk it's kind of a hunch, not really a proof at all

I see, so the issue is in principle $I(X \cap D(g))$ might be something smaller than $(f_1, \cdots, f_r, x_{n+1} g - 1)$.

So I guess you need to argue by hand, yeah

@BalarkaSen You mean bigger, but yea

Yeah, sorry about that.