Mathematics - 2019-10-30 (page 1 of 2)

manooooh

00:00

So $(2)$ must be incorrect. But how do we prove it i.e. $(1)\not\equiv(2)$? I hope to explain myself well

Shine On You Crazy Diamond

Let $P(x) = x \mid yz \implies x \mid y $ or $x \mid z, \forall y,z \in \Bbb{Z}$. It is false

for non-prime $x$

manooooh

What would be $Q(x)$?

Shine On You Crazy Diamond

Oh

$P(x) = x \mid yz$

$Q(x) = x\mid y$ or $x \mid z$.

That $\forall x \in \Bbb{Z}, P(x) \implies Q(x)$.

It's true if and only if $x$ is prime.

or $0$

just call $0$ a prime, since $(0)$ is a prime ideal

manooooh

@ShineOnYouCrazyDiamond ok. I don't understand how to interpret it to say that it is not equivalent to $(2)$

Shine On You Crazy Diamond

Oh, try inverting it

$\neg Q(x) = x \nmid y $ and $x \nmid z$.

So for all $x \in \Bbb{Z}$ there exists $y, z$ such that $x \nmid y, z$.

This is confusing me already

manooooh

00:06

Yeah lol

Shine On You Crazy Diamond

perhaps you want a symbolic counterexample

manooooh

I thought you gave a counterxample. What is a "symbolic counterexample"?

Shine On You Crazy Diamond

Suppose that $\exists x, p(x)$

and $\exists x, \neg q(x)$.

MadSpaceMemer

@AkivaWeinberger My good dude I wanted to tell you I actually read about them and spend some time since we talked and did more exercises and now I have a very better understanding . :D

Shine On You Crazy Diamond

Then $\exists x $ such that $p(x) \wedge \neg q(x)$.

That's true

I think they're equivalent!

Akiva Weinberger

00:08

@MadSpaceMemer Yay :D

Shine On You Crazy Diamond

Oh, wait

MadSpaceMemer

no

they are not equivalent

manooooh

No they aren't. It is taken from an exercise that ask to check the ONLY correct option

MadSpaceMemer

In the first case this x could be different and the second you are saying it is the same x for both cases

Shine On You Crazy Diamond

All you need is a different $\exists x, p(x)$ and $\exists y, \neg q(y)$. If $x \neq y$ then...

Akiva Weinberger

00:09

Mhm

Shine On You Crazy Diamond

So you have an infinite number of counterexamples described abstractly

manooooh

@MadSpaceMemer so, $(2)$ can be rewritten as $\exists x(p(x))\wedge\exists y(\neg q(y))$?

Shine On You Crazy Diamond

Yeah

Akiva Weinberger

On the other hand, you can conclude $\exists x,p(x)\lor\lnot q(x)$

though it's weaker

manooooh

Don't make it hard lol. I want to show $(1)\not\equiv(2)$, only that

Akiva Weinberger

00:10

I mean I guess that's true even from only one of the hypotheses

MadSpaceMemer

what is the problem ?

manooooh

I can't understand how to make a counterexample

MadSpaceMemer

Never mind the chat loaded for me I see the problem

well you can suppose that it is correct and then draw a contradiction

manooooh

I have taken $p(x)=\text{$x$ is multiple of $4$}$ and $q(x)=\text{$x$ is multiple of $2$}$. Then for all $x$ it is true that $p(x)\to q(x)$, so negating that shows that ...??

MadSpaceMemer

I would not use numbers and instead solve it abstract.

Akiva Weinberger

00:14

$\forall x,p\rightarrow q$ is true, so $\lnot(\forall x,p\rightarrow q)$ is false, right?

So (1) is false

manooooh

@AkivaWeinberger oh ok, right

Akiva Weinberger

and if (2) is true then you've won

'cause you've found an example where they give different truth-values

manooooh

@AkivaWeinberger the problem is: how can we shot that (2) is true knowing that (1) is false?

Akiva Weinberger

Oh sorry I thought the problem was showing (1) and (2) are not equivalent

manooooh

@AkivaWeinberger I think I did not find any example

MadSpaceMemer

00:16

or you can show that the inclusion only works the one direction and since 2 $\Rightarrow $ 3 but 3 does not go in the other direction it is also a good proof

manooooh

@AkivaWeinberger yes, we want to show that $\neg(\forall x(p(x)\to q(x)))\not\equiv\exists x(p(x))\wedge\exists x(\neg(q(x)))$, or the same: $\exists x(p(x)\wedge\neg(q(x)))\not\equiv\exists x(p(x))\wedge\exists x(\neg(q(x)))$

Akiva Weinberger

OK

Let p and q be as you said

manooooh

Ok, then?

Akiva Weinberger

What is the truth value for (1)? What is the truth value for (2)?

manooooh

@AkivaWeinberger well, I do not know how to make truth tables for predicates. They can be true or false depending on the different values of $x$!!!

Akiva Weinberger

00:18

You have quantifiers!

$\exists$ means it's true if you've got at least one $x$ where it's true!

So once you have a quantifier (like "$\exists$") then it doesn't depend on $x$ anymore

Leaky Nun

!

manooooh

@AkivaWeinberger sorry I still do not understand :(

Akiva Weinberger

What did you have? P is "x is a multiple of 4" and q is "x is a multiple of 2"?

So (1) (using the second version) is "there exists an x which is a multiple of 4 and not even"

Is that true or false?

(2) is "there exists an x which is a multiple of 4, and there exists an x (not necessarily the same) which is not even"

Is that true of false?

manooooh

@AkivaWeinberger Oh it is false

Akiva Weinberger

The first formulation of (1), for the record, reads "not all multiples of four are even"

OK so (1) is false

What about (2)?

manooooh

00:29

@AkivaWeinberger well we can pick $x=4$ so $x$ is a multiple of $4$, and we can pick $x=3$ so $3$ is not even. "True and true" is true. So (2) is true, meanwhile (1) is false. Right?

Akiva Weinberger

Yup

That means (1) and (2) are not equivalent

manooooh

@AkivaWeinberger oh nice!! But I have a question: why the "not necessarily the same"? Why do you say that?

Akiva Weinberger

I'm just expressing the fact that you can choose different $x$s for the different halves of (2)

manooooh

I mean if I read "$\exists x(p(x))\wedge\exists x(\neg(q(x)))$" I can say "I want to replace $x$ by $m$. So I do the following: $\exists m(p(m))\wedge\exists m(\neg(q(m)))$". Why am I wrong?

Akiva Weinberger

(as you did - you chose $x=4$ for the first half and $x=3$ for the second half)

manooooh

00:32

@AkivaWeinberger why?

Akiva Weinberger

You're not wrong - those are equivalent

manooooh

Oh ok

So why I can write $\exists x(p(x))\wedge\exists y(\neg(q(y)))$?

Akiva Weinberger

4 mins ago, by manooooh

@AkivaWeinberger well we can pick $x=4$ so $x$ is a multiple of $4$, and we can pick $x=3$ so $3$ is not even. "True and true" is true. So (2) is true, meanwhile (1) is false. Right?

For the same reason you could choose two different values of $x$ there^

manooooh

Sorry I forgot to say $\exists x(p(x))\wedge\exists y(\neg(q(y)))$ with not necessarily $x=y$?

Akiva Weinberger

"Not necessarily" doesn't really add anything, it's just a reminder

that these are two separate quantifiers, with two separate bound variables (even though they're given the same name)

so they don't need to be related to each other

Like - that first $x$ only has meaning inside the quantifier $\exists x(\cdots)$

manooooh

00:36

@AkivaWeinberger yes it adds something important: if not I could say $x=y$ holds so I can always write $\exists x(p(x))\wedge\exists x(\neg(q(x)))$ or $\exists y(p(y))\wedge\exists y(\neg(q(y)))$ because $x=y$. But we don't want that, we want to say that $x$ can be different wrt $y$

Akiva Weinberger

so it's a "bound variable"

@manooooh If we required $x=y$ then it would be the same as (1)

manooooh

@AkivaWeinberger but the first existence has the same letter for the second, and vice versa

@AkivaWeinberger ok. So my question is: why we don't require $x=y$?

@AkivaWeinberger but the first existence has the same letter for the second, and vice versa

@AkivaWeinberger they are separate quantifiers but with the same letter on each quantifier. That confuses me

Akiva Weinberger

Whenever you have a sentence that looks like $[\cdots]\land[\cdots]$ you can treat the parts independently

I should point out: while you're allowed to have the same name for the bound variables of multiple quantifiers, it's considered bad practice and should be avoided

because it's confusing

So working mathematicians will go out of their way to write things like that, usually, and try to make it clearer what they're actually trying to say

*not to write things like that

manooooh

Ok, thank you so much! You've helped me a lot!

Now I want to prove that (1) i.e. $\neg(\forall x(p(x)\to q(x)))$ IS NOT equivalent to $\forall x(\neg p(x))\wedge\forall x(q(x))$

I don't think that the $p(x)$ and $q(x)$ that we have taken would be a counterxample for this exercise

I mean (1) is still false because for all $x$ if $x$ is multiple of $4$ then is multiple of $2$ is true, so the negation is false. But what about $\forall x(\neg p(x))$? Is it true or false?

I think "For all $x$, $\neg p(x)$" is sometimes true and sometimes false: take $x=8$, then it is false. Take $x=9$, then it is true

Akiva Weinberger

00:52

$\forall x$ means there's no x where it's false

"Every x satisfies p"

manooooh

So?

Akiva Weinberger

"For all x, not p(x)" is false

manooooh

I don't understand. Why $\forall x$ cannot mean an $x$ where it is true?

Akiva Weinberger

That's what $\exists$ means

manooooh

Oh

Ok now I understand it

So $\forall x(\neg p(x))$ is always false, and false $\wedge$ something = False

So the counterexample of the multiples are wrong. What counterexample can we take?

Oh I think we can take: $p(x)=\text{$x$ is an integer}$ and $q(x)=\text{$x$ is a positive integer}$. Then for all $x$, $p(x)\to q(x)$ is false, because there exists an $-1$ where $p(-1)$ is true but $q(-1)$ is false. So the negation of that is always true

However, for all $x$, $\neg p(x)$ is always false. So false $\wedge$ something = false. Am I right?

Stan Shunpike

02:14

Does showing $A\mathbf{x}=\mathbf{0}$ has a nontrivial solution mean showing that the kernel of $A$ has more than just the zero vector and the empty set in it?

Leaky Nun

02:44

@StanShunpike the kernel of $A$ contains vectors and shouldn't contain the empty set

but "$Ax=0$ has a non-trivial solution" means that the kernel of $A$ has more than just the zero vector in it

nothing to do with the empty set

Stan Shunpike

@LeakyNun does the image of A contain the empty set?

Leaky Nun

nope

the empty set is usually not a vector

Stan Shunpike

I thought $\mathbb{R}^n$ contained the empty set, is that correct?

@LeakyNun

Leaky Nun

no it doesn't

$\Bbb R^n$ contains things of the form $(x_1, x_2, \cdots, x_n)$ where $x_i \in \Bbb R$

those things don't look like the empty set to me

Stan Shunpike

ok

cool that's interesting

@LeakyNun so does that mean rings and fields do not contain the empty set?

Leaky Nun

02:59

well they can but they don't need to

Ryan Unger

how is this interesting

are you confusing elements of a set and subsets of a set

Stan Shunpike

@LeakyNun what kind of math do u do?

/ interests u

Leaky Nun

number theory

Stan Shunpike

what drew u to it?

Leaky Nun

numbers

tigre

03:21

19 hours ago, by tigre

Serre linear representations of finite groups, page $163$: Let $A$ be a ring, and let $\mathcal{F}$ be a category of left $A$-modules. The Grothendieck group of $\mathcal{F}$, denoted $K(\mathcal{F})$, is the abelian group defined by generators and relations as follows: Generators. A generator $[E]$ is associated with each $E\in\mathcal{F}$. Relations. The relation $[E]=[E']+[E'']$ is associated with each exact sequence:
$$0\to E\to E'\to E''\to 0,$$
where $E,E',E"'\in\mathcal{F}$.

just to check, there is an error in the above right this isn't equivalent to taking the relation as $[E']=[E]…

@StanShunpike The emptyset is not going to be an element of most things you encounter, but it's a subset of every set

it won't be an element of any ring or field you've encountered ever probably

Stan Shunpike

@tigre but the empty is in the set of events for a probability space?

or is that wrong too lol

Akiva Weinberger

@StanShunpike I agree with @RyanUnger, I think you're confusing the elements of a set with the subsets of a set

tigre

@StanShunpike I don't know probability theory. But I did say 'most things you encounter' admittedly this is false, since topologies contain the empty set

and sigma algebras contain the empty set

@StanShunpike Are events the things that you can measure with a probability measure?

Stan Shunpike

I'm not sure. I'm more applied. I come here mainly to learn what I don't know so I can fix my understanding.

@AkivaWeinberger ok that makes sense. thanks for clarifying

tigre

okay, yeah, the events belong to the sigma algebra, so the empty set is an event there

Leaky Nun

03:36

@StanShunpike that's because events are sets

@tigre the exact sequence should be $0 \to E' \to E \to E'' \to 0$

this is the convention

tigre

@LeakyNun okay that's good. otherwise it seems to break everything

there should be an errata page for serre, but i couldn't find one

also, he writes that a discrete valuation of $K$ is a surjective homomorphism $v:K^*\to \Bbb Z$ such that $v(x+y)\geq \inf(v(x),v(y))$ for $x,y\in K^*$

why not just write min?

Leaky Nun

well that's just style

tigre

ok

just making sure i'm not missing anything there

Leaky Nun

btw $\inf$ \inf is the LaTeX command

tigre

i always just write $\text{inf}$ lol

`\text{inf}'

Leaky Nun

03:43

now it looks better

The East Wind

05:35

I have to express this in the form of beta/gamma functions.

Anyone any ideas?

maths student

05:59

Let A ⊆ N + and B ⊆ N + be nonempty sets of positive integers. Define A + B def = {a + b : a ∈ A, b ∈ B}. Show that A + B is finite if and only if both A and B are finite.

I guess I have to use well ordering principle

so A is finite then max element exist and B finite max element exist (say a1,b1) hence A+B has max element is a1+b1 and this will imply A+b is finite

Is it correct approach

1010011010

10:14

Howdy! Can limits be taken individually?

So for example if I have two products $\prod (1- \prod(1+a_{ij}))$ can I say I have some sort of product of double exponentials?

Sir Cumference

12:56

Is it true that all inner products on $\mathbb{R}^n$ induce the Euclidean topology?

Thorgott

13:17

Yes. In fact, all norms on $\mathbb{R}^n$ induce the Euclidean topology.

Sir Cumference

@Thorgott huh, i'd read on an answer that the 2-norm was "special", in that it's equivalent to all other norms induced by the inner product. but i struggle to see why

looking up a proof for "all inner products induce euclidean norm" isn't getting any results

Leaky Nun

@SirCumference I'm very confused as to what your question is

Sir Cumference

yeah i am very sleepy so i realized i misread him

lemme start over

basically do all inner products on $\mathbb{R}^n$ induce the Euclidean norm (up to scaling)?

Leaky Nun

no

$\|(x,y)\|_1 := \sqrt{x^2+y^2}$ and $\|(x,y)\|_2 := \sqrt{x^2+4y^2}$ for example

Sir Cumference

how about the euclidean metric then, up to scaling?

Semiclassical

13:23

@TheEastWind I think the substitution u=1/x^4 is a step in the right direction, tho I don’t think it’s enough

Leaky Nun

the eulicdean metric is just $d(\mathbf{u},\mathbf{v}) := \|\mathbf{u}-\mathbf{v}\|$

so no

Sir Cumference

hmm, i may be misunderstanding this answer then

25

A: Why do we use the Euclidean metric on $\mathbb{R}^2$?

The Euclidean metric is special because it comes from what is called an inner product, and up to scaling it is the only metric that does so. This allows you to talk about angles between vectors in a sensible way, which you cannot do with other metrics. So really we don't choose to use the Euclid...

does it mean that it's the only one that comes from an inner product, or specifically the dot product?

Leaky Nun

ah ok

the meaning of "scaling" is confused

the answer there permits scaling each axis independently

so going from $\|\cdot\|_1$ to $\|\cdot\|_2$ I scale the x-axis by 1 and the y-axis by 2

Secret

new weird idea

Capping cardinals

Sir Cumference

@LeakyNun i see. so in this sense, the 2-norm is still "special", right?

Leaky Nun

13:28

yes

Sir Cumference

e.g. the 1-norm wouldn't come from an inner product

Leaky Nun

an inner product

Thorgott

There are multiple inner products on $\mathbb{R}^n$. There is - up to scaling axes independently and orthogonal transformations - exactly one norm that gets induced by an inner product, which is the Euclidean norm. All norms on $\mathbb{R}^n$ are equivalent and induce the same Euclidean topology.

Leaky Nun

@SirCumference yes

Sir Cumference

hmm, i'm just not sure how to follow the reasoning though. i've looked up "inner product induces euclidean metric" in a lot of different ways but found no results for a proof

Leaky Nun

13:29

I still don't know what your question is

Thorgott

it is a straightforward calculation

If you know how your inner product is defined, calculate $\lVert x\rVert=\sqrt{\langle x,x\rangle}$ and then check that $\lVert x-y\rVert=|x-y|$, where the latter is the usual Euclidean metric.

Secret

Let $S$ be a directed set that is strictly increasing. Then the capping cardinal $\nu$ of $S$ is the smallest cardinal such that $S$ will repeat itself with some symmetry, within every interval $[\nu,\nu^+]$ defined recursively as $[\nu_{\alpha+1},\nu^+]^+ = [\nu_{\alpha},\nu^+]$

where ${}^+$ is understood to be a successor operation of cardinals defined by some large cardinal axiom $T$

Thorgott

In this context, it can also be noted that the $2$-norm is the only one of the $p$-norms that comes from an inner product.

Sir Cumference

Wait, is it the only norm that comes from an inner product?

or just $p$-norm?

Thorgott

Just $p$-norm

Secret

13:32

capping cardinals thus measures the periodicity of a rapidly increasing infinite cardinal sequence

Sir Cumference

@Thorgott OK, so all of them induce the Euclidean metric up to scaling of axes. Thanks

Semiclassical

I’m forgetting. Are all inner products on R^n of the form Q(x,y)=x^T A y where A is positive-definite?

Secret

As an example, the sequence s = 2,4,6,8,10,... has a capping cardinal of 2 because $2\Bbb{N}$ is cofinal relative to s

and the set of naturals have a capping cardinal of 1

The geometric sequence 2,4,8,16,32,..., I think, has a capping cardinal of $\omega$

Thorgott

@SirCumference as a comment points out, the post you linked seems to forget something. it is up to scaling of axes and up to orthogonal transformations

MatheinBoulomenos

@Semiclassical yes

Thorgott

13:38

@Semiclassical if your definition of positive-definite requires symmetry, yes

Semiclassical

Yeah

In which case one can write the square root of A as C, ie A=C^2 where C itself is PD

And the Q(x,y)=x^T C^2 y = (Cx)^T (Cy)

Hence, up to a linear transformation on the inputs, every inner product is indeed Euclidean

Thorgott

Neat. I wasn't aware of that

Semiclassical

(The phrasing of “up to linear transformation” is a bit too strong since it’s not arbitrary C: it has to be PD. That’s where “up to scaling and orthogonal transformation” properly comes in)

The East Wind

@Semiclassical I found the exact same question later on. It turns out the correct substitution is t = (1-x^4)/(1+x^4)

Oh and by the way figured out that transformation one too.

Thorgott

Don't you mean "too weak"?

Semiclassical

13:49

@TheEastWind ew

Thorgott

"up to scaling and orthogonal transformation" comes from looking at it via the spectral theorem + the fact that the characteristic polynomial of a real symmetric matrix splits over $\mathbb{R}$

Semiclassical

@Thorgott I guess so

The East Wind

@Semiclassical Yup. I wonder how people think of such weird substitutions though

Thorgott

more concisely, the fact that every real symmetric matrix can be orthogonally diagonalized. not sure if that theorem has a proper name

Semiclassical

I’m blanking as to which linear transformations correspond to PD matrices if I’m honest

Scaling/rotation seems plausible but I don’t rightly recall

Thorgott

13:58

I'm not sure if there is an easy answer since they are characterized by their nature as bilinear forms. You can orthogonally diagonalize a PD matrix and get a diagonal matrix with only positive entries, but that might not be the best intuition

Semiclassical

@TheEastWind best I can come up with is this: for that t, you have 1+t = 2/(1+x^4))

So you could possibly arrive at that t as follows : first, pick u=x^4 to get rid of the fourth powers, and then decide that u’=1+u makes progress

The East Wind

Hmm seems a bit of a stretch though.

Semiclassical

Then u’’=2/u’ and u’’’=1+u’’

Yeah. One might be able to stumble upon it but it’s indeed a stretch

The East Wind

That transformation one turned out to be nice though

No long calculations needed

Semiclassical

At a minimum, u=x^4 is a good first step

It’s just that seeing t=(1-u)/(1+u) isn’t obvious

The East Wind

14:04

@Semiclassical Yea I did that and fumbled about for a bit without much progress though.

Semiclassical

Same

Sir Cumference

@Semiclassical Huh, so I guess the dot product isn't all that arbitrary

Thorgott

Being completely arbitrary is out of the question, since it is the inner product corresponding to the identity matrix, which certainly is not just any matrix.

Anonymous

14:28

Is the product of two involuntary matrices (like Householder transformations) necessarily of finite order? It would be helpful if someone knows a reference regarding this topic.

Anonymous

(They do not necessarily commute.)

Leaky Nun

we're transforming household items now?

Secret

involuntarily

a reflection of a reflection is a rotation

Capping cardinals, better:

Let $S$ be a directed set

Then the capping cardinal $\nu$ is the smallest cardinal such that the generating set $<\nu,f>$ is cofinal to $S$ and $f$ is least increasing

Capping cardinals ensures $S$ will be closed under extensions that are described by increasing self map of $S$

Thus given all possible order preserving morphism of $S$, and given any extension $T$, then $\nu$ is cofinal to $T$

A cardinal is called transcendent if it is cofinal to any capping cardinals $\nu$

Gabriel Romon

15:43

Can you guys tell me if the Latex in this answer renders correctly on your side ? With the renderer "HTML-CSS" and Chrome on macOS I get a lot of blank formulas

Leaky Nun

@GabrielRomon it renders correctly for me

Gabriel Romon

I see this. It used to render normally not long ago... and it renders correctly with the renderer "Common HTML"

The East Wind

16:33

Why does the surface integral come out the same irrespective of which plane we take the projection of the surface in?

Is this immediately obvious, or does it involve a somewhat lengthy proof?

Akiva Weinberger

16:47

Had a nice after-class conversation with my math teacher today :)

Showed him some cool stuff

Shine On You Crazy Diamond

17:18

Nice, you riffed on some algebraics

bad ass

@AkivaWeinberger

;)

I have a question for the room.

Suppose that $\phi : M \to M$ is a monoid hom

such that $\phi^k(S) = s \in \Sigma^*$

I.e. for some generator $S$ of $M$, $\phi^k(S) = $ a string $s$ over $\Sigma$.

Akiva Weinberger

What's a monoid again? Group without inverses?

Shine On You Crazy Diamond

How can I make the identification of $S$ with $s$

yes

@Akiva

SO that we can indeed say that such "grammar homomorphism" (because each one is indeed describing a CFG for $s$) form a group?

I.e. because $S \equiv s$

under this identification

Akiva Weinberger

CFG?

Oh, context free grammar

Shine On You Crazy Diamond

Context-free grammar

You read my mind

For instance take $s = ababab$

$g = \{S \to AAA, A \to ab \}$ is both a CFG and also a homomorphism, that "fixes everything else".

So I'm taking the homomorphism view of the grammar.

$A, a,b \in M = \{M, a, b\}^*$ for now

Akiva Weinberger

I think I'm out of my depth here but I'll think about it

Shine On You Crazy Diamond

17:23

I.e. it's the free monoid on those letters

You're indeed not out of your depth

if you know what a monoid hom is

It's a group hom same thing!

Every monoid hom on a group is a group hom is what I meant to say

So $\phi$ must map the alphabet to some strings

Which you then concatenate together

$\phi(ab) = \phi(a)\phi(b) = \dots$

We could be also working with quotients of the free group $\text{FreeGroup}(\Sigma)$.

But I'm not finding that fruitful yet

I.e. you just add in formal symbol inverses to the mix

Anyway, since $\phi^k(S) = s$ and $\phi^k(A) = $ some substring of $s$.

for some $k \in \Bbb{N}$ which is well-defined and depends on $\phi$

Can we identify all variables with their expanded-to strings

So that

$\phi^{-1} = \phi^{k-1}$ and the grammars form a group?

$\phi^k = \phi \circ \phi \circ \dots \circ \phi$ ($k$ times).

?

That's the question, last part

Above is a typo. Should be $M = \{a,b, A\}^*$ no $M$!

user193319

18:00

0

Q: Killing Form for $\mathfrak{sl}_{n}(\Bbb{C})$

Show that for $\frak{g} = \frak{sl}_{n}(\Bbb{C})$, the Killing form is given by $K(x,y) = 2n tr(xy)$. This is problem 5.2 in Kirillov's book on Lie Algebras. Recall that $K(x,y) = tr (\text{ad } x \text{ ad } y)$, where $\text{ad } x \text{ ad } y$ is a composition of two operators acting o...

Shine On You Crazy Diamond

I didn't know the site let on violence math questions.

:D

@user193319

No one laughs

MadSpaceMemer

Hello good people. Can someone help me proof the following ? I can not come up with a quick solution and your advice is welcome. Let $K$ be a real number then follows that $K.K = -K.-K$

Semiclassical

18:17

(-K).(-K)=(-1).K.(-1).K=K.(-1).(-1).K

@user193319 having a basis for sl(n,CC) seems like a starting point

MadSpaceMemer

We have not defined yet that -1.-1 = 1

So this solution is not acceptable sofar.

Semiclassical

Then what defines -1?

That 1+(-1)=0?

MadSpaceMemer

-K is the additive inverse of K

Semiclassical

fair enough

MadSpaceMemer

Do you have another approach perhaps?

Semiclassical

18:21

K+(-K)=0, so K.(K+(-K))=K.K+K.(-K)=0

And also (-K).(K+(-K))=(-K).K+(-K).(-K)=0

So K.K and (-K).(-K) are both additive inverses of K.(-K)=(-K).K

MadSpaceMemer

Amazing :) thanks :)

Thorgott

Sanity check: $[\mathbb{Q}(\sqrt{2},i)\colon\mathbb{Q}]=4$, right?

Semiclassical

@user193319 actually, wouldn’t tr(E^{ij}E^{ji})=1?

MatheinBoulomenos

@Semiclassical ad(E_ij) is a (n^2)x(n^2)-matrix

you're not taking the trace of the product of elements of sl_n themselves

Ted Shifrin

@MadSpaceMemer: So the main lemma you have to prove is that $-K = (-1)K$. This follows from the fact that $0\cdot K = 0$ for every $K$.

I guess Semi gave you a different argument; I haven't looked at it.

Semiclassical

18:29

@MatheinBoulomenos oof

Ted Shifrin

@Thorgott Yes.

Thorgott

Thanks

Semiclassical

@TedShifrin based on what I did above, I think my approach for that lemma would be to consider $1(K+(-K))$ and $(-1)(K+(-K)$

Ted Shifrin

Meh.

Semiclassical

I guess I should only need the second one

Ted Shifrin

18:33

Somehow you should have to use $0\cdot K = 0$. I don't think I know an independent proof.

Semiclassical

That’s the approach I used

Ted Shifrin

OK. That's always the first thing I tell students to prove. They often assume what they're trying to prove.

But then it's just $(1+(-1))K = K + (-1)K = 0$, so $(-1)K = -K$.

Semiclassical

Though above I used $0=K+(-K)$ when it was probably better to use $0=1+(-1)$.

Ted Shifrin

Well, my proof works in vector spaces or rings.

Does yours?

Semiclassical

If K times K makes sense, yes. Otherwise no

But the original question was to show $KK=(-K)(-K).$

So I’m okay with needing that assumption

Ted Shifrin

18:37

Right. But Mad should understand what the fundamental important principle is.

Of course, there is no need to have $K$ and $K$, rather than any two general numbers, etc.

Semiclassical

That’s fair. My argument is more direct but also not as generic

Ted Shifrin

Not to beat a dead horse, but I wanted to say what the important fundamental principle is ... and it's applicable all over the place.

Sometimes direct arguments miss the point.

Semiclassical

Well, that’s what I mean about not being as generic

Ted Shifrin

OK, I'll quit :)

I shouldn't be discussing algebra, anyhow.

MatheinBoulomenos

19:26

@user193319 I have posted an elementary solution to your question about lie algebras

Leaky Nun

19:55

@MatheinBoulomenos "elementary"

MatheinBoulomenos

@LeakyNun I'm literally only using the definition of the objects that are needed to define the question and easy linear algebra results such as $\mathrm{Tr}(AB)=\mathrm{Tr}(BA)$

Leaky Nun

what's your best proof of tr(AB) = tr(BA)

MatheinBoulomenos

idk

Leaky Nun

@MatheinBoulomenos what's the trace of a (compact) linear operator on a Hilbert space?

is it just $\int \langle Tx, x \rangle \ \mathrm dx$

no that doesn't make sense

Trace class

In mathematics, a trace-class operator is a compact operator for which a trace may be defined, such that the trace is finite and independent of the choice of basis. Trace-class operators are essentially the same as nuclear operators, though many authors reserve the term "trace-class operator" for the special case of nuclear operators on Hilbert spaces and reserve "nuclear operator" for usage in more general Banach spaces. == Definition == A bounded linear operator A over a separable Hilbert space H is said to be in the trace class if for some (and hence all) orthonormal bases {ek}k of H, the sum...

looks like they also use a basis

(hopefully it's basis-independent)

MatheinBoulomenos

@LeakyNun every compact operator on a Hilbert space is a limit of operators with finite rank. Take the limit of the traces

Leaky Nun

20:04

oh nice

Jacob McMichael

Hello is this the best way to communicate with fellow learners fast?

Thorgott

20:24

$AB$ and $BA$ have the same characteristic polynomial. This is obvious when one of them is invertible. Using that, you can get the result for matrices whose entries are indeterminates and then evaluate to get the general case.

Ted Shifrin

As I recall, @Thorgott, there's a less sophisticated way, too. I think I have a hint in my book.

You do have to use stuff with determinants of block matrices.

@Leaky: I have no issue with interchanging the order of summation in a finite sum.

CroCo

20:48

At which field of mathematics the definiteness of a function is discussed? Can you suggest a book? I've searched about it in Calculus but couldn't find a single topic about positive definiteness.

Thorgott

Yeah, a proof like that is possible as well.

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