Mathematics - 2018-06-24 (page 2 of 3)

Leaky Nun

5:00 PM

@MatheinBoulomenos how do you represent the identity as an element of that?

MatheinBoulomenos

@LeakyNun the identity is an element of $\operatorname{Hom}_K(V,V)$

Leaky Nun

and then?

MatheinBoulomenos

If $V$ is finite-dimensional, that's isomorphic to $V^* \otimes_K V$

Nicholas Roberts

@MatheinBoulomenos so youre taking the identity element in Hom(V,V) and pushing it through to K using the universal property? How does this give dimension exactly

Leaky Nun

can you prove it?

MatheinBoulomenos

5:01 PM

yes

Leaky Nun

prove it

I'm not convinced that your proof is not circular

I'm convinced that your proof is circular

Ted Shifrin

runs to the kitchen to pop popcorn

?

lol hi @Ted

:45313958

5:04 PM

hi @Ale

MatheinBoulomenos

Let $V$ and $W$ be finite-dimensional vector spaces.
Define a map $V^* \times W \to \operatorname{Hom}_K(V,W)$ by sending the pair $(\xi,w)$ to the linear map $v \mapsto \xi(v)w$

Nicholas Roberts

@JasonKim what would you like to know?

Ted Shifrin

hi demonic @Alessandro

Leaky Nun

@MatheinBoulomenos ok fair enough

Jason Kim

You stated that you're not convinced and after you're convinced.

Leaky Nun

5:05 PM

@Ted can drop his popcorn :P

Jason Kim

$\LaTeX$

Leaky Nun

@JasonKim I had two "not"s in the first sentence

MatheinBoulomenos

this is bilinear, so we get an induced map $V^* \otimes_K W \to \operatorname{Hom}_K(V,W)$

note that this map doesn't depend on a choice of basis at all

Nicholas Roberts

yup

Daminark

Hey everyone!

Ted Shifrin

5:09 PM

Hi Demonark

Alessandro Codenotti

hi @Dami

Leaky Nun

@JasonKim hanguginiyeyo?

MatheinBoulomenos

If $f \in \operatorname{Hom}_K(V,W)$, then choose a basis $b_1, \dots b_n$ for $W$ (I know you're going to say that this is circular, but it's not, I'm only using the existence of a basis)
We can write $f(v)=\xi_1(v)b_1 + \dots x_n(v)b_n$ (where $\xi_i \in V^*$)
Then map $f$ to $\xi_1 \otimes b_1 + \dots + \xi_n \otimes b_n \in V^* \otimes_K W$

mercio

how do you define the trace of a map without knowing that vector spaces have dimensions ?

Leaky Nun

@mercio he just did

MatheinBoulomenos

5:11 PM

That's an inverse for this guy above

Leaky Nun

by UMP of tensor

@MatheinBoulomenos no I'm not, I already said "fair enough"

MatheinBoulomenos

this actually doesn't depend on a basis either, surprisingly

Leaky Nun

58 mins ago, by MatheinBoulomenos

Consider the identity as an element of $V^* \otimes_K V$, then apply the natural bilinear pairing $V^* \times V \to K, (\xi,v) \mapsto \xi(v)$, which gives us the trace of the identity matrix, i.e. the dimension of the vector space

@mercio apply UMP of tensor to that map ^

mercio

what's an UMP ?

Leaky Nun

universal mapping property

@MatheinBoulomenos happens if $V$ and $W$ are infinite-dimensional?

Ted Shifrin

5:12 PM

Hmm, so how do we get existence of a basis? Zorn's lemma just in case we don't know dimension?

MatheinBoulomenos

@LeakyNun the isomorphism doesn't work

Leaky Nun

@TedShifrin basis reduction theorem

(our space is finitely generated to begin with)

@MatheinBoulomenos is it surjective or injective?

Ted Shifrin

Oh, we know finitely generated even without dimension :P

MatheinBoulomenos

@TedShifrin for finitely generated vector spaces we don't need Zorn

mercio

so we have a map from $V^* \otimes V$ into $K$ and we apply it to the identity ?

Leaky Nun

5:13 PM

but to be fair, MB said finite dimension instead of finitely generated

@mercio right

MatheinBoulomenos

@LeakyNun the map $V^* \otimes_K W \to \operatorname{Hom}_K(V,W)$ is always injective and the image are precisely the linear maps with finite rank

Leaky Nun

fair enough

and if I replace $K$ with a ring?

Nicholas Roberts

i thought that map is an isomorphism?

Leaky Nun

(every ring is commutative with unity)

@NicholasRoberts if $V$ or $W$ are finite dimensional, then it is

Nicholas Roberts

also, can anyone explain for a vector space to be finitely generated? I know finitely generated means the generating set is finite. But isnt the generating set for a vector space, the basis elements?

MatheinBoulomenos

5:16 PM

I want only to prove that if there is a finite basis, then all finite bases have the same cardinality. not having a basis is not an issue

Ted Shifrin

No, no, Nicholas ... just a finite spanning set ... no need for linear independence.

Nicholas Roberts

Ah, so we have a set that spans our vector space. meaning every vector can be written as a linear combination of these guys. but we dont require linear independence. so its a weaker notion of a basis?

Ted Shifrin

LOL, it's the notion of span :P

Leaky Nun

@MatheinBoulomenos if $R$ is a ring and $M$ and $N$ are $R$-modules, then what can we say about $M^\ast \otimes_R N \to \operatorname{Hom}_R(M,N)$?

lol

MatheinBoulomenos

@LeakyNun there's always a natural map $\operatorname{Hom}_R(M,N) \otimes_S F \to \operatorname{Hom}_R(M,N \otimes_S F)$ where $M$ is a left $R$-module $N$ is a $(R,S)$-bimodule and $F$ is a left $S$-module

The Great Duck

5:18 PM

@Secret interesting

perhaps that might be useful

or perhaps not

Leaky Nun

@MatheinBoulomenos is that relevant?

The Great Duck

it might be better after all to go with sacrificing the chain rule if it actually gets that complicated to force it back in

MatheinBoulomenos

@LeakyNun take $R=S=N$

Leaky Nun

N?????

Nicholas Roberts

@TedShifrin so am i right or what?

MatheinBoulomenos

5:20 PM

@LeakyNun it's my name for an $(R,S)$-bimodule

Ted Shifrin

About what, @Nicholas?

Leaky Nun

@NicholasRoberts it's not a weaker notion of a basis, it's the notion of span

Linear span

In linear algebra, the linear span (also called the linear hull or just span) of a set of vectors in a vector space is the intersection of all subspaces containing that set. The linear span of a set of vectors is therefore a vector space. Spans can be generalized to matroids and modules. == Definition == Given a vector space V over a field K, the span of a set S of vectors (not necessarily infinite) is defined to be the intersection W of all subspaces of V that contain S. W is referred to as the subspace spanned by S, or by the vectors in S. Conversely, S is called a spanning set of W, and we say...

MatheinBoulomenos

@LeakyNun I meant, consider $R=S$ and take $N=R$ with the usual $(R,R)$-bimodule structure, is that better?

Ted Shifrin

I already said that ... grrr.

Nicholas Roberts

I dont see how a finite generating set is any different from a finite basis without the condition of linear independence

Leaky Nun

5:21 PM

it isn't

Nicholas Roberts

so yes, a span i suppose

Leaky Nun

but we don't call it "a weaker notion of basis"

we call it "span"

Secret

@TheGreatDuck Well nobody said that a differential algebra cannot have a chain rule

Ted Shifrin

Ordinarily a basis is defined to be a set that spans and is linearly independent.

Nicholas Roberts

ok, but in terms of describing things in english, it is a weaker notin of a basis

Secret

5:21 PM

As long product rule is intact, you should be fine

Ted Shifrin

So you're removing one of the criteria for basis.

Leaky Nun

@NicholasRoberts do you have a different definition of basis?

Nicholas Roberts

Yes, so its a generalization of a basis

Leaky Nun

oh godforsaken

Ted Shifrin

@Nicholas: It's like saying a ring is a weaker notion of field. I don't like it.

MatheinBoulomenos

5:22 PM

@LeakyNun there are some sufficient conditions for it to be an isomorphism: if $M$ is finitely generated projective it works. If $M$ is finitely presented and $F$ is flat, it also is an isomorphism

The Great Duck

@Secret differential algebra?

explain

Ted Shifrin

Forget about it, Nicholas.

The Great Duck

please

:-)

Nicholas Roberts

Lol

Leaky Nun

@MatheinBoulomenos is that always injective?

@NicholasRoberts no, I don't forget about it

MatheinBoulomenos

5:23 PM

@LeakyNun not sure

Secret

@TheGreatDuck Are you still developing that piecewise continous functon derivative thing?

Leaky Nun

I'll debate with you until you stop saying that it's a weaker notion of a basis

Ted Shifrin

throws away the popcorn and leaves in disgust

The Great Duck

working with it yes

i have a solid definition now

Leaky Nun

What do you want me to do? LEAVE? Then they'll keep being wrong!

Secret

5:23 PM

because if so, you are dealing with a differential algevra

Differential algebra

In mathematics, differential rings, differential fields, and differential algebras are rings, fields, and algebras equipped with finitely many derivations, which are unary functions that are linear and satisfy the Leibniz product rule. A natural example of a differential field is the field of rational functions C(t) in one variable, over the complex numbers, where the derivation is the differentiation with respect to t. Differential algebra refers also to the area of mathematics consisting in the study of these algebraic objects and their use for an algebraic study of the differential equations...

The Great Duck

i mean other than the definition of step funtion not being quite right maybe

Nicholas Roberts

@LeakyNun what is wrong with that?

MatheinBoulomenos

a linearly independent set is like a basis without the condition of a generating set.
Similarly, a subset of a vector space is like a basis without the conditions of generating set or linear independence

Leaky Nun

@NicholasRoberts how do you define a basis?

The Great Duck

@Secret flips table. arrrgh. someone told me it wasnt a differential algebra

it is?

Ted Shifrin

5:24 PM

won't feed the troll any more

The Great Duck

@TedShifrin are you calling me a troll?

Secret

How is it not a differential algebra when a derivative operator is involved?

Nicholas Roberts

A basis a set of vectors, linear independent, such that every vector in your VS can be described as a linear combination of the basis elements over the field the VS is defined over

@LeakyNun

The Great Duck

i was told something along the lines of a differential algebra being a type of ring or something closed under a derivative type operation

Leaky Nun

@NicholasRoberts so we agree that a basis is a spanning set that is linearly independent

The Great Duck

5:25 PM

i.e. that "differential" is a misnomer

Nicholas Roberts

@LeakyNun yea!

The Great Duck

now i suspect they were just lying

or confused

Leaky Nun

@NicholasRoberts would you agree that it's nonsense to say that a linearly independent set is a basis without the condition of being a spanning set?

Secret

@TheGreatDuck It is usually defined on a ring or a field

The Great Duck

ah ok

Nicholas Roberts

5:26 PM

@LeakyNun Haha, yeah it seems silly

Secret

@TheGreatDuck He is talking about Nicholas

The Great Duck

oh ok

Leaky Nun

@NicholasRoberts then how is it not nonsense to do the other thing?

Nicholas Roberts

@LeakyNun fair point!

The Great Duck

sorry @TedShifrin. I thought you were implying my confusion was insincere.

Leaky Nun

5:27 PM

@TedShifrin you can come back

Nicholas Roberts

Also, not trolling. not sure why someone said that

The Great Duck

well thanks @Secret

Ted Shifrin

@Leaky: It's still too scary.

Secret

lol

The Great Duck

at least now I can say i work with "differential algebra" instead of just blindly saying "I don't know what this thing counts as."

Leaky Nun

5:27 PM

@Ted at least you should clean up the confusion brought to the innocent duck

Ted Shifrin

I didn't say a word to poor innocent duck ... although in the past he frustrated me plenty.

The Great Duck

lets not call me innocent

thats a bit harsh

Secret

@TheGreatDuck However, as I have said earlier, your system involves not just step functions, but something like constant function with many removable singularities, and these isolated points will pose challenge to any notion of derivative

Nicholas Roberts

Not sure how someone asking genuine questions is called a troll...

The Great Duck

the derivative isnt unique in that sense

that's why

Secret

5:29 PM

You have to focus on working out e.g. the composition of two heaviside functions

Leaky Nun

@MatheinBoulomenos a monoid is a local field that doesn't need to be equipped with a locally compact totally disconnected non-discrete topology, that doesn't need multiplicative inverses, that doesn't need commutativity of addition, that doesn't need additive inverses, and that doesn't need a multiplication at all

The Great Duck

@Secret already understand that issue

it handles itself

@Secret honestly the operation is much more closely more and more to an anti-integral.

MatheinBoulomenos

@LeakyNun I think this discussion is over

Leaky Nun

ok

The Great Duck

when the input is continuous anyways

MatheinBoulomenos

5:30 PM

but fair point

Secret

But anyway, without seeing all the axioms involved, I cannot say much more, especially you said you have thrown away whatever I have read in your document a year ago

MatheinBoulomenos

Also a set is like a scheme that doesn't come with a structure sheaf or a topology

The Great Duck

it;s simple secret

Leaky Nun

@MatheinBoulomenos brilliant

a topological space is a hausdorff space whose diagonal doesn't need to be closed

MatheinBoulomenos

@LeakyNun we're being mean, it was an honest question

Leaky Nun

5:33 PM

ok

The Great Duck

@Secret for some function F(x) let $g(y,z)$ and $C(x)$ be a pair of functions such that $g$ has a partial derivative with respect to $y$ and $C(x)$ is piecewise constant. Then $f(x) = \partial_y g(x,C(x))$ is an implied derivative of $F$. Neither $g$ nor $C$ is unique so there are infinitely many such derivatives.

the current issue right now is that the chain rule doesnt exist for this operation

but if that is normal, then meh

so be it

Leaky Nun

@MatheinBoulomenos if $v \in \Bbb H\setminus \Bbb R$ then $qvq^{-1} \in \Bbb H \setminus \Bbb R$?

oh no, stupid question

@MatheinBoulomenos oh no that isn't what I wanted to ask

if $v \in \Bbb H$ with $\Re(v) = 0$, then $\Re(qvq^{-1}) = 0$?

MatheinBoulomenos

@LeakyNun yes. The conjugation action of $\Bbb H^\times$ is orthogonal because of the multiplicativity of the absolute value wrt. quaternionic multiplication and conjugation leaves $\Bbb R$ invariant, so it also leaves the orthogonal complement invariant which consists of purely imginary quaternions

Leaky Nun

ich folge nicht

The Great Duck

@MatheinBoulomenos it wasn't an honest query. The user was asking the same thing yesterday as well.

Leaky Nun

5:44 PM

@TheGreatDuck ted wasn't referring to you

The Great Duck

@LeakyNun i know

Leaky Nun

neither is MB

The Great Duck

i know

im referring to nathan

i mean nicholas

nicholas was very much trolling

Leaky Nun

he wasn't either

The Great Duck

@LeakyNun ted was correct about nicholas trolling

nicholas asked the exact same things yesterday

he's just looking for attention

i remember him here

MatheinBoulomenos

5:47 PM

@LeakyNun the statement is just that the map $v \mapsto qvq^{-1}$ preserves the standard norm, because $\|qvq^{-1}\|=\|q\|\|v\|\|q\|^{-1}=\|v\|$ it's also $\Bbb R$-linear, so it's an orthogonal map.So $\Bbb R$ is invariant under this map, thus it's orthogonal complement will be invariant as well

And the orthogonal complement of $\Bbb R$ in $\Bbb H$ are those quaternions with real part $0$

@TheGreatDuck oh that's good to know

Leaky Nun

@MatheinBoulomenos got it

@MatheinBoulomenos I'm proud of myself

The Great Duck

I mean I might be wrong, but I recall seeing them here yesterday being annoying.

MatheinBoulomenos

@LeakyNun lol, that was just a misunderstanding

Leaky Nun

what is

Liad

hi , $F$ is a field with $char(F) = p$ , $K/F$ is galios extention of order $p$. given $\sigma \in Gal(K/F)$ i need to show that $ker(\sigma -id) = F$. the inclusion $F \subset ker(\sigma -id)$ is easy, im not sure how to show the reversed one.
someone can help ?

MatheinBoulomenos

5:50 PM

our discussion about möbius pedagogy

Leaky Nun

ok

is that relevant?

The Great Duck

@Secret that doc was written 3 months ago

btw

MatheinBoulomenos

@LeakyNun I thought you were refering to that

Leaky Nun

no, I was referring to what I linked

MatheinBoulomenos

@Liad what does it mean for an element to be in the kernel of $\sigma - \operatorname{id}$? Write it out and see if a rearrangement helps

Liad

5:51 PM

$\sigma(x) = x$ @MatheinBoulomenos

why this would imply $x\in F$

MatheinBoulomenos

okay, so it's fixed by $\sigma$. What do you know from Galois theory about fixed things?

Liad

hmm

The Great Duck

@Secret Did you read the section on differential forks in my original paper?

Leaky Nun

@Liad that's false

Liad

about an element that is fixed im not sure if i know something .. @MatheinBoulomenos

MatheinBoulomenos

5:53 PM

@Liad you need $\sigma \neq id$ by the way

Liad

@MatheinBoulomenos yea it is a generator of the group

sorry for not mentioning that..

MatheinBoulomenos

okay, so it will be fixed by the whole group

Liad

right

but why it cant be in $K - F$ ?

MatheinBoulomenos

Are you sure you don't know something about elements fixed by the Galois group?

Liad

maybe its related to a statement/theorem i did study but i cant see it now

is it related to galios mapping @MatheinBoulomenos ?

MatheinBoulomenos

5:56 PM

@Liad do you know Galois theory?

Educ

Hi

Leaky Nun

@MatheinBoulomenos ooooooh

burntttttt

Liad

@MatheinBoulomenos i know the mapping theorem

if that's what you mean

Leaky Nun

@Educ hi

MatheinBoulomenos

@Educ hello

The Great Duck

5:57 PM

I don't know algebra of that sort

:-(

Educ

reading how to build Z, I didn't understand this sentence : "To visualize this, draw (in Quadrant I only) the set of lines y = x -c, where c is an integer. All points on the line y = x - c are in the same equivalence class; this equivalence class is the integer c."

MatheinBoulomenos

@Liad tell me what you know about the mapping theorem, I'm not sure what you mean

Educ

could someone please draw me this please ?

Liad

alright

Leaky Nun

@Educ desmos.com/calculator/og0elqtfon

Liad

5:58 PM

assuming $L/K$ is a finite galios extention then $[L:K] = |Gal(L/K)|$ and there is a one-to-one correspondence between subgroups of $Gal(L/K)$ and subfields of $L/K$. @MatheinBoulomenos

also

Secret

@TheGreatDuck I read I think 2 pages. If I recall it basically determines what functions will behave like constant function under that fork

Leaky Nun

the idea is that Z is built to solve the problem that you can't subtract a large thing from a small thing @Educ

so you say, pretend we can

Liad

order of subgroup = index in big extension, degree over base field = index of subgroup@MatheinBoulomenos

Leaky Nun

then denote the result of 2-4 as (2,4) formally @Educ

Liad

and the mapping reverse the inclusion

Leaky Nun

5:59 PM

and then you say, oh 2-4 should be the same as 1-3, so you relate (2,4) and (1,3) with an equivalence relation @Educ

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$Mathematics$ Mathematics