Mathematics - 2017-07-23 (page 4 of 4)

Eric Silva

21:01

@Steamy what is your kind of algtop

SteamyRoot

Lefschetz & Nielsen fixed point theory, for example

Eric Silva

Cool

Daminark

Hmm, what's that about?

Alessandro Codenotti

fixed points

Astyx

and theory

Eric Silva

21:11

Lol

Daminark

Wao

Alessandro Codenotti

Regardless of the misleading name though the object of study is neither Lefschetz nor Nielsen

Daminark

I'm like, surprised

Oh I thought it was literally fixed points of Nielsen automorphisms of Lefschetz

Alessandro Codenotti

Firefox tells me that the connection is not safe because the website is bady configured if I click on your Master's thesis link @Steamy

SteamyRoot

o.O

doesn't do so for me

Astyx

21:15

Me neither

Alessandro Codenotti

Hm, very weird

Mike Miller

@AlessandroCodenotti See if it tells you that on mozilla.org

SteamyRoot

Well, Firefox can be weird with its certificates

Try kuleuven-kulak.be/~u0092325/Masterpaper.pdf

Alessandro Codenotti

I rebooted on the ubuntu partition of my laptop and it works perfectly from here

it was probably just firefox acting strange and not a problem with the site

SteamyRoot

Well, in all honesty, my university's IT crew is... pretty bad.

Lucas Henrique

21:34

Yo!

Alessandro Codenotti

hi

Lucas Henrique

Can you guys help me with a proof?

Doesn't seem to advance.

"Prove that every non-constant polynomial $g: \Bbb Z \to \Bbb Z, g(x) = a_n x^n + \ldots + a_1 x + 1$ has infinite prime divisors"

Akiva Weinberger

I'm not even sure I understand the statement

@Daminark @Astyx Re: bijective holomorphic functions have holomorphic inverses, I think the idea is that,

the obvious near-counterexample is $x^3$, but that fails to be bijective in a neighborhood of the origin due to the branching

and there aren't any other obvious ways it could fail

so it seems pretty plausible

Astyx

What exactly does holomorphic mean again ?

Lucas Henrique

Given the polynomial, you can find infinite primes $p$ such that $\exists n$ s.t.
$p|g(n)$

Akiva Weinberger

21:40

(I was scrolling up and reading what the chat was up to while I was gone)

Complex differentiable @Astyx

@LucasHenrique Ahh

PVAL-inactive

This is the characterization of zeroes stuff.

Astyx

Is that it ? I thought there was more

Akiva Weinberger

Nah

It's literally, you can take the derivative of this

i.e., locally it looks like a linear map (which, on the complex plane, means a rotation-scaling mix-y thing)

Astyx

"Let's call it holomorphic. - Why ? - To confuse people. - Oh, sure !"

Akiva Weinberger

and just from that you can prove everything.

Somehow

I think the idea behind the name is that, away from spots with zero derivative, they preserve angles

(which makes sense from the "locally scalerotation" thing)

Astyx

21:43

Do you need continuously differentiable, or just differentiable ?

Akiva Weinberger

and you can somehow go from "preserves angles" to "holomorphic" if you take a detour through Greek or Latin or something

@Astyx Nah, just differentiable I think.

You don't want to think about differentiability on a point, though, you want differentiability on open sets

Daminark

Holomorphic functions are analytic

Astyx

Sure

Dattier

Hi, $$(1) n,k\in \mathbb{N}, 29+2^n=3^k \\ (2) n,k\in \mathbb{N}, 1+2^n=3^k \\ (3) a,b,c\in \mathbb{N}, 2^a+3^b=5^c$$

Solve

Akiva Weinberger

$x^2\sin(1/x)$ probably takes on all values near zero (because $\sin$ has image the entire complex plane—$-1\le\sin\le1$ fails), so our prime example of differentiable-but-not-continuously-so doesn't exist @Astyx

('cause you can't continuously extend the function to zero)

Astyx

21:46

Cool

Akiva Weinberger

@Dattier For (2), that's $1+2^3=3^2$, right?

And $2^4+3^2=5^2$ for (3)

Dattier

there are more solutions

(1) :0

Astyx

I remember having done 2 and 3

Dattier

(2) :2

(3):3

Akiva Weinberger

Oh, $1+2^1=3^1$

$2^1+3^1=5^1$

Dattier

21:48

yes

Lucas Henrique

It would be nice to find some factorization to limit the numbers

Akiva Weinberger

How would you even prove $x^2+1$ has infinitely many prime divisors of things in its image? @LucasHenrique

Dattier

why are there not more solutions ?

Akiva Weinberger

@Dattier What's the third for (3)?

And as for why there aren't more, no idea

Kirill

$\varphi: V \to V$ is normal, $\varphi^{\ast}$ is an adjoint operator to $\varphi$, $v$ is an eigenvector. What happens here: $\Vert \varphi^{\ast}(v) \Vert^2 = \langle v, \varphi(\varphi^{\ast}(v))\rangle$ ?

Akiva Weinberger

21:51

@LucasHenrique Oh, wait,

Astyx

@Akiva That's easy isn't it ?

Just like Euler

Akiva Weinberger

that's same as asking for infinitely many primes that satisfy $x^2+1\equiv0\pmod p$

which is a very famous special case

arctic tern

@Kirill $\|\varphi^{\ast}(v)\|^2=\langle \varphi^{\ast}(v),\varphi^{\ast}(v)\rangle=\langle v,\varphi(\varphi^{\ast}(v))\rangle$

Akiva Weinberger

'cause it's true for all primes that are $1$ mod $4$

and there's infinitely many of those

arctic tern

indeed, $\langle \varphi^{\ast}(v),u\rangle=\langle v,\varphi(u)\rangle$ for any $u,v,\varphi$, by definition of adjoint

Astyx

21:53

Can't you just suppose there are finitely many, and take the image of the product of them ?

Dattier

(3) 2^4+3^2=5^2 @Akiva

triplet de pythagore

Kirill

@arctictern thinking

Akiva Weinberger

@Dattier I gave that one already

("Pythagorean triple" in English)

arctic tern

@Kirill do you agree $\langle \varphi^{\ast}(v),u\rangle=\langle v,\varphi(u)\rangle$ for any $u,v,\varphi$ by definition of adjoint?

Akiva Weinberger

@Astyx I'm not sure I see how…

Dattier

21:55

2^1+3^1=5^1

Kirill

sure, I am not sure about the right argument in the product

Dattier

3^0+2^2=5^1

arctic tern

just set $u=\varphi^{\ast}(v)$ then

Akiva Weinberger

@Dattier Ahh

Thanks, makes sense

Secret

[My typical abstract algebra element convention]
a,b = generic element
c = zero divisor
d = differential operator
e = additive identity
f = multiplicative identity
g, h = generic functions and maps
i,j,k = quaternions
l,m,n = generic integers or natural numbers
p = prime
q = zero inverse
r = right zero inverse or remainder
s = successor operator
t, $\lambda$, $\mu$ parameters
u,v = unit elements (in the sense of rings or vector fields) or generic vectors
w,x,y,z = unique elements (for division algorithm and others) or generic coordinates

Akiva Weinberger

21:56

Yeah, I don't know how you'd show there aren't more

Astyx

Like, take $E = \{p\in\mathcal P\mid \exists n\in\Bbb N, p|n^2+1\}$

Dattier

I use a PC

Astyx

Suppose that set is finite

Dattier

computer

Astyx

And take $\left(\prod_{p\in E}p\right)^2+1$

Akiva Weinberger

21:56

Right @Dattier

Maybe ask on MSE

Kirill

$\varphi^{\ast}(v) = \varphi(\varphi^{\ast}(v))$

Dattier

I can't

@AkivaWeinberger

Akiva Weinberger

Why not?

Dattier

Because when I want do that : I have a message of error

Kirill

@arctictern gOt it, thank you!

Dattier

21:59

11

A: Why I can ask question in MathOverflow and not here (MSE)?

You cannot ask questions on Mathematics Stack Exchange because you are currently question banned. This ban is automatic and kicks in when some algorithm decides that your question contributions are of poor quality, and are not of positive value to the community. In contrast, you are not question...

they don't like my enigma

@AkivaWeinberger

A lot that put on hold

They say there no context

even I says it's a personnal production

They don't like this answer

user193319

Suppose I know that both $A$ and $B$ are convex sets, each generated from their collection of extreme points. What can I say about the extreme points of $A \times B$? How are they related to the extreme points of each individual factor?

Secret

A = Abelian group or algebra over a set
B, C = (unassigned)
D = Division algebra
$\mathcal{D}$ = differential algebra
E = Eigenspace
F = (forgot)
$\Bbb{F}$ = Field
$\mathcal{F}$ fourier transform
G = Group
H = subgroup
I, J = interval
K, L = (forgot)
$\mathcal{L}$ = language
M = generic mathematical object
$M_{mn}(\Bbb{F})$ matrix
N = (forgot)
O = Big O notation
P, Q = (forgot)
R = Ring
S = generic set or generic algebraic structure
T,U = (forgot)
V = vector space
W,X,Y,Z = (forgot)

Kirill

So, two eigenvectors from different eigenspaces to different eigenvalues are orthogonal, if the linear map $V \to V$ is normal, right?

Akiva Weinberger

@Dattier I found this math.stackexchange.com/questions/226415/…

Dattier

good, I have a another solution

Akiva Weinberger

22:06

@Astyx And then?

Kirill

Is there any evidence why for the bilinear form should be $\langle u+a \cdot v,w \rangle = \langle u,w \rangle + a \cdot \langle v,w \rangle$, and for the sesquilinear form $\langle u+a \cdot v,w \rangle = \langle u,w \rangle + \overline{a} \cdot \langle v,w \rangle$ ? I do not see any difference according to the other argument of the product in the definitions.

Daminark

math.stackexchange.com/questions/2369456/…

Oh for fuck's sake that example

arctic tern

@Kirill I don't understand what you're asking.

With a bilinear form, <a,a> may not be real. On the other hand, with a sesquilinear form, <a,a> is always real. You can use the standard inner product to measure size, after all.

So, they are indeed different.

for example, f(z,w)=zw is bilinear on C^1, whereas $\overline{z}w$ is sesquilinear

Kirill

22:23

@arctictern A form is bilinear, if it is linear in both arguments. The difference between the definition of a bilinear form and of a sesquilinear form is that you have $\overline{a}$ instead of $a$ in the left argument. I do not understand, why only in the left one.

Secret

Actually, I made a mistake here, I have unknowingly assumed 1x=b chat.stackexchange.com/transcript/message/38961782#38961782

arctic tern

@Kirill If you had it in both arguments it'd be more or less the same as in no arguments and just took the conjugate of the output, since $\overline{z}\,\overline{w}=\overline{zw}$

Kirill

@arctictern for $\mathbb{R}$ I see that $a\cdot \langle u,v\rangle = \langle u,v \rangle \cdot a = \langle a \cdot u,v \rangle = \langle u, a \cdot v \rangle.$ Is that the same for $\overline{a}$?

Lucas Henrique

what does "large" mean?

arctic tern

$a\cdot\langle u,v\rangle=\langle u,v\rangle$ is not true unless $a=1$ or $\langle u,v\rangle$. and in any case $a=\overline{a}$ if and only if $a\in\Bbb R$

Lucas Henrique

22:28

too abstract for me

4

A: Primes dividing a polynomial

Outline: If the constant term of the polynomial is $0$, the result is obvious. The rest of the proof imitates the standard Euclid-style proof that there are infinitely many primes. So let the constant term be $a\ne 0$. It follows that the polynomial $g(n)$ has the shape $$g(n)=nq(n)+a,$$ where...

Semiclassical

This problem annoys me just a little: math.stackexchange.com/questions/2369439/…

arctic tern

@LucasHenrique big

Kirill

@arctictern it was mistyped

Semiclassical

To my mind the 'right' way to do it is to check the case of $\beta=-\alpha=1$.

Daminark

Yeah it's just like, as your numbers go to infinity, blah is the asymptotic behavior

Semiclassical

22:29

The rest is just understanding how the central moments of one uniform distribution compare to another.

Kirill

@arctictern ok. Does it play a role for $\overline{a} \in \mathbb{C}$ if I multiply it with the right or with the left argument?

Semiclassical

But the intended solution is probably just to crunch things out brute-force.

arctic tern

@Kirill don't you already know $\langle au,v\rangle=\overline{a}\langle u,v\rangle$ and $\langle u,av\rangle=a\langle u,v\rangle$? (well, that's the physicist convention, the mathematician's convention is often the reverse)

Lucas Henrique

@arctictern :(

Kirill

@arctictern dot product is an integer for me, so I can multiply it from the left and from the right. So, according to your message I can do it with any argument from any side.

Semiclassical

22:33

huzzah physics :)

arctic tern

@Kirill I can't understand what you're saying.

Semiclassical

In defense of the physics way, in the case of $\langle u,v\rangle = u^\dagger v$ those pop out immediately.

(with $v$ understood as a column vector)

Kirill

What is $\langle u,v \rangle \cdot \overline{a}$ according to this convention?

Semiclassical

if $a$ is a complex scalar, then there's no difference between that and $\overline{a}\langle u,v\rangle$.

Kirill

but it is not $\langle u,v \cdot \overline{a} \rangle$?

Semiclassical

22:38

Sure? But that's already implied by what arctic said.

Dattier

@AkivaWeinberger I have send my solution

Akiva Weinberger

@LucasHenrique Oh that's a clever solution

Semiclassical

It's just doing $a\langle u,v\rangle = \langle u,av\rangle$ with $a$ replaced by $\overline{a}$.

Akiva Weinberger

It's simpler when $a=1$, which is what I think you had

So your polynomial is $a_nx^n+a_{n-1}x^{n-1}+\dotsb+a_1x+1$, right? @LucasHenrique

So, $p(n)$ is $1$ more than a multiple of $n$.

That means that $p(k!)$ is $1$ more than a multiple of $k!$.

Lucas Henrique

actually I was trying to prove the general thing but used a general polynomial $p(x)$ on $x = a_0 x'$

Kirill

22:41

$\langle au,v \rangle = \langle \overline{v, au} \rangle = \langle \overline{v}, \overline{au} \rangle = \langle v,u \rangle \cdot \overline{a} = \overline{a} \cdot \langle v,u\rangle$. That way? @Semiclassical

Lucas Henrique

then you get $p(a_0 x) = a_0(1+g(x))$ for some polynomial

Akiva Weinberger

There are only finitely many solutions to $p(n)=1$, right? @LucasHenrique

Lucas Henrique

yes

Akiva Weinberger

So, take $k$ such that $k!$ is larger than the largest such solution

If $p(k!)$ is $1$ more than a multiple of $k!$, and it doesn't equal $1$, then what could its prime factors be?

They cant be anything less than $k$

so they must all be larger

Semiclassical

sure, that works.

wait.

Akiva Weinberger

22:43

And then we're essentially done. If there were finitely many possible primes that were factors of a $p(n)$, then just take $k$ larger than the largest prime to obtain a contradiction.

(It doesn't matter that $p(k!)$ could be negative, right?)

Semiclassical

it'd be $\langle \overline{v},\overline{au}\rangle = \langle \overline{v},\overline{u}\rangle \overline{a}$.

And then $\langle \overline{v},\overline{u}\rangle = \langle u,v\rangle$.

Kirill

yeap, Ive twisted them up sorry

Secret

$\textbf{Revised version}$

$\textbf{Theorem}$: Let $A$ be a magma with the absorber $0$, and satisfying one of the following: $\forall a,b\in A,a\neq 0,\exists ! x: ax=b$ or $\forall a,b\in A,a\neq 0,\exists ! y: ya=b$. If $A$ contains a zero divisor, then the zero divisors can only have one sided inverses.

$\textbf{Proof}:$ Suppose we take the division rule to be $ax=b$. If $A$ has a zero divisor, that is $\exists c, ↄ\in A,c,ↄ \neq 0, cↄ=0$, then by definition, the division rule also holds for $c,ↄ$. Pick some $a \in A$ such that

Kirill

my point is to understand when I have to put the line over $a$, as I see no reason why it should be done for the left, and not for the right argument

Semiclassical

$\langle u,av\rangle=\langle \overline{a}u,v\rangle$

That's how scalars transfer from one argument to the other.

Akiva Weinberger

22:49

@Secret /wↄ/

Secret

$/wↄ/$ ?

Akiva Weinberger

$\rm/wↄ{:}/$

Woah

You used IPA as a variable

Secret

well, it seems fitting, since zero divisors kinda acts like chopping zero into halves

Kirill

@Semiclassical I got it. You take the $a$ out, rotate the product, put the $a$ in, rotate the product again.

Secret

you have two nonzero elements that when combined, they become nothing

Semiclassical

22:52

I don't know what you mean by 'rotate'

Kirill

@Semiclassical using symmetrie

Semiclassical

ah.

wasn't sure which scalar you meant (a is itself a scalar)

yeah, that'll work

Kirill

@Semiclassical sorry, it is a scalar product in German and a dot product in English. I mix them up.

Semiclassical

gotcha.

Akiva Weinberger

Hi

The dot product is also called the scalar product in English sometimes

Faust7

22:58

depends on the book

it actually makes a bit more sense to me but ima bit odd so...

Kirill

Wikipedia says we also use "Punktprodukt" for that. Never heard that.

Faust7

dunno most german technical stuff is relatively easy to read in context...

Semiclassical

When it's with the angle bracket I tend to think 'inner product'

Faust7

i always think area

Akiva Weinberger

Isn't area the cross product?

Magnitude of

Kirill

23:02

@Semiclassical inner product for me is a product of two elements of the set that gives an element of the set, again. Don't know, if it is a usual usage.

Faust7

...

Semiclassical

I have this in mind when I say that: en.wikipedia.org/wiki/Inner_product_space

Faust7

@AkivaWeinberger not sure cross product gives u a new vector

Akiva Weinberger

Magnitude of the cross product, I mean

Semiclassical

$|\vec{a}\times \vec{b}|=ab\sin \theta$

Akiva Weinberger

23:03

Like, note that the dot product of perpendicular vectors is zero, which doesn't make sense with area

Secret

$\textbf{Corollary}$: $A$ is an associative division algebra iff it has no zero divisors.

$\textbf{Proof}$: Suppose $A$ has a pair of zero divisors $c,ↄ$. $A$ is a divison algebra if it satisfies both the left and right division rules, i.e.

$$ax=b=ya$$

Since the division rule holds for all elements except zero (NB the case where it holds also for zero is a completely alien structure which is outside of the scope of this proof), we can pick a $b \in A$ such that:

$$cx=b=yc\text{ or }ↄx=b=yↄ$$

Semiclassical

so the magnitude of the cross product is the area of the parallelogram generated by the two vectors.

Faust7

whats the dot product give you

i always though it was a scaling factor for a linear map

Kirill

and what is the right name for inner operations I mentioned in the last post?

Akiva Weinberger

Project $x$ onto $y$ and multiply by the magnitude of $y$

Semiclassical

23:05

^

Akiva Weinberger

which admittedly is not all that obvious of a definition, and it doesn't help explain why it's symmetrical

Semiclassical

so a dot product with a unit vector gives the component along that unit vector.

Akiva Weinberger

3Blue1Brown has a good video on it

Semiclassical

to get to the symmetry, it's best to write it as $\vec{a}\cdot\vec{b}=ab\cos \theta$.

Secret

Typo in corollary: $A$ is an associative division algebra if it has no zero divisors.

The converse does not hold obviously, there are many nonassociative algebras with no zero divisors such as the octonions

Akiva Weinberger

23:07

You can think of it as a measure of how much $a$ and $b$ point in the same direction

That's how it's used in physics

A measure of how much they point in the same direction, with the sizes factored in as well

Faust7

gmm i almost remembered it then it flew away

Secret

NB: The inner product is the generalisation of the dot product to infinite number of dimensions

Faust7

the dot product of two vectors reprsents something important though

Akiva Weinberger

@Kirill The word you want is "binary operation" I think

You could also think of dot products as a measure of how much they fail to satisfy the Pythagorean theorem

since $|x+y|^2=|x|^2+|y|^2+2x\cdot y$

Faust7

that sounds more familiar

Akiva Weinberger

23:11

Also I'm 90% sure I'm right about this but not 100%:

Faust7

lol seems right

Akiva Weinberger

(cont'd) if you have an object with a force on it with direction and magnitude $\vec a$, and it moves with direction and magnitude $\vec b$, then $\vec a\cdot\vec b$ is the amount of work you've done on the object

@Faust7 I hadn't said it yet…

Secret

Yes this is correct

user193319

I am trying to show that $\Bbb{Q}$ does not form a locally compact set, and I am wondering whether the following is heading in the right direction: Suppose that $\Bbb{Q}$ is locally compact. Since it is also Hausdorff, there must exist a compact space $Y$ such that $\Bbb{Q} \subseteq Y$ and $Y - \Bbb{Q}$ is a singleton. Note that we cannot have $\Bbb{Q} \subseteq \Bbb{R} \subseteq Y$, since the sets would no longer differ by a singleton point.

Faust7

thats alot more familiar

user193319

23:13

Hence $\Bbb{Q} \subseteq Y \subseteq \Bbb{R}$ which implies $\overline{\Bbb{Q}} \subseteq \overline{Y} \subseteq \Bbb{R}$ or $\Bbb{R} \subseteq \overline{R} \subseteq \Bbb{R}$. Hence $\overline{Y} = \Bbb{R}$...

I feel like this should lead to a contradiction, but I cannot see it.

Akiva Weinberger

@Faust7 So like if the angle between the vectors is obtuse, then it's essentially going in the opposite direction of the force so it's negative

Faust7

@AkivaWeinberger thanks for the great explanations =)

Akiva Weinberger

and if they're directly opposite, it's very negative

and if they're perpendicular, it's literally orthogonal to the force you're doing so it's 0

Faust7

dot product is also a good example of a special ring ^^

Akiva Weinberger

@user193319 So you want to show that there's an open set in $\Bbb Q$ that's not contained in any compact set?

Well, in fact every open set in $\Bbb Q$ has that property

No compact set contains an open set

Kirill

23:16

@AkivaWeinberger of "how much they fail"? :)

user193319

@AkivaWeinberger Well, I was going in a slightly different direction, but I could try that. Can you see whether my ideas are leading in the right direction?

Daminark

@user193319 Why is that true?

user193319

@Daminark What specifically are you referring to?

Daminark

Oh wait a second

I think I get it, this is the one-point compactification, right?

user193319

@Daminark I am not sure. I haven't learned about that yet.

Daminark

23:21

OH WAIT

I have a really sneaky way of doing this

Akiva Weinberger

@Kirill Yeah exactly

Daminark

Have you heard of the Baire Category Theorem?

Akiva Weinberger

Just use $\Bbb Q=(-\infty,a)\cup(a,b)\cup(b,\infty)$ for $a,b$ irrational

That's a decomposition into open sets

Oh wait

Wait ignore me I'm confused

Daminark

It states that a locally compact Hausdorff space is Baire

Baire meaning, the countable intersection of open, dense sets is dense, or at least non-empty

Akiva Weinberger

I was thinking of connectedness for some reason

Say a compact set contains an open set in $\Bbb Q$. Think about any irrational number that's in the closure of that set in $\Bbb R$

Daminark

23:24

Since $\mathbb{Q}$ is Hausdorff, singletons are closed, so then $\bigcap_{q\in\mathbb{Q}} \mathbb{Q}\setminus\{q\} = \emptyset$

Kirill

Where do I fail again? $(\lambda - \mu)\langle v,w\rangle$ should be equal to $\langle \overline{\lambda}v,w \rangle - \langle v, \mu w \rangle.$ I get: $(\lambda - \mu)\langle v,w\rangle = \langle (\lambda - \mu)v,w \rangle = \langle \lambda v - \mu v, w \rangle = \langle \lambda v,w \rangle - \langle \mu v,w \rangle \ne \langle \overline{\lambda}v,w \rangle - \langle v, \mu w \rangle.$

Daminark

That's a countable intersection of open dense sets which is empty

So $\mathbb{Q}$ cannot be locally compact

Akiva Weinberger

Oh there's even a kind-of sneaky way to do this with the extreme value theorem EDIT: Ignore that that's unnecessarily complicated

But yeah I think the hint with the irrational number in the open set is what you need @user193319

Daminark

I like it since it extends to all countable Hausdorff spaces :P

Semiclassical

@Kirill $(\lambda-\mu)\langle v,w\rangle \neq \langle (\lambda-\mu)v,w\rangle.$

user193319

23:26

@Daminark and @AkivaWeinberger Okay. Let me think about everything you have both said. Thanks!

Akiva Weinberger

@Semiclassical Do you need to conjugate that one

Daminark

Alright. Do you know BCT though? If not, it'd probably count as way illegal machinery

user193319

@Daminark No. I don't know that yet.

Daminark

Okay, work more with what Akiva was saying then

Semiclassical

Right.

Kirill

23:33

@Semiclassical then how?

Semiclassical

then keep going.

You've got the tools. Use them.

@daminark Just got 100% with the nickname "universal cover" :P

Daminark

Nice

Kirill

@Semiclassical not really. I have finally read in wiki that the product is linear in the second argument and semilinear in the first one. That makes difference. My notes say nothing about that.

so, $\mu \langle v,w \rangle = \langle \overline{\mu}v,w \rangle = \langle v,\mu w \rangle$, right?

Semiclassical

right.

Kirill

thank you very much

Akiva Weinberger

23:45

So for $\Bbb C^1$, what's $\langle v,w\rangle$? $~\bar vw$?

Kirill

if I knew what $\mathbb{C}^1$ means...

Akiva Weinberger

(AKA, what's $\langle v,w\rangle$? $\sum \bar{v_i}w_i$?)

@Kirill I just mean for 1D

Kirill

$\langle v,w \rangle = f(v,w)$ which goes from $V \times V \to \mathbb{C}$.

the second question looks familiar to the product in $\mathbb{R}$. If so, it is $\langle \overline{v},w \rangle$. If not, it is something else.

@AkivaWeinberger

Akiva Weinberger

The prototypical example would be $V=\Bbb C^n$, no?

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