Mathematics

@s.harp I can't do my averaging trick if my group is not abelian right

because I find out I need to use right invariance...

maybe I defined things wrongly

s.harp

@Leaky I'm not sure what your averaging trick is, but usually you can't do averaging tricks if the group is not compact

non-abelian is not a problem

compact groups are unimodular

12:05 AM

my group is compact

@MatheinBoulomenos say what

why

lol

and how exactly do we detect unimodular from the group

the modular function is a continuous group homomorphism $G \to \Bbb{R}_{>0}$

ok

thanks

the image is a compact subgroup, but $\Bbb R_{>0}$ only has the trivial one

maths is beautiful

so we have compact hausdorff abelian group G acting unitarily on banach spaces V and W

and then L(V,W)^G = L_G(V,W) right?

G also acts unitarily on L(V,W) right

chandx

k, one more, is it done well?
linear isometry $F$ maps $\binom{1}{0}$ to $\binom{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}$ and changes orientation, find $F(\binom{0}{1}$

so let $F\binom{0}{1} = \binom{a}{b}$ and $E_1 = \binom{1}{0}$, $E_2 = \binom{0}{1}$
so we have $\det(E_1, E_2) = 1$ and $-1=\det(F(E_1),F(E_2)) = -\frac{b}{2} - \frac{a\sqrt{3}}{2}$, so $2 = a\sqrt{3} + b$
next $\langle F(E_1), F(E_2) \rangle = 0$, so $-\frac{a}{2} + \frac{b\sqrt{3}}{2} = 0$, so $b = \frac{1}{2}$ and $a = \frac{\sqrt{3}}{2}$, so $F\binom{0}{1} = \binom{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$

s.harp

12:10 AM

Unitary only makes sense on Hilbert spaces, what is $L(V,W)^G$ and $L_G(V,W)$?

hmm

right

hilbert spaces then

and we can find orthogonal complements using hilbert spaces I suppose

though why it is a complement is beyond me

maybe Riesz

@MatheinBoulomenos so does Maschke apply for unitary representations of compact groups?

yes, that's part of the Peter-Weyl theorem

cool

1:42 AM

Hey there guys.

Q: A surface geometry such that all lines only and always intersect at right angles.

0

I'm looking for any 3D surface such that given any two lines on the surface, they intersect and only intersect in right angles.

2:01 AM

Can someone please justify this? (or give a counterexample)

@LucasHenrique ?

Euclidean distance is the only rotation-invariant metric on $\Bbb R^n$

@LucasHenrique that link requires an account

please do not advertise for Quora

LOL, sorry then (?). I'll post the answer here.

> Euclidean distance is the only metric that is the same in all direction, that is, rotation invariant. This fits very nicely with the general qualities of our universe, which is also rotation invariant.

That is really all there is to it. All other metrics are dependent on how the coordinate system is rotated to be meaningful.

:thinking:

@LucasHenrique heck no

dist(x,y)/5

2:08 AM

Maybe $k \mathrm{dist}(a,b)$, then.

In fact, $\Bbb R^n$ is a model to Euclidean geometry axioms - and if you take points $A$, $B$ and map their distance to any arbitrary real number, then any other segment will be proportional to this distance.

true

We do also have that $a\mathbb{v} \mapsto \mathbb{v}, a > 0$ preserves all the metric space properties of $\Bbb R^n$.

Or more explictly, $\sqrt{a}v \mapsto v$.

ok

user330477

3:00 AM

Why is the closure of the set of matrices of rank $k$, the set of matrices of rank less than equal to $k$?

3:48 AM

Why can't there be a continuous function $f : [ 0,1 ] \rightarrow \mathbb { R } \text { such that } f ( [ 0,1 ] ) = ( 0,1 )$

@SharathZotis so a continuous function that only maps [0,1] to (0,1)

hmm

@SharathZotis because if 0 and 1 are the sup and inf of the range of the function then there is a sequence to each and therefore a limit to each. But that would imply the limit does not equal the value of the function

contradiction

as for constructing the sequence refer to the theorems you know

it's not that hard

Abcd

4:11 AM

In probability does OR include AND?

Will Hunting

In mathematics, or always includes and unless otherwise specified.

Abcd

thanks.

Will Hunting

A or B allows for the case of A and B.

That's why in a Venn diagram, A or B is represented by the union and A and B by the intersection.

4:31 AM

Im having a little bit of trouble understanding why the rationals are not connected

Two nonempty sets $ A . B \subset \mathbf { R } \text { are separated if } \overline { A } \cap B$ and $A \cap \overline { B }$ are both empty

This is the definition of seperated in my text.

Basically what I have so far is Let $A = \mathbb{Q} \cap (- \infty, \sqrt(2))$ and let $B = \mathbb{Q} \cap (\sqrt(2), \infty)$

Why would A complement intersect B be the empty set?

Doesn't A complement = B?

and so $\overline { A } \cap B = B$

@SharathZotis line over A is the complement of A correct?

yes

the complement of A does not equal B

because there are irrationals on that range

or any range for that matter

but B is still a subset of the complement of A

@SharathZotis in fact your definition of separation is equivalent to saying that A = B

because if the complement of A contains nothing in B

then the complement of the complement of A must contain everything in B

the complements cancel of course

same for B

meaning that everything in A is in B and everything in B is in A

A = B

4:47 AM

Still the complement of A interesect B is nonempty right? which contradicts the definition of seperate

@SharathZotis what do you mean?

?

basically in my example

A complement is the set of reals from $(\sqrt(2), \infty)$

and the intersection of this with B is not empty

@SharathZotis dont even need that

A not equal B

therefore not separate

your definition of separate is literally just another way of writing set equality

there's definitely a typo or mistake somewhere

@SharathZotis can you double or triple check that you wrote the definition correctly?

hmm I will check give me a moment

imgur.com/a/t2BbFgL

this is the definition

5:08 AM

@SharathZotis A = C and B = C satisfies that. Don't you agree?

where C is some other set

yup

but then by that reasoning

all sets

are disconnected

@SharathZotis so by that reasoning the rationals are disconnected

I see

cause C union C = C

no im pointing out how this definition makes no sense at all

there's nothing "separated" here

that definition is either made up on a whim to fool with a student for an assignment to see if they'll be tripped up into looking too deeply at the name of the term

(which would make sense if this is a class about learning proofs and logic)

or this book has a typo

@SharathZotis yer book has a typo

definitely

ok

makes sense

This is from Abbot's Understanding Analysis

5:12 AM

:-/

yeah definitely talk to your teacher if you have one

even if im wrong about equal sets being the only separated sets

the fact that equal sets satisfy that definition breaks the definition of connected

because then everything is disconnected

which is not right

Thank you for your help

@SharathZotis no problem. I'm just trying to think of how one might define separated

if I presume that it means there are two sets such that everything in one set is greater than everything in the other set

then it becomes an issue of whether the greater set contains the min

and the larger set contains its max

or inf and sup

sorry whether they have a max or min

@SharathZotis you've learned of max and min and sup and inf right? You know what I mean.

Yup

granted I'm sure if there's a way to define that without those

one the book was trying to use

cause basically the issue there

with (-inf, sqrt(2)) and (sqrt(2), inf)

regardless of rationals

is that sqrt(2) is in neither

it's between them

but it would take some thinking to make a definition that captures that without using max or min or sup or inf

or that only uses basic set operations

(arguably you cannot because some ordering information would be needed in the definition)

so now I wonder what the heck the author meant to intend by separated

(other than useless fluff)

Will Hunting

@Blue One of the reasons I don't like Collins Dictionary is that its definitions are often unclear. =)

Q: Where does Google get its dictionary data?

5:26 AM

@WillHunting who are you talking to?

Anonymous

@WillHunting Ah, that may be true :) I very rarely check definitions from Collins. Usually, just googling suffices (their primary source for definitions is "Oxford Dictionaries" afaik).

Anonymous

10

Use define: in Google and you will get dictionary definitions and etymologies. For instance: define: dictionary. Where is Google getting the definitions, etymologies, and pronunciations? I doubt they are compiling it themselves—they probably license it from somewhere.

google-search

@SharathZotis let me know if that definition ends up being used for something in the book. It might just be the books example of a poorly thought out definition. I'm not sure what it is trying to convey as a concept.

I will let you know

thanks

im heading off

Nick

5:54 AM

What does floor($\log_2(n!)$) mean?

I'll have that many bits for n! keys :/

Alessandro Codenotti

6:23 AM

@Sharath overline means closure

(and the definition in the book is fine)

Gaurang Tandon

8:03 AM

Hey all, in this question on rings how did they arrive at $abna^{-1} = c$? I know a is a unit implies that its multiplicative inverse exists. So, shouldn't it be $aa^{-1}bn=c$? How do we know that the multiplication commutes in that ring, as it wasn't explicitly mentioned?