Mathematics - 2024-12-25

hbghlyj

00:17

Suppose $f:Y\to X$ is a continuous map. Since $f^{-1}(\{f(0)\})$ is open in $Y$ and contains 0, it must be of the form $\{0\}\cup\{\frac1n:n\ge k\}$ for some $k$.

Thorgott

so what does that mean in terms of the connected components?

hbghlyj

01:07

$f:Y\to X$ and $g:X\to Y$ induce $f_:\pi_0(Y) \to \pi_0(X)$ and $g_:\pi_0(X)\to\pi_0(Y)$ and suppose $g\circ f\simeq \text{id}_Y$ then $g_*\circ f_*=\text{id}_*$ but $f_*$ is not injective, contradiction.

$f:Y\to X$ and $g:X\to Y$ induce $f _ *:\pi _ 0(Y) \to \pi _ 0(X)$ and $g _
*:\pi_0(X)\to\pi _ 0(Y)$ and suppose $g\circ f\simeq \text{id}_Y$ then $g_* \circ f _* =\text{id}_*$ but $f_*$ is not injective, contradiction.

So $X$ and $Y$ are not homotopy equivalent.

Jakobian

03:04

Shouldn't it be $f_0, g_0, \text{id}_0$

Notationally the asterisk looks like something you can plug anything into so e.g. $f_\ast:\pi_\ast(Y)\to\pi_\ast(X)$ is how I'd think the notation works here

leslie townes

07:08

jakobian while that makes sense as a convention it is not one i have seen in use, and it is pretty common to use a subscript asterisk for the induced map, at least in a context where one is considering only one "level" of homotopy group (i.e. just pi_k for one value of k, where it varies i have seen pi_k(f) but not, i think, f_k)

e.g. hatcher does it (not that i would defer to him as an authority on anything)

which is weird because it's definitely in a tiny amount of tension with the use of * as a wildcard or placeholder, which i have seen in at least the names of homology and cohomology groups

which i think hatcher also does

Jakobian

07:25

that's interesting, well, the asterisk now that you mention it, I think I have seen it as a map induced between fundamental groups

also interesting how $\pi_0$ doesn't actually depend on the base point

leslie townes

it's a christmas miracle

think_meaning_buildß

07:49

it was a chistmas 🎄 miracle that it only grazed his ear👂

Jakobian

I don't like where this is going

mo-_-

10:07

Merry Christmas 🥳🥳

Binky

10:31

Merry Christmas! 🎅🎄✨

mo-_-

10:45

Merry Christmas 🍕🍍

@Pizza Are you from Italy?

@Binky What are your Christmas traditions?

Pizza

@mo-_- Yes

@mo-_- Merry Christmas

Merry Christmas to everyone 😁

human

15

A: Prove that the additive groups $\mathbb{R}$ and $\mathbb{Q}$ are not isomorphic.

Let $\Phi: \mathbb{Q} \rightarrow \mathbb{R}$ homomorphism of additive groups. Then $\Phi$ is already determined by $\Phi(1)$ (as $\Phi(\frac{a}{b})= \Phi(\frac{1}{b})+ ... + \Phi(\frac{1}{b})$ ($a$ summands) and $\Phi(1)= \Phi(\frac{1}{b}) + ... + \Phi(\frac{1}{b}) $, ($b$ summands)) Now, say ...

ok I am really stupid can someone explain this answer

why can't I understand it :(

mo-_-

@Pizza Pizza + pineapple is illegal

Sine of the Time

@mo-_- you may get arrested in Italy

mo-_-

11:00

@SineoftheTime 🙀

Sine of the Time

@human which part it's not clear?

mo-_-

It’s been 3 hours without math, and I’m officially losing it. Someone pass me a derivative before I start integrating the number of ornaments on the tree

human

the last step
how $\sqrt{2}\cdot \Phi (1)$ does not have a preimage

Sine of the Time

what does it mean that $\sqrt 2 \Phi(1)$ has a preimage?

Jakobian

@human because $\Phi$ maps $x$ to a rational multiple of $\Phi(1)$ and $\Phi(1)\neq 0$

human

11:07

oh yeah oh yeah

thanks

Jakobian

the more obvious way to see that $\mathbb{Q}$ and $\mathbb{R}$ are not isomorphic as groups is to see they are not even of the same cardinality

$|\mathbb{R}| = 2^{\aleph_0} > \aleph_0 = |\mathbb{Q}|$

human

yeah I knew that, I was trying to understand this one
and I feel stupid now xd

mo-_-

11:26

@Binky did you solve yesterday's problem?

Binky

No

mo-_-

Why

Where are you stuck?

@Binky hello?

Binky

11:41

Hello

mo-_-

Show your wok

Binky

There isn't

mo-_-

Send the problem

Binky

But then why did you randomly start writing if I had solved it

@mo-_- Why ?

mo-_-

yesterday, by Binky

So there is no basis for the image?

What is your answer?

Binky

11:45

I told you there is no work

Tell me what answer I need to have

mo-_-

I just wanted to help you... :(

Binky

Okay

Can I help you?

@mo-_- I didn't ask for any help

At least for now

Anyway

💡

mo-_-

?

@Binky Why

Anyway do what you want...

Binky

Okay can you help me?

mo-_-

Ok

Binky

11:51

what should I do?

mo-_-

Show your work

Binky

Okay I can't show it, bye

mo-_-

Ok... Bye

Binky

no no wait

yesterday, by Binky

Okay they are not linearly independent

mo-_-

Nope, we're just spamming and you don't want to collaborate

Enough .

Binky

11:55

🤣

$$\begin{pmatrix} 1/2 \\ 1 \\ 1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$$

@mo-_-

the first vector can be write as a linear combination of the two others

would make sense that the other two vectors themselves formed a basis

since they are linearly independent

😭

mo-_-

6 mins ago, by Binky

🤣

👍

Binky

@mo-_- it's not true that's why

Pizza

What happened?

Sine of the Time

@Pizza nothing important :)

Pizza

12:22

Calculate:
$y(t) = e^{2t} + \int_{0}^{t} e^{2(t-\tau)} y(\tau) \, d\tau$

Anyone know the wolfram command for this?

I mean to find y(t)

@SineoftheTime 👍

Sine of the Time

@Pizza how do you usually solve these kind of problems?

you differentiate?

Pizza

I used the Leibniz rule for differentiation under the integral sign

But

I realized that it had to be solved by applying the Laplace transform

But still I have to find the same result using both methods

Sine of the Time

are you familiar with the convolution product?

@Pizza then you can differentiate first and then put it on wolfram

Pizza

@SineoftheTime $\mathcal{L}\{f(t) * g(t)\} = \mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\}$

Sine of the Time

does that integral look like a convolution product?

Pizza

12:31

the convolution $(f * g)(t)$ is defined as:
$(f * g)(t) = \int_0^t f(\tau) g(t - \tau) \, d\tau.$

In my integral equation the second term represents a convolution of $e^{2t}$ and $y(t)$

Sine of the Time

ok, so you have $y(t)=e^{2t} +e^{2t}*y(t)$

Pizza

Yes

Sine of the Time

now apply LT

Pizza

$\mathcal{L}\{y(t)\} = \mathcal{L}\{e^{2t}\} + \mathcal{L}\{e^{2t} * y(t)\}$

$\mathcal{L}\{e^{2t} * y(t)\} = \mathcal{L}\{e^{2t}\} \cdot \mathcal{L}\{y(t)\}$

I let $Y(s) = \mathcal{L}\{y(t)\}$, then:
$Y(s) = \mathcal{L}\{e^{2t}\} + \mathcal{L}\{e^{2t}\} \cdot Y(s)$

$\mathcal{L}\{e^{2t}\} = \frac{1}{s-2} \quad \text{for } s > 2$

$Y(s) = \frac{1}{s-2} + \frac{1}{s-2} Y(s)$

So

$Y(s) = \frac{1}{s-3}$

$y(t) = e^{3t}$

Sine of the Time

now try with Leibniz rule

Pizza

12:44

I used $e^{\alpha t} \quad \frac{1}{s-\alpha}, s > \alpha$

@SineoftheTime I had done it and I got the same result

Sine of the Time

to check, you can subsitute $e^{3t}$

In fact, $\int_0^{t} e^{2t-2\tau}e^{3\tau}d\tau=e^{2t}\int_0^{t}e^{\tau} d\tau=e^{2t}(e^t-1)$

So $e^{2t}+e^{2t}(e^t-1)=e^{3t}=y(t)$

Pizza

Yes

Sine of the Time

so it's correct

no need to use WA

Pizza

😃

Nice !

mo-_-

This seems interesting

@SineoftheTime What do you study?

Sine of the Time

12:54

maths

mo-_-

Do you do research?

Sine of the Time

no

user123234

does someone knows about path algebras?

Vladimir Lysikov

I know about path algebras

of quivers

not sure, maybe there are other ones

user123234

13:15

@VladimirLysikov Oh this would be nice, because I want to find the path algebra of the the following quiver

I know the following definition:

Vladimir Lysikov

So what's the problem?
Can you describe all paths in your quiver?

user123234

I don't see how to do this with Problem 2.8.6. I think it should be a nice task to compute this, and I think if I got it it is not difficult to solve different problems, but I really strugle.

I mean my paths are $\alpha_i$ for $2\leq i\leq n$ no?

Vladimir Lysikov

2.8.6 giver you description of an algebra in terms of generators, and if it is enough for you, you can just use it
But I think here it is more enlightening to use the definition 2.8.4 directly

paths are not just arrows $\alpha_i$, paths can have different lengths

Can you say how many paths there are from vertex $i$ to the vertex $j$?

user123234

ah so my paths are also all possible concatinations of $\alpha_i$?

Vladimir Lysikov

yes
For example, there is a path from 3 to 1 which is $\alpha_3 \alpha_2$

user123234

13:23

so from $i$ to $j$ are there $j-i$ paths?

Vladimir Lysikov

and there are also trivial paths $p_i$ that go from $i$ to itself

No
Let's take specific example. How many paths are from 3 to 1?

user123234

There is the one 3->2->1 but also 3->1 and with the self loops at each stage?

Vladimir Lysikov

There is no arrow from 3 to 1, only from 3 to 2 and from 2 to 1
This is a quiver, not a category

user123234

but then why are my paths not only $\alpha_i$

Vladimir Lysikov

Do you have a definition of a path before that?

user123234

13:27

I only have this before that

Vladimir Lysikov

Ah, that is unfortunate
A path from $i$ to $j$ is a sequence of arrows $a_1, \dots, a_k$ such that $h'(a_1) = i$, $h''(a_k) = j$ and $h''(a_i) = h'(a_{i + 1})$
So the idea is: a path is a sequence of arrows that go one after another

We also have a trivial path $p_i$ from $i$ to $i$, which is an empty sequence, and we postulate that trivial paths at each vertex are not equal to each other

user123234

ahaa, so there is only the path $\alpha_3\alpha_2$ from $3$ to $1$?

Vladimir Lysikov

Yes

user123234

okey so from $i$ to $j$ I have only $1$ path?

Vladimir Lysikov

Almost. We have a unique path from $i$ to $j$ as long as $i \geq j$, and no paths if $i < j$.

user123234

13:33

ah right, I agree until here and now?

Jakobian

@VladimirLysikov I think I've seen it before, but I can't say I know them

you know, outside of the definition

Thorgott

@hbghlyj yup, that's correct :)

Vladimir Lysikov

So by definition an element of a path algebra is just a linear combination of these paths
If we denote by $P_{ij}$ the path from $i$ to $j$ with $i \geq j$, then the path algebra as a linear space is generated by $P_{ij}$
It remains to understand how the multiplication works

Thorgott

@Jakobian nah, $(-)_{\ast}$ is completely standard for many kinds of induced maps

Vladimir Lysikov

What would be the product $P_{ij} P_{kl}$ by definition?

Thorgott

13:35

you'd write $f_{\ast}$ for most operations that are given by post-composing with $f$ or applying $f$, unless there is a need to distinguish

user123234

@VladimirLysikov the product would be the path from $i$ to $j$ concatinated with the one from $k$ to $l$ but this would only be nonzero if $k=j$?

Vladimir Lysikov

Yes

user123234

okey and then?

Vladimir Lysikov

So the path algebra of $A_n$ is generated as a linear space by paths $P_{ij}$ with $i \geq j$, and we have $P_{ij} P_{jk} = P_{ik}$ and $P_{ij} P_{kl} = 0$ when $j \neq k$
Do you recognize this algebra?

Jakobian

I believe path algebras appear in Algebra: Chapter 0 by Aluffi

user123234

13:38

@VladimirLysikov I mean it seems like some multiplication with matricies so there should be some matrices involved?

Vladimir Lysikov

Exactly
This is the algebra of lower triangular matrices

user123234

and one should see this immediately?

and why not upper one?

Vladimir Lysikov

@user123234 the algebra of upper triangular matrices and lower triangular are isomorphic. I just used the notation $P_{ij}$ with $i \geq j$, so the connection to lower matrices is more immediate

@Jakobian Aluffi's Algebra: Chapter 0 is a great book

user123234

ah I see, so one should see this immediate? are there maybe other exercises with solutions to train this?

I have found one:

which I would like to solve, can I try it and write it to you?

because I have not found solutions and I think it might be a good practice exercise

Vladimir Lysikov

@user123234 ok

user123234

13:45

thanks a lot

Vladimir Lysikov

:66877735 I mean that was probably in the linear algebra course in some form
The matrix space is generated by the spandard basis $E_{ij}$, and the multiplication works as $E_{ij} E_{jk} = E_{ik}$ and $E_{ij} E_{kl} = 0$ when $j \neq k$
Here we just restrict to $i \geq j$

user123234

oh okey I see

Vladimir Lysikov

And we can also see the description of 2.8.6 more clearly now
it says that the path algebra is generated by the trivial paths, which are our $P_{ii}$, and the arrows $\alpha_i$, which are our $P_{i, i-1}$
Subject to relations which are essentially $P_{ii}^2 = P_{ii}$, $P_{ii} P_{i,i-1} = P_{i,i-1} P_{i-1,i-1} = P_{i,i-1}$ and $P_{ii} P_{jj} = P_{ii} P_{j,j-1} = 0$ when $i\neq j$
And indeed, the unique path $P_{ij}$ with $i>j$ can be represented as a product $\alpha_i \alpha_{i-1} \dots \alpha_{j + 1} = P_{i,i-1} P_{i-1,i-2} \dots P_{j+1,j}$

user123234

@VladimirLysikov I mean in the quiver $S_n$ we have the trivial paths $p_i$ from. From $i$ to $j$ there is one path if $j=0$ and zero paths else i.e. $P_{i0}=1$ and $P_{ij}=0$ for $j\neq 0$. But there is no multiplication since I can't concatinate paths.

@VladimirLysikov I see thanks a lot!

Vladimir Lysikov

Good
You can actually concatenate with the trivial paths
So you have $P_{ii}$ and $P_{i0}$ and the multiplication is $P_{ii}^2 = P_{ii}$, $P_{ii}P_{i0} = P_{i0} P_{00} = P_{i0}$, everything else is 0

user123234

13:59

ah okey but then how do I get my algebra? So I see you use problem 2.8.6

I think the algebra is generated by $P_{i0}$ for $0\leq i\leq n$

Vladimir Lysikov

It is generated by $P_{ii}$ and $P_{i0}$
I don't think in this case there is an easier description
Technically, it is the algebra of matrices of the following form
$\begin{bmatrix}
* & 0 & 0 & \dots & 0\\
* & * & 0 & \dots & 0\\
* & 0 & * & \dots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
* & 0 & 0 & \dots & *
\end{bmatrix}$

But I don't think it easier than the description from 2.8.6

user123234

ah so $P_Q=\langle P_{ii},P_{i0}: 0\leq i\leq n\rangle$.

Vladimir Lysikov

Yes

user123234

And is it true that in my notation of the book $p_i=P_{ii}$ and $a_h=P_{ij}$

Vladimir Lysikov

Yes

user123234

14:06

okey thanks a lot you really helped me a lot!

Vladimir Lysikov

One thing which is not covered by these examples is quivers with loops
For these the path algebra is infinitely dimensional
For example, if we take a quiver with one vertex and one loop arrow $\alpha$, then every path is $\alpha^k$ (we can repeat the loop $k$ times), and the path algebra is the algebra of polynomials in $\alpha$

There are some theorems that only work for quivers without oriented loops, or are much easier to prove for quivers without oriented loops

Hopefully we didn't hog the chat too much, sorry

user123234

14:24

oh thanks a lot!

user123234

15:10

@VladimirLysikov sorry to disturb you again, if I have the quiver $1\rightarrow 2\rightarrow 3\rightarrow 4\rightarrow 5$ and also an arrow $3\rightarrow 6$ then I would say that the path algebra is the algebra of lower triangular matricies where also the 6 column is nonzero, i.e.:

@VladimirLysikov is this true because I would say that if $j\neq 6$ then $P_{ij}P_{kl}=P_{il}$ if and only if $j=k$ else it is zero and if $j=6$ then $P{ij}P_{kl}=P_{i6}$ if $k=l=6$ and else it is zero because if your target is $6$ you can't concatinate further paths so you end there. If this is wrong can you maybe help me furhter?

Vladimir Lysikov

@user123234 so here arrows are in the opposite direction to our $A_n$ example before, so these are a bit easier to see as upper triangular
And in the sixth column there is only entries $16$, $26$, $36$ and $66$
Because there are no paths from 4 and 5 to 6
So
$\begin{bmatrix}
* & * & * & * & * & * \\
0 & * & * & * & * & * \\
0 & 0 & * & * & * & * \\
0 & 0 & 0 & * & * & 0 \\
0 & 0 & 0 & 0 & * & 0 \\
0 & 0 & 0 & 0 & 0 & *
\end{bmatrix}$

Simd

15:45

I was hoping you were solving my challenge for a moment!

Jakobian

18:25

@VladimirLysikov no such thing as hogging the chat

user123234

@VladimirLysikov When would my matrix be correct then?

Vladimir Lysikov

Yours is not an algebra because it is not closed under multiplication
For 5->4->3->2->1 and additional arrows from each of 1...5 to 6 we have
$\begin{bmatrix}
* & 0 & 0 & 0 & 0 & *\\
* & * & 0 & 0 & 0 & *\\
* & * & * & 0 & 0 & *\\
* & * & * & * & 0 & *\\
* & * & * & * & * & *\\
0 & 0 & 0 & 0 & 0 & *
\end{bmatrix}

user193319

20:55

-1

Q: Normal Subgroups of Virtually Simple Group

I have a very elementary question, but I cannot answer it. My question is, If $G$ is a virtually simple group, does $G$ have finitely many normal subgroup?

leslie townes

23:14

haha yeah, i recommend editing in some of that comment, maybe also the observation from this chat, about "direct product with a simple factor" probably not being a way to produce an example.

maybe also add the definiton of 'virtually simple' the world of people who would understand what that means is much wider than the world of people who are used to that use of the term 'virtually'

anyway, merry christmas to you :)

Arjun Raghavan

23:55

Call a number despicable if it cannot be represented in the form p+n³ or p-n³ where p is prime and n an integer. Prove all despicable numbers are cubes. Hi I am stuck on this problem, could anyone help?

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$Mathematics$ Mathematics