Well one of the key thing seemed was (on a cursory look) to take the grading of $k[x,y,z]$ as $p,q,r$ respectively and then have the ideal generated by just homogenous elements, and then you start looking at various $\Bbb{N}$ linear combinations which come up as homogenous elements

Yeah, this I figured

But how do you link that to parity of the largest blah blah

Shrug

The moral is in algebra there is no reason why something is true or false. It just is.

This looks like combinatorics tho

But these intersection problems are weird problems. I think it is an open problem to show that every space curve in $\Bbb{P}^3$ is the intersection of two surfaces

Yeah these are pure algebra problems

Now I'm not going to be able to sleep

goddamn algebra

@LSS not sure what you mean by that, but the answer is probably no

Your sleep is too easily troubled then :)

Is it fine if the same basic variable reoccur in another iteration of simplex table?

Look at $\mathbb Z/4\mathbb Z$ and $(\mathbb Z/2\mathbb Z)^2$ for instance

Balarka: meh, it's just that I feel I'm close to understanding it, so my mind won't think about something else

can mods vote to reopen questions without it being absolute?

@kunalCh. Yes. For example, cycling can happen, in which one gets stuck at a vertex (bfs) in the polyhedron but the entering and exiting variables keep permuting around.

In any case, even without cycling, I don't think reappearance of the same basic variable is an issue.

In this video —> (youtu.be/aMLl6jUlpqA), around 5:22, the author says that y hat spans a 2D vector space by the vector equation $\hat{Y}=a(1,2,4)+b$ for a scalar $a$ and a constant vector $b$. Does this equation really span a plane?

i am not going to watch the vid, but it is a line through $b$ in the given direction. It depends on what the author means by spanning the space.

let $B \le A$ be finite abelian groups. how do I show that every irreducible character $\phi: B \to \mathbb{C}$ lifts to precisely $[A:B]$ irreducible characters $\psi: A \to \mathbb{C}$?

plz watch the video it's only got 900 views

(this is from a problem set - I can get tips but not the full answer :) )

@copper.hat Right, a line comes to mind

If the author means the span of all vectors of that form then itis a 2d space as long as $b$ does not lie on the line through $(1,2,4)$.

ambiguity makes good poetry but poor mathematics

I did something hipsted and hosted my own wordpress site

because I hadn't realized all the pros used wordpress.com

all the pro whats?

all the pro mathematicians

do you have a blog copper.hat?

no, just my ramblings that are spread all over :-)

@copper.hat what do you mean by "...all vectors of that form..."?

all vectors are a straight line with an arrow at the end

Each $a$ has a particular vector associated with it.

they all have the same shape

except the 0 vector

for the zero vector the arrow should point wherever the thing you most want is

I mean $\operatorname{sp} \{a(1,2,4)+b\}_{a \in \mathbb{R}}$.

like in pirates of the caribbean

Thanks for the quick replies

Processing the information

no problem

@schn in fact, you only need to pick two particular values of $a$ then the span of those two vectors will be 2d.

@LucasHenrique I would think of the induced representation

span means all linear combinations.

but i do not know if that is what the author intended

I found that out 2 years ago :/

I started a math blog once but I have since not posted anything :L

I always want to sit there and polish all my expository articles, and I do this in LaTeX since it's a lot more flexible than mathjax or katex in html I find.

And so I just opt for hosting my articles.

seems like too much baggage to have lying around and to have to maintain.

you just host the pdf articles?

or do you display them in some way on the site?

also, do you have a system for taking comments?

@copper.hat But for a fixed $b$, the two vectors associated with different values of $a$ lie on the same line, right?

This is how I interpreted the notation $\operatorname{sp} \{a(1,2,4)+b\}_{a \in \mathbb{R}}$.

yes, the two vectors lie on the same line. take the plane, the points $(1,0), (0,1)$ lie on a line but their span in the entire space.

there is a difference between the set of points and the span (unless the set of points is a linear space).

@copper.hat You also wrote "...as long as $b$ does not lie on the line through $(1,2,4).$" Did you mean the line that also goes through $(0,0,0)$?

yes, the line formed by the points $t(1,2,4)$ with $t$ a scalar

Got it.

look at examples in the plane, it will be fairly clear

of course, 3d and above are harder to visualise

@BalarkaSen I thought about that - there's a unique induced representation and its dimension is $[A:B]$ times the irrep's dimension

i'll think about it for a while, thanks

Remember your groups are abelian

so irreps have dimension 1

any deeper fact I should have in mind?

Nah :)

thanks Balarka!

@Ladiesandgentlemen just host, and no system for taking comments though I welcome anyone to comment with hypothes.is! It's a great tool!

@copper.hat thanks for sharing some insight

Is it correct that $\operatorname{sp} \{a(1,2,4)+b\}_{a \in \mathbb{R}, b \in \mathbb{R}^3}$ would span $\mathbb{R}^3$?

@schn this looks weird... since arbitrary b already gives you $\mathbb{R}^3$

^ this

@LucasHenrique are you saying that $\operatorname{sp} \{a(1,2,4)+b\}_{a \in \mathbb{R}}$ spans $\mathbb{R}^3$?

that is, $b$ is fixed

doesn't seem like what he was saying

depending on what b is, that's not going to be a vector subspace of R^3, and even when it is, it's incapable of being R^3

$b$ is a constant vector

geometrically that's a line through b in the direction of (1,2,4) regarded as a direction vector. no hope of being R^3. it's a line.

not clear what work, if any, "sp" { } is doing outside of that

@schn $\operatorname{sp} \{a(1,2,4)+b\}_{a \in \mathbb{R}}$ spans at most a 2 dimensional subspace. (It is not hard to show that $\operatorname{sp} \{a(1,2,4)+b\}_{a \in \mathbb{R}} = \operatorname{sp} \{(1,2,4),b\}$.)

linear span in the legalese :-)

but really, you would have to figure out what the youtube person meant by span

i agree with the irishman