somebody I know is taking Calc II

the homework is multiply choice with no partial credit

is that fair? is that good?

the same is true for exams

@MikeMiller ah, I see. Thanks!

@Bob let's just say it's not the height of mathematical pedagogy.

I don't see what would be unfair about that, but it's certainly not good

Cause that means you're grading by results, which misses the arguably most important aspect of doing mathematics yourself, the actual derivation

Is Martin Hairer here on MSE? If he is, CONGRATULATIONS!!

@Alessandro apparently our modular forms course won't be online

also, morning

1-1/2^n has limit point 1 ? How to write it formally ?

why 1 is only limit point ?

@TheTestosteroneFanatic If it's a research level question, ask it in math.overflow. However, if your question amounts to "is this result from x paper true?" then don't post it since it is not a very well-formed question.

@anakhro it's of the form "how do we prove x", where x is something I found in a research paper, without any citation or explanation (those cunning authors just wrote "We know that"). I'm just not sure if that proof will be research level or will have a more trivial proof

@TheTestosteroneFanatic are all the mathematical objects in the question things that you'd find in an undergrad curriculum? If so, post on stackexchange. If not and the objects are specialized, ask in math.overflow.

If you are unsure, post it in math.purgatory and wait the requisite number of years to absolve any mathematical sins you have committed before posting it in math.overflow.

very quick question, if e is idempotent, and R is a ring, to show eRe is a ring, can I just show x \to exe is a homomorphism?

makes little sense to show something is a homomorphism when you don't know that domain and codomain are rings

but you can of course observe it's a bijection and transport the structure+

but then i don't have to justify the introduction of the + structure? how do we deal with the abelian group structure?

I'm not sure what you mean

actually how do i even see it is a bijective? R might not have 1.

actually nvm

the additive identity is 0

but what i meant was

how do i know eRe also has the multiplicative operation

(exe)(eye) = e(xey)e

yes but how do u know u can multiply in the first place? don't we have to justify that?

You're defining a multiplication

The only thing you have to justify is that it is well-defined and satisfies the ring axioms

@Thorgott ? if $e$ is a nonidentity idempotent then the map $x \mapsto exe$ is not a bijection $R \to eRe$; eg $1-e$ is in the kernel

yeah, I mixed things up

how do i actually show the operation is well-defined...?

assume $exe=ex^{\prime}e$ and $eye=ey^{\prime}e$ and show that $e(xey)e=e(x^{\prime}ey^{\prime})e$

Sanity check: For $(W^{\omega})^{\omega} = W$ where $\omega$ denotes a symplectic form on a vector space $V$, we can use the same argument that we use for $(W^{\perp})^{\perp}$ right? I mean that there is an exact sequence just like before of the form $0 \to W^{\omega} \to V \to W^{*} \to 0$, so $dim(W^\omega) + dim(W) = dim(V)$ and then substituting that $W$ as $W^{\omega}$

@SayanChattopadhyay That's correct.

I didn't understand what you did at the end, but after the dimension argument you observe that $W \subseteq (W^\omega)^\omega$, so by dimension it must be an equality.

but of course it's well-defined

it's inherited from the multiplication of the original ring

for substructures you don't need to check well-defined

you need to check closure

(inb4 it's not a subring)

it isn't necessarily

@BalarkaSen I was just saying that in the dimension argument you get that $dim((W^{\omega})^{\omega}) + dim(W^{\omega}) = dim(V)$ which implies the dimension of the vector space under double perp is the same as $W$, and then it follows as you said

@SayanChattopadhyay Oh fine.

It's interesting that Lagrangian subspaces, i.e., those for which $W^\omega = W$, are half-dimensional. Follows from your argument.

Yeah, I was thinking though, why are the called Lagrangian?

I don't know, some physics thing

Maybe, though I haven't seen any explicit physics connection yet.

I dont particularly care haha

Using this can you prove that if $M$ is a oriented $3$-manifold with boundary, then the image of the snake map $H_2(M, \partial M) \to H_1(\partial M)$ is half-dimensional?

This is called the half-lives-half-dies fact

Oh lol, I don't see how this is implied from that though

Think about it

A better way to phrase it is to say the inclusion induced map $H^1(M) \to H^1(\partial M)$ has half-dimension image which is related to the homology formulation by PD

Sure

This is a very useful fact

I'll say more after you're done

Gotta run for now

Let A be a 9×7 matrix. Determine the number of solutions to the equation Ax = 0.

I think answer is it can not be determined I tried to come up with example

with in 1st row and first two colum I have 1 and everywhere 0 then system have infinitely many solution

If 1 in 7 pivot position then system have unique solution

hello

I'm a little behind in a class due to getting COVID; could someone help me understand how to do this homework problem about partially ordered sets? pretty lost rn but here's the question:

Let S be a set. Consider the partially ordered set (2^S, (subspace/equal symbol)). Given any U subspace 2^S show that U is bounded and has a least upper bound

I understand what the words mean but any attempt at a solution fails pretty quickly

@MadSpaces You mean like making them array variables?

@TedShifrin: good afternoon

Here is another rotating geometric shape. I bet you can identify it.