Here's a question

What's the simplest way to get Desmos to output this

0 is a bilinear alternating form

Except the zero

What do you think?

I dunno

What does it mean for a bilinear map to not be 0?

What's the image of $b$

It means it is not zero.

Wow !! i never thought of that!! what a great reply

well the image has to a subset of the reals, i dont see why one needs to be an element of it

A subset of what kind

Can the image of a linear map $V\to W$ between vector spaces be an arbitrary subset of $W$? Or does it have some extra structure

it is not linear, it is bilinear., if it were linear, then the zero is contained i know that

If you fix one of the entries it's a linear map

$b(-,w)$ is linear

yea man i dont know i am dumb, this is why i am asking, i dont remember why is one contained?

Fix a bilinear map $b$ other than the zero map. Since $b$ is not the zero map, we can find vectors $v$ and $w$ such that $b(w,v)$ is nonzero

Hint

Call this $C:=b(w,v)$

What is $b(w/C,v)$?

= b( w+ [b(w,v)], v) , where [] are equivalence classes.

@MadSpaces What's the type of $C$? Scalar, or vector space?

You are building modulo right?

No.

Oh okay its scalar ,who writes scalars as big letters?

usually used for sets

Sorry. $c:=b(w,v)$. What is $b(w/c,v)$?

much better

thats one

jesus ..

okay.. i am going to commit seppuku, thanks Akiva

(and the thing I was trying to get you to tell me is that the image of a linear map is a vector subspace)

Indeed. And since R is a one-dimensional space, the only vector subspaces are R itself (dim 1) and {0} (dim 0)

And that a bilinear map is nonzero iff there exists v and w s.t. b(v,w) is nonzero

(so not only you can find $v,w$ such that $b(v,w)=1$, but as long as $b$ is not zero, for any $v\neq 0$ you can find $w$ with $b(v,w)=1$, and the statement is symmetric in $v,w$ of course)

Okay guys, thanks for the help, this was embarassing

@AlessandroCodenotti Hm? No, $b(v,-)$ can be the zero map even if $b$ and $v$ are nonzero

Let $b(v,w)=p(v)p(w)$ where $p$ is projection onto first coordinate, say

As long as $b(v,-)$ is not zero, fine

Fun fact: while the image of a linear map is a vector subspace, the image of a bilinear map is not necessarily a vector subspace

because $b(u_1,u_2)+b(v_1,v_2)$ has no reason to be in the image

Basically because $V\times W\neq V\otimes W$

I guess I mean $V\times W\not\twoheadrightarrow V\otimes W$, i.e., it doesn't surject onto it

What is the importance of probability vector fields?

Is ChatJax being weird for anyone else? For me, when I type $\not\to$, the slash goes through the arrow, but when I type $\not\twoheadrightarrow$ it misses

(In retrospect I could've just said "$V\times W$ doesn't surject onto $V\otimes W$" in words)

I guess it's a list of probabilities for $n$ events, and if it's a vector field then those probabilities depend on where you are?

I guess weather maybe? You have a two-vector of "will rain" "won't rain" and it depends on location

@geocalc33 $\le$, surely

yes @AkivaWeinberger of course, thanks

I guess statistical manifolds generalize this

I've been rate limited on stack exchange. Never gotten that before

Is there some kind of marking for dead users

Passed last year juni, rest in peace.

@MadSpaces See the answer here

Interesting thanks

This is a lovely book review: maa.org/press/maa-reviews/…

I will have to haunt MSE spookily after my death.

So for example, $ P(x; \xi) \sim \exp \left ( - \frac{(x-\mu)^{2}}{2\sigma^{2}} \right ) $ the $ \xi = (\mu, \sigma^{2}) $ parameters vector can be considered a coordinate system for the corresponding manifold

@TedShifrin I aspire to haunt while I'm alive

@AkivaWeinberger Well, there's no denying you do that effectively.

this summer in Germany you can travel anywhere you want

for $10 dollars

a month transit pass

what's an example of two different math objects that are each representations of some other possibly more fundamental structure?

Maybe lattice, torus, and elliptic curve is the best I can come up with

@Ted Shifrin It should be correct now: $\frac{\partial \phi}{\partial x}=F(\frac{y}{x})+x\frac{d(F\circ g)}{dx_1}\cdot\frac{\partial g}{\partial x}=F(\frac{y}{x})+xF'(\frac{y}{x})\cdot (-\frac{y}{x^2})=F(\frac{y}{x})-\frac{y}{x}\cdot F'(\frac{y}{x})$ where $g:\mathbb{R}^2\to\mathbb{R},\ g\begin{pmatrix}x\\y\end{pmatrix}=\frac{y}{x}\equiv x_1$.

I wonder if I should buy Rudin xor spivak for real analysis.

@lorenzo OK. It might be good practice for you to set this up with the chain rule in matrix form, as in the earlier section.

Does anyone here know a good source that explains representability of cohomology, that is, the natural isomorphism $H^n(X;G)\cong [X,K(G,n)]_*$?

That was certainly in Spanier and other standard algebraic topology texts. Mosher/Tangora, too, probably. Does Hatcher not do that eventually?

Thanks I'll have a look in Spanier's book

The part I'm confused about wasn't clear for me in Hatcher (but other parts were nice)

@TedShifrin and also $\frac{\partial\phi}{\partial y}=\frac{\partial}{\partial y}\left(xF(\frac{y}{x})\right)=x\frac{d(F\circ g)}{dx_1}\cdot\frac{\partial g}{\partial y}=xF'(\frac{y}{x})\cdot (\frac{1}{x})=F'(\frac{y}{x}).$ Thanks for the advice. I will certainly try to set this up using the chain rule in matrix form tomorrow. Bye!

hatcher does it via abstract nonsense, observing the RHS defines a cohomology theory and then comparing in zeroth degree

theres a nice obstruction-theoretic explanation here, though ive never worked out the details

right, yea, my notes give this cell-by-cell approach, and I was indeed hoping to find additional literature that covered this

"I think that a nice write up can be found in the first chapter of Mosher and Tangora (a very nice book)."

I guess I'll have a look there