I'm not stopping you, but I'm not responsible for the consequences either.
Conventions definitely have their uses
Not abiding by them should never be considered a mistake, but of course, it's up to the person who doesn't follow them to then make clear what he means (for example, by properly labelling his axes using his own notation)
@BalarkaSen So if I have both the cohomology rings in real/rational and $\Bbb Z_2$ coefficients of some manifold, can I get it in $\Bbb Z$-coefficients?
To add a detail to SteamyRoot's definition, which is correct, if you have a function $(X,\mathcal{A},\mu)\to\Bbb R$ then it's assumed you're using the Borel $\sigma$-algebra on $\Bbb R$ when discussing measurability of functions unless otherwise specified
By the way, aren't Fourier transforms the same cool stuff that is used in signal processing? Looking forward to learn more soon. I've just started twelfth grade.
Yikes, do you guys remember all of those numbers, that link looks scary? Suppose I have two quadratic functions,
Back to my doubt,
Suppose I have two quadratic functions and I need to keep them both as low as possible altogether and ensure that it fits a criterion something like an equality constraint.
I um want to sort of know what a good strategy is to do that kind of thing, I know my description is most likely vague since I do not know the proper technical terminology for it.
Hey there @AlessandroCodenotti. Yeah, I've been trying to be active on main in recent days. My activity on here waxes and wanes...I've been spending my free time more over on chess.com haha
I haven't done much math of late. I'll look at some representation theory from time to time, and I want to start delving into applied stuff right now: coding theory and implementations of cryptosystems
I have a decent knowledge of the theory behind some public-key algorithms, but putting them into practice has a whole slew of challenges to overcome that aren't presented in the theory
I've been playing chess and teaching myself to play trumpet in recent weeks instead of math
@BalarkaSen So if I take the s.e.s. $\Bbb Z\to \Bbb Z \to \Bbb Z_2$ I should get a l.e.s. on the level of (Cech, and therefore singular) cohomology, right?
It's pretty interesting @BalarkaSen. One of the big ideas that appeals to me is the computational side of things. By finding faithful morphisms of some weird group into a group of matrices, we can work with the group a lot easier since the group multiplication can be computed as matrix multiplication, which is very well understood with super-fast algorithms
Ali was in here a few months ago telling me about algorithms that can multiply matrices close to $n^2$ time, which blew my mind
@Alessandro BTW, different way to see Hopf link has link group $\Bbb Z^2$ is to note that the complement of the Hopf link deformation retracts to a torus.
The first step in doing this is to realize $S^3$ as union of two solid torii glued along a diffeomorphism $T^2 \to T^2$ of their boundary. There are formal ways to do it but you should convince yourself that if you take a torus in $\Bbb R^3$, the complement is a punctured torus.
Is there some things to look at about the geometric properties of normed vector spaces? Like for example curvature of $\Bbb R^n$ with $\|\cdot\|_p$ norm, $0<p≤\infty$.
you have to be aware that when you look at those pictures you are viewing them as subsets of $\Bbb R^n$ with $2$-norm, so the unit spheres "think of themselves" in a totally different way than those pictures make them seem
I'm not sure what you mean, do you want to look at the $p$-sphere and give it $2$-norm, look at the $2$-sphere and give it $p$-norm or look at the $p$-sphere and give it $p$-norm?
Because in the last case the pictures you draw have almost nothing to do with the geometry of the sphere in my opinion
Hey guys, anyone good at tensor products in hilbert spaces. Specifically, I want to understand whether that $\vec{s}=\vec{v} \otimes \vec{u}$ can be decomposed into tensor products and $\vec{t}=\vec{v} \otimes \vec{u} - \vec{u} \otimes \vec{v}$ cannot is purely an algebraic phenomenon since both $\vec{s}$ and $\vec{t}$ are just vectors in the larger hilbert space geometrically. Where is this notion of separability to tensor products being stored in the geometric picture?
But these pictures, when I want to get geometric meaning from them I assume they have the $2$-norm, because I have other "intrinsic" way of looking at pictures in $\Bbb R^n$ in my brain
What I mean is look at the infty norm picture, it looks flat, but this flatness only comes from the fact that I am treating the pictures as a subset of $\Bbb R^n$ with $2$ norm
@Secret I'd relegate that to an algebraic phenomenon. in $\Lambda^k V$ for example, pure wedges represent oriented $k$-subspaces, but there are things that aren't pure wedges which must be formal combinations of them. one can depict distributivity of the wedge geometrically, but I don't find that helps grok the idea. and tensor products I don't even have much geometric intuition for.
so when I get the geometric idea "this sphere is flat almost everywhere" its not a statement about sphere$_p$ with $p$-norm but rather sphere$_p$ with $2$-norm
Notice that $\dim(V\otimes W)=\dim(V)\dim(W)$ but pairs of vectors in $V,W$ have form a space of dimension $\dim(V\oplus W)=\dim(V)+\dim(W)$, so it's definitely not just pairs of vectors.
The only thing you can really talk about in these vector spaces are geodesics. It can also be that you are of the opinion that the Lebesgue measure has nothing to do with euclidean space so you also get that in such a case
I need to go somewhere that has faster internet, I'll come back soon
If I have a finite Abelian group and I know that the homomorphism $2\cdot$ has kernel $\Bbb Z_2$, what can I conclude? That there is only one summand $\Bbb Z_p$ with $p$ even? Anything more?
Hey the coefficients $a_n$ of the Laurent series have this $(\xi - z)^{n+1}$ term is this n supposed to be the same as the n of coefficient? As of the proofs I've seen I strongly believe so but in my script they are suddenly noted with different indexes, e.g. k and a but might be an error??
@Danu if you have any finite abelian group $A$ that does not have a $\Bbb Z_{2^n}$ coefficient, then (I think) $2\cdot$ is an invertible homomorphism, and then on $A\oplus \Bbb Z_{2^n}$ the homomorphism should have kernel $\Bbb Z_2$. So it seems no other information about hte group can be gotten from that condition
I think you show if $p^n\mid\, |G|$ that then $G$ contains elements of order $p^n$, then you take all these and show they are a subgroup
then a subgroup of order $p^n$ has to be $ (Z_{p^n})^K$ and what you do is you divide this guy out to get a smaller group where you do the thing iteratively or something
I figured it wasn't going to be true, but given how dense primes are for small 2-digit odd integers, I thought going with 3-digit numbers was likely to be more succesfull
By the way about my question with curvature on normed spaces, there are really many geodesics on $\|\cdot\|_\infty$ which makes it super hard for me to look at geodesic flows
An algebraic structure is a set S which has one or more binary operations ∘1,∘2,…,∘n defined on all the elements of S×S, and is denoted (S,∘1,∘2,…,∘n).
if anyone wants a stab at this, the professor originally gave it as homework but decided it was too hard after he couldn't figure it out himself
"construct a surjective group homomorphism $\{e^{i2\pi q}|q \in \mathbb Q\} \to \{ 1, -1, i, -i \}$"
I tried playing around with treating points on a circle and picking the point that's "closest to", in some sense, the arc of between the two points, but it didnt have the homomorphism property
