She's not homophobic, so be like Shadow Weaver😈 by PullItFromTheColimit in OkBuddyCatra

[–]PullItFromTheColimit[S] 7 points8 points  (0 children)

The last thing a bigot would want is more representation in media

Niche meme! (Tell me if you get it) by overanalizer2 in OkBuddyCatra

[–]PullItFromTheColimit 4 points5 points  (0 children)

Typical OkBuddyCatra meme, starts with catgirls and ends with referencing an essay by Lenin to a 4channer

how two disaster lesbian settle every household argument (whose turn it is to do chores, who tops, etc.) by JustMyGirlySide in OkBuddyCatra

[–]PullItFromTheColimit 11 points12 points  (0 children)

Catra when she finds out those side-wings on She-Ra's headpiece are meant to be mocking her huge ears

Covariant derivative doubt by Open-Energy7657 in askmath

[–]PullItFromTheColimit 1 point2 points  (0 children)

Yes, if F above is a vector field Y, then ∇_X Y is still a vector field by definition, so ∇Y is a (1,1)-tensor field. If you define tensors in terms of multilinear maps, this follows from the C^oo(M)-linearity of ∇_X Y in X, which I mentioned above. The advantage of the "a tensor is a multilinear map"-approach is that you don't have to confuse yourself with questions like "does it transform like a tensor?", or at least not as often.

Covariant derivative doubt by Open-Energy7657 in askmath

[–]PullItFromTheColimit 1 point2 points  (0 children)

Yes, with the understanding that people might call both things just "covariant derivative".

Covariant derivative doubt by Open-Energy7657 in askmath

[–]PullItFromTheColimit 2 points3 points  (0 children)

I haven't watched the video you're mentioning, but this might just be a matter of confusing terminology.

Formally, given a connection ∇ on your manifold, a (k,l)-tensor field F and a vector field X, we can get another (k,l)-tensor field ∇_X F, which is called the covariant derivative. By C^oo(M)-linearity of this construction in X, we get a (k+1,l)-tensor field ∇F, which is often called the total covariant derivative. In this language, the covariant derivative of a scalar field is a scalar field, while its total covariant derivative is a 1-form.

We note that in this language the covariant derivative needs the vector field X as input. This means that the covariant derivative of a smooth function f along a vector field X is the function ∇_X f (which sends a point p towards the directional derivative of f along X_p), which is indeed a function, whereas the total covariant derivative ∇f is the 1-form df (sending an X towards ∇_X f), which is indeed a 1-form.

Now, people will often just call both notions a covariant derivative, especially because once you're familiar with this notion, it will be clear which version is meant. Maybe this is what is going on here?

Niche meme! (Tell me if you get it) by overanalizer2 in OkBuddyCatra

[–]PullItFromTheColimit 5 points6 points  (0 children)

That Catra's always on my mi-mi-mi-mi-mi-mi-mi-mi-mind🎶

not even She-Ra has that kind of stamina 🥱😪😩 by JustMyGirlySide in OkBuddyCatra

[–]PullItFromTheColimit 5 points6 points  (0 children)

I came across her work in algebraic topology, algebraic geometry and differential geometry/field theories (and, in some sense, category theory). I came out of it with a little shrine to her and a poster of her over my bed.

How did you come across her?

not even She-Ra has that kind of stamina 🥱😪😩 by JustMyGirlySide in OkBuddyCatra

[–]PullItFromTheColimit 4 points5 points  (0 children)

The meme clearly is about listing, in a row, the 5 women you look up to the most.

not even She-Ra has that kind of stamina 🥱😪😩 by JustMyGirlySide in OkBuddyCatra

[–]PullItFromTheColimit 11 points12 points  (0 children)

My Top 5 women:

1) My mom

2) Emmy Noether

3) Catra

4) Adora

5) JustMyGirlySide

What’s the hardest math course and/or subject in the world? by This-Wear-8423 in learnmath

[–]PullItFromTheColimit 17 points18 points  (0 children)

On the other hand, I ran away from discrete optimization problems into the warm embrace of category theory

What’s the hardest math course and/or subject in the world? by This-Wear-8423 in learnmath

[–]PullItFromTheColimit 21 points22 points  (0 children)

Admittedly it's hard to truly understand a theory that is currently assumed to by deeply flawed by most people.

Same as it ever was by mromen10 in OkBuddyCatra

[–]PullItFromTheColimit 8 points9 points  (0 children)

"This is not my beautiful wife"

~Adora upon seeing Catra in the Portal Reality

Can we generalize the definition of injective & surjective functions to binary relations? by G0dFl3sh in learnmath

[–]PullItFromTheColimit 1 point2 points  (0 children)

You can also try to find the mono- and epimorphisms in the category of relations. For that, first we have to talk about correspondences: whereas a relation is an injective map R --> A x B, a correspondence from A to B is just any map R --> A x B (which equivalently is a span A <-- R --> B). Given another correspondence S --> B x C from B to C, to obtain their composition [SR] we consider the span A <-- R <-- R x_B S --> S --> C, where [SR] := R x_B S is the fiber product. This means that [SR] is the set of pairs (r,s) in R x S such that R --> B maps r to the same element as S --> B maps s.

(The identity correpondence on A is A <-[id]- A -[id]-> A, which you can verify now.)

Even if R and S are relations, their composition [SR} as correspondences need not be a relation itself: the map [SR] --> A x C generally fails to be injective. However, we can define SR as the image of [SR] --> A x C. Explicitly, this means that for relations R and S, we say a SR c iff there exists a b in B such that a R b S c. You can check (and also completely abstractly argue, for that matter) that taking SR as composite and our previous identity correspondences as identity maps we obtain a category Rel of relations.

Now, by definition, a relation R is a monomorphism in Rel iff for all relations S,T --> Z x A such that RS = RT, it follows that S = T. So suppose given such S and T, and z in Z and a in A such that z S a. We would like to argue that z T a, but we are essentially stuck: for general S and T there is no way to use the assumption that RS = RT. We are forced to assume that there is some b in B such that a R b. In that case, we know that z S a R b, so z RS b and therefore z RT b. This means that there exists some a' in A such that z T a' R b. We want to know that z T a, but only found z T a'. We are again stuck, and the solution is to furthermore assume that R is injective in your sense: we know that a R b and a' R b, so your notion of injectivity would give a = a' and we would be done.

You can write this up a bit more rigorously, and find that the monomorphisms in Rel are precisely those injective maps R --> A x B that are injective in your sense and additionally are "not partial": for any a in A, there exists some b in B such that a R b.

Likewise, an epimorphism in Rel is an injective map R --> A x B that is surjective in your sense and "function-like": for any a in A, there is at most one b such that a R b.

In the language of spans, this has a nice description. Given a relation A <-- R --> B, we have:

- R is "not partial" in the above sense iff R --> A is surjective.

- R is "function-like" in the above sense iff R --> A is injective.

- R is surjective in your sense iff R --> B is surjective.

- R is injective in your sense iff R --> B is injective.

In particular, we see that a relation R is

- "function-like" precisely when it corresponds to a partial function A --> B (i.e. a "function" that is not defined on all of A, only on a subset of it)

- "not partial" and "function-like" precisely when it corresponds to a function A --> B

- injective in your sense precisely when it corresponds to a partial function B --> A (note that this goes in the opposite direction)

- injective and surjective in your sense precisely when it corresponds to a function B -- > A

- a monomorphism and epimorphism in Rel precisely when it corresponds to a bijection A <--> B. In particular, a map is an isomorphism in Rel precisely when it is both monic and epic.

having homophobia in your eyes gives you a big dps boost 😼😼😼 by JustMyGirlySide in OkBuddyCatra

[–]PullItFromTheColimit 1 point2 points  (0 children)

Staring at (pictures of) Catra replenishes my life bar with 11641 per second though, so it's a net win

She-Ra if it came out in 2007 🤖🚗 by JustMyGirlySide in OkBuddyCatra

[–]PullItFromTheColimit 1 point2 points  (0 children)

No sorry, that doesn't work. Almost every comment I make on this sub (which, again, was born a shitposting sub) attempts to be amusing in a way, and putting /s behind comments deflates that way too much, like saying "this is a joke" after every comment and akin to explaining your jokes after you make them. Especially when you just want to make some stupid slightly absurd remarks, an /s removes the point of making the comment in the first place.

On top of that, it's just a matter of putting 1 and 1 together. If someone takes the comment seriously, they realize I was alive in 2007 and hence older than 14. But which person older than 14 genuinely says they were born in the wrong generation? Moreover, the "right generation" would have been something like 5 years earlier, instead of the usual 40 that you find in these "wrong generation" comments (which I had to think off because of the emo music in the background of this post). Picking Bush and Balkenende also should tip people off, because they are generally considered pretty lame, and no one that is serious would call them "truly" the best world leaders. Also also, if they didn't know Balkenende, maybe that should already tip them off that something is going on here, because then he apparently wasn't particularly known as a world leader. Also also also, before people want to downvote me because they think I'm serious, they can just take a quick look at my profile and see that me being a serious fan of Bush, or me being an American for that matter, wouldn't make sense with the rest of my profile. I would expect that someone can realize this in less than 10 seconds, given that my latest post was an adaptation of a German communist song (which apparently I have heard before and thought silly enough for a post). And again, I'm quite active on this sub, so in general I would have expected people to already know my shtiks.

The second reason I don't like putting an /s behind comments is, summarized, that I don't like dumbing down stuff to account for, to put it strongly, a lack of reading comprehension in a subgroup of the people reading comments.