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Comments:
<0> He's been banned here several times. He just refuses to learn.
<1> how old is spx2?
<2> 4
<3> So I'm looking at the categorical definition of a group, and it seems to me that it uses Cayley's theorem implicitly.
<3> This doesn't help me, as I'm trying to explore means of categorifying Cayley's theorem.
<4> which definition?
<4> a 1-object groupoid?
<3> A one-object category where every morphism is an isomorphism
<0> A one object category where every morphism has an inverse?
<0> How is that using Cayley's?
<5> those two lines looked suprisingly similar
<0> Isn't that that every finite group of order n is a subgroup of S_n?
<4> (what is cayley's theorem anyway? that a group embeds into S_n?)
<3> Yeah.
<3> Because permutations are just isomorphisms from a set into itself.
<4> yeah, dont see why cayley comes into it
<4> but the morphisms in your group-as-category are not functions
<4> they are morphisms
<3> Well, if I forget it's a group, they're functions.
<4> no
<0> No, the morphisms are the elements of your group.
<0> The category has one object, call it 37.
<4> your category has one object *, the morphisms in the category are *-->* with one distinct morphism for each element in the group
<0> The elements of the group are the morphisms from 37 to 37.
<3> Ah, I read too much into that def.
<0> The object doesn't have elements.
<0> It's just some opaque object.
<3> Yeah, okay, that works.
<6> there's another ct definition of groups .. over a monoidal category
<4> i guess you mean groups internal to that monoidal category?
<0> So, if A and B are abelian groups, then Hom_Ab(A,B) is an End(B) module because if f in Hom(A,B) and g in End(B), gf is in Hom(A,B) plus checking a few axioms?
<4> should be
<0> This week's homework has been too easy.
<4> you dont really need to check the axioms, it's just composition in Cat
<7> |Steve|, Hom_Ab stands for Hom_Z?
<4> or maybe in "AbCat"
<8> huh, I can direct mbot to identify manually, but when I put the appropriate command into the rc file, it doesn't work.
<0> delta: Well, abelian groups are Z-modules.
<4> doesnt mbot just send the p***word as a server p***word?
<7> So what does mean Hom_Ab? :)
<4> maps in the category of abelian-categories
<8> Server p***word?
<0> Homomorphisms from A to B in the category abelian-groups.
<0> Well, that's what Hom_Ab(A,B) means.
<4> when you connect, you can identify by sending your nickserv p***word to the server
<0> Really?
<4> rather than having to send a privmsg yourself
<7> Never reached this notation then. I don't see why it shouldn't be Hom_Z :)
<0> I just use /ns identify p***word
<0> Why Z?
<8> what's the raw protocol for that?
<4> yeah, it all turns out to be the same thing
<7> |Steve|, because they're morphisms of Z-modules.
<4> i think it's just an argument to USER but i dont know
<0> But I'm thinking of them as abelian groups.
<0> I realize the two are one in the same.
<0> one and the same?
<0> Stupid English.
<2> it's P***
<2> not USER
<6> _llll_ : i guess you could say so
<8> okay, that works
<4> Hom_Ab(A,B) means "the maps from A to B in the category Ab"
<4> Hom_Z(A,B) means "maps in the category of Z-modules", so would be Hom_{Z-Mod}(A,B) in that notation
<0> I don't like how the -Mod is omitted in that case.
<0> But apparently that's common notation.
<4> becasue you have a category MOD whose objects are pairs (R,M) with M and R-Mod
<4> and R-Mod is a sub-category of MOD
<4> and the maps of R-modules are maps "over R"
<0> Don't I need to check that g(f+f')=gf+gf' and (g+g')f = gf+g'f to show that it's a module though?
<4> yes, but that;s obvious
<4> everything is essentially just compositon isnt it?
<0> Well, I need to show that the composition distributes over +.
<4> you've got End(B)=Hom(B,B), and the module structure is the map Hom(A,B)xHom(B,B)-->Hom(A,B) which si just composition
<0> But Ab is an additive category.
<4> so of course it is ***ociative, linear etc
<0> You mean Hom(B,B) x Hom(A,B), right?
<4> depends if you are doing left modules or right modules
<4> all the best people do right-modules :)
<0> Left.
<0> Heh.
<0> Hom(A,B) is a left End(B) module and a left End(A)^op module.
<4> an Ab-category is a category enriched over Ab, and so it comes with Hom(X,Y)xHome(Y,Z)-->Hom(X,Z) a map of categories
<0> (Which, I think, makes it a right End(A) module.)
<0> I don't know what enriched means here, but wikipedia used that term as well.
<4> how did you define Ab-category?
<0> I don't think we did.
<4> oh, well you wrote "additive category" above
<0> I think the additive category we defined is a pre-additive category according to wikipedia.
<0> If that's the same as Ab-category, he didn't mention it in cl***.
<4> a categoy is something with a load of objects, and sets Hom(A,B) of maps from A to B. and composition is a function Hom(A,B)xHom(B,C)-->Hom(A,C), plue "***ociativity" which is a condition on those composition functions
<4> and identitie are functions {*}-->Hom(A,A)
<0> Right.
<4> so now replace "set" with "object in C", and "function" with "map from C" and you have the notion of "category enriched over C"
<4> so if C=Ab, that's an additive category
<8> % 1
<9> Cale: 1
<4> if C=Set that's an ordinary cateory, if C=Cat that's a 2-category, etc
<0> Oh, okay.
<10> id_A
<4> so a 1-object Ab-category is an abelian group
<4> and a module is just an Ab-functor from that Ab-category to Ab, where Ab is considered enriched over itself
<4> (any Cartesian closed category is enriched over itself)
<4> or you want left modules, so you might want A^op-->Ab, but it depends how you are considereing your abelian group as a 1-object category
<11> Can someone help me integrate this expression: Int[(3x^2)/2 - 1/x^2] between -2 (bottom limit) and 3, I have done it myself and got 50/3 but my calculator gives me infinity
<0> It doesn't converge.
<11> This may be a load of crap (new to integration) but don't you just integrate it (got: 1/2x^3 - x^-1)
<0> % Integrate[3x^2/2-1/x^2,{x,-2,3}]
<9> |Steve|:
<9> 2
<9> -2 3 x
<9> Integrate::idiv: Integral of -x + ---- does not converge on {-2, 3}.
<9> 2
<9> Integrate[-x^(-2) + (3*x^2)/2, {x, -2, 3}]
<11> % Integrate[3x^2/2-1/x^2,x]
<0> EvilGuru: Not quite. Your function isn't continuous on (-2,3).
<0> Also, mbot just quit. Cale is playing with it.
<11> What do you mean by converge
<0> I mean that when considered as a Riemann sum, the sum does not converge.
<12> it's worse than that
<0> Catfive: Which "that"?
<11> aha
<11> it doesn't make any sense does it
<11> as if you plotted it between -2 and 3 when x = 0 you would get 1/0^2
<12> your function isn't even defined everywhere on that interval; even if it were, it isn't bounded. the definition of Riemann integrability requires both.
<0> Well, I said continuous because likely the form of the Riemann integral he learned required it.
<11> as the function is not defined for when x = 0
<0> Improper integrals don't require boundedness, though.
<0> I realize that admits a larger cl*** of functions to be integrated.
<13> Emmm, i want to read on what actually is happening with the argument of the function sin(x) inside it :)
<13> I want to write func in a programming language, anyone got i clue where i could read on this?
<13> a clue*
<0> What do you mean on the inside?
<14> steve, he means the definition of the function
<14> what the actual algorithm is
<14> to compute sin(X)
<0> Ah.
<13> I wanna know what happens to x in the function, i don't know what operation to do on it to return a number.
<13> Yeah !
<10> font: substitution.
<0> % TrigToExp[Sin[x]]
<9> |Steve|: (I/2)/E^(I*x) - (I/2)*E^(I*x)
<0> There you go.
<0> Just perform that complex exponentiation and you're all set.
<13> :))
<0> Cale: Is my using mbot killing it?
<8> no
<8> It's me
<0> Okay, good.
<8> Please ignore mbot for a while
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