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4:06 PM
how do I add a vector to a scalar like this : (2,3,2) + 4
 
cfp
4:18 PM
Is there any procedure for moving questions from math.se to mathoverflow? For example, this question math.stackexchange.com/questions/2763434/… has received zero answers, despite having had a bounty and multiple up votes. I wonder then if mathoverflow might be a better place for it.
 
can I add a vector to a scalar
 
@TheMaskedRebel No, you can't.
 
@TheMaskedRebel No you cannot
 
@Fargle @ÍgjøgnumMeg why
 
For the same reason that you can't compute $2 + \text{fish}$
3
 
4:23 PM
Do you want the result of the addition to be a vector or a scalar?
 
Because addition is only defined between pairs of vectors or pairs of scalars. Trying to add a vector to a scalar is nonsense, because + doesn't have any meaning between a vector and a scalar.
 
^a finer explanation
 
2+fish is the best explanation
 
@AkivaWeinberger A vector
 
4:24 PM
@Fargle does this have a meaning $(2, 3, 4) + 1$ = $(2 + 1, 3 + 1,4 + 1) then
 
Then you're doing $(2, 3, 4) + (1, 1, 1)$
 
@ÍgjøgnumMeg I see
 
2 + fish = I'm hungry
I really am
 
Yeah. You could just make up an addition law between vectors and scalars, but for it to be consistent, it would really just boil down to adding a multiple of some vector, and vector addition already covers that. There's no reason to add more structure.
@Balarka eet food
 
@TheMaskedRebel Something that that definition lacks is "basis independence"
Suppose I have two arrows in 3-space. You don't know how my coordinate system is set up; you don't know in what direction the x-axis is or in what direction the y-axis is. But you can compute the vector sum anyway, using the parallelogram rule.
 
4:36 PM
@Fargle I will do that
for convenience sake, of course
 
On the other hand, say I tell you to compute $(0,0,0)+1$. You have no way of figuring out what $(1,1,1)$ is without knowing anything about the coordinate system.
 
If I'm letting $m : G \times G \to G$ be the multiplication in a group, does showing that $G$ is a topological group amount to showing that $m^{-1}(U) \subset G \times G$ is open for all open sets $U \subset G$ (and for $i : G \to G$ by $g \mapsto g^{-1}$)? Is this a particularly practical way of doing things or is there some more workable trick that is more useful?
 
@ÍgjøgnumMeg depends on context, as always
 
4:40 PM
sometimes you have a subbase at the identity
what a coincidence, i'm working on topological groups in lean
 
I'm just learning a minimal amount about topological groups/rings/fields/<insert structure> to read some stuff about local fields
oh
what a coincidence lol
a bit beyond me tho :P
 
@ÍgjøgnumMeg Yeah, you need $m$ and $i$ continuous
Hm, is there any case where multiplication is continuous but inversion isn't?
 
Does anyone here know anything about getting the collision normal of two aabbs
 
I don't even know what an aabbs is... is it anything like the Swedish rock band ABBA, only in correct alphabetical order and made lowercase?
 
is ABBA rock?
 
4:45 PM
In mathematics, a paratopological group is a topological semigroup that is algebraically a group. In other words, it is a group G with a topology such that the group's product operation is a continuous function from G × G to G. This differs from the definition of a topological group in that the group inverse is not required to be continuous. As with topological groups, some authors require the topology to be Hausdorff. Compact paratopological groups are automatically topological groups. == References... ==
 
@ÍgjøgnumMeg Pop, maybe? Pop-rock?
 
4
A: A paratopological group $T_1$ that is not $T_2$?

Henno BrandsmaIn section 2.1 of this paper there is an explicit example: take $G= (\mathbb{Z},+)$ as the group, and a basic neighbourhood of $0$ looks like $U_n(0) = \{0\} \cup \{k \in \mathbb{Z}: k \ge n\}$, $ n \in \mathbb{N}$. At $p \in \mathbb{Z}$ we of course get the shifted neighbourhoods $U_n(p) = \{p\...

example here @AkivaWeinberger
 
"Euro-shite" perhaps
tho I like it
hahah
 
"Euro-shite?!"
Kids these days...
 
what an irony that the EU banning memes has become a meme
 
4:47 PM
:(
 
haha I mean
in the context of Eurovision
because everything from Eurovision is automatically awful
even if it's good
 
@ÍgjøgnumMeg I also did local fields in the same repo
a bit of them
 
To be fair, I suppose, there are very few Eurovision acts that have gone on to do much of anything of note.
So perhaps "Euro-shite" is appropriate. ;)
You just have to update your priors in light of ABBA
 
@Leaky Nice, I just wrote a big report on like.. classical ANT and now that that's finished I'd like to look at local fields which I avoided due to having no topology
@Xander Conchita Wurst was on every Austrian billboard for about a year after Austria won Eurovision
that's noteworthy!
 
4:51 PM
In geometry, the minimum or smallest bounding or enclosing box for a point set (S) in N dimensions is the box with the smallest measure (area, volume, or hypervolume in higher dimensions) within which all the points lie. When other kinds of measure are used, the minimum box is usually called accordingly, e.g., "minimum-perimeter bounding box". The minimum bounding box of a point set is the same as the minimum bounding box of its convex hull, a fact which may be used heuristically to speed up computation. The term "box"/"hyperrectangle" comes from its usage in the Cartesian coordinate system, where...
 
If M is an additive commutative monoid, then Spec(R[M]) is an affine group R-scheme for any commutative ring R
what should I call Spec(R[M])?
 
@ÍgjøgnumMeg I did say "very few," not "none"
 
@XanderHenderson that is what an aabb(see wiki post) is
 
Any excuse to advertise the bearded lady
 
I'm gonna leave. I've run out of questions :P
 
4:56 PM
sorry, M is an additive abelian group
 
Hi @Parth
 
@BalarkaSen hi
 
o/
 
@BalarkaSen you dont do alg geom?
 
5:00 PM
Go away bot
Begone bot
 
:(
 
@LeakyNun I do not
 
hi @loch
 
Hi @LeakyNun
Look up toric varieties
 
y tho
 
5:03 PM
If G is an additive abelian group, then Spec(R[G]) is an affine group R-scheme for any commutative ring R
 
what is up with tropical geometry
 
what should I call Spec(R[G])?
 
@LeakyNun Wait, are you sure? If Spec A is an affine group scheme, doesn't A have a coproduct?
 
R[G] tensor R[G] = R[GxG]
@BalarkaSen I just proved it in Lean B)
 
A --> A otimes A translates to Spec A x Spec A --> Spec A
 
5:05 PM
instance group_ring.cogroup (M : Type v) [add_comm_group M] :
  cogroup R (monoid_ring R M)
"if M is an add_comm_group, then monoid_ring R M is a cogroup over R"
 
@AkivaWeinberger please
the complement is ${(x,y),x2+y2>1}∩{(x,y);|x|>1 or y>0}$
or ${(x,y),x2+y2>1}∩{(x,y);|x|>1 and y>0}$ please?
 
@BalarkaSen to construct R[G] -> R[GxG], via G -> GxG, via diagonal
 
Ahh
Cool. I like that.
 
algebra is cool isn't it
B)
 
Only when there is geometry
 
5:07 PM
sure
i find that i get way more satisfaction typing up torus than typing up weil group
just look at the definition of cogroup
class cogroup :=
(comul : alg_hom A (tensor_a R A A))
(comul_assoc : (alg_hom.tensor_assoc R A A A).comp
    ((tensor_a.map R _ _ _ _ comul (alg_hom.id A)).comp comul)
  = (tensor_a.map R _ _ _ _ (alg_hom.id A) comul).comp comul)
(coone : alg_hom A R)
(comul_coone : (alg_hom.tensor_ring _ _).comp
    ((tensor_a.map R _ _ _ _ (alg_hom.id A) coone).comp comul)
  = alg_hom.id A)
(coone_comul : (alg_hom.ring_tensor _ _).comp
    ((tensor_a.map R _ _ _ _ coone (alg_hom.id A)).comp comul)
  = alg_hom.id A)
comul and coone and coinv :P
they just make me happy
 
oh ginore what i said about toric varieties
but what you said can be found here mathoverflow.net/questions/122394/…
 
who calls it spectrum
 
pretty sure that's what spec stands for lol
 
still
 
In abstract algebra and algebraic geometry, the spectrum of a commutative ring R, denoted by Spec ⁡ ( R ) {\displaystyle \operatorname {Spec} (R)} , is the set of all prime ideals of R. It is commonly augmented with the Zariski topology and with a structure sheaf, turning it into a locally ringed space. A locally ringed space of this form is called an affine scheme. == Zariski topology == For any ideal I of R, define V I {\...
oh it seems like this did not link to wha ti wanted it to link to
but anyway there's some motivation from linear algebra/functional analysis
im not sure if this is the historic reason why it's called spectrum though
 
5:21 PM
oh, that spectrum
 
@LeakyNun please the complement of $$\{(x,y):x^2+y^2\le1\}\cup\\\{(x,y):-1\le x\le1\text{ and }y\le0\}$$
is
${(x,y),x2+y2>1}∩{(x,y);|x|>1 or y>0}$
or ${(x,y),x2+y2>1}∩{(x,y);|x|>1 and y>0}$
please
 
@LeakyNun if you want a group scheme such that the "group scheme multiplication" reflects the multiplication in G, you should take the dual of R[G], not R[G] itself, then G doesnt need to be abelian either
 
dual?
 
The dual as a module, i.e. Hom_R(R[G],R)
 
hmm
 
5:28 PM
Because R[G] is a Hopf algebra, this will be a Hopf algebra as well
 
@PolineSandra Have you tried drawing the two sets?
 
so if $G$ and $H$ are groups and $f:G\to H$ preserve multiplication, then $f$ preserves identity
 
But basically the roles of comultiplication and multiplication get reversed
 
the argument is that $f(1) = f(1) 1= f(1) f(1) f(1)^{-1}= f(1) f(1) f(1)^{-1} = f(1) f(1)^{-1} = 1$
but good luck translating this argument to group objects
you used associativity, inverse, identity, basically everything
after that, I need to translate it to cogroup objects
 
The comultiplication induced from the diagonal G -> GxG doesnt really carry information about G, you have the diagonal map for every set
The multiplication in R[G] has information about multplication in G and is more interesting
 
5:31 PM
mul_one, mul_inv, mul_assoc, map_mul, mul_inv
so we used 5 commutative diagrams to prove that f preserves identity
what the hell
 
Und because R[G] is cocommutative, the dual is commutative
 
is there no shorter way @MatheinBoulomenos
und :P
 
@LeakyNun Yoneda lemma
 
was?
 
It's enough to do it for groups
 
5:32 PM
well
let's get our hands dirtier
 
"Yo, need a lemma?"
4
 
@Fargle smacks Fargle
 
I am immune to all smacks but @Ted smacks
check my stats
 
@MatheinBoulomenos but seriously, 5 godforsaken diagrams to prove f(1)=1
 
4
Q: Equivalence of definitions of "group object" using the Yoneda lemma

BenBarrowsI have just started learning category theory and I am trying to get an understanding of how to think about the Yoneda lemma. Obvious applications are clear to me (Yoneda embedding is full and faithful, for example), but I want to understand how one uses it in practical mathematics. In particular...

 
5:36 PM
@Akiva do you prefer wine or burger?
 
Burger
I actually hate wine
 
lol
 
@LeakyNun you ask for a shorter way and want to get your hands dirtier at the same time
 
shorter as in fewer steps
i.e. less than 5
 
"If the covariance function is isotropic it is invariant to all rigid motion". What does rigid motion actually mean here?
 
5:43 PM
Is "y shares at least one digit with x" a symmetric & reflexive relation (but not transitive) ?
nvm yes it is
 
@MatheinBoulomenos for god sake
16 mins ago, by Leaky Nun
the argument is that $f(1) = f(1) 1= f(1) f(1) f(1)^{-1}= f(1) f(1) f(1)^{-1} = f(1) f(1)^{-1} = 1$
the second step $f(1) 1 = f(1) f(1) f(1)^{-1}$ means applying product to a whole forsaken diagram
so is the fourth step $f(1) f(1) f(1)^{-1} = f(1) f(1)^{-1}$
 
Let $X = C[0, 1]$ be the space of all real-valued continuous functions on $[0, 1]$. Let me know a Cauchy sequence from $X$ that does not converge in $(X,d)$ where $d(f,g)=\int_0^1|f(x)-g(x)|\,dx$.
 
and then I skipped a step
 
@LeakyNun
 
Zee
6:05 PM
Algebraic topology is dead
 
@AkivaWeinberger hello
 
6:16 PM
@Zee just because it has become a subfield of category theory does not mean that it is dead
 
@PolineSandra Isn't a basis element a V-shaped set?
 
@MatheinBoulomenos, how to find last two digits of $9^{9^9}$? I can find last two digits of something like $9^{105}$, but not this
 
@PolineSandra
Here's an example of a basis set completely contained in the complement
 
@Silent $\varphi(100)=40$, so for the exponent, only the residue class mod $40$ matters, so compute $9^9 \pmod{40}$ first and then 9 to the power of that result mod 100
 
9^2 = 1
 
6:29 PM
I have seen the following:

Let $G\subset \mathbb{R}^n$ be open, bounded and $f:\overline{G}\rightarrow \mathbb{R}^n$ a continuous and open map. Then $\|f\|$ gets its maximum on the bounder of $G$.

Why does this holds?
 
Zee
@MatheinBoulomenos that’s exactly what it means
Catagory theory is a useful tool but a shallow one
 
@AkivaWeinberger what is it's name please
 
6:46 PM
@Hi all, I have here Vinberg and Fulton&Harris (on rep. theory) which one I should read first?
 
I think that particular V-shape was $D_{2,2.4,-1}$
 
what do you think about D_{a,1,1} with a>1
 
Those'll also be contained in the complement
 
why
we just take the part supperior
or with a=5
or a=10
@AkivaWeinberger
 
The upper part?
I am confused
 
7:00 PM
like your pic we take y>ax+1 and y>-ax+1
 
@MatheinBoulomenos group hom preserves 1
@Daminark
 
@MatheinBoulomenos i don't think it's fair to say algebraic topology is a subfield of category theory lol (and of course algebraic topology is not dead)
 
@AkivaWeinberger
 
@LeakyNun Gross.
 
@Fargle yes, it's big
hi @loch
 
@LeakyNun hi
 
@loch have you read the pic above?
 
Zee
@loch algebraic topology is just the study of (infinite , 1) categories
 
Is someone here good at number theory in gcd problems?
 
Zee
Does anybody have a good recipe for beef testicles ?
 
7:14 PM
I just want to be sure that gcd(3k-1, 14k+1)=2 is not possible for k integer.
 
gcd(3k-1,14k+1) = gcd(3k-1, 2k+5) = gcd(k-6, 2k+5) = gcd(k-6, 17)
which divides 17
 
@valer Can 2 divide 14k + 1?
 
oh right lmao
 
@LeakyNun no
 
@LeakyNun i am exactly in that point gcd(k-6,17) is never 2 correct?
 
7:17 PM
correct
 
Thank you very much @LeakyNun
 
@Zee im not sure if people would agree with that - eg im not sure if someone cares about homotopy groups of spheres would consider their field as only about infinite categories
 
@loch I think Zee is trolling, so I was trolling back
@LeakyNun glorious
 
@MatheinBoulomenos :D
@MatheinBoulomenos do you have a definition of separable extension that avoids talking about the elements or mentioning cardinality?
 
@Zee No idea about beef, but mutton ones are usually smoked (at least in Iceland).
 
7:30 PM
@MatheinBoulomenos cardinality: $K/F$ is separable iff the number of embeddings $K \to \overline F$ equals $[K:F]$
is that right?
 
7:42 PM
@MatheinBoulomenos If $T$ is an affine group $F$-scheme such that $T \times_F \overline F = GL(n,\overline F)$, then is it true that $T \times_F F^{sep} = GL(n, F^{sep})$?
 
@LeakyNun I'm 90% sure this works, but I need to check again: $K/F$ is separable iff for every reduced commutative $F$-algebra $A$, $K \otimes_F A$ is reduced
 
eh... do you have a simpler criterion?
 
you can always take $A$ an algebraic closure
 
what is reduced?
 
$K/F$ is separable iff $K \otimes_F \overline{F}$ is reduced
 
7:45 PM
what is reduced?
 
oh, no nilpotents
expect 0
 
how the hell can a field tensor a field over a field have nilpotent
what is the nilpotent element in $\Bbb F_2(\sqrt X) \otimes_{\Bbb F_2(X)} \overline{\Bbb F_2(X)}$?
 
$\Bbb{F}_2(\sqrt{X}) \cong \Bbb{F}_2(X)[T]/(T^2-X)$
 
@MatheinBoulomenos oh and you should know what I mean by simple criterion: basically 0% usability and 100% simplicity to state
we don't need to verify that anything is separable
we just need to state it
oh and would you have a simple criterion (in the above sense) for a Galois extension
 
Another question: Do other algebraic structures (like monoids or rings and so on) also have a representation theory? And if whats the peculiarity with groups there?
 
7:49 PM
I don't really know what kind of criterion you're looking for
 
@MatheinBoulomenos what you just stated is great
I'm just wondering if you have any more stored in your brain
 
@LeakyNun $K/F$ is separable iff $\Omega^1_{K/F} = 0$ (this is the module of Kähler differentials)
 
that isn't really simple to state...
 
at least that works for finite extension, not sure if it works for infinite ones
 
@MatheinBoulomenos so $T-\sqrt X$ is nilpotent?
 
7:54 PM
@LeakyNun yes
 
9 mins ago, by MatheinBoulomenos
$K/F$ is separable iff $K \otimes_F \overline{F}$ is reduced
might you have any references?
 
uhhhm
I think this is in Bourbaki - Algebra, but I don't know which chapter
@Rudi_Birnbaum there is representation theory for Lie algebras, algebras over a field and some objects which aren't even algebraic structures like quivers (basically graphs)
and then there's representation theory for groups with additional strucure, like Lie groups, topological groups or algebraic groups
 
@MatheinBoulomenos: What is a Lie Algebra, & any take on my question about Vinberg vs Fulton&Harris?
 

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