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12:29 AM
So, I’ve got this puny little calculator on hand: ernst.mulder.com/calculators/sharp-elsimateel326s-index.html
And I just noticed a little quirk of it
Suppose I enter the ops 1,+,2,=
I get 3, of course
If I now do = repeatedly, I get 5,7,9,11,...
So it’s doing +2 over and over. That’s what I’d expect it to do, and that’s also what it does for division and subtraction
But: if I do 1,x,3 and then repeatedly hit =
I get 3,3,3,3...
So, for multiplication only, it interprets repeated multiplication as -from the left-
 
have you tried 2x3=====?
 
Yeah. 6,12,24,48...
Looking up the data sheet for it, the main application they have in mind is for taxes
So I wouldn’t be shocked if that’s somehow behind it
I’m tempted to make an SE question about it. Not sure what site tho
 
12:50 AM
confuzled
 
1:02 AM
Hi @loch
 
... a thousand dollars?
 
I doubt it.
So, Leaky, have you met loch in person?
 
yeah
 
Good.
Have you introduced yourself to Artin and said hi for me? :P
 
not yet
 
1:10 AM
I'd tell you other names, but you're less likely to want to talk to them.
 
what are the names?
 
Mattuck, Guillemin, Melrose, Mrowka
 
1:52 AM
hi @TedShifrin
 
2:06 AM
hi chat
 
trying to think of how to word something. I've got two version of an inequality, one as $A^2\leq B^2$ and the other as $|A|\leq |B|$. Obviously these are equivalent, but how would I describe the latter in relation to the former?
 
@Semiclassical take the square root of both sides
 
Yeah. My inclination was something like "Aside from an overall square root, Eq. (2) is equivalent to Eq. (1)"
 
how should I go about proving that a strictly increasing function is injective? I am not quite sure about it since the definition of a strictly increasing function negates the definition of an injective function.
 
2:15 AM
What?
 
@HarryBattersby how does it negate it?
 
I mean, strictly increasing means $x>y\implies f(x)>f(y)\implies f(x)\neq f(y)$.
 
Yes, so isn't it the negation of the definition of injectivity which suggests that there exists $x_1$ and $x_2$ such that if $f(x_1)$ = $f(x_2)$ ⟹ $x_1$ = $x_2$?
 
No. The negation of $f(x_1)=f(x_2)\implies x_1=x_2$ would be $f(x_1)\neq f(x_2)\implies x_1\neq x_2$.
the contrapositive of injectivity, however, would be $x_1\neq x_2\implies f(x_1)\neq f(x_2)$.
fixed
And contrapositive is equivalent to the original statement, so strictly increasing implies injective (which is what you're trying to show)
 
But the thing that confuses me is that since the function is strictly increasing, then how can two points be equal to each other and produce $x_1$ = $x_2$?
 
2:23 AM
$\lambda \ell$ looks like $\mathcal{M}$
4 mins ago, by Harry Battersby
Yes, so isn't it the negation of the definition of injectivity which suggests that there exists $x_1$ and $x_2$ such that if $f(x_1)$ = $f(x_2)$ ⟹ $x_1$ = $x_2$?
no it doesn't say "there exists"
injectivity means that whenever we have $x_1$ and $x_2$ such that $f(x_1) = f(x_2)$, then $x_1 = x_2$
 
I understand. I totally forgot about the contrapositive. So using the contrapositive we can basically show that since there doesn't exist $x_1$ and $x_2$ such that $x_1$ = $x_2$, therefore $f(x_1)$ won't be equal to $f(x_2)$
which is essentially the contrapositive of the definition.
But I'm still finding it hard to visualize how the function is injective. If the function is always increasing and doesn't have an instantaneous point where the slope is 0, then how can the definition of injectivity hold?
I understand that you can use the contrapositive, but I don't quite get the logic behind it..
 
injectivity can also be stated as: whenever we have $x_1$ and $x_2$ that are distinct, i.e. $x_1 \ne x_2$, then $f(x_1) \ne f(x_2)$
@HarryBattersby if it rains, then the grass is wet
if the grass isn't wet, then it isn't raining
(because if it were, the grass would be wet)
 
"Distinct inputs give distinct outputs"
 
oh that makes sense! That's quite an abstract idea to think about.
 
Hi
 
2:32 AM
By comparison, strictly increasing means: "If one input is bigger than another, then its output is also bigger than the other's output."
 
So would directly proving that a strictly increasing function is injective be possible? (Without using the contrapositive)
 
But if one input is bigger than another, then those inputs are distinct
and if one output is bigger than the other, then the outputs are distinct
So strictly increasing implies injective.
 
@HarryBattersby I mean that depends on what you consider to be a direct proof
can I use trichotomy?
 
What is the reason behind the coboundary sign convention in Bredon or May or Milnor&Stasheff?
 
i.e. for any $x_1$ and $x_2$ we have either $x_1 < x_2$ or $x_1 = x_2$ or $x_1 > x_2$
 
2:34 AM
@Semiclassical @LeakyNun Thanks a lot. That really cleared up a lot.
 
It's probably a good idea to write this out as an explicit proof, of course.
 
@Leaky I haven't really learned about trichotomy, so I'm not exactly sure how that would work
 
I just stated it
 
Namely, show that: If $f$ is strictly increasing and $x\neq y$, then $f(x)\neq f(y)$.
 
@LeakyNun so you basically divided it into different cases?
 
2:35 AM
three cases
 
Trichotomy just says: "Given two real numbers, one of three things has to be true: the first number is bigger than the second, the first number is equal to the second, or the first number is smaller than the second."
 
@TedShifrin I'll hand it over to you :P
 
In particular, trichotomy implies: If two numbers aren't equal, then one is bigger than the other.
 
Ahh that sounds interesting. So far I have only touched negations, contrapositive and direct proof (Some statement, P that implies some other statement, Q). So, I still have much to cover :p
 
Note that trichotomy isn't a property of logic (as negation/contrapositive/converse are)
it's a property of the real numbers
 
2:40 AM
That's really interesting indeed. Are there any specific circumstances under which we can use this strategy?
 
If you're dealing with real numbers and their orderings, it's an obvious thing to remember.
so if you've got real numbers $x,y$ which are distinct, then you know that $x>y$ or $y>x$.
 
I see. That actually sounds really interesting and useful. I will definitely check it out at a deeper level.
So, just out of curiosity, would proving that an increasing function is injective using a direct proof (P implies Q) would be possible in this case? Are there any mathematical problems that can only be solved using a certain way due to their complexity?
 
3:07 AM
51
Q: An easy example of a non-constructive proof without an obvious "fix"?

Valentin GolevI wanted to give an easy example of a non-constructive proof, or, more precisely, of a proof which states that an object exists, but gives no obvious recipe to create/find it. Euclid's proof of the infinitude of primes came to mind, however there is an obvious way to "fix" it: just try all the n...

@HarryBattersby ^
 
3:30 AM
@loch why isn't H^2(Gal(C/R),C*) = 0?
isn't C* divisible
 
3:58 AM
dunno
idk galois cohomology
 
just treat it as group cohomology
it's just H^2(Z/2Z, C*)
where the non-trivial element of Z/2Z acts on C* by conjugation
 
ah
then i can say idk group cohomology instead!

but at least i can confirm that C* is divisible
 
lol
hmm what happened to cyclic cohomology
$\widehat{H}^{n+2}(\Bbb Z/n\Bbb Z,M) = \widehat{H}^n(\Bbb Z/n\Bbb Z,M)$
and $\widehat{H}^0(\Bbb Z/2\Bbb Z,\Bbb C^\times)$ is the cokernel of $\Bbb C^\times/\langle z \div \overline{z} \rangle \to \Bbb R^\times$
$z \div \overline{z}$ has magnitude $1$
it looks like the circle
so it's the cokernel of $\Bbb R_{>0} \to \Bbb R^\times$
where $r \mapsto r \overline{r} = r^2$
aha so it is $\Bbb R^\times / \Bbb R_{>0} = \Bbb Z/2\Bbb Z$
and therefore $H^2(\operatorname{Gal}(\Bbb C/\Bbb R),\Bbb C^\times) = \Bbb Z/2\Bbb Z$
@MatheinBoulomenos hej
 
4:34 AM
Hello
 
hi
 
5:29 AM
oh wow it follows from L C F T that $H^2(\operatorname{Gal}(\Bbb C/\Bbb R),\Bbb C^\times) = \Bbb Z/2\Bbb Z$
 
5:50 AM
@LeakyNun LCFT in the Archimedean case is obvious
 
"obvious"
 
 
3 hours later…
8:26 AM
Morning y’all
 
We didn't die!
 
Well maybe you didn't
 
 
2 hours later…
10:28 AM
If I have a connected manifold $M$, are all manifolds of the form $M \setminus U$ for some open set $U$ diffeomorphic?
Do they need to beeeee... homogeneous?
 
No and you can easily come up with counterexamples
 
Can I
Hm
lemme think
 
You're literally just asking if every closed submanifold of a connected manifold is diffeomorphic, and connectedness is more or less irrelevant here
 
Ah I think I can see one
Double torus minus $S \times [0,1]$
 
Not sure I understand tbh, that's not an open set
 
10:33 AM
$(0,1)$ yes
 
I guess I was expecting you wanted the complement to be a manifold without boundary but if you allow boundary it's even easier to come by yeah
 
Boundaries are fine
Hm
 
So for this example the complement is a disjoint union of two punctured tori and you just need to come up with absolutely any open set with non-diffeo complement
 
Is this true if $U$ is a disk tho?
 
lol yes but that changes the question dramatically, and you should probably assume U is the interior of a closed disc
 
10:35 AM
Just trying to prove things relating to a connected sum of a manifold to itself
So it will involve a lot of disks
 
at which point you prove this by showing that any two discs oriented the same in a connected manifold are isotopic (this is a lemma of Palais I think but it's straightforward)
It's also true topologically but much harder
 
Fortunately all my manifolds are smooth manifolds
I am pretty sure that the connected sum of a manifold to itself is equivalent to the connected sum of that manifold to an $(n-1)$-sphere bundle over $S^1$
or something close to that
 
Language way too fancy. It's either the connected sum with $S^{n-1} \times S^1$ or with the $n$-dimensional Klein bottle, where you take $S^{n-1} \times I /(v, 0) \sim (r(v), 1)$, $r$ being some reflection
Depending on whether or not you do the gluing in an or-preserving or or-reversing fashion
References for the claim about discs being isotopic
6
A: smooth homotopy 4-balls with sphere boundary in dimension 4

Mike MillerYes, you can perform that ambient isotopy: any oriented embedding $i: B^n \to M^n$ is isotopic to any other. (This is a lemma proven independently by Cerf and Palais1, but the idea is quite clear: shrink the image of $i$ until it's contained in the chart, then take the limit that defines the deri...

 
I once discussed this with a math people who told me that things could get fancy due to the whole spheres that could be homeomorphic but not diffeomorphic
Not sure if this is correct or relevent
Or if it doesn't matter if I assume everything smooth
 
Yeah they're wrong
It's not that it's irrelevant they're just making mistakes because they don't know the definition of connected sum
 
10:46 AM
Well to be fair I don't use a typical connected sum
It's more of a general cut and paste procedure
 
Define your terms dude this is getting frustrating
 
Sorry
Take a manifold $M$, remove two open disks from it, quotient the resulting manifold with boundaries along some gluing function $h$
So $$M' = (M \setminus (S_1 \cup S_2)) / \sim_h $$
Not assuming orientation-preserving for $h$ though it is smooth at least
 
Yeah that's indeed not a connected sum and indeed you can't say anything better about this than "connected sum with a sphere bundle" like you did above
 
Alas
This is physics-related so I'm not 100% sure what the exotic smooth structure business would be like
Exotic smooth structures are rather poorly studied in general
 
The connected sum procedure is carried out as follows. Choose an oriented embedding $i_1, i_2$ of the disc into a manifold $M$, $M'$. Delete the interior of the images. Then the connected sum is what you get when you identify $i_1(x) = i_2(x)$ for points $x$ in the unit sphere.
I assume you mean they're poorly studied in physics which fine
 
10:54 AM
Hello. Is the submersion $\mathbb R^2\to \mathbb R$ given by $(x,y)\mapsto xe^y$ a fiber bundle? It seems the gradient flow is complete and gives parallel transport. On the other hand, the Palais-Smale condition is not satisfied, so maybe I'm missing something...
 
@MikeMiller mostly yes
There's some papers on the topic tho
unfortunate that our universe happens to be exactly the correct dimension for that to matter
 
For that to matter to someone pathological in a particular way, sure
 
heh
You never know!
 
What's the difference between a knot and a loop?
 
What is a knot and what is a loop?
 
11:02 AM
I'd say a knot is a simple closed curve
 
And you're looking to know what a loop is?
 
Drop the simple condition
 
@MikeMiller hey, could you help me with some confusion about complete connections?
 
11:17 AM
@Arrow Not right now sorry. I think your thing is a fiber bundle by direct calculation ie write down local trivializations
 
@MikeMiller (I also think so. My question is about the Reeb foliation $(x^2-1)e^y$.) If you happen to have a couple of minutes later, I'd love to bug you. Thanks anyway!
 
12:15 PM
Who can point me to good material on Matsubara summation?
Actually no, let's break it down further
So I have a summation that i'm representing as a sum of residues of a function
How does the contour deformation work precisely?
There is no pole at infinity, it's a zero instead
So if I have a sum of poles on the real axis
And two poles in the lower half plane
(say)
I'm saying my summation is over the real axis poles, can I deform the clockwise contours around these real axis poles to a contour enclosing the complete lower half plane only ?
Or option 2: do I need to also have a contour in the upper half plane?
The imaginary part of the weighting function is positive in the upper half plane
The imaginary part of the two lower half plane poles is negative after the transformation x -> x+iepsilon
 
 
2 hours later…
2:23 PM
The power set operator is injective, but not surjective
Any inverse of power set operator is not total
e.g. there are no sets such that its power set is the natural numbers
 
2:41 PM
@Secret to speak of "surjective" you must first define its codomain.
 
well, the problem here is the "codomain" is the proper class, V
I don't know how functions that maps to proper class works
The powerset operator can be described as mapping between one set $S$ to another of cardinality $2^{|S|}$ along with the subset properties this new set has to obey
 
yeah, I think that will work
so $\mathcal{P}$ is not surjective because e.g. no set can map by a power set to the naturals
and $\mathcal{P} : V \to V$, as a definable function on classes
 
3:03 PM
$\bigcup$ is a left inverse of $\mathcal P$
 
$\bigcup 16 = 15 \neq 4$?
 
16 isn't the power set of 4
 
ok nvm
 
Every commutative boolean monoid has an associated commutative boolean ring
The finite monoid I'm working with is the set of subintervals in $\Bbb{N}$ of $[0, |s|)$ where $s$ is a finite string.
You can also limit the length of the subintervals and get a subring of that.
I'm actually interested in subintervals $[i,j)$ only up to length $j - i = \lfloor |s|/2\rfloor$.
Since any larger interval cannot result in a compression of $s$.
 
3:53 PM
@Ultradark
 
4:15 PM
So, how would you find the automorphism group of the semidirect product $Q_8\rtimes Q_8$?
 
I wouldn't
 
(Where the homomorphic image from $Q_8$ to $\text{Aut}(Q_8)$ is the klein four group)
Also, that is fair, Leaky
 
4:40 PM
@Arrow I can't stay consistently but if you ask your question I might see it later
 
Dear @MikeMiller, here is my question(s).
 
5:20 PM
hello, if $M_n=\max(u_n, v_n)$ why $u_n^3+v_3\geq M_n^3$
 
5:33 PM
have you an idea ? because by definition $u_n\leq M_n$
 
Take $u=1,v=2$, then $M=\max(u,v)=2$, but $u^3+v=1+2=3\not\ge 8=2^3=M^3$.
 
$u_n^3+v_n^3\geq M_n^3$
@Thorgott
 
ah, that is indeed true as long as $u,v\ge0$, because $M^3$ is either $u^3$ or $v^3$
 
5:49 PM
@Ultradark you could probably grasp my post, since you know what a ring is
I haven't brought in any theory yet, but i'll most likely be using basic properties of rings
 
okay I'll take a look!
@ShineOnYouCrazyDiamond can you have a colored (morphisms) category theory diagram
 
Yes, I do in BananaCats
though the color is just a user preference
Though it's not ready for use yet
 
oh cool
 
6:04 PM
@Arrow I don't really have anything to say, sorry
 
@MikeMiller no problem. Thanks anyway!
 
6:23 PM
-1
Q: Associative $+$ operation for a certain multiplicative subsets of a finite, commutative, boolean monoid?

Shine On You Crazy DiamondLet $X$ be a finite, commutative, boolean monoid with a zero element $0$. Define the set $C$ of subsets of $X$ to be $C = \{ x \subset X : 0 \in x, $ and $ \forall a,b \in x, \ ab \in \{a, b, 0\}\}$. Define for $x, y \in C$: $$ x + y = \{0\}\cup (x \Delta y) \setminus \{u \in x \Delta y: \exis...

anyone good at associativity proofs?
 
6:49 PM
Can anybody explain the sub manifold bit?
 
7:38 PM
@JakeRose If you're trying to figure out what a manifold (in Euclidean space) is, you might want to take a look at my YouTube lectures (linked in my profile) 3500 day 40 and 3510 day 23. The parametric definition you copied above appears as one of three equivalent ways of understanding manifolds.
 
a manifold is just [insert abstract and convoluted definition]
 
How can i determine $\lim\limits_{t \rightarrow 0}{\frac{v_1^2 \sin(tv_2)}{t^3v_1^4+tv_2}}$ where $||(v_1, v_2)|| = 1$ ?
 
Taylor is your friend
 
In what way could I use Taylor here?
 
@LeakyNun the category of smooth manifolds embeds fully faithfully into the category locally ringed spaces over $\mathrm{Spec}(\Bbb R)$
 
7:44 PM
@MatheinBoulomenos can you classify all the subspaces $V$ of $\Bbb R^\Bbb N$ closed under $\delta$ satisfying the property that there is a linear map $f: V \to \Bbb R$ such that $f(v) = v(0) + f(\delta(v))$ for all $v$?
$\Bbb R^\infty$ is such a subspace
 
what is $\delta$?
 
and no such subspaces can contain $(1,1,1,\cdots)$
the shifting operator
 
no idea
 
ok thanks
 
@LeakyNun I guess you did not meant me, right? Uh, maybe L'Hospital works?
 
7:49 PM
@T_01 you would probably need to case on whether $v_2 = 0$
or $v_1$
 
and if none of both is zero?
 
@MatheinBoulomenos how about $C^k$ manifolds
@T_01 sure
 
@LeakyNun sure what?
 
maybe it's just $v_1^2$ afterall
 
@LeakyNun same argument should work
 
7:50 PM
unless $v_2=0$ in which case it is $0$
@MatheinBoulomenos how about $C^0$ manifolds
 
yeah, should work as well
 
and PL manifolds?
 
idk
the argument for $C^k$ $k \in [0,\infty]$ does not work for PL manifolds (because I don't think PL manifolds admit PL partitions of unity)
 
I'm not sure whether for my presentation tomorrow I should cite the Albert--Brauer--Hasse--Noether--theorem without proof or define $\Bbb C \otimes_\Bbb R V$ as $V + iV$
 
why do you need Albert-Brauer-Hasse-Noether?
 
7:53 PM
maybe I don't
doesn't it talk about Br(C/R)
or is that just LCFT
 
@LeakyNun sorry, i am a bit confused - so is L'Hospital the right way to go (as i can interprete enumator and denominator as functions in $t$) , or not?
 
@T_01 yes, but I like Taylor more than L'Hopital
 
that's just LCFT and in the Archimedean case, it's just obvious
 
how about the A-structure thing
it can be generalized
 
Albert-Brauer-Hasse-Noether is about global Brauer group
 
7:54 PM
@LeakyNun Could you explain, how i could use the taylor series to get the limes? I really do not now much about analysis, sorry
 
@LeakyNun sure include that if you want, you don't need to cite me, I'm sure that's well-known to people who thought about it
 
I'm saying, I want to generalize things, but I don't know if I should
 
you should
 
it hurts me to state things that I know can be generalized :P
but it hurts the audience to generalize
 
no pain no gain
 
7:57 PM
I'm failing to figure out which language does mathematics in Latin
 
I don't understand that sentence
 
Serbo-Croatian :o
Limes je jedan od osnovnih pojmova u matematičkoj analizi. == Limes niza == Neka je ( a n ) {\displaystyle (a_{n})} niz realnih ili kompleksnih brojeva. Reći ćemo da niz ( a n ) {\displaystyle (a_{n})} konvergira broju L (realan ili kompleksan broj) ako vrijedi ( ∀ ϵ > 0 ( ∃ n…
Croatian:
Limes je jedan od osnovnih pojmova u matematičkoj analizi. == Limes niza == Neka je ( a n ) {\displaystyle (a_{n})} niz realnih ili kompleksnih brojeva. Reći ćemo da niz ( a n ) {\displaystyle (a_{n})} konvergira broju L (realan ili kompleksan broj) ako vrijedi ( ∀ ϵ > 0 ( ∃ n...
anyway @T_01 taylor series gives you sin(tv2) = tv2 + t^3 v2^3/6 + ...
 
Limes, in mathematica, est quantitas ad quam alia quantitas adpropinquit. Definitio haec est: lim x → a f ( x ) = b {\displaystyle \lim _{x\rightarrow a}f(x)=b} significat ∀ ϵ > 0 ∃ δ > 0 ut | x − a | < δ ⇒...
 
well that's Latin
did Euler use the word limes
@MatheinBoulomenos $\Bbb C \otimes_\Bbb R V = V + \overline{V}$
so in general $L \otimes_K V = \sum_{\sigma \in G} \sigma(V)$?
and this is the content of Galois descent?
 
@LeakyNun it doesn't appear in his "introductio in analysin infinitorum"
 
8:11 PM
yeah he just summed a series by writing 1+1/1+1/1.2+1/1.2.3+&c.
 
@LeakyNun yes
well, that follows from the normal basis theorem
 
well Galois descent also needs normal basis theorem oder
 
no, I don't think you need it
 
what's the content of Galois descent?
 
the version for finite Galois extension is: if $L/K$ is a finite Galois extension with Galois group $G$, then if we consider the category $L^*[G]$-Mod which consists of $L$-vector spaces together with a semilinear $G$-action, then $L^*[G]$-Mod is equivalent to k-Mod via thefunctors $V \mapsto V^G$ and $W \mapsto L\otimes_K W$
that's actually an equivalence of symmetric monoidal categories if we equip $L^*[G]$-Mod with the tensor product over $L$ and $K$-Mod with the tensor product over $K$
thus by abstract nonsense, you also get an equivalence of the corresponding categories of monoid objects and commutative monoid objects
 
8:17 PM
@Thorgott thank you
 
which are $K$-algebras and $L$-algebras with a semilinear $G$-action by ring automorphisms (respectively the commutative ones for commutative monoid objects)
and then continuing by abstract nonsense, you get a category equivalence of the corresponding cogroup objects in commutative $K$-algebras which are dual to affine algebraic groups
there's also some Galois descent for projective varieties, but that's harder
 
I'm still trying to prove $L \otimes_K V = \sum \sigma(V)$ lol
 
wlog $V=K$, then apply the normal basis theorem
 
no $V$ is an $L$-module and $=$ is an L-linear isomorphism
so ok wlog $V=L$
 
oh sorry
yeah that's what I wanted
 
8:20 PM
$L \otimes_K L$
 
you don't even need the normal basis theorem
primitive element is enough
 
$L \otimes_K L = L \otimes_K K[x]/(f) = L[x]/(f)$
 
yeah
$f$ splits over $L$, Chinese remainder theorem, Galois group acts transitively on the roots
 
where's the $\sigma$
 
Let $\alpha$ be a root of $f$, then $\prod(x-\sigma(\alpha))=f$
 
8:24 PM
wait
aha
no we have too little structure
I want $V$ to be a $L[H]$-module
where $H$ is an arbitrary group
 
you probably want to replace the left factor in the tensor product by $K[x]/(f)$
 
now $V$ and $\sigma(V)$ are not isomorphic
I thought it would work by naturality
afterall a $L[H]$-module is a functor from $H$ to $L$-Mod
 
why should it not work out?
 
well i can't even tell the difference between $L$ and $\sigma(L)$
ok so you say we replace the left factor instead of the right factor
 
you have a $\mathrm{Gal}(L/K)$ action on $L$ (the left factor)
you can actually chase the Chinese remainder theorem isomorphism to see that is just permutes the factors
 
8:31 PM
$L \otimes_K V = K[x]/(f) \otimes_K V = \bigoplus K[x]/(x-\sigma(\alpha)) \otimes_K V$
 
according to the action on the roots of $f$
 
ok let's chase the CRT
$K[x]/(f) \to \bigoplus K[x]/(x-\sigma(a))$
we send $p$ to $(p(\sigma(a)))_\sigma$
$L \to K[x]/(f)$ with $\alpha \mapsto x$
$L \to K[x]/(f) \to \bigoplus_\sigma K[x]/(x-\sigma(a))$
$\alpha \mapsto x \mapsto (\sigma(\alpha))_\sigma$
wait this is nonsense lemme start again
 
you want $L[x]/(f)$
 
$L \otimes_K L \to K[x]/(f) \otimes_K L \to L[x]/(f) \to \bigoplus_\sigma L$
$\alpha \otimes 1 \mapsto x \otimes 1 \mapsto x \mapsto (\sigma(\alpha))_\sigma$
suppose $p(\alpha) = \sigma(\alpha)$ for some $p \in K[x]$
then $\sigma(\alpha) \otimes 1 \mapsto p \otimes 1 \mapsto p \mapsto (p(\tau(\alpha)))_\tau = \tau(p(\alpha)) = \tau(\sigma(\alpha))$??
the composition is reversed
@MatheinBoulomenos I'm confused
 
8:46 PM
let $f : V \to \Bbb C \otimes_\Bbb R V$ be given by $f(v) := i \otimes v - 1 \otimes iv$
then $f(iv) = i \otimes iv + 1 \otimes v = i(1 \otimes iv - i \otimes v) = -if(v)$
so $f(zv) = \overline{z} f(v)$
@loch help
$g(v) := i \otimes v + 1 \otimes iv$
$g(iv) = i \otimes iv - 1 \otimes v = i(i \otimes v + 1 \otimes iv) = ig(v)$
$g(zv) = zg(v)$
 
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