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12:00 AM
(A hint here is that you'll need every vertex to have even valence.)
Another hint is that you can have multiple edges between any two nodes.
 
@D.ZackGarza you are right: we have more than two vertex of odd valence (the $v_1,v_2,v_3,v_4$)
 
Whoops, yeah, correction to my hint: you need zero or exactly two odd valence vertices.
 
It has Euler path!! We start from $v_4$: $v_4\to e_4\to v_2\to e_1\to v_1\to e_2\to v_2\to e_3\to v_2\to e_7\to v_3\to e_6\to v_2\to e_5\to v_4$
Well, that's probably an Euler cycle...
 
Unfortunately, this re-introduces Hamiltonian paths as well -- just travel around the outer edge.
 
Yeah... I'm almost done; it seems too difficult for me!
 
12:06 AM
The key is thinking about what proved the previous graph didn't have a Hamiltonian path - at some point, you get "stuck" at a vertex.
 
@D.ZackGarza yes, that "stuck vertex" is a vertex such that if we remove it, we disconnect the graph. I understand this, but I can't figure out how to meet the 2 conditions simultaneously
 
Think about something like this: you can traverse every edge by starting at the bottom, going up, looping the sides, then going up again.
But can you visit every vertex without backtracking?
No matter what choice you take in the middle, you get stuck at a dead-end.
 
This has Euler path: $v_5\to e_6\to v_1\to e_1\to v_2\to e_2\to v_1\to e_5\to v_4\to e_4\to v_1\to e_3\to v_3$: we passed through all edges just once
@D.ZackGarza this explains why this graph hasn't Hamiltonian path
 
Right - there is at least 1 Euler path, so the graph is Eulerian. But the graph isn't Hamiltonian, which requires some proof.
But instead of checking all possible paths, you can reduce to checking two cases: what happens when a potential path passes through $v_1$?
Case 1, you go to $v_2$ or $v_4$. But now your path must end, otherwise you are forced to take the remaining edge back to revisit $v_1$.
But if your path ends at $v_2$, you must have visited $v_3$, for example. But once you enter $v_3$, you can't leave, so your path can only have started at (say) $v_3$.
But now your path can not visit $v_5$, a contradiction. But the same argument goes through if I'd chosen $v_5$ instead of $v_3$ in that last sentence.
And the same argument again goes through if I replace $v_2$ by $v_4$ in the sentence before that.
 
@D.ZackGarza if we don't start at $v_1$, we need at some point to pass through $v_1$. When we have to visit another vertex we must pass through $v_1$, repeating $v_1$. The other case is if we start at $v_1$: then the same occurs, since if we go to any other vertex, to go back, we need to pass through $v_1$, repeating this vertex. Is this correct?
Hence, this graph does not have Hamiltonian graph since we must pass through $v_1$ more than once
 
12:23 AM
For case 1, if you don't start at $v_1$, then yes I agree -- but you'd want to say why you're forced to revisit $v_1$. For a slick example, you can note that removing $v_1$ disconnects the graph completely.
You have to pass through $v_1$ as your first move, but then you have 3 more vertices to visit, and the only way to get there is through $v_1$.
 
@D.ZackGarza yes, that's an excellent example!!
 
For case 2, yep, very similar. If you start at $v_1$, whatever choice you make, you've only visited 1 vertex and you need 2 more. But every path to them goes through $v_1$.
 
Can you help me about finding a graph that has Hamiltonian path but not Euler path?
 
Is an open subset of a quasi-compact space necessarily quasi-compact aswell (with the subspace topology)? I thought it was true originally, but then couldn't write down a proof yet (although I proved that closed subsets are quasi-compact along the way)
 
@manooooh So the idea is to cook up a graph where you can visit each vertex once, but are forced to re-use an edge.
 
12:31 AM
@D.ZackGarza exactly
 
Oh wait, an open subset of a hausdorff space is hausdorff, so I can just use euclidean space + Heine Borel to give a counterexample (meaning the failure of compactness of an open bounded subset of euclidean space is due to lack of quasi-compactness, and not hausdorffness)
 
@manooooh That one's slightly trickier. Try something like two closed cycles, then connect them up with one single edge.
@tigre Probably false, because [0, 1] is compact (and thus quasicompact), but (0,1) is neither.
 
@D.ZackGarza Yeah that's what I realised in the above paragraph
 
$(0,1)$ isn't quasi-compact, giving a counterexample. But are you sure you're right when you say it's not Hausdorff? Or you meant neither to compactness and quasicompactness, in which case I obviously agree
 
12:41 AM
@D.ZackGarza ^^^^^^ I think I followed your steps but came up with no edge forcing
 
@manooooh So it's worth checking - is there a Hamiltonian path? I think so, right, because you could just go from B to D. Now try to make a path that uses every edge.
 
@D.ZackGarza it has Hamiltonian path: $B\to1\to A\to3\to C\to4\to D$
I am not sure if I understand you correctly
 
$B\stackrel{1}{\to}A\stackrel{3}{\to}C\stackrel{4}{\to}D$
 
@tigre Right, just referring to the two versions of compactness.
(Hausdorff restricts to subsets just fine I think, you only have to worry about quotients.)
 
@tigre it is the same as I wrote! But this graph has Euler path
 
12:47 AM
$A\stackrel{1}{\to}B\stackrel{2}{\to}A\stackrel{3}{\to}C\stackrel{4}{\to}D\stackrel{5}{\to}C$
I was just suggesting a better way to type set
 
@manooooh So you'll want to modify this graph a little - you want to make it so there's no way to use every edge exactly once.
 
Oh, thank you!
 
So what about deleting edges $2, 4$ and making a new edge from $B\to D$?
It's almost immediate that this will just be a cycle, so that won't quite work.
So a further modification would be putting a new vertex $E$ in the middle, with 1 edge to everything else.
(The idea here is that you somehow want to force every choice of edges to "strand" you somewhere)
 
This has Hamiltonian path: $A\stackrel{1}{\to}B\stackrel{2}{\to}D\stackrel{3}{\to}C\stackrel{7}{\to}E$
I am not able to prove that it does not have Euler path
 
1:07 AM
1
Q: Is there a name for this mapping? $(x,\phi(x)) \mapsto \big(\frac{1}{\phi(x)},\frac{1}{x}\big)$

UltradarkIs there a name for this mapping? Does it appear in the literature, and where can I read about it more? How can I understand it best? $$ (x,\phi(x)) \mapsto \big(\frac{1}{\phi(x)},\frac{1}{x}\big) $$ If $f(x)=e^x$ and $g(x)=\frac{1}{\ln(x)}$ then let $\phi(x)=f(g(x)).$ I think it is some sort ...

any improvements I can make?
is the map an automorphism?
 
1:35 AM
@Ultradark Educate me. Is a map another name for a function? I read somewhere that there are 300 names for the concept of a function in mathematics.
 
@MatsGranvik yeah I'm mapping points to different points
 
@manooooh Just use the standard result: A eulerian path exists in a connected graph if and only if the number of odd degree vectices is zero or two.
 
Apparently: ${F_k}^2\mid F_m\implies F_k\mid m$
I have no idea why
Any ideas?
 
Is $F_k^2$ denoting $\Bbb F_{p^r}\times\Bbb F_{p^r}$ for $k=p^r$ some prime power?
 
$F_n$ is the $n$th Fibonacci number
 
1:47 AM
Oh
 
$F_0=0$, $F_1=1$
so $(F_n)_{n=1\dots6}=(1,1,2,3,5,8)$
 
2:22 AM
@AkivaWeinberger Sounds ripe for a proof by induction. :)
 
lol, too much algebra for @tigre
 
2:35 AM
Rule 30 is condensed down to this now:
B2=MOD(A1+B1+(1+B1)*C1,2)
 
I’m about to leverage a useful property in mathematics
 
@LeakyNun Can you shed some light on what the discriminant of an algebraic number field $K$ is? Is it exactly the volume of the "fundamental domain" of $\mathcal{O}_K$, squared?
 
@BalarkaSen indeed it is
it's also the norm of the different ideal, which encodes ramification data
 
By fundamental domain, I suppose I mean the following: Choose an integral basis $(\alpha_1, \cdots, \alpha_n)$ of $\mathcal{O}_K$, which sits canonically inside $K \otimes \Bbb R = \Bbb R^n$, so those elements span a parallelopiped.
What's the different ideal?
 
I don't like "$K \otimes_\Bbb Q \Bbb R = \Bbb R^n$" but oh well
 
2:41 AM
Haha why not
 
the different ideal is $\{ x \mid \operatorname{Tr}(x \mathcal O_K) \subseteq \Bbb Z \}^{-1}$
well because $K \otimes_\Bbb Q \Bbb R = \Bbb R^{r_1} \oplus \Bbb C^{r_2}$ as "etale algebras"
 
yeah, yeah :P Oh also I was working with a purely real field.
 
oh cool
the absolute of the discriminant of $\Bbb Z[a]$ is $N(f'(a))$
 
Let me take a simple example. Ring of integers of $\Bbb Q(\sqrt{d})$ has the integral basis $(1, \sqrt{d})$ if $d \not\equiv 1\pmod{4}$, and the discriminant is $\text{det}(1,1|\sqrt{d},-\sqrt{d})^2$ which is $4d$.
Can't quite see volume of what exactly is $2\sqrt{d}$
 
the covolume of the lattice generated by $(1,1)$ and $(\sqrt d, -\sqrt d)$
 
2:56 AM
The lattice of interest is $\Bbb Z + \sqrt{d}\Bbb Z$ in $\Bbb R^2$, yes?
 
@ÍgjøgnumMeg And what about the other divisors it might have? We needed to sum over all of them, right?
 
@BalarkaSen or more precisely $(\sigma_1(\Bbb Z[\sqrt d]), \sigma_2(\Bbb Z[\sqrt d]))$
no it isn't that
 
I'm confused then.
 
what's the map $K \otimes_\Bbb Q \Bbb R = \Bbb R^n$?
 
I'd imagine map comes from choice of a basis of $K$ over $\Bbb Q$. In our case $\{1, \sqrt{d}\}$ is a basis of $\Bbb Q(\sqrt{d})$, so define $1 \mapsto (1, 0)$ and $\sqrt{d} \mapsto (0, 1)$. But I guess then $\Bbb Z[\sqrt{d}]$ maps onto $\Bbb Z^2 \subset \Bbb R^2$, so something is off.
All in all a little confused. I guess my map isn't good because it won't map the ring of integers to an integral lattice in general.
 
3:02 AM
the map is $(\sigma_1, \sigma_2, \cdots, \sigma_n)$
 
I see, but why?
To make the image of $\mathcal{O}_K$ a sublattice of $\Bbb Z^n$?
 
let's say $K$ is simple (which it is)
then can you find a "natural" map $K \otimes_\Bbb Q \Bbb R \to \Bbb R^n$
 
I am looking for a reference for the following kind of result. For a commutative rng $A$ and a finitely generated $A$-module every endomorphism $\phi$ of $M$ is vanished by some polynomial in $A[x]$ acting on $\mathrm{End}_A(M)$ by evaluating.
This is Cayley-Hamilton's theorem in general, and from this one deduces that:
For any rng $A$ and a finitely generated $A$-module $M$ there is an element of $A$ that acts identically on $M$, that is, $a\in A$ such that $am=m$ for all $m\in M$.
 
Take $a = 1$?
 
rng
 
3:07 AM
It is a commutative rng, that is, a commutative ring possibly without identity.
 
Oh hah.
No idea then
 
The proof is just setting $\phi$ to be identity in the Cayley-Hamilton above.
 
I don't think I understand. What about $2\Bbb Z$ as a module over itself?
 
I am looking for a reference that covers results of this taste.
Is $2\Bbb Z$ finitely generated over itself?
 
I guess not! I can't get $2$
 
3:12 AM
Right.
 
@WilliamSun I have usually seen the usual Cayley-Hamilton for commutating rings to prove if $IM = M$ then there is some $i \in I$ such that for all $m \in M$, $im = m$.
But this a weird application
Never thought about this before
 
Yes but the version I wrote above is actually cleaner and more general.
 
I am skeptical there are interesting examples of finitely generated non-unital modules over non-unital rings.
 
Note that an ideal is a subrng.
And $IM=I$ is saying that if $M$ is f.g. over $A$ then it is f.g. over $I$.
 
Fair point
Can you not prove this general result by unitalization, from the standard result?
 
3:15 AM
Actually one can just prove it by induction or yes, unitalization.
 
Interesting though
 
I am looking for a reference that covers, for example, the module $M$ acts $A[x]$ through an endomorphism. Maybe I should look into some representation book.
 
There's a different proof of Nakayama which completely avoids Cayley-Hamilton, the induction argument you said. I think there you can even drop commutativity.
But I could be wrong
@LeakyNun I see the point. This is actually the canonical decomposition $K \otimes_{\Bbb Q} \Bbb R \to \Bbb R^n$ coming from the Galois embeddings $K \to \overline{\Bbb Q} \subset \Bbb R$; this is the more natural idea.
 
yeah
 
hey
can you have a rotation that is not circular and not hyperbolic?
 
3:21 AM
Yo Winterbash is here
 
and would this correspond to some other geometry that is undiscovered?
by not circular and not hyperbolic I mean that this roation would be neither circular nor hyperbolic
 
3:51 AM
@Ultradark By hyperbolic rotation do you mean something like $(x,y)\mapsto(ax,y/a)$?
i.e. there is a pseudo-norm preserved by the map
(in this case $\|(x,y)\|=xy$)
 
@BalarkaSen there is something more general here
$\Bbb R$ is just one completion of $\Bbb Q$
 
What's that
 
what's what
 
You said there's something more general; what's that more general thing
Milne defines discriminant of a finite extension $L/K$ to be the discriminant of the trace form $L \times L \to K$, $(\alpha, \beta) = \text{tr}(\alpha\beta)$.
 
@BalarkaSen $L \otimes_K K_v = \prod_{w \mid v} L_w$
@BalarkaSen yeah that's equivalent
I forgot that
 
4:03 AM
Trace is sum of all the Galois conjugates (inside $L$), right?
 
no not just inside $L$
 
Ah OK
Yes, of course, otherwise it won't map to $K$
 
and from that isomorphism one obtains $\operatorname{Hom}(L, \overline{K_v}) / \operatorname{Gal}(\overline{K_v} / K_v) \leftrightarrow \{ w \mid v \}$
 
I dunno anything about local fields yet
I'll understand that someday
 
$\Bbb R$ is just $\Bbb Q$ completed at $\infty$
I guess you know $\Bbb Q_p$?
 
4:05 AM
Yeah
I get the vague idea; you can complete wrt different norms (for $\Bbb Q$, $\nu_p$ are the $p$-adic norms, and $\nu_\infty$ is the plain vanilla guy)
 
yeah
 
Cool stuff.
 
do you see why that definition of discriminant is the same?
it's Nakamura 16.5 - 5.5 Naroditsky now
 
@AkivaWeinberger yep
 
Over the complex numbers these are the same I think
 
4:10 AM
@LeakyNun I think so. After choosing an integral basis $(\alpha_1, \cdots, \alpha_n)$ the matrix of the trace form becomes $(\text{tr}(\alpha_i \alpha_j))$ which is $(\sigma_i(\alpha_j))^T (\sigma_i(\alpha_j))$.
 
yep
precisely
 
@LeakyNun Oh shit
 
@AkivaWeinberger Consider the following matrix which transforms points in $\Bbb R \cap (0,1)^2,$ according to a parameter ${\mathfrak S}.$ $$ \mathfrak T_{\mathfrak S} = \begin{bmatrix} (e^{e^{\mathfrak S}})^{-1} & 0 \\\ 0 & (e^{e^{\mathfrak{ -S}}})^{-1} \end{bmatrix}$$
 
Naka crushing again
 
yep
 
4:11 AM
I think this works
 
that's more mathfrak than I have never seen
 
but I don't know what it preserves
 
What's it look like?
What are the trajectories of points?
 
If the set of trajectories are the level sets of some function, it preserves that function
 
4:12 AM
I can't figure out how to plot more trajects
 
You can biject your space to an easier-to-deal-with one
by saying $e^{\mathfrak S}\leftrightarrow s$
 
doesn't that biject to a hyperbola
 
Probably
 
yeah it looks like $f(x)=1/x$
 
So this is a weird image of a hyperbolic rotation
and so intrinsically they're identical
 
4:16 AM
yeah
 
I think $y=1/x$ is a geodesic on the pseudo-Riemannian metric $\|x,y\|=xy$?
 
does that mean the function that these points flow across is hyperbolic
 
@BalarkaSen ^Is that right
If so, your thing is geodesics on an isomorphic pseudo-Riemannian surface
@Ultradark What is "flow across a function"
 
you said level sets of some function above
 
Oh right
 
4:19 AM
so is the function then hyperbolic
or at least isomorphic?
 
I think it's level sets of $\ln(x)\ln(y)$?
 
@Akiva Oh holy shit I dunno, maybe.
 
@akiv
 
'Cause the original was level sets of $xy$ and then you did a bijection involving exp
 
I've know the function for at least 8 years
 
4:22 AM
@BalarkaSen Thank
 
$T(x, y) = (ax, y/a)$ are isometries wrt your pseudo-Riemannian metric right
 
So if those curves are preserved I think it means they are geodesic indeed
 
yes!
 
But I can't be sure
 
4:23 AM
Hm actually nah I don't think they are
Like the metric should change depending on location
 
are you still using $\ln(x)\ln(y)?$
 
Maybe $g_{ij}=xy\delta_{ij}$ or something weird like that
@Ultradark I'm talking about $(x,y)\mapsto(xt,y/t)$
 
I thought you meant that the matrix of the metric is (0, 1|1, 0)
 
But if it's geodesics on some weird metric then so is yours 'cause we have a diffeomorphism
 
That's what people usually mean when they write the metric as a bilinear form
 
4:26 AM
@BalarkaSen Yeah that's what I meant originally but it's probably wrong
(Well, half that)
 
Yeah half that. Why's it wrong
 
'Cause the geodesics of that are straight lines
 
Are they
 
Should be
 
Maybe
I have to compute, can't imagine shit
 
4:27 AM
I think that's what you get with a constant metric
hence my changing it to (xy,0|0,xy)
 
Yeah should be correct
Good call
 
I mean could you not just use the same constant lines to
be geodesics for my example
maybe I'm not following, but the rays that intersect hyperbolas of the form $f(x)=k/x$
define hyperbolic coordinates
so could you do the same here?
 
What are hyperbolic coordinates
Oh these
In mathematics, hyperbolic coordinates are a method of locating points in quadrant I of the Cartesian plane { ( x , y ) : x > 0 , y > 0 } = Q {\displaystyle \{(x,y)\ :\ x>0,\ y>0\ \}=Q} .Hyperbolic coordinates take values in the hyperbolic plane defined as: H P = { ( u , v ) : u...
Give me a minute
So what's your end goal here? I'm confused
Are you trying to see where this fits in with the spherical, Euclidean, and hyperbolic geometries?
 
oh I was just asking if this was neither a hyperbolic nor circular rotation
yeah
 
It's not a "rotation" in any sense I know, so I guess yeah it's neither of those things
If you can use this to turn the plane into a (Riemannian) metric space, and it has constant curvature, then it must be either spherical ($K>0$), Euclidean ($K=0$), or hyperbolic ($K<0$)
and it's probably not spherical, 'cause there, geodesics are loops
Does the parallel postulate hold? Is that a question we can ask?
Hm
All geodesics in the hyperbolic plane tend towards the idealized boundary at their edges
I guess so do these? (They go to $(1,0)$ and $(0,1)$ I think)
But in a hyperbolic plane, all geodesics are uniquely determined by their ideal points
 
4:40 AM
So this is a weird image of a hyperbolic rotation
and so intrinsically they're identical -Akiva
is this still accurate?
 
I'm honestly not sure how related the words "hyperbolic rotation" and "hyperbolic geometry" are
Oh wait
 
yeah I don't know either
 
Why would we even want them to be geodesics?
With a circular rotation, you don't get geodesics (lines), you get circles
In any case, the hyperbolic rotation is an isometry of the pseudo-Riemannian space with constant metric (0,1|1,0) I think, which is Euclidean (or at least has constant zero curvature)
(I dunno if you can call pseudo-Riemannian spaces Euclidean)
Your thing is diffeomorphic to that, so I think the same applies
 
I wonder what if you hyperbolify this weird Euclidean space, by (0, 1/y^2|1/y^2, 0) :P
 
I have my copy of Do Carmo within arm's reach, as it turns out
but it's 11:45 and I don't really want to go through those equations right now
@Ultradark Did you ever study Riemannian geometry? I feel like I might have seen you talk about it on here before but I'm not sure
 
4:46 AM
@Akiva This is actually isometric to Minkowksi plane R^{1, 1} I think. Because consider the change of coordintes (x, y) -> (x + y, x - y), which changes the metric from xy to x^2 - y^2
 
Right yes
Is that called "Euclidean"? It's a definitional thing I think 'cause it still has 0 curvature everywhere
 
It's just called the Minkowski space to my knowledge. Yeah, it is 0 curvature.
 
In mathematics and theoretical physics, a pseudo-Euclidean space is a finite-dimensional real n-space together with a non-degenerate quadratic form q. Such a quadratic form can, given a suitable choice of basis (e1, ..., en), be applied to a vector x = x1e1 + ... + xnen, giving q ( x ) = ( x 1 2 + ⋯ + x...
 
yeah Minkowski space has 0 curvature
 
It's pseudo-Euclidean. Wonderful
 
4:48 AM
$K \otimes_\Bbb Q \Bbb R$ is also called the Minkowski space
 
There is some connection between Minkowskian geometry and hyperbolic geometry
Some projective model or some shit
@LeakyNun lol
point
 
The hyperbolic plane is a subset of Minkowski 2+1-space I think
It's a paraboloid
 
Yeah
 
what Minkowski contributed to NT is just magical
 
I think the video game Zeno Rogue uses that to calculate its coordinates
like every point on the plane is stored as a $(x,y,t)$ value
*HyperRogue
That's what it's called, I think Zeno Rogue is the developer?
 
4:50 AM
Should play that game once
 
@BalarkaSen it's just magical how the geometry works
 
I've never actually played it either
but it's a rogue-like on a hyperbolic plane
 
like finiteness of class number
 
Ya @Leaky. I'll see if I get the chance to understand it someday
 
4:52 AM
@AkivaWeinberger I've talked about Riemannian geometry but never studied it
read a little about it
 
You should. I struggled a lot with Do Carmo but it was worth it. You seem to know more than me on lots of things so I think you'll struggle less
I think I'm technically like 2 pages from the end but never got through the last proof and put it down
lol
 
damn
i did like first 7 chapters i think
im still pretty bad at riemannian geometry
 
The metric should be visualized as Tissot ellipses
In cartography, a Tissot's indicatrix (Tissot indicatrix, Tissot's ellipse, Tissot ellipse, ellipse of distortion) (plural: "Tissot's indicatrices") is a mathematical contrivance presented by French mathematician Nicolas Auguste Tissot in 1859 and 1871 in order to characterize local distortions due to map projection. It is the geometry that results from projecting a circle of infinitesimal radius from a curved geometric model, such as a globe, onto a map. Tissot proved that the resulting diagram is an ellipse whose axes indicate the two principal directions along which scale is maximal and minimal...
 
@BalarkaSen and Hikaru is bad at chess
 
(or, for Minkowski space, Tissot hyperbolas!)
 
4:54 AM
yeah Ted taught me that
 
(and pseudo-Riemannian spaces in general)
 
norm 1 points on T_x M right?
 
@LeakyNun lol, of course
 
And norm $\pm1$ points give you a pair of conjugate hyperbolas if you're pseudo-Riemannian
 
4:55 AM
I gave up when there were ten million curvatures coming in I think
 
I should learn Ricci flow
 
I understand Riemann curvature tensor and sectional curvature and I'm pretty satisfied with that
@AkivaWeinberger same
 
'cause I saw the equation once and thought "Wait I understand all the parts of that"
It was like $dg_{ij}/dt=R_{ij}$ I think
 
I want to read
someday
we can try togather if you want
 
I have 97 tabs open. These are all my "read later" or "watch later" lists
 
4:57 AM
I started to bookmark them after a point because it was getting out of hand
 
Interesting chapter numbering system
 
When it crosses like 100 I bookmark them in a folder
with the date
@Akiva Gromov is crazy yes
 
just realize that you will never read them
 
(Re chapter numbering) A demonstration that all countable linear orders can be embedded in $\Bbb Q$
(proof: build the function one at a time)
 
@LeakyNun its true
but still
 
4:59 AM
@LeakyNun You mean to tell me I will never read this paper? arxiv.org/pdf/1810.00492.pdf
I'll prove you wrong! ...Eventually!
 
lol
 
wait there's more
this is only half of the space
 
Some day I oughta learn what an orbifold is
 
I usually refer to the little space as $\zeta_{\Bbb R^2}$ and the larger copy as $T_{\Bbb R^2}$
$T-$space haha
 
(Cont'd) My incredibly vague understand is, it's like when you take the universal cover of the plane minus a point, but then put the point back in
 
5:12 AM
where everyone drinks tea
 
and keep track of the differential information
 
It's a pseudomanifold with neighborhoods of singular points modelled on $\Bbb R^n/G$ as far I know, where $G$ is a finite group acting by homemorphisms on $\Bbb R^n$
 
Ah
That seems different to what I just said
but also it feels like it might not be
if you massage it a bit
 
I'd say finite covers of $\Bbb R^2 \setminus \{(0, 0)\}$ and then putting the point back in.
It's like $\Bbb R^2$ with $\Bbb Z_n$ acting by rotations
 
Right
I feel like I wanna make that work for infinity but it might not
In any case
I oughta sleep
 
5:15 AM
Good night
 
good bishop
 
Hahah
So trace of an element of $L$ for a finite extension of $K$ is also trace of the $K$-linear map $L \to L$ given by multiplication by that element, huh.
 
Yis
And Norm of an element is given by the determinant of the same map
 
That's clear if $L = K(\alpha)$ is a simple extension, and I am computing $\text{tr}(\alpha)$. The matrix of multiplication by $\alpha$ under the natural basis is just the companion matrix of the minimal polynomial of $\alpha$
The trace is just the coefficient of $x^{n-1}$, which is sum of all the Galois conjugates of $\alpha$
 
5:28 AM
aye, it’s slightly more subtle for the trace/norm of an element which isn’t the primitive element
 
Hm
Let me ponder on this, don't tell me
 
No worrieeeez
 
tnx
 
@tigre a necessary a sufficient condition for the existence of a Euler path is (from my textbook) that the number of odd degree vertices is zero, one, or two ("at most two odd degree vertices"). However this graph has more than two so it does not have Euler path. Thank you so much!!
 
welp it can't be one
because every edge connects two vertices
 
5:49 AM
@LeakyNun be careful: we are not talking about edges but vertices
 
I'm very careful thanks for reminding me
 
So we can have a vertex of odd degree: this can have a Euler path
 
6:04 AM
@manoooh Leaky is correct; the number of odd degree vertices cannot be one.
I played two 5+3 and both were missed wins.
I bungle the endgame so bad lmao
Need to read endgame theory yo
 
I love
it
When my psets have problems that are basically free points
 
Who doesn't?
 
Just has to classify the prime ideals of a direct product of rings
And that’s worth as many points as like.. showing that a pair of field extensions are determined by the primes that split
Or smth like that lol can’t remember the exercise
 
Spec sends products to coproducts boom
 
6:14 AM
a b s t r a c t nonsense
 
@BalarkaSen I recommend silman
 
I'll look into it
 

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