@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
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
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 $v_1$.
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 $v_1$.
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)
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)
$(0,1)$ 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
@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.
@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.)
Is 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 ...
@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?
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.
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 $2\sqrt{d}$
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.
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 $m\in M$.
@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 $m \in M$, $im = m$.
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.
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 $L \times L \to K$, $(\alpha, \beta) = \text{tr}(\alpha\beta)$.
and from that isomorphism one obtains $\operatorname{Hom}(L, \overline{K_v}) / \operatorname{Gal}(\overline{K_v} / K_v) \leftrightarrow \{ w \mid v \}$
@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 $(\sigma_i(\alpha_j))^T (\sigma_i(\alpha_j))$.
@AkivaWeinberger Consider the following matrix which transforms points in $\Bbb R \cap (0,1)^2,$ according to a parameter ${\mathfrak S}.$ $$ \mathfrak T_{\mathfrak S} = \begin{bmatrix} (e^{e^{\mathfrak S}})^{-1} & 0 \\\ 0 & (e^{e^{\mathfrak{ -S}}})^{-1} \end{bmatrix}$$
In mathematics, hyperbolic coordinates are a method of locating points in quadrant I of the Cartesian plane
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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?
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
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
@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
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
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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...
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 $\Bbb R^n$
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.
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 $\alpha$
@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!!