"Injective linear maps between finite-dimensional fields of the same dimension are bijective" is basically linear algebra's version of the pigeonhole principle
If you're familiar with the definition of $L^2(X,\Sigma,\mu)$ for a generic measure space this is the same as taking $X=\Bbb Z$, $\Sigma$ the full powerset and $\mu$ the counting measure
Hi @AlessandroCodenotti, unfort. I am not that familiar with it, but I can guess and look it up.
I am interested in probability currents for quite a while and now I came across that en.wikipedia.org/wiki/Probability_current#Discrete_definition which is facinating, as it seems to suggest that there is quantum mechanical world on discrete spaces...
@AlessandroCodenotti, @ÍgjøgnumMeg I try to imagine what a Hilbertspace vector (on that space?) looks like? Is it an infinte sequence of integers, or rather a sequence of field elements, like $\in\Bbb C$?
Alright I have a factor of $n$ in there so I'm not sure where that came from, but otherwise I have what I want
hehe
but what I did was, as you said, to look at the trace of the multiplication by $\alpha$ matrix (where $L = K(\alpha)$) and ended up with a companion matrix, and that guy has minimal polynomial equal to the minimal polynomial of $\alpha$, factored that guy over $\overline{K}$, and thus have that the multiplication by $\alpha$ matrix is similar to a matrix with $\sigma_i(\alpha)$ along the main diagonal
Okay so if you consider a general element of x of L, you want to choose a K basis for K(x) (just the standard one by powers of x) and a K(x) basis for L, then multiply those to get a nice basis for L over K
You get a block diagonal matrix with a bunch of companion matrices on the diagonal
@ÍgjøgnumMeg I don't think that's all that useful, since you may know the trace of $\alpha$, but not of powers of $\alpha$
You get $\mathrm{Tr}_{L/K} (x) = [L:K(x)] \mathrm{Tr}_{K(x)/K} $ if you construct a basis as I described above
Now you have to consider the question "for every K-algebra homomorphism $K(x) \to \overline{K} $ in how many different ways can I extend it to $L \to \overline{K} $?"
@Ultradark Abelian groups are all amenable, while hyperbolic groups (which are the generalization of non positive curvature) are amenable iff they are virtually cyclic, so an Abelian group needs to be essentially $\Bbb Z$ (to be precise it needs to be quasi-isometric to $\Bbb Z$) to also be non positively curved
@Ultradark I'm not sure why you consider the Cayley graph of a group to be two dimensional here
@AlessandroCodenotti yeah, nevermind about the dimensional thing. Do you know if there is a possible Cayley graph with two generators whose arrows only point in one direction.
The main theorem of Galois theory is that the étale site of Spec(K) is equivalent to the site of finite discrete G-sets where G is the absolute Galois group of k
@AlessandroCodenotti I guess I'm asking if you can simply look at a Cayley graph and see whether the elements pairwise commute, or is it not that simple
Start from the vertex corresponding to $e$, follow the arrow label $s_1$, then the one labeled $s_2$. Go back to the vertex corresponding to $e$, follow the arrow labeled $s_2$, then the one labeled $s_1$. If you end up in the same place then $s_1$ and $s_2$ commute, otherwise they don't
"The introduction of the digit 0 or the group concept was general nonsense too, and mathematics was more or less stagnating for thousands of years because nobody was around to take such childish steps..." - our Lord and saviour
Do you mean in the sense that you can recover an $\Bbb R$-form of $\Bbb{C}[G]$ by taking fixed points of that conjugate-linear map, which is acting as $Gal(\Bbb C/\Bbb R)$?
@GaloisintheField the equivalence of complex representations with a equivariant real structure and real representations follows immediately from Galois descent
For a distance to define a norm it is sufficient to have $\text{dist}(\alpha x, \alpha y)=\alpha \text{dist}(x, y)$ but is this the only way we can define a norm using a distance?
@Fuzzy: So you're on a vector space to start with, of course. The formula you just wrote down is the standard way to get a metric out of the norm, but of course you could take any positive multiple of that, so there's nothing unique.
I mean if you want a norm then it should be scale-invariant. It seems to me the only interesting question is the following. Say a subset of R^n is a unit sphere if it intersects every ray from the origin exactly once. Under what conditions is this the unit sphere of a norm?
As Ted says the corresponding unit ball must be convex; and the unit sphere must be a closed set.
I'm just saying I don't understand how to extract a question from what you asked beyond something unrelated and then I got kind of interested in that other question
But isn't it supposed to be in the form P implies Q and Q implies P since it states that this statement only holds true if and only if a-b is divisible by 4?
let $U$ be an open convex neighborhood of $0$ in $V=\Bbb R^n$ such that $v \in U \Rightarrow -v \in U$ and such that $\partial U$ has the property that every ray from the origin interesects $\partial U$ exactly once. Define $\|0\|=0$ and for any $v \in \Bbb R^n \setminus 0$, let $\|v\|=1/\alpha$ where $\alpha \in \Bbb R_{>0}$ is the unique constant such that $\alpha v \in \partial U$. Then clearly $\| \lambda v \|=\lambda \|v \|$ and since we have $-v \in U \Rightarrow v \in U$, we also have $v \in \partial U \Rightarrow -v \in \partial U$ which implies $\|-v\|=\|v\|$.
Okay, sorry about that. So we're pretty much given in that question that: $a (mod 4) = b (mod 4)$ with which we can prove that this relation is indeed an equivalence relation?
@TobiasKildetoft Okay. I understand. So, to prove that this relation is reflexive, would it be valid to say "For any m ∈ Z, m ≡ m (mod 4), as m − m = 0 is divisible by 4 as given." Therefore, this relation is reflexive?
and for symmetrical would that also be valid: "If a ≡ b (mod 4), then a−b is divisible by 4. But then b−a = −(a−b) is also divisible by 4, which implies that b ≡ a (mod 4)", as needed to show symmetry?
Oh, okay. I understand. That makes sense. And for the last part (proving that this relation is transitive), would this be acceptable? "Suppose that a-b and b-c are both divisible by 4 by definition. Consequently, their sum, (a − b) + (b − c) = a − c is also divisible by 4".
Hey! Are the following conclusions correct? For part 1: the number of elements of the largest transversal subset of that list is $7$. For part 2: the number of elements of the largest equivalence class that is a subset of that list is $3$ since the equivalence class subsets are ${1/2, 5/10}, {0/1, 0/2}, {2/1, -10/-5, 10/5}$.
Sorry for the second part, the sublists are: {$1/2, 5/10$}, {$0/1, 0/2$}, {$2/1, -10/-5, 10/5$}. Therefore, the last list contains the most elements ($3$).
If I draw a naive parallel curve to the trefoil I get an embedded curve which hits the Seifert surface of the trefoil at 3 points, so I put three twists in the opposite direction to make it the "true parallel" - intersection of the unit normal bundle of the trefoil with the Seifert surface.
If $a, b$ are the Wirtinger generators, then the knot group is $\langle a, b | aba = bab \rangle$ and the parallel comes out to be $ba^2ba^{-4}$ I think.
Which does have zero abelianization, as it should (the Seifert surface has that as boundary)
Yeah I am not sure I quite see it. But see the fundamental group: The meridian is $a$ and the longitude is $ba^2ba^{-4}$ (hoping I got the signs right). The +1 curve is $ba^2ba^{-4}a = ba^2ba^{-3}$
I don't understand the +1 curves though. If I just do a 0-surgery on the trefoil I glue a disk to the boundary of the Seifert surface, which gives me a torus bundle over the circle?
Yeah I seem to get fundamental group $\langle x, y | x^2 = y^3 = (xy)^4 \rangle$ (is that correct?)
Hm maybe not
$(xy)^6$
Meh there's some sign mistake in my calculations
Wish I could do these things fast but also correctly
OK, it has to be $\langle a, b | aba = bab, a^4 = ba^2b \rangle = \langle x, y | x^2 = y^3 = (y^{-1} x)^6 \rangle$. The abelianization matches at least, given what you said
@MatheinBoulomenos what's the generalization of the statement that an H-module is the same thing as a C-module with a negative-involutive conjugate-linear map?
I understand your comment about the Brieskorn sphere is that if I look at $x^2 + y^3 + z^5 = 1$ in $\Bbb C^3$ and a link of the $(0, 0, 0)$ singularity, that's a $3$-manifold inside $S^5$. What is the fibration over $S^2$?
$A_5$ acts on the $3$-manifold right (pi rotation on first component, 2pi/3 rotation on second, 2pi/5 rotation on the last)? It preserves the equation.
I guess I wasn't making sense above; I just produced $\Bbb Z/2 \times \Bbb Z/3 \times \Bbb Z/5 \leq S^1 \times S^1 \times S^1$ which acts on the hypersurface $x^2 + y^3 + z^5 = 1$ by $(x, y, z) \mapsto (-x, y, z)$, $(x, y, z) \mapsto (x, \zeta_3 y, z)$, $(x, y, z) \mapsto (x, y, \zeta_5 z)$ respectively.
@LeakyNun let $L/K$ a Galois extension with $G=\mathrm{Gal}(L/K)$ and let $f \in H^2(G,L^\times)$ a 2-cocycle, then we can construct a central simple algebra over $K$ like this. Take as a set the group algebra $L[G]$ and define the multiplication based on the relations $\lambda \cdot \sigma = \sigma \cdot \sigma(\lambda)$ and $\sigma \cdot \tau = (\sigma \circ \tau) f(\sigma,\tau)$
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