Actually I think this boils down to me not understanding tensor products with duals, for example I don't see the identification $\mathrm{End}(E)\simeq E\otimes E^\ast$ either
So going back to connections there's a natural identification of $E\otimes T^\ast M$ with $\mathrm{Hom}(TM,E)$, which is great since I have an element of $TM$ lying around and I need to produce an element of $E$
namely, a tangent vector is very obviously a derivation because you take the directional derivative along it. a derivation isn't obviously a tangent vector, and in fact you need a crucial "Taylor's theorem without reminders" for it, which says any smooth function $f$ can be written as $f = \sum x^i g_i$ in local coordinates where $g_i$ are smooth
this is false in $C^k$ because the regularity of $g_i$ goes one down to $C^{k-1}$
I don't know, just made that up. Feels like it should be
$M$ is compact manifold for simplicity
There are issues if it's not compact, because then $\text{Diff}(M)$ will not be locally contractible under the weak Whitney topology, in particular it's not a Banach manifold. Then you have to deal with strong Whitney topology which is ugh
@Adam Ein sof is super weird. It basically bleds God's essence from its domain all the way down to the physical world. I have not read much Kabbalah before get pulled into reading Zen Buddhism, so I have not done much study on the nature of Ein Sof yet
@MatsGranvik well that's why I'm in a bad mood this weekend after reading that a lot of the authors for the algorithms I'll be studying next claim themselves as ultrafinitists and their computers are coauthors, Like yes its very funny and I too liked to jest that maple is the best girlfriend ever but yeah I guess it just demonstrates how deep to the core the paradox is and how easy it is to raise the assumption levels to the danger zones of saying garbage
Like ok, yes I understand they may have their reasoning for having the belief but bringing that into the picture is ruining the best thing about mathematics
@Secret everything becomes normal if you are exposed to it for long enough that's my bubble wrap principle for imminent trauma, it's the bubble wrap for something worse to experience lol but no you don't need to be freaked out of course it reads like an episode of game of thrones the dialogue was written drawing from it I can only assume, apart from the other infinitude of reasons it sucks
all olden timey translations are written in that dramatic style
I'm guessing Pythagoras noticed the popularity of the dramatic amongst the chuds and that's when he started to announce "I see dead people" and formed a cult
All these packages are so awesome but I cant help but feel as if it would be neat if there was one that allowed you to import a circuit board layout from a datasheet into a project. @ÍgjøgnumMeg no megan you don't get stomach ulcers from pineapples unless you let it ferment and drink it ever day, even still ive never had stomach ulcers. i probably should have
I can really appreciate this one because it's one of my old favourite go to approaches, using the logical conjunction in anything other that the study of formal logic usually has the twits complaining but it really does become meaningless if you are not going to use principles of logic to make algebraic conclusions
but I guess I would need a respect costume to start talking like that
here is a challenge: 1) Estimate the reduction in government revenue and increase in public housing expenditure in the hypothetical scenario that all illicit drugs are decriminalised in your country 2) Estimate the historical expectation in civilian lives saved p.a and plot this against military and law enforcement expenditure p.a
3) Will the rate of home invasions and crimes of violence be likely to increase or decrease in the hypothetical scenario outlined in (1)? 4) What is likely to happen to the number of people inflicted with addictions to these substances? 5) What is likely to happen to the profits of criminal syndicates? 6) Which pharmaceutical company will be allowed to distribute these substances and monopolize on the market for them?
yeah I mean if you listen to rage against the machine you are likely to get accused of being a socialist or a communist, but their views are actually pretty libertarian, when they criticise the free market in some of their lyrics its because there is nothing free about it in many regards
if $f_n \to f$ pointwise and ${f_n}' \overset{\text{unif}}{\to}g$ on an unbounded interval, we can say $f' = g$, right?
I know that on a bounded interval it's enough
followup question: on a bounded interval, it's enough for $f_n(x_0) \to f(x_0)$ for a point $x_0$ in the bounded interval. Is there a way to extend that hypothesis to need only one point of convergence on an unbounded interval?
something like it holding for a point on every bounded subinterval or something
Blum, Blum, and Shub generator doesn't differ very much from the standard if there is a method for determining $m$ knowing $x_i$ that could be potentially not cool $x_{k+1}=x_k^{m} \operatorname{mod} n$
well it wouldn't affect me im trailer trash lol
but this seems like fun i mean opposed to doing nothing
@GFauxPas: Just restrict to bounded intervals $[x_0,b]$ and let $b$ increase. If you know convergence is uniform on $[x_0,\infty)$, then you certainly know it's uniform on any such compact subinterval.
I have a point in 3d space and am looking to find a unit quaternion that describes the direction of the point related to the origin. Any ideas on how to do this?
@Ted My $O(n)$ construction earlier actually realizes every group as a symmetry group of some nonregular polytope in general. Symmetry groups of regular polytopes are quite restrictive; they are reflection groups.
I have been on a streak thinking about polyhedra since :P
Here's a group theoretic Gauss-Bonnet for you: $\langle a, b, c | a^2, b^2, c^2, (ab)^p, (bc)^q, (ac)^r \rangle$ is finite if $1/p + 1/q + 1/r > 1$ and infinite if $1/p + 1/q + 1/r < 1$.
Still infinite but distinctly different. Here's the general statement: if $> 1$ it's a subgroup of $SO(3)$, if $< 1$ it's a subgroup of $\text{SL}_2(\Bbb R)$, if $= 1$ it's a subgroup of $\Bbb R^2 \rtimes SO(2)$
The isometry groups of the model geometries in 2 dimension
I was oversimplifying with the finite/infinite thing
I like how it's totally nonobvious from the presentation itself that it's going to be finite if $1/p + 1/q + 1/r > 1$ but once you realize it as a triangle group like that, it's a discrete subgroup of $O(3)$, so finite.
@TedShifrin I'm really struggling to find a way to describe the question I have well, so here goes: I want to define a coordinate system so that the x direction is facing a specified point. In ROS, coordinate transforms are defined using unit quaternions, which is why I mentioned that.
OK, @Mason, but you can think of the unit $\Bbb R^3$ vector as a unit quaternion. Or are you trying to do something fancy like represent a rotation in $\Bbb R^3$ by quaternionic multiplication?
@Ted Tile the hyperbolic plane by $(p, q, r)$ triangles if $1/p + 1/q + 1/r < 1$. The translates of a single triangle by this group gives all the triangles, so you can keep composing reflections in a way that will take one edge in the fundamental triangle to corresponding edges in triangles far off towards the ideal boundary
That gives an infinite order element. Less cleaner than the elliptic case
You can cook the elliptic ones by taking an arrangement of three 2-planes in $\Bbb R^3$ and looking at the group generated by them. These will actually be full symmetry groups of various regular polyhedra, except the degenerate cases (which is indirectly true because that's what discrete subgroups of $O(3)$ are).
@Ted A useful presentation for $A_5$ is $\langle x, y | x^2 = y^3 = (xy)^5 = 1 \rangle$ (this is the "von Dyck subgroup" of the triangle group), and the binary icosahedral guy is $\langle x, y | x^2 = y^3 = (xy)^5 \rangle$ if I am not too off.
My handy example is always the preimage of $V_4$ under $SU(2) \to SO(3)$. That's $Q_8$, group of coordinate quarternions, the central $\Bbb Z_2$-extension of $V_4$
In particular that means since $A_5$ contains $V_4$, the binary icosahedral group contains $Q_8$ as a subgroup of index $15$.
@TedShifrin Ah, but when I am thinking of $Q_8$ as a subgroup of $S^3$ I am really thinking of Hatcher's twist-glue cube picture. A modification of that gives the $3$-manifold $S^3/Q_8$. Do you still want to call me an algebraist? :)
Hm the formula $\sum (1 - 1/r_i) = 2(1 - 1/n)$ for a subgroup $G$ of $SO(3)$ acting on it's poles on $S^2$ can be derived just from Riemann-Hurwitzing on $S^2 \to S^2/G$, which is a branched cover, right?
To be clear I know it comes from orbit-stabilizer and some counting, but I think you can package that with R-H
I think $S^2/G$ is homeomorphic to $S^2$ but I don't know how I would quite see it without apriori knowing these are polyhedral groups. If we trust it then RH says $2 = 2|G| - \sum_p (r_p - 1)$ where $p$ varies over the poles, and $r_p$ is the order of the stabilizer subgroup of $p$, the ramification index
That's the good old $\sum_p (r_p - 1) = 2(n - 1)$
You just choose representatives $p_1, \cdots, p_k$, use orbit-stabilizer $r_p |\mathcal{O}(p)| = n$ and rearrange to get there now
