I like Coric, but he's not the most scintillating player.
Re your question, @Danu, of course you're supposed to have a self-adjoint linear map from $\Lambda^2$ to $\Lambda^2$. I guess I'd have to sort that out in your situation.
That's not sufficient though, because one needs Bianchi. In fact, the Kulkarni-Nomizu product of two two-forms already lies in $S^2(\Lambda^2 T^*)$, but is not in general an algebraic curvature tensor
There should be some nice way to understand the possible ways of making algebraic curvature tensors out of two two-forms, purely in algebraic terms, but idk how to do that
@Danu: One of my math friends has run an REU (undergraduate research experience) numerous summers on these sorts of questions. Would you like his name/email to contact?
I don't know what surface that is, @Ultra. Noncompact surfaces of revolution with two ends will of course have a foliation by circles, but in general you need Euler characteristic zero (homeo to a torus) to get a foliation of a compact surface.
@TedShifrin what about non-contractible geodesics on a compact surface with two singular endpoints. Would this be more in the realm of systolic geometry
I always advised my students to say that they were giving up their chance to see what was written. People reading the rec will trust it more if it's confidential, for obvious reasons.
Usually I told students I'd let them read it if they wanted, but almost no one asked. If I felt I couldn't write something good, I tried to get out of writing altogether.
Below is a problem that I did. I would like somebody to check it for me.
Thanks,
Bob
Problem:
Suppose that $z_1$ and $z_2$ are two independent random variables that are normally distributed with
mean $2$ and standard deviation of $4$. Now if $z = |z_1 - z_2|$, what is the probability that $z$ ...
I like how every question with Balarka starts as someone asking him something he knows nothing about but he just figures out the whole theory on the spot
Except set theory, which is not really important for most of maths anyway
But again, the knowledge of infinite topos means he will knew how to derive pretty much anything in maths, since everything can be expressed in terms of topos and categories
What’s the order type of the set of all integer polynomials, where $f \leq g \leftrightarrow \exists x \in \mathbb{Z} : \forall y \in \mathbb{Z} : f(y) \leq g(y)$?
@Secret Since you seem interested in infinities what are your thoughts on this?
Yeah I cannot work this out on mobile. Might need to sit down with paper. I guess the only things we can say is its linear order type has something to do with $\Bbb{Z}$
I am working out something relating to fixed point division, and I have an expression that's (AM×2¹⁶ + AL) / (BM×2¹⁶ + BL). Assuming the result is between -1 and 1, is there a well known expansion of this kind of binomial division in terms of the single terms?
(Apologies if I'm using the word "binomial" incorrectly there.)
So I'm asked to find whether the following integral is either convergent, divergent or indefinite. $$\int_1^2 \frac {\sqrt{x-1}}{\abs{\sin(1-x)}\tan(2-x)} dx$$ I've simplified it as: $$\int_0^2 \frac {\sqrtx}{\sinx \tan(1-x)} dx$$
now the Idea I had was to find a function which is greater [or smaller] at each point than the integrand, easier to integrate and which also converges [or diverges]. But I'm having a hard time finding any such functions. Is there something that escapes me? maybe an easier method?
the limits of the second integral are 0 and 1, not 0 and 2 as it is displayed
Let $V$ be a finite dimensional vector space over $\mathbb R$ and let $f$ and $g$ be two non-zero linear functionals on V such that whenever $f(x) ≥ 0$, we also have that $g(x) ≥ 0.$ Which of the following statements are true? a. $ker(f) ⊂ ker(g).$ b. $ker(f) = ker(g).$ c.$ f = αg$ for some $α > 0.$
This is what Artin in his book Algebra says about Gaussian integers:
I was wondering if the comment 'as many as four choices for $\gamma$' is for $\alpha=2+i$, or any nonzero Gaussian integer $\alpha$? Also, can we see that 'as many as four choices for $\gamma$' from this kind of geometric argument?
@MartinSleziak, Wow! Thank you so much! Those comments of yours made me think hard and allowed me to see what simple algebraic manipulations can do geometrically! (Sorry for my late appreciation, I am a slow learner.)
So, there can be one, two, three or four choices for $\gamma$, but there has to be at least one.
@Silent you can think of it in terms of (rational) integers; in the integers clearly $a$ and $-a$ have the same divisibility properties (for example they have the same prime factorisation, though this description only makes sense when your ring of integers is a UFD)
in fact two ring elements $a$ and $b$ are associates iff there exists a unit $u$ for which $a = ub$
Let $K$ be a compact Hausdorff space, and let $C(K)$ be the space of continuous complex valued functions equipped with the supremum norm. If one speaks of strong convergence in this context, does one mean convergence with respect to the supremum norm?
@user193319 My guess would be that if we're working with some normed space, strong convergence means convergence w.r.t. the given norm. (As opposed to weak* convergence, for example)
Of course, it's possible that this term is used also in some meanings I am not familiar with.
In Rudin's Functional Analysis, he states that the continuous dual of $C(K)$ equipped with the sup. norm can be identified with the space of all regular complex Borel measures on $K$. Does anyone have reference for the proof of this?
@user193319 This is equivalent to stating the pairing $C(K) \times M(K) \to \Bbb R$ taking a pair $(f, \mu)$ where $f \in C(K)$ and $\mu \in M(K)$ is a regular Borel measure, and spitting out $\int_K f d\mu$, is a nondegenerate pairing. This is the Riesz-Markov-Kakutani theorem
I believe the idea to show that $M(K) \to C(K)^*$ is surjective is, given any continuous functional $\phi : C(K) \to \Bbb R$, define $\mu(U)$ to be the supremum of $\phi(f)$ where $f$ varies over continuous functions with values in $[0, 1]$ which vanish outside of $U$ (so these approximate $\chi_U$, roughly speaking)
Then you extend $\mu$ over all Borel subsets to get the corresponding measure to $\phi$
It seems the information of the coefficient of the polynomial can be partially recovered somewhat from the resulting curve:
For example, the asymptotic behaviour of the polynomial is basically the same as when you truncate all the lower order terms. This can be easily explained as $x \to \infty$ then $cx^3+ax^2+bx \to cx^3$
which means, the coefficient of the highest order term can be recovered by inspecting the graph
Likewise, near the neighbourhood of the origin, the behaviour of the polynomial matches that of its corresponding linear term
meaning that the slope of the tangent of the polynomial at the origin will give you the coefficient for $x$
This means, if the polynomial has no linear term, then it will be flat locally at the origin. If in addition the polynomial has no constant term, then the origin will be a root
Now if the same logic holds for all other $x^n$, we might just have a way to predict where the roots of an arbitrary polynomial will be
The behaviour of the linear term can be proved as for all $n > 1$, pick any $x \in (-1,1)$ and the higher order terms will quickly go to zero, and only the linear and constant term survives
This desmos also showed how polynomials form a vector space. Since we knew the behaviour of the linear and highest degree term, note how that behaviour remains untouched when we add other terms of $x^n$
If we have some algorithm that can pin down the condition for $x$ such that $x^n$ is invariant, then one can in theory deconvolute any polynomial by inspection
If an integrand is unbounded on a given interval, can we say that its corresponding improper integral diverges? are there exceptions? (like odd functions on a [-a,a] interval?)
when can I be confident enough to say that it diverges?
@Secret Since you seem interested in infinities what are your thoughts on this?
so it is definitely not a well order, but what kind of linear order is to be investigated. My suspicion might be the product order $\prod_{i=1}^n \text{ord}(\Bbb{Z})$
@Simone I was responding to your question "If an integrand is unbounded on a given interval, can we say that its corresponding improper integral diverges?"
Would you say that a good strategy to find the convergence [divergence] of a given, maybe tough, integral is to find an integrand which is greater [smaller], and easier, at all points than the original one to then evaluate its convergence [divergence]?
how many subgroups of order $3$ does a nonabelian group of order $39$ have?
Attempt: $3 \times 13=39$. Let the number of such subgroups be $s$. So, $s \mid 13$ and $s=3k+1$ for some $k \in \mathbb{Z}$. $13$ is a prime and the only possible value that $s$ can take is $1$.
how many subgroups of order $3$ does a nonabelian group of order $39$ have?
Attempt: $3 \times 13=39$. Let the number of such subgroups be $s$. So, $s \mid 13$ and $s=3k+1$ for some $k \in \mathbb{Z}$. $13$ is a prime and the only possible value that $s$ can take is $1$.
It might not help much, but all I can say is that nearly everyone feels that way and it only gets worse the more you go down the rabbit hole. Try thinking about the progress you've already made rather than the progress you still have to make.
here's a question. A continuous function maps a compact metric spaces to another compact metric space. So, is the preimage of a compact metric space under a continuous function always compact?
My answer is "No". Take $f: \mathbb{R} \to [-1,1]$, where $f(x)=\sin(x)$.
Not sure if that's a good idea. You can get a lot of fundamentals about linear transformations essentially for free from corresponding facts in group theory. Not sure how Artin does it though.
However, the preimage of a compact set under a continuous mapping is compact when the domain is compact (in the context of metric spaces). Though, of course, this is not particularly revelatory.
Recall a space is totally disconnected if the only connected subsets are singletons (one-point subsets). Is a totally disconnected space, Hausdorff?
I think it is true since if $a $ and $b $ are two distinct points, they can be separated two disjoint open sets, since the main space is total...
Non T0: Anything can happen T0: At least one point can be separated by open sets T1: Points can be separated by open sets T2: There exists two disjoint open sets which points can be separated T2.5: There exists two disjoint compact set which points can be separated Complete T2: No open sets containing points overlap with each other T3: A closed set and a point can be separated by open sets T3.5: A closed set and a point can be separated by disjoint open sets T4: Closed sets can be separated by open sets
@RyanUnger If the Cantor space were discrete, it would be homeomorphic to $\mathbb{R}$ with the discrete topology (since any two discrete spaces with the same cardinality are homeomorphic). However, the Cantor space is compact whereas $\mathbb{R}$ with the discrete topology is not (consider the open cover of singletons), so this can't be.
which, kinda showing the limitation of the diagram as I tried to recover the Tx axioms by looking at the diagrams without reading the actual definitions, and get the wrong interpretations, it is suffice to say that my diagrams on T5, T6 are misleading
“Anything that happens, happens. Anything that, in happening, causes something else to happen, causes something else to happen. Anything that, in happening, causes itself to happen again, happens again. It doesn’t necessarily do it in chronological order, though.”
Statistical events: Anything that happens, has some chance of happening Quantum events: There are things that happens that follow the rules of linear algebra with interesting correlations Cauchy demons: There are things that happens without a source
@Mathein cool :) ich bekomme so 850€ im Monat vom DAAD und wenn ich mehr als 30% davon für Miete ausgeben muss zahlen sie so 250€ dazu, also such ich mir a schöne wohnung... hahaha
Below is a problem that I did. I would like somebody to check it for me.
Thanks,
Bob
Problem:
Suppose that $z_1$ and $z_2$ are two independent random variables that are normally distributed with
mean $2$ and standard deviation of $4$. Now if $z = |z_1 - z_2|$, what is the probability that $z$ ...
Recall a space is totally disconnected if the only connected subsets are singletons (one-point subsets). Is a totally disconnected space, Hausdorff?
I think it is true since if $a $ and $b $ are two distinct points, they can be separated two disjoint open sets, since the main space is total...
