Recent MathOverflow Questions

Continuous dual of a topological chain complex is chain equivalent to dual of a toplogical chain complex?

Math Overflow Recent Questions - Thu, 04/04/2019 - 17:18

I apologize in advance if this is a naive question.

Def: A topological chain complex is a chain complex of topological $\mathbb{R}$-vector spaces such that the boundary maps are continuous.

Let $C$ be a topological chain complex.

one can in a natural way consider the dual $C$* of the chain complex $C$ (ignoring the topology).

One can also consider the continuous dual of the topological chain complex $C^{*c}$ (because the boundary maps of $C$ are continuous)

Are $C$*, $C^{*c}$ chain equivalent? I would also like to know if the answer is yes after possibly adding some more conditions to ensure nothing pathological happens. For example, is the answer yes if one assumes further that all homologies of C are finite dimensional ?

Thank you,

How to apply fourier transform to this problem?

Math Overflow Recent Questions - Thu, 04/04/2019 - 17:10

I am struggling to figure out how to approach this problem. I've only solved a homogenous heat equation $u_t = u_{xx}$ using a fourier transform, where I can take the fourier transform of both sides then solve the general solution in fourier terms then inverse transform. However, since this question has extra terms I'm a little confused.

$$ u_{tt}(x,t) + 2u_t(x,t) = -u(x,t), -\infty < x < \infty, t> 0$$ $$u(x,t), as |x| \rightarrow \infty, t>0$$ $$u(x,0) = f(x), u_t(x,0)= g(x) , -\infty < x < \infty $$ Any advice and or guidance would be greatly appreciated. All I would do is take the fourier transform of all the terms but from there I don't think I know what to do.`

Seeking proof of Cuckoo Cycle Conjecture

Math Overflow Recent Questions - Thu, 04/04/2019 - 16:42

Allow me to introduce the Cuckoo Cycle Proof of Work, and conclude with a closely related Conjecture about random graphs, which I hope someone can find a proof for.

Cuckoo Cycle is named after the Cuckoo Hashtable, in which each data item can be stored at two possible locations, one in each of two arrays. A hash function maps the pair of item and choice of array to its location in that array. When a newly stored item finds both its locations already occupied, one is kicked out and replaced by the new item. This forces the kicked-out item to move to its alternate location, possibly kicking out yet another item. Problems arise if the corresponding Cuckoo graph, a bipartite graph with array locations as nodes and item location pairs as edges, has cycles. While n items, whose edges form a cycle, could barely be stored in the table, any n+1st item mapping within the same set of locations (a chord in the cycle) would not fit, as the pigeonhole principle tells us.

In the Proof-of-Work problem, we typically set n ≥ 29, and use edge indices (items) 0 .. N-1, where N = 2n. The endpoints of edge i are (siphash(i|0) % N, siphash(i|1) % N), with siphash being a popular keyed hash function. A solution is a cycle of length L in this Cuckoo graph, where typically L = 42.

Cuckoo Cycle solvers spend nearly all cycles on edge trimming; identifying and removing edges that end in a leaf node (of degree 1), as such edges cannot be part of a cycle. Trimming rounds alternate between the two node partitions.

The fraction fi of remaining edges after i trimming rounds (in the limit as N goes to infinity) appears to obey the

Cuckoo Cycle Conjecture: fi = ai-1 * ai, where a-1 = a0 = 1, and ai+1 = 1 - e-ai

fi could equivalently be defined as the fraction of edges whose first endpoint is the middle of a (not necessarily simple) path of length 2i. So far I have only been able to prove the conjecture for i ≤ 3. For instance, for i = 1, the probability that an edge endpoint is not the endpoint of any other edge is (1-1/N)N-1 ~ 1/e.

Here's hoping someone finds an elegant proof...

A Backtrack as a Single Word in a Group Presentation yields a Complex that isn't of the Same Homotopy Type?

Math Overflow Recent Questions - Thu, 04/04/2019 - 16:20

By "backtrack" I mean a subword of a relator in a group presentation of the form $x x^{-1}$.

Let $X = \langle a \rangle$ as a presentation complex.

Let $Y = \langle a$ | $aa^{-1} \rangle$ as a presentation complex.

Now we see that $X$ is a circle and $Y$ is a pinched torus, and these two spaces clearly do not have the same Homotopy Type as $\pi_2(Y)$ is nontrivial.

However it was said in "A Covering Space With no Compact Core" (MSN) by Daniel Wise that:

$\langle a, b, t $ | $ [a,b]^t = [a,b][b,a] \rangle$ is homotopy equivalent to $\langle a, b, t $ | $[a,b] \rangle$.

Is this always true when the backtrack is a proper subword of a relator? Is the above case with $X$ and $Y$ the only real nonexample?

Matroidal simplicial posets?

Math Overflow Recent Questions - Thu, 04/04/2019 - 15:32

A simplicial poset is a finite poset $P$ with minimial element $\hat{0}$ such that every interval $[\hat{0},x]$ is isomorphic to a Boolean lattice. Simplicial posets are generalizations of simplicial complexes (see, e.g., Now, one way to define a matroid is as a simplicial complex for which the restriction to any subset of vertices is a pure complex. We could then naively define a simplicial poset $P$ to be matroidal if $P\setminus F$ is always graded, where $F$ is the order filter generated by any subset of atoms of $P$.

Have these matroidal simplicial posets been studied at all? Are there conjectures (e.g., about face numbers) for them?

Centralizers of Cartan subgroups

Math Overflow Recent Questions - Thu, 04/04/2019 - 15:13

Let $E$ be an elliptic curve with CM by an order $\mathcal O$ in an imaginary quadratic field $K$. Choose a basis for $E[N]$ to get an isomorphism $\operatorname{Aut}(E[N])\cong \operatorname{GL}_2(\mathbb Z/N\mathbb Z) $. Complex multiplication on $E$ induces a homomorphism $(\mathcal O/ N\mathcal O )^\times\rightarrow \operatorname{GL}_2(\mathbb Z/N\mathbb Z)$. Let $C_N$ be the image.

Is $C_N$ equal to its centralizer in $\operatorname{GL}_2(\mathbb Z/N\mathbb Z)$?

This should occur in some cases, and should give rise to a rational point on certain modular curves. See Burcu Baran: Normalizers of non-split Cartan subgroups, modular curves, and the class number one problem, especially Proposition 4.1.

Lower bound for integral of a negativ part of a Brownian motion with respect to time

Math Overflow Recent Questions - Thu, 04/04/2019 - 14:38

My source states that $\int_0^1 (B_t)_{-} dt \geq |Z|$ with $Z = \int_0^1 B_t dt \sim N(0,1/3)$ and $(\cdot)_{-}$ ist the negativ part of $B_t$. If I think of a possible path of $(B_t)$ on $[0,1]$ where the Brownian Motion only takes positive values, the left hand side of the equation is zero whereas the right hand side is strictly positiv, hence the equation above doesn't hold for all possible $\omega$'s. Any ideas where my problem lies?

Furthermore they deduce that $\mathbb{P}\left( \int_0^1 (B_t)_{-}dt \leq \varepsilon\right) \leq \varepsilon$ holds for $\varepsilon >0$. That statement should readily follow from the above equation, meaning same argument schould hold for $|Z|$, since

$\mathbb{P}\left( \int_0^1 (B_t)_{-}dt \leq \varepsilon\right) \leq \mathbb{P}\left( | \int_0^1 B_t dt | \leq \varepsilon\right) = 2\Phi(\varepsilon/\sqrt(1/3)) -1 \overset{!}{\leq} \varepsilon$.

But the last equation doesn't hold for any $\varepsilon>0$. But $2|Z|$ seems to do the trick. The above equation can be found in link page 32 line 4

What exactly is and is not a concentration inequality?

Math Overflow Recent Questions - Thu, 04/04/2019 - 14:33

Hoeffding's inequality is surely a concentration inequality. It can be written in the form:

$\Pr(\bar X + f(n,\delta) \geq \mu) \geq 1-\delta,$

for some function $f$, where $X$ is a set of i.i.d. samples of a random variable with mean $\mu$, $\bar X$ is the sample mean, and $\delta \in [0,1]$. Maurer and Pontil presented an empirical Bernstein bound that uses the sample mean and sample standard deviation. It can be written in the form:

$\Pr(\bar X + f(X, n, \delta) \geq \mu) \geq 1-\delta,$

for some other function f that now depends on the sample $X$ (specifically, on the sample standard deviation of $X$).

Interestingly, Maurer and Pontil do not call their inequality a "concentration inequality". Notice that the width of the confidence interval it would provide on the mean depends on the samples, X. Similarly, Anderson (1969) presented a high-confidence bound on the mean that cannot (immediately) be formulated as a deviation from the sample mean. That is, its most natural form would be:

$\Pr(f(X, n, \delta) \geq \mu) \geq 1-\delta,$

for some other function f that one could (with work) phrase as the sample mean plus another term that depends on the sample, but which is not naturally in this form.

Which of these is a "concentration inequality"? Is Anderson's inequality not a concentration inequality because it is not clearly showing how the sample mean deviates from the expected value? Is even Maurer and Pontil's inequality not a concentration inequality because its confidence interval around the sample mean depends on the sample (and hence it cannot be written in the form Hoeffding's inequality is usually presented in)?

If Maurer & Pontil's inequality and/or Anderson's inequality are not concentration inequalities, what would you call them? Is the term "concentration inequality" sufficiently vague that it is up to personal interpretation whether each of these is a concentration inequality?

How to show if $X$ is Killing field then it is tangent to the geodesic spheres centred at a point $p$?

Math Overflow Recent Questions - Thu, 04/04/2019 - 14:31

Let $M$ be a Riemannianiam manifold with Levi-Civita connection and $X $ be a smooth vector field on $M$. Let $\phi : (-\epsilon, \epsilon) × V \to M$ be the local flow of $X$ in $M$.

Problem is- if $X$ is a killing field on $M $, $p \in M$ and $U$ be a normal neighborhood of $p$. If $p$ is the unique point satisfying $X(p)= 0$ then in $U$, $X$ is tangent to the geodesic spheres centred at $p$.

My attempt: With some hints, I've got from here and there, I know that first, we need to have that $\phi(t,p) = p \forall t\in(-\epsilon, \epsilon)$, which I am able to show.

Next, I think to show the required, it suffices to show that for any $q \in exp_{p}(S(0,\epsilon) = S_({p}(\epsilon))$, a geodesic sphere contained in $U$ which is of the form $q= exp_{p}(v)$ for some $v$ such that $|v|=\epsilon$, the radial vector joining $p$ to $q$ is orthogonal to $X(q)$.

But I don't know how to proceed to show this. Also, where do we use the uniqueness of point $p$ as the only zero of $X$?

It would be lovely if anyone could provide me some hint to solve this. Thank you!!

Compactifications and open sets with discrete boundaries

Math Overflow Recent Questions - Thu, 04/04/2019 - 14:30

A result here shows that every separable metric space having a basis of open sets with scattered boundaries has a rational metrizable compactification. Rational means having a basis of open sets with countable boundaries.

In my new paper I construct an unusual plane set which has a basis of open sets with discrete (but infinite) boundaries. I am wondering if such spaces have compactifications that are "very rational" in some sense. For instance, if $X$ is such a space, then is it possible to compactify $X$ so that the closure of any given basic discrete boundary set is a convergent sequence? Is a rational compactification of $X$ easy to construct?

Triangle perimeter & area [on hold]

Math Overflow Recent Questions - Thu, 04/04/2019 - 14:15

We have a triangle ABC:
a = 25cm
c = 30cm
γ = 2α


What is the perimeter & the area of the triangle?

Simultaneous diagonalization of the tensor products of Dirac gamma matrices

Math Overflow Recent Questions - Thu, 04/04/2019 - 14:07

Let $\gamma_i\ (i=1,2,\ldots N)$ be the Dirac gamma matrices satisfying the Clifford algebra $$\gamma_i\gamma_j+\gamma_j\gamma_i=2\delta_{ij} I\ \ (i,j=1,2,\ldots,N).$$ Then the tensor products $\gamma_i\otimes\gamma_i$ commute with each other: $$[\gamma_i\otimes\gamma_i, \gamma_j\otimes\gamma_j]=0.$$ This means all of $\gamma_i\otimes\gamma_i$ can be simultaneously diagonalized by some similarity transformation $S$ (probably not unique). I wonder what the characterizations of such diagonalization are, such as the arrangement of eigenvalues of $\gamma_i\otimes\gamma_i$ after diagonalization, the (fast) computational algorithm for $S$ etc. I find surprisingly little about this elementary construction online, any reference will also be appreciated.

Intersection of two quadrics in $\mathbb{R} P^5$

Math Overflow Recent Questions - Thu, 04/04/2019 - 13:08

Is there an ''algorithmic'' way to get that intersection of two quadrics $$x_1 y_1 -x_2 y_2 - z_1^2+z_2^2=0$$ and $$x_2 y_1 + x_1 y_2 -2z_1z_2=0$$ inside $\mathbb{R}P^5[x_1:x_2:y_1:y_2:z_1:z_2]$ is isomorphic $\mathbb{R}P^3?$ By ''algorithmic'' I mean similarily to statements like ''degree d curve in $\mathbb{C}P^2$ has genus (d-1)(d-2)/2.''

Do we really know that the axiom of regularity is true? [on hold]

Math Overflow Recent Questions - Thu, 04/04/2019 - 12:28

This website is for math professionals but I do not have a job of doing mathematical research so I think that means that on this website, it's very easy for me to figure out how to ask a question that belongs but super hard for me to figure out how to write an answer that belongs, so can you please write an answer that I can actually understand even though I don't have that kind of job.

According to the way I think of it, a formal system is not just a set of rules of how to deduce some strings of characters from other strings of characters. It also includes the information about the meaning of each string of characters that represents a statement. I think the intended meaning of the formal system of ZF - Regularity is pretty clear in this answer. If we just accept that that meaning of ZF - Regularity is a true model of set theory, then what that answer really claims we can prove is that if you add as an assumption, the axiom of regularity, then you can prove statements with that assumption that you cannot prove without it, giving you a supertheory of ZF - Regularity which is therefore a dubious model of set theory, but you can also have a subtheory of ZF - Regularity whose formal system is isomorphic to the formal system of the supertheory. What I mean by the axiom of regularity is the statement of the axiom of regularity in ZF the the meaning of the axiom the way it's written, if the intended meaning of the formal system of ZF is the supertheory of ZF - Regularity and not the subtheory. By that definition, if the supertheory is dubious, then then maybe the axiom of regularity which it proves is also dubious.

Just because the formal system of ZF is written like a supertheory of ZF - Regularity doesn't mean the actual supertheory is a true model of set theory. Maybe it's written that way because some people actually do strongly believe the supertheory itself because they find Quine atoms so counter-intuitive and we can't disprove the axiom of regularity either. That doesn't mean the supertheory is actually true. Some people think of the meaning of the formal system of ZF as being the subtheory of ZF - Regularity and not the supertheory so maybe the supertheory actually is a false model of set theory. What's not dubious is the theorem of the subtheory of ZF - Regularity that the axiom of regularity plays the role of.

$\lambda$-invariants in cyclotomic $\mathbb{Z}_p$ extensions

Math Overflow Recent Questions - Thu, 04/04/2019 - 12:09

The idea that Selmer groups and class groups are related is not new. More recently, we understand that the growth patterns of fine Selmer groups are very similar to that of class groups in cyclotomic $\mathbb{Z}_p$-extensions of number fields. There are results where the classical $\mu$-invariant has been compared with the $\mu$-invariant of the (fine) Selmer group (eg Coates-Sujatha 2005, Kestutis Cesnavcius 2015).

Has there been any work done in comparing the $\lambda$-invariants? My understanding is that Greenberg conjectured that for totally real fields $\lambda=0$. This I think is still an open problem. Are there similar conjectures for $\lambda$-invariants for (fine) Selmer groups?

Hachimori-Matsuno had worked out an analogue of Kida's formula for Selmer groups, and therefore can give an explicit formula for $\lambda(F_{cyc}/F)$ in terms of $\lambda(K_{cyc}/K)$ where $F/K$ is a $p$-power extension. They showed that in some cases, the growth $\lambda$-invariant of Selmer groups and the class group have a similar pattern. Are there any other such results?

How to "cite" MathOverflow answer? [migrated]

Math Overflow Recent Questions - Thu, 04/04/2019 - 11:22

I am writing a paper at the moment and for one proof I use an idea which I got from a MathOverflow answer (at the time I was asking idle questions but later of course found a way to use it). I of course feel uncomfortable just giving the idea without credit but I'm not really sure how to do it. What's the convention here? You ask if it's ok to just thank them informally in the paper or....???

Finding the average distance between all points [on hold]

Math Overflow Recent Questions - Thu, 04/04/2019 - 10:54

We have area NxN size. (Ex: For N=8 we have standard Chess Board)

What is the average of all shortest distances between all fields? For simplicity, we accept the floor of the result. What is the result if we measure in chebyshev distance, and what if it is manhattan distance.

I have no idea how to start even, or Where I should search for a clue. (Besides brute-force of course). I tried already with randomly picking two points N^2 times, but the approximation is too big.

Graded commutativity of the $n$th Browder bracket

Math Overflow Recent Questions - Thu, 04/04/2019 - 08:46

Let $\mathcal{O}$ be a topological operad and $X$ an algebra over $\mathcal{O}$. Then $H_*(X)$ is an algebra (in the category of $\mathbb{Z}$-graded $R$-modules) over $H_*(\mathcal{O})$. Each $e\in H_n(\mathcal{O}(2))$ gives us a graded product $$H_p(X)\otimes H_q(X)\to H_{p+q+n}(X),a\otimes b\mapsto e(a\otimes b).$$ Now $\mathfrak{S}_2$ acts on $H_*(\mathcal{O}(2))$ on the right and on $H_*(X)^{\otimes 2}$ by $\tau_*(a\otimes b) = (-1)^{ab}(b\otimes a)$. By equivariance, we get $$e(b\otimes a) = (-1)^{ab} (\tau^*e)(a\otimes b).$$ Now consider the special operad $\mathcal{C}_{n+1}$ of little $(n+1)$-cubes. Here, $\mathcal{C}_{n+1}(2)\simeq \mathbb{S}^n$, so its $n$th homology is generated by the fundamental class $s$ of the submanifold $$\mathbb{S}^n\to \mathrm{Conf}_2(\mathbb{R}^{n+1}), x\mapsto (0,x).$$ Now it is clear that $\tau^*s=(-1)^{n+1}s$ as $x\mapsto (x,0)$ is homotopic to $x\mapsto (0,-x)$. Cohen defines the Browder bracket by $[a,b]:=s(a\otimes b)$. By the above argumentation, we would get $$[b,a]=s(b\otimes a)=(-1)^{ab}(\tau^*s)(a\otimes b)= (-1)^{ab+n+1}[a,b].$$ However, in both Cohen’s paper The homology of $\mathcal{C}_{n+1}$-spaces (1976) and in May’s Operads, algebras and modules, the sign is different, namely $$[b,a] = (-1)^{ab+1+n(a+b+1)}[a,b].$$ It is clear that they don’t coincide for $|a|+|b|$ odd. What am I missing?

hook-length formula: "Fibonaccized": Part II

Math Overflow Recent Questions - Tue, 04/02/2019 - 19:55

This is a natural follow-up to my previous MO question, which I share with Brian Hopkins.

Consider the Young diagram of a partition $\lambda = (\lambda_1,\ldots,\lambda_k)$. For a square $(i,j) \in \lambda$, define the hook numbers $h_{(i,j)} = \lambda_i + \lambda_j' -i - j +1$ where $\lambda'$ is the conjugate of $\lambda$.

The hook-length formula shows that if $\lambda\vdash n$ then $$n!\prod_{\square\,\in\,\lambda}\frac1{h_{\square}}$$ counts standard Young tableaux whose shape is the Young diagram of $\lambda$.

Recall the Fibonacci numbers $F(0)=0, \, F(1)=1$ with $F(n)=F(n-1)+F(n-2)$. Define $[0]!_F=1$ and $[n]!_F=F(1)\cdot F(2)\cdots F(n)$ for $n\geq1$.

QUESTION. What do these integers count? $$[n]!_F\prod_{\square\,\in\,\lambda}\frac1{F(h_{\square})}.$$

A Matrix Inequality for positive definite matrices

Math Overflow Recent Questions - Tue, 04/02/2019 - 03:31

Let $X$ and $Y$ be positive semi-definite self-adjoint complex matrices of same finite order. The, is it true that $|X-Y|\leq X+Y$ where for any matrix $A$, $|A|$ is defined to be $|A|:=(A^*A)^{\frac{1}{2}}$ ?

PS. I think the answer is No. But I could not find any counterexample!


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