# Recent MathOverflow Questions

### When is a totally bounded set of an inductive limit contained in a component of this limit?

Math Overflow Recent Questions - Mon, 04/01/2019 - 23:37

A. P. Robertson and W. Robertson in their "Topological Vector Spaces" VII, 1.4, (and H.Jarchow in "Locally convex spaces", 4.6, Theorem 2) prove the following proposition:

Let $E=\lim_{n\to\infty}E_n$ be an inductive limit of a sequence of locally convex spaces $\{E_n;\ n\in {\mathbb N}\}$, and suppose that for each $n\in{\mathbb N}$

1) the topology of $E_n$ is induced from $E_{n+1}$ (in this case $E$ is called a strict inductive limit of $E_n$), and

2) $E_n$ is closed in $E_{n+1}$.

Then each bounded set $B$ in $E$ is contained in some $E_n$.

I wonder if the following modification of this statement is true:

Let $E=\lim_{\nu\to\infty}E_\nu$ be an inductive limit of a net of locally convex spaces $\{E_\nu;\ \nu\in I\}$ (where $I$ is directed, but not necessarily countable), and suppose that for each $\mu\le\nu\in I$

1) the topology of $E_\mu$ is induced from $E_\nu$, and

2) $E_\mu$ is closed in $E_\nu$.

Then each totally bounded set $B$ in $E$ is contained in some $E_\nu$.

Or perhaps some other modifications hold (where "bounded" is replaced by "totally bounded" and the other conditions can be weakened instead)?

### Is this a new Fibonacci Identity?

Math Overflow Recent Questions - Mon, 04/01/2019 - 16:34

I have found the following Fibonacci Identity (and proved it).

If $F_n$ denotes the nth Fibonacci Number, we have the following identity $$F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1}$$ where $F_1 = F_2 = 1$, $r \leq n$, $h \leq g$, and $n, g, k \in \mathbb{N}$.

It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few.

I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input!

### Percentages question [on hold]

Math Overflow Recent Questions - Mon, 04/01/2019 - 16:33

An alloy of brass is composed of 84.6% copper, 5.20% tin, 6.70% lead, and zinc. The number of pounds of zinc required to make 520.0 pounds of alloy is "_____"

Obviously this is a homework question for my trade school. What I don't get about this is the wording. The zinc equals 3.5% of an alloy from what I put together. I don't understand though. Do these percentages amount to the percentages of a pound of the named material? The way it's worded doesn't seem to make sense to me. If I did this the way my brain is trying to put it together, the answer would be 148.5714285714286. Can somebody fact check this for me?

### augmentation ideal and $\rho$-isotopic spaces

Math Overflow Recent Questions - Mon, 04/01/2019 - 15:57

These are two questions regarding Rubins paper - Global units and Ideal class groups in Inventiones 1987. The two questions are from section 3.

Let $p$ be a rational prime and $K \subset E \subset F$, where $F$ is an abelian extension of $K$ containing the Hilbert Class Field of $K$, with $G=Gal(F/K)$. Let $\rho$ be an irreducible $\mathbb{Z}_p$-representation of $\Delta = Gal(F/E)$. For any $\mathbb{Z}[\Delta]$-module $M$, write $M^{\rho}$ for $\rho$-eigenspace of $\hat{M}= \lim M/p^n M$. Explicitly, $M^{\rho}= e_{\rho}\hat{M}$, where $e_{\rho}$ is the $\rho$-idempotent. Let $W$ be a submodule of $(O_F^*)^{\rho}$ such that $(O_F^*)^{\rho}$ has no $\mathbb{Z}_p$ torsion. Let $N=p^n$.

Since $\rho \ne 1$, I have two questions:

1) Why is $(\mathbb{Z}/N\mathbb{Z})[G]^{\rho}$ contained in the augmentation ideal of $(\mathbb{Z}/N\mathbb{Z})[G]$? I couldnt understand the action of $\rho$ on a $(\mathbb{Z}/N\mathbb{Z})[G]$-module $M$.

2) Let $V=W/W^N$. Why is $O_K^* \cap V = 0$?

### How to calculate gross annual income from net monthly income [on hold]

Math Overflow Recent Questions - Mon, 04/01/2019 - 14:50

Unsure if this is the appropriate forum, but hoping someone can help. I have been able to calculate net monthly income from gross annual, but struggling to reverse it so I can calculate gross annual from net monthly. Given the variables below, does anyone know how the equation should look?

• taxRate1 applied to amounts up between 0 and taxBand1.
• taxRate2 applied to amounts between band 1 and 2.
• taxRate3 applied to amounts between band 2 and 3.
• taxRate 4 applied to amounts above band 3.

taxBand1 = 12500 taxBand2 = 45000 taxBand3 = 150000 - taxRate1 = 0% taxRate2 = 20% taxRate3 = 40% taxRate4 = 45%

### Monogenic suborders of irreducible cubic orders

Math Overflow Recent Questions - Mon, 04/01/2019 - 14:44

By an order of rank $n$ we mean a unital ring $R$ which is isomorphic to $\mathbb{Z}^n$. An order $R$ is irreducible if it is isomorphic to a subring of a degree $n$ number field $K$ and the fraction field of $R$ is equal to $K$. An order $R$ is monogenic if it is of the form $\mathbb{Z}[\alpha]$ for some $\alpha \in R$.

Let $R$ be an irreducible cubic order. Then the famous Delone-Faddeev correspondence states that $R$ is canonically associated with the $\text{GL}_2(\mathbb{Z})$-equivalence class an integral binary cubic form $F_R$.

We then have the following theorem: Let $p$ be a prime. Then an irreducible cubic order $R$ contains a monogenic suborder of index $p$ if and only if $p$ is representable by the form $F_R$.

What is the corresponding statement for composite $m$? Is it still true that representation by $F_R$ characterized the existence of monogenic suborders of index $m$?

### Story of Grothendieck's Prime Number

Math Overflow Recent Questions - Mon, 04/01/2019 - 14:31

I asked this question earlier, at hsm.stackexchange.com without much luck. Maybe somebody can answer it here.

There is a story about Alexander Grothendieck and the "Grothendieck Prime" 57, which goes roughly as follows (cf. this wikipedia article):

In a mathematical conversation, someone suggested to Grothendieck that they should consider a particular prime number. “You mean an actual number?” Grothendieck asked. The other person replied, yes, an actual prime number. Grothendieck suggested, “All right, take 57.”

This quote is taken from Allyn Jackson's article "Comme Appelé du Néant— As If Summoned from the Void: The Life of Alexandre Grothendieck". Jackson refers to the story as a "legend". One can argue that the story is quite believable given Grothendieck's way of thinking (David Mumford: "He (Grothendieck) doesn’t think concretely").

Question. What (if any) is the factual basis of the story?

For instance, when/where did it happen? (In different versions it is said to have happened during or after a Grothendieck's talk.) Did anybody hear this story from somebody present at Grothendieck's talk?

My guess is that the story is just a legend, but I could be mistaken.

The story is not made up: Grothendieck did make that silly blunder, in an exchange after a talk, after being asked to be more concrete by a member of the audience. Of course this doesn't change anything to the fact that Grothendieck was one of the most profound arithmeticians of the 20th century. And indeed 57 looks a bit prime for some psychological reason :-) . Conversely many mathematicians think I'm pulling their leg when I tell them that 4999 is prime! ... I've heard this story a long time ago. I think it is true but I can't prove it since, alas, most of the protagonists are dead. Anyway, this is just an amusing but quite meaningless anecdote: a genius made a lapsus linguae. So what? On the other hand I'm quite sure that Allyn Jackson can't disprove what he (strangely) calls a legend...

In contrast, here is what Franz Lemmermeyer said at HSM:

There is a similarly inane story about Kummer not being able to work out 7⋅9, or about Gauss on the number of needles on a Christmas tree. I have no idea why people continue to disseminate such stories without giving a reference - this is a very idiotic behavior even if you call the story a legend.

### Gauss' Posthumous Publications?

Math Overflow Recent Questions - Mon, 04/01/2019 - 13:56

I'm looking for any information about the posthumous publication of Gauss' mathematical correspondence and notebooks.

When did these become widely available, and how did it affect progress in mathematics?

### Function fitting rounded values

Math Overflow Recent Questions - Mon, 04/01/2019 - 13:41

I have a problem, I have a set of points in R2, and I want to find a function that can fit those points (f such that f(x) =y for all (x,y)), the problem is that the y values are rounded, and therefore it's impossible to do it with a standard curve fitting technique. The solution I've reached is to make a plot that shows the (x, y) and (x, y+1) (I would like to post a picture, but for the moment I'm not allowed to do it) And try to find a curve within those boundaries, therefore find a curve such that f(x) is between y and y+1. And basically my question is, Is there a way to find such function?

### Finitely generated matrix groups whose eigenvalues are all algebraic

Math Overflow Recent Questions - Mon, 04/01/2019 - 13:06

Let $G$ be a finitely generated subgroup of $GL(n,\mathbb{C})$. Assume that there exists a number field $k$ (i.e. a finite extension of $\mathbb{Q}$) such that for all $g \in G$, the eigenvalues of $g$ all lie in $k$. This implies that $g$ is conjugate to an element of $GL(n,k)$.

Question: must it be the case that some conjugate of $G$ lies in $GL(n,k)$? Or at least $GL(n,k')$ for some finite extension $k'$ of $k$? If this is not true, what kinds of assumptions can I put on $G$ to ensure that it is?

### if A is an invertible matrix and B is an invertible matrix, is A+B also an invertible matrix? [migrated]

Math Overflow Recent Questions - Mon, 04/01/2019 - 12:07

the answer is obviously false. for example: $$A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1/2 \\ \end{pmatrix} A^{-1}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1/2 & 0 \\ 0 & 0 & 2 \\ \end{pmatrix} \\ B=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1/2 \\ \end{pmatrix} B^{-1}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1/2 & 0 \\ 0 & 0 & -2 \\ \end{pmatrix} \\ A+B=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}$$ A+B dose not have an inverse matrix(rank is not n). is there a better way to prove it other then an example?

### Do kissing numbers with distance $d$ always grow polynomially or exponentially in dimension?

Math Overflow Recent Questions - Mon, 04/01/2019 - 11:38

Let $A_d(n)$ be the largest number of points that can be packed on the $n$-unit sphere, such that every point is at least $d$ apart. Compare with, for instance, https://arxiv.org/abs/1507.03631

When d=1, these are the standard kissing numbers, and they are known to grow exponentially in n. For any smaller d we can see that it must be exponential as well, because there is a simple exponential upper bound based on solid angles subtended by $A$-many circles on the surface.

For d=√2, these are vectors that are "at least" orthogonal (nonpositive inner product), and $A_d(n)$ is just 2n, in an octahedral packing. This implies that for any d>√2, the growth in n is at most linear.

On a smaller note, for d≥√3, A_d(n) is a constant in n.

Question: What happens for values of $d \in (1,\sqrt 2)$? Is it true that $A_d(n)$ is always either asymptotically linear, or at least exponential? If so, is the transition at d=√2 exactly?

And a side question, is $A_d(n)$ always either asymptotically linear or constant on $d\in(\sqrt 2,\sqrt 3)$?

### Derived maximal separated quotient

Math Overflow Recent Questions - Mon, 04/01/2019 - 11:11

Let $R$ be a Noetherian ring, $I \subset R$ a (finitely generated) ideal.

Let $M$ be an $R$-module equipped with the $I$-adic topology.

One can form its maximal Hausdorff quotient, which amounts to taking the quotient of $M$ by the $R$-submodule given by the closure of $(0)$ in the $I$-adic topology.

Is there an analog of this operation that works for complexes, or complexes up to quasi-isomorphism?

Namely, let $K^{\bullet}$ be a bounded complex of $R$-modules, with each $K^p$ equipped with the $I$-adic topology, and with continuous open differentials.

Is there a complex $K^{\bullet}_{\rm sep}$, functorial in $K^{\bullet}$, and such that its homology is the maximal Hausdorff quotient of the homology of $K^{\bullet}$?

### Essential Supremum and Supremum inside Expectation

Math Overflow Recent Questions - Mon, 04/01/2019 - 10:25

Suppose that $\{Z_i\}_{i \in I}$ are a family of densities in $L^2(\Omega,\mathcal{F},\mathbb{P})$, and $X=L^2(\Omega,\mathcal{F},\mathbb{P})$. When is it true that $$\sup_{i \in I} \mathbb{E}\left[Z_i\cdot (X- \mathbb{E}[Z_i\cdot X|\mathcal{G}])^2 \right] = \sup_{i \in I} \mathbb{E}\left[Z_i\cdot (X- \operatorname{esssup}_{i \in I}\mathbb{E}[Z_i\cdot X|\mathcal{G}])^2 \right] ?$$

I've seen similar questions asked, but I haven't come across something like this on here yet.

### When do Kan extensions preserve colimits?

Math Overflow Recent Questions - Mon, 04/01/2019 - 10:12

Assume that we have a pair of functors $Y:A \to B$ and $F:A \to C$ where $A$ is an essentially small category, $B,C$ are cocomplete categories and $Y,F$ preserve colimits. Assume also that for some reasons we know that both the left and the right Kan extensions of $F$ along $Y$ exist.

Question: When are they cocontinuous?

I know that if $Y$ is the Yoneda embedding, then the left Kan extension is cocontinuous, but this is not a case I am interested in, since the Yoneda embedding is not cocontinuous in general.

I also know from this question that $Lan_Y(F)$ preserves any colimit preserved by each representable functor $B(Yx,−): B \to \mathsf{Set}$ for $x \in A$, but it seems to me that in the context I am interested in, this is not the case.

I am not familiar with these constructions, can anyone give me some hints, please?

In the particular situation I am interested in $Y:A\to \left(\mathsf{Set}^A\right)^{op}$ is the "covariant Yoneda embedding".

### Reference request: conditions for the cardinality of the kernel of a linear map from $\mathbb{Z}_m^n \to \mathbb{Z}_m^k$ is a power of $m$

Math Overflow Recent Questions - Mon, 04/01/2019 - 09:19

Let $\mathbb{Z}_m = \mathbb{Z}/m\mathbb{Z}$. Let $A$ be an $k \times n$ matrix over $\mathbb{Z}_m$. Let $f: \mathbb{Z}_m^n \to \mathbb{Z}_m^k$ be a linear map defined by $f(x) = Ax$, $x \in \mathbb{Z}_m^n$. Are there some references about the condition such that $|\ker(f)|=m^p$ for some positive integer $p$?

For example, the kernel of the map $f: \mathbb{Z}_4 \to \mathbb{Z}_4$ given by $f(x) = 2x$ is $\ker(f) = \{0, 2\}$. Therefore $|\ker(f)|$ is not a positive integer power of $4$.

Let $f: \mathbb{Z}_m^4 \to \mathbb{Z}_m^2$ be a linear map given by $f(x) = (2x_1-x_3-x_4, x_2-x_3-x_4)^T$. In this case, $|\ker(f)| = m^2$. Thank you very much.

### Highest weight of a representation of a Lie Algebra

Math Overflow Recent Questions - Mon, 04/01/2019 - 08:54

Given a Lie Algebra $\mathfrak{g}$, its Cartan Matrix $A$ and a finite representation $R$, is there a way of determining its highest weight $\Lambda$ in a simple way?

In my course, we consider $\mathfrak{g}=A_2= \mathfrak{L}_{\mathbb{C}}(SU(3))$. It is stated that the highest weight of the fundamental representation has Dynkin labels $\Lambda = (1,0)$ and the highest weight of the adjoint representation has Dynkin labels $\Lambda = (1,1)$. Why is it so? From there, I can work out the other roots by removing weights given by the Cartan Matrix but it is of no use if I can't compute the highest weight in the first place.

Taking an example, let $\mathfrak{g}=B_2= \mathfrak{L}_{\mathbb{C}}(SO(5))$. How do I work out the highest weight for the fundamental and adjoint representation?

### On the determinants $\det\left[(i\pm j)\left(\frac{i\pm j}p\right)\right]_{1\le i,j\le(p-1)/2}$

Math Overflow Recent Questions - Mon, 04/01/2019 - 07:35

Let $p$ be an odd prime and define $$D_p^+:=\det\left[(i+j)\left(\frac{i+j}p\right)\right]_{1\le i,j\le(p-1)/2}$$ and $$D_p^{-}:=\det\left[(i-j)\left(\frac{i-j}p\right)\right]_{1\le i,j\le(p-1)/2},$$ where $(\frac{\cdot}p)$ is the Legendre symbol.

QUESTION. Is my following conjecture true? How to solve it?

Conjecture. Let $p>5$ be a prime. If $p\equiv1\pmod4$, then $$\left(\frac{D_p^+}p\right)=1=\left(\frac{D_p^-}p\right).$$ If $p\equiv3\pmod4$, then $p\nmid D_p^+D_p^-$. Moreover, when $p\equiv7\pmod8$ we have $$\left(\frac{D_p^-}p\right)=(-1)^{(h(-p)-1)/2},$$ where $h(-p)$ denotes the class number of the imaginary quadrtic field $\mathbb Q(\sqrt{-p})$.

Remark. (i) I have verified the conjecture for all primes $5<p<2000$.

(ii) For any prime $p\equiv1\pmod4$, clearly $D_p^-$ is a skew-symmetric determinant and hence it is an integer square by a result of Cayley. But I'm unable to show that $p\nmid D_p^-$ for all primes $p>5$ with $p\equiv 1\pmod4$.

In my opinion, the conjecture looks not so difficult. Your comments towards its solution are welcome!

### Is the Perron-Frobenius dimension of a G-Set given by its cardinality?

Math Overflow Recent Questions - Mon, 04/01/2019 - 05:51

Given a ring $R$ with finite additive basis $\{e_i\}_{i=1}^{n}$, such that $e_i e_j=\sum c_{ijk}e_k$ with $c_{ijk}\in \mathbb{N}$, we define the Perron-Frobenius dimension $FPDim(e_i)$ of a basis element $e_i$ to be the maximal positive real eigenvalue of matrix $M_{e_i}$, multiplication by $e_i$. This exists by the Perron-Frobenius theorem, and we extend by linearity to all of $R$.

Now take $B(G)$ to be the burnside ring of a finite group $G$, with basis given by isomorphism classes of transitive actions of $G$. One can directly check for $G\cong C_p$ that $FPDim(X)=|X|$. Does this hold in general for finite $G$-sets?

Note that our ring $B(G)$ is not necessarily transitive in the sense of Etingof's Tensor Categories (Definition 3.3.1), so this doesn't seem to follow immediately from the results in there.

### If $A$ is a dissipative self-adjoint operator with spectral decomposition $(H_λ)$, then $e^{tA}x$ tends to the projection of $x$ onto $H_0$ as $t→∞$

Math Overflow Recent Questions - Mon, 04/01/2019 - 04:50

Let $(T(t))_{t\ge0}$ be a strongly continuous contraction semigroup on a $\mathbb R$-Hilbert space $H$ with dissipative self-adjoint generator $(\mathcal D(A),A)$. In particular, $T(t)$ is self-adjoint for all $t>0$. By the spectral theorem, $$T(t)=e^{tA}\;\;\;\text{for all }t\ge0.\tag1$$ Let $(H_\lambda)_{\lambda\ge0}$ be the spectral decomposition related to $(\mathcal D(A),-A)$ (see, for example, Definition 1.8.1 and Theorem 1.8.2 on page 23 here) and $E_\lambda$ denote the orthogonal projection of $H$ onto $H_\lambda$. Using the spectral theorem, I was able to show that $$\lim_{t\to\infty}\left\|T(t)x\right\|_H=\left\|E_0x\right\|_H\;\;\;\text{for all }x\in H\tag2.$$

How can we conclude that we even got $$\left\|T(t)x-E_0x\right\|_H\xrightarrow{t\to\infty}0\tag3$$ for all $x\in H$?

I know that in a Hilbert space, convergence is equivalent to weak convergence together with convergence of the norms. So, we would be done if we could show that $$\langle T(t)x,y\rangle_H\xrightarrow{t\to\infty}\langle E_0x,y\rangle\tag4\;\;\;\text{for all }x,y\in H.$$ If this is the correct approach, how can we show that?