Recent MathOverflow Questions

On Euler's polynomial $x^2+x+41$

Math Overflow Recent Questions - Mon, 06/10/2019 - 08:28

This is an elementary question about something way outside my area of expertise. A well-known observation due to Euler is that the polynomial $P(x)=x^2+x+41$ takes on only prime values for the first 40 integer values of $x$ starting with $x=0$, namely the values $41,43,47,53,61,71,83,\cdots,1601$. In particular this gives a rather long sequence of primes such that the differences between successive terms form an arithmetic progression, namely $2,4,6,\cdots$, which is a consequence of $P(x)$ being quadratic. All this is related to the fact that the discriminant of $x^2+x+41$ is $-163$ and the field ${\mathbb Q}(\sqrt{-163})$ has class number 1.

Suppose one asks about the next 40 values of $P(x)$ after the value $P(40)=41^2$. We have $P(41)=1763=41\cdot 43$, also not a prime. After this the next two values $P(42)=1847$ and $P(43)=1933 $ are primes. Then comes $P(44)=2021=43\cdot 47$, then four primes, then $P(49)=2491=47\cdot 53$, then six primes, then $P(56)=3233=53\cdot 61$, then eight primes, then $P(65)=4331=61\cdot 71$, then ten primes, then $P(76)=5893=71\cdot 83$. The next four values are prime as well for $x=77,\ 78,\ 79,\ 80$, completing the second forty values. But then the pattern breaks down and one has $P(81)=6683=41\cdot 163$. Thus, before the breakdown, not only do we get sequences of $2,\ 4,\ 6,\ 8,\ 10$ primes but the non-prime values are the products of two successive terms in the original sequence of prime values $41,\ 43,\ 47,\ 53,\ 61,\ \cdots$. There is a simple explanation for this last fact, the easily-verified identity $P(40+n^2)=P(n-1)P(n)$, so when $n=1,2,3,\cdots$ we get $P(41)=P(0)P(1)=41\cdot 43$, $P(44)=P(1)P(2)=43\cdot47$, etc. However, this does not explain why the intervening values of $P(x)$ should be prime.

I've done an online search to find where this might be discussed, without success, so my question is, what is a reference for this curious behavior of the second 40 values of $P(x)$ (or anything related)? I'm also a little puzzled by the identity for $P(40+x^2)$, though perhaps it comes from the norm function in ${\mathbb Q}(\sqrt{-163})$.

Space derivative of flow of ODE with monotone source

Math Overflow Recent Questions - Mon, 06/10/2019 - 06:49

Consider the ODE $$ \begin{cases} \partial_t\Phi(t,x) = f(t,\Phi(t,x)), &\ t>0, \ x \in \mathbb R \\ \Phi(0,x) = x, & x \in \mathbb R \end{cases} $$ where $f$ is function which is a non-increasing in the second variable (without other assumptions on regularity).

Then $\Phi$ exists and is Lipschitz with respect to space (Flow of ODE with monotone source).

How can one compute the a.e. space derivative of this Lipschitz flow $\Phi(t, \cdot)$?

Remark. Note that, if $f$ was Lipschitz, we would get that the space derivative of the flow $\partial_x \Phi$ satisfies

$$\partial_t \partial_x \Phi = \partial_x f(t,\Phi(t,x))\partial_x \Phi.$$

To reiterate, the question of this post is the following:

In general, how can we compute $\partial_x \Phi(t,\cdot)$ if we only assume that $f$ is function which is a non-increasing in the second variable (without other assumptions on regularity of $f$)?

$k$-schemes isomorphic as abstract schemes

Math Overflow Recent Questions - Mon, 06/10/2019 - 05:46

Let $k$ be a perfect field. Given a connected scheme $X$, denote by $\mathrm{Aut}(X)$ the group of automorphisms of $X$ and denote by $\mathrm{Map}_k(X)$ the set of smooth proper morphisms $X\rightarrow \mathrm{Spec}\:k$. The former acts on the latter. Suppose the latter is non-empty, under what additional assumptions is the action transitive?

If $k=\mathbb{C}$ and $X$ is an elliptic curve that has $j$-invariant $i$ under one of the morphisms $X\rightarrow \mathrm{Spec}\:k$, then I think it is not transitive.

The number of rational semisimple conjugacy class/the Arthur-Selberg trace formula

Math Overflow Recent Questions - Sun, 06/09/2019 - 21:52

I was trying to understand a statement in Theorem 1.5 of this where the author seems to imply that if $G$ is a reductive group over $\mathbb{Q}$ such that $G/Z(G)$ is anisotropic, then for any function $f\in C^\infty_c(G(\mathbb{A}_\mathbb{Q}))$ the number of conjugacy classes $[\gamma]\in G(\mathbb{Q})/\sim$ such that $O_\gamma(f)\ne 0$ is finite. This led me to suspect that maybe the following is true/was what was meant:

Question: Let $G$ be a reductive group over $\mathbb{Q}$ which is anisotropic. Let $C\subseteq G(\mathbb{A}_\mathbb{Q})$ be a compact subset. Is the set of semi-simple $G(F)$-conjugacy classes which intersect $C$ finite?

If this is not true, can anyone provide insight into what is meant by `that sum is finite'?

Also, if anyone can clarify what happens when only $G/Z(G)$ is assumed to be anisotropic, that would also be super helpful.


EDIT: It was suggested (in a now deleted comment) that there might be a super easy solution that assumes nothing about $G$ (just that it's an algebraic group). The deletion of that comment has me worried. The suggested proof (as far as I understood it) is as follows:

Proof: Note that $G(\mathbb{Q})$ is closed in $G(\mathbb{A}_\mathbb{Q})$. This follows, I believe, from observing that if $G$ embeds into $\mathbf{A}^n_\mathbb{Q}$ (affine space) for some $n$, which gives a closed embedding of $G(\mathbb{A}_\mathbb{Q})$ in to $\mathbb{A}_\mathbb{Q}^n$. But, $\mathbb{Q}^n$ is closed in $\mathbb{A}_\mathbb{Q}^n$ and so $\mathbb{Q}^n\cap G(\mathbb{A}_\mathbb{Q})=G(\mathbb{Q})$ is closed in $G(\mathbb{A}_\mathbb{Q})$. Note then that if $C\subseteq G(\mathbb{A})$ is compact, then $C\cap G(\mathbb{Q})$ is a closed subset of $C$ and thus compact (since $C$ is Hausdorff). But, we also have that $C\cap G(\mathbb{Q})$ is a subspace of $G(\mathbb{Q})$, but $G(\mathbb{Q})$ is discrete in $G(\mathbb{A}_\mathbb{Q})$ so that $C\cap G(\mathbb{Q})$ is discrete. This implies that $G(\mathbb{Q})\cap C$ is finite. Thus, of course, the number of rational conjugacy classes that meet $C$ is finite.

Is there an issue with this proof? Any comment for/against it would be greatly appreciated!

Find the cocycle condition in fppf descent implies Galois descent

Math Overflow Recent Questions - Sun, 06/09/2019 - 11:56

I tried to give a proof that fppf (faithfully flat) descent implies Galois descent purely at the level of modules and I stumble to obtain the Galois cocycle condition. I'm interested to consider some questions of twisted sheaves with a Galois cohomological description and understanding how to obtain the former would be useful to me.

I obtained the following the following conditions: Given a finite Galois extension $L/K$ of Galois group $G$ and $M$ an $L$-vector space $M$, we have for each $\sigma \in G$ an isomorphism of $L$-vector spaces $\psi_\sigma : M \to M^\sigma$ satisfying $\psi_\sigma(am) = \sigma(a) \psi_\sigma(m)$, where $a \in L$ and $m \in M$ and such that for every pair $(\sigma, \tau) \in G \times G$ we have

$$ \psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma } \circ \psi_{ ( \sigma, \tau), (\sigma, \tau), \tau } = \psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma \tau } $$

as isomorphisms of $L$-modules. The $L$-module structure of $M^\sigma$ is twisted by $\sigma$, i.e given by $a \cdot m:= \sigma(a)m$.

My issue is that because of what one obtains for sheaves of modules ( I would expect the cocycle condition for modules to also have a twisting in the formula.

Translating what is done in stacks project into modules is certainly possible but I wasn't able to do so.

Edit: If someone can give me the details of how it is done at the level of modules (or a least some clear sketch), I would accept this as a correct answer.

Instead I will present a (somehow long) sketch of what I did. I can provide more details upon request and I apologize if there are too many. My difficulty is on Step 5. You can skip directly to this step if you want, the rest explains how I got there.

The context: Let $L/K$ be a finite Galois extension and let $M$ be an $L$-module together we an isomorphism of $L \otimes_K L$-modules $\phi: M \otimes_K L \to L \otimes_K M$ satisfying the cocyle condition $p_{13}^* \phi = p_{23}^* \phi \circ p_{12}^* \phi$ as isomorphisms of $L \otimes_K L \otimes_K L$-modules.

What I did:

Step 1: Describing some isomorphisms We have an isomorphism of $K$-algebras $L \otimes_K L \to \prod_{\sigma \in G} L$ given by $a \otimes 1 \mapsto ( a )_{\sigma \in G}$ and $1 \otimes a \mapsto ( \sigma(a) )_{\sigma \in G}$

and another one $ L \otimes_K L \otimes_K L \to \prod_{\sigma \in G} \Big( \prod_{\tau \in G} L \Big)$ given by

$$ a \otimes 1 \otimes 1 \mapsto \Big( (a)_{\tau \in G} \Big)_{\sigma \in G}, $$

$$ 1 \otimes a \otimes 1 \mapsto \Big( (\tau(a)_{\tau \in G} \Big)_{\sigma \in G}, $$

$$ 1 \otimes 1 \otimes a \mapsto \Big( \tau\sigma(a)_{\tau \in G} \Big)_{\sigma \in G}. $$

We can then describe the above isomorphisms as

$$ (1_L \coprod \sigma) : L \otimes_K L \to \prod_{\sigma \in G} L $$


$$ (1_L \coprod \tau \coprod \tau \sigma): L \otimes_K L \otimes_K L \to \prod_{\sigma \in G} \Big( \prod_{\tau \in G} L \Big). $$

Step 2: Obtain some $\prod_{\sigma \in G} L$-module structures

We then have a commutative diagram of modules.

$$ \begin{array}{ccccc} M \otimes_K L & \xrightarrow{} & \prod_{\sigma \in G} \Big( M \otimes_K L \Big) & \xrightarrow{} & \prod_{\sigma \in G} M\\ \downarrow & & \downarrow & & \downarrow \\ L \otimes_K M & \xrightarrow{} & \prod_{\sigma \in G} \Big( L\otimes_K M \Big) & \xrightarrow{} & \prod_{\sigma \in G} M^\sigma \end{array} $$

where the left most vertical arrow is $\phi$ and we denote by $\psi$ the induced the right most vertical arrow.

Using Step 1 we have ring morphisms $L \xrightarrow{1_L} L$ and $L \xrightarrow{\sigma} L$ for each $\sigma \in G$. Tensoring $M$ with these morphisms give $L$-module structures for $M \otimes_K L$ by $a \cdot ( m \otimes c ) = m \otimes ac$ and for $L \otimes_{L,\sigma} M$ by $a \cdot (c \otimes m) = \sigma(a)c \otimes m$. Now the isomorphism of $L$-modules $\mu: M \otimes_L L \to M: m \otimes c \mapsto cm$ then gives to $M$ the $L$-module structure $a \otimes m = am$. We also have the composite diagram

$$ L \otimes_{L,\sigma} M \xrightarrow{ \sigma^{-1} \otimes 1_M } L \otimes_L M \xrightarrow{\mu'} M : c \otimes m \mapsto \sigma^{-1}(c) \otimes m \mapsto \sigma^{-1}(c)m. $$

Then this $M$ has $L$-module structure given by $a \cdot m := \sigma(a)m$ and we denote it by $M^\sigma$ (We can relabel to get $M^\sigma$ instead of $M^{\sigma^{-1}}$.).

Now a structure of $\prod_{\sigma \in G} L$-module on $\prod_{\sigma \in G} M$ (resp. on $\prod_{\sigma \in G} M^\sigma$) is determined by an $L$-module structure on $M$ (resp. on $M^\sigma$) for each $\sigma \in G$. Therefore, if $(a_\sigma)_{\sigma \in G} \in \prod_{\sigma \in G}$ and $(m_\sigma)_{\sigma \in G} \in \prod_{\sigma \in G} M$ (resp. in $\prod_{\sigma \in G} M^\sigma$), then

$$ (a_\sigma)_{\sigma \in G} \bullet (m_\sigma)_{\sigma \in G} = (~a_\sigma m_\sigma)_{\sigma \in G} ( \text{ resp. } (a_\sigma)_{\sigma \in G} \circ (m_\sigma)_{\sigma \in G} = ( \sigma(a)_\sigma m_\sigma)_{\sigma \in G}~). $$

Step 3: Determine for each $\sigma \in G$ the isomorphisms of $L$-modules $\psi_\sigma$.

The isomorphism $\psi$ induced by $\phi$ must then satisfy

$$ \psi \big( a_\sigma m_\sigma)_{\sigma \in G} ) = (a_\sigma)_{\sigma \in G} \circ \psi( ( m_\sigma)_{\sigma \in G} ). $$

For each $\sigma \in G$ we have an isomorphism of $L$-modules

$$ M \xrightarrow{\iota_\sigma} \prod_{\sigma \in G} M \xrightarrow{\psi} \prod_{\sigma \in G} M^\sigma \xrightarrow{\pi_\sigma} M^\sigma $$

given by

$$ am \mapsto (0, \cdots, 0, am, 0, \cdots, 0) \mapsto \big( \sigma(a) \pi_\sigma \Big( \psi( \iota_\sigma(m) \Big) \big)_{\sigma \in G} \mapsto \sigma(a)\pi_\sigma \Big( \psi(m) \Big). $$

So for each $\sigma \in G$ we have an isomorphism $\psi_\sigma : M \to M^\sigma$ defined by $\psi_\sigma(m):=\pi_\sigma( \psi(m) )$ and such that $\psi_\sigma(am) = \sigma(a)\psi_\sigma(m)$.

Step 4: Determine some $\prod_{\sigma \in G} \prod_{\tau \in G} L$-module structures

I will skip some details, which I can provide upon request. I use the cocycle condition to determine three $\prod_{\sigma \in G} \prod_{\tau \in G} L$-module structures.

Consider $p_{12}^* p_1^* M$ (or equivalently $p_{13}^*p_1^*M$).

The $\prod_{(\sigma,\tau) \in G \times G } L$-module $\prod_{(\sigma,\tau) \in G \times G } M$ is

$$ (a_{g,h}) \cdot ( m_{g,h} ) = ( a_{g,h} m_{g,h} ). $$

Consider $p_{12}^* p_2^* M$ (or equivalently $p_{23}^*p_1^*M$).

The $\prod_{(\sigma,\tau) \in G \times G } L$-module $\prod_{(\sigma,\tau) \in G \times G } M^\tau$ is

$$ (a_{\sigma,\tau}) \cdot ( m_{\sigma,\tau} ) = ( \tau(a_{\sigma,\tau}) m_{\sigma,\tau} ). $$

Consider $p_{13}^* p_2^* M$ (or equivalently $p_{23}^*p_2^*M$).

The $\prod_{(\sigma,\tau) \in G \times G } L$-module $\prod_{(\sigma,\tau) \in G \times G } M^{\sigma \tau}$ is

$$ (a_{\sigma,\tau}) \cdot ( m_{\sigma,\tau} ) = ( (\sigma \circ \tau)(a_{\sigma,\tau}) m_{\sigma,\tau} ). $$

Step 5: Determine the cocycle condition

Finally, for each pair $(\sigma, \tau) \in G \times G$ we have three composite maps given as follows:

$$ M_{(\sigma, \tau)} \xrightarrow{ \iota_{(\sigma,\tau)}} \prod_{(\sigma,\tau) \in G \times G} M \xrightarrow{ p_{12}^* \psi } \prod_{(\sigma,\tau) \in G \times G} M^\tau \xrightarrow{ \pi_{\sigma,\tau}} M_{(\sigma, \tau)}^\tau $$

defining an $L$-module isomorphism $ \psi_{ ( \sigma, \tau), (\sigma, \tau), \tau } : M_{(\sigma,\tau)} \to M_{(\sigma,\tau)}^\tau$ satisfying

$$ \psi_{ ( \sigma, \tau), (\sigma, \tau), \tau }(am) = \tau(a)\psi_{ ( \sigma, \tau), (\sigma, \tau), \tau }(m) $$

$$ M_{(\sigma, \tau)} \xrightarrow{ \iota_{(\sigma,\tau)}} \prod_{(\sigma,\tau) \in G \times G} M \xrightarrow{ p_{13}^* \psi } \prod_{(\sigma,\tau) \in G \times G} M^{\sigma \tau} \xrightarrow{ \pi_{\sigma,\tau}} M_{(\sigma, \tau)}^{\sigma\tau} $$

defining an $L$-module isomorphism $ \psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma \tau } : M_{(\sigma,\tau)} \to M_{(\sigma,\tau)}^{\sigma \tau}$ satisfying

$$ \psi_{ ( \sigma, \tau), (\sigma, \tau), \tau }(am) = (\sigma \circ \tau)(a)\psi_{ ( \sigma, \tau), (\sigma, \tau), \tau }(m) $$


$$ \psi_{ ( \sigma, \tau), (\sigma, \tau), \tau }(am) = \tau(a)\psi_{ ( \sigma, \tau), (\sigma, \tau), \tau }(m) $$

$$ M_{(\sigma, \tau)}^\tau \xrightarrow{ \iota_{(\sigma,\tau)}} \prod_{(\sigma,\tau) \in G \times G} M^\tau \xrightarrow{ p_{23}^* \psi } \prod_{(\sigma,\tau) \in G \times G} M^{\sigma \tau} \xrightarrow{ \pi_{\sigma,\tau}} M_{(\sigma, \tau)}^{\sigma \tau} $$

defining an $L$-module isomorphism $ \psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma } : M_{(\sigma,\tau)}^\tau \to M_{(\sigma,\tau)}^{\sigma \tau}$ satisfying

$$ \psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma }(am) = \sigma(a)\psi_{ ( \sigma, \tau), (\sigma, \tau), \tau }(m). $$

Indeed, $\psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma }$ sends $\tau(a)m$ to $(\sigma \circ \tau)(a)m$ and so sends $am = \tau(\tau^{-1}(a))m$ to $\sigma( am )$.

Since $p_{23}^* \phi \circ p_{12}^*\phi = p_{13}^* \phi$, we have $p_{23}^* \psi \circ p_{12}^*\psi = p_{13}^* \psi$ and therefore for each pair $(\sigma, \tau) \in G \times G$ we have

$$ \psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma } \circ \psi_{ ( \sigma, \tau), (\sigma, \tau), \tau } = \psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma \tau } $$

as isomorphisms of $L$-modules.

Remarks: The map $\psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma }$ is from $M^\tau$ to $M^{\sigma \tau}$ and it is not clear to me how to re-express it as starting from $M$ and twisting it by $\tau$. Another problem is that the maps I found on Step 3 are not in an obvious way related to those of Step 5 and there might be a need to twick something here as well.

Combinatorics problem : find maximum gcd of subsequences [on hold]

Math Overflow Recent Questions - Sat, 06/08/2019 - 08:11

I have been given a sequence. I have to divide this sequence in into two subsequences (not necessarily contiguous) so that sum of their gcd is maximum possible. What should be the approach ?

What does it mean for two natural numbers to be *approximately equal*?

Math Overflow Recent Questions - Fri, 06/07/2019 - 17:37

This is related to this other question of mine about a paper of Colin and Honda.

I'm trying to follow the proofs line by line. I found the following piece of notation that is not explained in the text so i ask for its meaning or a reference where it is used other than this. I quote the beginning of page 6 in the cited paper

Since $d(h(\gamma_0), \gamma_0) = N$, it follows that $d(h(\alpha), h(\gamma_0)) \approx N$ for every $\alpha$. (Here $\approx$ means approximately equal to.)

For some context: $\gamma_0, \alpha$ are simple closed curves in a surface, $h$ is an automorphism of the surface, $d(\cdot, \cdot)$ is the distance in the complex of curves and $N$ is a natural number. Hence, we are saying that two natural numbers are approximately equal?

Since $N$ is a number that, in that context, is arbitrarily large one could think that it means that $\lim_{N\to \infty} d(h(\gamma_0), \gamma_0) - N = 0$? But later on the text a similar exppresion appears changing $N$ by $0$ (end of page $7$).

So I would appreciate someone explaining to me what it means.

Riccati equations for dealing with stiffness of ODEs

Math Overflow Recent Questions - Fri, 06/07/2019 - 17:10

Early texts [Forman Acton, Numerical Methods that Work; Numerical Recipes, first edition] suggested that for integrating a stiff linear ODE it was very helpful to integrate the associated Riccati equation instead. This point was made specifically for using shooting methods, i.e. algorithms that use an initial value method to solve a boundary value problem. I have not seen any followup on this point in the literature -- the statement does not appear in later editions of Numerical Recipes, for instance -- and I'm wondering whether this Riccati methodology is no longer considered a useful way to deal with stiffness.

Entropy of moments of integers

Math Overflow Recent Questions - Fri, 06/07/2019 - 15:23

Let $x \in \{0,1\}^n$ be uniformly at random. What is an estimate for the entropy of moments, $H(\sum_i x_i, \sum_i i\cdot x_i, \sum_i i^2\cdot x_i)$ ?

Reference request: Gauge natural bundles, and calculus of variation via the equivariant bundle approach

Math Overflow Recent Questions - Fri, 06/07/2019 - 14:56

Let $P\rightarrow M$ be a principal fibre bundle with structure group $G$, $F$ a manifold and $\alpha: G\times F\rightarrow F$ a smooth left action.

There is an associated fibre bundle $E\rightarrow M$ with $E=P\times_\alpha F=(P\times F)/G$.

As it is well known, one may either treat sections of the associated fibre bundle "directly", or consider maps $\psi:P\rightarrow F$ which satisfy the equivariance property $\psi(pg)=g^{-1}\cdot\psi(p)$, where $\cdot$ denotes the left action. Let us refer to this latter method as the "equivariant bundle approach".

I am interested in describing the gauge field theories of physics using global language with appropriate rigour. However, most references I know treat this topic using the "direct approach", and not with the equivariant approach, the chief exception being Gauge Theory and Variational Principles by David Bleecker.

Bleecker's book however doesn't go far enough for my present needs.

  • Bleecker only uses linear matter fields, eg. the case where $F$ is a vector space and $\alpha$ is a linear representation. Some things are easy to generalize, others appear to be highly nontrivial to me.
  • Bleecker treats only first-order Lagrangians. The connection between a higher-order variational calculus based on the equivariant bundle approach and between the more "standard" one built on the jet manifolds $J^k(E)$ of the associated bundle is highly unclear to me. Example: If $\bar\psi:M\rightarrow E$ is a section of an associated bundle, its $k$-th order behaviour is represented by the jet prolongation $j^k\bar\psi:M\rightarrow J^k(E)$, but if instead I use the equivariant map $\psi:P\rightarrow F$, what represents its $k$-th order behaviour? I assume it is related to something like $J^k(P\times F)/G$, but the specifics are unclear to me.

  • In Bleecker's approach, connections are $\mathfrak g$-valued, $\text{Ad}$-equivariant 1-forms on $P$, however I am interested in treating them on the same footing as matter fields. Connections however are higher order associated objects in the sense that they are associated to $J^1P$. Bleecker absolutely doesn't treat higher order principal bundles.

In short, I am interested in references that consider gauge theories, gauge natural bundles, including nonlinear and higher-order associated bundles and calculus of variations/Lagrangian field theory from the point of view where fields are fixed space-valued objects defined on the principal bundle (equivariant bundle approach), rather than using associated bundles directly.

Regularity of a shrunken domain

Math Overflow Recent Questions - Fri, 06/07/2019 - 14:52

I am encountering a geometrical question that intuitively seems obvious but I have a lack of argument to prove or disprove it in a rigorous manner.

Let $\Omega\subset\Bbb R^d$ be an open bounded (may be connected just to make it simpler)

For $\delta>0$ small enough we define the shrunken version of $\Omega$ as by $$\Omega_\delta= \{x\in \Omega~: \operatorname{dist}(x,\partial \Omega)>\delta\}$$

Basically, for $\delta$ small enough, $\Omega$ and $\Omega_\delta$ have a similar shape. For instance if $\Omega= B(0,1)$ is the unit ball then, $\Omega_\delta=B(0,\delta) =\delta B(0,1)$ is the ball centered at 0 with radius $\delta$.

Definition An open set $\Omega$ is said to satisfy the Volum Density Condition(VDC) if there exists a constant $\kappa>0$ such that for all $x\in \partial \Omega$ and $r\in (0,1)$ $$|\Omega\cap B(x,r)|\geq \kappa r^d.$$

An open set $\Omega$ is said to be of class $C^k$ if for every $x\in \partial \Omega$ there exists $r>0$ and a mapping $\gamma: \Bbb R^{d-1}\to\Bbb R$ such that

$$ \Omega\cap B(x,r)= \{x=(x',x_d)\in B(x,r)~: x_d>\gamma(x')\}$$

A natural question would be: Does $\Omega_\delta$ inherit the regularity properties of $\Omega$? Precisely,

1)If $\Omega$ is $C^k$ do we have that $\Omega_\delta$ is also $C^k$?

2) If $\Omega$ satisfies the Volum Density Condition(VDC) does $\Omega_\delta$ also satisfies the Volum Density Condition(VDC)?

Equivariant sheafs and $G$ actions on modules

Math Overflow Recent Questions - Fri, 06/07/2019 - 13:55

I am reading Simpson's paper on The Hodge filtration on nonabelian cohomology. In particular Chapter 5 (p.24) and I am confused about the notion of a group acting on an equivariant sheaf.

The set up is as follows:

$\mathbb{A}^1=Spec(\mathbb{C}[z])$ admits an obvious action by the multiplicative group $\mathbb{G}_m=Spec(\mathbb{C}[z,z^{-1}])$ \begin{equation} a:\mathbb{G}_m\times \mathbb{A}^1\to \mathbb{A}^1. \end{equation} which is equivalent to giving the canonical grading on $\mathbb{C}[z]$ (cf. here).

Now what is a $\mathbb{G}_m$-equivariant vector bundle over $\mathbb{A}^1$?

For me this is a $\mathbb{G}_m$-equivariant locally free sheaf $W$. In the affine case this reduces to a locally free module $B=W(\mathbb{A}^1)$ with a grading that respects the grading of $\mathbb{C}[z]$ or equivalently a comodule map $B\to B\otimes_\mathbb{C} \mathbb{C}[z,z^{-1}]$.

So I was quite surprised when I found the following comment in a more written out version of the paper*:

It seems like there is an identification $\mathbb{G}_m$ with $\mathbb{C}^*$, which is confusing to me.

  • How does this action of $\mathbb{G}_m$ (or rather $\mathbb{C}^*$ ?) on $\mathbb{C}[z,z^{-1}]\otimes_\mathbb{C} V$ induce an equivariant structure on the sheaf or equivalently a comodule structure?
  • Is my approach for the definition correct to begin with?

This question is cross-posted from stackexchange

Conceptual (operadic?) reason for the generalized EHP fiber sequence $J_{q-1}(S^{2n}) \to J S^{2n} \to JS^{2nq}$?

Math Overflow Recent Questions - Fri, 06/07/2019 - 13:31

Let $q$ be a prime and $q=p^r$ a power. Then there is a $p$-local fiber sequence from the $q-1$st stage of the James construction on $S^{2n}$, to $J(S^{2n}) = \Omega \Sigma S^{2n}$, to $J(S^{2nq}) = \Omega \Sigma S^{2nq}$. Here the first map is the natural inclusion map for the James filtration, and the second map is the James-Hopf map, adjoint (under $\Sigma \dashv \Omega$) to the projection coming from the Snaith splitting $\Sigma J(X) = \Sigma \vee_{q\geq 1} X^{\wedge q}$ (with $X = S^{2n}$). The EHP sequence arises when $p=q=2$.

What I'd like to understand is the fact that this is a fiber sequence. The proof I'm familiar with (from lectures 3-5 of notes by Akhil Mathew on a course by Mike Hopkins) can be seen by computing what it does on $\mathbb F_p$ homology and considering the Serre spectral sequence. The homology picture is rather suggestive:

  • Recall that the James construction is the free $E_1$-space on an $E_0$-space (i.e. a pointed space), and the Snaith splitting tells us that correspondingly $H_\ast(J(X))$ is the free associative algebra on the augmented vector space $H_\ast(X)$: $H_\ast(J(X)) = T \tilde H_\ast(X)$.

  • Moreover, the James filtration corresponds to the filtration of the tensor algebra by tensor rank: $H_\ast(J_k(X)) = T_{\leq k} \tilde H_\ast(X)$.

  • The James-Hopf map is rather complicated, but one works out using the comultiplication (and computing some mod $p$ multinomials) that $p$-locally, and when $X = S^{2n}$, it looks additively like the obvious projection $H_\ast(J(S^{2n})) = \mathbb F_p[x_{2n}] \to \mathbb F_p[y_{2nq}] = H_\ast(J(S^{2nq}))$.

So the fiber sequence boils down to the additive decomposition $\mathbb F_p[x_{2n}] \cong (\mathbb F_p[x_{2n}] / x_{2n}^q) \otimes \mathbb F_p[x_{2n}^q]$. This closely mirrors the universal properties of $J_{q-1} S^{2n}$ and $J(S^{2nq})$. So in some sense,

The fiber sequence $J_{q-1} S^{2n} \to J S^{2n} \to J S^{2nq}$ says something about decomposing operations in the $E_1$ operad into operations of arity $<q$ and operations of arity divisible by $q$.

But what exactly it says, I'm not sure. So I suppose my question is:


  1. Is there a conceptual explanation for the $p$-local fiber sequence $J_{q-1} S^{2n} \to J(S^{2n}) \to J(S^{2nq})$?

  2. In particular, is there such an explanation which either circumvents or else better explains the seeming contingency that the James-Hopf map is a $H_\ast(-,\mathbb F_p)$-surjection?

  3. As an alternative desideratum, is there such an explanation which follows from operadic considerations?

To flesh out this operadic perspective a bit more, here's a description of a "James-Hopf sequence"in a more general setting: one might formalize (3) above as asking under what conditions the following sequence

$$J^O_{q-1}(X) \to J^O(X) \to J^O(O(q)_+ \wedge_{\Sigma_q} X^{\wedge q})$$

is a fiber sequence.

Claim: ("Destabilization of the stable Snaith splitting"): Let $O$ be an operad with $O(0) = \ast$ and admitting a map from the $A_2$ operad (the latter condition means that every $O$-space is an $H$-space). Let $J^O$ be the free functor from $E_0$-spaces to $O$-spaces. Then the natural map

$$J^O(\vee_{n \geq 1} O(n)_+ \wedge_{\Sigma_n} X^{\wedge n}) \xrightarrow{J^O\varphi} (J^O)^2(X)$$

is an equivalence for every connected space $X$.

Proof: Use the stable Snaith splitting, the fact that $J^O$ preserves stable equivalences, and the fact that a stable equivalence between connected $H$-spaces is an equivalence.

Corollary: ("Operadic James-Hopf map") Let $O$ and $X$ be as above. Then for any $q \geq 1$, there is a "James Hopf" map

$$ J^O(X) \xrightarrow{\eta_{J^O(X)}} (J^O)^2(X) \overset{(J^O\varphi)^{-1}}{\simeq} J^O(\vee_{n \geq 1} O(n)_+ \wedge_{\Sigma_n} X^{\wedge n}) \xrightarrow{J^O(\pi_q)} J^O(O(q)_+ \wedge_{\Sigma_q} X^{\wedge q}) $$

which kills the subspace $J^O_{q-1}(X) \subseteq J^O(X)$. Here the filtration $\dots \subseteq J^O_k(X) \subseteq \dots J^O(X)$ is defined by arity of operations in $O$ as in the James filtration, and $\eta_Y: Y \to J^O(Y)$ is the unit map.

This is analogous to the usual James-Hopf map; note that the "adjointing over" that occurs with the unit map is analogous to the $\Sigma \dashv \Omega$ adjointing that occurs in the usual James-Hopf map.

Ratio of integrals with increasing dimension over Euclidean balls

Math Overflow Recent Questions - Fri, 06/07/2019 - 12:13

Let $f_n(x)\geq0$ be any sequence of nonnegative $L^1(\mathbb{R}^n)$ functions such that $\int_{\mathbb{R}^{n}}f_n(x)dx=1$ where $dx$ is the Lebesgue measure on $\mathbb{R}^n$. For any $a>1,\epsilon>0$, does there exist a sequence $x_n\in\mathbb{R}^n$ such that $$\lim_{n\to+\infty}a^n\frac{\int_{\|x-x_n\|^2\leq n^{1-\epsilon}}f_n(x)dx}{\int_{\|x-x_n\|^2\leq n}f_n(x)dx}=0$$?

If it does not hold, is there any counter example of the sequence $f_n$? What additional conditions do we need on the sequence $f_n$?

Inverse limit and graded functor commute?

Math Overflow Recent Questions - Fri, 06/07/2019 - 12:06

I am trying to understand a proof where there are graded algebras and inverse limit involved.

In one of the steps it seems to commute this two elements. Is there any reference where this is stated.

$$\varprojlim (gr(\Lambda^*_n))=gr(\varprojlim \Lambda_n^*).$$

Where $\Lambda^*_n$ stands for the shifted symmetric polynomials of $n$ indeterminates.

Note that the inverse limit $\varprojlim \Lambda_n^* = \Lambda^*$ is taken in the category of filtered algebras.


Is a smooth family of vector spaces always locally trivial?

Math Overflow Recent Questions - Fri, 06/07/2019 - 09:58

Let's define a smooth family of vector spaces to be the following data $(E,M,p,a,s,z)$, where

(1) $p:E\to M$ is a smooth submersion of smooth manifolds

(2) There is some fixed integer $n\ge 0$ such that the fibre $E_m = p^{-1}(m)$ has the structure of an $n$-dimensional vector space over $\mathbb R$ for each point $m\in M$. [The smooth structure of $E_m$ as a vector space coincides with its smooth structure as a submanifold of $E$.]

(3) For this fibrewise vector space structure, the addition map $a:E\times_ME\to M$ is smooth, the scaling map $s:\mathbb R\times E\to E$ is smooth and the zero section $z:M\to E$ is a smooth embedding.

Then, is $p:E\to M$ a locally trivial vector bundle (with $a,s,z$ being its addition, scaling and zero section maps)?

Just by virtue of $p:E\to M$ being a smooth submersion, we have a locally trivial vector bundle $T_{E/M}$ on $E$ (the "vertical tangent bundle" of $p$), and we can pull it back to $M$ to get a locally trivial vector bundle $z^*T_{E/M}$.

Now, define a map $\phi:E\to z^*T_{E/M}$ given by $e\mapsto\frac{\partial}{\partial t}|_{t=0}s(t,e)$. This is a smooth map (since $s,z$ are smooth) that commutes with the projections to $M$ and is a (fibrewise linear) bijection.

Finally, if we look at the map $\phi$ along the image $z(M)$ of the zero section, it quite clearly has full rank. Thus, it has full rank on a neighborhood of $z(M)\subset E$. But now, since $\phi$ commutes with scaling, it follows that $\phi$ has full rank on all of $E$ and thus, it is a diffeomorphism by the inverse function theorem. Thus, $E$ being isomorphic (as a smooth family of vector spaces) to $z^*T_{E/M}$ means it is locally trivial.

My question is: Is there any (possibly subtle) mistake in this proof (and if so, is local triviality true nevertheless)?

Curiously, this argument does not seem to need even the smoothness of $a$ (it follows as a consequence of the isomorphism $\phi$, which itself depends only on the smoothness of $s$ and $z$). Also, the same argument seems to work if we replace all the objects involved by complex manifolds, holomorphic maps and complex vector spaces. The bracketed assumption in (2) seems to be used when we show that $\phi$ is a fibrewise linear bijection and that it has full rank along the zero section. I think it can be removed and that the proof should still go through - is this true?

If you think the answer to the first question is NO, then I'd still like to hear about it in the comments. My actual motivation is to apply this in the setting of infinite dimensional Banach manifolds, and an answer to the second question covering this case would also be appreciated.

Proving the representability of a functor that is covered by open subfunctors

Math Overflow Recent Questions - Fri, 06/07/2019 - 09:10

I already posted this on math.stackexchange some while ago, but haven't received any answers yet. (

I want to proof Theorem 8.9 from Algebraic Geometry I ( U.Görtz, T.Wedhorn), which reads as follows:

Let $S$ be a scheme $F: Sch/S°\rightarrow Set $ a functor such that:

  1. F is a sheaf for the Zariski topology
  2. F has a cover by open subfunctors $\alpha_i:F_i\rightarrow F$, such that every $F_i$ is representable by a scheme $X_i$

Then F is representable.

A cover by open subfunctors means, that for every scheme $T$ and for every morphism $h_T\rightarrow F$, the pullback $F_i\times_F h_T$ is representable, say by $Y_i$ and the morphism of schemes $Y_i\rightarrow T$ corresponding to the projection $F_i\times_F h_T\rightarrow h_T$ is an open immersion. In addition the images of $Y_i\rightarrow T$ form an open covering of $T$

Let me explain what I have done so far and where I am stuck:

The $X_i$ can be glued to a scheme $X$. The morphisms $\tilde{\alpha_i}:h_{X_i}\cong F_i\rightarrow F$ correspond via the yoneda lemma to elements $f_i\in F(X_i)$. Using the sheaf property of F, the $f_i$ glue together to an element $f\in F(X)$ which gives us a natural transformation $\alpha: h_X\rightarrow F$. For a scheme $T$ and a morphism $g\in\mathrm{Hom}(T,X)$ this is given by $\alpha(T)(g)=F(g)(f)$. The last step is to show that this assignment is bijective. I managed to show the surjectivity, but can't find a proof for the injectivity. It would suffice to show that the following diagram is a pullback $\require{AMScd}$ \begin{CD} h_{X_i} @>>> h_X\\@VidVV @VV\alpha V\\ h_{X_i} @>>\tilde{\alpha_i}> F \end{CD} where the morphism $h_{X_i}\rightarrow h_X$ is induced from the open immersion $X_i\rightarrow X$. The commutativity of this diagram is clear to me. To proof that this is a pullback we could try the following: Since $h_{X_i}$ is an open subfunctor, there is an open subscheme $U_i$ of $X$, such that the following square is cartesian: \begin{CD} h_{U_i} @>>> h_X\\@VVV @VV\alpha V\\ h_{X_i} @>>\tilde{\alpha_i}> F \end{CD} By the commutativity of the first square the open immersion $X_i\rightarrow X$ factors through $U_i\rightarrow X$, so that $X_i$ is an open subscheme of $U_i$. But I couldn't find a way to show $X=U_i$. Another way would simply be checking that, $h_{X_i}$ satisfies the universal property. However, when trying to do so one needs that $\alpha(T)$ is injective for every scheme $T$.

I also looked at the proof in EGA I (Springer edition 1971), where this is Proposition 4.5.4 in chapter 0. There Grothendieck uses (without comment) that the fiber product $F_i\times_F h_X$ is represented by $X_i$, which I think is equivalent to saying that the square from above (the one with $X_i$) is cartesian.

I am thankful for any thoughts on this.

Density of primes in the set $\{p_1+2a,...,p_n+2a,...\}$ for every natural $a$, any conjectures?

Math Overflow Recent Questions - Fri, 06/07/2019 - 08:49

If we shift the set $\mathbb P=\{p_1,...,p_n,...\}$ of all prime numbers by some natural number $2a$ to obtain a set $\mathbb P+2a=\{p_1+2a,...,p_n+2a,...\}$ then I expect that $\mathbb P +2a$ contains an infinite number of prime numbers and that it contains an infinite number of composite numbers.

I would like to know are there any conjectures about density of primes in the set $\mathbb P+2a$, that is, what is known about the limit $$\lim_{n \to + \infty} \dfrac{nop\{p_1+2a,...,p_n+2a\}}{n}$$, where $nop$ stands for the "number of primes", that is, if $S$ is any set then $nop(S)$ gives as a number of primes in the set $S$.

We can denote $\lim_{n \to + \infty} \dfrac{nop\{p_1+2a,...,p_n+2a\}}{n}=f(a)$, and, I would also like to know what is known about $f$, for example, is it reasonable to expext that $f$ is a constant function?

More particularly, is there any evidence that we could have $f(a)=0$ for every $a \in \mathbb N$, that is, that shifting of the set of primes by some even number $2a$ gives us a set where "almost all" numbers are composite numbers?

More generally, I am also interested in every conjecture about this topic that you know of.

Cardinal exponentiations inequality

Math Overflow Recent Questions - Fri, 06/07/2019 - 08:30

Let $\kappa < \beth_2$ and $\lambda<\beth_1$ be cardinals. What can we say about $\kappa^{\lambda}$ without assuming CH? Is it true that $\kappa^{\lambda} < \beth_2$ or $\kappa^{<\beth_1} < \beth_2$?

Is the least ordinal containing all countable ordinals defined by a formula an element of itself?

Math Overflow Recent Questions - Fri, 06/07/2019 - 08:24

The following argument supports a "yes" answer; is it convincing?

Let $\Omega$ be the least uncountable ordinal. A finite ($L_{\omega \omega}$) formula of signature $\{\in\}$ of one free variable $\varphi(x)$ defines an element of $\Omega$ if $$(\Omega,\in) \models \exists!_x \varphi(x).$$ Let $X$ be the set of all $\varphi(x)$ such that $\varphi$ defines an element of $\Omega$. Since $L_{\omega \omega}$ is countable, so is $X$. Let $F\colon X\rightarrow \Omega$ be such that $F(\varphi)$ is the element of $\Omega$ such that $\varphi(F(\varphi))$. Let $F[X]$ be the range of $F$. Since $F[X]$ is a countable subset of $\Omega$, and since no countable subset of $\Omega$ is cofinal in $\Omega$, there exists a least $\beta\in\Omega$ such that $F[X]\subseteq\beta$. Since the preceding remarks can be formalized in ZFC and define $\beta$ uniquely, there exists a formula $\psi(x)$ such that $$(\Omega,\in)\models\forall_x \psi(x) \iff x = \beta.$$ Since $\psi\in X$ and $F(\psi)=\beta$, we have $\beta\in F[X]\subseteq \beta$, hence $\beta\in\beta$.


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