Fix a totally real field $K$, a level $\mathfrak{n}$, a (parallel) weight $k\geq 2$ and a primitive ray class character $\chi$ modulo $\mathfrak{n}$.

Then one can form the space $S_k(\mathfrak{n},\chi)$ of Hilbert cusp forms (as in Shimura). This is the full Hecke module so can be considered as $h_K^{+}$ tuples of modular forms that transform for various “Gamma_0” style groups.

Consider the module of forms which have $\mathfrak{p}$-integral Shimura coefficients. Then there is a well defined mod $\mathfrak{p}$ reduction map into the space of mod $\mathfrak{p}$ HMF’s of weight $k$, level $\mathfrak{n}$ and character $\bar{\chi}$ (a la Katz).

Is it known explicitly when this map is surjective, i.e. is there an explicit analogue of Carayol’s lemma? (as in Edixhoven’s Serre’s conjecture chapter in “Modular Forms and FLT”). Could the answer be $k\geq 2$ as in the rational case?

If the result is only known for large enough $k$ then that is fine too, but would prefer to know an explicit absolute bound.

Let A is a $n\times n$ matrix given by \begin{align*} a_{ij} = [\Gamma(\lambda_{i}+\mu_{j})] \end{align*} where $0 < \lambda_{1} < \ldots < \lambda_{n}$ and $0 < \mu_{1} < \ldots < \mu_{n}$ are real positive numbers and $\Gamma$ denotes the Gamma function given by $\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt$ \, for Re(z)>0.

We need to show that matrix A is non-singular.

I have no idea how to start. Any hint or solution will be appreciated.

Thanks in advance.

Elliptic curves have a group structure over the rational points. Why is this impossible for curves having genus greater than 1? I read somewhere that this impossibility is implied by Faltings theorem, which states that the number of rationals is finite. I don't see the implication. The group over the rational points could be finite.

If, $u: \mathbb{R}^n \to \mathbb{R}$ be a non-negative test function, i.e., $u \in \mathcal{D}(\mathbb{R}^n)$ and $u \ge 0$, then does it follow that,

$$\mathcal{H}^{n}(\{s < u \le t \} \cap \{ \nabla u = 0\}) \ge \mathcal{H}^{n}(\{s < u^{\ast} \le t \} \cap \{ \nabla u^{\ast} = 0\}), \text{ for } 0 < s < t$$

where, $u^{\ast}$ denotes the symmetric decreasing rearrangement of $u$?

**Context:** Using the above fact I wish to conclude the inequality, $$\int_{\{u = t\}} \frac{1}{|\nabla u|}\,d\sigma \le \int_{\{u^{\ast} = t\}} \frac{1}{|\nabla u^{\ast}|}\,d\sigma \tag{2}$$ for almost every regular value $t$ of $u$. Since, we are supposed to have equality in $(2)$, both quantities equal to the derivative of the distribution function of $u$, I am trying to find a direct argument for $(1)$ that does not use $(2)$.

Any help is appreciated. Thanks!

Consider the following commutative diagram of the fiber bundles $% F\rightarrow E\rightarrow B$ and $F^{\prime }\rightarrow E^{\prime }\rightarrow B^{\prime }$ where $B^{\prime }$ is simply connected space (but $B$ is not simply connected space) and all spaces are path-connected spaces. $\require{AMScd}$ \begin{CD} F @>{}>> E @>{}>> B \\ @VVV @VVV @VVV\\ F' @>{}>> E' @>{}>> B' \end{CD} Suppose that \begin{equation*} H^{\ast }\left( B^{\prime };% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) \rightarrow H^{\ast }\left( B;% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) \end{equation*} and \begin{equation*} H^{\ast }\left( E^{\prime };% %TCIMACRO{\U{211a} } %BeginExpansion \mathbb{Q} %EndExpansion \right) \rightarrow H^{\ast }\left( E;% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) \end{equation*} are isomorphisms.

If \begin{equation*} H^{i }\left( F;% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right)=0 \end{equation*} for all $i \geq n$ ($n$ fixed), then \begin{equation*} H^{i }\left( F^{\prime };% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right)=0 \end{equation*} for all $i \geq n$?

I am a physics student and are interesting in the study of invariant metrics. I have searched several textbooks, including those fat books of Krantz, but the following concern seems not to be mentioned in these books.

Let $\Omega\subset\mathbb{C}^{n}$ be an open, bounded domain. In literature, the Bergman distance between two points $z_{0}$ and $z_{1}$ (with $z_{0},z_{1}\in\Omega$) is defined by $$ d\left(z_{0},z_{1}\right):=\inf\left\{ l\left(\gamma\right):\gamma\in C^{1}\left(\left[0,1\right],\Omega\right),\gamma\left(0\right)=z_{0},\gamma\left(1\right)=z_{1}\right\} . $$ Here $$ l\left(\gamma\right):=\intop_{0}^{1}\sqrt{\sum_{j,k=1}^{n}\dfrac{\partial^{2}\left(\log K\left(\gamma\left(t\right),\gamma\left(t\right)\right)\right)}{\partial z_{j}\partial\overline{z}_{k}}\gamma'_{j}\left(t\right)\overline{\gamma'_{k}}\left(t\right)}dt, $$ and $K$ denotes the Bergman kernel of $\Omega$.

My question:

Is $\left(\Omega,d\right)$ a metric space? If so, is the topology induced by $d$ the standard topology on $\Omega$?

Let $P_1$ be a set of 4 points in the Euclidean plane. Formally, $P_1$ determines a set $L_1$ of 6 lines, which then determine only 3 points not already in $P_1.$ Let $P_2$ be the set of 7 points thus far determined. Formally, $P_2$ determines only 3 lines not already in $L_1.$ Let $L_2$ be the set of 9 lines thus far determined. Continuing in this manner determines sets $P_n$ and $L_n.$ What can be said about the cardinalities $|P_n|$ and $|L_n|$?

I am trying to evaluate an integral in a $3$d space (where the coordinates are given as $x=(x_1, x_2, x_3)$) but am having difficulty.

For convex polyhedra I am trying to evaluate the below:

$\int_{E}(\pi-k)dT_3$

Where $k$ is the angular defect at each vertex of the polyhedron, E is the set of edges of the polyhedron, and $T_3$ is the unit vector tangent to E in the positive direction of $x_3$.

There are a few things I know about this integral:

- If you take the polyhedron and sandwich it tightly between two parallel planes, this integral should come out to the average perpendicular distance between the two planes over all rotations of the polyhedron.
- For a regular polyhedron this is equal to the sum of the lengths of all edges times the angular defect of an angle (e.g. for a cube it's $3\pi l$ where $l$ is the length of an edge).
- It should be independent of the orientation of the shape.

I'm struggling to work out how to evaluate it for non-regular polyhedra - e.g. a square based pyramid or something like that. Does anyone know how I could do this (thank you in advance!)

I believe that in various fields of mathematics there are general principles which might lead to good questions and good results in concrete situations. I would like to have a list of such principles. Below is my own list.

1) Given a Banach space, one may ask what is its dual. In many concrete situations this question has interesting and important answers. (For example the dual space of the space of continuous functions on a compact space is the space of measures on it.)

2) Given a functor between two categories, one may ask whether is has right/ left adjoint. One may also ask whether it has right/ left derived functors. (For example the simple operation of push-forward of sheaves of abelian groups has left adjoint - pull back- which is somewhat less obvious.)

3) Given a metric space, one may ask what is its completion. (Concrete example which I like: consider the set of isometry classes of $n$-dimensional closed Riemannian manifolds of diameter at most $D$ and sectional curvature at least $\kappa$ equipped with the Gromov-Hausdorff metric. Points of its completion are compact metric spaces which are so called Alexandrov spaces with curvature bounded below.)

Let $V$ be an $\mathbb{R}$-vector space of finite dimension, let $N$ be a $\mathbb{Z}$-structure on $V$, and let $M$ be its dual $\mathbb{Z}$-structure on the dual space $V^*$.

Let $\sigma\subseteq V$ be an $N$-rational pointed polyhedral cone and let $R$ be a ring. The intersection $\sigma^\vee_M$ of the dual cone of $\sigma$ with $M$ is a monoid, and thus gives rise to the $R$-algebra $R[\sigma^\vee_M]$. Its spectrum $X_\sigma(R)$ is known as *the affine toric scheme over $R$ associated with $\sigma$.*

Let $\tau\subseteq V$ be a further $N$-rational pointed polyhedral cone, and suppose that $\tau$ is a *subset* of $\sigma$. The above procedure then yields an injection of monoids $\sigma^\vee_M\hookrightarrow\tau^\vee_M$ and thus a morphism of $R$-schemes $h_{\tau,\sigma}(R)\colon X_\tau(R)\rightarrow X_\sigma(R)$.

It is a well-known and basic fact in the theory of toric varieties that if $\tau$ is a *face* of $\sigma$, then the above morphism of $R$-schemes $h_{\tau,\sigma}(R)$ is an open immersion. I wonder about the converse, and I would be surprised if the following is not true:

**Conjecture:** If the canonical morphism $h_{\tau,\sigma}(R)\colon X_\tau(R)\rightarrow X_\sigma(R)$ is an open immersion *for some non-zero ring $R$,* then $\tau$ is a face of $\sigma$.

I tried to attack this in several ways but did not succeed so far. (One idea led to this more general question.) So, my question is whether this conjecture is known to be true, and of course about any source proving or disproving it.

(If this is of any use, note that by base change we can suppose that $R$ is nice (e.g., noetherian, a field, an algebraically closed field...), and that by faithfully flat descent we can even suppose that $R$ is a prime field.)

The degree of the Todd class of the tangent bundle of a variety $Y$ gives the holomorphic genus of $Y$, which is a birational invariant. We also know that the $\hat{A}$-roof is closely related to the Todd class. I will call the degree of the $\hat{A}$-roof the A-genus. On a spin manifold, the $\hat{A}$-genus gives the index of the Dirac operator by the Atiyah-Singer theorem.

I would like to know what is the invariance of the $\hat{A}$-genus.

Because the $\hat{A}$-genus can be expressed by Pontryagin numbers, the $\hat{A}$-genus is a homotopy invariant and an oriented diffeomorphic invariant. The $\hat{A}$-genus is even an oriented homeomorphic invariant since Novikov proved in the 1960s that Pontryagin numbers are oriented homeomorphic invariants.

But is the $\hat{A}$-genus also a birational invariant as the Todd-genus?

In this generality, the answer will be negative by looking at the case of complex surfaces. For surfaces, the $\hat{A}$-genus is $\frac{(-c_1^2+2c_2)}{24}$ while the holomorphic Euler characteristic is proportional to $\frac{c_1^2+c_2}{12}$. Thus, if both were birational invariant, so would be $c_1^2$ and $c_2$, which is not the case by considering their behavior under the blowup of a smooth point.

But if we restrict to crepant birational maps, I suspect that we might get an invariant. (A well-known example of invariants of crepant birational maps between non-singular varieties are the Hodge numbers.) So my question is:

**Is the $\hat{A}$-genus preserved under a crepant birational map between two smooth algebraic varieties?**

(I work over the complex numbers and if it makes a difference, I am particularly interested in the case of projective fourfolds. )

Let $(\alpha,g)$ be a Morse-Smale pair on a closed smooth manifold $M$, i.e. $\alpha$ is a Morse form and $g$ a Riemannian metric on $M$ such that stable and unstable manifolds of the gradient vector field $X$ intersect transversally.

Let $\pi \colon \tilde{M} \to M$ be the associated cover to $\ker [\alpha]$. Denote by $\tilde{f} \colon \tilde{M} \to \mathbb{R}$ a primitive of the exact pullback $\pi^* \alpha$ and set $\tilde{g}=\pi^*g$.

It is clear that $(\tilde{f},\tilde{g})$ is a Morse-Smale pair on the cover $\tilde{M}$ and it is also clear that $\tilde{M}$ is non-compact whenever the Morse form $\alpha$ is non-exact. This leads to the natural question:

Does $\tilde{f} \colon \tilde{M} \to \mathbb{R}$ satisfy the Palais-Smale condition in the non-exact case?

For various reasons I'd like this to be true, but I'm not able to find a proof for this (neither by looking in the literature, nor by trying myself). Is anyone aware of some place in the literature that answers this (or vaguely related) question?

I wonder whether one could reduce the following integral to some combination of special (e.g. Bessel) functions: $$\int_{m}^{\infty}d\epsilon \exp\left(-s\epsilon\right)\left( \epsilon^{2}-m^{2}\right)^{1/2}\left( \epsilon^{2}-\lambda m^{2}\right)^{1/2}$$

The integral without the last term in the integrand reduces to a well know integral representation of $K_{1/2}$, but I can not figure out if there is a clever variable substitution which brings the integral above to something better known.

Every vector $v$ in a finite-dimensional vector space space $V$ of dimension $n$ over a field $F$ has a unique representation in terms of a basis ${\frak B} \subseteq V$, where a basis for $V$ is a set of $n$ linearly independents vectors in $V$. Intuitively, a vector $v \in V$ doesn't have an "absolute representation" (except perhaps as the symbol "$v$"), but only "relative representations" in terms of other vectors in $V$ together with scalars in $F$.

A vector $v \in V$ and a basis $\frak B := \{b_0, b_1, \cdots, b_{n-1}\}$ yield a unique $n$-nuple of scalars $(s_0, s_1, \cdots, s_{n-1}) \subseteq F^n$ with the property that $v$ is the linear combination

$$ v = s_0 b_0 + s_1 b_1 + \cdots s_{n-1} b_{n-1}. $$

The $n$-tuple $(s_0, s_1, \cdots, s_{n-1})$ is sometimes called the coordinate vector of $v \in V$, although I like to think of it as $(s_0, s_1, \cdots, s_{n-1})_{\frak B}$ to emphasize that its only relationship to $v$ is through the basis $\frak B$.

Every integer $d$ in the commutative ring of integers ${\mathbb Z}$ has a unique representation in terms of a base $b \in {\mathbb Z_{>1}}$ where a base for ${\mathbb Z}$ is an integer larger than $1$. Intuitively, an integer $d \in {\mathbb Z}$ doesn't have an "absolute representation" (except perhaps as the symbol $d$), but only "relative representations" in terms of other integers in ${\mathbb Z}_{>1}$ together with integers in ${\mathbb Z}$.

An integer $d$ and a base $b$ yield a unique $n$-tuple of integers $(s_0, s_1, \cdots, s_{n-1})$ with the property that $d$ is the "non-linear combination"

$$ d = s_0 b^0 + s_1 b^1 + s_2 b^2 + \cdots + s_{n-1} b^{n-1}. $$

The integer $d$ is then written down as $s_{n-1} \ldots s_2s_1s_0$, or $s_{n-1} \ldots s_2s_1s{_0}_{b}$ to clarify that $b$ is the base.

Now, I know that thinking of the integers as a vector space (or an algebra, or a curve, or something) over a field (or ring) is absurd (since they don't satisfy the definitions), but are these similarities just a coincidence, or is there something to say about it?

(Wikipedia says that it'd be really convenient if the integers could be constructed as a curve over $F_{un}$.)

We have a group of $n$ people who must make a journey of length $d$. They are to start together, and their goal is to arrive at the destination at same time. They have a single bicycle, which they ride in turns. Each time a rider dismounts he leaves the bike by the side of the road, and walks on, while one of the other ones eventually arrives at the bike and jumps on it (if a group member sees the bike by the road, she can use the bike, but doesn't have to).

Group member $i$ has constant walking speed $w_i\in\mathbb{Q}$ and constant bicycling speed $b_i\in\mathbb{Q}$ such that $b_i \geq w_i$.

Is there an algorithm that takes $n$, $d$, $(w_i)_{i\in\{1,\ldots,n\}}$ and $(b_i)_{i\in\{1,\ldots,n\}}$ and decides whether it is possible that the $n$ people reach the destination at the same time?

Does there exist a bounded set $A \subset \mathbb{R}^ n$ (for some $n$) and a function $f:A\to A$ (not necessarily onto) such that $\forall x,y \in A$ then $\|x-y\| < \|f(x)-f(y)\|$?

If such a set $A$ (and a function $f$) exists, does there even exist such a set $A$ and a function $f$ such that $f$ is continuous? (Or - can it be proved that such a continuous $f$ doesn't exist?)

How can you find a divisor of the numerator of the sum: 1-1/2+1/3-1/4-...+1/47

there are alternatives given such as 59, 61, 67, 71 and 79 using a calculator you get the answer is 71 but how can you do it without it?

The Bennequin bound [1] says that, for a transverse knot (or later link) $K$ in $S^3$, $$\mathrm{sl}(K) \le - \chi(\Sigma)$$ for any Seifert surface $\Sigma$ for $K$, where $\mathrm{sl}$ is the self-linking number.

Lisca and Matic proved [2, Theorem 3.4] that for a Legendrian knot $K$ $$|r(K)| + \mathrm{tb}(K) \le 2g(\Sigma) - 1$$ for any smooth surface $\Sigma \subset B^4$ with boundary $K$, where $\mathrm{tb}$ and $r$ are the Thurston-Bennequin and rotation numbers.

EDIT: This was actually proved by Rudolph [3].

I have several questions here.

(a) Why is the Lisca-Matic bound always stated in terms of Legendrian knots? The combination $|r| + \mathrm{tb}$ is exactly the self-linking number of the transverse push-off, with one of the two orientations. Wouldn't it be simpler to just state it in terms of transverse knots?

(b) Isn't the Lisca-Matic bound strictly stronger than the Bennequin bound? Why isn't it more popular to quote it? Maybe it's about the extension to links (which, in the case of the Bennequin bound, is due to Eliashberg)? It seems that Lisca and Matic only state the result for knots in their paper, but it seems to me that the proof should carry over.

For knots/links in more general 3-manifolds, the Bennequin bound applies as long as you have a tight contact structure, while to state Lisca-Matic you need some sort of filling.

[1] Bennequin, "Entrelacements et équations de Pfaff", In Third Schnepfenried Geometry Conference, Vol. 1, Schnepfenried, 1982, 87–161. Astérisque 107. Paris: Société Mathématique de France, 1983.

[2] Lisca, P., and G. Matić. “Stein 4-manifolds with Boundary and Contact Structures.” 55–66. Topology and Its Applications 88, nos. 1–2, 1998.

[3] Rudolph, Lee, "Quasipositivity as an obstruction to sliceness." Bull. Amer. Math. Soc. (N.S.) 29 (1993), no. 1, 51–59.

My question is about the regularity of a metric whose curvature is bounded in $L^2$. Of course, this question doesn't really make sense since the regularity of the metric depends on the coordinates choice.

But, thanks to Deturck-Kazdan, we know that harmonic coordinates are the best choice, more precisely, if $g\in C^2$ and $Ric\in C^{k,\alpha}$ for $k>0$ in harmonic coordinates then $g\in C^{k+2,\alpha}$.

The proof use some "classical elliptic theory" and the fact that in harmonic coordinates we get roughly speaking $\Delta g=Ric$.

My question is two folds.

1) Assuming the existence of harmonic coordinates, do we have an $L^p$-estimate, like $Ric \in L^p$ then $g\in W^{2,p}$ for $p>1$. It seems reasonable with respect to regularity theory but I didn't find any reference.

2) let assume 1) with $p=2$ and $n=4$(dimension of the manifold). Then $g\in W^{2,2}$ but this space doesn't not belongs to $C^0$, hence it seems difficult to give sense to any harmonic coordinates since the coefficients of the Laplace-Beltrami won't be regular enough to apply the regularity theory. Can we give some sense to this assuming only $Rm$ or $Ric$ in $L^2$.

My motivation comes from the regularity theory for critical point of quadratic functional with respect to the curvature in dimension $4$, like $$\int_M \vert Ric_g\vert^2\, dv_g$$, for instance Einstein metric are critical point of this functional and they are known to be analytic (theorem 5.26 of Besse) assuming that the manifold is $C^1$ and I guess the metric at least $C^0$ before bootstraping. Can we give a sense to all this theory working only in energy space?

The only statement I'm sure of is that any hyperbolic or Euclidean manifold is a $K(G,1)$ (i.e. its higher homotopy groups vanish), since its universal cover must be $\mathbb H^n$ or $\mathbb E^n$. But for example, if a complete Riemannian manifold $M$ satisfies one of the following, can I conclude that $M$ is a $K(G,1)$?

$M$ has sectional curvature bounded above by some negative number.

$M$ has negative sectional curvature.

$M$ has nonpositive sectional curvature.

$M$ has sectional curvature bounded above by $f(\operatorname{vol}(M))$ (where $f: \mathbb R \to \mathbb R$ is some function depending only on the dimension of $M$ that I don't know).

$M$ has scalar curvature bounded above by some negative number.

$M$ has negative scalar curvature.

$M$ has nonpositive scalar curvature.

$M$ has scalar curvature bounded above by $f(\operatorname{vol}(M))$.

Do the answers change if I assume that $M$ is compact? Have I left out a relevant condition of some kind?