I am reading the paper "Two Theorems on Extensions of Holomorphic Mappings" by PHILLIP A. GRIFFITHS. Proposition 2.9 of the paper is: If $\Psi$ is a pluri-sub-harmonic on the punctured ball $B_n^{*}$ and $n\ge 2$, then $\Psi$ extends to a pluri-sub-harmonic function on the whole ball $B_n$. Then, the paper gives a proof in the special case for $n=2$, where I am stuck.

By defining $$\Psi(0)=\lim\sup \Psi(z)$$ , it attempts to show that $\Psi(0)<+\infty$. When $z_1\not= 0$, we have the estimate $$\Psi(z_1,z_2)\le \frac{1}{2\pi}\int_{0}^{2\pi}\Psi(z_1,z_2+\epsilon e^{i\theta})d\theta$$ for $\epsilon>0$ small enough so that the integrand does not pass through $z=0$.

My question is: in the paper, it follows immediately that $\Psi(0)<\infty$. I do not know why it follows, since it seems that we do not have a uniform bound through this estimate.

Any hint is welcome. Thanks in advance!

- Let $c$ be a positive integer, $G$ a finite group with at most $c$ conjugacy classes of maximal subgroup. What can we say about $G$?
- Same question, but this time $G$ is a finite group with at most $c$ conjugacy classes of
**core-free**maximal subgroup.

Notes:

- Question 1 is equivalent to asking about groups with at most $c$ primitive permutation representations (up to permutation equivalence).
- Question 2 is equivalent to asking about groups with at most $c$ faithful primitive permutation representations (up to permutation equivalence).
- I'm interested in the same idea, but instead of permutation equivalence, one considers
**permutation isomorphism**. This would change the original questions so that one considers $\textrm{Aut}(G)$-conjugacy classes, instead of just $G$-conjugacy classes.

What might an answer look like:

- If $c=1$, then one can prove that $G$ must be cyclic of prime power order. Core-free will require $G$ is of prime order. Can one give a full classification for $c=2,3,4,\dots$? Non-solvable groups enter at $c=3$ (e.g. $A_5$) -- I'm presuming that $c=2$ will still imply solvability, although I haven't written down a proof.
- If one considers the variant mentioned in the third bullet point of the notes above -- referring to $\textrm{Aut}(G)$-conjugacy classes -- then one also obtains elementary-abelian groups for $c=1$.
- I'm presuming that this stuff has been studied before so this question is also a reference-request. There is a conjecture about upper bounds for the number of maximal subgroups -- see my answer here... But my interest in the current question is specifically about very small values of $c$, so that conjecture is not so relevant...

Let $N$ be a big integer number and consider the equation :

$$ x^{N} + a_{N-1} x^{N-1} + ...+a_{1} x + o(\frac{1}{N})=0,$$ where $o(h)$ is by definition a term such that $\lim_{h \to 0} o(h)/h =0$. Assume that all coefficients $a_j$ have a big norm : $|| a_j||\approx N^{j-1}$ except $a_{1}$ which is asymptotically a nonzero constant.

Let $x_s$ be a complex root of above polynomial with the smallest norm. I want to show something similar to $|| x_s || \approx o(\frac{1}{N}) $ or any upper bound like $|| x_s || \leq \frac{1}{N} $ or even smaller than that. It's obvious that if we don't have the term $o(\frac{1}{N})$ then smallest norm root of the polynomial $$ x^{N} + a_{N-1} x^{N-1} + ...+a_{1} x=0$$ is $x_s=0$. So intuitively makes sense to say that for the first polynomial $|| x_s ||$ is also small but I can't show how small it is .

I do not expect someone gives me the exact solution of this problem because I understand we need more details, but please let me know how you tackle with these kind of questions.

Is there way to round number to decimal places using just simple math operations in one formula?

No converting to text allowed.

No considerations (like "if point is..then") allowed.
No conversions like INT, FLOOR allowed.

Something like this: (2344.334*100)/100 ...

TL;DR.

Some local root numbers of the Hecke character associated with our specific CM elliptic curve by $\mathbf{Q}(i)$ seem to have value in $\mu_4$. But apparently our computation via Rohrlich's local root number formula says otherwise. What am I missing here?

Settings.Let $K = \mathbf{Q}(\sqrt{-1})$ the quadratic ground field and $E$ an elliptic curve over $K$ having complex multiplication by the ring of integers $O_K$ of $K$ defined by an affine Weierstrass equation \begin{equation*} y^2 = x^3 - Dx \end{equation*} for some $D \in O_K$. To $E/K$ we attach a Hecke character $\psi=\psi_{E/K} = \prod' \psi_v: \mathbf{I}_K/K^\times \to \mathbf{C}^\times$ as usual (c.f. e.g. [Sil], sections II.9--10), of conductor $\mathfrak{f}$ which have the same prime factors as the conductor of the elliptic curve $E/K$.

The local root number at $v$ of the Hecke character ($\psi$ in our case) is defined as the signature of its epsilon-factor of the Weil--Deligne representation induced from $\psi_v$, by the local Artin map. The formulae for the local root number are also well-known thanks to Rohrlich [Rohr], pp. 32--33. In particular, when $v$ is a finite and ramified place (i.e.~$v$ divides the conductor $\mathfrak{f}$) not dividing $2$ (i.e.~$v$ is relatively prime to the different ideal of $K/\mathbf{Q}$), the formula for $\chi = \psi_v$ can read: \begin{equation*} W(\chi) = \chi_{u}(\beta) \cdot q^{-a(\chi)/2} \cdot \sum_{x \in (O_v/\mathfrak{f}_v)^\times} \chi^{-1} (x) \exp (2 \pi i \operatorname{tr}^{K_v}_{\mathbf{Q}_p}(x/ \beta) ), \end{equation*} where

- $\beta \in K_v^\times$ is any element with $v(\beta) = a(\chi)$,
- $a(\chi)$ is the conductor exponent of $\psi$ at the place $v$,
- $q$ is the size of the residue field of $K$ modulo $v$,
- $\chi_v$ denotes the unitary part of the character $\chi$.

Let $\ell$ be a rational prime that splits completely in $K$ and $v \not\mid \ell$ be any finite prime of $K$. If $V_\ell E$ is the $\ell$-adic Tate module of $E$ over $\mathbf{Q}_\ell$, then we have $V_\ell E \otimes_{\mathbf{Q}_\ell} \mathbf{C} \cong \psi_v \oplus \psi_v$. See e.g. [Rubin], Proposition 5.4 and Theorem 5.15(ii).

Property 2.It is well-known that the infinity type $\psi_\infty: \mathbf{C}^\times \to \mathbf{C}^\times$ of $\psi$ is given by $z \mapsto z^{-1}$, so that for any $\alpha \in K_{\mathfrak{f},1}$ (the trivial ray), we have $\psi(\alpha O_K) = \psi_\infty^{-1}(\alpha)$ (of course with a fixed embedding $K \hookrightarrow \mathbf{C}$).

From now on, we assume $p$ is a rational prime $\equiv 1 \pmod{4}$ so that $p O_K = \mathfrak{p} \overline{\mathfrak{p}}$. Since $K$ has class number $1$, we can find generators $\pi$ and $\overline\pi$ in $O_K$ of $\mathfrak{p}$ and of $\overline{\mathfrak{p}}$, respectively, so that $p = \pi \overline\pi$. We may and do assume that $\pi$ and $\overline\pi$ are primary, i.e. $\equiv 1 \pmod{2(1+i)}$. Put $D = \pi$ and we may assume $E$ has good reduction modulo $(1+i)$. The last condition is surely satisfied when $D = 1 + 2i$, etc. This implies that our Hecke character $\psi$ is of conductor $\mathfrak{f} = \mathfrak{p}$. We also take $\beta = p$.

Property 3.Recall that $\chi = \psi_v: K_v^\times \to \mathbf{C}^\times$ is the local part of the Hecke character $\psi$. From [Rohr], Proposition 2.1, we can see that \begin{align*} \chi | O_K^\times &= \epsilon^{-1} \\ \chi(\pi) &= \psi_\infty^{-1}(\pi) \end{align*} since there are no other primes dividing $\mathfrak{f}$ except $\mathfrak{p}$. As values of the $\epsilon$-type lie in the group of 4th roots of unity $\mu_4$ (recall $O_K/\mathfrak{f} \cong O_K/\mathfrak{p} \cong \mathbf{F}_5$), we see that \begin{equation*} \chi_u(\beta) = \chi_u(\pi) \chi_u(\bar{\pi}) \equiv \psi_\infty^{-1}(\pi)/|\psi_\infty^{-1}(\pi)| = \pi/|\pi| \pmod{\mu_4}. \end{equation*}

Property 4.Write the ``Gauss sum part'' in the formula of $W(\chi)$ as $G$, i.e. \begin{equation*} G = \sum_{x \in (O_v/\mathfrak{f}_v)^\times} \chi^{-1} (x) \exp (2 \pi i \operatorname{tr}^{K_v}_{\mathbf{Q}_p}(x/ \beta) ) \end{equation*} In our case, this is just \begin{equation*} G = \sum_{x \in \mathbf{F}_5^\times} \epsilon(x) \exp ( 2 \pi i x / p). \end{equation*} Note that there are only four characters $\epsilon: \mathbf{F}_5^\times \to \mathbf{C}^\times$. When $\epsilon$ is the quartic residue symbol, then it is well-known that $G^2 = \pm\pi \sqrt{p}$.

Question.From Property 1, since the root number of elliptic curves is in $\{ \pm 1 \}$, it follows that the root number of $\chi = \psi_v$ lies in the group $\mu_4$. However, from the properties 3--4, we see that \begin{equation*} W(\chi) \equiv \frac{\pi}{\sqrt{p}} \cdot \frac{1}{\sqrt{p}} \cdot \sqrt{\pi} \sqrt[4]{p} \pmod{\mu_4} \neq 1 \pmod{\mu_4} \end{equation*} and hence $W(\chi) \not\in \mu_4$. Why do such kinds of inconsistency occur? We will be very gratitude if someone find out any flaw/mistake in our consideration.

References.[Rohr] Rohrlich, D. E., *Root numbers*.

[Rubin] Rubin, K., *Elliptic curves with complex multiplication and the conjecture of Birch and Swinnerton-Dyer*.

[Sil] Silverman, *Advanced topics in the arithmetic of elliptic curves*.

I'm interested in affinely connected spaces, on which a metric is not necessarily defined, i.e. $(\mathcal{M},\Gamma)$. Since (as a physicist) my goal is to consider a generalized model of gravity, I restrict myself to the space of possible connections compatible with a symmetry group, by requiring that the Lie derivative of the connection---along the vectors associated with the generators of the symmetry group---vanishes: $$\mathcal{L}_\xi \Gamma^a_{bc} = \xi^m \partial_m \Gamma^a_{bc} - \Gamma^m_{bc} \partial_m \xi^a + \Gamma^a_{mc} \partial_b \xi^m + \Gamma^a_{bm} \partial_c \xi^m + \frac{\partial^2 \xi^a}{\partial x^b \partial x^c} = 0.$$

In this space I've been able of define a symmetric tensor of type $\binom{0}{2}$, $T_{ab}$, which is parallel under the action of the torsion-free connection compatible with the symmetries (as defined above), i.e. $\nabla^{\Gamma} T = 0$.

In General relativity, the metric, $g$, is a symmetric $\binom{0}{2}$-tensor, that is required to satisfy the *metricity* condition, $\nabla g = 0$, i.e. it is parallel, under the (Levi-Civita) connection.

Given the similarities a natural question is: Can the tensor $T_{ab}$ be interpreted like a metric? Is there a theorem ensuring that a parallel tensor of $\binom{0}{2}$-type is a metric?

Side noteA while ago, I found a set of notes about holonomy groups saying that (if I remember well) the existence of a parallel $\binom{0}{2}$-tensor was equivalent to restricting the holonomy group to $O(N)$---where $N = \dim(\mathcal{M})$.

It sound to me like the action of the parallel transportation preserves the length of the vectors. Am I right?

Does it mean that starting from an affinely connected space I've end up in a Riemannian space? If so, Where did I make the transition? Since I've pulled the tensor $T$ out from a hat!!!

In Kleene's "On Notation for Ordinal Numbers", Journal of Symbolic Logic, Volume 2, Number 4, December 1938, he says that a function of natural numbers is taken to be effective if it is Herbrand-Goedel recursive, and then cites a 1934 set of notes by Goedel. Does this mean primitive recursive or general recursive?

I have a surjective smooth map with surjective differential between two balls $\phi:B^{2n}\rightarrow B^{2k}$. Fix an integrable almost complex structure $J$ on $B^{2n}$. Assume that $\mathrm{Ker}\:d\phi$ is preserved by the action of $J$.

For any point $q \in B^{2k}$, I can find a point $p\in B^{2n}$ satisfying $\phi(p)=q$ and I can pushforward the complex structure from $T_{B^{2n}, p}$ to $T_{B^{2k}, q}$. Is it true that the resulting complex structure on $T_{B^{2k}, q}$ does not depend on the lift? If so, do I get an integrable almost complex structure on $B^{2k}$?

I think that the answer to the first question is positive if $\phi^{-1}(q)$ is connected as for sufficiently small open sets in the fiber we have local normal form (and independence from the lift becomes self-evident). However, as Mike Miller mentions, the fibers don't have to be connected.

In a dg-category $\mathcal{C}$, the *$n$-translation* of an object $C$ is an object $C[n]$ representing the functor
$$
{\rm Hom}(-,C)[n].
$$
The *cone* of a closed morphism $f\colon C \to D$ of degree zero is an object
${\rm Cone}(f)$ representing the functor
$$
{\rm Cofiber}\big({\rm Hom}(-,C)
\stackrel{{f_{\ast}}}{\longrightarrow}{\rm Hom}(-,D)\big).
$$
On the other hand, the *homotopy cofiber* of $f$ is an object
${\rm Cofiber}(f)$ representing the functor
$$
{\rm Fiber}\big({\rm Hom}(D,-)
\stackrel{{f^{\ast}}}{\longrightarrow}{\rm Hom}(C,-)\big).
$$

Now, suppose $\mathcal{C}$ has zero object, all translations of all objects, and all cones of all morphisms.

My question: **Is there any easy way to show ${\rm Cofiber}(f)$ and ${\rm Cone}(f)$ are isomorphic?**

To be more efficiently, I know that there are canonical closed morphism $\iota\colon D\to{\rm Cone}(f)$ of degree $0$ and morphism $h\colon C\to{\rm Cone}{f}$ of degree $-1$. They induce a natural cochain map $$ {\rm Hom}\big({\rm Cofiber}(f),-\big) \longrightarrow {\rm Fiber}\big({\rm Hom}(D,-) \stackrel{{f^{\ast}}}{\longrightarrow}{\rm Hom}(C,-)\big) $$ which sends any $x\colon{\rm Cofiber}(f)\to X$ to the pair $(x\circ\iota,x\circ h)$.

However, I don't know how to finish the proof, i.e. show this is an isomorphism.

Prove or disprove: Let $ABCD$ be a quadrilateral in Pasch Geometry for which the following holds $AB\cap CD=\{E\}$, $AC\cap BD=\{F\}$ and $AD\cap BC=\{G\}$. Then the points $E,F,G$ are not collinear.

I tried to prove by contradiction defining a graph on set $\{A,B,C,D\}$ with edge between two points if they lie on different sides of the line $EFG$ but couldn't arrive to a contradiction. Maybe there is a counterexample in Poincaré Geometry.

The following observation appears to be implicit in this paper but I am having trouble seeing it.

Let $H \cup_\Sigma H'$ be the Heegaard splitting of $S^3$ of genus $g$. Let $L$ be a $g$ component framed link on $\Sigma$ where the framing coefficients are given by the pushoffs on the surface $\Sigma$. Also assume that the union of the components of $L$ is nonseparating on $\Sigma$. Let $M$ be the 3-manifold obtained by surgery on the framed link $L$.

Let $x_1,...,x_g$ be a collection of simple closed curves on $\Sigma$ that bound disks in $H$. Why is $(\Sigma, \{x_1,...,x_g\}, L)$ a Heegaard diagram for $M$?

Let $G$ be a connected simple Lie group, and $K$ its maximal compact subgroup. Denote by $\tilde{G}$ the universal covering group of $G$, and the covering map $p:\tilde{G}\rightarrow G$ gives a maximal compact subgroup $\tilde{K}=p^{-1}(K)$ of $\tilde{G}$. By polar decomposition, it is known that a simple Lie group has the same fundamental group with its maximal compact subgroup.

Now let $G=\mathrm{E}_{6(-14)}$, and its maximal compact subgroup is $K=\mathrm{SO}(2)\times\mathrm{Spin}(10)$/(center). Thus, the Lie algebra of $K$ is $\mathfrak{so}(2)\oplus\mathfrak{so}(10)$, which is also the Lie algebra of $\tilde{K}$. However, since $\tilde{G}$ is simply connected, so is $\tilde{K}$. But there ought not to be a simpy connected compact Lie group with the Lie algebra $\mathfrak{so}(2)\oplus\mathfrak{so}(10)$ because the simply connected Lie group corresponding to $\mathfrak{so}(2)\oplus\mathfrak{so}(10)$ is $\mathbb{R}\times\mathrm{Spin}(10)$ which is not compact.

I am confused at this point, and I think that I must be get something wrong here. I shall be grateful if experts here may give any comments.

I want to consider the sheaf of an arbitrary category (not only of sets, groups, modules and so on) on a topological space, using the language of etale space.

In all references I am reading (Hartshorne's book, Harris' book, Vakil's note), only the sheaf of sets is considered in this way, although it is not hard to show that the same discussion works for sheaf of groups or modules. But how about the sheaf of an arbitrary category?

For example, let $\mathcal{F}$ be a presheaf of sets on a topological space $X$. We can construct a topology on the disjoint union $F$ of $\mathcal{F}_p$ for all point $p \in X$ as well as a continuous map $\pi: F \rightarrow X$. Then the sheaf $\mathcal{F}'$ of sections of $\pi$ is the sheafification of $\mathcal{F}$.

But I cannot do the same discussion as above in the case where $\mathcal{F}$ is a presheaf of an arbitrary $\mathbf{C}$, because the construction of $\mathcal{F}'$ is based on topology but not on category.

That is, since $\mathcal{F}$ is a presheaf of the category $\mathbf{C}$, every $\mathcal{F}(U)$ and every $\mathcal{F}_p$ is an object of $\mathbf{C}$. Then if we assume that $\mathbf{C}$ has coproduct, the topological space $F =\coprod_{p \in X} \mathcal{F}_p$ can be also regarded an object of $\mathbf{C}$.

But $\mathcal{F}' (U)$ is defined to be the collection of sections of the map $\pi: F \rightarrow X$ on $U$, and none of categorical information is involved in this step. Then how can I claim that $\mathcal{F}'$ is a sheaf not only of sets but also of $\mathbf{C}$?

Of course, for the case of category of groups, rings, or $R$-modules, we can define addtions, productions, or $R$-actions of the sections, then I can show that $\mathcal{F}'$ is what I want. But what is the correct construction for an arbitrary $\mathbf{C}$?

This question is motivated by one of the problem set from this year's Putnam Examination. That is,

**Problem.** Let $S_1, S_2, \dots, S_{2^n-1}$ be the nonempty subsets of $\{1,2,\dots,n\}$ in some order, and let
$M$ be the $(2^n-1) \times (2^n-1)$ matrix whose $(i,j)$ entry is
$$M_{ij} = \begin{cases} 0 & \mbox{if }S_i \cap S_j = \emptyset; \\
1 & \mbox{otherwise.}
\end{cases}$$
Calculate the determinant of $M$. **Answer:** If $n=1$ then $\det M=1$; else $\det(M)=-1$.

I like to consider the following variant which got me puzzled.

**Question.** Preserve the notation from above, let
$A$ be the matrix whose $(i,j)$ entry is
$$A_{ij} = \begin{cases} 1 & \# (S_i \cap S_j) =1; \\
0 & \mbox{otherwise.}
\end{cases}$$
If $n>1$, is this true?
$$\det(A)=-\prod_{k=1}^nk^{\binom{n}k}.$$

*Remark.* Amusingly, the same number counts "product of sizes of all the nonempty subsets of $[n]$" according to OEIS.

**POSTSCRIPT.**

If $B$ is the matrix whose $(i,j)$ entry is $B_{ij} = q^{\#(S_i\cap S_j)}$ then does this hold? $$\det(B)=q^n(q-1)^{n(2^{n-1}-1)}.$$

The Minkowski functional on a normed linear space $E$ is usually defined for convex (or sometimes even non convex) subsets $C$ of $E$ such that $0 \in \operatorname{int}(C)$. Is there any standard reference for the definition on convex sets such that $0 \in C$ but not necessarily $0 \in \operatorname{int}(C)$? For Minkowski functionals of convex sets convexity and positive homogeneity still hold on a convex cone after all.

so Silverman in his 1986 book mentioned about approximating distributions with Gaussian mixture models but he didn't go much further into the topic...I'm just wondering, say I'm given a N-dimensional uniform box (and as a further extension to this, any arbitrary distribution) $u(\bar{x})$ , is there a neat way to approximately it with a (truncated) k-kernel Gaussian mixture model $g(\bar{x}) = \sum_{i} \omega_i N(\mu_i,\sigma^2_i)$?

I've attempted this by trying to minimise some sort of measurable divergence between the two, say the Hellinger distance or the Kullback–Leibler divergence, but analytical solutions don't seem to be plausible and in terms of numerical approaches, the the computational costs climb up way too rapidly as the number of kernels/dimension goes up.

I wonder if any previous studies have been done on this topic before but haven't been quite successful in finding anything useful... if someone can kindly give me some leads or perhaps point me to the relevant literature on this it'd be greatly appreciated!

Cheers!

The $pq$ analog of Serre's conjecture (see "Mod pq Galois representations and Serre's Conjecture"- Khare, Kiming) states that if $\bar{\rho}_1:G_{\mathbb{Q}}\rightarrow \text{GL}_2(\mathbb{F}_p)$ is a mod $p$ Galois representation and $\bar{\rho_2}:G_{\mathbb{Q}}\rightarrow \text{GL}_2(\mathbb{F}_q)$ is a mod $q$ Galois representation, then if $\bar{\rho}_1$ and $\bar{\rho}_2$ are odd, irreducible, ramified at only finitely many primes and of the same "weight", then under some further hypotheses, $\bar{\rho}_1$ and $\bar{\rho}_2$ both lift to Galois representations coming from the same eigenform.

Before Serre made his conjecture (for one prime) there was certainly a lot of computational evidence for it, is there some computational evidence for this conjecture?

It seems that the easiest way to come up with examples of $\bar{\rho}_1$ and $\bar{\rho}_2$ as described would be to take the mod $p$ and mod $q$ representations associated a newform, but such representations don't serve the purpose of producing evidence for the conjecture. One would instead seek to contrive to find $\bar{\rho}_1$ and $\bar{\rho}_2$ that do not arise in this way to find computational evidence.

Let $(M,g)$ be a Riemannian compact manifold without boundary, and $\Delta$ is the Laplace-Beltrami operator on $M$. Is there any result on the elliptic regularity like this:

For any $u\in H^1(M)$, and $f\in L^2(M)$ such that $\Delta u = f$ (in the sens of distributions), Then $u \in H^2(M)$. If there is a nice reference for such regularity result It would be good.

For nearly two years, I have been struggling with the next task I have already published on MSE, but unfortunately with no respond.

Let $M$ be a non-empty and finite set, $S_1,...,S_k$ subsets of $M$, satisfying:

(1) $|S_i|\leq 3,i=1,2,...,k$

(2) Any element of $M$ is an element of at least $4$ sets among $S_1,....,S_k$.

Show that one can select $[\frac{3k}{7}] $ sets from $S_1,...,S_k$ such that their union is $M$.

Partial solution with probabilistic method: I can find a family of ${13\over 25}k$ such sets that no element in $X$ is in more then 3 set from that family. Thus we have a family of the size ${13\over 25}k$ instead of ${4\over 7}k$.

Let' s take any set independently with a probability $p$. Let's mark with $X$ a number of a chosen sets and with $Y$ a number of elements that are ''bad'' i.e. elements which are in at least 4 sets among a chosen sets. Note that $4n\leq 3k$. Then we have $$E(X-Y)=E(X)-E(Y) \geq kp-np^4 \geq kp (1-3p^3/4)$$Since a function $x \mapsto x(1-3x^3/4)$ achives a maximum at $x=\sqrt[3]{1/3}$ we have $E(X-Y)\geq {\sqrt[3]{9}\over 4}k> {13\over 25}k$.

So with the method of alteration we find constant ${\sqrt[3]{9}\over 4}$ which is about $0,051$ worse then ${4\over 7}$.

How far away is

$$\max_{x: x \in \{0, \ldots, N\}} \left|W\left(\frac{x}{N}\right)\right|$$

from

$$\max_{0 \leq t \leq 1} |W(t)|$$

In other words, if you simulate a Wiener process over a finite set of equidistant rational numbers in $[0,1]$ and take the maximum of the Wiener process at those points, how far away is that maximum away from the true maximum of the Wiener process (in any sense you want, be it probability, distribution, expectation, almost surely, etc.)?

This matters because I need to simulate this maximum and since I cannot do so over a continuous set on a computer I would like to know how much of an error I make by working with the next-best thing: a very fine discretization of $[0,1]$.

Another approach to simulating the maximum would be to find the maximum over the discrete set, take the two closest neighboring points, then simulate Brownian bridges connecting the points and computing a new maximum. This seems like a lot of computational work, though, and also seems to just move the problem elsewhere.

Any results appreciated.