Mathematically the definitions are as follows : if $H_n$ is a $n-$dimensional complex Hilbert space then its two different corresponding ``Fock space"(s) are often denoted as $F_{1}$ and $F_{-1}$ defined as, $F_1 = \oplus_{k=0}^{\infty} Sym^k(H_n)$ and $F_{-1}= \oplus_{k=0}^{\infty} \Lambda^k(H_n)$.

Physically for a Quantum Field Theory one "defines" its so-called Hilbert space as the dual of an implicit vector space over $\mathbb{C}$ whose basis is in bijective correspondence to the set of all possible values for all the classical fields that occur in the underlying Lagrangian.

Now my question is two fold,

Does this physical notion of a "Hilbert space of a QFT" correspond to the $H_n$ or some ``total Fock space" that can be defined from the first mathematical definition as, $\otimes_{i \in Fields} F^i_{p_i}$ where $p_i=$1 if the $i^{th}$ field is Bosonic or $-1$ if it is Fermionic? (..I guess this tensoring is needed because the QFT can have both Fermionic as well as Bosonic fields..)

If we agree as above that the states of a QFT live in such a "total Fock space" and not in the Hilbert space defined in the second paragraph then shouldn't the "Quantum Field" be mapping into a space of Hermitian operators on this total Fock space and not just the Hilbert space?

We are studying the behavior of families of curves inside stable families of surfaces. The non-existance of the following configurations of curves in a non-normal surface would be sufficient to prove our result.

Let $X$ be a non-normal surface over $\mathbb{C}$ with non-normal locus (i.e. double locus) $X_{dl}$.

Let $D \subset X$ be a nodal curve, with node $p$ so that $p \subset X_{dl}$, the pair $(X,D)$ has semi-log canonical singularities, and $K_X + D$ is ample.

Let $C = C_g \cup C_0$ be a genus $g(C) = g$ irreducible curve, which is the union of a genus $g$ curve $C_g$ and a rational curve $C_0$. Further suppose that $C_g \cap C_0 = p$ (the node of $D$), and that $C_0 \subseteq X_{dl}$.

Finally, suppose that $C_0 \cap D = p$ and the normalization of $X$ is irreducible.

Is such a configuration possible? For instance, is $C_0 \cap D$ forced to be more than one point?

I am looking at the first proof of the existence of a fundamental solution for Linear partial differential equations with constant coefficient (The 3.1.1, Linear Partial Differential Operators, Hormander, Springer Verlag 1964), using Hahn-Banach. In order to construct a fundamental solution such that $$ P(D) E = \delta $$ Hormander uses a lemma giving for $u$ a smooth function with compact support : $$| u(0) | \leq C || P(D)u || .$$ He constructs a function $E \ast \delta_0$ which is defined as $P(D) u \mapsto u(0)$ on the vector space of functions $\{g | g = P(D)u \text{ for some } u \}$ and then extended to $E \ast \delta_0 : \mathcal{D}(\mathbb{R}^n) \to \mathbb{R} $ by Hahn-Banach. My naive question is how is $E \ast \delta_0$ well defined in the first place, as when considering a function $f = Du$, $u$ and thus $u(0)$ may vary ? I agree that once $E$ is fixed then $u$ and $u(0)$ are unique, but I don't see how this can be at the beginning of the proof. Is any of you familiar with this proof ? Am I missing something obvious ?

Thanks.

The finite field of order $p^n$ is isomorphic to $(\mathbb Z/p \mathbb Z)[X]/(P)$, where $P$ is an irreducible polynomial in $(\mathbb Z/p \mathbb Z)[X]$ of degree $n$. This describes every finite field up to isomorphism.

My question is, what are alternate ways of describing these finite fields?

For example, Conway gave an alternate way of describing the fields $GF(2^{2^n})$ (see this example answer).

Say $L^2(M_1,\mathbb{R})$ is the space of square-integrable functions on $M_1$ which is a compact manifold with a measure defined on it. Let $L^2(\mathbb{R}^n,\mathbb{R})$ be the space of all square-integrable functions mapping $\mathbb{R}^n \rightarrow \mathbb{R}$.

Now I consider the space of all functions $F$ mapping, $M_1 \rightarrow \mathbb{R}$ of the form, $\{ f(g_1,..,g_n) \vert f \in L^2(\mathbb{R}^n,\mathbb{R}) \text{ and } g_i \in L^2(M_1,\mathbb{R})\}$

Now given a (countable) basis for $L^2(M_1,\mathbb{R})$ and $L^2(\mathbb{R}^n,\mathbb{R})$ can one write down a countable basis for $F$?

Are there general theorems known about when one can write down a basis for the composed space in terms of a given set of bases for the spaces being composed?

Task question:

Please solve and visualize movement using winbgi.h graphic library (available from faculty/AeroDiv/Courses website).

• dx/dt = -y - z • dy/dt = x + ay • dz/dt = b + z(x - c) • Usual parameters: a = b = 0.2, c = 5.7

Visualize z=z(t,x,y). Below is the code for the above problem :

include include include include include include"winbgi2.h" define n 3 define dist 0.1 define max 50void rungekutta(double x, double y[], double step);
double fun(double x, double y[], int i);
void main()
{
graphics(50,30);
double t, z[n];
int j;
FILE *f;
fopen_s(&f, "padela.txt", "w+");
t = 0.0;
z[0] = 0;
z[1] = 0;
z[2] = 0;
fprintf(f,"time\t\tZ-axis\n");
fprintf(f, "%lf\t\t%lf\n",t,z[2]);
for (j = 1; j*dist <= max; j++)
{
t = j*dist;
rungekutta(t,z,dist);
fprintf(f, "%lf\t\t%lf\n",t,z[2]);
circle(t+200,z[2]+200,7);
}

fclose(f); getch(); }

void rungekutta(double x, double z[], double step)
{
double h = step / 2.0,
t1[n], t2[n], t3[n],
k1[n], k2[n], k3[n], k4[n];
int i;
for (i = 0; ifun(x, z, i));
}
for (i = 0; i(k2[i] = step*fun(x + h, t1, i));
}
for (i = 0; ifun(x + h, t2, i));
}
for (i = 0; i
*

*double fun(double x, double z[], int i)
{
double a=0.2,b=0.2,c=5.7;
if (i == 0)
{
return -z[1]-z[2];
}
if (i == 1)
{
return z[0]+(az[1]);
}
if (i == 2)
{
return b+(z[2]*(z[0]-c));
}
}
I need the continuation code for the program which states below question:
1) Find the peak value of Z from Z-t graph at 43 second. (Since the h is changing, so find the peak value at 43 second) consider the peak value as ZR.
2) Calculate value of Z (h) where h= 1/100000. The values of h is multiplied by 10 every time till h=1.
3) Calculate the error. Error = (Z (h) - ZR) / ZR
4) If the value of error in the range 2*10-3 < …….<4*10-3. Then print the value of error and also h (step).

Let $E$ be an infinite-dimensional complex Hilbert space.

For $T = (T_1,\cdots,T_n)\in\mathcal{L}(E)^n$, the algebraic spectral radius of $T$ was given by $$ r_a(T_1,\cdots,T_n)=\lim_{k\to+\infty}\left\|\sum_{f\in F(k,n)} T_f^* T_f\right\|^{\frac{1}{2k}} , $$ where $F(k,n):=\{f:\,\{1,\cdots,k\}\longrightarrow \{1,\cdots,n\}\}$ and $T_f:=T_{f(1)}\cdots T_{f(k)}$, for $f\in F(k,n)$.

Gelu Popescu has a paper (Memoirs of the AMS, arXiv). In page 8 of this memoirs I see the following paragraph:

Are the two definitions equivalents?

A question I was asked recently lead me to the following question. For every closed first order formula $\theta$ in the group signature consider the set $N_\theta$ of natural numbers $n$ such that the symmetric group $S_n$ satisfies $\theta$.

**Question.** Can $N_\theta$ always be defined by a first order formula in the signature of natural numbers?

**Update.** The original question has been answered below by Noah Schweber, but it occurred to me that I am mostly interested in the converse translation. So here is

**Converse question.** Given a first order formula that defines a set $M$ of natural numbers, is there always a first order formula in the group signature defining the set of symmetric groups $\{S_n\mid n\in M\}$?

Suppose we have an M$\times$N complex matrix $H$ and its singular value decomposition $H=U\Lambda V^*$ and an N$\times$N covariance matrix $R_s$ with its eigendecomposition $R_s = U_s\Lambda_sU_s^*$. We also have the eigendecomposition of $HR_sH^*$ as $U_A\Lambda_AU_A^*$. In my research problem setting, I know $U_A$, $H$ but not $R_s$ and $\Lambda_A$. I want to find $U_s$ using $H$ and $U_A$. I was wondering if there exists any connection between them.

I am trying to prove Proposition 2.1.1 of Gauduchon's note on Kähler extremal metrics (page 67). In order to show that, for compact Kähler manifolds, the complex Lie algebra of real holomorphic vector fields $\mathfrak{h}$ is finite-dimensional, he argues the following.

Firstly $\mathfrak{h}$ is identified with the space of holomorphic vector fields of $T^{1,0}$, which is equipped with an obvious $\overline{\partial}$ operator. Then he claims that holomorphic vector fields belong to $Ker(\overline{\partial})$ and $Ker(\overline{\partial}^*)$, being $\overline{\partial}^*$ the adjoint of $\overline{\partial}$, with respect to any compatible metric. Finally, since the kernel of the elliptic operator $\overline{\partial}^*$ + $\overline{\partial}$ is clearly finite dimensional, the result follows.

I am not able to prove that an holomorphic vector field $X$ belongs to $Ker(\overline{\partial}^*)$. When I identify $X$ with a (0,1)-form $\omega$, using the Kähler metric, it is not hard to prove that $\omega$ is $\overline{\partial}$-closed. However, expressing $\omega$ in local coordinates, I am not able to derive also the $\overline{\partial}^*$-closedness.

Do you have any suggestion?

If $Y\subset X^*$ is a closed subspace (where $X$ is a separable Banach space), the preannihilator of $Y$ in $X$ is $Y_{\perp}:=\{x\in X : y^*(x)=0, \forall y^*\in Y \}$. If $Y$ is a proper subspace of $X^*$, and $X$ is reflexive then it can be proved that $Y_{\perp}\neq\{0\}$. On the other hand if $X=l_1$ and $Y=c_0\subset l_{\infty}$, then $Y_{\perp}=\{0\}$.

- 1) If $X^*$ is separable, is it still possible that $Y_{\perp}=\{0\}$?
- 2) Can $c_0$ be a subspace of a separable dual? This question is probably unrelated, and most likely well known, but I realized I do not know an answer to it.

Let be $S$ a separable(non compact) metric space and $X=C_b(S)$ the set of all bounded continuous functions, then it's topological dual $X^{\star}=rba(S)$ is the set of all regular Borel additive measures endowed with the variation norm. Denote by $\mathscr{P}(S)$ the subset of $rba(S)$ of all additive probability measures, endowed with the weak$^{*}$ topology.

Fix $\mu$ a extremal point of $ \mathscr{P}(S)$ and a closed set $\mathcal{F}$ in $\mathscr{P}(S)$ such that the convex hull $co(\mathcal{F})$ do not contains $\mu,$ then it is possible to show that there is an affine functional $\ell_{\mu,\mathcal{F}}:\mathscr{P}(S)\to \mathbb{R}$ such that $\ell_{\mu,\mathcal{F}}(\mu)=0$ and $\ell_{\mu,\mathcal{F}}$ is strictly positive in $\mathcal{F}$

**Question 1:** Is there some way to use the above stated to get a global result, that is, I want to show that there is a affine functional $\ell_{\mu}:\mathscr{P}(S)\to \mathbb{R}$ such that $\ell_\mu(\mu)=0$ and $\ell_\mu(\nu)>0$ for all $\nu\in \mathscr{P}(S)\setminus \{\mu\}$ ?

**Question 2:** If not, is there some other approach to show that there is a affine functional $\ell_{\mu}:\mathscr{P}(S)\to \mathbb{R}$ such that $\ell_\mu(\mu)=0$ and $\ell_\mu(\nu)>0$ for all $\nu\in \mathscr{P}(S)\setminus \{\mu\}$ ?

**Edit:** Following the suggestions of Jochen Glueck I had made some edits in the second paragrapher

For any integer $m>2$, let $P_m$ be the set of primes less than $m$, and let $$ f(m) = \sum\limits_{p \in P_m} \frac{1}{m-p}. $$ For example, $f(3)=\frac{1}{3-2}=1$, $f(4)=\frac{1}{4-2}+\frac{1}{4-3}=\frac{3}{2}$, and so on.

The question is to estimate $I=\inf\limits_{m>2} f(m)$.

A simple Mathematica calculation shows that $f(m)\geq f(223)\approx 0.60178$ for all $m$ up to $10,000$. It is true that $I>0$? Is $I>0.5$? Is $I=f(223)$?

What is the finite description of Rayo's number? And how does it compare in theoretical size to transfinite numbers? How many operator levels are required to comprehend Rayo's number?

Let x be a random n-bit string, and let $I ={i_1,i_2,...,i_n}$ be the starting indexes of the longest 0-runs of x, sorted in decreasing order (so $i_1$ is the starting index of the longest (~$\log n$) 0-run, and ties are broken by lexicographic order). Note that the entropy of $I$ is at most $H(I) <= n$ bits (since $I$ is determined by $x$ and $H(x)=n$). What I'm trying to show is that "most" of this entropy is actually concentrated on a rather *small* number of indices with "large" entropy. More precisely, is it true that there is a subset $S \subseteq I$, $|S|=n/poly\log(n)$, s.t $H(S) > |S|*h$ , where $h >= \log^{\epsilon}n$ or even $h =\Omega(\log(\log(n)$)).

Note $H(i_1) = \log n$ by symmetry. The intuition is that $i_2,\ldots, ,i_{\Omega(\sqrt{\log n)}}$ are *determined* by $i_1$ since there's a gap of at least ~$\Omega(\sqrt{\log n)}$ between the first and second longest 0-runs. Then there's another discontinuous "random jump" to the next non-overlapping run, etc.

Would appreciate if anyone can point out relevant references.. Thanks!

Let $(R, \mathfrak m)$ be a valuation ring such that $\mathfrak m$ and every ideal of finite height is principal. Then is $R$ Noetherian , i.e. a discrete valuation ring ?

Let $d(n)$ denote the number of positive divisors of $n$. Is it known how to evaluate the sum

$$\displaystyle \sum_{1 \leq m < n \leq X} d(m) d(n) d(n-m)?$$

A slightly more difficult question is if we change the height condition in the summation, to obtain the sum

$$\displaystyle \sum_{\substack{1 \leq mn(n-m) \leq X \\ 1 \leq m < n}} d(m) d(n) d(n-m).$$

This is a generalization of the single variable case, where it is known how to evaluate sums of the form

$$\displaystyle \sum_{1 \leq n \leq X} d(an + b) d(cn+d)$$

for fixed positive integers $a,b,c,d$.

The *density* of a set
$X\subseteq\omega$ refers to:
$\limsup\limits_{n\rightarrow\infty}\dfrac{C\cap n}{n}$.

Given a set of positive integers $F= \{m_0<\cdots<m_{k-1}\}$, let $C\subseteq \omega$ be such that for every $x$ there exists $y\in (\{x\}\cup x+F)\cap C$.

Q1. Given $F$, what is the smallest possible density of $C$? Is it always $1/(1+k)$?

Q2. Given $F$, if one build $C$ stochastically as following, what is the density of $C$? For every $x$, if there is not any element in the current $(\{x\}\cup x+F )\cap C $, then select an element $y$ uniformly randomly from $ \{x\}\cup x+F$ and add $y$ into $C$.

Let $G$ be locally compact group and let $H$ be a open subgroup in $G$. Then the full group $C^*$-algebra of $H$, $C^*(H)$, is a subalgebra of $C^*(G)$ and there is a conditional expectation $$E\colon C^*(G)\to C^*(H),$$ which is induced by restriction $f\in L^1(G) \mapsto f_{|H}\in L^1(H)$ of functions which are integrable w.r.t. the left Haar measure on $G$, see Rieffel, induced representations of $C^*$-algebras, Proposition 1.2.

Note that $E$ is'n faithful in general, there is an example in this paper in the section above definition 3.: Consider $G$ a nonamenable discrete group and $H$ an open subgroup consisting of the identity element of $G$. Then there are nonzero elements $c$ in the kernel of the left regular representation of $G$ (since $G$ is not amenable) and they satisfy $E(c^*c)=0$.

My Question: Now, let $G$ be a locally compact amenable group and $H$ be an open compact (amenable) subgroup.

Is then $E$ faithful?

I think yes (I have considered some examples), but I am stuck with a proof.

If I additionally assume $G$ (and $H$) to be discrete I can prove it considering $E$ as a conditional expectation $C_r^*(G)\to C_r^*(H)$ and then it is $\tau_G=\tau_H\circ E$, where $\tau_G$ and $\tau_H$ are the canonical faithful tracial states on $C_r^*(G)$, $C_r^*(H)$ respectively. It follows that $E$ must be faithful.

For the more general case, I thought about trying a similar strategy using the fact that $C_r^*(G_1)$ of a locally compact group $G_1$ which contains a non-trivial amenable open subgroup $H_1$ has a tracial state $\tau^{G_1}$ satisfying $$\tau^{G_1}( \lambda_{G_1}(f))=\int_{H_1}f(s)d\mu(s),$$ see corollary 4.1 in 'embedding theorems in group $C^*$-algebra' by Lee for this fact. If one can check that the tracial state $\tau^{H_1}$ is faithful, then this together with $\tau^{G_1}=\tau^{H_1}\circ E$ implies that $E$ is faithful. But I am stuck with proving faithfulness of the trace. Probably I am on the wrong track..Other strategies regarding my question are welcome.

I am interested in the generating function of $SO(N)$ random matrix, that is, I want to compute $$ Z_N[J]=\int dM e^{{\rm Tr} (J^T M)}, $$ where $dM$ is the $SO(N)$ Haar measure, and $J$ is an arbitrary $N\times N$ matrix. From this generating function, I can generate all correlations $\langle M_{ij}M_{kl}\cdots\rangle$ by taking derivatives with respect to the elements of $J$.

Due to the invariance of the measure, one sees that $Z[J]=Z[U^TJV]$ with $U,V\in SO(N)$, and thus $Z$ only depends on the singular values of $J$. (Stated otherwise, $Z$ only depends on the $N$ invariants ${\rm Tr}((J^TJ)^n)$, $n=1,...,N$).

Finally, at least for $N=2$ and $N=3$, one can show that $Z$ is also invariant under permutations of the singular values of $J$ (maybe it can be generalized for all $N$ ?).

It is not too hard to compute explicitly $Z_2[J]$, which is given in terms of a Bessel function of the sum of the two singular values of $J$.

Is there a way (or has it been done in the literature) to compute $Z_N$ for any $N$ ? I would already be happy with $Z_3$, which I cannot manage to compute explicitly.

EDIT : Here is an attempt, which kind of works for $N=2$, but for which I am stuck for $N=3$. If we define a Laplacian $\Delta=\sum_{ij}\frac{\partial^2}{\partial J_{ij}^2}$ (with $J_{ij}$ the elements of $J$), one shows easily that $$ \Delta Z[J]=N Z[J]. \tag{1} $$ If we call $\lambda_i$ the singular values of $J$ (with $\lambda_1>\lambda_2>\ldots$), using the fact that $Z[J]=Z[\lambda_1,\lambda_2,\ldots]$, one shows (at least for $N=2$ and $N=3$, but it might be generalizable to $N\geq4$) that $$ \Delta Z=\frac{1}{D}\sum_{i}\frac{\partial}{\partial \lambda_i}\left(D\frac{\partial}{\partial \lambda_i}Z\right), $$ where $D=\prod_{i< j}(\lambda_i^2-\lambda_j^2)$ is related to the Jacobian to go from $J_{ij}$ to $\lambda_i$. This equation looks nice enough, so my hope is that a solution exists, I am not quite sure how to find it for $N=3$.

In the case $N=2$, we can compute $Z_2$ exactly via its definition, and it reads $Z_2[\lambda_1,\lambda_2]=I_0(\lambda_1+\lambda_2)$, with $I_\nu$ the modified Bessel function of the first kind. One checks that this is indeed a solution of Eq. (1).

Unfortunately, even in that case, it is not clear to me how to find this solution starting from Eq. (1) only. Given that $D=\lambda_1^2-\lambda_2^2$, it is tempting to defined $u=(\lambda_1+\lambda_2)/2$ and $v=(\lambda_1-\lambda_2)/2$. Then Eq. (1) is solved by separation of variables and we find a family of solution $Z_{2,\mu}$ (where I have already use the fact that $Z_N[0]=1$) : $$ Z_{2,\mu}[u,v]=I_0(\sqrt{\mu}u)I_0(\sqrt{4-\mu}v). $$ Clearly the solution to my problem corresponds to $\mu=4$, but it is not clear to me what is the rigorous argument to pick this value of $\mu$ (since $Z_N>0$ $\forall J$, we must have $\mu\geq 4$ as $I_0$ can be negative for imaginary variables; but how to select $\mu=4$ as the only viable solution ?).

One way to solve this issue of $\mu$ is to use the fact that $Z_2[J=\lambda Id_2]$ can be computed explicitly: $Z_2[J=\lambda Id_2]=I_0(2\lambda)$, which unambiguously selects $\mu=4$.

Since we can always compute the expansion of $Z_N[\lambda Id_N]$ explicitly at least for small $\lambda$, this kind of argument might be enough to fix the constant also for $N>2$. For $N=3$, a few special cases can be computed explicitly, which might help too.

Any insight for the solution of Eq. (1) would be greatly appreciated.