Denote by $W$ the first Weyl algebra over a field $K$ of characteristic zero, $W := \langle X,Y | YX-XY=1 \rangle$.

Based on Guccione, Guccione, and Valqui - The Dixmier conjecture and the shape of possible counterexamples (denote it by GGV. Notice that I refer to the first version), I would like to ask the following question:

It seems that **the Dixmier Conjecture** follows quite immediately from the results in GGV, isn't it? Or please where is my error? (high chances that I have an error somewhere, I guess).

**My proof:**
It seems that the following argument is a proof for the Dixmier Conjecture:

By Theorem 3.3[GGV], we would like to show that there is no irreducible endomorphism of $W$.

We assume that there exists an irreducible endomorphism of $W$ and reach a contradiction, hence there is no irreducible endomorphism of $W$.

Let $f: (X,Y) \mapsto (P,Q)$ be an *irreducible* endomorphism of $W$.

**Claim:** If $h$ is an automorphism of $W$, then $hf$ is also an irreducible endomorphism of $W$.
**Proof of Claim:** Follows from Definition 3.1[GGV].

There always exists (regardless of being irreducible) a change of variables, $g$, such that $(gf)(x)$ is of the following form: $y^N+r_{N-1}y^{N-1}+\cdots+r_1y+r_0$, where $r_{N-1},\ldots,r_1,r_0 \in K[X]$; indeed, we can take $g: (X,Y) \mapsto (X+Y^M,Y)$, for large enough $M \in \mathbb{N}$.

Let $\tilde{f}:= gf$. On the one hand, our claim shows that $\tilde{f}$ is an irreducible endomorphism of $W$. On the other hand, by Lemma 5.10[GGV], $\tilde{f}$ cannot be an irreducible endomorphism of $W$.

(I presented a short proof; if one wishes that I will present a proof with more details, then please let me know and I will).

Moreover, if the results in GGV are also valid for the polynomial ring $K[x,y]$ (I have not yet tried to find proofs for the analogous commutative results to the non-commutative results), then my proof also serves as a proof for **the two-dimensional Jacobian Conjecture**.

Any comments are welcome! Thank you!

Suppose that $\phi(t,x):[0,\infty)\times \mathbb{R}^d\rightarrow \mathbb{R}^d$ is a flow. Is it possible to extend $\phi$ to a family of stochastic flows $\{\Phi(t,x,\sigma)\}_{\sigma \in [0,1]}$ such that

- For every $\sigma_0 \in (0,1]$, $\Phi(t,x,\sigma_0):[0,\infty)\times \mathbb{R}^d\rightarrow \mathbb{R}^d$ is a stochastic flow,
- $\Phi(t,x,0)=\phi(t,x)$,
- For every $\sigma_0 \in[0,1]$, $\lim\limits{\sigma \downarrow \sigma_0} \Phi(t,x,\sigma)=\Phi(t,x,\sigma_0)$ almost-surely?
- For $\sigma>0,x \in \mathbb{R}^d$, $\lim\limits_{t \mapsto \infty}\Phi(t,x,\sigma_0) \in \mathbb{R}^d$?

Let $\alpha$ be a root of a polynomial $a_nx^n + \ldots + a_1x$ with integral coefficients.

I would like to determine $\varepsilon > 0$ depending on $a_1, \ldots, a_n$ so that $|\alpha| < \varepsilon$ implies $\alpha = 0$.

Is it possible to give a "formula" for such an $\varepsilon$ without refering to the complete list of roots?

Consider the linear operator $\mathbb{L} : L^2([0,1])\to L^2([0,1])$ defined by $$ (\mathbb{L}f)(x) = \int_0^1 xy(f(x)-f(y)) \mathrm{d}y $$ for all $f\in L^2([0,1])$ and $x \in [0,1]$. Is $\mathbb{L}$ diagonalizable and why?

**Definition:** $\mathbb{L}$ diagonalizable means that there exists eigenvalues $\{\lambda_k\}_{k\in\mathbb{N}}$ and an orthonormal basis $\{f_k\}_{k\in\mathbb{N}}$ such that

$$ \mathbb{L}f = \sum_{k\in\mathbb{N}} \lambda_k\langle f, f_k \rangle f_k $$ for all $f\in L^2([0,1])$.

$\zeta(3)$ has at least two well-known representations of the form $$\zeta(3)=\cfrac{k}{p(1) - \cfrac{1^6}{p(2)- \cfrac{2^6}{ p(3)- \cfrac{3^6}{p(4)-\ddots } }}},$$

where $k\in\mathbb Q$ and $p$ is a cubic polynomial with integer coefficients. Indeed, we can take $k=1$ and$$ p(n) =n^3+(n-1)^3=(2n-1)(n^2-n+1)=1,9,35,91,\dots \qquad $$ (this one generalizes in the obvious way to the odd zeta values $\zeta(5),\zeta(7),...$) or, as shown by Apéry, $k=6$ and
$$ p(n) = (2n-1)(17n^2-17n+5)= 5,117,535,1463,\dots
. $$ **Numerically, I have found that $k=\dfrac87$ and $p(n) = (2n-1)(3n^2-3n+1)$ also works.** (Is that known? Maybe Ramanujan obtained that as some by-product?)

The question:

- Are there other values of $k$ where such a polynomial exists?
- Must all those polynomials have a zero at $\dfrac12$ for some reason?

I have the following trivariate ($\rho_{11}, \rho_{22}, \mu$) function \begin{equation} 4 \mu ^{3 \beta +1} \rho_{11}^{3 \beta +1} \left(-\rho_{11}-\rho_{22}+1\right){}^{3 \beta +1} \rho_{22}^{3 \beta +1} \left(\mu ^2 \rho_{22}+\rho_{11}\right){}^{-3 \beta -2}, \end{equation} the (three-fold) integral (for $\beta$ nonnegative integer) of which over $\mu \in [0,1]$, $\rho_{11} \in [0,1]$,$\rho_{22} \in [0,1- \rho_{11}]$ is \begin{equation} \frac{\Gamma \left(\frac{3 \beta }{2}+1\right)^4}{\Gamma (6 \beta +4)}. \end{equation} I would like to know the (two-fold) integral, say $f(\mu,\beta)$, over $\rho_{11} \in [0,1]$,$\rho_{22} \in [0,1- \rho_{11}]$.

A reference for the (quantum-information-theoretic) background of this problem is sec. II.A.1 of Slater - Extended Studies of Separability Functions and Probabilities and the Relevance of Dyson Indices, in particular eq. (8) there. (For later related work, see secs. III,IV of Slater - Master Lovas-Andai and Equivalent Formulas Verifying the 8/33 Two-Qubit Hilbert-Schmidt Separability Probability and Companion Rational-Valued Conjectures, where I try to relate Lovas-Andai and Slater "separability functions".)

Mathematica computes the integral for specific nonnegative integer $\beta$, but apparently not for general $\beta$. (I can compute for $\beta=1,2,\ldots$ and then try to employ the Mathematica command FindSequenceFunction to obtain the general [two-fold] rule--which is, I think, how I got the three-fold result.)

For example, for $\beta=1$, we have \begin{equation} \frac{\mu ^4 \left(12 \left(\mu^8+16\mu ^6+36 \mu ^4+16\mu ^2+1\right) \log (\mu )-5 \left(5 \mu ^8+32 \mu ^6-32 \mu ^2-5\right)\right)}{945 \left(\mu ^2-1\right)^9} \end{equation} and for $\beta=2$, \begin{equation} \frac{\mu ^7 \left(140 \left(\mu^2+1\right)\left(\mu ^{12}+48 \mu ^{10}+393 \mu ^8+832 \mu ^6+393 \mu ^4+48 \mu ^2+1\right)\log(\mu ) -(\mu^2-1) \left(363 \mu ^{12}+10310 \mu ^{10}+58673 \mu^8+101548 \mu ^6+58673 \mu ^4+10310 \mu ^2+363\right)\right) }{900900 \left(\mu ^2-1\right)^{15}}. \end{equation}

A matrix $A \in \mathbb{R}^{n \times n}$ is called completely positive if there exists an entrywise nonnegative matrix $B \in \mathbb{R}^{n \times r}$ such that $A = BB^{T}$. All eigenvalues of $A$ are real and nonnegative.

My question is when will a completely positive matrix have all positive eigenvalues?

The only completely positive matrix I know so far have zero eigenvalues is \begin{equation} A = \begin{pmatrix} 41 & 43 & 80 & 56 & 50 \\ 43 & 62 & 89 & 78 & 51 \\ 80 & 89 & 162 & 120 & 93 \\ 56 & 78 & 120 & 104 & 62 \\ 50 & 51 & 93 & 62 & 65 \end{pmatrix} . \end{equation} So probably that the case completely positive matrix has zero eigevalues are rare. But I could not find any documment on this.

**Update:** due to @Robert Israel answer.

Usually, we do not know $B$ in general and indeed the composition may not unique. Therefore the condition depends only in $A$ would be easier to verify. For example with \begin{equation} A = \begin{pmatrix} 18 & 9 & 9 \\ 9 & 18 & 9 \\ 9 & 9 & 18 \end{pmatrix} \end{equation} there are at least three decompose satisfies $B \geq 0$. In particular \begin{equation} B_{1} = \begin{pmatrix} 4 & 1 & 1 \\ 1 & 4 & 1 \\ 1 & 1 & 4 \end{pmatrix} , \quad B_{2} = \begin{pmatrix} 3 & 3 & 0 & 0 \\ 3 & 0 & 3 & 0 \\ 3 & 0 & 0 & 3 \end{pmatrix} , \quad B_{3} = \begin{pmatrix} 3 & 3 & 0 \\ 3 & 0 & 3 \\ 0 & 3 & 3 \end{pmatrix} . \end{equation}

$\require{AMScd}$Let $F:\mathcal{C}^{op}\times \mathcal{C} \to \mathcal{D}$ be a functor and $\mathcal{C}' \subseteq \mathcal{C}$ a full subcategory. Assume that the coends $C$ over $F$ and $C'$ over $F|_{\mathcal{C}'}$ exist. By the universal property of $C'$, we obtain a map $i:C' \to C$.

I am looking for a necessary and sufficient condition for this to be a monomorphism.

I think we can consider $C$ as a pushout of the diagram $$ \begin{CD} \coprod_{c\in\mathcal C'} F(c,c) @>>> C' \\ @VVV \\ \coprod_{c\in\mathcal C} F(c,c) \end{CD} $$ in the category of cocones $Cocone(F)$ of $F$. Hence, if coprojections are monos in $\mathcal{D}$ and pushouts preserve monos in $Cocone(F)$, $i$ is a monomorphism. However, the last condition seems to be a bit strong, so is there another argument I'm not seeing? In particular, is it true for abelian categories?

It is well known that for a discrete random walk on the integers with a fair coin, the expected distance of the walker from the origin after $N$ time steps is $\sqrt{\frac{2N}{\pi}}$ if $N$ is large. For example, Wolfram Mathworld has a thorough explanation here.

Consider instead the following situation. We have two types of weighted coins. Type A coins have a probability $1/3$ to land heads, and a probability $2/3$ to land tails. Type B coins have a probability $2/3$ to land heads, and a probability $1/3$ to land tails. At each integer location, we place one of the coin types there at random, so that each integer location has exactly a 50% chance to get a type A coin, and a 50% chance to get a type B coin (coins are independently chosen at each integer location).

When the walker is at a particular integer location, she flips the coin placed there and goes one step right if it lands heads, and one step left if it lands tails. I believe in the literature this is known as a "Random Walk in a Random Environment" (RWRE)

What then is the expected value of the distance of the walker from the origin after $N$ steps for large $N$? I have done numerics that suggest that this value is still $\sim \sqrt N$, but the constant term is probably not $\sqrt{\frac{2}{\pi}}$.

$$\partial_t f(x,t)= f(x,t)+f(x-a,t)$$

Consider the determinantal function $$F(a,b,c):=\det\left[\binom{i+j+a+b}{i+a}\right]_{i,j=0}^{c-1}.$$ I would like to ask:

**QUESTION.** Can you provide an argument, combinatorial or otherwise, to prove the symmetry
$$F(a,b,c)=F(c,a,b)=F(b,c,a)=F(a,c,b)=F(c,b,a)=F(b,a,c)$$
without actually computing the its values? I know how to justify this based on direct evaluation.

**Example.** The following are all equal: $F(2,3,4)=F(4,2,3)=F(3,4,2)$ or
$$\det\begin{pmatrix} 10&15&21&28\\20&35&56&84\\35&70&126&210\\56&126&252&462
\end{pmatrix}=
\det\begin{pmatrix} 15&35&70\\21&56&126\\28&84&210
\end{pmatrix}=\det\begin{pmatrix} 35&56\\70&126
\end{pmatrix}.$$

Let $E$, $F$ be Banach spaces, $D$ be open in $E$, and $K=[0,1]$. Given $\varphi\colon K\times D\to F$ I call
$$
\varphi^\sharp\colon D^K\to F^K,\quad u\mapsto \varphi(\cdot,u(\cdot))
$$
the *superposition operator*.

I am interested in a run-down on how $\varphi^\sharp$ as a mapping between some vector spaces $V\subseteq D^K$ and $W\subseteq F^K$, both endowed with appropriate norms, inherits regularity from $\varphi$.

E.g., I can show that $\varphi^\sharp \in C^k\left(C\left(K,D\right),C\left(K,F\right)\right)%\tag{$\star$}$ provided $\varphi$ and $\partial_2^k\varphi$ are continuous.

I am especially interested in finding $V$ to be some Hilbert space that contains all piecewise smooth functions $K\to D$.

Many thanks in advance!

We consider the f-divergence, which takes the form $$ D_f(P \| Q) = \int_\Omega f\left(\frac{dP}{dQ}\right) dQ. $$ For example, when $f(t) = t \log t$, we obtain the KL-divergence.

My question is that, as a function of $P$, under what regularity conditions on $f$ and $Q$ is the f-divergence $D_f(P \| Q)$ (geodesic) smooth with respect to the Wasserstein metric $W_2$?

For any function $g(P)$, the (geodesic) $\mu$-smoothness of $g(P)$ is defined as $$ g(P') \leq g(P) + \langle\text{grad}\,g(P), \exp_P^{-1} g(P') \rangle_P + \frac{\mu}{2}\cdot W_2^2(P, P') $$ or equivalently that the Hessian $\text{Hess}(f)$ is bounded from the above by $\mu$ for all $P$.

A 1991 paper of Lewis, title “Is there a convenient category of spectra?” proves that there is no *category* $\mathrm{Sp}$ satisfying the following desiderata$^1$:

- There is a symmetric monoidal smash product $\wedge$;
- There is an adjunction $\Sigma^\infty\colon\mathrm{Top}_*^\mathrm{CGWHaus}\rightleftarrows\mathrm{Sp}:\Omega^\infty$;
- The sphere spectrum $\mathbb{S}$ is a unit for $\wedge$;
- There is either a natural transformation $$(\Omega^\infty E)\wedge(\Omega^\infty F)\Rightarrow\Omega^\infty(E\wedge F)$$ or a natural transformation $$\Sigma^\infty(E\wedge F)\Rightarrow(\Sigma^\infty E)\wedge(\Sigma^\infty F).$$ Furthermore, these natural transformations are required to commute wit]h unity, commutativity, and associativity isomorphisms of $\mathrm{Top}_*^\mathrm{CGWHaus}$ and $\mathrm{Sp}$;
- There is a natural weak equivalence $\Omega^\infty\Sigma^\infty X\xrightarrow{\cong}\varinjlim(\Omega^n\Sigma^nX)$.

**Question:** The theorem is proved for (ordinary) categories. But what happens for $\infty$-categories? Is the $\infty$-category of spectra defined in Higher Algebra “convenient”?

The $\infty$-category $\mathrm{Sp}$ satisfies (in an appropriate sense) [1] and [3] (see HA 4.8.2.19), and [2]$^2$ (see HA 1.1.2.8). What about [4] and [5]?

$^1$Which I am partly paraphrasing from the 2017 Talbot notes and partly paraphrasing from Lewis.
$^2$More is true: they are in fact *homotopy inverses* to each other.

Let $M$ be a $\mathbb{Z}$-module and $p$ a prime number such that $M\otimes_{\mathbb{Z}}\mathbb{Z}/p\mathbb{Z}=0$ and $Tor_{1}^{\mathbb{Z}}(M,\mathbb{Z}/p\mathbb{Z})=0$. Is it true that $M$ is an $\mathbb{Z}[\frac{1}{p}]$-module ?

Suppose we have a group $\Gamma$ acting on an abelian group $V$. Then it is well-known that the second cohomology group $H^2(\Gamma,V)$ corresponds to equivalence classes of central extensions of $\Gamma$, or equivalently, equivalences classes of short exact sequences of the form $$1 \longrightarrow V \longrightarrow E \longrightarrow \Gamma \longrightarrow 1$$ This interpretation can also be stated in terms of a lifting property: $H^2(\Gamma,V)$ comprises "obstacles" to the lifting of a homomorphism from $\Gamma$ to a quotient $W/V$ to a homomorphism from $\Gamma$ to $W$, where $W$ is another abelian group containing $V$.

Is there any such natural interpretation of the **bounded cohomology** group $H_b^2(\Gamma,V)$ in terms of obstacles to lifting?

I know that when $V=\mathbb{R}$ and the action of $\Gamma$ is trivial, then the kernel of the comparison homomorphism $$c: H_b^2(\Gamma,\mathbb{R}) \to H^2(\Gamma,\mathbb{R})$$ comprises the space of non-trivial quasi-homomorphisms.

But what do the elements of the group $H_b^2(\Gamma,V)$ themselves represent for arbitrary $V$? I have been unable to find any satisfactory interpretation in existing literature so far.

Ravenel defined the notion of a harmonic spectrum (see, for example, the Definition 7 here. A theorem of Ravenel shows that the Brown-Peterson spectrum is harmonic.

Can one deduce from here that the cobordism spectrum is harmonic?

What other examples of harmonic spectrum there are? For example, is the connective complex $K$-theory, $ku$, harmonic?

Let

- $h\in C^2(\mathbb R)$ with $$h''\ge\rho\tag1$$ for some $\rho>0$ and $$\int\underbrace{e^{-h}}_{=:\:\varrho}\:{\rm d}\lambda=1$$
- $\mu$ be the measure with density $\varrho$ with respect to the Lebesgue measure $\lambda$ on $(\mathbb R,\mathcal B(\mathbb R))$
- $(X^x_t)_{(t,\:x)\in[0,\:\infty)\times\mathbb R}$ be a continuous process on a probability space $(\Omega,\mathcal A,\operatorname P)$ with $$X^x_t=x+\int_0^t\frac12(\ln\varrho)'(X^x_s)\:{\rm d}s+W_t\;\;\;\text{for all }t\ge0\text{ almost surely for all }x\in\mathbb R\tag2$$ for some Brownian motion $(W_t)_{t\ge0}$ on $(\Omega,\mathcal A,\operatorname P)$,$$\kappa_t(x,B):=\operatorname P\left[X^x_t\in B\right]\;\;\;\text{for }(x,B)\in\mathbb R\times\mathcal B(\mathbb R)$$ and $$\kappa_tf:=\int\kappa_t(\;\cdot\;,{\rm d}y)f(y)$$ for Borel measurable $f:\mathbb R\to\mathbb R$ with $\kappa_t|f|<\infty$ for $t\ge0$

Now, let $$\Gamma(f):=\frac12|f'|^2\;\;\;\text{for }f\in C^1(\mathbb R)$$ and $$Lf:=-\frac12h'f'+\frac12f''\;\;\;\text{for }f\in C^2(\mathbb R).$$

How can we show that $$\Gamma(\kappa_tf)\le e^{-\rho t}\kappa_t(\Gamma(f))\tag3\;\;\;\text{for all }f\in C_c^\infty(\mathbb R)$$ for all $t\ge0$?

We know that $(\kappa_t)_{t\ge0}$ is a strongly continuous contraction semigroup on $C_0(\mathbb R)$. If $(\mathcal D(A),A)$ is the corresponding generator, then $$\tilde{\mathcal D}(A):=\left\{f\in C_0(\mathbb R)\cap C_b^2(\mathbb R):Lf\in C_0(\mathbb R)\right\}\subseteq\mathcal D(A)$$ and $$\left.A\right|_{\tilde{\mathcal D}(A)}=\left.L\right|_{\tilde{\mathcal D}(A)}\tag4.$$ Moreover, $\mu$ is reversible (and hence invariant) with respect to $(\kappa_t)_{t\ge0}$. By $(1)$, $$\Gamma_2(f):=\frac14(|f''|^2+h''|f'|^2)\ge\frac\rho2\Gamma(f)\;\;\;\text{for all }f\in C_c^\infty(\mathbb R)\tag5.$$

In order for $(3)$ to make sense, we need to prove that $\kappa_tf\in C^1(\mathbb R)$ (otherwise, the left-hand side would be undefined) for all $f\in C_c^\infty(\mathbb R)$ and $t\ge0$.

We know that $$(\kappa_tf)(x)=f(x)+\int_0^t\left(\kappa_s\left(Lf\right)\right)(x)\:{\rm d}s\;\;\;\text{for all }(t,x)\in[0,\infty)\times\mathbb R\tag6$$ for all $f\in C_b^2(\mathbb R)$.

Let $X,Y,Z$ be random variables jointly distributed on the same probability space then we know the conditional probability $P(X=a,Y=b\mid Z)$ is a random variable which is $Z$ measurable. Can we say

$$P(X=a,Y=b\mid Z)=P(X=a\mid Y=b,Z)P(Y=b\mid Z)?$$

Moreover is it true to say

$$\operatorname E[P(x=a,Y=b \mid Z)]=P(x=a,Y=b)?$$

Question 35996 asks about the Ehrhart polynomial $i_d(n)$ of the standard regular cross-polytope. It can be defined equivalently by $$ \sum_{n\geq 0}i_d(n)x^n = \frac{(1+x)^d}{(1-x)^{d+1}}. $$ It can be shown that the coefficients of $i_d(n)$ are positive, using Theorem 3.2 of http://math.mit.edu/~rstan/papers/cycles.pdf to show that all zeros of $i_d(n)$ have real part $-1/2$. Is there some "positive" formula for $i_d(n)$ that makes it transparent that the coefficients are positive? Or at least, is there another proof that doesn't involve the zeros of $i_d(n)$?