Let $A=k[x_1,\cdots,x_r]/I$ for some prime ideal $I$ and field $k$. Consider the free $A$-module $A^n$. Given an element $e\in A^n$ is there a method to tell whether $e$ can be admitted as an element of some $A$-basis for $A^n$? Obviously components of $e$ need to generate $A$ but I doubt this is enough. What about instead of one element $e$ we have a set of $A$ linear independent elements $e_1,\cdots , e_k$ for $k<n$. Is there a method to tell whether these can be admitted as basis elements.

I'm looking for more practical methods than theoretical ones. For example you can consider the determinant polynomial with one column equal to $e$ and equate it to 1 and say whenever this has a solution in $A$, then $e$ is an element of some basis. but this method is not practical.

This was my main question. Let $S$ be the set of elements in $A^n$ that can be part of some basis. Now define an equivalence relation on $S$ by setting two elements to be equivalent iff there is some basis containing both of them as basis elements. After taking the quotient under this equivalence is $S$ connected? or what are its connected components?

In the article *Centralizers in R. Thompson group $V_n$*, the following question is asked:

**Question:** Is the centraliser of an element of $V_n$ always finitely presented?

I am wondering: is this question still open? I have a plausible argument in mind, but I would like to be sure that the answer is not already somewhere in the literature.

I'm little bit puzzled with the following question, I was wondering if someone could help with a suggestion or a hint.

Let start with some notations. Suppose that $G$ is a Hausdorff topological group such that

- As a space $G$ is contractible.
- As a group $G$ is isomorphic to the free product $\mathrm{Q}\ast F(X) $ where $\mathrm{Q}$ is the additive abelian group of rational numbers and $F(X)$ is a free group generated by a set $X$.
- We assume that $\mathrm{Q}$ is a topological subgroup of $G$ in an obvious way and that $\{1\}\ast_{\mathrm{Q}} G$ is a retract of a free topological group.

Let $\mathbf{Ab}: \mathsf{Tgr}\rightarrow \mathsf{AbTgr}$ be the abelianization functor from the category of topological groups to abelian topological groups.

Since $G$ is a contractible group, for any natural number $n>0$ the power maps $p^{n}: G\rightarrow G$, $t\mapsto t^n$ are weak homotopy equivalences of underlying topological spaces.

My question is the following: is there a hope that the induced power maps: $p^{n}: \mathbf{Ab}(G)\rightarrow \mathbf{Ab}(G)$ induce weak homotopy equivalences of underlying topological spaces?

Thank you in advance for any help!

lemma: For every list of powers of consecutive numbers, prove that at least one of the numbers is co=prime to their sum.

Please prove this lemma.

So we are given an infinite supply of bulbs, following Exp($\lambda$)

You arrive at time t, note the bulb that is burning at time t.

$L_t$ = Lifetime of the bulb we noted

$A_t$= Lifetime of the noted bulb spent until time t

$B_t$ = Remaining lifetime of the noted bulb

I was able to calculate the distribution of the $L_t$, which came out as

P[$L_t \leq x$] = $1-((x\lambda + 1)* e^{-\lambda x}$ for $0<x\leq t$ and 1-(($t\lambda + 1)e^{-\lambda x}$ for $x \geq t$

I know $B_t$ follows Exp($\lambda$), which is the paradox.

How to calculate the distribution of $A_t$?

I was trying to use change of variable but because of dependent variable I was unable to proceed

Find each of the following quantities exactly using csc(a) = -3 and tan(b) = -7 such that b is a quadrant 11 angle and that -pi < a < -pi/2

Is it true that every tournament is contained in some vertex-transitive tournament? If not, is it known which tournaments satisfy this property? This seems like a basic question, but I have not been able to find a relevant reference.

The following extension of the Jordan Curve Theorem is well known: every closed connected hypersurface of the sphere $\mathbb S^N$ separates $S^N$ into exactly two connected components. As a consequence, every compact connected hypersurface of the euclidean space $\mathbb R^N$ separates $\mathbb R^N$ into exactly two connected components.

Does this result keep its validity if the word `compact' is replaced by *closed*? With other words, is it true that every closed connected hypersurface of the euclidean space $\mathbb R^N$ separates $\mathbb R^N$ into exactly two connected components?

I am interested in (as explicit as possible) descriptions of non-Gorenstein integral projective curves. Most of the literature on singular curves appears to be focused around the Gorenstein case, with a notable exception being the paper of Kleiman, *The Canonical Model of a Singular Curve*, and a handful of others. Any other references would be appreciated. Specifically I am interested in knowing what has been understood about their derived categories of coherent sheaves (or the subcategory of perfect complexes), so I would very interested in any developments along those lines.

There are some estimates about initial value problem for example Kato's $H$-smoothness and Strichartz's estimate. These estimates roughly say the solution $u$ of some PDE is bounded for initial value, say $u_0$, in the $L^p$-norm sense. That is, $$\||\nabla|^s u\|_{L^p} \le \|u_0\|_{L^q} \quad and \quad \|u\|_{L^p}\le \|u_0\|_{L^q} $$ for some $s>0$ and $1/p +1/q = 1$ for $p>1$.

I hope know that meaning of bounding for norm of initial value. In addition, meaning of regularity, $|\nabla|^s$ also.

I saw this result in *A Model Category Structure for Differential Graded Coalgebras* by Getzler-Goerss, but when the coalgebra is non-negatively graded, is this property also satisfied when the dg coalgebra is $\mathbb{Z}$-graded?.

Thanks.

Let $(X,\mathcal T)$ be a topological space. About the subsets $A,B,C$ of $X$ it is known that $$\mathrm {cl} (C)= A \cup B\,, \quad \mathrm {cl} (A) = A\,, \quad \mathrm {cl} (B) = B\,.$$

Does it always imply that there exist two subsets of $X$, $C_A$ and $C_B$ such that $$C = C_A \cup C_B\,, \quad \mathrm {cl} (C_A) = A\,, \quad \mathrm {cl} (C_B) = B \; ?$$

If this is not true in general, is it true at least for Euclidean spaces?

**Preliminaries**

Let $(\Omega, \mathcal{G}, \mathbb{P})$ be a complete probability space.

Let $D$ be a complete, separable, metrizable topological space with Borel $\sigma$-algebra $\mathcal{B}(D)$ (such as $D = \mathbb{R}^q$ with $\sigma$-algebra $\mathcal{B}(D) = \mathcal{B}(\mathbb{R}^d)$).

Let $\mathbb{R}$ be equipped with its canonical Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$.

Let $g: \Omega \times D \rightarrow \mathbb{R}$ be a bounded $(\mathcal{G} \otimes \mathcal{B}(D) ) / \mathcal{B}(\mathbb{R})$-measurable function.

Let $\Pi: \Omega \rightarrow D$ be a $\mathcal{G}/\mathcal{B}(D)$-measurable random variable.

Let $H : \Omega \rightarrow \mathbb{R}$ be a $\mathcal{G}/\mathcal{B}(\mathbb{R})$-measurable random variable, defined by $$ H(\omega) := g(\omega, \Pi(\omega)).$$ Note, that, since $g$ is bounded, we have $H \in \mathcal{L}^2(\Omega, \mathcal{G}, \mathbb{P})$.

Let $j: D \rightarrow \mathcal{L}^2(\Omega, \mathcal{G}, \mathbb{P}) $ be defined by $$ j(\pi)(\omega) := g(\omega, \pi) $$

For all $\pi \in D$, let $j(\pi)$ be independent of $\Pi$.

**Question**

I am interested in the conditional expectation $$\mathbb{E}[H \mid \Pi] :\Omega \rightarrow \mathbb{R}$$ of $H$ with respect to $\Pi$. More specifically, I suspect that (a $\mathbb{P}$-unique version of) this condititional expectation is given by

$$ \mathbb{E}[H \mid \Pi] (\omega) = \mathbb{E}[j(\Pi(\omega))], \quad (\dagger) $$ whereby $\mathbb{E}[j(\Pi(\omega))]$ can of course also be written as $$\mathbb{E}[j(\Pi(\omega))] = \int_{\Omega} j(\Pi(\omega))(\tilde{\omega}) d\mathbb{P}(\tilde{\omega}) . $$

How can I prove, that $(\dagger)$ is the case? I have tried, tracking the definition of conditional expectation and using Fubini, but with little success so far.

Thanks for any advice!

Consider the extension

$$1\to SU(2)\to X\to O\to1,$$

there are 4 possibilities for $X$: $X=O\times SU(2)$ or $E\times_{\mathbb{Z}_2}SU(2)$ or $Pin^+\times_{\mathbb{Z}_2}SU(2)$ or $Pin^-\times_{\mathbb{Z}_2}SU(2)$ where $E$ is defined in Freed-Hopkins's work1 as the colimit of the group $E(d)$, the group $E(d)$ is defined to be the subgroup of $O(d)\times\mathbb{Z}_4$ consisting of the pairs $(A,j)$ such that $\det A=j^2$, where $\mathbb{Z}_4=\{\pm1,\pm\sqrt{-1}\}$ is the multiplicative group of order 4.

Here the notation $G_1\times_{\mathbb{Z}_2} G_2 :=\frac{G_1\times G_2}{\mathbb{Z}_2} $ is defined as mod out the common $\mathbb{Z}_2$ of $G_1\times G_2$.

The question is about computing $MT(E(d)\times_{\mathbb Z_2} SU(2))$ and the bordism group $\Omega_d^{E \times_{\mathbb Z_2}SU(2)}$.

(1) There is a short exact sequence of groups: $1\to SO(d)\to E(d)\to\mathbb{Z}_4\to 1$. So naively, people may suspect that $$MT(E(d)\times_{\mathbb Z_2} SU(2))=MT E(d)\wedge\Sigma^{-3}M SO(3)=MSO(d)\wedge\Sigma^{-2}M\mathbb Z_4\wedge\Sigma^{-3}M SO(3).$$ However, this is likely to be incorrect.

(2) The space $B(E \times_{\mathbb Z_2}SU(2))$ sits in a homotopy pullback square: a map $M \to B(E \times_{\mathbb Z_2}SU(2))$ is determined by two maps $M \to BO$ and $M\to BSO(3)$ which correspond to bundles $TM$ and $V_{SO(3)}$ such that $w_1(TM)^2=w_2(V_{SO(3)})$.

To compute the bordism group $\Omega_d^{E \times_{\mathbb Z_2}SU(2)}$, we need to know the Madsen-Tillmann spectrum $MT(E \times_{\mathbb Z_2}SU(2))$ and decompose it as the wedge sum or smash product of some familiar spectra.

The attachment is the author's attempt,

but the map $f$ is not a homotopy equivalence. I actually obtain an identification $$ \text{Thom$(B(E \times_{\mathbb Z_2}SU(2)),-2V)=MT(Pin^+ \times_{\mathbb Z_2}SU(2))$} $$ which is already known in 1604.06527 paper, but we need to know $$ \text{Thom$(B(E \times_{\mathbb Z_2}SU(2)),-V)=MT(E \times_{\mathbb Z_2}SU(2))$,}$$

where $V$ is the induced virtual bundle of dimension 0 by $B(E \times_{\mathbb Z_2}SU(2)) \to BO$.

- Is Thom$(B(E \times_{\mathbb Z_2}SU(2)),-2V)$=smash product of Thom$(B(E \times_{\mathbb Z_2}SU(2)),-V)$ and Thom$(B(E \times_{\mathbb Z_2}SU(2)),-V)$? If so, how to obtain Thom$(B(E \times_{\mathbb Z_2}SU(2)),-V)$ as the "square root" of Thom$(B(E \times_{\mathbb Z_2}SU(2)),-2V)$?

1 Reflection positivity and invertible topological phases Daniel S. Freed, Michael J. Hopkins, arXiv:1604.06527

Supposing I have complex square matrices $B_i$ and $C_i$ ($i = 1,\dots,N$) of dimension $4 \times 4$.

- Is there an effective algorithm for solving the following problem?

$$\begin{align} A=&\underset{A\in\mathbb{C}^{4\times 4}}{\text{argmin}}\sum_{i=1}^{N}{\left\|AB_iA^{-1}-C_i\right\|_2}\\ \,\\ &\text{subject to }\left\{\begin{matrix}A_{12}=A_{13}=A_{42}=A_{43}=0\\A_{11}+A_{41}=1\\ A_{14}+A_{44}=1\end{matrix}\right. \end{align} $$

- How can I make sure that there exists only one unique solution for $A$?

Let $(X,d)$ be a metric space. Suppose that $\{A^n\}_{n \in \mathbb{N}}$ is a sequence of closed, non-empty subsets of $X$.

Is there a Hausdorff topology on the space of closed subsets of $X$, guaranteeing that if $A^n$ converges in this space to a $A\subseteq X$, then for any continuous function $f:X \rightarrow \mathbb{R}$, we have that $$ \sup_{x \in A}f(x)\leq \sup_{n \in \mathbb{N}}\sup_{x \in A^n}f(x) ? $$

Did someone develop ZFC by means of ZF plus axioms for a binary well-ordering constant, say $\blacktriangleleft$? Are there results that suggested accounts are conservative extensions of ZFC?

Let $F: C \to D$ be an infinity functor. Is it true that the homotopy fiber at $y$ can be described as $C \times_D D^{\simeq}_{/y}$? If not, is there a simple formula resembling this one?

Beside the infinity structure, its points are pairs $(x \in C, s: F(x) \to y \text{ equivalence})$. I heard this in a class but never seen a proof. I came now with a weird fact that seems a consequence of this fact and I would like to have a confirm.

Thanks, Andrea

Let $k$ be a field of characteristic zero and $R$ a local $k$-algebra. By Stacks \tag 00TX, if we assume that $R$ is of finite type, then the $R$ is smooth over $k$ if and only if $\Omega_{R/k}$ is free. I'm wondering if the analogous statement is true for formal smoothness, i.e is the formal smoothness of $R$ over $k$ equivalent to the freeness of $\Omega_{R/k}$?

My solution would be to use Stacks project 031I. We choose a polynomial ring P and a surjection $P\rightarrow R$ and denote by $J$ the kernel of $P\rightarrow R$. Then we have to show that the sequence $ (*) \ J/J^2 \rightarrow \Omega_{P/k}\otimes_P R \rightarrow \Omega_{R/k} \rightarrow 0$ is left exact and splits. If we can show that it is left exact, then by the projectivity of $\Omega_{R/k}$, we are done. By Stacks 00S2, we know that $ 0 \rightarrow H_1(L_{R/k}) \rightarrow J/J^2 \rightarrow \Omega_{P/k}\otimes_P R$. In http://www1.mat.uniroma1.it/people/manetti/dispense/defosing.pdf, it is writen (page 20, Def 9.2), that if $R$ is dg-algebra, and $M$ an $R$-dg-module ( and $k$ of characteristic zero!), then $H_i(L_{R/k}\otimes_R M)=\text{Tor}_i^R(L_{R/k},M)$. If we assume that $R$ is a dg-algebra, then for $i=1$ and $M=R$ we get that $H_1(L_{R/k})=0$. Hence the sequence (*) is left exact and we are done. However I'm not confident in this solution.

I have two question:

1) Are there any examples of complete manifold with strictly positive and bounded section curvature which has zero injectivity radius?

2) Is there a sequence of non-compact complete manifolds with strictly positive and bounded section curvature with injectivity radius approach to zero?

I think one may construct these examples from Beger's spheres, but I cannot do it rigorously.