gn.general topology – On Čech-complete space

I’m reading an article of topology and i came across a Properties :

Properties :

Closed subspaces and arbitrary products of Čech-complete spaces are Čech-complete

Every Čech-complete space is a Baire space

Where the author just explain the concept. I am stuck on what actually does it mean and how to prove it. Is there an easy way to prove this propeties?

general topology – A topological detail in the definition of Lie group

A Lie group $ G $ is an $r$-times-differentiable manifold endowed with a group structure, i.e. with an associative binary operation

mu:quad Gtimes G longrightarrow G :qquadleft{x,, yright} longmapsto xcenterdot y

and an inversion operation

zeta:quad G longrightarrow G :qquad g longmapsto g^{-1}~~,

both of which are $r$-times-differentiable.

Does this definition restrict in any way the topology wherewith the manifold is equipped?

More specifically, if I take the inverses of all points residing in an open set $U$,

will the resulting set $tilde{U}$ be open in that topology?

gn.general topology – The space of skew-symmetric orthogonal matrices

Let $M_n subseteq SO(2n)$ be the set of real $2n times 2n$ matrices $J$ satisfying $J + J^{T} = 0$ and $J J^T = I$. Equivalently, these are the linear transformations such that, for all $x in mathbb{R}^{2n}$, we have $langle Jx, Jx rangle = langle x, x rangle$ and $langle Jx, x rangle = 0$. They can also be viewed as the linear complex structures on $mathbb{R}^{2n}$ which preserve the inner product.

I’d like to understand $M_n$ better as a topological space, namely an $(n^2-n)$-manifold.

$M_1$ is just a discrete space consisting of two matrices: the anticlockwise and clockwise rotations by $pi/2$.

For $n geq 2$, we can see that $M_n$ is an $M_{n-1}$-bundle over $S^{2n-2}$. Specifically, given an arbitrary unit vector $x$, the image $y := Jx$ must lie in the intersection $S^{2n-2}$ of the orthogonal complement of $x$ with the unit sphere $S^{2n-1}$. Then the orthogonal complement of the space spanned by $x$ and $y$ is isomorphic to $mathbb{R}^{2n-2}$, and the restriction of $J$ to this space can be any element of $M_{n-1}$.

Since the even-dimensional spheres are all simply-connected, it follows (by induction) that $M_n$ has two connected components for all $n in mathbb{N}$, each of which is simply-connected. For instance, $M_2$ is the union of two disjoint 2-spheres: the left- and right-isoclinic rotations by $pi/2$. The two connected components of $M_n$ are two conjugacy classes in $SO(2n)$; they are interchanged by conjugating with an arbitrary reflection in $O(2n)$.

Is (each connected component of) $M_n$ homeomorphic to a known well-studied space? They’re each an:

$S^2$-bundle over an $S^4$-bundle over $dots$ an $S^{2n-4}$-bundle over $S^{2n-2}$

but that’s not really very much information; can we say anything more specific about their topology?

at.algebraic topology – $p$-completeness of the function spectrum $F(Sigma^{infty} BS, Sigma^{infty} BK)$

Let $S$ be a finite $p$-group and $K$ a compact Lie group, in the paper A Segal conjecture for $p$-completed classifying spaces, it is said that the function spectrum $F(Sigma^{infty} BS, Sigma^{infty} BK)$ is $p$-complete, but I have not succeeded in proving it. I hope this remains true when, more generally, $S$ is a $p$-toral group. Any suggestion or idea?, please.

differential topology – May this slice disk for the unknot be pushed into the boundary?

Write the 4-ball as $mathbb{D}^4=mathbb{D}^2times mathbb{D}^2$.
Then its boundary $mathbb{S}^3simeq mathbb{S}^1times mathbb{D}^2cup mathbb{D}^2times mathbb{S}^1$. We will use implicitely this homeomorphism.

Consider the unknot $K=mathbb{S}^1times{0}$ in the boundary $partial mathbb{D}^4simeqmathbb{S}^3$ of the four ball.

The disk $D= mathbb{D}^2times {0}$ is a smooth slice disk for $K$.

Is $D$ boundary parallel? I.e. is it obtained by pushing an unknotting disk $D’subset mathbb{S}^3,$ $partial D’ = K$ inside $int(mathbb{D}^4)?$

A possible approach: take as $D’subset partial mathbb{D}^4$ the PL disk obtained by gluing the annulus $mathbb{S}^1times ((-1,1)times {0})in mathbb{S}^1times mathbb{D}^2subsetpartial mathbb{D}^4$ to the disk $mathbb{D}^2times ({-1}times {0})in mathbb{D}^2times mathbb{S}^1subsetpartial mathbb{D}^4$.
Then gluing $D’$ to $D$ we get an embedded 2-sphere. If we manage to prove that this sphere is unknotted, i.e. it bounds a 3-ball then we can use the latter to push $D’$ to $D$.

Relevance of this problem
In studying Kirby calculus, one finds often the claim that when you attach a 2-handle, the cocore of a 2-handle is an unknotted 2-disk, i.e. boundary parallel. How to prove this?
The above problem is a possible way.

gn.general topology – irreducible compact set vs prime compact set

Let X be a topological space.
A compact set K is called irreducible if for any two compact subsets K1,K2 of K with K is equal to the union of K1 and K2, then K is equal to K1 or K2.
A compact set K is called prime if for any two compact sets K1,K2 of X with K is included in the union of K1 and K2, then K is included in K1 or K2.

Are these two properties equivalent?
I guess that they are not the same. But I could not come up with an example.

ag.algebraic geometry – Topology of the set of polynomials with bounded real algebraic varieties (inside the v. s. of polynomials in $n$ variables and up to degree $d$)

Set $x=(x_{1}, dots, x_{n}).$ Consider the set $mathbb{R}(x)_{d}$ of polynomials with coef. in $mathbb{R}$ in $n$ variables up to degree $d.$ This set can be seen as a finite-dimensional vector space with basis formed by all the monomials up to degree $d$ in $n$ variables and thus endowed accordingly with a(n Euclidean) topology coming naturally from this vector space structure. Now consider particularly the subset $mathbb{R}0(x)_{d}$ of polynomials in $mathbb{R}(x)_{d}$ that, when restricted to a real line, have only real zeros (see Section 2.1 here where RZ polynomials are defined). Finally, take the subset $bmathbb{R}0(x)_{d}subsetneqmathbb{R}0(x)_{d}$ of polynomials defining bounded real algebraic varieties. Does $bmathbb{R}0(x)_{d}$ has non-empty interior in $mathbb{R}(x)_{d}$ with respect to the Euclidean topology previously mentioned?

I really intuit that this question has to be easily answerable and that $bmathbb{R}0(x)_{d}$ has in fact non-empty interior in $mathbb{R}(x)_{d}$ but I cannot see how to prove it right now. Thanks for any insight!

general topology – Discussing the definition of continuity of linear functionals on the space $mathcal{D} (mathbb{R}^{n})$

$mathcal{D} (mathbb{R}^{n})$ the vector space over the field $mathbb{C}$ such that its elements are “nice” functions $varphi :mathbb{R}^n to mathbb{R}$. We say that a function is “nice” if:

  1. It’s smooth
  2. It has compact support

This is what I’ve studied.

The functional $F_{f} (varphi)$ is continuous in the following sense: Consider a sequence of functions ${ varphi_{k}(x) }$ in $mathcal{D} (mathbb{R}^{n})$ with the following two properties:

  1. There exists a compact set $K subset mathbb{R}^{n}$ such that for all $k$, $mbox{supp} varphi_{k} subset K$.

  2. $lim_{k to infty} partial^{m}varphi_{k}(x) = 0$ uniformly for all $m = 0, 1, 2, cdots$.

Such a sequence is said to go to $0$ in $mathcal{D}$ and is written $varphi_{k} to 0, k to infty$ in $mathcal{D}$.

We then say that the functional $F_{f}(varphi)$ is continuous if $F_{f}(varphi_{k}) to 0$ for $varphi_{k} to 0, k to infty$ in $mathcal{D}$.

Continuity is a topological property. The space $mathcal{D}$ is a linear vector space. It is made into a topological vector space by defining the neighbourhood of the “point” $varphi (x) = 0$ by a sequence of semi-norms.

I’ve read that the two conditions required above in the definition of $varphi_{k} to 0$ in $mathcal{D}$ follow from the conditions used to define the neighbourhood of $varphi(x) = 0$. Why is this true though?

at.algebraic topology – unlinking when relaxing the homeomorphism condition

Say that we have two knots $K_1$ and $K_2$ in $S^3$ linked together in $S^3$ and forming the Hopf link. Usually, we can prove that we cannot unlink them by using a link invariant that shows that the “two-component unlink” that consists of two separate circles in $S^3$ have a different value (with respect to the invariant) in comparison to its value on the Hopf link. This effectively shows that there is no homomorphism from $S^3$ to itself that separates the two links. I want to relax the condition of homomorphism a little bit and ask: is there a continuous function that separates the images of the two links? in other words, is there a continuous function $f:S^3to S^3$ such that $f(K_1)$ is contained in an closed disk $D_1$ and $f(K_2)$ is contained in another closed disk $D_2$ and $D_1$ and $D_2$ are disjoint? it seems that the answer is no but I am not sure how to show something like this. Any pointer is appreciated.

at.algebraic topology – Homotopy groups of certain geometric fixed point spectrum

Let $G$ be a finite group and $E$ be a genuine $H$-spectrum for $Hleq G.$ Then for any subgroup $K$ of $G$, consider the $K$-spectrum $X=Res^G_K Ind^G_H(E).$

Is there any reference for computing the homotopy groups of the geometric fixed point $Phi^K(X): = (widetilde{Emathcal{P}}wedge X)^K?$ Here $widetilde{Emathcal{P}}$ is a $K$-space given the following fixed point data:

$$widetilde{Emathcal{P}}^L= begin{cases} S^0, & text{ if }L=K\ ast, & text
{ if } L text{ is any proper subgroup of } K.end{cases}

Maybe the double coset formula could be useful.

Thank you so much in advance. Any help will be appreciated.