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2 votes
1 answer
99 views

Stabilizing conjugacy classes of integer matrices

$\DeclareMathOperator{\Conj}{Conj} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\id}{id} \newcommand\Z{\mathbb{Z}}$ For an $n \times n$ integer matrix $A \in \GL_n(\Z)$, let $\Conj(A)$ be the ...
Ingrid's user avatar
  • 23
4 votes
1 answer
186 views

The upper bounds on rank $ 2 $ real matrices

Let $ A_{n}(F) $ be the collection of all skew-symmetric matrices over the field $ F $ ($\operatorname{char} F \neq 2 $). Let M be a subspace of $ A_{n}(F) $ such that all non zero elements have rank ...
Sky's user avatar
  • 923
1 vote
1 answer
94 views

Problem concerning about an $n$-subspace of $ A_{n}(F) $

Let $A_{n}(F) $ denote the $n \times n$ skew symmetric matrices over a finite field $F$. Suppose $n$ be even and $N$ be a subspace of $A_{n}(F) $. Now if all the non-zero matrices in $N$ are ...
Sky's user avatar
  • 923
8 votes
1 answer
690 views

Subspaces of $ A_{n}(\mathbb {Q})$ in which all nonzero matrices are invertible

Let $A_{n}(\mathbb{Q}) $ denote the $n$ times $n$ skew symmetric matrices over the rational number field. Let $N$ be a subspace of $A_{n}(\mathbb{Q}) $. If all the non-zero matrices in $N$ are ...
Sky's user avatar
  • 923
11 votes
1 answer
520 views

Problems concerning subspaces of $M_{n}(\mathbb{Q}) $

Let $M_{n}(\mathbb{Q}) $ denote the $n$ times $n$ matrices over the rational number field. $N$ be a subspace of $M_{n}(\mathbb{Q}) $.Then if all the non-zero matrices in $N$ are invertible, what is ...
Sky's user avatar
  • 923
8 votes
1 answer
234 views

What is the inverse in K-theory represented by Clifford module extensions?

I am working on a model for topological KO-theory which is represented by explicit spaces of orthogonal Clifford module extensions. That is, assuming $M$ compact, $KO^{-n+1}(M) := [M,X_n]$ where the ...
Matt C's user avatar
  • 151
5 votes
1 answer
142 views

On the width of the Catalan monoid and the rank of K-groups of the Furstenberg transformation group

The semigroup algebra of the Catalan monoid is isomorphic to the incidence algebra of $P_n$, where $P_n$ is the poset consisting of subsets of { 1,...,n } where for two subsets $X \leq Y$ if and only ...
Mare's user avatar
  • 26.5k
5 votes
0 answers
91 views

elementary matrices over a regular ring

Let $n\in\mathbb{N}$ with $n\geq 3$. Let $A$ be a regular ring and $\gamma\in GL_{n}(A)$. We suppose that $\gamma$ is homotopic to the identity, i.e. there exists $\alpha\in GL_{n}(A[X])$ such that $\...
prochet's user avatar
  • 3,472
18 votes
0 answers
540 views

A curious switch in infinite dimensions

Let $V$ be a finite dimensional real vector space. Let $GL(V)$ be the set of invertible linear transformations, and $\Phi(V)$ be the set of all linear transformations. We can also characterize $\Phi(V)...
Thomas Rot's user avatar
  • 7,583
8 votes
1 answer
353 views

$E_n(\ell^\infty)=SL_n(\ell^\infty)$?

Let $R$ be a commutative unital ring $R$ with unit element $1$. For $n\in \mathbb{N}=\{1,2,3,\cdots\}$, let $SL_n(R)$ be the group of all $n\times n$ matrices with entries from $R$ having ...
KevinC's user avatar
  • 81
1 vote
1 answer
322 views

A (possible) equivalent relation on the space of vector bundles

Edit: According to the essential comment of Alex Degtyarev, we revise the question as follows; Assume that $\alpha$ and $\beta$ are two oriention preserving automorphism of Lie groups $O(n)$ and $...
Ali Taghavi's user avatar
18 votes
2 answers
1k views

Karoubi versus Kasparov K-theory

I have the following, probably very elementary question: Let $Cl^{p,q}$ be the Clifford algebra on generators $e_i$, $i=1, \ldots, p+q$ with $e_i e_j = -e_j e_i$ and $e_{i}^{2}=-1$ for $i=1,\ldots,p$, ...
Johannes Ebert's user avatar
2 votes
1 answer
186 views

Can I bound the degree of a contracting homotopy in an exact filtered complex?

Suppose that I have given you a bigraded vector space $V = \bigoplus_{i,j} V_{i,j}$. The first grading is a "homological" $\mathbb Z$-grading, and the second is an independent $\mathbb Z$-grading. ...
Theo Johnson-Freyd's user avatar