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Algebraic and topological K-theory, relations with topology, commutative algebra, and operator algebras

1 vote
0 answers
237 views

Smooth version of the splitting principle

Inspried by this MO question A manifold whose tangent space is a sum of line bundles and higher rank vector bundles we pose the following question as a possible smooth version of the splitting p …
2 votes
0 answers
199 views

The trigonometric $C^*$-algebra

The trigonometric $C^*$-algebra is the universal $C^*$-algebra generated by $\mathcal{G}=\{x,y,z\}$ subject to relations \begin{align}x^2=x=x^*, &\quad y^2=y=y^*\\ [x, z]=y, &\quad [y,z]=-x \\ …
2 votes
0 answers
90 views

The group of quasi unitary elements of a (simple) Banach algebra

For a Banach algebra $A$ with invertible group $G(A)$ we define the following group: $$QG(A)=\{u\in G(A)\mid \;\text{the mapping}\; a\mapsto u^{-1} a u \;\text{is an isometry}\}$$ What is an exampl …
0 votes
0 answers
119 views

Subalgebras of $B(H)$ consisting of all operators leaving a given finite dimensional space i...

Let $H$ be an infinite dimensional separable Hilbert space. Let $V$ be a finite dimensional subspace of $H$. Put $$A=\{T\in B(H)\mid T(V)\subseteq V\}.$$ So $A$ is a Banach algebra. Can we equip $A$ w …
1 vote
0 answers
132 views

A possible kind of $K$ theory via comparison of sphere bundles associated to given vector bu...

Let $E$ be a vector bundle on a topological space $X$.Thanks to Allen Hatcher's book "Vector Bundles and K theory", the construction of sphere bundle $S(E)$ can be done without any inner product on fi …
4 votes
0 answers
332 views

Other kinds of equivalence relations on the set of idempotents of a Banach or $C^*$-algebra ...

The standard equivalent relations on idempotents of a $C^*$ algebra or a Banach algebra are Murray von Neumann, similarity and homotopy equivalent. In this post we consider two other kinds of equivale …
3 votes
0 answers
166 views

"Somewhat connected" spaces or algebras

Before we state our question, we give a motivational simple example: Put $X$ for disjoint union of two circles. However $X$ is not a connected space but it has an open dense subset $U$ such that $U$ h …
5 votes
0 answers
132 views

Riemannian version of topological $K$-theory

Let $X$ be a compact Hausdorff space.Put $Vec(X)$, the space of all real (or complex) vector bundles over $X$.We put also $Vec_g(X)$, the space of all Riemannian vector bundles over $X$, that is the s …
3 votes
0 answers
155 views

Left and right topological K-theory of Banach algebras

Let us consider the topological $K$-functor on the category of Banach algebras as described in page $18$ of "Introduction to the Baum–Connes conjecture" by Alain Valette. The definition is based on in …
3 votes
0 answers
286 views

(Non trivial) coidempotents(Co-$K$-theory)

I was interested to know about coalgebraic version of "Idempotents". So I seached the web and I found the following interesting post : https://math.stackexchange.com/questions/689322/co-idempotents-al …
1 vote
1 answer
141 views

A kind of isomorphicity of vector bundles

Let $X$ be a connected topological space. Let $E$ be a $k$ dimensional sub vector bundle of the trivial vector bundle $X\times \mathbb{R}^n$. Then $E$ defines an idempotent with trace $k$ in $M_n(C(X) …
2 votes
2 answers
124 views

Invertible (isometric) sections of certain hom bundles over sphere

Assume that we have a vector bundle $E$ over $S^n$. Is there a continuous family of invertible linear maps $T_x:E_x \to E_{-x}$? Here continuity has the obvious meaning as soon …
2 votes
1 answer
263 views

Are these vector bundles, trivial bundle?

We identify the vector space tensor product $\mathbb{R}^{m} \otimes \mathbb{R}^{n}$ with $\mathbb{R}^{mn}$ Let $X$ be the space of all non zero simple tensors $X=\{a\otimes b \mid a\in \math …
6 votes
1 answer
547 views

A generalized K- theory via generalized idempotents

Edit After the answer by Neil Strickland, I add the word "a ring" in this new version. In the literature, there is a concept of generalized idempotent: an n- idempotent is an element $a$ of a Banach …
2 votes
0 answers
208 views

A functor on the category of rings, algebras or compact Hausdorff topological space

Assume that $R$ is a unital ring or a complex or real (Banach or $C^{*}$) algebra. We define a relation $M$ on $R$ as follows: $$a\;M b \;\;\; \text{iff}\;\; a=xy,\;b=yx \;\; \text{for so …

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