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This is a result in algebraic topology, where we study the structure of topological spaces $X$. One early way to do this is to calculate a thing called $H_*(X)$, the ordinary homology of $X$. Later …

Freed, Hopkins and Teleman will be using the homotopical definition $K^0(X)=[X,\mathbb{Z}\times BU]$. For many spaces $X$ this is the same as the Grothendieck group of vector bundles on $X$; in parti …

Usually you are computing $H^*(X)$ or $K^*(X)$ for a reason; for example if $H^*(X)\not\simeq H^*(Y)$ then you know that $X$ and $Y$ are not homotopy equivalent, but if $H^*(X)\simeq H^*(Y)$ and this …

Let $E_n(A)$ be the set of $n$-idempotents in $A$, and let $u_1,\dotsc,u_n$ be the elements of $E_n(\mathbb{C})$. Let $E'_n(A)$ be the set of $n$-tuples $e_1,\dotsc,e_n\in E_2(A)$ with $e_ie_j=0$ for …

The cobordism groups can be calculated using the Adams spectral sequence, which is based on the homological algebra of modules over the Steenrod algebra. This works most nicely for the unoriented cob …

I'll just give an answer for finite complexes $X$. The condition $\widetilde{K}^*(X)=0$ only depends on the suspension spectrum of $X$ so this is naturally regarded as a question in stable homotopy th …

Let $L$ denote $\mathbb{C}$ with $S^1$ acting by multiplication, and let $\mathbb{C}$ denote $\mathbb{C}$ with trivial $S^1$-action. Then the projective space $P(L\oplus\mathbb{C})$ is homeomorphic t …

Put $u_\alpha=[\mathbb{C}]-[L_\alpha]$. It is a standard fact, known as the projective bundle theorem, that $$K(\mathbb{P}(E))=K(X)[t]/\prod_{\alpha}(t-u_\alpha)$$ One can express $Fl(E)$ as the top o …

Some comments: You might want to look at 'Vector bundles over classifying spaces of compact Lie groups' by Jackowski and Oliver. They discuss a situation in which you can understand $K(V(X))$ quite …

You have given yourself an invertible element $u\in\pi_2(R)$ and a coordinate $x\in R^2(\mathbb{C}P^\infty)$ with $\psi(x)=x\otimes 1 + 1\otimes x + ux\otimes x$. This means that the class $m=1+ux\in …

First, I do not think that strict associativity makes a difference, so I will ignore it. Next, let $G$ be a finite group, and let $\mathcal{C}G$ be the category of finite $G$-sets and equivariant bij …

Using the Hopf fibration you can show that $\Omega^2S^2\simeq\mathbb{Z}\times\Omega^2S^3$. Also $\pi_1(\Omega^2S^3)=\pi_3(S^3)=\mathbb{Z}$, so the Universal Coefficient Theorem gives $[\Omega^2S^3,S^ …

There is a paper by Milnor called On the homology of Lie groups made discrete which contains many results on this kind of problem, as well as references to related literature. However, I don't think …

This seems to be an answer, based on discussion with Maxime Ramzi in the comments. The space of $E_\infty$ $KU$-algebra maps from $KU\otimes S^1_+\to F(S^1_+,KU)$ is the same as the space of $E_\infty …

Consider a map $f\colon X\to Y$ of spaces (such as $B(G^{\text{disc}})\to B(G)$). Say that $f$ is a $K(n)$-equivalence if $K(n)^*(f)\colon K(n)^*(Y)\to K(n)^*(X)$ is an isomorphism. We will allow th …