MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a pointed simplicial set. For each $n$, you are given a pointed subsimplicial set $A_n$ of $X^{\wedge n}$, the $n$-fold smash product of $X$. With this data, we can construct a bigraded ${\mathbb{Z}}/2$-vector space $E:=\oplus_n\oplus_i \tilde{H}_i (X^{\wedge n}/A_n; {\mathbb{Z}}/2)$, consisting of the reduced homologies of the quotient spaces $X^{\wedge n}/A_n$.

Suppose that $A_{n+m}$ is a subsimplicial set of $A_n\wedge X^{\wedge m} \vee_{A_n\wedge A_m} X^{\wedge n}\wedge A_m$. This condition gives us a morphism $f_{n,m}:X^{\wedge n}/A_n \wedge X^{\wedge m}/A_m\to X^{\wedge n+m}/A_{n+m}$. We now have a multiplication map on the homologies $\tilde{H}_i (X^{\wedge n}/A_n)\otimes \tilde{H}_j (X^{\wedge m}/A_m)\to $ $\tilde{H}_{i+j} (X^{\wedge n}/A_n \wedge X^{\wedge m}/A_m)$
$\stackrel{H_{i+j}(f_{n+m})}{\to}$ $H_{i+j}(X^{n+m}/A_{n+m})$. This gives $E$ the structure of a graded algebra.

Suppose further that $X\simeq *$ is contractible. The short exact sequence $A_n\to X^{\wedge n}\to X^{\wedge n}/A_n$ induces a long exact sequence. From this long exact sequence we read off that for $i\ge 1$, we have the isomorphism $\tilde{H}_i(X^{\wedge n}/ A_n)\cong \tilde{H}_{i-1}(A_n)$. Supposing that $X^{\wedge n}/A_n$ is connected, we have an isomorphism of vector spaces $A \cong \oplus_n \oplus_{i=1}^\infty \tilde{H}_{i - 1}(A_n;{\mathbb{Z}}/2)$.

The multiplication on $E$ induces via this isomorphism a multiplication on the right-hand space. Is there a simple description of the multiplication map $\tilde{H}_{i-1}(A_n)\otimes \tilde{H}_{j-1}(A_m) $ $\to H_{i+j-1} (A_{n+m})$?

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.