15
$\begingroup$

In the usual singular homology of a topological space $X$, one consider the free abelian group generated by all continuous maps from the standard simplex $\Delta^{n}$ to $X$.

Now we can replace $\Delta^{n}$ by the orthonormal group $O(n)$. There are $n+1$ (both topological and group theoretical) obvious embedding $\epsilon_{i}:O(n)\to O(n+1),\;\;i=1,2,\ldots,n+1$. These obvious embeddings are as follows:

$$\epsilon_{1}(A)=1 \oplus A,\;\;\\ \epsilon_{i}(A)=\lambda_{i}\epsilon_{1}(A)\lambda_{i}^{-1}$$ where $\lambda_{i}$ is the elementary matrix obtaining from the identity matrix by replacing first row by $i_{th}$ row.

For example, for $n=2$, the matrix $\epsilon_{i} \left (\begin{pmatrix} a&b\\c&d \end{pmatrix} \right)$ is $\begin{pmatrix}1&0&0\\0&a&b\\0&c&d \end{pmatrix}$, $\begin{pmatrix}a&0&b\\0&1&0\\c&0&d \end{pmatrix}$ and $\begin{pmatrix} a&b&0\\c&d&0\\0&0&1\end{pmatrix}$, for $i=1,2,3,\;$ respectively.

For a topological space $X$, one can define $\overline{C_{n}(X)}$ as the free abelian group generated by all continuous maps from $O(n)$ to $X$.

The boundary maps $\delta: \overline{C_{n}(X)} \to \overline{C_{n-1}(X)}$ is defined by $\delta(\phi)=\sum (-1)^i \phi \circ \epsilon_i$. Then we have $\delta \circ \delta =0$.

So we obtain a kind of "homology" as a functor on the category of topological spaces. Of course homeomorphic spaces have isomorphic homology.

(However this functor is not necessarily a homotopoic invariant functor but it impose an equivalent relation on the space of all continuous maps $f,g:X\to Y:\; f\simeq g$ iff $f_{*}=g_{*}$.

Is this equivalent relation, stronger than the homotopy equivalent?

On the other hand, for a group $G$, one can define the free abelian group generated by all group homomorphism from $O(n)$ to $G$. The same processes as above, gives us a homology of groups.

Are these type of homologies studied already?

$\endgroup$

This question has an open bounty worth +50 reputation from Ali Taghavi ending in 2 days.

This question has not received enough attention.

  • $\begingroup$ Why is $d \circ d = 0$? $\endgroup$ – Ryan Budney Jan 1 '17 at 19:14
  • $\begingroup$ @RyanBudney The reason is similar to the singular homology case. Please write it for n=2, or 3. Then you would be convinced that it works for arbitrary n. $\endgroup$ – Ali Taghavi Jan 1 '17 at 19:16
  • $\begingroup$ Perhaps I don't understand what the "obvious" boundary maps are. Are they precomposition with your inclusions? There is no canonical inclusion and your objects are not defined to be homotopy-classes of maps, so it is unclear how $d$ is defined. $\endgroup$ – Ryan Budney Jan 1 '17 at 19:18
  • 1
    $\begingroup$ For the space $\mathrm{Map}(O(n),X)$ you consider the free $\mathbb{Z}$-module generated by this space, say $\mathbb{Z}\{\mathrm{Map}(O(n),X)\}$. If instead you were asking about $\mathrm{Map}(X,O(n))$ then I think the question could be related to $KO$ of $\Sigma X$ if $X$ was paracompact, after passing to homotopy of course! So, something like a free resolution of the $KO(X)$ might appear somewhere; this is just a very vague guess to the ``mirror'' of the construction you suggest!!! But, I am not sure such an example has been studied in May's book. $\endgroup$ – user51223 Jan 2 '17 at 6:14
  • 1
    $\begingroup$ My first guess is this is singular cohomology with a funny coefficient group. The coefficient group would have something to do with the bundle structure $O_n \to O_{n+1} \to S^n$. In the $n=1$ case this is a trivial bundle so I believe you get $H^0(X; \mathbb Z^2)$, for your object derived from maps $O_1 \to X$. $\endgroup$ – Ryan Budney Oct 11 at 18:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.