most recent 30 from http://mathoverflow.net 2013-06-19T22:20:04Z http://mathoverflow.net/feeds/user/22156 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93958/simple-equivariant-homology-no-borel-moore laerne 2012-04-13T13:39:42Z 2012-04-13T15:47:45Z

<p>Hey. I'm working with Bredon's equivariant cohomology. At some point I need to compute the $4$th equivariant cohomology group of $S^1 \x D^3$ relatively to its boundary for the antipodal action of $\mathbb{Z}_2$.</p> <p>I found a paper that just use "equiviriant poincaré duality" to bring the problem to compute the $0$th equivariant homology which is said to be $\mathbb{Z}_2$. This sounds trivial, but Bredon does not define what equivariant homology is, just cohomology, hence he does not talk of Poincare duality either (At least not in the 30 first pages of his paper "Equivariant Cohomology Theories" I use.).</p> <p>I'm trying to search for a nice definition of equivariant homology, but every paper I find use concept I don't master well or at all (Vector bundles, groupoids, Borel-Moore Homology). I also tried to design my own definition of equivariant homology, simply letting $H_0$ be the quotient of equivariant chains $H_0 = C_0^{\mathbb Z_2}/\partial(C_1^{\mathbb Z_2})$ with $C_i^{\mathbb Z_2} = \lbrace c \in C_i| (1+\mathbb Z) \curvearrowright c = c \rbrace$. But using the relative exact sequence of chain, one finds $C_0^{\mathbb Z_2}(S^1\times D^3, S^1 \times S^2)$ is a subgroup of $C_0(S^1\times D^3, S^1 \times S^2)$, which is $0$, so $H_0^{\mathbb Z_2}(S^1\times D^3, S^1 \times S^2) \approx 0$, not $\mathbb Z_2$ !</p> <p>Many thanks for any help !</p>

http://mathoverflow.net/questions/91277/equivalence-of-definitions-of-carmichael-numbers laerne 2012-03-15T13:39:20Z 2012-03-16T14:14:51Z

<p>I would like to prove the equivalence of the two most common definitions of a composite integer $n > 1$ being a Carmichael number: $a^n \equiv a \mod n $ for all $a$ $\iff a^{n-1} \equiv 1 \mod n$ for all $a$ such that $\mathrm{gcd}(a,n)=1$. </p> <p>I do not see how to prove the right-to-left statement (that is, why if the congruence on the right holds whenever $\mathrm{gcd}(a,n)=1$ then the congruence on the left holds for all $a$). Of course if $n$ divides $a$, the congruence on the left is obvious since both terms are 0.</p> <p>I would like to use the Chinese remainder theorem to try to reduce the problem to the case of a prime-power modulus $n = p^e$ (since I don't know yet $n$ must be square-free), but $a^{n-1} \equiv 1 \mod{p^e}$ is not a very helpful equation.</p> <p>Every article on the web says it is obvious, but not for me. Can you help me?</p>

http://mathoverflow.net/questions/93958/simple-equivariant-homology-no-borel-moore laerne 2012-04-13T16:33:39Z 2012-04-13T16:33:39Z

Yeah, I've already found it. But I'm rather asking &quot;what is equivariant homology ?&quot;.

http://mathoverflow.net/questions/93958/simple-equivariant-homology-no-borel-moore/93968#93968 laerne 2012-04-13T16:31:52Z 2012-04-13T16:31:52Z

Huuu... This is not encouraging, especially since I don't know what a Mackey functor and most of those concepts are...

http://mathoverflow.net/questions/91277/equivalence-of-definitions-of-carmichael-numbers/91372#91372 laerne 2012-03-17T18:19:00Z 2012-03-17T18:19:00Z

So you add a constraint ($a \equiv 1 (n/p^k)$) to find an element. Quite clever, I didn't see that one. Thank you very much !

http://mathoverflow.net/questions/91277/equivalence-of-definitions-of-carmichael-numbers/91372#91372 laerne 2012-03-17T09:42:07Z 2012-03-17T09:42:07Z

Thank you very, I begin to see the pattern of the proof. Though I do not need Korselt Criterion, I shall prove it on the run, so it's done. But one big problem remains for me : why do $(a,n)=1$ ? for instance if $n$ would be $p^k(1+p^{k-1})$, the statement would be wrong since $a|n$. Of course such a $n$ is not a caermichael number for $k \ge 2$ since it is not square-free, but that's what I want to prove ! Finding coprimes is a recurring issue I cannot get rid of. PS: Sorry for the answer, I use another account to answer. I didn't understand you can recover an account.