most recent 30 from http://mathoverflow.net 2013-06-19T10:24:32Z http://mathoverflow.net/feeds/user/19731 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88842/when-is-a-finite-cw-complex-a-compact-topological-manifold/88852#88852 Adam 2012-02-18T19:42:35Z 2012-02-18T19:42:35Z

<p>Here is one necessary condition for the space being locally Euclidean: If $X$ is an $n$-dimensional manifold, then for every $p\in X$ the relative homology groups $H_*(X,X\setminus p;\mathbb{Z})$ are trivial except in dimension $n$ where they are equal to $\mathbb{Z}$. It's an easy check that this detects some of the obvious types of singularities. For example, if $p$ is the singular point in $X$, the wedge of two $S^n$s, then $H_n(X,X\setminus p;\mathbb{Z})=\mathbb{Z}^2$.</p> <p>I'm pretty sure it isn't a sufficient condition, however, since one should be able to cook up a singular space where a neighborhood of some point looks is a cone on a homology sphere.</p>

http://mathoverflow.net/questions/87308/homology-related-to-e1-e2/87351#87351 Adam 2012-02-02T17:52:14Z 2012-02-02T17:52:14Z

<p>For the audience, $E(1)$ is the underlying smooth 4-manifold for a rational elliptic surface. i.e. $\mathbb{C}P^2$ blown up 9 times, while $E(2)$ is the underlying smooth manifold for a $K3$ surface and is diffeomorphic to the symplectic fiber sum of two copies of $E(1)$ along an elliptic fiber. ($E(n)$ is the fiber sum of $n$-copies of $E(1)$)</p> <p>I assume that you don't find the change of basis in $H_2(E(n))$ method in Gompf and Stipsciz sufficient for your purposes? An alternative way of finding the intersection forms is to look at Kodira's classification of singular fibers in an elliptic fibration. (See for example Barth-Hulek-Peters-Van de ven.) From this we can obtain an elliptic fibration with two or three singular fibers one of which has intersection form $-E8$, the others being either a cusp or two nodes. The regular neighborhood $W$ of the $-E8$ singular fiber is diffeomorphic to a plumbing of $-2$-spheres in a configuration given by the Dynkin diagram for $E8$ and has boundary equal to the Poincare homology sphere $\Sigma(2,3,5)$. The intersection for the other side, called the nucleus $N(1)$, is $\left(\begin{matrix} 0 &amp; 1 \ 1 &amp; -1 \end{matrix}\right)$.</p> <p>The homology of $E(1)$ is generated by $h,e_1,\ldots, e_9$ (the generators of $\mathbb{C}P^2$ and the $\overline{\mathbb{C}P}^2$s. Then $H_2(N(1))$ is generated by $[F]=3h-\sum_{i=1}^8 e_i$ and $e_9$ where $F$ is the elliptic fiber and $H_2(W)$ is generated by $e_1-e_2,e_2-e_3,\ldots,e_7-e_8,-h+e_6+e_7+e_8$. </p> <p>You can now reconstruct your statement using the Meyer-Vietoris and relative homology sequences. The three square $0$ classes end up being the 3 "rim tori" in the $T^3$ boundary of $E(n)\setminus N(F)$.</p>

http://mathoverflow.net/questions/87071/is-this-manifold-orientable/87077#87077 Adam 2012-01-30T22:04:01Z 2012-01-30T22:04:01Z

<p>Identifying your points with matrices $M$ with column vectors $(a,b)^T$ and $(c,d)^T$, your equations come from the components of $M M^\dagger=I$ where $M^\dagger$ denotes the conjugate transpose. So $C$ is $U(2)$ and $S$ is $SU(2)$. Since $SU(2)$ is diffeomorphic to $S^3$, it is orientable.</p>

http://mathoverflow.net/questions/84388/presentation-of-the-clifford-group-by-generators-and-relations/84400#84400 Adam 2011-12-27T18:10:35Z 2011-12-27T18:10:35Z

<p>Yes, from Lawson and Michelson's "Spin Geometry': If $e_1,\ldots, e_n$ is an orthonormal basis for $\mathbb{R}^n$ then the Clifford group can be presented by the abstract elements $-1,e_1,\ldots, e_n$ subject to the relation that $-1$ is central and $(-1)^2=1$, $(e_i)^2=-1$ , and $e_i e_j = (-1) e_j e_i$ for all $i\neq j$. </p> <p>You can obtain your matrices through the complex spin representation.</p>

http://mathoverflow.net/questions/84084/is-the-double-twisted-moebius-strip-isotopic-to-the-trivial-strip/84111#84111 Adam 2011-12-22T19:13:20Z 2011-12-22T19:13:20Z

<p>You could look at the knot type of the boundary of these twisted bands. (Substitute the unit sphere bundle if you want the noncompact version.) For each of your $\mu(k)$ you get a $T(2,k)$ torus link as the boundary. Each of these is non-isotopic. For example, $k=2$ gives the Hopf link and $k=3$ the trefoil. This implies that no two would be ambiently isotopic.</p>

http://mathoverflow.net/questions/88103/comparables-to-journal-of-algebra-journal-of-pure-and-applied-algebra Adam 2012-02-10T15:20:50Z 2012-02-10T15:20:50Z

@Donu They are both Elsevier journals. If you are not aware, there is a boycott of Elsevier being organized. See <a href="http://thecostofknowledge.com/" rel="nofollow">thecostofknowledge.com</a>

http://mathoverflow.net/questions/87343/is-the-lipschitz-unique-for-close-alexandrov-space-other-than-dimension-4/87353#87353 Adam 2012-02-03T02:27:17Z 2012-02-03T02:27:17Z

The Fintushel-Stern paper you reference sightly predates Freedman's proof of the topological $h$-cobordism theorem in dimension $4$. Hence the weaker statement.

http://mathoverflow.net/questions/87071/is-this-manifold-orientable/87077#87077 Adam 2012-01-31T16:51:51Z 2012-01-31T16:51:51Z

@Mohammad Muro's comment wasn't visible to me when I posted.

http://mathoverflow.net/questions/86757/can-we-explain-why-1-998001-or-1-9801-yields-to-magic-results/86758#86758 Adam 2012-01-26T21:57:12Z 2012-01-26T21:57:12Z

@JohnJohnGa Let $x=10^3$ and expand the first few terms of martin's formula. Also, $1/998001 = 1.0020030040050060070080090100110120130140150160170\ldots$

http://mathoverflow.net/questions/86112/unsociability-of-the-quintic-over-the-quaternion-and-octinion-extension-of-the-ra Adam 2012-01-19T17:58:08Z 2012-01-19T17:58:08Z

What does solvability mean in the context of the quaternions? I ask since polynomials in $\mathbf{H}$ do not uniquely factor. For example: $x^2+1=(x+i)(x-i)=(x+j)(x-j)=(x+k)(x-k)$