User eric - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T04:56:10Z http://mathoverflow.net/feeds/user/24646 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102051/tessellating-mathbbrn-by-bricks/102196#102196 Answer by Eric for Tessellating $\mathbb{R}^n$ by bricks. Eric 2012-07-14T00:39:39Z 2012-07-14T05:41:33Z <p>If I undderstand the problem correctly one case would be that all bricks are translates of each other in which case a tessalation can be summarized by the lattice of centers of bricks. Up to reordering coordinates, generators for this lattice can be chosen to be the columns of a triangular matrix $M$ and two which differ by an integer triangular basis change are equivalent. In this case $s(M)=dt(M)/L(M)$ where $t(M)$ is the trace of the matrix $M$ and $L(M)$ is the minimum 1-norm of a nonzero lattice element. </p> <p>Claim: $s(d+1)/(d+1) \geq s(d)/d + 1/2$.</p> <p>Thus $s(d) \geq d(d+1)/2$.</p> <p>Proof: If M is obtained by adding a new vector v with diagonal entry r (and dimension) to N then up to translation by the lattice spanned by columns of N, the projection of v to the space spanned by N has 1-norm at least $L(N)/2$ and $s(M)/(d+1)=(t(N)+r)/\min(L(N),L(N)/2 + r) \geq s(N)/d + 1/2$, with equality if $r=L(N)/2$. </p> <p>There is a large space of lattices achieving this bound.<br> One simple solution has diagonal entries (brick edge lengths) of (d/2)(2,1,1,...,1) and first off diagonal of (d/2)(1,1,...,1). </p> http://mathoverflow.net/questions/101634/discrete-version-of-some-topological-object/101720#101720 Answer by Eric for Discrete version of some topological object. Eric 2012-07-09T00:06:05Z 2012-07-09T00:06:05Z <p>This should be a comment. </p> <p>Since the image is Abelian $\phi$ factors through the universal Abelian cover (much smaller than the universal cover) and your objects are equivalent to a map $\psi$ from the integer first homology of your surface to $Z^n$ along with the data of an equivariant map from the one skeleton of this cover to $R^n$ linear on edges and with vertices going to integer points.<br> Since this is equivariant this descends to a map $\tau$ from the surface to the torus $R^n$ mod $Z^n$. </p> <p>In terms of flat connections $\psi$ is the monodromy action of the (Abelianization of) the fundamental group on the fiber ($R^n$). </p> <p>In the case n=2 one characteristic class interpretation of your associated 2-cocycle class is as the pullback of the fundamental class of the torus $R^2$ mod $Z^2$ (classifying space of $Z$ x $Z$) via $\tau$. </p> http://mathoverflow.net/questions/99924/effect-on-homology-of-decorating-vertices-of-a-simplicial-complex/100385#100385 Answer by Eric for Effect on homology of decorating vertices of a simplicial complex Eric 2012-06-22T18:58:27Z 2012-06-22T20:21:23Z <p>If I understand your question correctly $C_i(X[m])=\oplus_P C_i(X)\otimes U^{i,m}_P\otimes V_P$ where $U^{i,m}_P$ has dimension $d^{i,m}_P$ which is the number of semistandard fillings of the shape $P$ with $0, 1, 2, \ldots, i, \ldots i$ and $V_P$ is an irreducible representation of $S_m$. </p> <p>For some representations this is easy to compute. For instance if $P_1=m-d-1$ then $d^{d-1,m}_P= 0$, but $d^{d,m}<em>P > 0$, so $H</em>{d,P}(X[m])=C_d(X)\otimes U^{d,m}_P\otimes V_P$. </p>