User eric - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-21T04:56:10Zhttp://mathoverflow.net/feeds/user/24646http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/102051/tessellating-mathbbrn-by-bricks/102196#102196Answer by Eric for Tessellating $\mathbb{R}^n$ by bricks.Eric2012-07-14T00:39:39Z2012-07-14T05:41:33Z<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#101720Answer by Eric for Discrete version of some topological object.Eric2012-07-09T00:06:05Z2012-07-09T00:06:05Z<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#100385Answer by Eric for Effect on homology of decorating vertices of a simplicial complexEric2012-06-22T18:58:27Z2012-06-22T20:21:23Z<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>