User antonio - MathOverflow

Cone over the Join of two topological spaces

2012-03-21T04:05:27Z

Suppose $X$ and $Y$ are topological spaces. Let's define the join $X\ast Y$ as the quotient space $X\times Y\times [0,1]/\sim$, where $\sim$ is the equivalence relation generated by $(x,y,0)\sim(x,y',0)$ and $(x,y,1)\sim(x',y,1)$. In particular, define the cone over $X$, $Cone(X)$, as the join of $X$ with a point. Is it true that $Cone(X\ast Y)$ is homeomorphic to $Cone(X)\times Cone(Y)$? If not, when does this happen?

Neighbourhood of a point in the relative interior of a cell in a convex cell complex

2012-03-21T05:34:28Z

Hi, Im trying to understand the proof of the next proposition:

$Proposition.$

If a point $x$ in a locally finite convex cell complex $\Lambda$ lies in the interior of some cell $F$, then there is an open neighbourhood of $x$ in $\Lambda$ of the form $int(F)\times N_\varepsilon$, where $int(F)$ is the relative interior of $F$ and $N_\varepsilon$ is an $\varepsilon-$neighbourhood of the cone point in $Cone(Lk(F,\Lambda))$. Equivalently, there is a neighbourhood of $x$ homeomorphic to the cone on the $k-fold$ suspension of $Lk(F,\Lambda)$, where $k=dimF$.

Here, $Lk(F,\Lambda)$ is the link of $F$ in $\Lambda$.

The idea of the proof is to show that $St(F,\Lambda)$, the star of $F$ in $\Lambda$, i.e., the union of all the relative interiors $int(F´)$, where $F´$ is a cell in $\Lambda$ such that $F$ is a face of $F´$, is homeomorphic to $int(F)\times N_\varepsilon$. For this we use that $int(F´)\cong int(F\ast \sigma_{F´})$.

Here $\sigma_{F´}=Lk(F,F´)$ and $F\ast\sigma_{F'}$ is the join of the two convex cells.

The book where I'm studying this says that from the homeomorphisms $int(F´)\cong int(F\ast \sigma_{F´})$, it follows that $St(F,\Lambda)\cong int(F)\times N_\varepsilon$ but I can't see this.

For the statement of the suspension, the book says that if we take a disk neighbourhood in the relative interior of the cell $F$ around $x$, then $D^{k}\times Cone(Lk(F,\Lambda))\cong Cone(S^{k-1}\ast Lk(F,\Lambda))$. The thing is that I can't see this either.

Hereditary algebras

2011-12-01T18:37:34Z

I have the following problem: If $\Lambda$ is a hereditary, basic and connected algebra and $e$ is an idempotent of $\Lambda$, how can I prove that $e\Lambda e$ is also hereditary?

Comment by Antonio

2012-03-21T13:30:49Z

This is the Proposition I need and where I began to ask myself if this question is true mathoverflow.net/questions/91796/…

Comment by Antonio

2012-03-21T13:29:08Z

It´s not a homework I came uo with this problem when I was traying to understand the proof of a Proposition.

Comment by Antonio

2011-12-01T23:47:04Z

Ok, to be honest i have no idea of how to solve this problem... so if you could give a simple proof of it... im really new in this area

Comment by Antonio

2011-12-01T23:20:40Z

So i think it is equivalent to what you call split basic

Comment by Antonio

2011-12-01T23:18:48Z

The definition of basic that Im considered is the next one: If $\Lambda$ is a $k$-algebra with a complete set ${e_{1},...,e_{n}}$ of primitive orthogonal idempotents, then $\Lambda$ is basic if $e_{i}\Lambda$ is not isomorphic to $e_{j}\Lambda$ for all $i\noteq j$

Comment by Antonio

2011-12-01T20:41:25Z

Is not a homework is just that Im interested in studying this things and I found that problem. Yes I assume $\Lambda$ is finite dimensional and is over any field.