most recent 30 from http://mathoverflow.net 2013-06-19T22:40:06Z http://mathoverflow.net/feeds/user/7726 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56245/graphs-embedded-on-fractals Sergeib 2011-02-22T05:10:19Z 2011-02-23T11:50:51Z

<p>It is fairly well know when graphs can be embedded on various surfaces. Also, it is not hard to see that any graph can be embedded in 3-dimensional space. Has anyone ever studied the embeddability of graphs on various fractals? If so, are there any interesting results?</p> <p>For example, I made some quick deductions about graphs on Sierpinski's Gasket. No graph that has vertices of degree greater than 4 can be embedded on Sierpinski's Gasket. Also, I conjecture that one can not embed $K_4$ into Sierpinski's Gasket either. </p> <p>Also, one might ask questions like, "Is it true that one can embed any graph on any space with Hausdorff dimension that is greater than or equal to 3?"</p> <p>These are just some curiosities that came to me and a friend of mine earlier today, and I am curious if the Math Overflow community knows anything about such things.</p>

http://mathoverflow.net/questions/32397/vector-spaces-without-natural-bases Sergeib 2010-07-18T20:34:25Z 2010-08-03T20:32:38Z

<p>Does anyone know any nice examples of vector spaces without a basis that is in some sense "natural".</p> <p>To clarify what I mean, suppose we look at $\mathbb{R}^2$. We define $\mathbb{R}^2$ as pairs of real numbers. In some sense, what we are doing is expressing vectors in terms of a natural basis : (1,0) and (0,1). This is not what I want. </p> <p>An example that I thought of is a tangent space to a manifold. When one picks a tangent space to a manifold, there is no natural basis that one can pick. </p> <p>Are there other nice examples?</p>

http://mathoverflow.net/questions/32734/jordan-curve-homotopy Sergeib 2010-07-21T02:29:33Z 2010-07-21T15:14:56Z

<p>Does there exist a notion of Jordan curve homotopy? </p> <p>In particular, suppose we have two Jordan curves $C_0 : S^1 \rightarrow \mathbb{R}^2$ and $C_1 : S^1 \rightarrow \mathbb{R}^2$. When does there exist a continuous function $f: S^1 \times [0,1] \rightarrow \mathbb{R}^2$ such that:</p> <p>$f(x,0) = C_0(x)$, $f(x,1) = C_1(x)$, and for all $t \in [0,1]$, the function $C_t: S^1 \rightarrow \mathbb{R}^2$ defined by $C_t(x) = f(x,t)$ is a Jordan curve. </p> <p>My intuition tells me that such a function always exists, but I'm unsure about how to go about proving this. Also, if this is a known result, are there similar results for manifolds other than $\mathbb{R}^2$?</p>