# simplicial complex equipped with barycenric metric is complete [closed]

Consider a simplicial complex $C$. On its support $$|C|=\lbrace \alpha = \sum_{v\in C}\alpha_{v}v \mid 0\leq \alpha_{v} \leq 1 , \sum_{v\in C}\alpha_{v} =1\mbox{ and }v|{\alpha_{v}} \neq 0\mbox{ is a simplex in }C \rbrace$$ there is a metric, the barycentric metric, defined by $d(\alpha,\beta)=\sqrt{\sum_{v\in C}(\alpha_{v} - \beta_{v})^{2}}$. Is this metric complete ?

Greetings Ben

-

## closed as off topic by Misha, Benoît Kloeckner, Lee Mosher, Oscar Randal-Williams, Chris GerigJun 11 '13 at 16:21

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

This doesn't look like research level… Please try another forum (e.g. MathStackExchange). – Loïc Teyssier Jun 10 '13 at 7:48
Benjamin - some extra context might be helpful. Do you want to place any restrictions on $C$? Is it finite-dimensional, for instance? – HJRW Jun 10 '13 at 12:21

No, this metric need not be complete. Let $C$ denote the complex with vertex set $\mathbb{N}$, where the simplices are the nonempty finite subsets. Thus, the points of $|C|$ are sequences of nonnegative real numbers that are eventually zero and sum to one. Consider the sequences \begin{align*} a_0 &= (1/2,1/2,0,0,0,0,\dotsc) \\\\ a_1 &= (1/2,1/4,1/4,0,0,0,\dotsc) \\\\ a_2 &= (1/2,1/4,1/8,1/8,0,0,\dotsc) \\\\ a_3 &= (1/2,1/4,1/8,1/16,1/16,0,\dotsc) \end{align*} and so on. Then $(a_n)_{n=0}^\infty$ is a Cauchy sequence in $|C|$ that does not converge.