Let $K$ be a simplicial complex. Is there a way to calculate the number of k-simlices in the barycentric subdivision $K'$ of $K$? Given the number of $l$-simplices in $K$, for any $l$, of course.

(I am suposed to show that the Euler characteristic does not change, directly from the definition. I can not see how to do this without calculating the number of k-simplices...)


1 Answer 1


I suppose this should be tagged as homework, since this is problem (1) on http://kurser.math.su.se/file.php/853/Top11.pdf.

The following might be of some help. I am fairly certain that it is correct. Let $K_n$ denote the simplicial complex consisting of a simplex of dimension $n$, together with all its faces. Let $K_n'$ denote its barycentric subdivision. Let $s(n)$ be the number of simplexes in $K_n$. Let $s(m,n)$ be the number of $m$-simplexes in $K_n$. Let $s'(m,n)$ be the number of $m$-simplexes in $K_n'$. Let for $n > 0$ $s''(m,n)$ be the number of $m$-simplexes in $K_n'$ not spanned by the barycenter of the $n$-simplex. Then

$$ s(n) = 2^{n+1}-1, $$


$$ s(m,n) = \binom{n+1}{m+1}, $$


$$ s'(m,n) = \begin{cases} s(n) &\text{if } m=0, \\ s''(m,n) + s''(m-1,n) &\text{otherwise}, \end{cases} $$


$$ s''(m,n) = \begin{cases} s(n)-1 &\text{if } m=0, \\ \sum\limits_{i=m}^{n-1}\left( s(i,n) \left[s'(m,i) - s''(m,i) \right] \right) &\text{if } 0 < m < n, \\ 0 &\text{if } m = n. \end{cases} $$

It is possible that the formulas could be considerably simplified. I can provide derivations of the above after homework deadline at May 13.

UPDATE: The expressions for $s(n)$ and $s(m,n)$ are stated at http://en.wikipedia.org/wiki/Simplex.

For $s'(m,n)$ reason as follows. The case $m = 0$ is clear. For $m > 0$, in addition to the $s''(m,n)$ $m$-simplexes not spanned by the barycenter of the $n$-simplex, there are $s''(m-1,n)$ $m$-simplexes spanned by the barycenter and the vertices of an $(m-1)$-simplex not spanned by the barycenter. There are no other $m$-simplexes.

For $s''(m,n)$ reason as follows. The cases $m = 0$ and $m = n$ are clear. For $0 < m < n$ and $m \leq i \leq n-1$, there are $s(i,n)$ $i$-complexes in $K_n$, each contributing to $K_n'$ with $s'(m,i)$ $m$-simplexes not spanned by the barycenter of the $n$-simplex. There are no other contributions of such simplexes. In the summation, discount for $m$-simplexes already contributed by simplexes of lower dimension: of the $s'(m,i)$ $m$-simplexes contributed by an an $i$-simplex, $s''(m,i)$ $m$-simplexes have already been counted (since these are not spanned by the barycenter of the $i$-simplex).

HOWEVER: Visualizing the complexes $K_n'$ for $n=0,1,2,3$, I get

$$ s'(0,0)=1 \\ s'(0,1)=3, \quad s''(0,1)=2, \\ s'(1,1)=2, \quad s''(1,1)=0, \\ s'(0,2)=7, \quad s''(0,2)=6, \\ s'(1,2)=12, \quad s''(1,2)=6, \\ s'(2,2)=6, \quad s''(2,2)=0, \\ s'(0,3)=15, \quad s''(0,3)=14, \\ s'(1,3)=50, \quad s''(1,3)=36, \\ s'(2,3)=?, \quad s''(2,3)=24, \\ s'(3,3)=24, \quad s''(3,3)=0. $$

Using the formulas above, I get differing results $s'(1,3)=68$ and $s''(1,3)=54$ (and $s'(2,3)=78$). It seems the expression for $s''$ above is counting some contributions more than once (or I am counting simplexes wrong when visualizing $K_3'$).

UPDATE 2: Formula updated, gives correct results at least for $n \leq 3$ now. I think it is correct for all $n$ now.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.