3
$\begingroup$

Let $g$ be a continuous function from the unit simplex $D(n)$ (with $n$ vertices) into itself, that leaves invariant its vertices, and such that $g$ is not onto: to fix ideas say that $g(D(n))$ does not contain $e=((1/n),(1/n),...,(1/n))$.

I believe that g cannot satisfy the following symmetry property:

Let $F_{i}$ be the face of $D(n)$ where the $i$-th coordinate is zero; and $s^{ij}$ be the permutation of $[n]$ exchanging $i$ and $j$. Then $g$ commutes with $s^{ij}$ on $F_{i}\cup F_{j}: g(s^{ij}(x))=s^{ij}(g(x))$.

This is trivial for $n=2$ and easy for $n=3$. By the symmetry assumption, the winding number in $D(3)$ of the image by $g$ of $F_{i}$ (viewed as a path between two vertices, taken clock-wise) with respect to $e$ does not depend on $i$, and is equal to $1/3+k$ (even though not all three symmetries are rotations). Hence, the image by g of $\Delta (3)$, the frontier of $D(3)$ viewed as a circular path, has a non zero winding number. This is impossible because $\Delta (3)$ is contractible in $D(3)$ whereas its image is not, as it avoids $e$. For higher dimensions, the notion of a winding number is more complicated and may or may not help; perhaps an argument can be made without this notion.

$\endgroup$
  • 1
    $\begingroup$ Is it true that you function $g$ does not move the point of the boundary of $D(n)$? If yes, then the map should be surjective, otherwise one can construct a retraction of the simplex onto its boundary (and it is well-known that such retraction does not exist). $\endgroup$ – Taras Banakh Oct 6 at 13:15
  • $\begingroup$ No we do not assume that the function is constant on the frontier, the image of the frontier can be anywhere in the simplex, except for the vertices which are invariant. Thanks for your interest. $\endgroup$ – herve Moulin Oct 6 at 17:28
  • $\begingroup$ Could you write your symmetry condition for the simplex $D(2)$ in the barycentric coordinates ? Is it $f(0,x,y)=f(x,0,y)=f(x,y,0)$ for any $x,y\in[0,1]$ with $x+y=1$? If yes, then for $x=1$ and $y=0$ we get $f(0,1,0)=f(1,0,0)$, which contradicts the condition that $f$ preserves vertices and hence $f(0,1,0)=(0,1,0)\ne (1,0,0)=f(1,0,0)$. $\endgroup$ – Taras Banakh Oct 6 at 22:23
  • $\begingroup$ What do you mean by saying that the functions $g^{-i}$ coincide? They are defined on distinct copies of the $(n-1)$-simplex. Does it follow from your definition that each vertex of $D(n)$ must be mapped to the same point, because the image of the $i$th vertex of the copy of $D(n-1)$ given by missing out vertex $n$ must be the same as the image of the $i$th vertex of the copy of $D(n-1)$ given by missing out vertex $0$? Doesn't this contradict your statement that vertices are fixed by the map? $\endgroup$ – IJL Oct 7 at 9:00
1
$\begingroup$

EDIT: I was wrong, and the actual answer is that there IS such a mapping. We need to specify it on the boundary, and check that the winding (i.e. image of the boundary in $H^n (D_n \ e)$ is 0). The image of the identity map has index 1.

Now, consider the simplex of dimension 6. Note that it has 6 five-dimensional facets, 6*5/2 = 15 four-dimensional facets, 6*5*4/6= 20 three-dimensional facets.

Note that gcd(6, 15, 20) = 6+15-20 = 1. It is possible to construct such a map that it is identity away from the small neighborhoods of the centers of 5, 4, 3 - dimensional facets. Near the centers of the 5 and 4-dimensional facets we define the map as follows:

Denote the center of the facets as $c$. Away from the circle of radius $R<<1$ it is identity. Inside it it is identity plus vector $(e-c) f(r)$, where $r$ is a distance to $c$, and $f(r) = (1+\varepsilon)(1-\frac{r}{R})$. $\varepsilon << 1$.

Each such facet will decrease winding number by $1$.

Near the 3-dimensional facets, the following reverse construction works:

Fix some $R_2 < R$. Out of the circle of radius $R$ the map F is identity). In the annulus $R_2 < r < R$ it is $F(x) = c + (x-c)\frac{r - R_2}{R-R_2}$. In the inner circle the map is $c - (x-c)\sin(\pi \frac{R_2-r}{R_2}) + (e-c)f(r)$.

Here, $f(r) = (1+\varepsilon)(1-\frac{r}{R_2})$

Each such facet will lower the winding by $1$.

Now, the resulting map will have winding $0$, hence can be prolonged to the interior of the simplex.

I think that it exists if and only if gcd($C_n^k$) over all $1<k<n$ equals 1.


Previous answer (WRONG):

I think you can prove this fact in a following way (I'd like to denote by $D_n$ the simplex of dimension $n$, so $D(n+1)$ in your notation) consider the image of a boundary. It is some homological cycle $s \in H^{n-1} (D_n \ e) \simeq \mathbb{Z}$.

It is clear that any two symmetric maps from $D_n$ to itself can be transformed to each other by a homotopy in a class of symmetric maps. One can also arrange this homotopy in such a way that only the images of the $n-1$-dimensional facets ever pass through $e$.

Then, it is clear that this homology element changes by some multiple of $n$ on each such occasion (because facets pass through $e$ simultaneously due to symmetry condition), and by exactly $n$ if the passing is transverse.

For the identity mapping, $s = 1$, so by the argument above for any mapping $s = 1+kn \neq 0$. Hence, the image of $D_n$ should intersect $e$ (because if it won't intersect $e$ the element $s$ would be homologous to $0$.~~

$\endgroup$
  • $\begingroup$ Many thanks for your answer. I will ask a colleague to explain it at my lower level and get back to you. $\endgroup$ – herve Moulin Oct 10 at 15:28
  • $\begingroup$ Lev's answer looks good! $\endgroup$ – Gael Meigniez Oct 11 at 16:29
  • $\begingroup$ This map actually exists! I've changed the answer. $\endgroup$ – Lev Soukhanov Oct 12 at 16:03
  • $\begingroup$ @GaelMeigniez it looked good but apparently was wrong :) $\endgroup$ – Lev Soukhanov Oct 12 at 18:40
0
$\begingroup$

The question is nice! provided that in the hypotheses you actually assume that the barycenter is not in the image.

Your other question, where you only assume that g is not onto, admits easy counterexamples: it is easy (eg by means of a partition of the unity) to make on your (n-1)-dimensional simplex Dn a smooth vector field V such that - V vanishes at each point of each face of dimension <=n-3; - V is transverse to each hyperface F (face of dimension n-2) at every point x interior to F, and enters Dn at this point.

Then, you consider the action of the symmetric group S_n on the simplex, and you change V to the sum of its images through all the symmetric trannsformations. This way, you have moreover that V is S_n-invariant. Then, the time 1 of the flow of V is a counterexample to your question.

$\endgroup$
  • $\begingroup$ You are absolutely right I started by the right question then got carried away. $\endgroup$ – herve Moulin Oct 10 at 15:07
0
$\begingroup$

To Taras Banakh and IJL Apologies to both of you for expressing the symmetry property in this cryptic way. Here is a clearer way, I hope:

Let $F_{i}$ be the face of $D(n)$ where the $i$-th coordinate is zero; and $s^{ij}$ be the permutation of $[n]$ exchanging $i$ and $j$. Then $g$ commutes with $s^{ij}$ on $F_{i}∪F_{j}$: $g(s^{ij}(x))=s^{ij}(g(x))$. In particular if $f(0,1,0)$ is the vertex of the second coordinate, then $f(1,0,0)$ is the vertex of the first.

$\endgroup$
  • $\begingroup$ you should probably post this stuff in comments and tag users so they notice your answer, for example @herveMoulin (actually doesn't work because you are OP but should work on other users). Answers are for actual answers or extensive discussion. Your question is very nice though. $\endgroup$ – Lev Soukhanov Oct 9 at 16:30

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.