Unimodal category and the monotonicity conjecture

Abstract

We completely characterize the unimodal category for functions $f:{\mathbb {R}}\rightarrow [0,\infty )$ using a decomposition theorem obtained by generalizing the sweeping algorithm of Baryshnikov and Ghrist. We also give a characterization of the unimodal category for functions $f:S^1\rightarrow [0,\infty )$ and provide an algorithm to compute the unimodal category of such a function in the case of finitely many critical points. We then turn to the monotonicity conjecture of Baryshnikov and Ghrist. We show that this conjecture is true for functions on ${\mathbb {R}}$ and $S^1$ using the above characterizations and that it is false on certain graphs and on the Euclidean plane by providing explicit counterexamples. We also show that it holds for functions on the Euclidean plane whose Morse–Smale graph is a tree using a result of Hickok, Villatoro and Wang.

Appendices

A Direct descriptions of the counterexamples in ${\mathbb {R}}^2$

1.1 A.1 First example

Piecewise linearity of the functions $u_1,u_2,F:{\mathbb {R}}^2\rightarrow [0,\infty )$ appearing in Sect. 4.4 follows from the properties of the $\infty$-distance. For concreteness, we explicitly describe the decomposition of the plane with respect to which the functions are piecewise linear. The main advantage of $u_1$ and F being piecewise linear is that other facts about these functions can be verified completely computationally.

Before we begin, we need some notation. Two points $p_1,p_2\in {\mathbb {R}}^2$ determine a segment

$$\begin{aligned} p_1p_2=\{p\in {\mathbb {R}}^2\mid \exists t\in [0,1]:p=(1-t)p_1+tp_2\}. \end{aligned}$$

We write $p_1p_2\ldots p_n$ for the union of segments $p_1p_2,p_2p_3,\ldots ,p_{n-1}p_n$. If $p_1p_2\ldots p_np_1$ is a topological circle in ${\mathbb {R}}^2$, it is the boundary of a uniquely determined compact set in ${\mathbb {R}}^2$, which we denote by $\overline{p_1p_2\ldots p_n}$.

We can describe $u_1,u_2$ and F as piecewise linear functions defined by their values on the vertices of a polygonal decomposition of $S=\mathrm{supp}f$ consisting of 44 vertices, 95 edges and 52 faces, namely triangles, trapezoids and two non-convex quadrilaterals (see Fig. 17). Note that we do not count the “face at infinity” and we consider parallelograms to be a special case of trapezoids. Some care must be taken as not every choice of values at the vertices of a quadrilateral can be extended to a linear function. First, we list the vertices (indexed lexicographically):

$$\begin{aligned} \begin{array}{l} x_1=(-4, 0),\\ x_2=(-4, 2),\\ x_3=(-4, 4),\\ x_4=(-3, 1),\\ x_5=(-3, 2),\\ x_6=(-3, 3),\\ x_7=(-2, -2),\\ x_8=(-2, 0),\\ x_9=(-2, 1),\\ x_{10}=(-2, 2),\\ x_{11}=(-2, 3), \end{array} \qquad \begin{array}{l} x_{12}=(-2, 4),\\ x_{13}=(-1, -1),\\ x_{14}=(-1, 1),\\ x_{15}=(-1, 2),\\ x_{16}=(-1, 3),\\ x_{17}=(0, -4),\\ x_{18}=(0, -2),\\ x_{19}=(0, 2),\\ x_{20}=(0, 4),\\ x_{21}=(1, -3),\\ x_{22}=(1, -2), \end{array} \qquad \begin{array}{l} x_{23}=(1, -1),\\ x_{24}=(1, 1),\\ x_{25}=(1, 2),\\ x_{26}=\left( 1, \tfrac{11}{5}\right) ,\\ x_{27}=(1, 3),\\ x_{28}=(2, -4),\\ x_{29}=(2, -3),\\ x_{30}=(2, -2),\\ x_{31}=(2, -1),\\ x_{32}=(2, 0),\\ x_{33}=(2, 1), \end{array} \qquad \begin{array}{l} x_{34}=(2, 2),\\ x_{35}=(2, 4),\\ x_{36}=\left( \tfrac{11}{5}, 3\right) ,\\ x_{37}=(3, -3),\\ x_{38}=(3, -2),\\ x_{39}=(3, -1),\\ x_{40}=(3, 1),\\ x_{41}=(4, -4),\\ x_{42}=(4, -2),\\ x_{43}=(4, 0),\\ x_{44}=(4, 2). \end{array} \end{aligned}$$

Fig. 17

A polygonal decomposition of $S=\mathrm{supp}f$ and the values of F at its vertices

We omit listing the edges $h_1,h_2,\ldots ,h_{95}$ (ordered lexicographically by the indices of the vertices) as they are simply the edges of the 52 polygons in the decomposition. Finally, we list the faces, using the notation defined above (ordered lexicographically by the corresponding sets of vertices):

$$\begin{aligned} \begin{array}{l} f_1=\overline{x_1x_4x_5x_2},\\ f_2=\overline{x_1x_8x_9x_4},\\ f_3=\overline{x_2x_6x_3},\\ f_4=\overline{x_2x_5x_6},\\ f_5=\overline{x_3x_6x_{12}},\\ f_6=\overline{x_4x_{10}x_5},\\ f_7=\overline{x_4x_9x_{10}},\\ f_8=\overline{x_5x_{10}x_{11}x_6},\\ f_9=\overline{x_6x_{11}x_{12}},\\ f_{10}=\overline{x_7x_{13}x_{14}x_8},\\ f_{11}=\overline{x_7x_{18}x_{23}x_{13}},\\ f_{12}=\overline{x_8x_{14}x_9},\\ f_{13}=\overline{x_9x_{14}x_{15}x_{10}}, \end{array} \qquad \begin{array}{l} f_{14}=\overline{x_{10}x_{16}x_{11}},\\ f_{15}=\overline{x_{10}x_{15}x_{16}},\\ f_{16}=\overline{x_{11}x_{16}x_{20}x_{12}},\\ f_{17}=\overline{x_{13}x_{23}x_{24}x_{14}},\\ f_{18}=\overline{x_{14}x_{19}x_{15}},\\ f_{19}=\overline{x_{14}x_{24}x_{19}},\\ f_{20}=\overline{x_{15}x_{19}x_{20}x_{16}},\\ f_{21}=\overline{x_{17}x_{21}x_{22}x_{18}},\\ f_{22}=\overline{x_{17}x_{28}x_{29}x_{21}},\\ f_{23}=\overline{x_{18}x_{22}x_{23}},\\ f_{24}=\overline{x_{19}x_{27}x_{20}},\\ f_{25}=\overline{x_{19}x_{24}x_{25}},\\ f_{26}=\overline{x_{19}x_{25}x_{26}}, \end{array} \qquad \begin{array}{l} f_{27}=\overline{x_{19}x_{26}x_{34}x_{27}},\\ f_{28}=\overline{x_{20}x_{27}x_{35}},\\ f_{29}=\overline{x_{21}x_{30}x_{22}},\\ f_{30}=\overline{x_{21}x_{29}x_{30}},\\ f_{31}=\overline{x_{22}x_{30}x_{31}x_{23}},\\ f_{32}=\overline{x_{23}x_{32}x_{24}},\\ f_{33}=\overline{x_{23}x_{31}x_{32}},\\ f_{34}=\overline{x_{24}x_{33}x_{34}x_{25}},\\ f_{35}=\overline{x_{24}x_{32}x_{33}},\\ f_{36}=\overline{x_{25}x_{34}x_{26}},\\ f_{37}=\overline{x_{27}x_{34}x_{35}},\\ f_{38}=\overline{x_{28}x_{37}x_{29}},\\ f_{39}=\overline{x_{28}x_{41}x_{37}}, \end{array} \qquad \begin{array}{l} f_{40}=\overline{x_{29}x_{37}x_{38}x_{30}},\\ f_{41}=\overline{x_{30}x_{39}x_{31}},\\ f_{42}=\overline{x_{30}x_{38}x_{39}},\\ f_{43}=\overline{x_{31}x_{39}x_{43}x_{32}},\\ f_{44}=\overline{x_{32}x_{36}x_{33}},\\ f_{45}=\overline{x_{32}x_{40}x_{34}x_{36}},\\ f_{46}=\overline{x_{32}x_{43}x_{40}},\\ f_{47}=\overline{x_{33}x_{36}x_{34}},\\ f_{48}=\overline{x_{34}x_{40}x_{44}},\\ f_{49}=\overline{x_{37}x_{42}x_{38}},\\ f_{50}=\overline{x_{37}x_{41}x_{42}},\\ f_{51}=\overline{x_{38}x_{42}x_{43}x_{39}},\\ f_{52}=\overline{x_{40}x_{43}x_{44}}. \end{array} \end{aligned}$$

We can now state the alternative descriptions of $u_1,u_2$ and F. The function $u_1$ can be defined on the vertices of the above decomposition:

$$\begin{aligned} u_1(x_i)={\left\{ \begin{array}{ll} 5;&{}i=40,\\ 1;&{}i=4,5,6,9,13,14,23,24,31,33,36,37,38,39,\\ 0;&{}\text {elsewhere}. \end{array}\right. } \end{aligned}$$

Similarly, we have

$$\begin{aligned} u_2(x_i)={\left\{ \begin{array}{ll} 5;&{}i=27,\\ 1;&{}i=6,11,13,14,15,16,21,22,23,24,25,26,29,37,\\ 0;&{}\text {elsewhere}. \end{array}\right. } \end{aligned}$$

Summing these, we obtain

$$\begin{aligned} F(x_i)={\left\{ \begin{array}{ll} 5;&{}i=27,40,\\ 2;&{}i=6,13,14,23,24,37,\\ 1;&{}i=4,5,9,11,15,16,21,22,25,26,29,31,33,36,38,39,\\ 0;&{}\text {elsewhere}. \end{array}\right. } \end{aligned}$$

The values of F (and the polygonal decomposition of $S=\mathrm{supp}f$) are pictured in Fig. 17.

1.2 A.2 Second example

For the same reason as our first example, $f:{\mathbb {R}}^2\rightarrow [0,\infty )$ is actually a piecewise linear function, so it can be described by its function values at the vertices of a polygonal decomposition of its support $S=\mathrm{supp}f$. This decomposition consists of 47 vertices, 94 edges and 46 faces, namely triangles, trapezoids and two non-convex quadrilaterals (see Fig. 18). Again note that not every choice of values at the vertices of a trapezoid can be extended to a linear function, but in our case such issues do not arise as the two function values on each of two parallel sides agree. We first list the vertices (indexed lexicographically):

$$\begin{aligned} \begin{array}{l} x_1=(-7, -7),\\ x_2=(-7, 7),\\ x_3=(-6, -6),\\ x_4=(-6, 6),\\ x_5=(-5, -5),\\ x_6=(-5, -1),\\ x_7=(-5, 1),\\ x_8=(-5, 5),\\ x_9=(-4, 0),\\ x_{10}=\left( -\tfrac{10}{3}, 0\right) ,\\ x_{11}=(-3, -5),\\ x_{12}=(-3, -1), \end{array} \qquad \begin{array}{l} x_{13}=(-3, 1),\\ x_{14}=(-3, 5),\\ x_{15}=(-2, -4),\\ x_{16}=(-2, 0),\\ x_{17}=(-2, 4),\\ x_{18}=(-1, -5),\\ x_{19}=(-1, -3),\\ x_{20}=(-1, -1),\\ x_{21}=(-1, 1),\\ x_{22}=(-1, 3),\\ x_{23}=(-1, 5),\\ x_{24}=(0, -6), \end{array} \qquad \begin{array}{l} x_{25}=(0, -4),\\ x_{26}=(0, 0),\\ x_{27}=(0, 4),\\ x_{28}=(0, 6),\\ x_{29}=(1, -7),\\ x_{30}=(1, -5),\\ x_{31}=(1, -3),\\ x_{32}=(1, -1),\\ x_{33}=(1, 1),\\ x_{34}=(1, 3),\\ x_{35}=(1, 5),\\ x_{36}=(1, 7), \end{array} \qquad \begin{array}{l} x_{37}=(2, -4),\\ x_{38}=(2, 0),\\ x_{39}=(2, 4),\\ x_{40}=(3, -5),\\ x_{41}=(3, -1),\\ x_{42}=(3, 1),\\ x_{43}=(3, 5),\\ x_{44}=\left( \tfrac{10}{3}, 0\right) ,\\ x_{45}=(4, 0),\\ x_{46}=(5, -1),\\ x_{47}=(5, 1). \end{array} \end{aligned}$$

We again omit listing the edges $h_1,h_2,\ldots ,h_{94}$ and proceed to the faces:

$$\begin{aligned} \begin{array}{l} f_1=\overline{x_1x_3x_4x_2},\\ f_2=\overline{x_1x_{29}x_{24}x_3},\\ f_3=\overline{x_2x_4x_{28}x_{36}},\\ f_4=\overline{x_3x_5x_8x_4},\\ f_5=\overline{x_3x_{24}x_{18}x_5},\\ f_6=\overline{x_4x_8x_{23}x_{28}},\\ f_7=\overline{x_6x_9x_7},\\ f_8=\overline{x_6x_{12}x_9},\\ f_9=\overline{x_7x_9x_{13}},\\ f_{10}=\overline{x_9x_{12}x_{10}x_{13}},\\ f_{11}=\overline{x_{10}x_{12}x_{16}},\\ f_{12}=\overline{x_{10}x_{16}x_{13}}, \end{array} \qquad \begin{array}{l} f_{13}=\overline{x_{11}x_{15}x_{16}x_{12}},\\ f_{14}=\overline{x_{11}x_{18}x_{25}x_{15}},\\ f_{15}=\overline{x_{13}x_{16}x_{17}x_{14}},\\ f_{16}=\overline{x_{14}x_{17}x_{27}x_{23}},\\ f_{17}=\overline{x_{15}x_{19}x_{20}x_{16}},\\ f_{18}=\overline{x_{15}x_{25}x_{19}},\\ f_{19}=\overline{x_{16}x_{21}x_{22}x_{17}},\\ f_{20}=\overline{x_{16}x_{20}x_{26}},\\ f_{21}=\overline{x_{16}x_{26}x_{21}},\\ f_{22}=\overline{x_{17}x_{22}x_{27}},\\ f_{23}=\overline{x_{18}x_{24}x_{25}},\\ f_{24}=\overline{x_{19}x_{25}x_{26}x_{20}}, \end{array} \qquad \begin{array}{l} f_{25}=\overline{x_{21}x_{26}x_{27}x_{22}},\\ f_{26}=\overline{x_{23}x_{27}x_{28}},\\ f_{27}=\overline{x_{24}x_{29}x_{30}x_{25}},\\ f_{28}=\overline{x_{25}x_{31}x_{32}x_{26}},\\ f_{29}=\overline{x_{25}x_{30}x_{40}x_{37}},\\ f_{30}=\overline{x_{25}x_{37}x_{31}},\\ f_{31}=\overline{x_{26}x_{33}x_{34}x_{27}},\\ f_{32}=\overline{x_{26}x_{32}x_{38}},\\ f_{33}=\overline{x_{26}x_{38}x_{33}},\\ f_{34}=\overline{x_{27}x_{35}x_{36}x_{28}},\\ f_{35}=\overline{x_{27}x_{34}x_{39}}, \end{array} \qquad \begin{array}{l} f_{36}=\overline{x_{27}x_{39}x_{43}x_{35}},\\ f_{37}=\overline{x_{31}x_{37}x_{38}x_{32}},\\ f_{38}=\overline{x_{33}x_{38}x_{39}x_{34}},\\ f_{39}=\overline{x_{37}x_{40}x_{41}x_{38}},\\ f_{40}=\overline{x_{38}x_{41}x_{44}},\\ f_{41}=\overline{x_{38}x_{42}x_{43}x_{39}},\\ f_{42}=\overline{x_{38}x_{44}x_{42}},\\ f_{43}=\overline{x_{41}x_{45}x_{42}x_{44}},\\ f_{44}=\overline{x_{41}x_{46}x_{45}},\\ f_{45}=\overline{x_{42}x_{45}x_{47}},\\ f_{46}=\overline{x_{45}x_{46}x_{47}}. \end{array} \end{aligned}$$

An alternative definition of the piecewise linear function $f:{\mathbb {R}}^2\rightarrow [0,\infty )$ is then given by specifying it on the vertices as follows (see Fig. 18):

$$\begin{aligned} f(x_i)={\left\{ \begin{array}{ll} 3;&{}i=9,45,\\ 1;&{}i=3,4,10,15,16,17,24,25,26,27,28,37,38,39,44,\\ 0;&{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

Fig. 18

A polygonal decomposition of $S=\mathrm{supp}f$ and the values of f at its vertices

Proposition 5.1

The function $f:{\mathbb {R}}^2\rightarrow [0,\infty )$ has a unimodal $\infty$-decomposition of length 2.

Proof

Define two piecewise linear functions $v_1,v_2:{\mathbb {R}}^2\rightarrow [0,\infty )$ on a polygonal decomposition of S by specifying their values at the vertices. Start with the decomposition of S defined above. Further subdivide it by adding the edges $x_{15}x_{18}, x_{17}x_{23}, x_{30}x_{37}, x_{35}x_{39}$. The function $v_1$ assumes the value 1 at the vertices $x_3,x_4,x_{10},x_{15},x_{16},x_{17},x_{24},x_{25},x_{26},x_{27},x_{28}$; the value 3 at the vertex $x_9$; and the value 0 at all the other vertices. The function $v_2$ assumes the value 1 at the vertices $x_3,x_4,x_{24},x_{25},x_{26},x_{27},x_{28},x_{37},x_{38},x_{39},x_{44}$; the value 3 at the vertex $x_{45}$; and the value 0 at all the other vertices. It is clear that $\max \{v_1,v_2\}=f$, however, these functions are not unimodal, as $x_3x_{24}x_{28}x_4x_3$ yields a non-trivial cycle in some of the superlevel sets. However, we can modify $v_1$ and $v_2$ to obtain unimodal functions $u_1,u_2:{\mathbb {R}}^2\rightarrow [0,\infty )$. To retain the property $\max \{u_1,u_2\}=f$, we modify them on sets with disjoint interiors, namely, $u_1$ is modified on the sets $R_1=[-1,1]\times [2,3]$, $R_2=[-1,1]\times [-2,-1]$ and $R_3=[-7,-5]\times [-2,-1]$, whereas $u_2$ is modified on the sets $R_4=[-1,1]\times [1,2]$, $R_5=[-1,1]\times [-3,-2]$ and $R_6=[-7,-5]\times [1,2]$. They are modified in the same way on each of these. Namely, if R is of one of these rectangles, subdivide it into four triangles using the center point of R. Then define a piecewise linear function on R that takes the value 1 at the center point of R and value 0 at the vertices of the rectangle, and extend it by 0 to the whole plane to obtain a function $\varphi _R:{\mathbb {R}}^2\rightarrow [0,\infty )$. Now, $u_1$ and $u_2$ are defined by putting $u_1=v_1-\varphi _{R_1}-\varphi _{R_2}-\varphi _{R_3}$ and $u_2=v_2-\varphi _{R_4}-\varphi _{R_5}-\varphi _{R_6}$. (These functions are piecewise linear w.r.t. a further subdivision of S that takes into account the six rectangles.) A straightforward verification shows that $u_1$ and $u_2$ are unimodal, but it is ultimately unenlightening, so instead of this, we provide the graphs of the two functions (Fig. 19). $\square$

Fig. 19

The graphs of $u_1$ and $u_2$

Fig. 20

The graphs of $u_1$, $u_2$ and $u_3$

Proposition 5.2

The function $f:{\mathbb {R}}^2\rightarrow [0,\infty )$ has a unimodal decomposition of length 3.

Proof

An explicit unimodal decomposition of $f:{\mathbb {R}}^2\rightarrow [0,\infty )$ can be described as follows: take the polygonal decomposition of S as defined in the beginning. Further subdivide it by adding the edges $x_1x_4, x_4x_5, x_{15}x_{18}, x_{20}x_{25}, x_{21}x_{27}, x_{25}x_{32}, x_{27}x_{33}, x_{35} x_{39}$. Now, define piecewise linear functions $u_1,u_2,u_3:{\mathbb {R}}^2\rightarrow [0,\infty )$ by specifying their values at the vertices of this decomposition. For the sake of brevity, we only specify the nonzero values. The function $u_1$ is defined by taking the value 3 at the vertex $x_9$ and the value 1 at the vertices $x_4,x_{10},x_{15},x_{16},x_{17},x_{27},x_{28}$. The function $u_2$ is defined by taking the value 3 at the vertex $x_{45}$ and the value 1 at the vertices $x_3,x_{24},x_{25}, x_{37},x_{38},x_{39},x_{44}$. Finally, $u_3$ takes the value 1 at the vertex $x_{26}$. It is straightforward to verify that these are all unimodal and that $f=u_1+u_2+u_3$. (See the graphs in Fig. 20.) $\square$

B Algorithm for the circle

Theorem 3.7 provides a generalization of the sweeping algorithm computing $\mathbf{ucat}(f)$ for $f:{\mathbb {R}}\rightarrow [0,\infty )$ with finitely many critical points, given in Baryshnikov and Ghrist (2011). Here we describe Algorithm 1 which computes $\mathbf{ucat}(f)$ for $f:S^1\rightarrow [0,\infty )$ in the case of finitely many critical points. Although this is possible in principle, as the proofs of the theorems leading to the algorithm are constructive in nature, we do not compute the explicit decomposition.