Proof. First, let $i=1$ . It follows from the assumption that r is odd and from Lemma 3.5, that $\operatorname {Z}$ is translation invariant. Let $Q\in \mathcal {P}(\mathbb {Z}^2)$ be a segment or a point. In either case, a vector $v\in \mathbb {Z}^2$ exists such that $-Q = Q+v$ . Since r is odd and $\operatorname {Z}$ is $\operatorname {GL}_2(\mathbb {Z})$ equivariant, the valuation $\operatorname {Z}$ is also odd. Hence, by translation invariance,

$$\begin{align*}-\operatorname{Z}(Q) = \operatorname{Z}(-Q)=\operatorname{Z}(Q+v) = \operatorname{Z}(Q) \end{align*}$$

and, thus, $\operatorname {Z}(Q)=0$ .

Now, let $i>1$ . By induction and Lemma 3.4, the associated functions $\operatorname {Z}_{i-j}^{r-j}$ of $\operatorname {Z}$ are simple for $0<j<i$ . This implies $\operatorname {Z}(Q+v)=\operatorname {Z}(Q)$ for any $v\in \mathbb {Z}^2$ and any lower-dimensional $Q\in \mathcal {P}(\mathbb {Z}^2)$ .

It follows from equation (3.1) that $\operatorname {Z}(\{0\})=0$ . For any $p\in \mathbb {Z}^2$ with $p\ne 0$ , the translation invariance on lower-dimensional polygons implies that $\operatorname {Z}(\{p\}) = \operatorname {Z}(p+\{0\})=0$ . So, let $S\in \mathcal {P}(\mathbb {Z}^2)$ be a segment. For $m\ge 1$ , we can decompose the dilate $mS$ into m translates $S_i$ with $1\leq i\leq m$ of S that intersect at lattice points. We just saw that our valuation vanishes on points. Hence, we obtain from the homogeneity of $\operatorname {Z}$ that

$$\begin{align*}m^i\operatorname{Z}(S) = \operatorname{Z}(mS) = \sum_{i=1}^m \operatorname{Z}(S_i) = m\operatorname{Z}(S). \end{align*}$$

Since $i>1$ , we conclude that $\operatorname {Z}(S)=0$ .

Proof. Since $\operatorname {Z}$ is translation invariant, all its associated valuations $\operatorname {Z}^{r-j}\colon \mathcal {P}(\mathbb {Z}^2)\to \mathbb {R}[x,y]_{r-j}$ vanish except for $\operatorname {Z}^r=\operatorname {Z}$ . Thus, Lemma 3.8 implies that it is enough to check $\operatorname {Z}(m\mathbf T) = m\operatorname {Z}(\mathbf T)$ for $m\in \mathbb {Z}_{>1}$ .

We observe that

$$\begin{align*}m\mathbf T = \Big(\bigsqcup_{v\in(m-2)\mathbf T\cap\mathbb{Z}^2} (v+[0,1]^2)\Big) \sqcup \Big( \bigsqcup_{a,b\in\mathbb{Z}_{\geq 0},~a+b=m-1} (ae_1 +be_2 + \mathbf T) \Big). \end{align*}$$

Since $\operatorname {Z}$ is odd and translation invariant and $[0,1]^2$ is centrally symmetric, it follows that $\operatorname {Z}(v+[0,1]^2) = 0$ for all $v\in \mathbb {Z}^2$ . Thus, by equation (3.2),

$$\begin{align*}\operatorname{Z}(m\mathbf T) = \sum_{a,b\in\mathbb{Z}_{\geq 0},~a+b=m-1} \operatorname{Z}(ae_1 + be_2 +\mathbf T) = m\operatorname{Z}(\mathbf T), \end{align*}$$

where we used again that $\operatorname {Z}$ is simple and translation invariant.

Proof. First, we show that $\Theta ^r(\operatorname {Z})\in \mathbb {R}[x,y]_r^{\mathbf {G}}$ for $\operatorname {Z}\in \mathrm {Val}_1^r$ . If $\phi \in \mathbf {G}$ , then there exists $v\in \mathbb {Z}^2$ such that $\mathbf T=\phi \mathbf T + v$ . Since $\operatorname {Z}$ is translation invariant by Lemma 3.5 and $\operatorname {\mathrm {GL}}_2(\mathbb {Z})$ equivariant, it follows that Z(T) = Z(T) ∘ ϕ ^⋆ for $\phi \in \mathbf {G}$ , which shows that $\operatorname {Z}(\mathbf T)\in \mathbb {R}[x,y]_r^{\mathbf {G}}$ as required.

The linearity of $\Theta ^r$ is obvious. The injectivity follows directly from Lemma 3.8. To show that $\Theta ^r$ is surjective, we show that for any $f\in \mathbb {R}[x,y]_r^{\mathbf {G}}$ the expression (4.1) can serve as a definition of a simple valuation $\operatorname {Z}_f\in \mathrm {Val}_1^r$ , which is then necessarily $(\Theta ^r)^{-1}(f)$ .

To this end, consider a unimodular triangle $T\in \mathcal {P}(\mathbb {Z}^2)$ represented in two ways as an affine image of the standard triangle $\mathbf T$ , that is, $T = \phi \mathbf T + v = \phi '\mathbf T+v'$ for certain $\phi ,\phi '\in \operatorname {GL}_2(\mathbb {Z})$ and $v,v'\in \mathbb {Z}^2$ . Rearranging yields

$$\begin{align*}\mathbf T = \phi^{-1}\phi'\,\mathbf T +\phi^{-1}(v'-v). \end{align*}$$

Hence, $\phi ^{-1}\phi '\in \mathbf {G}$ . Since f is $\mathbf {G}$ invariant, it follows that $f\circ (\phi ')^\star = f\circ \phi ^\star $ , that is, is a well-defined translation invariant, $\operatorname {\mathrm {GL}}_2(\mathbb {Z})$ equivariant function on the set of unimodular triangles.

Next, we note that the triangles $\rho _1\mathbf T$ and $\rho _2\mathbf T$ occurring in equation (3.3) are opposite in the sense that $\rho _1\mathbf T = -\rho _2\mathbf T$ . Thus, since the degree of f is odd, we have $\operatorname {Z}'(-\mathbf T) = -\operatorname {Z}'(\mathbf T)$ and $\operatorname {Z}'(\rho _1\mathbf T) = -\operatorname {Z}'(\rho _2\mathbf T)$ . Since $\operatorname {Z}'$ is translation invariant, equation (3.3) is satisfied. Hence, Corollary 3.7 yields a simple $\operatorname {GL}_2(\mathbb {Z})$ equivariant valuation $\operatorname {Z}\colon \mathcal {P}(\mathbb {Z}^2)\to \mathbb {R}[x,y]_r $ with $\operatorname {Z}(T) = \operatorname {Z}'(T)$ for all unimodular triangles T. Since $\operatorname {Z}'$ is translation invariant, it follows from equation (3.2) that $\operatorname {Z}$ is translation invariant as well. In particular, $\operatorname {Z}\in \mathrm {Val}^r$ and Lemma 3.8 shows that $\operatorname {Z}(P) = \operatorname {Z}_f(P)$ as given by equation (4.1).

Finally, Lemma 4.2 yields that $\operatorname {Z}_f$ is 1-homogeneous, that is, we have $\operatorname {Z}_f\in \mathrm {Val}_1^r$ as desired.

Proof. It is enough to apply the Reynolds operator $\frac {1}{\lvert \mathbf {G}\rvert }\sum _{\phi \in \mathbf {G}} \phi $ , which is a projection onto $\mathbb {R}[x,y]^{\mathbf {G}}$ , to the polynomials $x^2$ and $x^3$ and thus compute the generators $\mathbf {p}_2$ and $\mathbf {p}_3$ .

Proof of Lemma 5.1.

Let . By assumption, $\operatorname {X}$ is translation invariant, as well as $\operatorname {\mathrm {GL}}_2(\mathbb {Z})$ equivariant and i-homogeneous, and by Lemma 4.1, it is simple. If $i=2$ , then [Reference Ludwig and Silverstein19, Proposition 23] yields $\operatorname {X}=0$ and thus the claim. If $i>2$ , we have $X=0$ by [Reference McMullen20, Theorem 5].

Proof. First, we observe that condition (2) is by linearity equivalent to

1. (2’) There exists $h\in \mathbb {R}[x,y]_{r+1}^{\mathbf {G}}$ such that $2h-(\tfrac {x}{3}+\tfrac y3)\operatorname {Z}_1^r(\mathbf T)\in \mathbb {R}[x,y]_{r+1}^{\mathbf {D}}$ .

In the proof, we will work with (2’) rather than (2).

We start by showing that (1) implies (2’). By Lemma 4.1, the valuation $\operatorname {Z}_2^{r+1}$ is simple. So, it is determined by

by Lemma 3.8. Consider a unimodular transformation $\phi \in \mathbf {G}$ such that $\phi \mathbf T + v = \mathbf T$ for some $v\in \mathbb {Z}^2$ . Then, by equation (5.1) and since $f_1\in \mathbb {R}[x,y]_r^{\mathbf {G}}$ ,

(5.2) $$ \begin{align} f_2 = \operatorname{Z}_2^{r+1}(\phi \mathbf T + v) = f_2\circ\phi^\star + f_1 v. \end{align} $$

Suppose $\widetilde {f}_2$ is another solution of equation (5.2). Then $f_2-\widetilde {f}_2\in \mathbb {R}[x,y]_{r+1}^{\mathbf {G}}$ . Next, we make the guess that $(\tfrac {x}{3}+\tfrac {y}{3})f_1$ is a particular solution to equation (5.2). To see that the guess is correct, note that corresponds to the centroid of the standard triangle.

$$\begin{align*}\big(\!\operatorname{\mathrm{cen}}(\mathbf T)f_1\big)\circ \phi^\star + vf_1 = \operatorname{\mathrm{cen}}(\phi \mathbf T)(f_1\circ\phi^\star) +vf_1 = \operatorname{\mathrm{cen}}(\phi\mathbf T +v) f_1=\operatorname{\mathrm{cen}}(\mathbf T)f_1. \end{align*}$$

So $(\tfrac {x}{3}+\tfrac {y}{3})f_1$ is indeed a solution to equation (5.2) and $f_2 = h+(\tfrac {x}{3}+\tfrac {y}{3})f_1$ for a suitable $h\in \mathbb {R}[x,y]_{r+1}^{\mathbf {G}}$ .

Now consider the square $[0,1]^2$ . This is a centrally symmetric polygon, so we have $\operatorname {Z}_1^r([0,1]^2)=0$ because $\operatorname {Z}_1^r$ is odd and translation invariant. From equation (5.1), it follows that $\operatorname {Z}_2^{r+1}([0,1]^2+v)=\operatorname {Z}_2^{r+1}([0,1]^2)$ for any $v\in \mathbb {Z}^2$ , which implies that

(5.3) $$ \begin{align} \operatorname{Z}_2^{r+1}([0,1]^2)\in\mathbb{R}[x,y]_{r+1}^{\mathbf{D}}, \end{align} $$

since $\phi [0,1]^2$ is a translate of $[0,1]^2$ for $\phi \in \mathbf {D}$ . Using the triangulation $[0,1]^2 = \mathbf T\sqcup (-\mathbf T+e_1+e_2)$ , we obtain

$$ \begin{align*} \operatorname{Z}_2^{r+1}([0,1]^2) &= \operatorname{Z}_2^{r+1}(\mathbf T) + \operatorname{Z}_2^{r+1}(-\mathbf T+e_1+e_2)\\ &= f_2(x,y) + f_2(-x,-y) +(x+y)f_1(-x,-y)\\ &= 2f_2(x,y)-(x+y)f_1(x,y) \\ &= 2h-(\tfrac{x}{3}+\tfrac{y}{3})f_1. \end{align*} $$

Combined with equation (5.3), this proves the implication.

Second, we show that (2’) implies (1). We define . For a unimodular triangle $T=\phi \mathbf T +v$ with $\phi \in \operatorname {GL}_2(\mathbb {Z})$ and $v\in \mathbb {Z}^2$ , we set

We claim that this is independent of the representation of T as an affine image of $\mathbf T$ . As in the proof of Theorem 4.3, consider $\phi \mathbf T +v = \psi \mathbf T+w$ for suitable $\phi ,\psi \in \operatorname {\mathrm {GL}}_2(\mathbb {Z})$ and $v,w\in \mathbb {Z}^2$ . Then, $\mathbf T = \phi ^{-1} \psi \mathbf T + \phi ^{-1}(w-v)$ . This means $\phi ^{-1}\psi \in \mathbf {G}$ . Set . We already saw in the previous step that $f_0$ satisfies

$$\begin{align*}f_0 = f_0\circ (\phi^{-1}\psi)^\star + \phi^{-1}(w-v)\cdot f_1 = f_0\circ\psi^\star\circ\phi^{-\star} + \phi^{-1}(w-v)\cdot f_1. \end{align*}$$

Adding h to the equation, applying $\phi $ and rearranging gives

$$\begin{align*}\begin{aligned} f_2\circ\phi^\star + v\cdot f_1\circ\phi^\star &= h\circ\phi^\star + f_0\circ\phi^\star + v\cdot f_1\circ\phi^\star\\ &= h\circ \phi^\star + f_0\circ\psi^\star + w\cdot(f_1\circ \phi^\star)\\ &= h\circ (\phi^{-1}\psi)\circ\phi^\star + f_0\circ\psi^\star + w\cdot(f_1\circ (\phi^{-1}\psi)\circ \phi^\star)\\ &=h\circ \psi^\star + f_0\circ\psi^\star + w\cdot(f_1\circ \psi^\star)\\ &=f_2\circ\psi^\star + w\cdot(f_1\circ\psi^\star) \end{aligned}\end{align*}$$

Thus, $\operatorname {Z}'(T)$ is, in fact, well-defined. Moreover, it is easy to verify that $\operatorname {Z}'$ is $\operatorname {\mathrm {GL}}_2(\mathbb {Z})$ equivariant.

For $v\in \mathbb {Z}^2$ , consider the triangles $T_1 = v+\mathbf T$ and $T_2 = v+e_1+e_2-\mathbf T$ , as well as $T_3=v+e_1+\rho _2\mathbf T$ and $T_4=v+e_2+\rho _1\mathbf T$ , where again $\rho _i\in \mathbf {D}$ denotes the reflection in the ith coordinate axis. Using that r is odd, we have

$$\begin{align*}\begin{aligned} \operatorname{Z}'(T_1) + \operatorname{Z}'(T_2) &= f_2 + vf_1 + f_2 - (x+y)f_1 - vf_1\\ &= 2(h+(\tfrac x3+ \tfrac y3)f_1) - (x+y)f_1\\ &= 2h - (\tfrac x3 + \tfrac y3) f_1. \end{aligned} \end{align*}$$

Using that r is odd, we also obtain

$$\begin{align*}\begin{aligned} \operatorname{Z}'(T_3) + \operatorname{Z}'(T_4) &= f_2(-x,y) + (x+v)f_1(-x,y) + f_2(x,-y) + (y+v)f_1(x,-y)\\ &= 2f_2(-x,y) + (x-y)f_1(-x,y)\\ &= 2h(-x,y) - (-\tfrac x3 +\tfrac y3) f_1(-x,y)\\ &= 2h - (\tfrac x3 + \tfrac y3) f_1. \end{aligned} \end{align*}$$

We used the $\mathbf {D}$ invariance of $2h - (\tfrac x3 + \tfrac y3) f_1$ to obtain the last line. So the function $\operatorname {Z}'$ satisfies equation (3.3). By Corollary 3.7, we find a simple, $\operatorname {\mathrm {GL}}_2(\mathbb {Z})$ equivariant valuation $\operatorname {Z}\colon \mathcal {P}(\mathbb {Z}^2) \to \mathbb {R}[x,y]_{r+1}$ with $\operatorname {Z}(T) = \operatorname {Z}'(T)$ for all unimodular triangles T. In fact, $\operatorname {Z}$ satisfies equation (5.1): Since $\operatorname {Z}$ and $\operatorname {Z}_1^r$ are simple, it suffices to check equation (5.1) for a unimodular triangle $T = \phi \mathbf T + w$ and a translation vector $v\in \mathbb {Z}^2$ . We have

$$\begin{align*}\begin{aligned} \operatorname{Z}(T+v) &= \operatorname{Z}'(T+v) = f_2\circ \phi^\star + (v+w)\cdot f_1\circ\phi^\star\\ &= \operatorname{Z}'(T) + v\cdot \operatorname{Z}_1^r(\mathbf T) \circ\phi^\star = \operatorname{Z}(T) + v\operatorname{Z}_1^r(T), \end{aligned} \end{align*}$$

where in the last step we used that $\operatorname {Z}_1^r$ is translation invariant by Lemma 3.5. It remains to show that $\operatorname {Z}$ is 2-homogeneous. First, we show that $\operatorname {Z}(2^k T) = 2^{2k}\operatorname {Z}(T)$ holds all unimodular simplices T and for all $k\in \mathbb {Z}_{\geq 0}$ by induction on k. For $k=0$ , there is nothing to prove, so let $k>0$ . Multiplying the triangulation

$$ \begin{align*} 2\mathbf T = \mathbf T\sqcup (\mathbf T+e_1)\sqcup (\mathbf T+e_2) \sqcup (-\mathbf T+e_1+e_2). \end{align*} $$

with $2^{k-1}$ , we obtain a triangulation of $2^k\mathbf T$ into four copies of $\pm 2^{k-1}\mathbf T$ :

$$ \begin{align*} 2^k\mathbf T = 2^{k-1}\mathbf T \sqcup 2^{k-1}(\mathbf T + e_1) \sqcup 2^{k-1}(\mathbf T+e_2) \sqcup 2^{k-1}(-\mathbf T+e_1+e_2). \end{align*} $$

Using induction, we find

$$ \begin{align*} \operatorname{Z}&(2^k\mathbf T) \\ &= \operatorname{Z}(2^{k-1}\mathbf T) + \operatorname{Z}(2^{k-1}(\mathbf T + e_1))+\operatorname{Z}(2^{k-1}(\mathbf T+e_2)) + \operatorname{Z}(2^{k-1}(-\mathbf T+e_1+e_2))\\ &= 2^{2k-2}\big( \operatorname{Z}(\mathbf T) + \operatorname{Z}(\mathbf T+e_1) + \operatorname{Z}(\mathbf T+e_2) + \operatorname{Z}(-\mathbf T+e_1+e_2)\big)\\ &= 2^{2k}\operatorname{Z}(\mathbf T), \end{align*} $$

where we used equation (5.1) to obtain the last line. For any $\phi \in \operatorname {\mathrm {GL}}_2(\mathbb {Z})$ , we then also have $\operatorname {Z}(2^k\phi \mathbf T) = 2^{2k}\operatorname {Z}(\phi \mathbf T)$ . For an arbitrary unimodular triangle $\phi \mathbf T+v$ , it follows from equation (5.1) that

$$\begin{align*}\begin{aligned} \operatorname{Z}(2^k(\phi\mathbf T+v)) &= \operatorname{Z}(2^k \phi \mathbf T) + \operatorname{Z}_1^r(2^k\phi\mathbf T) 2^kv \\ &=2^{2k} (\operatorname{Z}(\phi\mathbf T) + \operatorname{Z}_1^r(\phi\mathbf T)v) = 2^{2k}\operatorname{Z}(\phi\mathbf T+v), \end{aligned}\end{align*}$$

where we used that $\operatorname {Z}_1^r$ is one-homogeneous. So we have $\operatorname {Z}(2T) = 2^{2k}\operatorname {Z}(T)$ for all unimodular triangles T and all $k\in \mathbb {Z}_{\geq 0}$ . By Theorem 3.2, this means that $\operatorname {Z}$ is 2-homogeneous on unimodular triangles. It follows from equation (3.2) that $\operatorname {Z}$ is 2-homogeneous on arbitrary $P\in \mathcal {P}(\mathbb {Z}^2)$ .

Finally, we establish the uniqueness of h. Note that in the construction above we have

$$ \begin{align*}\operatorname{Z}([0,1]^2) = 2{h}-(\tfrac{x}{3}+\tfrac y3)f_1.\end{align*} $$

Suppose that a second $\mathbf {D}$ invariant polynomial $\tilde {h}$ exists such that

$$ \begin{align*}2\tilde{h}-(\tfrac{x}{3}+\tfrac y3)f_1\in\mathbb{R}[x,y]_{r+1}^{\mathbf{D}}.\end{align*} $$

Then there exists a valuation $\operatorname {\tilde {Z}}\in \mathrm {Val}_2^{r+1}$ that satisfies equation (5.1) and has

$$ \begin{align*}\operatorname{\tilde{Z}}([0,1]^2) = 2\tilde{h}-(\tfrac{x}{3}+\tfrac y3)f_1 \neq \operatorname{Z}([0,1]^2).\end{align*} $$

However, by Lemma 5.1, we have $\operatorname {Z}=\operatorname {\tilde {Z}}$ , a contradiction.

Proof. Since the permutation of the coordinates x and y is in both $\mathbf {G}$ and $\mathbf {D}$ , it follows for $f,h\in \mathbb {R}[x,y]^{\mathbf {G}}$ that $h+(x+y)f\in \mathbb {R}[x,y]^{\mathbf {D}}$ if and only if $h+(x+y)f\in \ker (\delta )$ , the kernel of $\delta $ . Since the projection to the first factor $\ker (\delta \circ \rho )\to \mathbb {R}[x,y]_r^{\mathbf {G}}$ is injective by Proposition 5.2, the claim follows.

Proof. We observe that

$$ \begin{align*} \mathbf{p}_2 &= \tilde{\mathbf{q}} - \mathbf{q},\\ (x+y)\mathbf{p}_3 &= \frac{1}{2}(x+y)^2(2x^2 - 5xy + 2y^2) \\ &= \frac{1}{2}(\tilde{\mathbf{q}} + 2\mathbf{q})(2\tilde{\mathbf{q}} - 5\mathbf{q}),\\ \mathbf{p}_3^2 &= \frac{1}{4}(\tilde{\mathbf{q}} + 2\mathbf{q})(2\tilde{\mathbf{q}} - 5\mathbf{q})^2. \end{align*} $$

Therefore, any element in the image of $\rho $ is a linear combination of terms of the form $(\tilde {\mathbf {q}} - \mathbf {q})^k (2\tilde {\mathbf {q}} - 5\mathbf {q})^\ell (\tilde {\mathbf {q}} + 2\mathbf {q})^{\lfloor \frac {\ell +1}{2}\rfloor }$ such that $2\big (k+\ell +\lfloor \frac {\ell +1}{2}\rfloor \big ) = r+1$ .

To determine the range of $\ell $ , note that k must be non-negative, and therefore

$$\begin{align*}\ell + \left\lfloor\frac{\ell+1}{2}\right\rfloor \leq \frac{r+1}{2}. \end{align*}$$

By distinguishing cases, we see that this is equivalent to

$$\begin{align*}\ell\leq\left\lfloor\frac{r+1}{3}\right\rfloor, \end{align*}$$

provided that r is odd.

Proof. It suffices to observe that $\tilde {\mathbf {q}}$ is even in y and $\mathbf {q}$ is odd in y.

Proof of Lemma 5.4.

For $r>1$ odd, let $d(r)$ be the dimension of the image of $\delta \circ \rho $ . Set

with for $0\leq \ell \leq \lfloor \frac {r+1}{3}\rfloor $ . It follows from Lemma 5.6 that $d(r)$ is the number of linearly independent polynomials in

$$\begin{align*}\Big\{\delta(\mathbf{g}_\ell): 0\leq\ell\leq\Big\lfloor\frac{r+1}{3}\Big\rfloor\Big\}. \end{align*}$$

Let R be the matrix of coefficients of monomials of these polynomials, that is, $R_{\ell , i}$ is the coefficient of $\mathbf {q}^i\tilde {\mathbf {q}}^{\frac {r+1}{2}-i}$ in $\delta (\mathbf {g}_\ell )$ . Since $\delta (\mathbf {g}_\ell )$ has no monomials of even degree in $\mathbf {q}$ by Lemma 5.7, every second column of R vanishes. We will show that the remaining

$$\begin{align*}\Big\lfloor\frac{r+3}{4}\Big\rfloor = \Big\lfloor\frac{\frac{r+1}{2} + 1}{2}\Big\rfloor\end{align*}$$

columns of R are linearly independent. This determines the dimension $d(r)$ .

By Lemma 5.7, the entries of R are integers. Therefore, it suffices to show that the remaining columns of R are linearly independent modulo $2$ . This is easier, since we have

$$\begin{align*}\mathbf{g}_\ell = (\tilde{\mathbf{q}} + \mathbf{q})^k \mathbf{q}^\ell\tilde{\mathbf{q}}^{\left\lfloor\frac{\ell+1}{2}\right\rfloor}\end{align*}$$

in $\mathbb Z_2[\tilde {\mathbf {q}}, \mathbf {q}]$ . By the binomial theorem, we obtain that $R_{\ell , 2i+1}$ equals, modulo $2$ ,

$$\begin{align*}\binom{k}{2i+1-\ell} = \binom{\frac{r+1}{2}-\ell-\left\lfloor\frac{\ell+1}{2}\right\rfloor} {2i+1-\ell}. \end{align*}$$

We now show that the indices of the last odd entries in the remaining columns of R are all distinct. More precisely, we show that these indices are

In both cases, the binomial coefficient equals $1$ , which follows in the second case from $\ell _i$ being even. Since r is odd, the indices $\ell _i$ are distinct. It remains to show that $0\leq \ell _i\leq \left \lfloor \frac {r+1}{3}\right \rfloor $ and that subsequent entries in the same column vanish.

For the first case, we observe that $2i+1\leq \left \lfloor \frac {r+1}{3}\right \rfloor $ if and only if $i<\left \lfloor \frac {r+3}{6}\right \rfloor $ , and that the binomial coefficient vanishes when replacing $\ell _i$ with any larger value.

For the second case, we check the extremal values of i. On the one hand, for $i = \left \lfloor \frac {r+3}{6}\right \rfloor $ , we require

$$\begin{align*}\ell_i = r-1-4 \left\lfloor\frac{r+3}{6}\right\rfloor \leq \left\lfloor\frac{r+1}{3}\right\rfloor. \end{align*}$$

Indeed, this is the case for all three relevant congruence classes modulo $6$ , namely, for $r+1 = 6s$ , $r+1= 6s+2$ , and $r+1=6s+4$ . On the other hand, for $i = \left \lfloor \frac {r+3}{4}\right \rfloor -1 = \left \lfloor \frac {r-1}{4}\right \rfloor $ , we require

$$\begin{align*}\ell_i = r-1-4 \Big\lfloor\frac{r-1}{4}\Big\rfloor\geq 0, \end{align*}$$

which is obviously true. It remains to show that all subsequent entries in the same column vanish. This is certainly the case if, for $j> 0$ ,

$$\begin{align*}\frac{r+1}{2}-(\ell_i+j)-\left\lfloor\frac{\ell_i+j+1}{2}\right\rfloor < 2i+1-(\ell_i+j). \end{align*}$$

Using $\ell _i = r-1-4i$ , we obtain the condition

$$\begin{align*}\frac{r+1}{2}-\left\lfloor\frac{r-1-4i+j+1}{2}\right\rfloor < 2i+1, \end{align*}$$

which simplifies to

$$\begin{align*}-\left\lfloor\frac{j-1}{2}\right\rfloor < 1. \end{align*}$$

This concludes the proof.

Proof. If $\operatorname {\overline {Z}}$ is translatively exponential, then

$$\begin{align*}\sum_{r\geq 0} \operatorname{Z}^r(P+v) = \operatorname{\overline{Z}}(P+v) = \mathrm{e}^v \operatorname{\overline{Z}}(P) =\sum_{k\geq 0}\frac{v^k}{k!}\,\sum_{r\geq 0} \operatorname{Z}^r(P) = \sum_{r\geq 0}\sum_{j=0}^r \operatorname{Z}^{r-j}(P)\frac{v^j}{j!} \end{align*}$$

for every $P\in \mathcal {P}(\mathbb {Z}^2)$ and $v\in \mathbb {Z}^2$ . Comparing the r-summands on both sides shows that $\operatorname {Z}^r$ is translatively polynomial with associated valuations $\operatorname {Z}^{r-j}$ for $0\leq j\leq r$ . The reverse implication can be seen in the same way.

Proof. If $\operatorname {\overline {Z}}$ is d-dilative, then

$$ \begin{align*} \sum_{r\geq 0}\operatorname{Z}^r(mP)(x,y)&= \overline{\operatorname{Z}}(mP)(x,y) \\ &= m^{-d}\sum_{r\geq 0}\operatorname{Z}^r(P)(mx,my) = \sum_{r\geq 0}m^{r-d}\operatorname{Z}^r(P)(x,y) \end{align*} $$

for $m\in \mathbb {Z}_{>0}$ . Comparing the r-summands on both sides shows that $\operatorname {Z}^r$ is $(r-d)$ -homogeneous. The reverse implication can be seen in the same way. The last two statements immediately follow from Theorem 3.2.

Proof. We start by showing that the map is well-defined. For any $P\in \mathcal {P}(\mathbb {Z}^2)$ and $r\geq 0$ there are only finitely many $d\geq -2$ such that the r-summand of $\operatorname {\overline {Z}}_d$ is nonzero (cf. Lemma 6.2). Hence, $\sum _{d\geq -2}\operatorname {\overline {Z}}_d(P)$ is a formal power series. The valuation property, the $\operatorname {GL}_2(\mathbb {Z})$ equivariance and the fact that $\sum _{d\geq -2}\operatorname {\overline {Z}}_d$ is translatively exponential follow directly from $\operatorname {\overline {Z}}_d\in \overline {\mathrm {Val}}_d$ . So $\sum _{d\geq -2}\overline {\mathrm {Val}}_d$ is a well-defined element of $\overline {\mathrm {Val}}$ .

The linearity of the map is clear. To see that the map is injective, suppose that

$$\begin{align*}\sum\nolimits_{d\geq -2}\overline{\operatorname{Z}}_d = 0.\end{align*}$$

We write $\operatorname {\overline {Z}}_d = \sum _{r\geq 0} \operatorname {Z}^r_{r-d}$ . Note that the difference of upper and lower indices is d, so there is no conflict of notation for different d. Extracting the r-summand of the sum, we see that $\sum _{d\geq -2} \operatorname {Z}^r_{r-d} = 0$ . It follows from Lemma 6.2 that $\operatorname {Z}_{r-d}^r$ is $(r-d)$ -homogeneous. This implies $\operatorname {Z}^r_{r-d}=0$ for all $d\geq -2$ . Since r was arbitrary, we obtain $\overline {\operatorname {Z}}_d=0$ for all $d\geq -2$ .

For the surjectivity, let $\overline {\operatorname {Z}} \in \overline {\mathrm {Val}}$ be written as in equation (6.1). By Theorem 3.2, each $\operatorname {Z}^r$ has a homogeneous decomposition $\operatorname {Z}^r = \sum _{i=0}^{r+2}\operatorname {Z}^r_i$ with $\operatorname {Z}_i^r\in \mathrm {Val}_i^r$ . Since $\operatorname {\overline {Z}}$ is translatively exponential, Lemma 6.1 implies that $\operatorname {Z}^r$ is translatively polynomial with associated valuation $\operatorname {Z}^{r-j}$ for $0\leq j\leq r$ . Thus, it follows from Lemma 3.3 that $\operatorname {Z}^r_i$ is translatively polynomial with associated valuations $\operatorname {Z}^{r-j}_{i-j}$ for $0\leq j\leq r$ . Since the difference between rank and homogeneity degree is the same for every j, the reverse implication of Lemma 6.1 shows that the valuation

is translatively exponential for any $d\geq -2$ . By Lemma 6.2, it is d-dilative. Since the r-summands of $\overline {\operatorname {Z}}_d$ are zero for $r<d$ by Lemma 6.2, in the sum $\sum _{d\geq -2} \overline {\operatorname {Z}}_d(P)$ , we see only finitely many polynomial terms of each degree of homogeneity. Therefore, $\sum _{d\geq -2}\overline {\operatorname {Z}}_d(P)$ defines a formal power series which is equal to $\overline {\operatorname {Z}}(P)$ .

Proof. The fact that $\Lambda ^r=0$ for $r\leq d$ follows from Theorem 3.2 and Lemma 3.4. To see that $\Lambda ^r$ is injective for $r>d$ , consider $\overline {\operatorname {Z}}=\sum _{s\geq 0} \operatorname {Z}^s\in \overline {\mathrm {Val}}_d$ with $0=\Lambda ^r\operatorname {\overline {Z}} = \operatorname {Z}^r$ . It follows from Lemma 6.1 that $\operatorname {Z}^s = 0$ for all $s\leq r$ , since these are the associated valuations of $\operatorname {Z}^r$ . Again by Lemma 6.1, the associated valuations of $\operatorname {Z}^{r+1}\in \mathrm {Val}_{r-d+1}^{r+1}$ (other than $\operatorname {Z}^{r+1}$ itself) are all zero. Since by assumption $r-d+1\geq 2$ , Lemma 5.1 applies and we obtain $\operatorname {Z}^{r+1}=0$ . Using induction on j, we see that $\operatorname {Z}^{r+j}=0$ for all $j>0$ . Thus, $\overline {\operatorname {Z}}=0$ , as desired.

Proof. Since any unimodular triangulation of K is a refinement of ${\mathcal {H}}(K)$ , it suffices to prove the lemma for K being itself a unimodular cone, that is, ${\mathcal {H}}(K)=\{K\}$ and $K=\operatorname {\mathrm {pos}}\{u,v\}$ , where u and v generate $\mathbb {Z}^2$ . We use induction on . For $m=1$ , there is nothing to prove. So, let $m>1$ and suppose that the statement of the lemma is true for any unimodular cone K and any decomposition $\mathcal {T}$ of size less than m.

First, we show that the ray $\operatorname {\mathrm {pos}}\{u+v\}$ is contained in the boundary of two cones of $\mathcal {T}$ . To see this, we apply a unimodular transformation so that $K=\mathbb {R}^2_{\geq 0}$ and, thus, $u+v = (1,1)$ . If $\operatorname {\mathrm {pos}}\{(1,1)\}$ is not a boundary ray of a cone in $\mathcal {T}$ , there exists a unimodular cone $C\in \mathcal {C}(\mathbb {Q}^2)$ with primitive generators $(a,b), (c,d)\in \mathbb {Z}^2$ such that $a>b\geq 0$ and $d>c\geq 0$ and such that $(1,1)$ is contained in the interior C. Since C is unimodular, there exist $k,\ell \in \mathbb {Z}_{> 0}$ with $(1,1) = k(a,b) + \ell (c,d)$ . Using that the matrix $\big (\begin {smallmatrix} a & b\\ c & d \end {smallmatrix}\big )$ is in $\operatorname {\mathrm {SL}}_2(\mathbb {Z})$ and solving for k and $\ell $ gives $k=d-c\geq 1$ and $\ell = a-b\geq 1$ . This implies $(a,b)=(1,0)$ and $(c,d)=(0,1)$ . Hence, $C=K$ , which contradicts $m>1$ . Thus, the ray $\operatorname {\mathrm {pos}}\{u+v\}$ must be contained in the boundary of a cone of $\mathcal {T}$ .

Consequently, we obtain that $\mathcal {T}$ is the disjoint union of the two unimodular decompositions ${\{T\in \mathcal {T}\colon T\subseteq \operatorname {\mathrm {pos}}\{u,u+v\}\}}$ and $ \{T\in \mathcal {T}\colon T\subseteq \operatorname {\mathrm {pos}}\{u+v,v\}\}$ of the unimodular cones $\operatorname {\mathrm {pos}}\{u,u+v\}$ and $\operatorname {\mathrm {pos}}\{u+v,v\}$ , respectively. Since each of the two decompositions contains fewer cones than $\mathcal {T}$ , the induction hypothesis applies, and we see that $\mathcal {T}$ is obtained by taking the balanced decomposition of K first and then refining the two halves inductively.

Proof. We only have to show that $\zeta _0$ is a valuation. Let $K\in \mathcal {C}(\mathbb {Q}^2)$ and $L\subset \mathbb {R}^2$ be a rational one-dimensional subspace (L is generated by a rational vector) that intersects K. Let $K_+$ and $K_-$ denote the intersections of K with the two closed half-spaces determined by L. By [Reference Barvinok2, Lemma 19.5], it is sufficient to check that

(7.4) $$ \begin{align} \zeta_0(K) = \zeta_0(K_+) + \zeta_0(K_-) \end{align} $$

for all $K\in \mathcal {C}(\mathbb {Q}^2)$ and rational one-dimensional subspaces L.

If $\dim K<2$ , this is trivial. So, let K be two-dimensional. First, we assume that K is pointed. Let and for $\sigma \in \{-,+\}$ . Clearly, is a unimodular decomposition of K. As such, it is obtained from ${\mathcal {H}}$ by a finite sequence of elementary refinements using Lemma 7.3. Consider a unimodular cone $C\in \mathcal {T}$ . By $\operatorname {\mathrm {GL}}_2(\mathbb {Z})$ equivariance, we may assume that $C=\mathbb {R}^2_{\geq 0}$ . If we show that

$$\begin{align*}\zeta_0(\mathbb{R}^2_{\geq 0})=\zeta_0(\operatorname{\mathrm{pos}}\{e_1,e_1+e_2\}) + \zeta_0(\operatorname{\mathrm{pos}}\{e_2+e_1,e_2\}),\end{align*}$$

then we can use the $\operatorname {\mathrm {GL}}_2(\mathbb {Z})$ equivariance and apply Lemma 7.3 inductively to obtain equation (7.4). Hence, by definition of $\zeta _0$ , we need to verify that

(7.5) $$ \begin{align} R(x,y) = R(x,x+y) + R(x+y,y) \end{align} $$

for $x,y\in \mathbb {R}$ . Writing , , and , we have

$$\begin{align*}\begin{aligned} R(x,x+y) = \frac{\operatorname{Z}_2^{d+2}(P_0)(x,x+y)}{x(x+y)} = \frac{\operatorname{Z}_2^{d+2}(P_1)}{x(x+y)}. \end{aligned} \end{align*}$$

Similarly, we obtain

$$\begin{align*}R(x+y,y) = \frac{\operatorname{Z}_2^{d+2}(P_2)}{(x+y)y}.\end{align*}$$

Thus,

$$\begin{align*}R(x,x+y) + R(x+y,y) = \frac{x \operatorname{Z}_2^{d+2}(P_2) + y\operatorname{Z}_2^{d+2}(P_1)}{xy(x+y)}. \end{align*}$$

Let . Using the decompositions

$$\begin{align*}P_0=T\sqcup (-T+e_1+e_2), \,P_1 = T\sqcup (-T+2e_1+e_2), \, P_2=(-T+e_1+e_2)\sqcup(e_2+T)\end{align*}$$

together with the facts that $\operatorname {Z}_2^{d+2}$ is simple by Lemma 4.1 and that $\operatorname {Z}_1^{d+1}$ is odd and translation invariant by Lemma 3.5, we compute

$$ \begin{align*} x(\operatorname{Z}_2^{d+2}(P_2) - \operatorname{Z}_2^{d+2}(P_0)) &= x(\operatorname{Z}_2^{d+2}(T+e_2) - \operatorname{Z}_2^{d+2}(T)) \\ &= xy\operatorname{Z}_1^{d+1}(T)\\ &=y(-x\operatorname{Z}_1^{d+1}(-T+e_1+e_2)) \\ &= y(\operatorname{Z}_2^{d+2}(-T+e_1+e_2) - \operatorname{Z}_2^{d+2}(-T+2e_1+e_2))\\ &=y(\operatorname{Z}_2^{d+2}(P_0) - \operatorname{Z}_2^{d+2}(P_1)). \end{align*} $$

Hence,

$$\begin{align*}x\operatorname{Z}_2^{d+2}(P_2) + y\operatorname{Z}_2^{d+2}(P_1) = (x+y)\operatorname{Z}_2^{d+2}(P_0)\end{align*}$$

and equation (7.5) follows. So we obtain equation (7.4) if K is a pointed cone.

Next, let K be a halfspace. We may assume that L intersects the interior of K. Then, $K_+$ is unimodular if and only if $K_-$ is unimodular. First, assume that the cones $K_+$ and $K_-$ are unimodular. By $\operatorname {\mathrm {GL}}_2(\mathbb {Z})$ equivariance, it suffices to consider the case where K is the upper half-plane and L is the y-axis. In this case, equation (7.4) is equivalent to the functional equation

(7.6) $$ \begin{align} 0 = R(x,y) + R(-x,y) \end{align} $$

for $x,y\in \mathbb {R}$ . Since $\operatorname {Z}_2^{d+2}([0,1]^2)$ is $\mathbf {D}$ invariant, equation (7.6) is fulfilled.

Next, we assume that $K_+$ is a nonunimodular cone. We can choose a one-dimensional rational subspace $L^\prime $ that meets the interior of $K_+$ such that the cone $K_+^\prime $ bounded by the boundary of K and $L^\prime $ is unimodular. Let $K_{-}^\prime $ be the closure of $K\setminus K_+^\prime $ , which is a unimodular cone. Thus,

$$\begin{align*}\zeta_0(K) = \zeta_0(K_+^\prime) + \zeta_0(K_-^\prime). \end{align*}$$

Moreover, $K_{-}^\prime $ is decomposed by the subspace L into the cones $K_-$ and $K_+^{\prime \prime }$ , where $K_+^{\prime \prime }$ is the closure of $K_+\setminus K_+^\prime $ . Since we already proved that equation (7.4) holds for the pointed cone $K_-^\prime $ we deduce

$$\begin{align*}\zeta_0(K) = \zeta_0(K_+^\prime) + \zeta_0(K_+^{\prime\prime}) + \zeta_0(K_-) = \zeta_0(K_+) + \zeta_0(K_-), \end{align*}$$

which verifies equation (7.4) for half-spaces. Since the statement is trivial for $K=\mathbb {R}^2$ , the proof is complete.

Proof. We set $f_2: = \operatorname {Z}_2^{d+2}(\mathbf T)$ and . Since $\operatorname {Z}_{i}^{d+i}$ is simple for $i\in \{1,2\}$ by Lemma 4.1, considering the triangulation $[0,1]^2 = \mathbf T \sqcup (-\mathbf T+e_1+e_2)$ , we obtain

(7.8) $$ \begin{align} R(x,y) = \frac{2f_2(x,y) - (x+y)f_1(x,y)}{xy}. \end{align} $$

For the standard triangle, we have

$$\begin{align*}\operatorname{\mathrm{fcone}}(0;\mathbf T) = \mathbb{R}^2_{\geq 0},~\operatorname{\mathrm{fcone}}(e_1;\mathbf T) = \phi_1\mathbb{R}^2_{\geq 0},~\text{and}~\operatorname{\mathrm{fcone}}(e_2;\mathbf T) = \phi_2\mathbb{R}^2_{\geq 0}, \end{align*}$$

where $\phi _1 = \big (\begin {smallmatrix} -1 & -1\\ \phantom {-}1 & \phantom {-}0 \end {smallmatrix}\big )$ and $\phi _2 = \big (\begin {smallmatrix} \phantom {-}0 & \phantom {-}1\\ -1 & -1 \end {smallmatrix}\big )$ . Note that $\phi _i\mathbf T = \mathbf T-e_i$ for $i\in \{1,2\}$ . So, in particular, we have $\phi _1,\phi _2\in \mathbf {G}$ .

Using the representation (7.8) of R and the $\mathbf {G}$ invariance of $f_1$ , we compute

$$\begin{align*}\begin{aligned} R\circ \phi_2^\star &= \frac{1}{(x-y)(-y)}\left( 2f_2\circ\phi_2^\star - (x-y + (-y))f_1\circ\phi_2^\star \right)\\ &=\frac{1}{y(y-x)}\left( 2\operatorname{Z}_2^{d+2}(\mathbf T-e_2) - (x-2y)f_1 \right)\\ &=\frac{1}{y(y-x)}\left( 2(f_2 - yf_1) -(x-2y)f_1\right)\\ &=\frac{1}{y(y-x)}\left( 2f_2 -xf_1\right). \end{aligned} \end{align*}$$

Following the same lines, we compute

$$\begin{align*}R\circ \phi_1^\star = \frac{1}{x(x-y)}\left( 2f_2 -yf_1\right). \end{align*}$$

So we obtain

$$\begin{align*}\zeta(\mathbf T) &= \mathrm{e}^0\zeta_0(\mathbb{R}^2_{\geq 0}) + \mathrm{e}^x\zeta_0(\phi_1 \mathbb{R}^2_{\geq 0}) + \mathrm{e}^y\zeta_0(\phi_2\mathbb{R}^2_{\geq0}) \\&= R+\mathrm{e}^x\, R\circ\phi_1^\star + \mathrm{e}^y\,R\circ\phi_2^\star\\&=\frac{2f_2-(x+y)f_1}{xy} + \mathrm{e}^x\,\frac{2f_2 -yf_1}{x(x-y)} + \mathrm{e}^y\,\frac{2f_2 -xf_1}{y(y-x)} \\&=2f_2\cdot\left(\frac{\mathrm{e}^y-1}{y(y-x)} - \frac{\mathrm{e}^x-1}{x(y-x)}\right) - f_1\cdot\left( \frac{x(\mathrm{e}^y-1)}{y(y-x)} - \frac{y(\mathrm{e}^x-1)}{x(y-x)} \right). \end{align*}$$

This proves the first identity.

For the second identity, note that $\tfrac {e^x-1}{x} = \sum _{m\geq 0} \tfrac {x^m}{(m+1)!}$ is an analytic function and that $y^m-x^m = (y-x)[m]_{x,y}$ . For the first bracket term, we thus obtain

$$\begin{align*}\frac{\mathrm{e}^y-1}{y(y-x)} - \frac{\mathrm{e}^x-1}{x(y-x)} = \sum_{m\geq 0} \frac{y^m-x^m}{(m+1)!(y-x)}= \sum_{m>0} \frac{[m]_{x,y}}{(m+1)!}. \end{align*}$$

In particular, the left-hand side is an analytic function on all $\mathbb {R}^2$ . Similarly, we have for the second bracket term:

$$\begin{align*}\begin{aligned} \frac{x(\mathrm{e}^y-1)}{y(y-x)} - \frac{y(\mathrm{e}^x-1)}{x(y-x)} &= \sum_{m\geq 0} \frac{xy^m-yx^m}{(m+1)!(y-x)}\\ &= -1+\sum_{m> 0} \frac{xy(y^{m-1}-x^{m-1})}{(m+1)!(y-x)} = -1 + \sum_{m>0}\frac{xy[m-1]_{x,y}}{(m+1)!}\\ &= \sum_{m\geq 0}\frac{[m+1]_{x,y}-x^m-y^m}{(m+1)!} = \sum_{m>0} \frac{[m]_{x,y}-x^{m-1}-y^{m-1}}{m!}. \end{aligned}\end{align*}$$

Again, the left-hand side is an analytic function on all $\mathbb {R}^2$ .

Proof. By Proposition 5.5, we have $\dim \mathrm {Val}_2^{d+2} = p_{2,3}\left (\tfrac d2 + 1\right )$ . By Lemma 7.6, we have $\dim \overline {\mathrm {Val}}_d = p_{2,3}\left (\tfrac d2 + 1\right )$ . Now let $i\geq 2$ and let $A_i^{d+i} \colon \mathrm {Val}_i^{d+i} \to \mathrm {Val}_{i-1}^{d+i-1}$ be the linear operator that maps a valuation $\operatorname {Z}_i^{d+i}$ to its associated valuation of rank $d+i-1$ . Note that this associated valuation is $(i-1)$ -homogeneous by Lemma 3.3. Since $i\geq 2$ , applying Lemma 5.1 inductively shows that each $A_i^{d+i}$ with $i\geq 2$ is injective. So, the linear map $A_i^{d+i} \circ \cdots \circ A_3^{d+3}$ from $\mathrm {Val}_i^{d+i}$ to $\mathrm {Val}_2^{d+2}$ is injective. At the same time, the map $\Lambda ^{d+i}\colon \overline {\mathrm {Val}}_d\to \mathrm {Val}_i^{d+i}$ is injective by Lemma 7.1. This shows the claim for all $i\geq 2$ (cf. Figure 1).

Proof. We start by showing that $\operatorname {\mathrm {L}}_i^r$ for $r-i$ odd is not identically zero on $\mathcal {P}(\mathbb {Z}^1)$ and thus not identically zero on $\mathcal {P}(\mathbb {Z}^2)$ . Using Faulhaber’s formula (see [Reference Ludwig and Silverstein19] for more information), we obtain for $m\ge 1$ that

$$\begin{align*}\operatorname{\mathrm{L}}^r(m\,[0,e_1])(x,y) = \frac{1}{r!}\sum_{k=0}^m k^rx^r = \frac{x^r}{r!(r+1)}\sum_{\ell=0}^r (-1)^\ell \binom{r+1}{\ell}B_\ell\, m^{r+1-\ell}, \end{align*}$$

where $B_\ell $ denotes the $\ell $ th Bernoulli number. Since $B_\ell \neq 0$ for $\ell $ even, this shows that $\operatorname {\mathrm {L}}_i^r([0,e_1])\neq 0$ for $r-i$ odd. Since also $B_1\neq 0$ , we obtain the claim that $\operatorname {\mathrm {L}}_i^r([0,e_1])\ne 0$ holds for all i and r as in the proposition with $i\neq r+2$ . Moreover, $\operatorname {\mathrm {L}}_{r+2}^r$ is not identically zero by [Reference Ludwig and Silverstein19, Lemma 26].

Next, for $r=0$ , it is easy to see that $\operatorname {\mathrm {L}}^0_i$ is not identically zero on $\mathcal {P}(\mathbb {Z}^2)$ for $i\in \{0,1,2\}$ . Thus, for $r\leq i\leq r+2$ , using Lemma 3.3, we see that $\operatorname {\mathrm {L}}_i^r$ is not translation invariant since $\operatorname {\mathrm {L}}_{i-r}^0$ is among its associated functions. So, in particular, $\operatorname {\mathrm {L}}_i^r\ne 0$ . So, we have established $\operatorname {\mathrm {L}}_i^r\ne 0$ for the indices occurring in the proposition.

We note that $\mathrm {Val}_i^0 = \operatorname {\mathrm {span}}\{\operatorname {\mathrm {L}}_i^0\}$ by the Betke–Kneser theorem. For $r\geq 2$ even, consider the one-homogeneous valuation $\operatorname {Z}_1^r$ , which is translation invariant by Lemma 3.5. Define the valuation $\operatorname {Z}^\prime \colon \mathcal {P}(\mathbb {Z}^1)\to \mathbb {R}[x,y]_r$ by . By [Reference Ludwig and Silverstein19, Lemma 5], we have $\operatorname {Z}^\prime (S)\in \mathbb {R}[x]_r$ . It is elementary to check that there is only one even, one-homogeneous, translatively polynomial valuation from $\mathcal {P}(\mathbb {Z}^1)$ to $\mathbb {R}[x]_r$ . Thus, $\operatorname {Z}^\prime = c\operatorname {\mathrm {L}}_1^r$ for some $c\in \mathbb {R}$ . It follows that the valuation is simple. Lemma 24 from [Reference Ludwig and Silverstein19] yields that $\operatorname {Y}([0,1]^2)=0$ . Since we have $[0,1]^2 = \mathbf T \sqcup (-\mathbf T+e_1+e_2)$ , it follows from the assumption that r is even, and the fact that $\operatorname {Y}$ is simple, translation invariant, and $\operatorname {\mathrm {GL}}_2(\mathbb {Z})$ equivariant that $\operatorname {Y}(\mathbf T)=0$ . But this also implies $\operatorname {Y}=0$ . So, we have established the statement for one-homogeneous valuations of even rank.

Next, suppose that $i=r=1$ . Then, $\operatorname {\mathrm {L}}_1^1$ is called the discrete Steiner point and it follows from Theorem 5 of [Reference Böröczky and Ludwig8] that $\mathrm {Val}_1^1 = \operatorname {\mathrm {span}}\{\operatorname {\mathrm {L}}_1^1\}$ . Hence, we proved the statement for $i=1$ .

Finally, let $r>0$ and $i>1$ satisfy the proposition’s assumptions. Let $\operatorname {Z}_i^r\in \mathrm {Val}_i^r$ . Then, $r-1$ and $i-1$ also satisfy the assumptions. By induction on i, we have $\operatorname {Z}_{i-1}^{r-1}=c\operatorname {\mathrm {L}}_{i-1}^{r-1}$ . Since the associated functions are hereditary by Lemma 3.1, this means that

$$ \begin{align*}\operatorname{Z}_{i-j}^{r-j} = c\operatorname{\mathrm{L}}_{i-j}^{r-j}\end{align*} $$

for all $0<j\leq \min \{r,i\}$ . Consequently, $\operatorname {Z}_i^r-c\operatorname {\mathrm {L}}_i^r$ is translation invariant. Using the homogeneous decomposition theorem for translation invariant valuations from [Reference McMullen20], we obtain that the polynomial $m\mapsto (\operatorname {Z}_i^r -c_i\operatorname {\mathrm {L}}_i^r)(mP)$ has degree 2. This yields $\operatorname {Z}_i^r-c\operatorname {\mathrm {L}}_i^r=0$ as desired, except for $i=2$ . But then the claim follows from [Reference Ludwig and Silverstein19, Proposition 23].

Proof. The fact that ${\overline {L}}_d([0,e_1])\neq 0$ for $d=0$ or d odd follows from $\operatorname {\mathrm {L}}_1^{d+1}([0,e_1])\neq 0$ . To see that ${\overline {L}}_{-2}$ is not identically zero, it suffices to note that its constant part is $\operatorname {\mathrm {L}}_2^0$ , the two-dimensional volume.

Let $\operatorname {\overline {Z}}\in \overline {\mathrm {Val}}_d$ and consider for $r>d$ its r-summand $\operatorname {Z}^r$ . By Lemma 6.2, we have $\operatorname {Z}^r \in \mathrm {Val}_{r-d}^r$ . By the assumption on d, Proposition 8.1 applies, so there exists a constant $c_r\in \mathbb {R}$ with $\operatorname {Z}^r = c_r\operatorname {\mathrm {L}}_{r-d}^r$ . By Lemma 6.1, the valuations $\operatorname {Z}^r$ for $r\geq 0$ are associated functions of one another. Since also $\operatorname {\mathrm {L}}_{r-d}^r$ for $r\geq 0$ are associated functions of one another, we see that $c_r = c$ for all $r\geq 0$ and some $c\in \mathbb {R}$ that does not depend on r. Thus, $\operatorname {\overline {Z}} = c {\overline {L}}_d$ , as claimed.

Proof of Theorem 1.2.

The first case follows from Proposition 8.1 for $i\geq 1$ , and from Theorem 1.1 for $i=r=0$ . The second case is Corollary 4.5. The third case is Corollary 7.7. The fourth case consists of the two subcases, $i=0<r$ and $i>r+2$ . In the former case, the statement follows from Lemma 3.4, and in the latter case, it follows from Theorem 3.2.

Proof of Theorem 1.3.

The polynomials with $2k+3\ell =r$ form a basis of $\mathbb {R}[x,y]_r^{\mathbf {G}}$ by Lemma 4.4. By Theorem 4.3, the evaluation map $\Theta ^r$ is an isomorphism from $\mathrm {Val}_1^r$ to $\mathbb {R}[x,y]^{\mathbf {G}}_r$ , if $r>1$ is odd. Hence,

$$\begin{align*}\begin{aligned} \mathrm{Val}_1^r &= \operatorname{\mathrm{span}}\{ (\Theta^r)^{-1}(\mathbf{f}_{k,\ell}) \colon 2k+3\ell=r\}\\ &= \operatorname{\mathrm{span}}\{ \operatorname{Z}_{\mathbf{f}_{k,\ell}} \colon 2k+3\ell=r\}, \end{aligned} \end{align*}$$

where $\operatorname {Z}_{\mathbf {f}_{k,\ell }}$ is given by equation (4.1). Note that $ \operatorname {\mathrm {L}}^r_1(\mathbf T) \in \mathbb {R}[x,y]^{\mathbf {G}}_r $ also if r is even. From Lemma 4.4, we obtain $\mathbb {R}[x,y]_r^{\mathbf {G}} = \operatorname {\mathrm {span}}\{\mathbf {p}_r\}$ for $r\in \{2,3\}$ . Hence, $\mathbf {f}_{k,\ell }$ is a multiple of $\operatorname {\mathrm {L}}_1^2(\mathbf T)^k\operatorname {\mathrm {L}}_1^3(\mathbf T)^\ell $ . By [Reference Ludwig and Silverstein19, Lemma 28], this multiple is nonzero, which concludes the proof.

Proof of Theorem 1.4.

The statement on the dilative decomposition is proven in Proposition 6.3. For the determination of dimensions, the first case is proved in Corollary 8.2, the second case in Corollary 7.7, and the third case in Lemma 6.2.