Smooth toric variety which is a cube is a bott tower (reference request)

Question

According to Lee, Masuda and Park (page 3), the following result is "well-known in toric topology". I've found a proof, but I would like a published reference.

Let $X$ be a toric variety. Then the following are equivalent:

(1) $X$ is smooth and the poset of torus orbits (under closure) is the face lattice of the $n$-dimensional cube.

(2) There is a sequence of toric varieties and toric maps $X = X_n \to X_{n-1} \to X_{n-2} \to \cdots \to X_1 \to X_0 = \{ \text{point} \}$ such that each $X_i \to X_{i-1}$ is a $\mathbb{P}^1$ bundle. This is often called a "Bott tower".

The implication $(2) \implies (1)$ is clear. I can do the converse as well -- it's a clever bit of linear algebra -- but I'd rather just cite it. Does anyone know where this appears?

---

For the record, here is the combinatorially equivalent statement: Assume (1). Let $\Sigma$ be the corresponding fan, so $\Sigma$ is combinatorially the dual fan to a cube. In particular, it has $2n$ rays; call these rays $v_1^0$, $v_1^1$, $v_2^0$, $v_2^1$, ..., $v_n^0$, $v_n^1$ with $v_j^0$ and $v_j^1$ dual to antipodal faces. So the maximal cones of this fan are of the form $\text{Span}(v_1^{r_1}, v_2^{r_2}, ..., v_n^{r_n})$.

The condition that these cones form a fan says that $\det(v_1^{r_1}, v_2^{r_2}, \ldots, v_n^{r_n})$ has sign $(-1)^{\sum r_j}$, and the condition that the toric variety is smooth says that this determinant is $\pm 1$, so we can combine these as

(1') $\det(v_1^{r_1}, v_2^{r_2}, \ldots, v_n^{r_n}) = (-1)^{\sum r_j}$.

Meanwhile, (2) is equivalent to

(2') We can reorder the lower subscripts such that $\dim \text{Span}(v_1^0, v_1^1, \ldots, v_j^0, v_j^1) = j$, and $v_{j+1}^0+v_{j+1}^1 \in \text{Span}(v_1^0, v_1^1, \ldots, v_j^0, v_j^1)$.

The implication (1') to (2') is what I mean by "a clever bit of linear algebra"; I'll write it up if no one gives me a reference.

Answer 1

Thank you for the question. As is mentioned in the same paper (on page 6), you can find the statement in Corollary 3.5 in the following paper:

Semifree circle actions, Bott towers and quasitoric manifolds by M. Masuda, T. E. Panov, https://doi.org/10.1070/SM2008v199n08ABEH003959 (https://arxiv.org/pdf/math/0607094)

Answer 2

$\def\ZZ{\mathbb{Z}}$No answers after a week so, for the record, I'll write up my proof. Let $v_1^0$, $v_1^1$, $v_2^0$, $v_2^1$, ..., $v_n^0$, $v_n^1$ be as in the original post.

Rename $v_i^0$ to $e_i$ and $v_j^1$ to $f_j$. By hypothesis, $(e_1, e_2, \ldots, e_n)$ is a basis of $\ZZ^n$ and without loss of generality, we can assume it is the standard one. Let $f_j = \sum_i a_{ij} e_i$.

Lemma 1 We have $a_{ii} = -1$.

Proof Use the assumption that $\det(e_1, e_2, \ldots, e_{i-1}, f_i, e_{i+1}, \ldots, e_n) = -1$. $\square$

Now, define a directed graph $\Gamma$. The vertex set is $\{ 1,2,\ldots, n \}$, and there is an edge $i \to j$ if $i \neq j$ and $a_{ij}$ is nonzero. There are no edges of the form $i \to i$.

Lemma 2 The directed graph $\Gamma$ is acyclic.

Proof Suppose, to the contrary that $\Gamma$ has a directed cycle. Let $j_1 \to j_2 \to \cdots \to j_k \to j_1$ be the shortest directed cycle. (Note that we could have $k=2$, but we can't have $k=1$, since $\Gamma$ has no loops.) Then there are no other edges between any $j_a$ and $j_b$ other than those in the cycle. Let $J = \{ j_1, j_2, \ldots, j_k \}$.

Now, consider $\det{\big(} \{ e_i : i \not\in J \} \cup \{ f_j : j \in J \} {\big)}$. On the diagonal, we get $(-1)^k$ (using Lemma 1). The only other permutation which can contribute to the determinant is the cycle $(j_1 j_2 \cdots j_k)$, and it contributes $\prod_{r=1}^k a_{j_r j_{r+1}}$. (The inner subscript is periodic modulo $k$.) So the determinant is $(-1)^k + (-1)^{k-1} \prod_{r=1}^k a_{j_r j_{r+1}}$.

Since $j_r \to j_{r+1}$ is an edge of $\Gamma$, we have $a_{j_r j_{r+1}} \neq 0$, so $\prod_{r=1}^k a_{j_r j_{r+1}} \neq 0$ and the determinant is not $(-1)^k$. But our hypothesis exactly is that the determinant should be $(-1)^k$, a contradiction. $\square$

So, we can reorder our subscripts so that $\Gamma$ only has edges $i \to j$ for $i<j$. Then $$v_j^0+v_j^1 = e_j + \left( - e_j + \sum_{i<j} a_{ij} e_i \right) = \sum_{i<j} a_{ij} v_i^0$$ as desired. QED