Lemma
Let \(X\) be a toric variety with torus \(T\). Suppose \([C,f,\vec x]\in \mathcal M\) is a stable map (logarithmic or otherwise) which is fixed by the induced action on \(\mathcal M\). If \(C'\) is a component of \(C\), then either
- \(f(C')\) is contracted to a fixed point of \(X\)
- \(C'\) is isomorphic to \(\mathbb P^1\) and the restriction of \(f\) to \(C'\) is a degree \(d\) map to a copy of \(\mathbb P^1\) in \(X\)given in local coordinates by \([z_0:z_1]\mapsto [z_0^d:z_1^d]\).
Proof
Suppose first that \(C'\) is contracted by \(f\). By the behavior of a fixed stable map at special points, if \(C'\) has a node, a mark or a ramification point then we immediately know \(f(C') = p\). If \(C'\) has none of these points, then \(C' =C\) since there are no nodes, and by the curve class data defining the moduli space we must then have that \(X = \text{pt}\).
Now suppose that \(C'\) is not contracted by \(f\). For convenience, we briefly assume \(C' = C\), so that we may write \(f\) instead of \(f|_{C'}\).
Claim 1: \(f\) covers a copy of \(\mathbb P^1\) in \(X\). Since \([C,f,\vec x]\) is fixed by the \(T\)-action we get for every \(t\) an isomorphism \(\varphi_t:C\to C\) making the diagram
commute. Since \(f(C) = t\cdot \varphi_t(f(C)) = t\cdot f(C)\), the closure of \(f(C)\) in \(X\) must be a dimension \(1\) subvariety which is fixed set-wise by the torus action on \(X\). In particular, \(\ell = \overline{f(C)}\) is a one-dimensional toric variety, and hence is either \(\mathbb C^*\), \(\mathbb C\) or \(\mathbb P^1\). Since \(C\) is necessarily projective, if \(V\) is affine then \(f\) must contract \(C\), hence the only possibility is that \(\ell \cong \mathbb P^1\) . Since any non-constant morphism of projective curves is surjective, \(f\) is a cover of \(\mathbb P^1\).
Claim 2: The branched cover \(f:C\to \mathbb P^1\) is totally ramified; that is, \(f^{-1}(0)\) and \(f^{-1}(\infty)\) consist of only one point each. To prove this, we unfortunately need to appeal to GAGA and move into the smooth category in order to make use of topology.
Consider the monodromy representation \(\rho\) of \(f\). Since this is a branched cover of Riemann surfaces, the monodromy representation is transitive. Since \(\mathbb P^1 \setminus \{0,\infty\}\) is just \(\mathbb C^*\), which is homotopic to a circle, \(\pi_1(\mathbb P^1 \setminus \{0,\infty\}, x_0)\) is generated by a single element \(\gamma\). This means that \(\langle \rho(\gamma)\rangle \) is a transitive subgroup of \(S_d\). There is only one of these up to conjugation, the cyclic permutation \((1 ~ 2 ~ 3 ~ ... ~ d)\). The order of the permutation corresponds to the ramification index of the branch point (analytic locally in a neighborhood of the branch point \(b\) the map \(f\) is \(z \mapsto z^{e_b}\)). Thus, the number of sheets at a point in \(f^{-1}(0)\) is \(|(1 ~ 2 ~ 3 ~ ... ~ d)| = d\), as is the case for \(f^{-1}(\infty)\). This implies that \(f\) is totally ramified.
Alternatively, since we’re in the topological category, we can appeal to homotopy. Once we remove the branch locus, the base is homotopic to \(S^1\), so we have a degree \(d\) cover of \(S^1\). This immediately implies \(X\setminus f^{-1}(0,\infty) \simeq S^1\) and that \(f\) is given in coordinates by the degree \(d\) power map. This skips claim 3, but feels worse since we’re applying homotopy to topological spaces which feels further from the algebraic category.
Claim 3: \(C\) has genus \(0\). Applying Riemann-Hurwitz gives us \[2g(C) - 2 = d(g(\ell) - 2) + \sum_{p\in X}(v_p - 1) \implies 2g(C) = 2 - 2d + \sum_{p\in X}(v_p - 1).\] Applying Claim 2 gives us \[2g(C) = 2 - 2d + (2d - 2) = 0.\] So \(g(C) = 0\).
Claim 4: There exist coordinates on \(C\cong \mathbb P^1\)so that \(f:[x:y] \mapsto [x^d : y^d]\). Simply apply an automorphism of \(\mathbb P^1\) so that \(f^{-1}([1:0]) = [1:0]\) and \(f^{-1}([0:1]) = [0:1]\). Done.