Smooth Euler-symmetric varieties generated by a single polynomial

Let $\operatorname{Sym}^{d}W$ be the vector space of degree $d$ homogeneous differential operator with constant coefficients. Consider the follow $j$-th partial derivative map of $P$:

$$P_{j,r-j}:\operatorname{Sym}^{j}W\to\operatorname{Sym}^{r-j}W^{*};\;D\mapsto D(P).$$

By the definition of the symbol system $\textbf{F}_{P}$, we have that $\operatorname{Im}(P_{j,r-j})=F_{P}^{r-j}$ for $j\neq r-1$. We claim that

$$\operatorname{rank}(P_{j,r-j})=\operatorname{rank}(P_{r-j,j}).$$

This was mentioned in [GL19] without explicit proof. To prove the claim, we write

$$P=\sum_{\begin{subarray}{c}I=(a_{1},\cdots,a_{m}),a_{i}\geq 0\\ a_{1}+\cdots+a_{m}=r\end{subarray}}A_{I}\dfrac{1}{a_{1}!\cdots a_{m}!}x^{a_{1}}_{1}\cdots x^{a_{m}}_{m}.$$

For $D^{j}\in\operatorname{Sym}^{j}W$, it can be written as

$$D^{j}=\sum_{\begin{subarray}{c}J=(b_{1},\cdots,b_{m}),b_{i}\geq 0\\ b_{1}+\cdots+b_{m}=j\end{subarray}}D^{j}_{J}\dfrac{\partial^{j}}{\partial x^{b_{1}}_{1}\cdots\partial x^{b_{m}}_{m}}.$$

Then we have

$$D^{j}(P)=\sum_{I,J,a_{i}\geq b_{i}}A_{I}D^{j}_{J}\dfrac{1}{(a_{1}-b_{1})!\cdots(a_{m}-b_{m})!}x^{a_{1}-b_{1}}_{1}\cdots x^{a_{m}-b_{m}}_{m}.$$

Define $\Phi:\operatorname{Im}(P_{j,r-j})\rightarrow\operatorname{Im}(P_{r-j,j})$ as

$$\Phi(D^{j}(P))=\sum_{I,J,a_{i}\geq b_{i}}A_{I}D^{j}_{J}\dfrac{1}{(b_{1})!\cdots(b_{m})!}x^{b_{1}}_{1}\cdots x^{b_{m}}_{m}.$$

It is easy to see that $\Phi$ is a bijection and hence the claim holds. Therefore, $\operatorname{dim}(F^{j}_{P})=\operatorname{dim}(F^{r-j}_{P})$ for any $j\neq 1$.

If $P$ is homaloidal, then $\operatorname{dim}(F^{r-1}_{P})=m=\operatorname{dim}(W^{*})$. ∎

Consider the terminal point $z=[0:\cdots:0:1]\in M_{P}\subset\mathbb{P}V_{P}$. Firstly, for any point $w\in W$ such that $P(w)\neq 0$, we have that the projective tangent cone at $z$ contains the direction $[0:\cdots:0:dP(w):0]$. We claim that the the map $w\mapsto dP(w)$ is not degenerate, namely the image is not contained in a hyperplane. If not, by choosing a suitable coordinate, we can assume that $\partial P/\partial w_{m}\equiv 0$. Hence $M_{P}$ is degenerate. It contradicts with the non-degeneracy of Euler-symmetric projective varieties. Therefore, the projective tangent cone must contain the projective subspace $L=\{[0:\cdots:0:*:*]\}$ of dimension $m$.

Consider the tangential projection of $M_{P}$ from $z$:

	$\displaystyle\psi:M_{P}$	$\displaystyle\dashrightarrow L$
	$\displaystyle[t^{r}:t^{r-1}w:\cdots:tdP(w):P(w)]$	$\displaystyle\mapsto[tdP(w):P(w)].$

Since the Euler-symmetric variety $M_{P}$ is smooth, the projective subspace $L$ coincides with projective tangent space of $M_{P}$ at point $z$. Then $\psi$ must be a local isomorphism in the complex topology. That is, the map $\psi$ must be a local isomorphism at $z$, in particular it must be locally injective.

On the other hand, suppose that we can find two non-colinear vectors $w_{1}$ and $w_{2}$ with $P(w_{1}),P(w_{2})\neq 0$ such that the vectors $dP(w_{1})$ and $dP(w_{2})$ are colinear. After multiplying them by a suitable constant, we may suppose that $P(w_{1})^{-1}dP(w_{1})=P(w_{2})^{-1}dP(w_{2})$. Then for $t$ small enough, $[t^{r}:t^{r-1}w_{1}:\cdots:tdP(w_{1}):P(w_{1})]$ and $[t^{r}:t^{r-1}w_{2}:\cdots:tdP(w_{2}):P(w_{2})]$ are two distinct points in $M_{P}$ close to $z$. But $[tdP(w_{1}):P(w_{1})]$ and $[tdP(w_{2}):P(w_{2})]$ coincide, a contradiction with the local injectivity of $\psi$. We conclude that the rational map $[w]\in\mathbb{P}W\mapsto[dP(w)]\in\mathbb{P}W^{*}$ must be injective on the open subset $P(w)\neq 0$. Hence, the polynomial $P$ is homaloidal. ∎

Let $\mathscr{L}_{w}$ be the image of the line $L_{w}=\{[a:bw]\mid[a:b]\in\mathbb{P}^{1},P(w)\neq 0\}$ under the rational map $\Phi_{P}$. The tangent direction of $\mathscr{L}_{w}$ at $z$ is

$$[0:\cdots:\frac{dP(w)}{P(w)}:0],$$

and the coordinate of the neighborhood of $z$ is

$$[\frac{t^{r}}{P(w)}:\frac{t^{r-1}w}{P(w)}:\frac{t^{r-2}\iota^{2}_{w}}{P(w)}:\cdots:\frac{tdP(w)}{P(w)}:1].$$

Therefore, the tangent space $T_{z}M_{P}$ can be identified with $W^{*}$. Hence by [FH20, Lemma 2.5] and the smoothness of $M_{P}$, we can take the coordinate

$$(z_{m_{r}}^{(r)},\cdots,z_{1}^{(r)},\cdots,z_{m_{2}}^{(2)},\cdots,z_{1}^{(2)},z_{m},\cdots,z_{1}),$$

such that $z_{j}=\frac{t\partial_{j}P(w)}{P(w)}$, and analytic functions $h^{k}_{j}(z_{m},\cdots,z_{1})$ satisfy

$$z_{j}^{(k)}=h^{k}_{j}(z_{m},\cdots,z_{1}),2\leq k\leq r,1\leq j\leq m_{i}.$$

By the Taylor’s expansion of $h^{k}_{j}$ and comparing the degree w.r.t $t$, we know that $h^{k}_{j}$ are homogeneous polynomials of degree $k$. Hence $r_{z}=r$ and the system of fundamental forms of $M_{P}$ at $z$ is $\textbf{F}_{z}=\oplus_{k=0}^{r}F^{k}_{z}$ with $F_{z}^{k}=\langle h^{k}_{j}\mid 1\leq j\leq m_{k}\rangle\subset\operatorname{Sym}^{k}W$ for $k\geq 2$. Together with Proposition 2.3 and Proposition 2.4 we know $m_{k}=\operatorname{dim}(F_{P}^{r-k})=\operatorname{dim}(F_{P}^{k})$, for any $0\leq k\leq r$. ∎

From the proof of Proposition 2.5, we know that $\frac{1}{P(w)}=h^{r}_{1}(\frac{dP(w)}{P(w)})$, namely $P_{*}=h^{r}_{1}$. Hence $P$ is an EKP-homaloidal polynomial. ∎

By definition of $h^{r-1}_{j}$, we have that $h^{r-1}_{m+1-j}(\frac{dP(w)}{P(w)})=\frac{w_{j}}{P(w)}$, for $w\in W$. Let $\Pi:W^{*}\to W,w\mapsto(h^{r-1}_{m}(w),\cdots,h^{r-1}_{1}(w))$. Identifying $E$ with $\mathbb{P}W^{*}$, the rational map $\overline{\pi}$ and $\Pi$ are same under the projectivization. Consider the morphisms

	$\displaystyle\Phi_{1}:$	$\displaystyle\;W\to W^{*},w\mapsto dP(w),$
	$\displaystyle\Phi_{1}^{*}:$	$\displaystyle\;W^{*}\to W,w\mapsto dP_{*}(w).$

By Proposition 2.7 and the definition of EKP-homaloidal polynomials, we know that

$$\Phi_{1}\circ\Phi_{1}^{*}(w)=P_{*}^{r-2}(w)w,\Phi_{1}^{*}\circ\Phi_{1}(w)=P^{r-2}(w)w.$$

On the other hand since $h^{r-1}_{j}$ are homogeneous polynomials of degree $r-1$, we also have

$$\Pi\circ\Phi_{1}(w)=P^{r-2}(w)w,$$

therefore $\Pi\circ\Phi_{1}=\Phi_{1}^{*}\circ\Phi_{1}$. Composed on the right by the morphsim $\Phi_{1}^{*}$, we have

	$\displaystyle\Phi_{1}^{*}\circ\Phi_{1}\circ\Phi_{1}^{*}(w)$	$\displaystyle=P^{(r-2)(r-1)}_{*}(w)\Phi_{1}^{*}(w),$
	$\displaystyle\Pi\circ\Phi_{1}\circ\Phi_{1}^{*}(w)$	$\displaystyle=P^{(r-2)(r-1)}_{*}(w)\Pi(w).$

Therefore, we know that $\Pi=\Phi_{1}^{*}$. ∎

By the definition of $\mathbb{G}_{m}$-action on the projective space $\mathbb{P}V_{P}$, the set of fixed points is the disjoint union $\coprod_{i=0}^{r}\mathbb{P}V_{F_{P}^{i}}$. Hence, $(M_{P})^{\mathbb{G}_{m}}=\coprod_{i=0}^{r}M_{i}$. ∎

From Theorem 1.6 we know $M_{P}$ is an equivariant compactification of the vector group $W$, it suffices to prove that at the original point $o\in M_{P}$, there exists a line in $\mathbb{P}V_{P}$ which lies on $M_{P}$. Also from Theorem 1.6, this can be reduced to prove that $V(F^{2}_{P})\neq\emptyset$.

Note that from Proposition 2.10, $F^{r-1}_{z}=F^{r-1}_{P_{*}}=\langle dP_{*}\rangle$. Let $Z$ be the image of $V(P_{*})$ of the following rational map:

$$\Phi_{F^{r-1}_{z}}:\mathbb{P}T_{z}M_{P}(\supset V(P_{*}))\dashrightarrow\mathbb{P}V_{F_{z}^{r-1}}(\subset\mathbb{P}V_{P}),[w]\mapsto[dP_{*}(w)].$$

Since $P$ is a reduced homogeneous polynomial, $P_{*}$ is also a reduced homogeneous polynomial. Clearly $Z$ is the dual variety of $V(P_{*})$, which is non-empty. From the proof of Proposition 2.5, $Z$ is fixed by $\mathbb{G}_{m}$ and hence $Z\subset M_{1}$.

Write the coordinates of $\mathbb{P}V_{P}$ as $[z_{0}:z_{1}:\cdots:z_{m}:w^{(2)}_{1}:\cdots:w^{(2)}_{m_{2}}:\cdots:w^{(r)}_{1}]$, and let $F^{2}_{P}=\langle Q^{(2)}_{1},\cdots,Q^{(2)}_{m_{2}}\rangle$ generated by $Q^{(2)}_{1},\cdots,Q^{(2)}_{m_{2}}$. Then the following homogeneous polynomials of degree two lie in the vanishing ideal of $M_{P}\subset\mathbb{P}V_{P}$:

	$\displaystyle f_{1}^{(2)}$	$\displaystyle=z_{0}w_{1}^{(2)}-Q_{1}^{(2)}(z_{1},\cdots,z_{m});$
		$\displaystyle\vdots$
	$\displaystyle f_{m_{2}}^{(2)}$	$\displaystyle=z_{0}w_{m_{2}}^{(2)}-Q_{m_{2}}^{(2)}(z_{1},\cdots,z_{m}).$

Therefore, $Z\subset M_{1}\subset V(F^{2}_{P})$. In particular, $V(F^{2}_{P})\neq\emptyset$. ∎

By Proposition 2.10, we know that $Z\subset\mathbb{P}V_{F_{P}^{2}}$ is the dual variety of $V(P_{*})\subset\mathbb{P}T_{z}M_{P}$. Therefore, from the proof of Proposition 3.2 and Remark 3.3, we only need to show that $M_{1}=Z$.

Let $L_{z}^{r}$ be the union of all $\mathbb{G}_{m}$-stable rational curves through $z$ whose tangent direction at $z$ lies in $\operatorname{Sm}(V(P_{*}))=V(P_{*})\cap(dP_{*}\neq 0)\subset\mathbb{P}T_{z}M_{P}$. Then we have

$$\operatorname{dim}L_{z}^{r}=\operatorname{dim}(V(P_{*}))+1=n-1.$$

From the definition of $L_{z}^{r}$, we know that $L_{z}^{r}\backslash\{z\}\subset M_{1}^{+}$, hence $Z$ is a union of some irreducible components of $M_{1}$.

Since $M_{1}$ is smooth, the irreducible components of $M_{1}$ do not intersect. Let $M_{1}=Z\cup Z_{1}$. Suppose that $Z_{1}\neq\emptyset$, then $\overline{Z_{1}^{+}}$ is also a divisor. By the definition of $Z$ (see the proof of Proposition 3.2), $Z$ is actually the set of source points of the $\mathbb{G}_{m}$-stable rational curves on $M_{P}$ through $z$ with degree $r-1$. Since $Z_{1}\cap Z=\emptyset$, we know $z\notin\overline{Z_{1}^{+}}$. Write the coordinates of $\mathbb{P}V_{P}$ as $[z_{0}:z_{1}:\cdots:z_{m}:w^{(2)}_{1}:\cdots:w^{(2)}_{m_{2}}:\cdots:w^{(r)}_{1}]$, then $\overline{Z_{1}^{+}}$ is contained in the hyperplane $\{w_{1}^{(r)}=0\}$. Also we know $\overline{Z_{1}^{+}}\subset\{z_{0}=0\}$. This implies that $\dim(\overline{Z^{+}_{1}})\leq m-2$, which contradicts with $\operatorname{dim}(\overline{Z_{1}^{+}})=m-1$. Therefore, $M_{1}=Z$. ∎

From [HT99, Proposition 2.3, Theorem 2.5], we know that the action $W$ on boundary divisor $D$ stabilizes all its irreducible components. Let $D_{1}$ be an irreducible component of $D$, and let $p=[0:\cdots:0:f^{j}:\cdots:f^{r}]$ be a point of $D_{1}$ where each $f^{i}(j\leq i\leq r)$ is viewed as an element in $(F^{i}_{P})^{*}$ and $f^{j}\neq 0$. Then $j\geq 1$.

For $v\in W$, define

$$\mathcal{L}_{v}^{p}=\overline{\{g_{tv}\cdot p\mid t\in\mathbb{C}\}}.$$

Therefore, $\mathcal{L}_{v}^{p}$ is a rational curve and $\mathcal{L}_{v}^{p}\subset D_{1}$ for any vector $v\in W$.

From the explicit expression of the action of $W$ on $\mathbb{P}V_{P}$ (see Remark 1.7), we know that

$$g_{tv}\cdot p=[0:\cdots:0:g_{tv}^{p}\cdot f^{j}:\cdots:g_{tv}^{p}\cdot f^{r}],$$

where for each $j\leq i\leq r$,

$$g_{tv}^{p}\cdot f^{i}=\sum_{\ell=j}^{i}C_{i}^{\ell}f^{\ell}\circ\iota_{tv}^{i-\ell}=\sum_{\ell=j}^{i}t^{i-\ell}C_{i}^{\ell}f^{\ell}\circ\iota_{v}^{i-\ell}.$$

Since $F^{j}_{P}$ consists of all $(r-j)$-th derivatives of $P$ and $j\geq 1$, for general $v\in W$, $f^{j}\circ\iota_{v}^{r-j}\neq 0$. Hence $g_{tv}^{p}\cdot f^{r}\neq 0$ and the rational curve $\mathcal{L}_{v}^{p}$ contains the terminal point $z$, then we have $z\in\mathcal{L}_{v}^{p}\subset D_{1}$. ∎