Proof. For $(a)$ , first of all, observe that $\Gamma _f\cap \Delta _{{\mathbb {P}}^n}=\emptyset $ since, otherwise, $V(f)$ would be singular by Proposition 2.1. Hence, by Remark 2.4, we have that $\Gamma _f$ has dimension at most $n-1$ . However, by definition, we have that $\Gamma _f$ is cut by $n+1$ divisors of ${\mathbb {P}}^n\times {\mathbb {P}}^n$ of bidegree $(1,1)$ so each component of $\Gamma _f$ has dimension at least $n-1$ .

Since $\Gamma _f$ is a complete intersection, its connectedness follows by the Fulton-Hansen-type theorem (see [Reference Fulton and Hansen14], [Reference Lazarsfeld21, Ch.3] or [Reference Martinelli, Naranjo and Pirola25]).

For $(b)$ , let us assume by contradiction that there exists an irreducible component W of ${\mathcal {D}}_k(f)$ of dimension $d\geq k$ . Over the general point $[w]$ of W, the fiber of the projection $\operatorname {\mathrm {pr}}_1$ from $\Gamma _f$ is a projective space $\iota ([x])\simeq {\mathbb {P}}^s$ with $s\geq n-k$ . Therefore, there exists a component of $\Gamma _f$ of dimension at least $d+s\geq n$ . This would mean, by Remark 2.4, that $\Gamma _f\cap \Delta _{{\mathbb {P}}^n}$ is not empty, giving a contradiction.

For $(c)$ , assume that W is as in $(b)$ and of dimension $k-1$ . Then the same reasoning as above implies the existence of an irreducible component of $\Gamma _f$ dominating W. For the converse, let G be an irreducible component of $\Gamma _f$ , set $G'=\operatorname {\mathrm {pr}}_1(G)$ and let m be the dimension of $G'$ . If $m=n-1$ , then $G'$ is a component of ${\mathcal {H}}_f={\mathcal {D}}_n(f)$ , so we are done. Otherwise, $m=n-1-a$ with $a>0$ so that $\operatorname {\mathrm {pr}}_1|_G$ has the general fiber F of dimension a. Since the whole fiber of $\operatorname {\mathrm {pr}}_1$ over a general point of $[x]\in G'$ is a projective space containing a fiber of dimension a, we have that $\operatorname {\mathrm {Rank}}(H_f(x))\leq n-a$ (i.e., the general point of $G'$ lies in ${\mathcal {D}}_{n-a}(f)$ ), as claimed.

Question 2.6. For which $[f]\in {\mathcal {U}}$ does $\operatorname {\mathrm {Sing}}({\mathcal {H}}_f)$ have dimension $n-2$ ?

Proof. In order to prove the claims, we will use extensively that $\operatorname {\mathrm {Sing}}(V(f))$ can be identified with $\{[x]\in {\mathbb {P}}(A^1)\,|\, x^2=0\}$ (see Proposition 2.1). More precisely, we will proceed by contradiction by proving that if the conclusion of $(a),(b)$ or $(c)$ is false, then there exists an element whose square is $0$ (i.e., a singular point for f), which is impossible by assumption.

Let us start by proving $(a)$ . For $P=([y_1],[y_2])\in \Gamma _f$ , assume, by contradiction, that $y_1^2$ and $y_2^2$ are linearly dependent. Then, there exists $\lambda \in {\mathbb {K}}$ such that $y_1^2=-\lambda ^2y_2^2$ . As $y_1y_2=0$ vanishes, we would have that $(y_1+\lambda y_2)^2=0$ , which is impossible.

By $(a)$ , in order to prove $(b)$ , it is enough to show that $z^2\not \in \left \langle x^2,y^2\right \rangle $ in $A^2_f$ : let us assume again by contradiction that it is the case (i.e., there exist $\alpha $ and $\beta $ in ${\mathbb {K}}^*$ such that $z^2=-\alpha ^2 x^2-\beta ^2 y^2$ ). In the same way as before, we can consider the square $(z+\alpha x+\beta y)^2$ , which is zero since $xy=xz=yz=0$ , leading a contradiction.

For $(c)$ , assume, by contradiction, that we could write $z=\alpha x+\beta y$ for some $\alpha ,\beta \in {\mathbb {K}}^*$ . Then, by definition of triangle, we would have that $x^2,y^2$ and $z^2$ are linearly dependent, which is impossible by $(b)$ .

Proof. The variety ${\mathcal {T}}$ can be described in ${\mathbb {P}}(A^1)^3$ as

$$ \begin{align*}\{([x],[y],[z])\in ({\mathbb{P}}(A^1))^3\,|\, xy=xz=yz=0\}\end{align*} $$

so that the forms $xy,xz$ and $yz$ vanish identically on ${\mathcal {T}}$ . If $\underline {v}=(v_1,v_2,v_3)\in T_{{\mathcal {T}},T}$ is a tangent vector and $v_l=[x_l']$ as in the statement, then $(x_1+tx_1',x_2+tx_2',x_2+tx_3')$ satisfies the equations $xy=xz=yz=0$ at first order:

$$ \begin{align*}0=(x_i+tx_i')(x_j+tx_j') \quad\mod t^2=x_ix_j+t(x_ix_j'+x_jx_i') \quad\mod t^2,\end{align*} $$

so $x_ix_j'+x_jx_i'=0$ , as claimed. Multiplying by $x_j$ , we get $x_j^2x_i'=0$ (i.e., we have $x_i'\in \mathrm {Ann}_{A^1}(x_j^2,x_k^2)/\left \langle x_i\right \rangle $ ).

Proof. Let us start by proving that the map $\psi $ is injective (up to the permutation of such vertices). Let $T=([x_1],[x_2],[x_3])$ and $T'=([y_1],[y_2],[y_3])$ be two triangles which are not equivalent via permutation of the vertices. Assume, by contradiction, that $\psi (T)=\psi (T')$ (i.e., the two triples of vertices span the same projective plane). We can then write $y_i=\sum _{j=1}^3a_{ij}x_j$ . Recall that $x_kx_l=\delta _{kl}x_k^2$ , since $x_kx_l=0$ for every $k\neq l$ , while $x_i^2\neq 0, y_i^2\neq 0$ for every i, by the smoothness of $V(f)$ . Hence, for $i\neq j$ , we get

$$ \begin{align*}0=y_iy_j=\sum_ka_{ik}x_k\cdot\sum_la_{jl}x_l=\sum_{k,l}a_{ik}a_{jl}x_kx_l=a_{i1}a_{j1}x_1^2+a_{i2}a_{j2}x_2^2+a_{i3}a_{j3}x_3^2.\end{align*} $$

From Lemma 2.10, we have then that $a_{ik}a_{jk}=0$ for every $k=1,2,3$ and $i\neq j$ . From this, one can easily see that, for every $i=1,2,3$ , at least (and at most, by construction) two coefficients among $a_{i1},a_{i2},a_{i3}$ are zero. Hence, the vertices of $T'$ and of T are the same up to a permutation.

Fix a triangle $T=([x_1],[x_2],[x_3])$ in ${\mathcal {T}}$ . We claim now that the differential

$$ \begin{align*}d_T\psi:T_{{\mathcal{T}},T}\rightarrow T_{\mathrm{Gr}(2,{\mathbb{P}}(A^1_f)), \left\langle x_1,x_2,x_3\right\rangle}\end{align*} $$

of $\psi $ at T is injective. Let us consider a nontrivial vector

$$ \begin{align*}\underline{v}=(x_1',x_2',x_3')\in T_{{\mathcal{T}},T}\subset T_{{\mathbb{P}}(A^1)^3,T}\simeq \bigoplus_{i=1}^3 A^1/\left\langle x_i\right\rangle.\end{align*} $$

Via the isomorphism $T_{\operatorname {\mathrm {Gr}}(k,V),W}\simeq \operatorname {\mathrm {Hom}}(W,V/W)$ , we have that $d_T\psi (\underline {v})$ is the homomorphism

$$ \begin{align*}d_T\psi(\underline{v})\in \operatorname{\mathrm{Hom}}\left(\left\langle x_1,x_2,x_3\right\rangle, A^1/\left\langle x_1,x_2,x_3\right\rangle\right)\qquad \mbox{such that }\quad x_i\mapsto x_i' \quad \mbox{ for } i\in \{1,2,3\}.\end{align*} $$

Hence, if we assume that $d_T\psi (\underline {v})\equiv 0$ , we have $x_i'\in \left \langle x_1,x_2,x_3\right \rangle $ , so we can write

(2.6) $$ \begin{align} x_i'=\sum_{m=1}^3a_{im}x_m \end{align} $$

for suitable $a_{im}\in {\mathbb {K}}$ . By Lemma 2.11, we have $x_ix_j'+x_i'x_j=0$ for $i\neq j$ , so using the relations in Equation (2.6), we obtain

$$ \begin{align*}0=x_i\sum_{m=1}^3a_{jm}x_m+x_j\sum_{m=1}^3a_{im}x_m=a_{ji}x_i^2+a_{ij}x_j^2 \qquad \mbox{ for } i\neq j.\end{align*} $$

By Lemma 2.10, squares of vertices of a triangle are independent, so we obtain $a_{ij}=0$ for $i\neq j$ and $x^{\prime }_i=a_{ii}x_i$ . This is impossible since $x_i'\in A^1/\langle x_i\rangle $ , and we would have $\underline {v}=0$ , whereas $\underline {v}$ is assumed to be nontrivial.

Proof. If ${\mathcal {D}}_{k}(f)\setminus {\mathcal {D}}_{k-1}(f)$ is a single point, $\varphi $ is clearly injective. Assume then that $z_1,z_2\in {\mathcal {D}}_{k}(f)\setminus {\mathcal {D}}_{k-1}(f)$ are distinct and that $\varphi ([z_1])=\varphi ([z_2])$ . Then $\iota ([z_1])=\iota ([z_2])$ , so $H_f(z_1)$ and $H_f(z_2)$ have the same kernel. Up to a change of coordinates, we can assume that $\ker (H_f(z_i))=\langle e_0,\dots , e_{n-k}\rangle $ , where $\{e_0,\dots ,e_n\}$ is the basis corresponding to the basis $\{y_0,\dots ,y_n\}$ under the identification ${\mathbb {P}}^n\simeq {\mathbb {P}}(A^1)$ . Hence, there exist two square matrices $A_1$ and $A_2$ of order k with coefficients in ${\mathbb {K}}$ and maximal rank such that

$$ \begin{align*}H_{f}(z_i)=\begin{bmatrix} 0 & 0\\ 0 & A_i \end{bmatrix}. \end{align*} $$

Being $x\mapsto H_f(x)$ linear, ${\mathbb {P}}(\langle z_1,z_2\rangle )\simeq {\mathbb {P}}^{1}$ is clearly contained in ${\mathcal {D}}_k(f)$ .

Set $p(\lambda ,\mu )$ to be the polynomial $\det (\lambda A_1+\mu A_2)$ . Since $\det (A_i)\neq 0$ by assumption, we have that p is homogeneous of degree k and nontrivial. Hence, there exists $[\lambda _0:\mu _0]$ such that $p(\lambda _0,\mu _0)=0$ (i.e., $H_f(\lambda _0 z_1+\mu _0 z_2)$ has rank at most $k-1$ ). Thus, ${\mathbb {P}}(\langle z_1,z_2\rangle )$ cuts ${\mathcal {D}}_{k-1}(f)$ , as claimed.

Proof. By assumption, one has $\operatorname {\mathrm {Rank}}(H_f(v))\leq n-k$ for all $v\in V\setminus \{0\}$ . One can see ([Reference Bricalli, Favale and Pirola6]) that the quadric of ${\mathbb {P}}^n$ given by the vanishing of the polynomial ${\partial _{v}}(f)$ is represented by the square symmetric matrix $H_f(v)$ . Then the singular locus of the quadric $V({\partial _{v}}(f))$ contains a ${\mathbb {P}}^{k}$ . By setting $W=\{{\partial _{v}}(f)\}_{v\in V}$ , we can then observe that $|W|$ is a linear subsystem of dimension $n-k-1$ , since the map $v\mapsto {\partial _{v}}(f)$ is injective as $V(f)$ is smooth (it would have been enough to ask that $V(f)$ is not a cone).

Let $J=J_f$ be the Jacobian ideal of f. Since $|W|\subset |J^2|$ and $J^2$ is spanned by a regular sequence, we have that $B:=\operatorname {\mathrm {BL}}(|W|)$ has pure dimension k. Indeed, being B cut by $n-k$ quadrics, we have that $\dim (B)\geq k$ . However, if there were a component of B with dimension at least $k+1$ , then we would be able to complete a basis of W in such a way that $J^2$ is not spanned by a regular sequence, against our assumptions. Indeed, if $q_0,\dots ,q_{n-k-1}$ is a basis of W, then for any choice of elements $q_{n-k},\dots ,q_n$ in $J^2$ , we would have $\bigcap _iV(q_i)\neq \emptyset $ . This cannot happen since f is smooth by assumption.

By Bertini’s theorem, the general element of $|W|$ is smooth away from B, which has pure dimension k. However, as observed before, all quadrics of $|W|$ have a ${\mathbb {P}}^k$ contained in the singular locus. Therefore, there exists a component of B which is a ${\mathbb {P}}^k$ . Since B has dimension k, it contains at most a finite number of ${\mathbb {P}}^k$ : there exists a component of B which is a ${\mathbb {P}}^k$ contained in the singular locus of all the elements of $|W|$ . Let ${\mathbb {P}}(U)\simeq {\mathbb {P}}^k$ be this linear space.

Consider $[u]\in {\mathbb {P}}(U)$ . Since all the quadrics $V({\partial _{v}}(f))$ parametrized by W are singular along ${\mathbb {P}}(U)$ , we have

$$ \begin{align*}{\partial_{u}}{\partial_{v}}(f)=0\qquad \mbox{ for all }[v] \in {\mathbb{P}}(V).\end{align*} $$

This implies that ${\partial _{v}}({\partial _{u}}(f))=0$ for all $[v]\in {\mathbb {P}}(V)$ : $V({\partial _{u}}(f))$ is a quadric whose singular locus contains the $(n-k-1)$ -plane $\Pi ={\mathbb {P}}(V)$ . Therefore, $H_f(u)$ has rank at most $k+1$ , and thus, ${\mathbb {P}}(U)\subseteq {\mathcal {D}}_{k+1}(f)$ .

Finally, notice that $\dim ({\mathcal {D}}_{k}(f))\leq k-1$ by Proposition 2.5, so ${\mathbb {P}}(U)\simeq {\mathbb {P}}^k$ cannot be contained in ${\mathcal {D}}_{k}(f)$ . Hence, for the general point $[u]\in {\mathbb {P}}(U)$ , the singular locus of $V({\partial _{u}}(f))$ is exactly the $(n-k-1)$ -plane $\Pi $ . In other terms, we have $\iota ([u])={\mathbb {P}}(V)$ for $[u]\in {\mathbb {P}}(U)$ general.

Proof. Assume that ${\mathbb {P}}(V)$ is a $(n-k-1)$ -plane in ${\mathcal {D}}_{n-k}(f)$ . By Proposition 3.6, we have that there exist ${\mathbb {P}}^k\simeq {\mathbb {P}}(U)\subseteq {\mathcal {D}}_{k+1}(f)$ such that ${\mathbb {P}}(V)\subseteq \iota ([u])$ for $[u]\in {\mathbb {P}}(U)$ with equality holding for $[u]$ general. Since, in this case, ${\mathcal {D}}_{k+1}(f)$ and ${\mathcal {D}}_k(f)$ do not coincide, for dimensional reason (by Proposition 2.5), we can define the map $\varphi :{\mathcal {D}}_{k+1}(f)\setminus {\mathcal {D}}_{k}(f)\to G(n-k-1,{\mathbb {P}}^n)$ . Then, the injectivity of $\varphi $ fails on two general points of ${\mathbb {P}}(U)$ . Finally, by Proposition 3.5, we have that ${\mathbb {P}}(U)\cap {\mathcal {D}}_{k}(f)\neq \emptyset $ , as claimed.

Proof. First of all, let us assume that for a fixed $k\geq 0$ , there exists ${\mathbb {P}}^k\simeq {\mathbb {P}}(V)\subseteq {\mathcal {D}}_{k+1}(f)$ . Then, by Proposition 3.6, there also exists ${\mathbb {P}}(U)\simeq {\mathbb {P}}^{n-k-1}$ contained in the locus ${\mathcal {D}}_{n-k}(f)$ . Moreover, we know that for all $[u]\in {\mathbb {P}}(U)$ , the projective space ${\mathbb {P}}(V)$ is contained in $\iota ([u])$ . This means that for every $[u]\in {\mathbb {P}}(U)$ and $[v]\in {\mathbb {P}}(V)$ , we have $uv=0$ , with the identification ${\mathbb {P}}^n={\mathbb {P}}(A^1)$ . Let us notice that the spaces ${\mathbb {P}}(U)$ and ${\mathbb {P}}(V)$ are skew in ${\mathbb {P}}^n$ and of complementary dimension. Indeed, if their intersection was nontrivial, we could find a point $[x]\in {\mathbb {P}}(V)\cap {\mathbb {P}}(U)$ : from above, we would obtain $x^2=0$ , against the smoothness of $V(f)$ . We can then consider for ${\mathbb {P}}^n$ a coordinate system $x_0,\dots ,x_n$ where ${\mathbb {P}}(V)=V(x_0,\dots ,x_{k})$ and ${\mathbb {P}}(U)=V(x_{k+1},\dots ,x_n)$ . Up to a change of coordinates, we can then write the polynomial f with respect to these variables: since, by construction, $x_ix_j=0$ in $A^2$ for every $i=0,\dots ,k$ and $j=k+1,\dots ,n$ , we get the claim.

Let us now assume that f is TS: as in Remark 3.3, we can write it (up to a possible change of coordinates) as $f=f_1(\underline {x})+f_2(\underline {x}')$ . As already observed, the Hessian matrix of f is of the form

(3.2) $$ \begin{align} H_f(\underline{x},\underline{x'})=\begin{bmatrix} H_{f_1}(\underline{x}) & 0 \\ 0 & H_{f_2}(\underline{x'}) \end{bmatrix}. \end{align} $$

Hence, by defining ${\mathbb {P}}(V):=V(x_0,\dots ,x_k)\simeq {\mathbb {P}}^k$ , one easily sees that $\operatorname {\mathrm {Rank}}(H_f(v))\leq k+1$ for every $[v]\in {\mathbb {P}}(V)$ (i.e., ${\mathbb {P}}(V)\subseteq {\mathcal {D}}_{k+1}(f)$ , as claimed).

Proof. Notice that Z is not contained in ${\mathcal {D}}_{k-1}(f)$ by Proposition 2.5, as we are assuming $\dim (Z)=k-1$ . Hence, the general point $z\in Z$ lies in ${\mathcal {D}}_k(f)\setminus {\mathcal {D}}_{k-1}(f)$ , and thus, $\iota ([z])\simeq {\mathbb {P}}^{n-k}$ . Since the general fiber of $\operatorname {\mathrm {pr}}_1|_{\tilde {Z}}:\tilde {Z}\to Z$ has dimension $n-k$ by construction, one has that the whole fiber $\operatorname {\mathrm {pr}}_1^{-1}([z])=\{[z]\}\times \iota ([z])$ is contained in $\tilde {Z}$ . However, $\operatorname {\mathrm {pr}}_1|_{\tilde {W}}:\tilde {W}\to W$ is surjective and $Z\subseteq W$ by assumption, so there exists at least a point $p=([z],[y])$ of the whole fiber $\pi _1^{-1}([z])$ with $p\in \tilde {W}$ . Then $p \in U=\tilde {W}\cap \tilde {Z}$ and $\operatorname {\mathrm {pr}}_1|_{U}:U\to Z$ is such that $\operatorname {\mathrm {pr}}_1|_{U}(p)=z$ . In particular, $\operatorname {\mathrm {pr}}_1|_{U}$ dominates Z.

By the above argument, we have that $\tilde {W}$ and $\tilde {Z}$ are irreducible components of $\Gamma _f$ which meet in a variety U of dimension at least $k-1$ . Then, we have a family of dimension $k-1$ since this variety is contained in $\operatorname {\mathrm {Sing}}(\Gamma _f)$ by construction and each point yields (at least) a triangle by Lemma 2.9.

Let $Z\subseteq {\mathcal {D}}_{n-1}(f)$ be an irreducible component of dimension $n-2$ and let ${\mathcal {F}}$ be a family of triangles dominating Z via $\pi _1$ , so that $\dim ({\mathcal {F}})\geq n-2$ . By construction, the general point $[x]$ of Z is such that $\iota ([x])\simeq {\mathbb {P}}^1$ so $\pi _1^{-1}([x])\subset \{[x]\}\times {\mathbb {P}}^1\times {\mathbb {P}}^1$ . If the general fiber $\pi _1^{-1}([x])$ has positive dimension, we would have that $\pi _1^{-1}([x])\cap \{[x]\}\times \Delta _{{\mathbb {P}}^1}$ is not empty, thus giving rise to a singular point of $V(f)$ . Then the general fiber of $\pi _1$ has dimension $0$ , and thus, $\dim ({\mathcal {F}})=n-2$ .

Proof. For $(a)$ , w.l.o.g. we can assume $\dim (Y_1)\geq 1$ and $Y_1\subset V(f)$ . If $T=([x],[y],[z])\in {\mathcal {F}}$ is a general triangle, then the differential $d\pi _{1,T}:T_{{\mathcal {F}},T}\rightarrow T_{Y_1,[x]}$ is surjective and it sends a tangent vector $(x',y',z')$ to $x'$ . By Lemma 2.11, $x'\in \operatorname {\mathrm {Ann}}_{A^1}(y^2,z^2)/\langle x\rangle $ . Moreover, since $Y_1\subset X$ , by Proposition 2.1, we have $X=\{[x]\,|\, x^3=0\}$ so $T_{X,[x]}=\operatorname {\mathrm {Ann}}_{A^1}(x^2)/\langle x\rangle $ . Hence,

$$ \begin{align*}T_{Y_1,[x]}\subseteq \operatorname{\mathrm{Ann}}_{A^1}(x^2,y^2,z^2)/\langle x\rangle.\end{align*} $$

Since T is a triangle, by Lemma 2.10, one has that $\langle x^2,y^2,z^2\rangle $ has dimension $3$ , and thus, by the Gorenstein duality, $\dim (\operatorname {\mathrm {Ann}}_{A^1}(x^2,y^2,z^2))=n+1-3=n-2$ . Moreover, being $[x]\in X$ , one has $x\in \operatorname {\mathrm {Ann}}(x^2,y^2,z^2)$ , so $\operatorname {\mathrm {Ann}}_{A^1}(x^2,y^2,z^2)/\langle x\rangle $ has dimension $n-3$ .

For $(b)$ , w.l.o.g. assume that $Y_3$ is of dimension $n-2$ . By contradiction, let us assume that $Y_1=\{[x]\}$ and that f is not of TS type. Hence, $Y_3\subseteq \iota ([x])$ , and this implies that $\iota (x)={\mathbb {P}}^s$ with $s\in \{n-2,n-1\}$ .

Then, we would get $\iota ([x])={\mathbb {P}}^{n-2}=Y_3\subseteq {\mathcal {D}}_{n-1}(f)$ and $[x]\in {\mathcal {D}}_1(f)$ , respectively. Both cases yield a contradiction by Theorem 3.4.

Proof. This follows from a more general fact: if $g:X\to Z$ is a surjective morphism between irreducible varieties and $f:X\to Y$ is a morphism, the locus

$$ \begin{align*}A=\{ z\in Z \,|\, \dim(f(g^{-1}(z)))=0\}\end{align*} $$

is an open Zariski set of Z. In order to prove this, let us consider the map $F=(f,g):X\rightarrow Y\times Z$ and denote by $X'\subseteq Y\times Z$ the image of X under the morphism F. Moreover, let $p_1$ and $p_2$ be the two projections from $X'$ to Y and Z, respectively. It is then clear that

$$ \begin{align*}A=\{z\in Z \ | \ \dim(p_1(p_2^{-1}(z)))=0\}=\{z\in Z \ | \ \dim(p_2^{-1}(z))=0\},\end{align*} $$

and thus, it is an open subset of Z. In our situation, if we assume that a fiber of $\pi _3$ is contracted to points via $\pi _1$ , then the same is true for the general one, contradicting the assumption $\dim (Y_1)=\dim ({\mathcal {F}})$ .

Proof. Notice that the proof follows easily if none of the three projections of ${\mathcal {F}}$ is contained in X. Indeed, in this case, $\pi _i^{-1}(Y_i\cap X)$ , the locus where the triangles of ${\mathcal {F}}$ have the i-th vertex on the cubic X, is a proper closed subset of ${\mathcal {F}}$ . We would like to show that also for $n\leq 5$ , this is indeed the only possible case (i.e., no projection of ${\mathcal {F}}$ can be contained in X). For $n=3$ , This is an easy consequence of Lemma 4.3, so we can assume $n\geq 4$ .

By hypothesis, $\pi _1$ is generically finite (and thus, by Lemma 4.3, $Y_1$ is not contained in X) and, by contradiction, $Y_3\subset X$ . Notice that, under these assumptions, $Y_2$ is not contained in X. Otherwise, $\pi _1^{-1}(Y_1\cap X)$ would be an $(n-3)$ -dimensional family of triangles with all the vertices contained in X. Then, we would have a contradiction as observed in Remark 4.4. For brevity, set ${\mathcal {F}}_c={\mathcal {F}}\cap (X\times X\times X)$ (i.e., ${\mathcal {F}}_c$ is the locus of the triangles of ${\mathcal {F}}$ with all $3$ vertices on the cubic hypersurface X).

Assume that $n=4$ . By Lemma 4.3, since $Y_3\subset X$ and $Y_1$ has dimension $2=n-2=\dim ({\mathcal {F}})$ , we have that $\dim (Y_3)=1$ . Then, since ${\mathcal {F}}$ is irreducible, all the fibers of $\pi _3$ have pure dimension $1$ . The dimension of $Y_2$ is either $1$ or $2$ . If the dimension of $Y_2$ is $1$ , all the fibers of $\pi _2$ are curves too and $Y_2\cap X$ is not empty. Consider $[y_0]\in Y_2\cap X$ and its fiber $C=\pi _2^{-1}([y_0])$ . By Lemma 4.5, C cannot be contracted by $\pi _1$ , so $\pi _1(C)$ is a curve. Then, $\pi _1(C)\cap X$ is not empty, and we can consider a point $[x_0]$ in this intersection. Hence, any element in $\pi _1^{-1}([x_0])\cap C\neq \emptyset $ is a triangle in ${\mathcal {F}}_c$ . This is impossible by Remark 4.4. Then we have necessarily $\dim (Y_2)=2$ .

Let Y be an irreducible component of $Y_2\cap X$ . Being $Y_2\not \subseteq X$ and of dimension $2$ , Y is a curve, and there exists an irreducible component C in $\pi _2^{-1}(Y)$ of dimension $1$ (since ${\mathcal {F}}$ is irreducible) dominating Y via the second projection. If either $\pi _1(C)$ is a curve or $\pi _1(C)=[x_0]$ with $[x_0]\in X$ , we have an element in ${\mathcal {F}}_c$ , so the only possible case is $\pi _1(C)=[x_0]$ with $[x_0]\not \in X$ .

Looking at the third projection, we have that $\pi _3(C)$ either is a point $[z]$ or it coincides with $Y_3$ . Moreover, let us observe that $\iota ([x_0])\simeq {\mathbb {P}}^s$ with $s\in \{1,2\}$ such that Y and $\pi _3(C)$ are contained in $\iota ([x_0])$ . To rule out the first case, namely $\pi _3(C)=[z]$ , first of all, observe that $[z]\not \in Y$ ; otherwise, we would have a singular point for the cubic X. Then s is forced to be $2$ . Moreover, by construction, we have $yz=0$ and $y^3=z^3=0$ for any $[y]\in Y$ : by reasoning as in Remark 4.4, all the lines $\langle [y],[z]\rangle $ lie in X. This implies that the whole $\iota ([x_0]) \simeq {\mathbb {P}}^2$ is contained in the smooth threefold X, but this is not possible, as observed in Remark 4.4.

For the remaining case, we have that $\pi _3(C)=Y_3$ , and by construction, both the curves Y and $Y_3$ are contained in $\iota ([x_0])$ . Then, if $s=1$ , we necessarily have $Y=Y_3\simeq {\mathbb {P}}^1$ and C is a family of triangles of dimension $1$ in $\{[x_0]\}\times Y\times Y$ . This yields a contradiction since C has to meet $\{[x_0]\}\times \Delta _{Y}$ , thus giving a singular point for X. Hence, we have necessarily $s=2$ , and we can assume $Y\neq Y_3$ or $Y=Y_3\neq {\mathbb {P}}^1$ . In both cases, as done above, considering the lines $\langle [y],[z]\rangle $ with $[y]\in Y, \ [z]\in Y_3$ and $yz=0$ , we get that the $2$ -plane $\iota ([x_0])$ is contained in the cubic threefold X, which is not possible.

Assume now that $n=5$ . We are working in the following framework: ${\mathcal {F}}$ is an irreducible $3$ -dimensional family of triangles with $\pi _1$ generically finite, $Y_1,Y_2$ not contained in X and $Y_3\subseteq X$ (this will lead to a contradiction). Being $Y_3\subseteq X$ by assumption, ${\mathcal {F}}_c$ is cut out from ${\mathcal {F}}$ by two divisors, so its expected dimension is $n-4=1$ . Then, either ${\mathcal {F}}_c$ is empty or $\dim ({\mathcal {F}}_c)\geq 1$ . However, as observed in Remark 4.4, under the above-mentioned hypotheses, we have that $\dim ({\mathcal {F}}_c)\leq 0$ , so ${\mathcal {F}}_c$ is necessarily empty. We will now prove that ${\mathcal {F}}_c$ is not empty, thus leading to a contradiction.

First of all, notice that $\dim (Y_2)\in \{1,2,3\}$ by Lemma 4.3. If $\dim (Y_2)\leq 2$ , the general fiber of $\pi _2$ has positive dimension and cannot be contracted to a point by $\pi _1$ by Lemma 4.5. Then, its image meets X, and thus, we produce a triangle in ${\mathcal {F}}_c$ as the analogous case for the threefold. We can then assume $\dim (Y_2)=3$ so that $\pi _2$ is generically finite as $\pi _1$ .

Denote by Y an irreducible component of $Y_2\cap X$ . The preimage $\pi _2^{-1}(Y)$ has dimension $2$ , and we can consider an irreducible component $S\subset \pi _2^{-1}(Y)$ dominating Y. If $\pi _1(S)$ is not a point or is a point on the cubic fourfold, as in the threefold case, one can easily construct an element in ${\mathcal {F}}_c$ : we can assume $\pi _1(S)=[x_0]\not \in X$ . As done in the previous case, we have that $\iota ([x_0])\simeq {\mathbb {P}}^s$ containing Y and $\pi _3(S)$ (thus, $s\in \{2,3\}$ since X is not of TS type), where $\pi _3(S)\subseteq Y_3$ can be either a point $[z]$ , a curve $C\not \subset Y_3$ (with $\dim (Y_3)=2$ ) or the whole $Y_3$ (with $\dim (Y_3)\in \{1,2\})$ . To conclude the proof, let us study these distinguished cases.

• • $\mathbf {\pi _3(S)=[z_0]}$ : since $[z_0]\not \in Y$ (otherwise we would have a singular point in X), s is forced to be equal to $3$ . Considering again the lines $\langle [y],[z_0]\rangle $ with $[y]$ varying in Y, we have that the whole $3$ -space $\iota ([x_0])$ is contained in the smooth cubic fourfold X, which is clearly not possible.

• • $\mathbf {\pi _3(S)=C}$ : In this case, one has $\iota ([x_0])\simeq {\mathbb {P}}^3$ since, otherwise, we would have $C\subseteq Y={\mathbb {P}}^2$ and $S\subseteq \{[x_0]\}\times Y\times Y$ and then a singular point for X as in a previous case. If we assume $C\not \subset Y\subset \iota ([x_0])={\mathbb {P}}^3$ , then one easily sees that the lines $\langle [y],[z]\rangle $ with $[y]\in Y, [z]\in C$ and $yz=0$ cover $\iota ([x_0])$ : we have a contradiction since we would have a projective $3-$ space in X. The only remaining case to analyse is then the one where $C\subset Y\subset {\mathbb {P}}^3$ with Y surface which is not a ${\mathbb {P}}^2$ . In this case, S is a surface in $\{[x_0]\}\times Y\times C$ with the projections $p_2=\pi _2|_S$ and $p_3=\pi _3|_S$ which are surjective. Then, for all $[z]\in C$ , $p_3^{-1}([z])$ has pure dimension $1$ . Let $[z]$ be a point in C and let D be an irreducible component of one of those fibers. For all $[y]\in p_2(D)$ , we have $yz=y^3=z^3=0$ and $[z]\not \in p_2(D)$ , so the joint variety $J(p_2(D),[z])$ has dimension $2$ , is a cone with vertex $[z]$ and is completely contained in $\iota ([x_0])\cap X$ . Since $\iota ([x_0])\simeq {\mathbb {P}}^3$ cannot be contained in X, these cones have to vary at most discretely, when $[z]$ moves in C. Notice that Y lies, by construction, in the union of these cones, so $[z]$ is in the vertex $\operatorname {\mathrm {Vert}}(Y)$ of Y. Then $C\subseteq \operatorname {\mathrm {Vert}}(Y)$ , and this forces Y to be a ${\mathbb {P}}^2$ , which is against our assumptions.

• • $\mathbf {\pi _3(S)=Y_3}$ : As in the previous case, we necessarily have $\iota ([x_0])\simeq {\mathbb {P}}^3$ containing both the surface Y and $\pi _3(S)$ , which can be either a curve or a surface. We can assume, moreover $Y_3=\pi _3(S)\subseteq Y$ , since otherwise, proceeding as above, we would have $\iota ([x_0])\subset X$ . In particular, Y is not a $2$ -plane. If $Y_3$ is a curve, we can obtain a contradiction as in the previous case by considering the cones with vertex $[z]\in Y_3$ spanned by the curves in Y whose elements annihilate $[z]$ . Hence, we can assume $Y=Y_3$ (and thus, $\pi _3$ is generically finite). By construction, for any element $[y]\in Y$ , there exists at least one element $[z]\in Y$ such that $yz=0$ ; hence, again, the line $\left \langle [y],[z]\right \rangle $ is contained in $\iota ([x_0])\cap X$ . If, for $[y]$ general, at least one of these lines is not contained in Y, one can see that the whole $3$ -space $\iota ([x_0])$ is contained in X, yielding a contradiction. We can then assume that for $[y]\in Y$ general, the above-mentioned lines are contained in Y. Our aim is now to show that $Y\simeq {\mathbb {P}}^2$ , against our assumptions. First of all, let us show that the union $\Lambda $ of these lines as subset in the Grassmannian $\operatorname {\mathrm {Gr}}(1,{\mathbb {P}}^3)$ has dimension $2$ . If, by contradiction, $\dim (\Lambda )=1$ , this would mean that for all $\ell \in \Lambda $ and for all $[y]\in \ell $ , there exists $[z]\in \ell $ with $yz=0$ . In other words, this would yield a correspondence in ${\mathbb {P}}^1\times {\mathbb {P}}^1$ , which intersects the diagonal $\Delta _{{\mathbb {P}}^1}$ nontrivially: the cubic X would be singular, which is not possible. Let us finally consider the incidence variety

  $$ \begin{align*}\Psi:=\{(y,\ell) \ | \ y\in\ell\in\Lambda\}\subset Y\times \operatorname{\mathrm{Gr}}(1,{\mathbb{P}}^3)\end{align*} $$
  and denote by $\psi _1$ and $\psi _2$ the two projections. We have just shown that $\operatorname {\mathrm {Im}}(\psi _2)=\Lambda $ , and moreover, it is clear that if $\ell \in \Lambda $ , then $\psi _2^{-1}(\ell )$ is described by $\ell $ itself; hence, such a fiber has dimension $1$ . Then $\dim (\Psi )=3$ , and looking at the first projection $\psi _1$ , we have that there exist infinitely many lines in $\Lambda $ contained in Y and passing through the general point $[y]\in Y$ . Hence, Y has to be a cone with $[y]\in \operatorname {\mathrm {Vert}}(Y)$ : from the generality of $[y]$ , it follows that $Y\simeq {\mathbb {P}}^2$ , as claimed.

Framework 5.4. Assume that f defines a smooth cubic $X=V(f)$ which is not of TS type (i.e., $[f]\in {\mathcal {V}}$ ) such that ${\mathcal {H}}_f$ is not normal. Then by Lemma 4.1 and Remark 4.2, we have a family ${\mathcal {F}}$ of dimension $n-2$ of triangles for ${\mathcal {H}}_f$ dominating, via the first projection, a component $Y_1$ of ${\mathcal {D}}_{n-1}(f)$ of the same dimension. As we have done in the previous sections, we denote by $Y_i$ the images of the projections $\pi _i$ . As just observed, $\pi _1:{\mathcal {F}}\to Y_1$ is generically finite.

If $T=([x],[y],[z])=([x_1],[x_2],[x_3])$ is a general point of ${\mathcal {F}}$ , by generic smoothness, we can assume that the differentials $d_T\pi _i:T_{{\mathcal {F}},T}\rightarrow T_{Y_i,[x_i]}$ are surjective; in particular, $d_T\pi _1$ is an isomorphism. Moreover, since we are assuming $n\leq 5$ , by Theorem 4.6, we have that none of the vertices of T belongs to the cubic $V(f)$ (i.e., $x_i^3\neq 0$ for $i\in \{1,2,3\}$ ). If we set

(5.1) $$ \begin{align} V_1=\left\langle x,y,z\right\rangle\subset A^1 \qquad \mbox{and}\qquad V_2=\operatorname{\mathrm{Ann}}_{A^1}(x^2,y^2,z^2)\subset A^1, \end{align} $$

by Lemma 2.10 and by Gorenstein duality, we have that $\dim _{{\mathbb {K}}}(V_1)=3$ and $\dim _{{\mathbb {K}}}(V_2)=n-2$ . Moreover, since $x^3,y^3,z^3\neq 0$ , we also have $V_1\cap V_2=\{0\}$ . Hence, by dimension reason, one has

(5.2) $$ \begin{align} A^1=V_1\oplus V_2. \end{align} $$

If $\{i,j,k\}=\{1,2,3\}$ , by Lemma 2.10 and Lemma 2.11, one has also

(5.3) $$ \begin{align} \dim_{{\mathbb{K}}}\operatorname{\mathrm{Ann}}_{A^1}(x_j^2,x_k^2)=n-1\qquad \dim_{{\mathbb{K}}}\operatorname{\mathrm{Ann}}_{A^1}(x_j^2,x_k^2)/\langle x_i\rangle=n-2. \end{align} $$

By dimension reason and since $x_i\not \in V_2$ , we have a canonical isomorphism $V_2\simeq \operatorname {\mathrm {Ann}}_{A^1}(x_j^2,x_k^2)/\langle x_i\rangle $ induced by the inclusion $V_2\hookrightarrow \operatorname {\mathrm {Ann}}_{A^1}(x_j^2,x_k^2)$ followed by the quotient by $\langle x_i\rangle $ . By Lemma 2.11 and since we have that $d_{T}\pi _i$ is surjective, we have

$$ \begin{align*}T_{Y_i,[x_i]}\subseteq \operatorname{\mathrm{Ann}}_{A^1}(x_j^2,x_k^2)/\langle x_i\rangle \simeq V_2,\end{align*} $$

so we can interpret $d_T{\pi _i}$ as maps $T_{{\mathcal {T}},T}\to V_2$ . Being $d_T\pi _1$ an isomorphism, we have then the endomorphisms

(5.4) $$ \begin{align} \psi_m=d_T\pi_m\circ (d_T\pi_1)^{-1}:V_2\to T_{Y_m,[x_m]}\hookrightarrow V_2\qquad \mbox{for } m\in \{2,3\}. \end{align} $$

Our approach is to analyse obstructions for such a configuration, by studying the Zariski tangent spaces. These information can be naturally codified by looking at the operators $\psi _m$ ; this strategy can be followed a priori in every dimension and in the cases of our interest yield to a direct and conclusive computation. In the proof of Theorem 5.1, we will start by analyzing these specific maps, ruling out both the cases where $\psi _2$ (or $\psi _3$ ) can be diagonalized or not and ultimately proving that a family of triangles of dimension $n-2$ cannot exist.

As we have done in the proof of Lemma 2.11, to a tangent vector $\underline {v}=(x',y',z')\in T_{{\mathcal {T}},T}$ , we can associate the ‘first-order deformation’ of T in the direction of $\underline {v}$ (for brevity, $\underline {v}$ -deformation of T), which we write in a compact way as

(5.5) $$ \begin{align} T+\underline{v}=(x+tx',y+ty',z+tz'). \end{align} $$

Proof. Since $Y_1$ has dimension $n-2$ , its general point $[x]$ is in ${\mathcal {D}}_{n-1}(f)\setminus {\mathcal {D}}_{n-2}(f)$ (i.e., the kernel of the multiplication map $x\cdot :A^1\to A^2$ has dimension $2$ ). The point $[x]$ is also the vertex of an element $T=([x],[y],[z])$ of ${\mathcal {F}}$ , and thus, $\ker (x\cdot )=\langle y,z\rangle \subseteq V_1$ . However, by the assumptions, one has $V_1\cap V_2={0}$ , as observed above.

Proof. By contradiction, let us assume that $[f]\in {\mathcal {V}}$ and $\dim (\operatorname {\mathrm {Sing}}({\mathcal {H}}_f))=n-2$ . Notice that if $n=2$ , since $\operatorname {\mathrm {Sing}}({\mathcal {H}}_f)={\mathcal {D}}_1(f)$ , we have a contradiction by Theorem 3.4. Hence, we can assume $n=3$ .

Fix the notation explained in the framework (see 5.4). Since $n=3$ , we have that

$$ \begin{align*}\operatorname{\mathrm{Ann}}_{A^1}(x^2,y^2,z^2)=V_2=\langle u\rangle\end{align*} $$

for suitable $u\in A^1\setminus \{0\}$ . All projections of ${\mathcal {F}}$ have dimension exactly $1$ by Lemma 4.3, so $T_{{\mathcal {F}},T}=\langle (Au,Bu,Cu)\rangle $ with $A,B,C\in {\mathbb {K}}^*$ . By Lemma 2.11, the associated first-order deformation $T+u(A,B,C)$ has to satisfy

$$ \begin{align*}Bux+Auy=0, \qquad Cux+Auz=0, \qquad Cuy+Buz=0.\end{align*} $$

We can then observe that the three independent points $Bx+Ay, \ Cx+Az, \ Cy+Bz$ belong to the kernel of the multiplication by u. In other words, $\iota (u)\supseteq {\mathbb {P}}^2$ , so that $[u]\in {\mathcal {D}}_1(f)$ . Hence, as before, we have a contradiction.

Proof. By contradiction, let us assume that $[f]\in {\mathcal {V}}$ and $\dim (\operatorname {\mathrm {Sing}}({\mathcal {H}}_f))=2$ . Then we are in the situation described in 5.4: $T=([x],[y],[z])$ will denote a general triangle in ${\mathcal {F}}$ (recall that by Theorem 4.6 none of its vertices belongs to the cubic $X=V(f)$ ). Since $n=4$ , we have that $\dim (V_2)=2$ . Recall that we have the endomorphisms

$$ \begin{align*}\psi_i=d_T\pi_i\circ (d_T\pi_1)^{-1}:V_2\simeq T_{Y_1,[x]}\to V_2\qquad i\in \{2,3\},\end{align*} $$

which have image $T_{Y_2,[y]}$ and $T_{Y_3,[z]}$ , respectively. We have that either one of the two is diagonalizable or that none is. We treat differently the two cases.

Case (I): Let us suppose that at least one of the two above endomorphisms is diagonalizable. W.l.o.g. we can assume that $\{u,w\}$ is a basis of $V_2$ whose elements are eigenvectors for $\psi _2$ . Then two independent tangent vectors to ${\mathcal {F}}$ in T are given as

(5.6) $$ \begin{align} \underline{v}=(u,Au,Cu+Dw) \qquad\mbox{ and }\qquad \underline{v}'=(w,Bw,Eu+Fw) \end{align} $$

for suitable $A,B,C,D,E,F$ scalars depending on the triangle, where, in particular, A and B are the eigenvalues corresponding respectively to u and w.

We recall that, for a given tangent vector $\underline {v}=(x',y',z')$ at $T=([x],[y],[z])$ , we have the relation $xy'+yx'=0$ by Lemma 2.11. For brevity, we refer to this relation with the notation $(\underline {v})_{xy}$ . We denote by $(\underline {v})_{xz}$ and $(\underline {v})_{yz}$ the analogous relations.

For example, using the vectors in (5.6), we have the corresponding first-order deformations

$$ \begin{align*}T+t\underline{v}=(x+tu,y+tAu,z+t(Cu+Dw))\qquad T+s\underline{v}'=(x+sw,y+sBw,z+s(Eu+Fw)),\end{align*} $$

which yield

(5.7) $$ \begin{align} \begin{array}{ll} (\underline{v})_{xy}: Aux+yu=0 \qquad &\quad (\underline{v}')_{xy}: Bwx+yw=0 \\ (\underline{v})_{xz}: Cux+Dwx+zu=0 \qquad &\quad (\underline{v}')_{xz}: Eux+Fwx+zw=0 \\ (\underline{v})_{yz}: Cuy+Dwy+Auz=0 \qquad &\quad (\underline{v}')_{yz}: Euy+Fwy+Bwz=0. \end{array} \end{align} $$

By elementary operations, one gets two new relations:

$$ \begin{align*} 2ACux+D(A+B)wx=0 \qquad E(A+B)ux+2FBwx=0. \end{align*} $$

For example, the former relation is obtained by substituting in $(\underline {v})_{yz}$ of Equation (5.7), the products $uy$ , $wy$ , and $uz$ obtained respectively from $(\underline {v})_{xy}$ , $(\underline {v}')_{xy}$ and $(\underline {v})_{xz}$ of Equation (5.7).

By Lemma 5.5, the multiplication map $x\cdot :V_2\to A_2$ is injective, so

(5.8) $$ \begin{align} AC=0 \qquad D(A+B)=0 \qquad E(A+B)=0\qquad FB=0. \end{align} $$

Let us now observe that A and B cannot be simultaneously zero when evaluated in a general triangle T; otherwise, the second projection of ${\mathcal {F}}$ would be zero-dimensional, contradicting Lemma 4.3.

Proof. Let $T=([x],[y],[z])$ be a general triangle of ${\mathcal {F}}$ . W.l.o.g. we can assume by contradiction that $A\neq 0$ and $B=0$ . We claim that $Y_2\subseteq \iota (z)$ and $Y_3\subseteq {\mathcal {D}}_2(f)$ .

Since $A\neq 0$ , we also get that $C=D=E=0$ by Equation (5.8). Hence, the first-order deformations of T are given by

(5.9) $$ \begin{align} T+t\underline{v}=(x+tu,y+tAu,z)\quad \mbox{ and }\quad T+s\underline{v}'=(x+sw,y,z+sFw). \end{align} $$

Since the differential maps of the projections from ${\mathcal {F}}$ are surjective, we have $\dim (Y_1)=2$ and $\dim (Y_2)=\dim (Y_3)=1$ .

Consider the curve $C_T=\pi _3^{-1}([z])$ . By construction, the tangent to $C_T$ in T is spanned by $\underline {v}$ which is projected to u and $Au$ via $d_T(\pi _1|_{C_T})$ and $d_T(\pi _2|_{C_T})$ , respectively. Hence, $\pi _1(C_T)$ is a curve in $Y_1$ and $\pi _2(C_T)=Y_2$ . In particular, $\pi _1(C_T)\cup Y_2\subseteq \iota ([z])$ , as claimed. Moreover, since ${\mathcal {D}}_1(f)=\emptyset $ , by assumption (by Theorem 3.4), we get that $\iota ([z])\simeq {\mathbb {P}}^2$ up to the case where it is a projective line coinciding both with $Y_2$ and $\pi _1(C_T)$ . But in this last case, we would have an involution on $Y_2\simeq {\mathbb {P}}^1$ , which yields a fixed point and then a singular point for $V(f)$ . Hence, $Y_3\subseteq {\mathcal {D}}_2(f)$ , as claimed.

Notice that the same argument can be used to prove that $Y_3\subseteq \iota ([y])$ and thus that $Y_2\subseteq {\mathcal {D}}_2(f)$ . By Proposition 3.5, one can see that for two general points $[z]$ and $[z']$ in $Y_3$ , we have that $\iota ([z])\neq \iota ([z'])$ . Hence, $Y_2\subseteq \iota ([z])\cap \iota ([z'])={\mathbb {P}}^1$ , and thus, $Y_2={\mathbb {P}}^1\subseteq {\mathcal {D}}_2(f)$ . This is impossible by Theorem 3.4 since f is not of TS type.

From the above Lemma 5.8, since T is general, we have that both A and B are not zero; hence, by Equation 5.8, we also get $B=-A$ . Indeed, if $A+B\neq 0$ , we would obtain $C=D=E=F=0$ , which is not possible by Lemma 4.3. Since $A=-B\neq 0$ , from Equations (5.8), we have $C=F=0$ . Then, the first-order deformation, in this case, can be written as

(5.10) $$ \begin{align} T+t\underline{v}=(x+tu,y+tAu,z+tDw)\qquad T+s\underline{v}'=(x+sw,y-sAw,z+sEu), \end{align} $$

and the conditions (5.7) are equivalent to

(5.11) $$ \begin{align} u(y+Ax)=0 \qquad zw+Exu=0 \qquad w(y-Ax)=0 \qquad zu+Dxw=0. \end{align} $$

Moreover, we cannot have $D=E=0$ as observed above, so we can assume $D\neq 0$ .

Since these equations hold, by assumption, for the general point $T\in {\mathcal {F}}$ , by deforming T at the first order in the direction of $\underline {v}$ , that is, by considering a curve

$$ \begin{align*}T(t)=([x(t)],[y(t)],[z(t)])=T+t\underline{v}+t^2(\cdots),\end{align*} $$

also the corresponding eigenvectors of $\psi _2$ ‘move’. More precisely, we have two curves

$$ \begin{align*}\gamma_u:U\to A^1\qquad \gamma_w:U\to A^1\end{align*} $$

defined in a neighbourhood of $0$ such that $\gamma _u(0)=u$ , $\gamma _w(0)=w$ and $\{\gamma _u(t),\gamma _w(t)\}$ is a basis of eigenvectors of $d_{T(t)}\pi _2\circ d_{T(t)}\pi _1^{-1}$ . These eigenvectors satisfy equations analogous to the ones in (5.7) where the coefficients depend on t. As observed above, the sum of the two eigenvalues of $\psi _2$ is $0$ also in a neighbourhood of T, so that the Equations (5.11) hold also locally.

We can then consider an expansion of the curves

$$ \begin{align*}\gamma_u(t)=u+tu'+t^2(\cdots)\qquad \gamma_w(t)=w+tw'+t^2(\cdots)\end{align*} $$

and substitute them in the Equations (5.11) in order to get new relations. We write $A(t)=A+A't+t^2(\cdots )$ for the curve following the eigenvalue relative to $\gamma _u(t)$ with an analogous notation for the coefficients that appear in Equations (5.11).

For example, from the condition $u(y+Ax)=0$ , one has

$$ \begin{align*}0\equiv \gamma_u(t)(y(t)+A(t)x(t))=(u+tu'+t^2(\cdots)) (y+Ax+t(2Au+A'x)+t^2(\cdots)),\end{align*} $$

so we get $2Au^2+A'xu+u'(Ax+y)=0$ . One can do the same reasoning for the $\underline {v}'$ -deformation, and we also can use two ‘parameters’ to take into account in a compact description the deformation of T in the direction of $t\underline {v}+s\underline {v}'$ . In this way, u and w ‘deform’ at first order as

$$ \begin{align*}u+tu'+su" \qquad\mbox{ and }\qquad w+tw'+sw",\end{align*} $$

respectively. Moreover, we can assume that $u',u"$ and $w',w"$ do not depend on u and w, respectively. This argument yields the relations

(5.12) $$ \begin{align} 2Au^2+A'xu+(Ax+y)u'=0\qquad u"(y+Ax)+A"xu=0 \end{align} $$

(5.13) $$ \begin{align} w'z+Dw^2+Eu^2+E'ux+Eu'x=0\qquad 2Euw+w"z+E"ux+Eu"x=0 \end{align} $$

(5.14) $$ \begin{align} w'(y-Ax)-A'xw=0\qquad w"(y-Ax)-A"xw-2Aw^2=0 \end{align} $$

(5.15) $$ \begin{align} 2Duw+D'xw+Dw'x+u'z=0\qquad Dw^2+Dxw"+D"xw+Eu^2+zu"=0. \end{align} $$

First of all, observe that multiplying by x Equation (5.14) ${}_I$ , since $xy=0=x^2w$ , we get

$$ \begin{align*}x^2w'=0.\end{align*} $$

Hence, multiplying by x Equation (5.15) ${}_I$ , we obtain

(5.16) $$ \begin{align} xuw=0. \end{align} $$

Since $xuw=0$ , from Equation (5.11), one can easily see that also

(5.17) $$ \begin{align} zu^2=zw^2=yuw=0. \end{align} $$

We claim now that $A'=A"=0$ and $xu^2,xw^2\neq 0$ . Indeed, let us observe that the product $xu$ vanishes if multiplied by $x, \ y, \ z, \ w$ . If $xu^2=0$ too, then by the Gorenstein duality in the apolar ring $A_f$ , we would get that $xu=0$ , which is not possible as observed with Lemma 5.5.

In the same way, one sees that $xw^2\neq 0$ . Then, recalling that $u(Ax+y)=0=w(y-Ax)$ , we can multiply by u and by w, respectively, Equations (5.12) ${}_{II}$ and (5.14) ${}_I$ , getting $A"xu^2=0$ and $A'xw^2=0$ , and so the claim:

(5.18) $$ \begin{align} A'=A"=0 \qquad xu^2,xw^2\neq 0. \end{align} $$

Proof. Let us prove it for $w'$ . Since we can assume that $w'$ does not depend on w, we can write it as $w'=\alpha x+\beta y+\gamma z+\delta u$ . Multiplying Equation (5.14) ${}_I$ by x and y, we get $x^2w'=0$ and $y^2w'=0$ respectively. Moreover, multiplying by z Equation (5.13) ${}_I$ , we have also $z^2w'=0$ . These last conditions yield

$$ \begin{align*}\alpha x^3=\beta y^3= \gamma z^3=0,\end{align*} $$

but since no vertex for the general triangle T belongs to $V(f)$ , we have that $\alpha =\beta =\gamma =0$ . Finally, since we have just shown that $A'=0$ , from Equation (5.14) ${}_I$ , we get

$$ \begin{align*}\delta u(y-Ax)=0.\end{align*} $$

Since, from Equations (5.11), we get $uy=-Aux$ , we would have $-2\delta Axu=0$ , which implies that $\delta =0$ , by Lemma 5.5.

Then $w'=0$ , as claimed. The same reasoning can also be used to prove that $u"=0$ .

We claim now that $wu^2=0$ and $w^2u=0$ . Multiplying by u Equation (5.15) ${}_I$ , one gets

(5.19) $$ \begin{align} 2Du^2w+zuu'=0. \end{align} $$

Since we have just shown that for T general, also the condition $zu^2=0$ is satisfied (see Equation (5.17)), we can deform it at the first order:

(5.20) $$ \begin{align} 0\equiv (z+tDw+sEu)(u+tu')^2 \quad\mod \langle t,s\rangle^2 \qquad \mbox{ and so }\qquad Du^2w+2zuu'=0. \end{align} $$

Putting together Equations (5.19) and (5.20), one gets

$$ \begin{align*}u^2w=0.\end{align*} $$

Let us now do the same for the second claim. Multiplying by w the Equation (5.13) ${}_{II}$ , we get $2Euw^2+zww"=0$ . Moreover, by deforming the condition $zw^2=0$ (see Equation (5.17)), we get ${Euw^2+2zww"=0}$ . As before, putting together these last conditions, one gets

$$ \begin{align*}uw^2=0\end{align*} $$

by using $E\neq 0$ (see Remark 5.10).

We claim now that $zuw=0$ . In order to show this last claim, let us deform the condition just obtained (i.e., $u^2w=0$ ):

$$ \begin{align*}0\equiv (u+tu')^2(w+sw") \quad\mod \langle t,s\rangle^2 \qquad\mbox{ and so }\qquad uwu'=0.\end{align*} $$

Write $u'$ as $\alpha x+\beta y+\gamma z+\delta w$ for simplicity. Since $xuw=yuw=uw^2=0$ and since $z^2w=0$ by definition of w, one has

$$ \begin{align*}0=uwu'=\gamma zuw\qquad z^2u'=\gamma z^3.\end{align*} $$

If $\gamma \neq 0$ we have done, let us then assume $\gamma =0$ : we get $z^2u'=0$ , and multiplying by z Equation (5.15) ${}_I$ we get $2Dzuw=0$ , as desired.

Finally, having $zuw=0$ yields a contradiction: in this case, from Equations (5.11), we would have $xw^2=0$ , which is not possible by Equation (5.18). Hence, neither $\psi _2$ nor $\psi _3$ can be diagonalizable for T general.