# On the quadratic equations for odeco tensors

Benjamin Biaggi  Mathematical Institute, University of Bern, Sidlerstrasse 5,
3012 Bern, Switzerland<EMAIL_ADDRESS>,  Jan Draisma  Mathematical
Instittue, University of Bern, Sidlerstrasse 5, 3012 Bern, Switzerland; and
Department of Mathematics and Computer Science, Eindhoven University of
Technology, P.O. Box 513, 5600 MB Eindhoven, The Netherlands
<EMAIL_ADDRESS>and  Tim Seynnaeve  Mathematical Institute, University of
Bern, Sidlerstrasse 5, 3012 Bern, Switzerland<EMAIL_ADDRESS>
###### Abstract.

Elina Robeva discovered quadratic equations satisfied by orthogonally
decomposable (“odeco”) tensors. Boralevi-Draisma-Horobeţ-Robeva then proved
that, over the real numbers, these equations characterise odeco tensors. This
raises the question to what extent they also characterise the Zariski-closure
of the set of odeco tensors over the complex numbers. In the current paper we
restrict ourselves to symmetric tensors of order three, i.e., of format
$n\times n\times n$. By providing an explicit counterexample to one of
Robeva’s conjectures, we show that for $n\geq 12$, these equations do not
suffice. Furthermore, in the open subset where the linear span of the slices
of the tensor contains an invertible matrix, we show that Robeva’s equations
cut out the limits of odeco tensors for dimension $n\leq 13$, and not for
$n\geq 14$ on. To this end, we show that Robeva’s equations essentially
capture the Gorenstein locus in the Hilbert scheme of $n$ points and we use
work by Casnati-Jelisiejew-Notari on the (ir)reducibility of this locus.

The authors were partly funded by Vici grant 639.033.514 from the Netherlands
Foundation for Scientific Research and by project grant 200021_191981 from the
Swiss National Science Foundation.

To Giorgio Ottaviani, on the occasion of his 60th birthday.

##  1\. Introduction

In [Rob16], Robeva discovered quadratic equations satisfied by orthogonally
decomposable (odeco) tensors. In [BDHR17], it is proved that over the real
numbers, these quadratic equations in fact characterise odeco tensors.

This raises the question whether Robeva’s equations also define (the Zariski
closure of) the set of complex odeco tensors. Indeed, Robeva conjectured that
they might even generate the prime ideal of this Zariski closure, at least in
the case of symmetric tensors [Rob16, Conjecture 3.2]. She proved this
stronger statement when the ambient space has dimension at most $3$ [Rob16,
Figure 2]. In general, however, the answer to the (weaker) question is no, as
already pointed out by Koiran in [Koi21]. In this short paper, based on the
first author’s Master’s thesis, we give an explicit symmetric tensor in
$({\mathbb{C}}^{12})^{\otimes 3}$ that satisfies Robeva’s equations but is not
approximable by complex odeco tensors. We do not know whether $12$ is the
minimal dimension for which this happens, but we show that if we impose a
natural, additional open condition on the tensor, then Robeva’s equations
characterise the Zariski closure of the odeco tensors precisely up to
dimension $13$.

A key idea in [BDHR17] is to associate an algebra $A$ to a symmetric three-
tensor $T$ and to realise that Robeva’s equations express the associativity of
that algebra. Furthermore, the symmetry of the tensor implies that $A$ is
commutative and that the bilinear form is an invariant form on $A$; see below
for definitions. If, in addition, $A$ contains a unit element—this turns out
to be an open condition on $T$—then $A$ is a Gorenstein algebra. Consequently,
we can use the results of [CJN15] on (ir)reducibility of the Gorenstein locus
in the Hilbert scheme of points in affine space to study the variety defined
by Robeva’s equations.

In the opposite direction, we use this relation between algebras and tensors
to give an elementary proof that the Gorenstein locus in the Hilbert scheme of
$n$ points in ${\mathbb{A}}^{n}$ has a dimension that grows as
$\Theta(n^{3})$. This seems surprising at first, since the component
containing the schemes consisting of $n$ distinct reduced points has dimension
only $n^{2}$; on the other hand, it is well-known that the dimension of the
Hilbert scheme itself does grow as a cubic function of $n$.

This relation between algebras and tensors is, of course, not new: a bilinear
multiplication on $V$ can be thought of as an element of $V^{*}\otimes
V^{*}\otimes V$, and in the presence of a bilinear form on $V$, the copies of
$V^{*}$ may be identified with $V$. Further properties of the algebra, such as
associativity, cut out subvarieties of the corresponding tensor space. A
classical reference for varieties of algebras is [Fla68], where the term
algebraic geography is coined. Another, more closely related paper is [Poo08],
whose Remark 4.5 is closely related to Lemma 9.2.2, and together with [Poo08,
Theorem 9.2] gives the cubic lower bound on the dimension of the Hilbert
scheme mentioned above. Finally, we note that the Zariski closure of the odeco
tensors consists of tensors of minimal border rank; equations for these are
studied in the recent paper [JLP22]. In particular, [JLP22, Proposition 1.4],
which states that, in the $1$-generic locus, the $A$-Strassen equations are
sufficient to characterise tensors of minimal border rank is closely related
to our Theorem 2.6.3.

###  1.1. Organisation of this paper

In Section 2, we introduce the fundamental notions of this paper, including
Robeva’s equations, of which we show that they are the only quadrics that
vanish on odeco tensors. We also state our main results (Theorems 2.5.2 and
2.6.3). In Section 3 we extend the well-known decomposition of finite-
dimensional unital algebras into products of local algebras to the non-unital
case. In Section 4 we recall that the Zariski closure of the odeco tensors is
a component of the variety cut out by Robeva’s equations; this was already
established in [Rob16, Lemma 3.7]. In Section 5 we show that there are many
weakly odeco tensors; combined with later results, this gives a lower bound on
the dimension of the Gorenstein locus in the Hilbert scheme of $n$ points in
${\mathbb{A}}^{n}$. In Section 6 we show how to unitalise algebras along with
an invariant bilinear form to turn them into a local Gorenstein algebra, and
vice versa. In Section 7 we use this construction to motivate the search for
nilpotent counterexamples to Robeva’s conjecture. Then, in Section 8 we find
such a counterexample for $n=12$. In Section 9 we make the connection with the
Gorenstein locus in the Hilbert scheme of $n$ points and prove our second main
result—that a version of Robeva’s conjecture holds in the (open) unital locus
precisely up to $n=13$. Finally, in Section 9.5 we show that the dimension
$d(n)$ of that Gorenstein locus is lower-bounded bounded by a cubic polynomial
in $n$. Since it is also trivially upper-bounded by such a polynomial, we have
that $d(n)=\Theta(n^{3})$.

##  2\. Set-up

###  2.1. Weakly and strongly odeco tensors

Let $V_{\mathbb{R}}$ be a finite-dimensional real vector space equipped with a
positive-definite inner product $(.|.)$.

######  Definition 2.1.1.

A symmetric tensor $T\in S^{3}V_{\mathbb{R}}\subseteq V_{\mathbb{R}}\otimes
V_{\mathbb{R}}\otimes V_{\mathbb{R}}$ is called orthogonally decomposable
(odeco, for short) if, for some integer $k\geq 0$, $T$ can be written as

$T=\sum_{i=1}^{k}v_{i}^{\otimes 3}$

where $v_{1},\ldots,v_{k}\in V_{\mathbb{R}}$ are nonzero, pairwise orthogonal
vectors. We write $Y(V_{\mathbb{R}})\subseteq S^{3}V_{\mathbb{R}}$ for the set
of odeco tensors.

Positive-definitiveness of the form implies that $(v_{i}|v_{i})>0$ for each
$i$, so that $v_{1},\ldots,v_{k}$ are linearly independent. Hence $k\leq
n:=\dim(V_{\mathbb{R}})$, and $Y(V_{\mathbb{R}})$ is a semi-algebraic set of
dimension at most $\binom{n}{2}+n$: the dimension of the orthogonal group plus
$n$ degrees of freedom for scaling. It turns out that $Y(V_{\mathbb{R}})$ is
in fact the set of real points of an algebraic variety defined by quadratic
equations that we will discuss below. Furthermore, it is easy to see that the
$k$ and the $v_{i}$ in the decomposition of an odeco tensor $T$ are unique,
which implies that the dimension of $Y(V_{\mathbb{R}})$ is precisely
$\binom{n}{2}+n$.

We now set $V:={\mathbb{C}}\otimes V_{\mathbb{R}}$ and extend $(.|.)$ to a
complex symmetric bilinear form (not a Hermitian form; that setting is studied
in [BDHR17] under the name udeco).

######  Definition 2.1.2.

A symmetric tensor $T\in S^{3}V\subseteq V\otimes V\otimes V$ (where the
tensor product is over ${\mathbb{C}}$) is weakly odeco if $T$ can be written
as

$T=\sum_{i=1}^{k}v_{i}^{\otimes 3}$

where the $v_{i}$ are nonzero pairwise orthogonal vectors. It is called
strongly odeco if $T$ admits such a decomposition where, in addition,
$(v_{i}|v_{i})\neq 0$. We write $Y(V)\subseteq S^{3}V$ for the Zariski closure
of the set of strongly odeco tensors.

######  Remark 2.1.3.

The set called $\mathrm{SODECO}_{n}({\mathbb{C}})$ in [Koi21] consists of the
strongly odeco tensors in $S^{3}{\mathbb{C}}^{n}$. Koiran proves that these
are precisely the set of symmetric tensors whose $n\times n$ slices are
diagonalisable and commute.

As pointed out above, every element of $Y(V_{\mathbb{R}})$, regarded as an
element of $S^{3}V$, is strongly odeco. Since the real orthogonal group
$O(V_{\mathbb{R}})$ is Zariski dense in the complex orthogonal group $O(V)$,
$Y(V)$ is the Zariski closure of $Y(V_{\mathbb{R}})$. On the other hand, due
to the presence of isotropic vectors and higher-dimensional spaces in $V$, the
set of weakly odeco tensors strictly contains the set of strongly odeco
tensors; we will return to this theme shortly. First we give the easiest
example that shows the need for a Zariski closure in the definition of $Y(V)$.

######  Example 2.1.4.

Consider the vector space $V={\mathbb{C}}^{2}$ equipped with the symmetric
bilinear form for which $(e_{1}|e_{2})=1$ and all other products are zero.
Then

$S=e_{1}\otimes e_{2}\otimes e_{2}+e_{2}\otimes e_{1}\otimes
e_{2}+e_{2}\otimes e_{2}\otimes e_{1}=\frac{1}{2}\lim_{t\to
0}[(t^{2}e_{1}+t^{-1}e_{2})^{\otimes 3}+(t^{2}e_{1}-t^{-1}e_{2})^{\otimes 3}]$

shows that $S$ is a limit of strongly odeco tensors. So $S\in Y(V)$, but $S$
is not strongly odeco, because its tensor rank is $3$ rather than $2$.

(Of course, since all nondegenerate symmetric bilinear forms on a finite-
dimensional complex vector space are equivalent, we could have changed
coordinates such that the bilinear form is the standard form.)

###  2.2. Commutative algebras from symmetric tensors

We now associate an algebra structure on $V$ to a tensor.

######  Definition 2.2.1.

We identify $V$ with $V^{*}$ via the map $v\mapsto(v|.)$. Then any element
$T\in V^{\otimes 3}$ can also be regarded as an element of $V^{*}\otimes
V^{*}\otimes V$, hence a bilinear map $\mu_{T}:V\times V\to V$. We call $V$
with $\mu_{T}$ the algebra associated to $T$.

If $T$ is symmetric, then, first, $\mu_{T}$ is commutative, and second,
$\mu_{T}$ satisfies

$(\mu_{T}(x,y)|z)=(x|\mu_{T}(y,z)),$

i.e., the bilinear form $(.|.)$ is invariant for the multiplication $\mu_{T}$.
When $T$ is fixed in the context, then we will often just write $xy$ instead
of $\mu_{T}(x,y)$.

###  2.3. Odeco implies associative

######  Proposition 2.3.1.

If $T\in S^{3}V$ is weakly odeco, then $\mu_{T}$ is associative.

This was observed in [BDHR17] in the real case, but the argument easily
generalises, as follows.

###### Proof.

Write

$T=\sum_{i=1}^{k}v_{i}^{\otimes 3}$

where the $v_{i}$ are pairwise orthogonal. Let $x,y,z\in V$. Then

$\displaystyle\mu_{T}(x,\mu_{T}(y,z))=$
$\displaystyle\mu_{T}(x,\sum_{i=1}^{k}(y|v_{i})(z|v_{i})v_{i})=\sum_{j=1}^{k}\sum_{i=1}^{k}(y|v_{i})(z|v_{i})(x|v_{j})(v_{i}|v_{j})v_{j}$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{k}(y|v_{i})(z|v_{i})(x|v_{i})(v_{i}|v_{i})v_{i}$

where the last equality uses that $(v_{i}|v_{j})=0$ whenever $i\neq j$. A
similar computation for $\mu_{T}(\mu_{T}(x,y),z)$ yields the exact same
result. ∎

###  2.4. Robeva’s equations

For any fixed $x,y,z$ (e.g. chosen from a basis of $V$), the condition that
$\mu_{T}(x,\mu_{T}(y,z))$ equals $\mu_{T}(\mu_{T}(x,y),z)$ translates into $n$
quadratic equations for $T$. All these equations together, with varying
$x,y,z$, are called Robeva’s equations.

######  Definition 2.4.1.

We denote by $X(V)\subseteq S^{3}V$ the affine variety defined by Robeva’s
equations, i.e.,

$X(V):=\\{T\in S^{3}V\mid\mu_{T}\text{ is associative}\\}.$

To prove that Robeva’s equations are all quadratic equations satisfied by
strongly odeco tensors, we reinterpret them as follows. The condition that
$\mu_{T}(x,\mu_{T}(y,z))$ equals $\mu_{T}(\mu_{T}(x,y),z)$ means that for
every $w\in V$, we have

$(\mu_{T}(x,\mu_{T}(y,z))|w)=(\mu_{T}(\mu_{T}(x,y),z)|w),$

which can be rewritten as

(1) $(\mu_{T}(y,z)|\mu_{T}(x,w))=(\mu_{T}(x,y)|\mu_{T}(z,w)).$

In other words: Robeva’s equations precisely express that the 4-linear map

$\displaystyle T\bullet T:V^{4}$ $\displaystyle\to{\mathbb{C}}$
$\displaystyle(x,y,z,w)$ $\displaystyle\mapsto(\mu_{T}(x,y)|\mu_{T}(z,w))$

is invariant under arbitrary permutations of $(x,y,z,w)$. This was, in fact,
Robeva’s original description of these quadratic equations in [Rob16].

######  Proposition 2.4.2.

The only quadratic equations vanishing on $Y(V)$ are Robeva’s equations.

###### Proof.

If we consider the natural map

$\displaystyle S^{2}(S^{3}V)$ $\displaystyle\to S^{2}(S^{2}V)$
$\displaystyle(v_{1}\otimes v_{2}\otimes v_{3})\otimes(w_{1}\otimes
w_{2}\otimes w_{3})$ $\displaystyle\mapsto(v_{1}|w_{1})(v_{2}\otimes
v_{3})\otimes(w_{2}\otimes w_{3}),$

then for any $T\in S^{3}V$, we can identify $T\bullet T$ with the image of $T$
under the composition

(2) $S^{3}V\xrightarrow{\nu_{2}}S^{2}S^{3}V\to S^{2}S^{2}V,$

where $\nu_{2}$ is the second Veronese embedding. Then $T$ satisfies Robeva’s
equations if and only if $T\bullet T\in S^{4}V\subseteq S^{2}S^{2}V$.

Since quadratic equations on $Y(V)$ correspond to linear equations on
$\nu_{2}(Y(V))$, we want to show that the image of (2), when restricted to
$Y(V)$, linearly spans $S^{4}V$. But (2) maps an odeco tensor
$\sum_{i=1}^{k}v_{i}^{\otimes 3}$ to the odeco tensor
$\sum_{i=1}^{k}(v_{i}|v_{i})v_{i}^{\otimes 4}$, and these tensors clearly span
$S^{4}V$. ∎

###  2.5. The main question

By Proposition 2.3.1, $X(V)$ contains weakly odeco tensors, hence in
particular the variety $Y(V)$. On the other hand, the results of [BDHR17]
imply that the set of real points of $X(V)$ equals $Y(V_{\mathbb{R}})$. This
raises the following question.

######  Question 2.5.1.

For which dimensions $n=\dim(V)$ is $X(V)$ equal to $Y(V)$?

Our partial answer to this question is as follows.

######  Theorem 2.5.2.

For $n=\dim(V)\leq 3$, $X(V)$ equals $Y(V)$. For $n\geq 12$, we have
$X(V)\subsetneq Y(V)$.

In fact, the result for $n\leq 3$ is due to Robeva [Rob16, Figure 2]. Our
contribution is a counterexample to [Rob16, Conjecture 3.2] for $n=12$.

###  2.6. The existence of a unit

The answer in Theorem 2.5.2 is unsatisfactory because of the large interval of
dimensions $n=\dim(V)$ for which we do not know whether $X(V)$ is strictly
larger than $Y(V)$. However, in a certain open subset of $S^{3}V$, we do know
precisely where the two stop being equal.

######  Lemma 2.6.1.

The condition on $T\in X(V)$ that $\mu_{T}$ has a multiplicative unit element
is equivalent to the condition that there exists a $v\in V$ such that the
multiplication map $L_{x}:v\mapsto\mu_{T}(x,v)$ is invertible. This is an open
condition on $T$.

In the case of ordinary tensors, the analoguous condition is called
$1$-genericity; see, e.g., [JLP22].

###### Proof.

We write $xv$ instead of $\mu_{T}(x,v)$. For the implication $\Rightarrow$
take $x$ to be the unit element. For the implication $\Leftarrow$, assume that
$L_{x}$ is invertible. Then in particular there exists an $e\in V$ such that
$ex=xe=x$. We then find, for any $y\in V$, that

$ey=exL_{x}^{-1}(y)=xL_{x}^{-1}(y)=y$

so that $e$ is a unit element.

It follows that $(V,\mu_{T})$ is not unital if and only if $\det(L_{x})=0$ for
all $x\in V$; this is a system of degree $n$ polynomial equations on $T\in
S^{3}(V)$ defining the complement of the unital locus. ∎

We will refer to the variety in $S^{3}(V)$ defined by the degree $n$ equations
in the proof above as the non-invertibility locus. If $T\in S^{3}(V)$ is not
in the non-invertibility locus, then $\mu_{T}$ needs not have a unit element;
but it does if furthermore $T$ lies in $X(V)$—we have used associativity of
$\mu_{T}$ in the proof above.

We define $X^{0}(V)$ as

$X^{0}(V):=\\{T\in X(V)\mid\mu_{T}\text{ is unital}\\},$

and similarly for $Y^{0}(V)$. Note that $\overline{Y^{0}(V)}=Y(V)$. This leads
to the following weakening of Question 2.5.1.

######  Question 2.6.2.

For which dimensions $n=\dim(V)$ is $X^{0}(V)$ equal to $Y^{0}(V)$? In other
words, for which dimensions do Robeva’s quadrics characterise the set of
limits of strongly odeco tensors in the complement of the non-invertibility
locus?

To answer this question, we will prove the following theorem.

######  Theorem 2.6.3.

The number of irreducible components of $X^{0}({\mathbb{C}}^{n})$ equals that
of the Gorenstein locus in the Hilbert scheme of $n$ points in
${\mathbb{A}}^{n}_{\mathbb{C}}$.

###### Proof.

See Section 9. ∎

We can now make use of the following result on the irreducibility of
Gorenstein loci of Hilbert schemes.

######  Theorem 2.6.4 ([CJN15]).

The Gorenstein locus of $n$ points in ${\mathbb{A}}^{d}_{\mathbb{C}}$

  * •

is irreducible if $n\leq 13$, or if $n=14$ and $d\leq 5$.

  * •

has 2 irreducible components if $n=14$ and $d\geq 6$.

Combining the two previous theorems gives complete answer to Question 2.6.2.

######  Corollary 2.6.5.

The locus $X^{0}({\mathbb{C}}^{n})$ is irreducible and equal to
$Y^{0}({\mathbb{C}}^{n})$ for $n\leq 13$, and is not irreducible and not equal
to $Y^{0}({\mathbb{C}}^{n})$ for $n\geq 14$.

##  3\. Decomposing (the algebra of) a tensor in $X(V)$

###  3.1. Motivation

Recall that if $B$ is a unital finite-dimensional algebra over ${\mathbb{C}}$,
then $B$ is isomorphic to a product $B_{1}\times\cdots\times B_{k}$ of local
algebras; here $k=|\operatorname{Spec}(B)|$. In this section we want to
establish a similar decomposition for not necessarily unital algebras.

###  3.2. The unital/nilpotent decomposition

######  Proposition 3.2.1.

Let $T\in X(V)$ and equip $V$ with the corresponding commutative, associative
multiplication $\mu_{T}$. Then $V$ has a unique decomposition as direct sum

$V_{1}\oplus\cdots\oplus V_{k}\oplus N$

where the $V_{i}$ are nonzero and $N$ is potentially zero, such that $N$ and
each $V_{i}$ is an ideal in $(V,\mu_{T})$, $(V_{i},\mu_{T}|_{V_{i}\times
V_{i}})$ is a local unital algebra for each $i$, and $N$ is a nilpotent
algebra. Furthermore, this unique direct sum decomposition is orthogonal.

Accordingly, $T$ decomposes as $T_{1}+\cdots T_{k}+T_{N}$ with $T_{i}\in
X(V_{i})$ and $T_{N}\in X(N)$; and we have $T\in Y(V)$ if and only if
$T_{i}\in Y(V_{i})$ for all $i$ and $T_{N}\in Y(N)$.

###### Proof.

Note that if $x,y\in V$ belong to different factors in any decomposition of
$V$ as a direct sum of ideals, and if the first factor has unit element $e$,
then

$(a|b)=(ae|b)=(e|ab)=(e|0)=0;$

this proves the orthogonality of the decomposition.

That a unital finite-dimensional commutative, associative algebra $V$ has a
unique product decomposition into local algebras is well-known—it is found by
taking a decomposition of $1$ into minimal idempotents $e_{i}$ satisfying
$e_{i}e_{j}=\delta_{ij}e_{i}$ and taking $V_{i}:=e_{i}V$.

To reduce to the unital case, we proceed as follows. If $V$ is nilpotent, then
we set $k:=0$ and $N:=V$. Otherwise, there exists an element $x\in V$ that is
not nilpotent. Let $L_{x}:V\to V$ be multiplication with $x$. Then there
exists an $m$ such that
$\operatorname{im}L_{x}^{m}=\operatorname{im}L_{x}^{m+1}=\ldots$. Set
$y:=x^{m}$ and $W:=yV$. Then $(L_{y})|_{yV}$ is invertible, so the ideal $yV$
is unital by Lemma 2.6.1. On the other hand, we have $V=yV\oplus\ker(L_{y})$:
indeed, the dimensions of $yV=\operatorname{im}(L_{y}),\ker(L_{y})$ add up to
$n$, and if $L_{y}(yv)=0$, then $L_{y}^{2}v=0$, so already $L_{y}v=0$, so
$yv=0$.

Now $\ker L_{y}$ has strictly lower dimension than $V$, and by induction we
know that $\ker L_{y}$ is the direct sum of a nilpotent ideal $N$ and an ideal
$I$ that is unital as an algebra. Then so is $V$: it equals the direct sum of
the ideals $N$ and $yV\oplus I$, where the latter is unital as an algebra. The
decomposition of $V$ into a unital ideal $V_{0}$ and a nilpotent ideal $V_{1}$
is unique, because $V_{0}$ is the space of elements $v\in V$ for which there
is an idempotent $e\in V$ with $v\in L_{e}$.

The statements about $T$ are straightforward from the fact that $\mu_{T}$ is
the sum of its restrictions to the ideals $V_{i}$ and $N$. Here we note that
the restrictions of $(.|.)$ to the $V_{i}$ and to $N$ are non-degenerate, so
that the notation $X(V_{i}),X(N),Y(V_{i}),Y(N)$ make sense.

##  4\. $Y(V)$ is a component of $X(V)$

###  4.1. Motivation

While, as we will see, $Y(V)$ is in general not equal to $X(V)$, at least it
is an irreducible component. This was already observed in [Rob16]; we
paraphrase the argument here.

###  4.2. A tangent space computation

Here, and later in the paper, we will write $e_{1},\ldots,e_{n}$ for the
standard basis of ${\mathbb{C}}^{n}$.

######  Proposition 4.2.1.

For each $V$ equipped with a nondegenerate symmetric bilinear form, the
variety $Y(V)$ of limits of strongly odeco tensors is an irreducible component
of the variety $X(V)$ defined by Robeva’s quadrics.

###### Proof.

It suffices to prove that for a suitable tensor $T_{0}\in Y(V)$, the tangent
space to $X(V)$ at $T_{0}$ has dimension equal to $\dim(Y(V))=\binom{n+1}{2}$.
We take $V={\mathbb{C}}^{n}$ and $T_{0}=E=\sum_{i=1}^{n}e_{i}^{\otimes 3}$.

Let us first write Robeva’s equations in coordinates: writing
$T=\sum_{i,j,k}T_{ijk}e_{i}\otimes e_{j}\otimes e_{k}$, equation (1) becomes

(3) $\sum_{r}T_{jkr}T_{i\ell r}=\sum_{r}T_{ijr}T_{k\ell r}$

The equations defining the tangent space at $E$ are given by substituting
$T=E+\varepsilon X$ in (3):

(4)
$\delta_{jk}X_{ij\ell}+\delta_{i\ell}X_{ijk}=\delta_{ij}X_{ik\ell}+\delta_{k\ell}X_{ijk}.$

By taking $i=\ell\neq j\neq k\neq i$ in (4) we find that $X_{ijk}=0$ for
$i,j,k$ pairwise distinct, and by taking $i=\ell\neq j=k$ we find that
$X_{iij}=-X_{ijj}$ for all $i\neq j$. But this implies that our tangent space
has dimension at most $n+\binom{n}{2}=\binom{n+1}{2}$. ∎

######  Remark 4.2.2.

In a similar manner, one finds that all strongly odeco tensors of tensor rank
$n$ are smooth points of $Y(V)$.

##  5\. Many weakly odeco tensors

###  5.1. Motivation

In this section, we give our first negative answer to Question 2.5.1 by
showing that, for $n$ sufficiently large, there are many more weakly odeco
tensors than strongly odeco tensors.

###  5.2. Weakly odeco tensors from isotropic spaces

Recall that $V$ is a complex vector space of dimension $n$ equipped with a
symmetric bilinear form $(.|.)$.

######  Proposition 5.2.1.

The variety $X(V)$ contains the union over all (maximal) isotropic subspaces
$U\subseteq V$ of $S^{3}U$. This union is an affine variety $Z(V)\subseteq
S^{3}V$ of dimension

$\binom{\lfloor n/2\rfloor+2}{3}+\binom{\lceil n/2\rceil}{2}.$

###### Proof.

For the first statement, note that if $u_{1},\ldots,u_{k}$ are elements of an
isotropic subspace $U$ of $V$, then $\sum_{i}u_{i}^{\otimes 3}$ is weakly
odeco, hence in $X(V)$ by Proposition 2.3.1.

There is no harm in restricting our attention to maximal isotropic subspaces,
i.e., those of dimension $\lfloor n/2\rfloor$. Hence $Z(V)$ is the projection
of the incidence variety

$\\{(U,T)\in\mathrm{Gr}_{\mathrm{iso}}(\lfloor n/2\rfloor,V)\times S^{3}V\mid
T\in S^{3}U\\}$

onto the second factor. Since the isotropic Grassmannian is a projective
variety, $Z(V)$ is closed. Furthermore, for $U\subseteq V$ isotropic of
dimension $\lfloor n/2\rfloor$ and $T\in S^{3}U$ concise, i.e. such that the
associated linear map $S^{2}U^{*}\to U$ is surjective, the fibre over $T$ is
the single point $(U,T)$, hence $\dim(Z(V))$ equals the dimension of the
isotropic Grassmannian, which is the second term above, plus the dimension of
$S^{3}U$ for a fixed isotropic $U\subseteq V$ of dimension $\lfloor
n/2\rfloor$, which is the first term. ∎

######  Remark 5.2.2.

Clearly, $\dim(Z(V))$ grows as a cubic (quasi-)polynomial in $n$, whereas
$\dim(Y(V))$ is a quadratic polynomial in $n$. Since $X(V)\supseteq Z(V)$,
this shows that $X(V)\supsetneq Y(V)$ for all $V$ of sufficiently high
dimension. In fact, $\dim(Z(V))>\dim(X(V))$ for $n\geq 16$. However, we will
show with an explicit example that $X(V)\supsetneq Y(V)$ holds already for
$n\geq 12$ (and possibly already for smaller $n$).

######  Remark 5.2.3.

The variety $Z(V)$ consists precisely of the tensors $T\in S^{3}V$ whose
algebra is $2$-step nilpotent:

$T\in Z(V)\iff\mu_{T}(x,\mu_{T}(y,z))=0\quad\forall x,y,z\in V.$

One implication is clear: if $T\in Z(V)$, we can write
$T=\sum_{i}{u_{i}^{\otimes 3}}$ with the $u_{i}$ isotropic and pairwise
orthogonal, and the computation from the proof of Proposition 2.3.1 gives that
$\mu_{T}(x,\mu_{T}(y,z))=0$. For the other direction we can work in
coordinates: write $V={\mathbb{C}}^{n}$, then the condition

$\mu_{T}(e_{i},\mu_{T}(e_{j},e_{k}))=0\quad\forall i,j,k\in\\{1,\ldots,n\\}$

is equivalent to

$\sum_{r=1}^{n}{T_{ijr}T_{k\ell r}}=0\quad\forall
i,j,k,\ell\in\\{1,\ldots,n\\}.$

But this means that that the space $U$ spanned by the columns of $T$ is
isotropic.

##  6\. Unitalisation and de-unitalisation

###  6.1. Motivation

It is well known that if an associative algebra $A$, say over ${\mathbb{C}}$,
has no multiplicative unit element, then one can turn $A$ into a unital
associative algebra by setting $A^{\prime}:={\mathbb{C}}1\oplus A$ and
extending the multiplication on $A$ to $A^{\prime}$ via
$1a^{\prime}:=a^{\prime}$ for all $a^{\prime}\in A^{\prime}$. In this section,
we describe a process that also extends an invariant bilinear form.

###  6.2. Unitalising algebras with invariant forms

Let $A$ be an associative algebra over $C$ equipped with a bilinear form
$(.|.)$ such that $(ab|c)=(a|bc)$ for all $a,b,c\in A$. We do not require $A$
to be commutative or $(.|.)$ to be symmetric.

We construct a new algebra

$\widetilde{A}:={\mathbb{C}}1\oplus A\oplus{\mathbb{C}}y$

with multiplication determined by

$\displaystyle 1*x:=x\text{ for all }x\in\widetilde{A},$ $\displaystyle
a*a^{\prime}:=aa^{\prime}+(a|a^{\prime})y\text{ for all }a,a^{\prime}\in A,$
$\displaystyle a*y:=0,y*a=0\text{ for all }a\in A,\text{ and}$ $\displaystyle
y*y:=0.$

We also extend the form $(.|.)$ to $\widetilde{A}$ by requiring that

$(1|1)=(a|1)=(1|a)=(a|y)=(y|a)=(y|y)=0\text{ for all }a\in A$

and $(1|y)=(y|1)=1$.

######  Remark 6.2.1.

Let us consider the special case where the multiplication on $A$ is
identically zero. If we let $a_{1},\ldots,a_{n}$ be an orthonormal basis of
$A$, then the tensor associated to $\tilde{A}$ is equal to

$\sum_{i=1}^{n}{(1\otimes a_{i}\otimes a_{i}+a_{i}\otimes 1\otimes
a_{i}+a_{i}\otimes a_{i}\otimes 1)}+1\otimes 1\otimes y+1\otimes y\otimes
1+y\otimes 1\otimes 1.$

This tensor is known as the Coppersmith-Winograd tensor [CW90]; it has played
in central role in the literature on the complexity of matrix multiplication.
We refer the reader to [Lan17] (in particular Chapter 3.4.9) for an overview.
In the notation of the latter reference, the above tensor is denoted
$T_{n,CW}$.

######  Proposition 6.2.2.

The algebra $\widetilde{A}$ is associative, and the form $(.|.)$ on
$\widetilde{A}$ is invariant. Furthermore, if $(.|.)$ is nondegenerate or
symmetric on $A$, then its extension to $\widetilde{A}$ has the same property;
and if $A$ is commutative and $(.|.)$ is symmetric, then $\widetilde{A}$ is
commutative.

###### Proof.

It suffices to prove the identity $a*(b*c)=(a*b)*c$ for $a,b,c$ ranging over a
spanning set of $\widetilde{A}$. If at least one of $a,b,c$ is $1$, then the
identity is immediate. If none of them is $1$ and at least one of them is $y$,
then both sides are zero. So the interesting case is the case where $a,b,c$
are all in $A$. Then we have

$a*(b*c)=a*(bc+(b|c)y)=a(bc)+(a|bc)y+0=a(bc)+(a|bc)y$

and

$(a*b)*c=(ab+(a|b)y)*c=(ab)c+(ab|c)y+0=(ab)c+(ab|c)y.$

These two expressions are equal by associativity of $A$ and invariance of
$(.|.)$ on $A$.

Now we turn to the identity $(a*b|c)=(a|b*c)$ for $a,b,c$ ranging over the
same spanning set. If $b=1$, then the identity is immediate. If $b\in
A\oplus{\mathbb{C}}y$ and $a=1$, then the identity reads

$(b|c)=(1|b*c).$

Now the right-hand side is the coefficient of $y$ in $b*c$. Write
$b=b^{\prime}+\beta y$ and $c=\gamma 1+c^{\prime}+\delta y$ with
$\beta,\gamma,\delta\in{\mathbb{C}}$ and $b^{\prime},c^{\prime}\in A$. Then
the coefficient of $y$ in $b*c$ equals $\beta\gamma+(b^{\prime}|c^{\prime})$,
and this also equals $(b|c)$. Since $a,c$ play symmetric roles, the identity
also holds when $c=1$. So we are left with the case where $a,b,c\in
A\oplus{\mathbb{C}}y$. But then, since $y$ is perpendicular to $A$, we have

$(a*b|c)=(ab|c)=(a|bc)=(a|b*c),$

as desired.

That the extension of $(.|.)$ inherits the properties of symmetry and non-
degeneracy is immediate, and so is the statement about the commutativity of
$\widetilde{A}$. ∎

######  Remark 6.2.3.

The space $M:=A\oplus{\mathbb{C}}y$ is a maximal ideal in $\widetilde{A}$, and
in particular a non-unital subalgebra of $\widetilde{A}$. This subalgebra has
an ideal ${\mathbb{C}}y$, and the natural map $A\to M/{\mathbb{C}}y$ is an
isomorphism of algebras. We will use this construction below to de-unitalise a
Gorenstein local algebra in a canonical manner.

######  Lemma 6.2.4.

Suppose that $A$ is commutative and that $(.|.)$ is symmetric. Then $A$ is
nilpotent if and only if $\widetilde{A}$ is (unital and) local.

###### Proof.

If $A$ is nilpotent, then $M$ consists of elements that are nilpotent in
$\widetilde{A}$, and hence any element not in $M$ is invertible. Conversely,
if $\widetilde{A}$ is local, then $M$ is the unique maximal ideal and its
elements are nilpotent. This implies that $A\cong M/{\mathbb{C}}y$ is
nilpotent. ∎

We now show that each local, unital algebra with an invariant bilinear form
arises as $\widetilde{A}$ for some $A$ equipped with a bilinear form.

######  Proposition 6.2.5.

Let $B$ be a commutative, local, unital algebra of dimension at least $2$
equipped with a nondegenerate invariant symmetric bilinear form. Then
$B\cong\widetilde{A}$ for some nilpotent algebra $A$ equipped with a
nondegenerate invariant symmetric bilinear form.

###### Proof.

Let $M$ be the maximal ideal of $B$, and let $d$ be maximal such that $M^{d}$
is nonzero. Then $d\geq 1$ since $\dim(B)\geq 2$.

We claim that $M^{d}$ is one-dimensional. Indeed, if it were at least two-
dimensional, then $1^{\perp}\cap M$ would contain a nonzero element $x$. This
element would satisfy $(x|1)=0$ and $(x|z)=(xz|1)=(0|1)=0$ for all $z\in M$,
contradicting the non-degeneracy of $(.|.)$.

Choose a spanning vector $z\in M^{d}$. Then $(1|z)\neq 1$, and hence we may
replace $z$ by a (unique) scalar multiple with $(1|z)=1$. Furthermore,
$z^{\perp}=M$. We define $A$ as the algebra $M/{\mathbb{C}}z$ equipped with
the induced symmetric bilinear form. We claim that $\widetilde{A}\cong B$ as
algebras with bilinear forms. The isomorphism $\varphi:\widetilde{A}\to B$
sends $1\in\widetilde{A}$ to $1\in B$, $y\in\widetilde{A}$ to $z\in B$ and
$m\in A$ to the unique element $m^{\prime}\in m+{\mathbb{C}}z\subseteq B$ that
satisfies $(1|m^{\prime})=0$ in $B$. All checks are then straightforward. ∎

Now let $V$ be an $n$-dimensional complex vector space equipped with a
nondegenerate symmetric bilinear form $(.|.)$. Define
$\widetilde{V}:={\mathbb{C}}1\oplus V\oplus{\mathbb{C}}y$, equipped with the
symmetric bilinear form as above.

Let $T\in X(V)$ and let $V=V_{1}\oplus\cdots\oplus V_{k}\oplus N$ be the
decomposition of Proposition 3.2.1. Let $e_{i}$ be the unit element in
$V_{i}$. Now $\widetilde{V}={\mathbb{C}}1\oplus V\oplus{\mathbb{C}}y$ is also
a commutative, associative algebra with invariant symmetric bilinear form
$(.|.)$, hence it corresponds to an element $\widetilde{T}\in
X(\widetilde{V})$, which in turn gives a decomposition of $\widetilde{V}$ as
in Proposition 3.2.1. The following proposition expresses the latter
decomposition into the former.

######  Proposition 6.2.6.

We have an orthogonal decomposition

$\widetilde{V}=V_{1}^{\prime}\oplus\cdots\oplus V_{k}^{\prime}\oplus
N^{\prime}\oplus 0$

into ideals, where $V_{i}^{\prime}\subseteq\widetilde{V}$ is isomorphic to
$V_{i}$ via the isomorphism

$\varphi_{i}:V_{i}\to V_{i}^{\prime},\ \varphi_{i}(v):=v+(v|e_{i})y$

and where $N^{\prime}$ is a local unital algebra spanned by $N$, $y$, and the
unit element $e_{k+1}:=1-e_{1}-\cdots-e_{k}$.

###### Proof.

First, $\varphi_{i}$ is clearly injective. It is also an algebra homomorphism
because

$\displaystyle\varphi_{i}(vw)=vw+(vw|e_{i})y=vw+(v|we_{i})y$
$\displaystyle=vw+(v|w)y=(v+(v|e_{i})y)*(w+(w|e_{i})y)=\varphi_{i}(v)*\varphi_{i}(w).$

Now note that if $v,w$ belong to $V_{i}\neq V_{j}$, respectively, then

$\varphi_{i}(v)*\varphi_{j}(w)=(v+(v|e_{i})y)*(w+(w|e_{j})y)=vw+(v|w)y=0.$

This shows that $V_{i}^{\prime}*V_{j}^{\prime}=\\{0\\}$. Similarly, we have
$N^{\prime}*V_{i}=\\{0\\}$ for all $i$—e.g., for $v\in V_{i}$ we have

$e_{k+1}*\varphi_{i}(v)=(1-e_{1}-\cdots-
e_{k})*(v+(v|e_{i})y)=v+(v|e_{i})y-(e_{i}v+(e_{i}|v)y)=0.$

Finally, $e_{k+1}$ is clearly a unit element in $N^{\prime}$. Indeed, we even
have an isomorphism $\widetilde{N}\to N^{\prime}$ of unital algebras with
invariant bilinear forms that sends $1$ to $1-e_{1}-\cdots-e_{k}$ and $y$ to
$y$. ∎

######  Proposition 6.2.7.

The map $T\mapsto\widetilde{T}$ is a morphism from $X(V)$ into
$X(\widetilde{V})$ that maps $Y(V)$ into $Y(\widetilde{V})$.

We call this morphism the unitalisation morphism.

###### Proof.

The first statement is immediate: the algebra structure on $\widetilde{V}$
depends in a polynomial manner on the algebra structure on $V$. For the last
statement, we note that if $T$ is strongly odeco of tensor rank $n$, then
$(V,\mu_{T})$ is an orthogonal direct sum of $n$ one-dimensional unital
ideals. By Proposition 6.2.6, $(\widetilde{V},\mu_{\widetilde{T}})$ is then an
orthogonal direct sum of $n$ one-dimensional unital ideals and one two-
dimensional ideal which, as an algebra with symmetric bilinear form, is
isomorphic to ${\mathbb{C}}[y]/(y^{2})$ with the blinear form determined by
$(1|y)=1$. The latter corresponds to the tensor

$S:=y\otimes y\otimes 1+y\otimes 1\otimes y+1\otimes y\otimes y$

from Example 2.1.4, hence it is a limit of strongly odeco tensors.
Consequently, by Proposition 3.2.1, $\widetilde{T}$ is in $Y(\widetilde{V})$.
Since the map $T\mapsto\widetilde{T}$ maps the dense subset of $Y(V)$ of
strongly odeco tensors of rank $n$ into $Y(\widetilde{V})$, it maps $Y(V)$
into $Y(\widetilde{V})$. ∎

######  Remark 6.2.8.

Unfortunately, we see no reason why, if $T\in X(V)$ satisfies
$\widetilde{T}\in Y(\widetilde{V})$, $T$ should be in $Y(V)$. Indeed, the
assumption says that $\widetilde{T}$ is a limit of sums with $n+2$ pairwise
orthogonal terms, and we do not see a natural construction that shows that $T$
is a limit of sums with $n$ pairwise orthogonal terms; we do not have a
counterexample, though.

##  7\. Nilpotent counterexamples

###  7.1. Motivation

When one studies the Hilbert scheme of $n$ points in a fixed space for
increasing $n$, and $n$ is taken minimal such that the scheme has more than
one irreducible component, then all components other than the main component
parameterise subschemes supported in a single point. We will establish a
similar result here.

###  7.2. First counterexamples are nilpotent

######  Theorem 7.2.1.

Let $n=\dim(V)$ be minimal such that $X(V)\neq Y(V)$. Then for all $T\in
X(V)\setminus Y(V)$ the algebra $(V,\mu_{T})$ is nilpotent.

###### Proof.

Let $T\in X(V)\setminus Y(V)$. Decompose $V=V_{1}\oplus\cdots\oplus
V_{k}\oplus N$ as in Proposition 3.2.1, and decompose
$T=T_{1}+\cdots+T_{k}+T_{N}$ accordingly. By Proposition 3.2.1, either some
$T_{i}$ does not lie in $Y(V_{i})$, or $T_{N}$ does not lie in $Y(N)$. By
minimality of $n$, we find that either $k=0$ and we are done, or else $k=1$
and $N=\\{0\\}$. In the latter case, by Proposition 6.2.5, the algebra
$(V,\mu_{T})$ equals $\widetilde{A}$ for some nilpotent algebra $A$ of
dimension $\dim(V)-2$ equipped with a nondegenerate symmetric bilinear form.
This means that $T=\widetilde{S}$ for some tensor $S\in X(A)$. By minimality
of $n$, $S$ lies in $Y(A)$. But then, by Proposition 6.2.7, $T=\widetilde{S}$
lies in $Y(V)$, a contradiction. Hence $(V,\mu_{T})$ is nilpotent, as claimed.
∎

##  8\. Proof of Theorem 2.5.2

Let $V$ be a finite-dimensional complex vector space of dimension $n$ and let
$(.|.)$ be a nondegenerate symmetric bilinear form on $V$. We first show that
$X(V)$ is not equal to $Y(V)$ when $n=12$.

In [Jel18] an explicit $14$-dimensional local Gorenstein algebra is
constructed which is not smoothable. Call this algebra $B$, and let $(.|.)$ be
a nondegenerate invariant symmetric bilinear form on $B$. Let $M$ be the
maximal ideal of $B$, and let $M^{d}$ be its minimal ideal. Then $A:=M/M^{d}$
is a nilpotent algebra and since $M^{d}$ is the radical of the restriction of
$(.|.)$ to $M$, $(.|.)$ induces a nondegenerate bilinear form on $A$. Note
that $\dim(A)=12$, so we may assume that $V$ (with its bilinear form) is the
underlying vector space of $A$ (with its bilinear form). Let $T\in X(V)$ be
the tensor corresponding to the algebra $A$. We claim that $T$ does not lie in
$Y(V)$. Indeed, if it does, then $T=\lim_{i\to\infty}T_{i}$ for a convergent
sequence of strongly odeco tensors $T_{i}$. Applying the unitalisation
morphism, we obtain $\widetilde{T}=\lim_{i\to\infty}\widetilde{T_{i}}$. Now
$\widetilde{T}$ is the structure tensor of the algebra $\tilde{A}$, which by
(the proof of) Proposition 6.2.5 is isomorphic to $B$.

However, by (the proof of) Proposition 6.2.7, each $\widetilde{T}_{i}$ is the
direct product of $12$ one-dimensional ideals and one copy of
${\mathbb{C}}[x]/(x^{2})$. In particular, each $\widetilde{T}_{i}$ corresponds
to a smoothable algebra, and $B$ is smoothable, as well. This contradicts the
choice of $B$. ∎

##  9\. Proof of Theorem 2.6.3

We set $V:={\mathbb{C}}^{n}$, equipped with the standard symmetric bilinear
form $\beta_{0}(u,v):=\sum_{i}u_{i}v_{i}$. Recall that $X^{0}(V)$ is the
variety of tensors corresponding to unital associative algebras on $V$ for
which $\beta_{0}$ is invariant. We want to show that $X^{0}(V)$ has the same
number of irreducible components as $\mathcal{H}^{\mathrm{Gor}}$.

###  9.1. Locating the unit element

######  Lemma 9.1.1.

The map $u:X^{0}(V)\to V$ that assigns to a tensor $T$ the unit element of
$(V,\mu_{T})$ is a morphism of quasi-affine varieties.

###### Proof.

For given $T$, the unit element $u=u(T)$ is the solution to the system of
linear equations $\mu_{T}(u,e_{i})=e_{i}$ for $i=1,\ldots,n$. For each $T\in
X^{0}(V)$, this system has a unique solution. This means that we can cover
$X^{0}(V)$ with open affine subsets in which some subdeterminant of the
coefficient matrix has nonzero determinant, and on such an open subset the map
$u$ is morphism with a formula in which that determinant appears in the
denominator. These morphisms glue to a global morphism $u$. ∎

###  9.2. A map from $X^{0}(V)$ to the Hilbert scheme

We write $R:={\mathbb{C}}[x_{1},\ldots,x_{n}]$, denote by $\mathcal{H}$ the
Hilbert scheme of $n$ points in ${\mathbb{A}}^{n}$, and by
$\mathcal{H}^{\mathrm{Gor}}$ the open subscheme of $\mathcal{H}$
parameterising Gorenstein schemes. In fact, since we care only about
irreducible components, we may and will replace both of these by the
corresponding reduced subvarieties, and we will only speak of
${\mathbb{C}}$-valued points of these varieties. Points in $\mathcal{H}$ will
be regarded as ideals in $R$ of codimension $n$. To define a morphism from an
affine variety $B$ over ${\mathbb{C}}$ to $\mathcal{H}$, it suffices to
indicate a subscheme of $B\times{\mathbb{A}}^{n}$ (product over
${\mathbb{C}}$), flat over $B$, such that the fibre over each $b\in B$ is
defined by such a codimension-$n$ ideal.

Take $B=X^{0}(V)$. A tensor $T\in X^{0}(V)$ gives rise to the ideal
$I_{T}:=\ker(\varphi_{T})$, where
$\varphi_{T}:K[x_{1},\ldots,x_{n}]\to(V,\mu_{T})$ is the kernel of the
homomorphism of associative algebras that maps $x_{i}$ to $e_{i}$ and $1$ to
the unit element $u(T)$ from Lemma 9.1.1. The ideals $I_{T}$ have vector space
codimension $n$ in $T$ and together define a subscheme of
$X^{0}(V)\times{\mathbb{A}}^{n}$ flat over $X^{0}(V)$. Hence we have described
a morphism $\Phi:X^{0}(V)\to\mathcal{H}$. Since any algebra corresponding to a
tensor in $X^{0}(V)$ has a nondegenerate invariant bilinear form,
$\Phi(T)\cong(V,\mu_{T})$ is Gorenstein for each $T\in X^{0}(V)$, so $\Phi$ is
a morphism $X^{0}(V)\to\mathcal{H}^{\mathrm{Gor}}$. We want to use $\Phi$ to
compare irreducible components of $\mathcal{H}^{\mathrm{Gor}}$ and $X^{0}(V)$.
However, the map $\Phi$ is not an isomorphism, so some care is needed for
this. We first describe the image of $\Phi$; the following is immediate.

######  Lemma 9.2.1.

The image of $\Phi$ consists of all codimension-$n$ ideals
$I\in\mathcal{H}^{\mathrm{Gor}}$ such that $x_{1},\ldots,x_{n}\in R$ map to a
basis of $R/I$ and moreover the bilinear form on $R/I$ for which this basis is
orthonormal is invariant for the multiplication in $R/I$. ∎

The following lemma shows that, as far as irreducible components are
concerned, it is no real restriction to consider ideals $I$ modulo which
$x_{1},\ldots,x_{n}$ is a basis.

######  Lemma 9.2.2.

The locus $\mathcal{H}^{\mathrm{Gor}}_{0}$ in $\mathcal{H}^{\mathrm{Gor}}$ of
ideals $I$ in $R$ for which $x_{1},\ldots,x_{n}$ maps to a vector space basis
of $R/I$ is open and dense in $\mathcal{H}^{\mathrm{Gor}}_{0}$. Consequently,
$\mathcal{H}^{\mathrm{Gor}}_{0}$ has the same number of irreducible components
as $\mathcal{H}^{\mathrm{Gor}}$.

This is well-known to the experts—see [Poo08, Remark 4.5]—but we include a
quick proof.

###### Proof.

The condition on $I$ can be expressed by the non-vanishing of certain
determinants; this shows that $\mathcal{H}^{\mathrm{Gor}}_{0}$ is open. For
density, suppose that some component $C$ of $\mathcal{H}^{\mathrm{Gor}}$ does
not meet $\mathcal{H}^{\mathrm{Gor}}_{0}$, and let $I_{0}$ be a point in $C$
such that the image of $\langle x_{1},\ldots,x_{n}\rangle_{\mathbb{C}}$ in
$R/I_{0}$ has maximal dimension, say $m<n$. After a linear change of
coordinates (which preserves all components of $\mathcal{H}^{\mathrm{Gor}}$
and hence in particular $C$), we may assume that $x_{m+1},\ldots,x_{n}$ are in
$I_{0}$, and there exists a monomial $r$ in $x_{1},\ldots,x_{m}$ of degree
$\neq 1$ such that $x_{1},\ldots,x_{m},r$ are linearly independent in
$R/I_{0}$.

Now, for $a\in{\mathbb{C}}$, consider the nonlinear automorphism
$\psi_{a}:R\to R$ that maps all $x_{i},i\neq n$ to themselves but $x_{n}$ to
$x_{n}+a\cdot r$. The map $(a,I)\mapsto\psi_{a}^{-1}(I)$ defines an action of
the additive group $({\mathbb{C}},+)$ on $\mathcal{H}^{\mathrm{Gor}}$, and
since the additive group is irreducible, this action preserves all components
of $\mathcal{H}^{\mathrm{Gor}}$. Since, for $a\neq 0$,
$x_{1},\ldots,x_{m},x_{n}+ar$ are linearly independent modulo $I_{0}$, their
pre-images $x_{1},\ldots,x_{m},x_{n}$ under $\psi_{a}$ are linearly
independent modulo $I_{a}:=\psi_{a}^{-1}(I_{0})$. Since $I_{a}$ is in $C$,
this contradicts the maximality assumption in the choice of $I_{0}$.

The last statement is now immediate. ∎

###  9.3. Varieties $Z_{2}\to Z_{1}\to\mathcal{H}^{\mathrm{Gor}}_{0}$ with the
same number of components

In what follows, we will identify $V$ with the space in $R$ spanned by the
variables $x_{1},\ldots,x_{n}$, via the identification $e_{i}\mapsto x_{i}$.
Each point in $\mathcal{H}^{\mathrm{Gor}}_{0}$ defines a unital, commutative,
associative algebra structure on $V$. The structure constant tensor in
$(S^{2}V^{*})\otimes V$ of this algebra does not necessarily lie in
$X^{0}(V)$, though, because the standard form $\beta_{0}$ may not be invariant
for it.

######  Lemma 9.3.1.

Let $Z_{1}$ be the subvariety

$\\{(I,[\beta])\in\mathcal{H}^{\mathrm{Gor}}_{0}\times{\mathbb{P}}(S^{2}V^{*})\mid\beta\text{
is invariant for
}R/I\\}\subseteq\mathcal{H}^{\mathrm{Gor}}_{0}\times{\mathbb{P}}(S^{2}V^{*}).$

Then the projection $Z_{1}\to\mathcal{H}^{\mathrm{Gor}}_{0}$ is surjective and
induces a bijection on irreducible components.

###### Proof.

Indeed, every (possibly degenerate) invariant bilinear form on $R/I$ is of the
form $\beta(r,s)=\ell(rs)$ for a unique linear form $\ell\in(R/I)^{*}$,
namely, the form $\ell(r):=\beta(1,v)$. Moreover, since for
$I\in\mathcal{H}^{\mathrm{Gor}}_{0}$ the space $V$ is a vector space
complement of $I$ in $R$, the natural map $(R/I)^{*}\to V^{*}$ is a linear
bijection. We conclude that, in fact, $Z_{1}$ is isomorphic to
$\mathcal{H}^{\mathrm{Gor}}_{0}\times{\mathbb{P}}(V^{*})$ via the map that
sends $(I,[\beta])$ to $(I,[v\mapsto\beta(1,v)])$. So each component of
$Z_{1}$ is just a component of $\mathcal{H}^{\mathrm{Gor}}_{0}$ times the
projective space ${\mathbb{P}}(V^{*})$. ∎

######  Lemma 9.3.2.

Let $Z_{2}$ be the subvariety

$\\{((I,[\beta]),g)\in Z_{1}\times\operatorname{GL}\nolimits(V)\mid
g[\beta]=[\beta_{0}]\\}\subseteq Z_{1}\times\operatorname{GL}\nolimits(V).$

Then the projection $Z_{2}\to Z_{1}$ has dense image and induces a bijection
between irreducible components.

###### Proof.

For the first statement, if $I\in\mathcal{H}^{\mathrm{Gor}}_{0}$, then by
definition there are _nondegenerate_ invariant bilinear forms on $R/I$. These
correspond to a dense open subset of ${\mathbb{P}}(V^{*})$ via the
correspondence in the proof above. This shows that $Z_{2}\to Z_{1}$ has dense
image. This image, $U$, is open in $Z_{1}$.

Next we claim that, in the analytic topology, $Z_{2}\to U$ is a fibre bundle
with fibre the group
${\mathbb{C}}^{*}\cdot\operatorname{O}\nolimits(\beta_{0})\subseteq\operatorname{GL}\nolimits(V)$;
here $\operatorname{O}\nolimits(\beta_{0})$ is the orthogonal group of the
form $\beta_{0}$. To see this, it consider a point $(I,[\beta_{1}])\in U$. By
definition of $U$, there exists a $g_{1}\in\operatorname{GL}\nolimits(V)$ such
that $g_{1}[\beta_{1}]=[\beta_{0}]$. Furthermore, there exists a holomorphic
map $\gamma$ defined in an open neighbourhood $\Omega$ in
${\mathbb{P}}S^{2}V^{*}$ of $[\beta_{0}]$ to $\operatorname{GL}\nolimits(V)$
such that $\gamma([\beta_{0}])=\operatorname{id}_{V}$ and
$\gamma([\beta])[\beta]=[\beta_{0}]$ for all $[\beta]\in\Omega$. Essentially,
$\gamma([\beta])$ is found by the Gram-Schmidt algorithm—note that in this
algorithm one has to divide by square roots of complex numbers, which, since
$[\beta]$ is close to $[\beta_{0}]$, are close to $1$; this can be done
holomorphically.

Now the map

$\displaystyle(U\cap(\mathcal{H}^{\mathrm{Gor}}_{0}\times
g_{1}^{-1}\Omega))\times({\mathbb{C}}^{*}\cdot\operatorname{O}\nolimits(\beta_{0}))\to
Z_{2},$
$\displaystyle((I,[\beta]),g)\mapsto((I,[\beta]),g\cdot\gamma(g_{1}[\beta])\cdot
g_{1})$

trivialises the map $Z_{2}\to Z_{1}$ over an open neighbourhood of
$(I,[\beta_{1}])$; here we use that
${\mathbb{C}}^{*}\cdot\operatorname{O}\nolimits(\beta_{0})$ is the stabiliser
of $[\beta_{0}]$ in $\operatorname{GL}\nolimits(V)$.

Now since $Z_{2}\to U$ is a fibre bundle with irreducible fibre
${\mathbb{C}}^{*}\cdot\operatorname{O}\nolimits(V)$—this is where it is
important that we work with the projective space ${\mathbb{P}}S^{2}V^{*}$
rather than $S^{2}V^{*}$; the orthogonal group $\operatorname{O}\nolimits(V)$
itself has two components!—that map induces a bijection between irreducible
components. ∎

###  9.4. Completing the proof

###### Proof of Theorem 2.6.3..

Recall that $X^{0}(V)$ parameterises the unital associative, commutative
algebra structures on $V$ such that $\beta_{0}$ is invariant for the
multiplication. Now consider the map

$\operatorname{GL}\nolimits(V)\times
X^{0}(V)\to\mathcal{H}^{\mathrm{Gor}}_{0},(g,T)\mapsto g\cdot\Phi(T)$

By Lemma 9.2.1, this map is surjective. Since $\operatorname{GL}\nolimits(V)$
is irreducible, the left-hand side has as many irreducible components as
$X^{0}(V)$. This shows that $X^{0}(V)$ has at least as many irreducible
components as $\mathcal{H}^{\mathrm{Gor}}_{0}$, hence as
$\mathcal{H}^{\mathrm{Gor}}$ by Lemma 9.2.2.

For the converse, by Lemmas 9.3.1, 9.3.2 $\mathcal{H}^{\mathrm{Gor}}_{0}$ has
as many irreducible components as $Z_{2}$. Now we claim that the morphism

$\displaystyle Z_{2}$ $\displaystyle\to(S^{2}V^{*})\otimes V$
$\displaystyle((I,[\beta]),g)$ $\displaystyle\mapsto\text{ the structure
constant tensor of }R/(g\cdot I)$

has as image the variety $X^{0}(V)$. Indeed, if $((I,[\beta]),g)$ lies in
$Z_{2}$, then $\beta$ is an invariant symmetric bilinear form for the
multiplication on $R/I$, and $g\beta\in{\mathbb{C}}^{*}\cdot\beta_{0}$ is an
invariant symmetric bilinear form for the multiplication on $R/(g\cdot I)$;
this therefore corresponds to an element in $X^{0}(V)$. We conclude that the
number of components of $X^{0}(V)$ is also at most that of
$\mathcal{H}^{\mathrm{Gor}}$. This concludes the proof. ∎

###  9.5. Cubic dimension growth for the Gorenstein locus

We conclude this paper with an observation on the dimension of
$\dim(\mathcal{H}^{\mathrm{Gor}})$.

######  Proposition 9.5.1.

The dimension of $\mathcal{H}^{\mathrm{Gor}}$, and hence that of the Hilbert
scheme $\mathcal{H}$ of $n$ points in ${\mathbb{A}}^{n}$, is lower-bounded by
a cubic polynomial in $n$ for $n\to\infty$.

Note that this was already known for $\mathcal{H}$ by [Poo08, Theorem 9.2].
The algebras constructed there are of the form
$A:={\mathbb{C}}[x_{1},\ldots,x_{d}]/(V+\mathfrak{m}^{3})$ where
$\mathfrak{m}$ is the maximal ideal $(x_{1},\ldots,x_{d})$ and where
$V\subseteq\mathfrak{m}^{2}/\mathfrak{m}^{3}$ has the correct codimension
$r=n-1-d$ for this quotient to have dimension $n$. Since any $1$-dimensional
subspace of $\mathfrak{m}^{2}/(V+\mathfrak{m}^{3})$ is a minimal ideal in $A$,
$A$ is not Gorenstein unless $\mathfrak{m}^{2}/(V+\mathfrak{m}^{3})$ is one-
dimensional, in which case $r=1$. However, to obtain cubic behaviour in $n$,
one needs $r$ to grow linearly with $d$. So the cubic-dimensional locus in
[Poo08] is a non-Gorenstein part of the Hilbert scheme.

###### Proof.

The unitalisation morphism sends $X({\mathbb{C}}^{n-2})$ into
$X^{0}({\mathbb{C}}^{n})$ by Proposition 6.2.7, and it does so injectively.
This means that the latter variety has dimension at least that of
$Z({\mathbb{C}}^{n-2})$, which is lower-bounded by a cubic polynomial by
Proposition 5.2.1. Furthermore, the morphism
$\Phi:X^{0}({\mathbb{C}}^{n})\to\mathcal{H}^{\mathrm{Gor}}$ is also injective.
∎

######  Remark 9.5.2.

The coefficient of $n^{3}$ in $\dim(Z({\mathbb{C}}^{n}))$ equals
$\frac{1}{48}$, which is considerably smaller than the coefficient
$\frac{2}{27}$ in [Poo08] for the lower bound on the dimension of the Hilbert
scheme of $n$ points in ${\mathbb{A}}^{n}$. We do not know whether the
$\frac{1}{48}$ can be improved.

## References

  * [BDHR17] Ada Boralevi, Jan Draisma, Emil Horobeţ, and Elina Robeva.  Orthogonal and unitary tensor decomposition from an algebraic perspective.  Israel J. Math., 222(1):223–260, 2017.
  * [CJN15] Gianfranco Casnati, Joachim Jelisiejew, and Roberto Notari.  Irreducibility of the Gorenstein loci of Hilbert schemes via ray families.  Algebra Number Theory, 9(7):1525–1570, 2015.
  * [CW90] Don Coppersmith and Shmuel Winograd.  Matrix multiplication via arithmetic progressions.  J. Symbolic Comput., 9(3):251–280, 1990.
  * [Fla68] Francis J. Flanigan.  Algebraic geography: Varieties of structure constants.  Pac. J. Math., 27:71–79, 1968.
  * [Jel18] Joachim Jelisiejew.  VSPs of cubic fourfolds and the Gorenstein locus of the Hilbert scheme of 14 points on A6.  Linear Algebra and its Applications, 557:265–286, 2018.
  * [JLP22] Joachim Jelisiejew, Joseph M. Landsberg, and Arpan Pal.  Concise tensors of minimal border rank.  2022\.  Preprint, `2205.05713`.
  * [Koi21] Pascal Koiran.  Orthogonal tensor decomposition and orbit closures from a linear algebraic perspective.  Linear Multilinear Algebra, 69(13):2353–2388, 2021.
  * [Lan17] Joseph M. Landsberg.  Geometry and complexity theory, volume 169 of Cambridge Studies in Advanced Mathematics.  Cambridge University Press, Cambridge, 2017.
  * [Poo08] Bjorn Poonen.  The moduli space of commutative algebras of finite rank.  Journal of the European Mathematical Society, page 817–836, 2008\.
  * [Rob16] Elina Robeva.  Orthogonal decomposition of symmetric tensors.  SIAM J. Matrix Anal. Appl., 37(1):86–102, 2016.