The Noether problem for spinor groups of small rank

TL;DR

This paper proves that the Noether Problem has a positive solution for spinor groups of rank up to 14 over any field with characteristic not equal to 2, extending previous results in the area.

Contribution

It extends the known cases where the Noether Problem is positively resolved for spinor groups up to rank 14, building on prior foundational work.

Findings

Positive solution for $ ext{Spin}_n$ with $n extless 15$ over arbitrary fields of characteristic not 2

Generalizes previous results to higher ranks

Provides new insights into the structure of spinor groups and their invariants

Abstract

Building on prior work of Bogomolov, Garibaldi, Guralnick, Igusa, Kordonskii, Merkurjev and others, we show that the Noether Problem for $\operatorname{Spin}_n$ has a positive solution for every $n\leq 14$ over an arbitrary field of characteristic $\neq 2$.

Equations

$$x\in X_0(\overline{k})$$

$$X\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$G\wr P\to \mathrm{GL}(V^n)\text{and}H\wr P\to \mathrm{GL}(W^n),$$

$$(V^n\times Z)/(G\wr P)\sim ((V/G{)}^n\times Z)/P\sim ((W/H{)}^n\times Z)/P\sim (W^n⋊Z)/(H\wr P),$$

$$k(V\times W{)}^G=k(V_0\times W_0{)}^G=k(W/H)=k(W{)}^H,$$

$$\pi (w_1)=\phi (v_0,w_1)=\phi (v_0,w_2)=\pi (w_2),$$

$$v\cdot (\lambda ,v^{\prime }):=(\lambda ,\lambda v+v^{\prime })$$

$$V_0/(V⋊G)\sim ({\mathbb{A}}^1\oplus W)/G\sim {\mathbb{A}}^1\times W/G,$$

$$A:(X,Y)\mapsto (AX,Y{A}^{-1})$$

$$\tau(A\cdot(X,Y))=\big{(}(A^{-1})^{T}Y^{T},X^{T}A^{T}\big{)}=(A^{-1})^{T}\cdot\tau(X,Y)$$

$$\pi :V\to M_{(n-1)\times (n-1)},$$

$$\pi (A\cdot (X,Y))=\pi (X,Y)$$

$$J=\left(\begin{array}{cccc}1& 0& \dots & 0\\ 0& 1& \dots & 0\\ 4& & & \\ 0& 0& \dots & 1\\ 0& 0& \dots & 0\end{array}\right)\in M_{n\times (n-1)}\text{and}{Y}^{\prime }\in M_{(n-1)\times n}.$$

$$Y=\left(\begin{array}{ccccc}{y}_{1,1}& y_{1,2}& \dots & y_{1,n-1}& y_{1,n}\\ y_{2,1}& y_{2,2}& \dots & y_{2,n-1}& y_{2,n}\\ 5& & & & \\ y_{n-1,1}& y_{n-1,2}& \dots & y_{n-1,n}& y_{n-1,n}\end{array}\right),$$

$$\pi (J,Y)=YJ=\left(\begin{array}{cccc}{y}_{1,1}& y_{1,2}& \dots & y_{1,n-1}\\ y_{2,1}& y_{2,2}& \dots & y_{2,n-1}\\ 4& & & \\ y_{n-1,1}& y_{n-1,2}& \dots & y_{n-1,n-1}\end{array}\right).$$

$$Z=\left(\begin{array}{ccccc}{y}_{1,1}& y_{1,2}& \dots & y_{1,n-1}& z_{1,n}\\ y_{2,1}& y_{2,2}& \dots & y_{2,n-1}& z_{2,n}\\ 5& & & & \\ y_{n-1,1}& y_{n-1,2}& \dots & y_{n-1,n}& z_{n-1,n}\end{array}\right)$$

$$\left\{\begin{array}[]{l}AJ=J\\ ZA=Y,\\ \end{array}\right.$$

$$A=\left(\begin{array}{ccccc}1& 0& \dots & 0& t_1\\ 0& 1& \dots & 0& t_2\\ 5& & & & \\ 0& 0& \dots & 1& t_{n-1}\\ 0& 0& \dots & 0& 1\end{array}\right)$$

$$\left\{\begin{array}[]{l}y_{1,1}t_{1}+\dots+y_{1,n-1}t_{n-1}+z_{1,n}=y_{1,n}\\ y_{2,1}t_{1}+\dots+y_{2,n-1}t_{n-1}+z_{2,n}=y_{2,n}\\ {1}\\ y_{n-1,1}t_{1}+\dots+y_{n-1,n-1}t_{n-1}+z_{n-1,n}=y_{n-1,n}.\end{array}\right.$$

$$B(\mathbb{Z}/2\mathbb{Z})$$

$$k(V\times W{)}^{{\mathrm{SL}}_n⋊(\mathbb{Z}/2\mathbb{Z})}$$

$$k(V\times W)^{\operatorname{SL}_{n}\rtimes(\mathbb{Z}/2{\mathbb{Z}})}=k\big{(}(V/\operatorname{SL}_{n})\times W\big{)}^{{\mathbb{Z}}/2{\mathbb{Z}}}=k(\operatorname{M}_{(n-1)\times(n-1)}\times W)^{{\mathbb{Z}}/2{\mathbb{Z}}}.$$

$$1\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$C:=⟨(\gamma ,-1)⟩\simeq {\mu }_2$$

$$1\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$S\simeq G_2$$

$$g(p)=g((h,\lambda )p)=g(\lambda hp)={\lambda }^{2}g(hp)={\lambda }^{2}g(p).$$

$$1\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

Full text

Zinovy Reichstein was partially supported by National Sciences and Engineering Research Council of Canada Discovery grant 253424-2017.

Federico Scavia was partially supported by a graduate fellowship from the University of British Columbia.

1. Introduction

Let $k$ be a field, $\overline{k}$ be an algebraic closure of $k$, $G$ be a linear algebraic group defined over $k$, and $\rho\colon G\hookrightarrow\operatorname{GL}(V)$ be a $k$-representation. Assume that $\rho$ is generically free; that is, the scheme-theoretic stabilizer $\operatorname{Stab}_{G}(v)$ is trivial for a point $v\in V(\overline{k})$ in general position. The Noether Problem asks whether the field of rational $G$-invariants $k(V)^{G}$ is a purely transcendental extension of $k$. Equivalently, it asks whether the Rosenlicht quotient $V/G$ is rational over $k$. (For the definition of the Rosenlicht quotient, see Section 2.) The following variants of the Noether Problem are also of interest: Is $V/G$ stably rational? Is $V/G$ retract rational? Recall that a $d$-dimensional algebraic variety $X$ is called rational if $X$ is birationally equivalent to the affine space ${\mathbb{A}}^{d}$, stably rational if $X\times{\mathbb{A}}^{r}$ is birationally equivalent to ${\mathbb{A}}^{d+r}$ for some $r\geqslant 0$ and retract rational if the identity morphism $\operatorname{id}\colon V/G\to V/G$, viewed as a rational map, can be factored through the affine space ${\mathbb{A}}^{m}$ for some $m\geqslant d$.

By the no-name lemma [RV06, Lemma 2.1] the answer to the Noether Problem for stable and retract rationality depends only on the group $G$ and not on the choice of the representation $V$. Following A. Merkurjev [Mer17], we will say that the classifying stack $BG$ is stably (respectively, retract) rational if $V/G$ is stably (respectively, retract) rational for some (and thus every) generically free representation $G\hookrightarrow\operatorname{GL}(V)$. We will also say that $BG$ and $BH$ are stably birationally equivalent if $V/G$ and $W/H$ are stably birationally equivalent, where $H\hookrightarrow\operatorname{GL}(W)$ is a generically free representation of $H$. This terminology is related to the fact that $V/G$ can be thought of as an approximation to $BG$. Note that for us “$BG$ is stably rational” will be a convenient short-hand for “the Noether Problem for stable rationality has a positive answer for $G$”; we will not actually work with stacks in this paper.

In the case where $G$ is a finite group, and $V$ is the regular representation of $G$, the question of rationality of $V/G$ was posed by E. Noether in the context of her work on the Inverse Galois Problem [Noe17]. For a finite group $G$, $V/G$ may not be stably (or even retract) rational. The first such examples over number fields $k$ were given by R. Swan [Swa69] and V. Voskresenskiĭ [Vos70] and over $k={\mathbb{C}}$ by D. Saltman [Sal84]. For many specific finite groups $G$ the Noether Problem remains open.

In the case where $G$ is a connected split semisimple groups over $k$, no counter-examples to the Noether Problem are known. It is known that $BG$ is stably rational for some $G$ (e.g., for $G=\operatorname{GL}_{n}$, $\operatorname{SL}_{n}$, $\operatorname{SO}_{n}$) but for many other connected split semisimple groups the Noether Problem remains open. Among these, projective linear groups $\operatorname{PGL}_{n}$ and spinor groups $\operatorname{Spin}_{n}$ have received the most attention.

For $G=\operatorname{PGL}_{n}$ the Noether Problem arose independently in ring theory in connection with universal division algebras; see [Pro67, p. 254]. It is known that $BG$ is stably rational for every $n$ dividing $420$; the remaining cases are open. See [LB91] for an overview.

In [Bog86], F. A. Bogomolov claimed that $BG$ is stably rational for every simply connected simple complex algebraic group $G$. However, there is a mistake in his argument; in particular, it breaks down for the spinor groups. V. Kordonskiĭ [Kor00] subsequently proved that $B\operatorname{Spin}_{7}$ and $B\operatorname{Spin}_{10}$ are stably rational (again, over the field of complex numbers). More recently, Merkurjev [Mer19, Section 4] showed that $B\operatorname{Spin}_{n}$ is retract rational for $n\leq 14$ over any field $k$ of characteristic $\neq 2$111Over a field of characteristic [math] retract rationality of $\operatorname{Spin}_{n}$ for $n\leqslant 12$ was proved earlier by J.-L. Colliot-Thélène and J.-J. Sansuc [CTS07]., and conjectured that $B\operatorname{Spin}_{n}$ is, in fact, not retract rational for $n\geq 15$. We strengthen Merkurjev’s result as follows.

Theorem 1.1**.**

Let $k$ be a field of characteristic $\neq 2$. Then $B\operatorname{Spin}_{n}$ is stably rational over $k$ for every $n\leq 14$.

The remainder of this paper will be devoted to proving Theorem 1.1. Our strategy will be as follows. For $n\leqslant 6$, it is well known that $\operatorname{Spin}_{n}$ is special; see [Gar09, Section 16.1]. Hence, $B\operatorname{Spin}_{n}$ is stably rational; see Lemma 3.1. Merkurjev [Mer19, Corollary 5.7] showed that $B\operatorname{Spin}_{2m+1}$ is stably birationally equivalent to $B\operatorname{Spin}_{2m+2}$ for every $m\geq 0$. Thus in order to prove Theorem 1.1 it suffices to show that $B\operatorname{Spin}_{n}$ is stably rational for $n=7$, $10$, $11$ and $14$. We will give self-contained proofs that work over an arbitrary field $k$ of characteristic $\neq 2$ in Propositions 6.2, 6.4, 7.1 and 8.1, respectively. Along the way we will show that $B(\operatorname{SL}_{n}\rtimes{\mathbb{Z}}/2{\mathbb{Z}})$ is stably rational; see Section 4. In the last section we give an alternative proof of Theorem 1.1 over $k=\mathbb{C}$ suggested to us by G. Schwarz.

2. Rosenlicht quotients

For the remainder of this paper $k$ will denote a field of arbitrary characteristic and $G$ will denote a smooth linear algebraic group defined over $k$. Starting from Section 6, we will assume that $\operatorname{char}(k)\neq 2$, but for now $k$ is arbitrary.

Let $X$ be a reduced and absolutely irreducible algebraic variety equipped with an action of $G$ over $k$. A rational map $\pi\colon X\dasharrow Y$ is called a Rosenlicht quotient map for the $G$-action on $X$ if

• •

$\pi\circ g=\pi$ for every $g\in G(\overline{k})$,

• •

$Y$ is reduced and irreducible,

• •

$k(Y)=k(X)^{G}$ and

• •

$\pi$ is induced by the inclusion of fields $k(X)^{G}\hookrightarrow k(X)$.

It is clear from this definition that a Rosenlicht quotient map exists for every $G$-action: just let $Y$ be any variety with function field $k(Y)=k(X)^{G}$, and $\pi\colon X\dasharrow Y$ be the rational map induced by the inclusion $k(X)^{G}\hookrightarrow k(X)$. Note that $Y$ (or $\pi$) is often called the rational quotient in the literature (see e.g. [PV94, 2.4]); we will refer to it as “the Rosenlicht quotient” in this paper in order not to overuse the term “rational”.

If $\pi\colon X\dasharrow Y$ is a Rosenlicht quotient map, then by a theorem of Rosenlicht [Ros56, Theorem 2] there exist a $G$-invariant open $k$-subvariety $X_{0}\subset X$ and an open $k$-subvariety $Y_{0}\subset Y$ such that $\pi$ restricts to a regular map $\pi_{0}\colon X_{0}\to Y_{0}$ and

[TABLE]

For a modern proof of Rosenlicht’s theorem, see [BGR17, Section 7]. The following Lemma is readily deduced from Rosenlicht’s Theorem.

Lemma 2.1**.**

Consider an action of $G$ on $X$ as above, and let $\pi\colon X\to Z$ be a $G$-equivariant morphism, where $G$ acts trivially on $Z$. Suppose that

(i) there exists a dense open subvariety $Z_{0}\subset Z$ such that for any $z\in Z_{0}(\overline{k})$ the fiber $\pi^{-1}(z)=G\cdot x$ for some $x\in X(\overline{k})$, and

(ii) $\pi$ is generically smooth (which is automatic in characteristic [math]).

Then $\pi$ is a Rosenlicht quotient for the $G$-action on $X$. In particular, $\pi$ induces an isomorphism between $k(Z)$ and $k(X)^{G}$.

Proof.

Let $\mu\colon X\dasharrow Y$ be a Rosenlicht quotient map. Since $\pi$ is $G$-equivariant, it factors through $\mu$ as follows:

[TABLE]

(This is called the universal property of Rosenlicht quotients.) It now suffices to show that $\alpha$ is a birational isomorphism. Choose open subvarieties $X_{0}\subset X$ and $Y_{0}\subset Y$ as in Rosenlicht’s theorem (2.1) and such that $\alpha$ is regular on $Y_{0}$. Then (i) tells us that $\alpha$ induces a bijection between $Y_{0}(\overline{k})$ and $Z_{0}(\overline{k})$ for some dense open subvariety $Z_{0}\subset Z$. In characteristic [math] this implies that $\alpha$ is a birational isomorphism, and we are done.

In finite characteristic the fact that $\alpha$ induces a bijection between $Y_{0}(\overline{k})$ and $Z_{0}(\overline{k})$ only tells us that the field extension $k(Y)/k(Z)$ induced by $\alpha$ is purely inseparable, so we need to use condition (ii) to finish the proof. By (ii), $\pi$ is generically smooth and hence, so is $\alpha$. This implies that $\alpha$ is a birational isomorphism, as desired. ∎

Remark 2.2*.*

Consider a generically free action of $G$ on a variety $X$ defined over $k$. By [BF03, Theorem 4.7], there exists a dense $G$-invariant open subvariety $X_{0}\subset X$ which is the total space of a $G$-torsor $\pi\colon X_{0}\to Y$ over some $k$-variety $Y$. Conditions (i) and (ii) of 2.1 are satisfied, because $G$ is smooth and $\pi$ is a $G$-torsor. It follows that $\pi\colon X_{0}\to Y$ (viewed as a rational map $X\dasharrow Y$) is a Rosenlicht quotient for the $G$-action on $X$.

3. Preliminaries on the Noether Problem

In this section we collect several known results on the Noether Problem for future use. Given two $k$-varieties, $X$ and $Y$, we will write $X\sim Y$ if $X$ and $Y$ are birationally isomorphic over $k$.

Recall that a smooth linear algebraic group $G$ is called special if $H^{1}(K,G)=\{1\}$ for every field $K$ containing $k$. Special groups were introduced by Serre [Ser58]; over an algebraically closed field of characteristic [math] they were classified by Grothendieck [Gro58].

Lemma 3.1**.**

If $G$ is special and stably rational, then $BG$ is stably rational.

Proof.

See [CTS07, Proposition 4.7] or [FR18, Remark 3.2]. ∎

Example 3.2**.**

It is known that the groups $G=\operatorname{GL}_{n},\operatorname{SL}_{n},\operatorname{Sp}_{2n}$ are special for every $n\geqslant 1$, and so are the groups $G=\operatorname{Spin}_{n}$ for $n\leqslant 6$. Thus $BG$ is stably rational for these $G$.

Let $P\to\operatorname{S}_{n}$ be a permutation representation of a linear algebraic group $P$. If $G$ is a another linear algebraic group, this representation gives rise to the wreath product $G\wr P$, which is defined as the semidirect product $G^{n}\rtimes P$ via the permutation action of $P$ on $G^{n}$.

Lemma 3.3**.**

Let $G$, $G_{1}$, $H$, and $P$ be linear algebraic groups over $k$, and $P\to\operatorname{S}_{n}$ be a permutation representation.

(a) If $BG$ and $BH$ are stably birational, then $B(G\wr P)$ and $B(H\wr P)$ are stably birational.

(b) If $BG$ is stably rational, then $B(G\wr P)$ is stably birational to $BP$.

(c) If $BG$ is stably rational, then $B(G\times G_{1})$ is stably birational to $B(G_{1})$.

Proof.

(a) Let $V$ be a generically free $G$-representation, and $W$ be a generically free $H$-representation. By our assumption the Rosenlicht quotients $V/G$ and $W/H$ are stably birationally equivalent, say $V/G\times\mathbb{A}^{r}\sim W/H\times\mathbb{A}^{s}$. After replacing $V$ by $V\oplus\mathbb{A}^{r}$ and $W$ by $W\oplus\mathbb{A}^{s}$, where $G$ acts trivially on $\mathbb{A}^{r}$ and $H$ acts trivially on $\mathbb{A}^{s}$, we may assume that $V/G\sim W/H$.

The product actions of $G^{n}$ on $V^{n}$ and of $H^{n}$ on $W^{n}$ naturally extend to linear representations

[TABLE]

respectively, where $P$ acts on $V^{n}$ and $W^{n}$ by permuting the factors. Now let $P\to\operatorname{GL}(Z)$ be some generically free linear representation of $P$. Then the representations $G^{n}\rtimes P$ on $V^{n}\times Z$ and of $H^{n}\rtimes P$ on $W^{n}\times Z$ are generically free. Comparing the Rosenlicht quotients $(V^{n}\times Z)/(G\wr P)$ and $(W^{n}\rtimes Z)/(H\wr P)$, we obtain

[TABLE]

as desired.

(b) Letting $H$ be the trivial group in part (a), we deduce that $B(G^{n}\rtimes P)$ is stably birational to $BP$.

(c) is a special case of (b) with $P=G_{1}$, equipped with the trivial permutation representation $P\to\operatorname{S}_{1}$. ∎

Lemma 3.4**.**

Let $G\to\operatorname{GL}(V)$ be a finite-dimensional representation defined over $k$. Suppose there exists a $k$-point $v_{0}\in V$ such that the scheme-theoretic stabilizer $H$ of $v_{0}$ in $G$ is smooth, and the $G$-orbit of $v_{0}$ is dense in $V$. Then $BG$ and $BH$ are stably birationally equivalent.

When $G$ is reductive and $\operatorname{char}k=0$, this lemma reduces to [CTS07, Proposition 3.13]. The following argument works in arbitrary characteristic.

Proof.

Let $G\to\operatorname{GL}(W)$ be a generically free representation of $G$. Denote the open orbit of $v_{0}$ in $V$ by $V_{0}$ and the Rosenlicht quotient map for the $H$-action on $W$ by $\pi\colon W\dasharrow W/H$. After possibly replacing $W/H$ by a dense open subvariety, we can choose an $H$-invariant dense open subvariety $W_{0}\subset W$ such that $\pi$ restricts to a morphism $W_{0}\to W/H$ whose fibers are exactly the $G$-orbits in $W_{0}$; see (2.1). In fact, by 2.2, we may assume that $\pi\colon W_{0}\to W/H$ is an $H$-torsor.

We claim that $\varphi\colon V_{0}\times W_{0}\to W/H$ given by $(v,w)\to\pi(w)$ is a Rosenlicht quotient map for the $G$-action on $V_{0}\times W_{0}$. If we establish this claim, then

[TABLE]

and the lemma will follow. By 2.1, it suffices to show that

(i) $\varphi^{-1}(\pi(w))$ is a single $G$-orbit for any $w\in W_{0}(\overline{k})$, and

(ii) $\varphi$ is generically smooth.

To prove (i), suppose $\varphi(v_{1},w_{1})=\varphi(v_{2},w_{2})$ for some $(v_{1},w_{1})$ and $(v_{2},w_{2})$ in $V_{0}\times W_{0}$. Our goal is to show that $(v_{1},w_{1})$ and $(v_{2},w_{2})$ lie in the same $G$-orbit. After translating these points by suitable elements of $G$, we may assume that $v_{1}=v_{2}=v_{0}$. Since $\pi$ is an $H$-torsor and

[TABLE]

we conclude that $w_{2}=h(w_{1})$ for some $h\in H(\overline{k})$. Since $v_{0}$ is stabilized by $H$, we have $h(v_{0},w_{1})=(v_{0},w_{2})$, as desired.

To prove (ii), note that $\pi$ is the composition of the projection map $p\colon V_{0}\times W_{0}\to W_{0}$ and the Rosenlicht quotient map $\pi\colon W_{0}\to W/H$. Clearly, $p$ is smooth. Moreover, $\pi$ is also smooth, because $H$ is smooth and $\pi$ is an $H$-torsor. Thus $\varphi=\pi\circ p$ is smooth, as desired. This completes the proof of (ii) and thus of Lemma 3.4. ∎

Example 3.5**.**

Consider the 1-dimensional representation $V$ of $G={\mathbb{G}}_{m}$, where $t\in{\mathbb{G}}_{m}$ acts on $V$ via scalar multiplication by $t^{n}$, where $n$ is not divisible by $\operatorname{char}(k)$. Taking $v_{0}$ to be any non-zero vector in $V$, we see that the stabilizer of $v_{0}$ in $G=\mathbb{G}_{m}$ is $H=\mu_{n}$. Since ${\mathbb{G}}_{m}=\operatorname{GL}_{1}$ is special, $B{\mathbb{G}}_{m}$ is stably rational over $k$. By Lemma 3.4, so is $B\mu_{n}$.

For more sophisticated applications of Lemma 3.4, see [CTS07, Section 4].

Remark 3.6*.*

Representations of connected groups which admit a dense open orbit have been studied by M. Sato and T. Kimura [SK77] (over ${\mathbb{C}}$). They referred to such representations as prehomogeneous vector spaces.

Lemma 3.7**.**

(cf. [Bog86, Corollary to Lemma 2.2])* Let $V$ be a linear representation of $G$, and consider $V$ as a vector group scheme over $k$. Then $B(V\rtimes G)$ is stably birational to $BG$.*

Proof.

Let $W$ be a generically free representation of $G$, and consider the $G$-representation $V_{0}:=\mathbb{A}^{1}\oplus V\oplus W$, where $G$ acts trivially on $\mathbb{A}^{1}$. We let the vector group scheme $V$ act linearly on $\mathbb{A}^{1}\oplus V$ by

[TABLE]

and trivially on $W$. This gives $V_{0}$ the structure of a $V\rtimes G$-representation. Since $G$ acts generically freely on $W$ and $V$ acts generically freely on $\mathbb{A}^{1}\oplus V$, $V_{0}$ is generically free as a $V\rtimes G$-representation. It suffices to show that $V_{0}/(V\rtimes G)$ is stably birational to $W/G$.

The projection map $\pi:V_{0}\to\mathbb{A}^{1}\oplus W$ is $V\rtimes G$-equivariant. Moreover, $V$ acts trivially on $\mathbb{A}^{1}\oplus W$ and simply transitively on the fibers of points in $(\mathbb{A}^{1}\setminus\{0\})\times W$. By 2.1, $\pi$ is the Rosenlicht quotient map for the $V$-action on $V_{0}$. Hence

[TABLE]

as desired. ∎

4. The Noether Problem for $\operatorname{SL}_{n}\rtimes\,({\mathbb{Z}}/{2}{\mathbb{Z}})$

Let $\mathbb{Z}/2\mathbb{Z}=\langle\tau\rangle$ be the cyclic group of order $2$. In this section we will study the Noether Problem for the group $\operatorname{SL}_{n}\rtimes(\mathbb{Z}/2{\mathbb{Z}})$, where $n\geqslant 1$ and $\tau$ acts on $\operatorname{SL}_{n}$ by $A\to(A^{-1})^{T}$. Our main result is as follows.

Proposition 4.1**.**

$B(\operatorname{SL}_{n}\rtimes(\mathbb{Z}/2{\mathbb{Z}}))$* is stably rational for every $n\geqslant 1$.*

In the sequel we will write $\operatorname{M}_{a\times b}$ for the space of rectangular matrices with $a$ rows and $b$ columns. Assume $n\geqslant 2$ and consider the linear representation of $\operatorname{SL}_{n}\rtimes(\mathbb{Z}/2{\mathbb{Z}})$ on $V=\operatorname{M}_{n\times(n-1)}\times\operatorname{M}_{(n-1)\times n}$ given by

[TABLE]

Here $X^{T}$ denotes the transpose of $X$ and similarly for $Y$. This action is well defined:

[TABLE]

for every $A\in\operatorname{SL}_{n}$, $X\in\operatorname{M}_{n\times(n-1)}$, and $Y\in\operatorname{M}_{(n-1)\times n}$. Set

[TABLE]

where $\pi(X,Y)=YX$. Clearly

[TABLE]

for any $A\in\operatorname{SL}_{n}$, $X\in\operatorname{M}_{n\times(n-1)}$ and $Y\in\operatorname{M}_{(n-1)\times n}$.

Lemma 4.2**.**

Suppose $\pi(X,Y)$ is a non-singular $(n-1)\times(n-1)$ matrix for some $n\geqslant 2$, $X\in\operatorname{M}_{n\times(n-1)}(k)$ and $Y\in\operatorname{M}_{(n-1)\times n}(k)$. Then

(a) the $\operatorname{SL}_{n}$-orbit of $(X,Y)$ in $V$ contains a point of the form $(J,Y^{\prime})$, where

[TABLE]

(b) The scheme-theoretic stabilizer $\operatorname{Stab}_{\operatorname{SL}_{n}}(X,Y)$ is trivial and $\pi^{-1}(\pi(X,Y))$ is a single $\operatorname{SL}_{n}$-orbit.

(c) $\pi$ is a Rosenlicht quotient map for the $\operatorname{SL}_{n}$-action on $V$. In particular, $\pi$ induces an isomorphism between $k(\operatorname{M}_{(n-1)\times(n-1)})$ and $k(V)^{\operatorname{SL}_{n}}$.

Proof.

(a) is a consequence of the fact that $\operatorname{SL}_{n}$ acts transitively on $(n-1)$-tuples of linearly independent vectors in $k^{n}$.

(b) In view of part (a), we may assume that $X=J$. If

[TABLE]

then

[TABLE]

Now suppose $(X^{\prime},Z)$ is another point in the fiber $\pi^{-1}(\pi(J,Y))$. We want to show that $(X^{\prime},Z)$ is an $\operatorname{SL}_{n}$-translate of $(J,Y)$. By part (a), we may assume without loss of generality that $X^{\prime}=J$. Since we are assuming that $\pi(J,Y)=\pi(J,Z)$, this tells us that

[TABLE]

for some $z_{1,n},\ldots,z_{n-1,n}\in k$.

We claim that the locus $\Lambda$ of solutions to the system $A\cdot(J,Y)=(J,Z)$, or equivalently

[TABLE]

is (scheme-theoretically) a single point $A\in\operatorname{SL}_{n}$. The fact that $\Lambda$ is non-empty implies that $(J,Y)$ and $(J,Z)$ lie in the same $\operatorname{SL}_{n}$-orbit. The fact that $\Lambda$ is a single point tells us that the scheme-theoretic stabilizer of $(J,Y)$ is trivial (just set $Z=Y$ in the claim).

It thus remains to prove the claim. Note that $AJ=J$ if and only if

[TABLE]

for some $(t_{1},\dots,t_{n-1})\in{\mathbb{A}}^{n-1}$. On the other hand, $ZA=Y$ translates to

[TABLE]

The matrix of this linear system is (4.3). This matrix is non-singular by our assumption. Hence, the system has a unique solution, $(t_{1},\dots,t_{n-1})$ in ${\mathbb{A}}^{n-1}$. This completes the proof of the claim and thus of part (b).

(c) In view of (4.1) and part (b), it suffices to show that $\pi$ is generically smooth; see 2.1. In other words, we need to check that the differential $d\pi\colon T_{v}(V)\to T_{\pi}(M_{(n-1)\times(n-1)})$ is surjective for $v\in V$ in general position. This is readily seen by restricting $\pi$ to the affine subspace $\{J\}\times\operatorname{M}_{(n-1)\times n}$ and using the formula (4.3). ∎

Proof of 4.1.

First let us settle the case, where $n=1$. Here $\operatorname{SL}_{n}\rtimes({\mathbb{Z}}/2{\mathbb{Z}})\simeq{\mathbb{Z}}/2{\mathbb{Z}}$. Examining the natural two-dimensional permutation representation $W$ of ${\mathbb{Z}}/2{\mathbb{Z}}\simeq\operatorname{S}_{2}$, we readily see that $k(W)^{{\mathbb{Z}}/2{\mathbb{Z}}}$ is rational over $k$. (Here $k$ is a field of arbitrary characteristic.) Consequently,

[TABLE]

Now suppose $n\geqslant 2$. Set $V=\operatorname{M}_{n\times(n-1)}\times\operatorname{M}_{(n-1)\times n}$ and consider the representation $V\times W$ of $\operatorname{SL}_{n}\rtimes({\mathbb{Z}}/2{\mathbb{Z}})$, where $\operatorname{SL}_{2}\rtimes({\mathbb{Z}}/2{\mathbb{Z}})$ acts on $W$ via $\mathbb{Z}/2\mathbb{Z}$. Since the $\operatorname{SL}_{n}$-action on $V$ is generically free (see 4.2(b)) and the ${\mathbb{Z}}/2{\mathbb{Z}}$-action on $W$ is generically free (obvious), we conclude that so is the $\operatorname{SL}_{n}\rtimes(\mathbb{Z}/2{\mathbb{Z}})$-action on $V\times W$. It remains to show that

[TABLE]

In view of 4.2(c), we have

[TABLE]

To see how ${\mathbb{Z}}/2{\mathbb{Z}}=\langle\tau\rangle$ acts on $V/\operatorname{SL}_{n}=\operatorname{M}_{(n-1)\times(n-1)}$, recall that $\tau$ sends $v=(X,Y)\in V$ to $(Y^{T},X^{T})$. Hence, the induced action of $\tau$ on $V/\operatorname{SL}_{n}=\operatorname{M}_{(n-1)\times(n-1)}$ takes $\pi(X,Y)=YX$ to $\pi(Y^{T},X^{T})=X^{T}Y^{T}=\pi(X,Y)^{T}$. In other words, the induced ${\mathbb{Z}}/2{\mathbb{Z}}$-action on $V/\operatorname{SL}_{n}=\operatorname{M}_{(n-1)\times(n-1)}$ is given by $\tau\colon Z\mapsto Z^{T}$. In particular, this action is linear.

Now (4.6), tells us that $k(\operatorname{M}_{(n-1)\times(n-1)}\times W)^{{\mathbb{Z}}/2{\mathbb{Z}}}$ is stably rational over $k$. This completes the proof of (4.7) and thus of 4.1. ∎

Remark 4.3*.*

For $n\geqslant 3$, the $\operatorname{SL}_{n}\rtimes({\mathbb{Z}}/2{\mathbb{Z}})$-action on $V$ is generically free, so we can work directly with $V$, rather than $V\times W$. The extra factor of $W$ is only needed when $n=2$. Note also that if $\operatorname{char}(k)\neq 2$, then ${\mathbb{Z}}/2{\mathbb{Z}}$ is isomorphic to $\mu_{2}$, and thus (4.6) is a special case of Example 3.5.

5. Group extensions

In the sequel we will apply 4.1 in combination with the following proposition.

Proposition 5.1**.**

Let $n$ be an odd integer and let $d\geq 1$. Consider a short exact sequence

[TABLE]

of algebraic groups. Then either $H\simeq\operatorname{SL}_{n}\times({\mathbb{Z}}/2{\mathbb{Z}})$ or $H\simeq\operatorname{SL}_{n}\rtimes({\mathbb{Z}}/2{\mathbb{Z}})$, where the generator $\tau$ of ${\mathbb{Z}}/2{\mathbb{Z}}$ acts by $\tau\colon A\mapsto(A^{-1})^{T}$, as in Section 4.

Proof.

Note that the fiber $\pi^{-1}(\tau)$ is an $\operatorname{SL}_{n}$-torsor. Since $\operatorname{SL}_{n}$ is a special group, this torsor is split. In other words, there exists an $x\in H(k)$ such that $\pi(x)=\tau$. Let $\varphi_{x}:\operatorname{SL}_{n}\to\operatorname{SL}_{n}$ denote conjugation by $x$: $\varphi_{x}(A)=xAx^{-1}$. Since $\varphi_{x}$ is a $k$-group automorphism of $\operatorname{SL}_{n}$, there exists $B\in\operatorname{SL}_{n}(k)$ such that either $\varphi_{x}(A)=BAB^{-1}$ for every $A\in\operatorname{SL}_{n}$, or $\varphi_{x}(A)=B(A^{-1})^{T}B^{-1}$. After replacing $x$ by $Bx$, we may assume that either $\varphi_{x}=\operatorname{Id}$ or $\varphi_{x}(A)=(A^{-1})^{T}$. It now suffices to show that there exists a $y\in H(k)$ such that $\pi(y)=\tau$ and $y^{2}=1$.

In both cases, $\varphi_{x^{2}}=(\varphi_{x})^{2}$ equals the identity, i.e., $x^{2}Ax^{-2}=A$ for every $A\in\operatorname{SL}_{n}({k})$. It follows $x^{2}$ lies in the center of $\operatorname{SL}_{n}$, i.e., $x^{2}\in\mu_{n}(k)\subset\operatorname{SL}_{n}(k)$ is a diagonal matrix.

Let $\left\langle{x}\right\rangle\subset H(k)$ be the subgroup generated by $x$. The restriction of $\pi$ to $\left\langle{x}\right\rangle$ is surjective and it sends $x^{m}$ to $\tau^{m}$. It follows that $\operatorname{Ker}(\pi)\cap\left\langle{x}\right\rangle=\left\langle{x^{2}}\right\rangle$, so that we have a short exact sequence

[TABLE]

The order of $\mu_{n}(k)$ divides $n$, hence $\mu_{n}(k)$ is cyclic of odd order. Since $\left\langle{x^{2}}\right\rangle$ is a subgroup of $\mu_{n}(k)$, it is also cyclic of odd order. On the other hand, since $\left\langle{x}\right\rangle$ surjects onto ${\mathbb{Z}}/2{\mathbb{Z}}$, $\left\langle{x}\right\rangle$ is of even order. We conclude that $\left\langle{x}\right\rangle$ contains an element $y$ of order $2$, and $\pi(y)=\tau$ as desired. ∎

6. The Noether Problem for $G_{2}$, $\operatorname{Spin}_{7}$ and $\operatorname{Spin}_{10}$

For the remainder of this paper we will assume that $k$ is a field of characteristic $\neq 2$. In particular, $\mu_{2}\simeq{\mathbb{Z}}/2{\mathbb{Z}}$ over $k$.

Proposition 6.1**.**

$BG_{2}$* is stably rational.*

Proof.

Let $V$ be the octonion representation of $G_{2}$. If we let $\mathbb{G}_{m}$ act on $V$ by scaling, the induced $G_{2}\times\mathbb{G}_{m}$-action on $V$ has an open orbit, and the stabilizer in general position is isomorphic to $\operatorname{SL}_{3}\rtimes\mu_{2}$, where $\mu_{2}$ acts on $\operatorname{SL}_{3}$ by $A\mapsto(A^{-1})^{T}$; see [Jac58, Theorem 4], [ST05, Proposition 3.3] or [PT19, Lemma 3.3].222In [PT19] it is assumed that $\sqrt{-1}\in k$. However, this assumption can be dropped if one works with the split form of $G_{2}$ throughout. By Lemma 3.4, $BG_{2}$ is stably birationally equivalent to $B(\operatorname{SL}_{3}\rtimes\mu_{2})$. By 4.1, $B(\operatorname{SL}_{3}\rtimes\mu_{2})$ is stably rational, and hence so is $BG_{2}$. ∎

For an alternative proof of Proposition 6.1, see [ZS18, Corollary 2, p. 568].

Proposition 6.2**.**

$B\operatorname{Spin}_{7}$* is stably rational.*

Proof.

Let $V$ be the spin representation of $\operatorname{Spin}_{7}$. Letting $\mathbb{G}_{m}$ act on $V$ by scalar multiplication, we obtain an action of $\operatorname{Spin}_{7}\times\mathbb{G}_{m}$ on $V$. If $\gamma$ is the generator of the center of $\operatorname{Spin}_{7}$, then $\gamma$ acts as $-\operatorname{Id}$ on $V$. It follows that the subgroup

[TABLE]

of $\operatorname{Spin}_{7}\times\mathbb{G}_{m}$ acts trivially on $V$. The quotient $(\operatorname{Spin}_{7}\times\mathbb{G}_{m})/C$ acts faithfully on $V$. This quotient group is usually called “the even Clifford group” and is denoted by $\Gamma_{7}^{+}$. The natural short exact sequence

[TABLE]

shows that $B\operatorname{Spin}_{7}$ and $B\Gamma_{7}^{+}$ are stably birational; see [Mer19, Lemma 4.3] or [Mer19, Corollary 4.4].

It remains to show that $B\Gamma_{7}^{+}$ is stably rational. By [Igu70, Proposition 4], there exists a quadratic form $g:V\to\mathbb{A}^{1}$ such that the orbits of the $\operatorname{Spin}_{7}$-action on $V$ are exactly the fibers of $g$. In particular, the $\Gamma_{7}^{+}$-action on $V$ has an open orbit. Furthermore, if $p\in V(k)$, $g(p)\neq 0$, and $S$ is the stabilizer of $p$ in $\operatorname{Spin}_{7}$, then

[TABLE]

see also [GG17, Table 1]. Note that Igusa only showed that $S\simeq G_{2}$ at the level of points. This settles (6.1) in characteristic [math]. To complete the proof of (6.1) in finite characteristic ($\neq 2$) it remains to show that $S$ is smooth, or equivalently, that $\operatorname{dim}\operatorname{Lie}(S)=\dim S$, where $\operatorname{Lie}(S)$ denotes the Lie algebra of $S$. This is done in [SK77, pp. 115-116], where it is shown that $\operatorname{Lie}(S)\simeq\mathfrak{g}_{2}$; see 6.3. 333This argument is implicit in [GG17]; we are grateful to S. Garibaldi for clarifying it to us.

Denote by $H$ the stabilizer of $p$ in $\operatorname{Spin}_{7}\times\mathbb{G}_{m}$. If $(h,t)\in{H}(\overline{k})$, then

[TABLE]

Since $g(p)\neq 0$, we deduce that $\lambda^{2}=1$. Thus the projection $\operatorname{Spin}_{7}\times{\mathbb{G}}_{m}\to{\mathbb{G}}_{m}$ to the second coordinate gives rise to a short exact sequence

[TABLE]

Since $S\simeq G_{2}$ has trivial center, $C$ intersects $S$ trivially. The inclusion $S\hookrightarrow{H}$ now induces an isomorphism between $S$ and $H/C$, which is the stabilizer of $p$ in $\Gamma_{7}^{+}$. By Lemma 3.4, $B\Gamma_{7}^{+}$ is stably birational to $B(H/C)=BS=BG_{2}$. By 6.1, $BG_{2}$ is stably rational; hence so is $B\Gamma_{7}^{+}$. ∎

Remark 6.3*.*

The base field in [SK77] is assumed to be ${\mathbb{C}}$. However, the Lie algebra calculation of [SK77, pp. 115-116] remains valid over any field $k$ of characteristic $\neq 2$. The same is true for the Lie algebra calculations from [SK77] which will be used in the proofs of Propositions 6.4 and 7.1.

Proposition 6.4**.**

$B\operatorname{Spin}_{10}$* is stably rational.*

Proof.

Let $V$ be the half-spin representation of $\operatorname{Spin}_{10}$. By [Igu70, Proposition 2] there are two non-zero orbits in $V$. Let $H$ be the stabilizer of a $k$-point in the open orbit. The subgroup $H$ is explicitly described in [Igu70, Lemma 3] (with $n=5$): we have $H=W\rtimes G_{0}$, where $W$ has the structure of an $8$-dimensional vector space and $G_{0}$ acts linearly on $W$. By [Igu70, Proposition 1], $G_{0}\simeq\operatorname{Spin}_{7}$. 444Here we could have cited [Igu70, Proposition 2] which asserts that $H=(\operatorname{Spin}_{7})\cdot(\mathbb{G}_{a})^{8}$. In [Igu70], the symbol $H=H_{1}\cdot H_{2}$ is used for the semidirect product $H=H_{2}\rtimes H_{1}$; see e.g. [Igu70, Lemma 1]. To avoid confusion, we spelled out the argument in detail. Note that $H$ is smooth by the Lie algebra computation in [SK77, p. 121]; once again, see 6.3. Hence $H\simeq W\rtimes\operatorname{Spin}_{7}$ as group schemes (and not just at the level of points).

By Lemma 3.4, $B\operatorname{Spin}_{10}$ is stably birational to $BH=B(W\rtimes\operatorname{Spin}_{7})$. By Lemma 3.7, $B(W\rtimes\operatorname{Spin}_{7})$ is stably birational to $B\operatorname{Spin}_{7}$, which is stably rational by 6.2. We conclude that $B\operatorname{Spin}_{10}$ is stably rational. ∎

7. The Noether Problem for $\operatorname{Spin}_{11}$

Proposition 7.1**.**

$B\operatorname{Spin}_{11}$* is stably rational.*

Our proof will follow the same pattern as the proof of stable rationality of $B\operatorname{Spin}_{7}$ in 6.2, except that the second half of the argument will be more involved.

Proof of 7.1.

Let $V$ be the spin representation of $\operatorname{Spin}_{11}$. Our starting point is the following result of J. Igusa [Igu70, Proposition 6].

(a) There exists a non-zero $\operatorname{Spin}_{11}$-invariant homogeneous form $J:V\to\mathbb{A}^{1}$ of degree $4$, such that the $\operatorname{Spin}_{11}$-orbits in $V\setminus\{0\}$ are the $J^{-1}(\lambda)$, $\lambda\in\mathbb{A}^{1}\setminus\{0\}$, together with four orbits inside $J^{-1}(0)$.

(b) If $-\lambda\in\mathbb{A}^{1}(k)$ is a non-zero square, the orbit $J^{-1}(\lambda)$ contains a rational point whose stabilizer is $k$-isomorphic to $\operatorname{SL}_{5}$. This is an isomorphism of group schemes; see the proof of [SK77, Proposition 39] or [GG17, Table 1].

Letting $\mathbb{G}_{m}$ act on $V$ by scalar multiplication, we obtain an action of $\operatorname{Spin}_{11}\times\mathbb{G}_{m}$ on $V$. If $\gamma$ is the generator of the center of $\operatorname{Spin}_{11}$, then $\gamma$ acts as $-\operatorname{Id}$ on $V$. It follows that the subgroup $C:=\left\langle{(\gamma,-1)}\right\rangle\simeq\mu_{2}$ of $\operatorname{Spin}_{11}\times\mathbb{G}_{m}$ acts trivially on $V$. The quotient $(\operatorname{Spin}_{11}\times\mathbb{G}_{m})/C$ acts faithfully on $V$. This quotient is usually called the even Clifford group and is denoted by $\Gamma_{11}^{+}$. Applying [Mer19, Lemma 4.3] to the short exact sequence

[TABLE]

we see that $B\operatorname{Spin}_{11}$ and $B\Gamma_{11}^{+}$ are stably birational; cf. [Mer19, Corollary 4.4]. It remains to show that $B\Gamma_{11}^{+}$ is stably birational.

By (a), $\operatorname{Spin}_{11}\times\mathbb{G}_{m}$ acts transitively on $U:=V\setminus J^{-1}(0)$. Let $p\in U(k)$, $H$ be the stabilizer of $p$ in $\operatorname{Spin}_{11}\times\mathbb{G}_{m}$, and $\overline{H}$ be the stabilizer of $p$ in $\Gamma_{11}^{+}=(\operatorname{Spin}_{11}\times\mathbb{G}_{m})/C$. Since $C$ acts trivially on $V$, $C$ is a central subgroup of $H$ and $\overline{H}=H/C$. Since $p$ belongs to the open orbit of the $\Gamma_{11}^{+}$-action, Lemma 3.4 tells us that $B\Gamma_{11}^{+}$ is stably birationally equivalent to $B\overline{H}$. Thus it suffices to show that $B\overline{H}$ is stably rational.

If $(h,\lambda)\in H(\overline{k})$, then

[TABLE]

Since $J(p)\neq 0$, we deduce that $\lambda^{4}=1$. This shows that $H\subseteq\operatorname{Spin}_{11}\times\mu_{4}$. The kernel of the projection $\pi:H\to\mu_{4}$ is the stabilizer of $p$ in $\operatorname{Spin}_{11}\times\{1\}$. By (b), this stabilizer is isomorphic to $\operatorname{SL}_{5}$. We thus obtain a short exact sequence

[TABLE]

where $\mu:=\operatorname{Im}(\pi)$. Note that $C\cap\operatorname{Spin}_{11}=1$ and thus $C\cap\operatorname{SL}_{5}=1$. Modding out by $C$, we obtain a short exact sequence

[TABLE]

Since $C\cap\operatorname{SL}_{5}=1$, we have $\pi(C)\simeq\mu_{2}$. If $\mu=\pi(C)$, then $\overline{H}\simeq\operatorname{SL}_{5}$ is special. In this case $B\overline{H}$ is stably rational by Lemma 3.1. Hence we may assume that $\mu=\mu_{4}$. In this case $\mu/\pi(C)\simeq\mu_{2}$ and our exact sequence reduces to

[TABLE]

By 5.1 either (i) $\overline{H}=\operatorname{SL}_{5}\times\mu_{2}$ or (ii) $\overline{H}=\operatorname{SL}_{5}\rtimes\mu_{2}$, where $\mu_{2}$ acts on $\operatorname{SL}_{5}$ by $A\mapsto(A^{-1})^{T}$.

In case (i), $B\overline{H}$ is stably birational to $B\mu_{2}$ by Lemma 3.3(c), and $B\mu_{2}$ is stably rational by Example 3.5. Thus $B\overline{H}$ is stably rational. In case (ii), $B\overline{H}$ is stably rational by 4.1. ∎

8. The Noether Problem for $\operatorname{Spin}_{14}$

Proposition 8.1**.**

$B\operatorname{Spin}_{14}$* is stably rational.*

Proof.

Let $V$ be the half-spin representation of $\operatorname{Spin}_{14}$, $v\in V$ be a $k$-point in general position, $S$ be the stabilizer of $v$, and $N$ be the normalizer of $S$. By [Gar09, Example 21.1],

[TABLE]

and the $\operatorname{Spin}_{14}$-orbit of $[v]$ is open in $\mathbb{P}(V)$; cf. also [GG17, §8]. The $\mu_{8}$-action on $G_{2}\times G_{2}$ factors through the surjection $\mu_{8}\to\mu_{2}$, where $\mu_{2}\simeq S_{2}$ acts on $G_{2}\times G_{2}$ by switching the two factors. Note that this action is well defined even if $k$ does not contain an $8$th root of unity. We will write $N\simeq G_{2}\wr\mu_{8}$, as in Section 3.

Letting $\mathbb{G}_{m}$ act on $V$ by scalar multiplication, we obtain an action of $\operatorname{Spin}_{14}\times\mathbb{G}_{m}$ on $V$. The orbit of $v$ under this action is open and dense in $V$. Let $H$ be the stabilizer of $v$ in $\operatorname{Spin}_{14}\times\mathbb{G}_{m}$. Consider the composition

[TABLE]

Note that $\varphi$ is injective. Indeed, its kernel, the stabilizer of $v$ in $\mathbb{G}_{m}$, is trivial. Thus $H\simeq\operatorname{Im}(\varphi)$. Moreover, clearly $S\times\{1\}\subset H$, and thus

[TABLE]

We claim that $\operatorname{Im}(\varphi)\subseteq N$. Here the inclusion should be understood scheme-theoretically. To prove this claim, let $R$ be a $k$-algebra and $g\in\operatorname{Spin}_{14}(R)$ be in the image of $\varphi$. Then $gv=\lambda v$ for some $\lambda\in R^{\times}$. For any $h\in S(R)$, we have

[TABLE]

This shows that $g^{-1}hg\in S(R)$ for any $h\in S(R)$. In other words, $g\in N(R)$, as claimed.

We have thus shown that $\operatorname{Im}(\varphi)$ is a subgroup of $N\simeq(G_{2}\times G_{2})\rtimes\mu_{8}$ containing $G_{2}\times G_{2}$. Thus $\operatorname{Im}(\varphi)$ is the preimage of a subgroup of $\mu_{8}$ under the natural projection $N\to N/(G_{2}\times G_{2})\simeq\mu_{8}$. We conclude that

[TABLE]

where $\mu_{m}$ is a subgroup of $\mu_{8}$, i.e., $m$ is a divisor of $8$. In particular, in view of our standing assumption that $\operatorname{char}(k)\neq 2$, this shows that $H$ is smooth. (As an aside, we remark that the semidirect product $(G_{2}\times G_{2})\rtimes\mu_{m}$ is direct if $m=1$, $2$ or $4$ and not direct if $m=8$.) To finish the proof, observe that,

(i) by 6.1, $BG_{2}$ is stably rational.

(ii) By Lemma 3.3(b), $B(G_{2}\wr\mu_{m})$ is stably birationally equivalent to $B\mu_{m}$. On the other hand, $B\mu_{m}$ is stably rational by Example 3.5.

(iii) By Lemma 3.4, applied to the representation of $\operatorname{Spin}_{14}\times\mathbb{G}_{m}$ on $V$, $B(\operatorname{Spin}_{14}\times\mathbb{G}_{m})$ is stably birationally equivalent to $BH$, where $H\simeq\operatorname{Im}(\varphi)\simeq G_{2}\wr\mu_{m}$. By (ii), $BH$ is stably rational, and hence, so is $B(\operatorname{Spin}_{14}\times\mathbb{G}_{m})$.

(iv) By Lemma 3.3(c), $B\operatorname{Spin}_{14}$ is stably birationally equivalent to $B(\operatorname{Spin}_{14}\times{\mathbb{G}}_{m})$. Thus $B\operatorname{Spin}_{14}$ is stably rational. ∎

Remark 8.2*.*

Assume that $-1$ is a square in $k$, and let $D_{2m+1},D^{\prime}_{2m+1}$ be the two non-isomorphic extraspecial $2$-groups of order $2^{2m+1}$. It is shown in [BB13] that $BD_{2m+1}$ and $BD^{\prime}_{2m+1}$ are stably birationally equivalent. By [Mer19, Corollary 6.2], $B\operatorname{Spin}_{n}$ is stably birational to $BD_{2m+1}$. Therefore, 1.1 has the following consequence: $BD_{2m+1}$ and $BD^{\prime}_{2m+1}$ are stably rational for any $m\leq 6$.

9. Coregular representations of spinor groups

In this section we present an alternative proof of 1.1 over $k={\mathbb{C}}$ suggested to us by G. Schwarz.

Let $G$ be a linear algebraic group over $k$. A linear representation $\rho\colon G\to\operatorname{GL}(V)$ is called coregular if $k[V]^{G}$ is a polynomial ring over $k$. Coregular representations of finite groups $G$, whose order is not divisible by $\operatorname{char}(k)$, are described by the celebrated theorem of Chevalley-Shephard-Todd: $\rho$ is coregular if and only if $\rho(G)$ is generated by pseudo-reflections. Now suppose that $G$ is a simple linear algebraic groups over $k=\mathbb{C}$. In this setting irreducible coregular representations were classified by V. Kac, V. Popov and E. Vinberg in [KPV76] and arbitrary (not necessarily irreducible) coregular representations by G. Schwarz in [Sch78] and (independently) by O. Adamovich and E. Golovina [AG83]. For an overview of this area of research we refer the readers to [PV94, §8]. We will not need the full classification here; we will only use the fact that certain specific representations of $\operatorname{Spin}_{n}$ ($n=7,10,11,14$) are coregular.

Coregular representations are related to the Noether Problem via the following simple observation.

Lemma 9.1**.**

Suppose a smooth algebraic $k$-group $G$ has no non-trivial characters (over $k$). If $G$ admits a generically free coregular representation over $k$, then $BG$ is stably rational over $k$.

Proof.

Let $\rho\colon G\to\operatorname{GL}(V)$ be a representation defined over $k$. Since $k[V]$ is a unique factorization domain, and $G$ has no non-trivial characters, $k(V)$ is the fraction field of $k[V]^{G}$; see [PV94, Theorem 3.3]. If $V$ is coregular, this shows that $k(V)^{G}$ is rational over $k$. If $V$ is also generically free, we conclude that $BG$ is stably rational. ∎

Now recall from the Introduction that in order to prove Theorem 1.1, it suffices to show that $B\operatorname{Spin}_{n}$ is stably rational for $n=7$, $10$, $11$ and $14$. Over the field $k=\mathbb{C}$ of complex numbers, Theorem 1.1 is now an immediate consequence of the following.

Proposition 9.2**.**

Let $k=\mathbb{C}$. Then $G:=\operatorname{Spin}_{n}$ admits a coregular generically free representation for $n=7,10,11,14$.

Proof.

Let $V_{n}$ be the natural representation of $\operatorname{Spin}_{n}$ (via the projection $\operatorname{Spin}_{n}\to\operatorname{SO}_{n}$), and $W_{n}$ be the spin representation. If $n$ is even, let $W_{n}^{1/2}$ be the half-spin representation. The following representations are shown to be coregular in [Sch78]:

(i) $V_{7}^{\oplus 3}\oplus W_{7}$, see Table 3a.9;

(ii) $V_{10}^{\oplus 5}\oplus W_{10}^{1/2}$, see Table 3a.20;

(iii) $V_{11}^{\oplus 4}\oplus W_{11}$, see Table 3a.25;

(iv) $V_{14}^{\oplus 3}\oplus W_{14}^{1/2}$, see Table 3a.31.

It thus remains to show that these representations are generically free.

(i) Let $w\in W_{7}$ be a point in general position. By [GG17, Table 1], the stabilizer of $w$ is isomorphic to $G_{2}$. We claim that

[TABLE]

Note that $\Lambda=\{x\in V_{7}^{\oplus 3}\oplus W_{7}\,|\,\operatorname{Stab}_{\operatorname{Spin}_{7}}(x)=1\}$ is a constructible subset of $V_{7}^{\oplus 3}\oplus W_{7}$. If the claim is established, then by the Fiber Dimension Theorem,

[TABLE]

Thus $\Lambda$ contains a dense open subset of $V_{7}^{3}\times W_{7}$, i.e., the $\operatorname{Spin}_{7}$-action on $V_{7}^{3}\times W_{7}$ is generically free.

It remains to prove the claim. Since $\operatorname{dim}(V_{7})=7$, the $G_{2}$-action on $V_{7}$ is the octonion representation of $G_{2}$. In other words, we may identify $V_{7}$ with the space of trace-zero octonions in such a way that $G_{2}$ acts via octonion automorphisms. If $x,y,z$ generate the octonions as a $\mathbb{C}$-algebra, then clearly the $G_{2}$-stabilizer of $(x,y,z)\in V_{7}^{\oplus 3}$ is trivial. The set $U\subset V_{7}^{\oplus 3}$ of generating triples is readily seen to be open. Note also that $U$ is non-empty, because $(i,j,k)\in U$, where $i$, $j$ and $k$ are the standard generators.

(ii) If $v:=(v_{1},v_{2},v_{3},v_{4},v_{5})\in V_{10}^{\oplus 5}$ is a general point, the stabilizer of $v$ is isomorphic to $\operatorname{Spin}_{5}$. Arguing as in (i), it remains to show that the restriction of $W_{10}^{1/2}$ to $\operatorname{Spin}_{5}$ is generically free. Recall that, for every $n\geq 1$,

• •

the restriction of the spin representation $W_{2n+1}$ to $\operatorname{Spin}_{2n}$ is $W_{2n}^{1/2}\oplus W_{2n}^{1/2}$, and

• •

the restriction of $W_{2n}^{1/2}$ to $\operatorname{Spin}_{2n-1}$ is the spin representation $W_{2n-1}$;

see [Igu70, p. 1000] or [Bum13, Ex. 31.2, Ex. 31.3]. Restricting $W_{10}^{1/2}$ from $\operatorname{Spin}_{10}$ to $\operatorname{Spin}_{9}$, then to $\operatorname{Spin}_{8}$, etc., we see that as a $\operatorname{Spin}_{5}$-representation, $W_{10}^{1/2}$ is isomorphic to $W_{5}^{\oplus 4}$. Note that $\dim(W_{5})=4$. There is an accidental isomorphism $\operatorname{Spin}_{5}\cong\operatorname{Sp}_{4}$, under which $W_{5}$ corresponds to the natural $4$-dimensional representation of $\operatorname{Sp}_{4}$; see [Ada96, Proposition 5.1]. Thus we can identify the $\operatorname{Spin}_{5}$-action on $W_{5}^{4}$ with the $\operatorname{Sp}_{4}$-action on $\operatorname{M}_{4\times 4}$ via left multiplication. The latter action is clearly generically free.

(iii) If $v:=(v_{1},v_{2},v_{3},v_{4})\in V_{11}^{\oplus 4}$ is a general point, the stabilizer of $v$ is isomorphic to $\operatorname{Spin}_{7}$. Once again, it suffices to show that when we restrict $W_{11}$ from $\operatorname{Spin}_{11}$ to $\operatorname{Spin}_{7}$, it becomes generically free. Arguing as in (ii) above, we see that as a $\operatorname{Spin}_{7}$-representation, $W_{11}$ is isomorphic to $W_{7}^{\oplus 4}$. If $w\in W_{7}$ is a general point, then the $\operatorname{Spin}_{7}$-stabilizer of $w$ is isomorphic to $G_{2}$; see (6.1). Thus it suffices to show that the $G_{2}$-action on $W_{7}^{\oplus 3}$ is generically free. As a $G_{2}$-reresentation, $W_{7}=\mathbb{C}\oplus V_{7}$, where $\mathbb{C}$ is the trivial $1$-dimensional representation generated by the vector stabilized by $G_{2}$, and $V_{7}$ is a $7$-dimensional representation, which can be identified with trace-zero octonions, as in (i). The desired conclusion now follows from (9.1).

(iv) If $w\in W_{14}^{1/2}$ is a general point, then by [GG17, Table 1] the stabilizer of $w$ in $\operatorname{Spin}_{14}$ is isomorphic to $G_{2}\times G_{2}$. It thus suffices to show that $G_{2}\times G_{2}$ acts generically freely on $V_{14}^{\oplus 3}$. Since $G_{2}\times G_{2}$ has trivial center, we may view it as a subgroup of $\operatorname{SO}(V_{14})$ via the projection $\operatorname{Spin}_{14}\to\operatorname{SO}(V_{14})$. By [GG17, §8], $V_{14}\cong V_{7}\oplus V_{7}$ as a $G_{2}\times G_{2}$-representation. Here $V_{7}$ is identified with trace-zero octonions, with the natural action of $G_{2}$, and $(g,g^{\prime})\in G_{2}\times G_{2}$ acts on $(v,v^{\prime})\in V_{7}\times V_{7}$ by $(v,v^{\prime})\mapsto(g(v),g^{\prime}(v^{\prime}))$. By (9.1), $G_{2}$ acts generically freely on $V_{7}^{3}$. Hence, $G_{2}\times G_{2}$ acts generically freely on $V_{14}^{\oplus 3}\simeq V_{7}^{\oplus 3}\oplus V_{7}^{\oplus 3}$, as desired. ∎

Acknowledgments

The authors are grateful to Gerald Schwarz for contributing Proposition 9.2, to Skip Garibaldi and Mattia Talpo for informative correspondence, and the anonymous referee for helpful comments.