2.1 The Clifford algebra

Let V be a finite-dimensional vector space over K endowed with a quadratic form q. The Clifford algebra $\operatorname{\mathrm{Cl}} (V,q)$ of V is the quotient of the tensor algebra $T(V) = \bigoplus _{d \geq 0 } V^{\otimes d}$ by the two-sided ideal generated by all elements

(2.1) $$ \begin{align} v \otimes v - q(v) \cdot 1, \: v \in V. \end{align} $$

This is also the two-sided ideal generated by

(2.2) $$ \begin{align} v \otimes w + w \otimes v - 2 (v|w) \cdot 1, \: v,w \in V, \end{align} $$

where $(\cdot |\cdot )$ denotes the bilinear form associated to q defined by $(v|w):=\frac {1}{2}(q(v+w)-q(v)-q(w))$ .

The Clifford algebra is a functor from the category of vector spaces equipped with a quadratic form to the category of (unital) associative algebras. That is, any linear map $\varphi \colon (V,q) \to (V',q')$ with $q'(\varphi (v))=q(v)$ for all $v \in V$ induces a homomorphism of associative algebras $\operatorname{\mathrm{Cl}} (\varphi ) \colon \operatorname{\mathrm{Cl}} (V,q) \to \operatorname{\mathrm{Cl}} (V',q')$ . If $\varphi $ is an inclusion $V \subseteq V'$ , then $\operatorname{\mathrm{Cl}} (\varphi )$ is injective, and hence, $\operatorname{\mathrm{Cl}} (V,q)$ is a subalgebra of $\operatorname{\mathrm{Cl}} (V',q')$ .

The decomposition of $T(V)$ into the even part $T^+(V):=\bigoplus _{d \text { even}} V^{\otimes d}$ and the odd part $T^-(V):=\bigoplus _{d \text { odd}} V^{\otimes d}$ induces a decomposition $\operatorname{\mathrm{Cl}} (V,q) = \operatorname{\mathrm{Cl}} ^+(V,q) \oplus \operatorname{\mathrm{Cl}} ^-(V,q)$ , turning $\operatorname{\mathrm{Cl}} (V,q)$ into a ${\mathbb {Z}}/2{\mathbb {Z}}$ -graded associative algebra. Note that, via the commutator on $\operatorname{\mathrm{Cl}} (V,q)$ , the even Clifford algebra $\operatorname{\mathrm{Cl}} ^+(V,q)$ is a Lie subalgebra of $\operatorname{\mathrm{Cl}} (V,q)$ .

The anti-automorphism of $T(V)$ determined by $v_1 \otimes \cdots \otimes v_d \mapsto v_d \otimes \cdots \otimes v_1$ preserves the ideal in the definition of $\operatorname{\mathrm{Cl}} (V,q)$ and therefore induces an anti-automorphism $x \mapsto x^*$ of $\operatorname{\mathrm{Cl}} (V,q)$ .

2.2 The Grassmann algebra as a $\operatorname{\mathrm{Cl}} (V)$ -module

From now on, we will write $\operatorname{\mathrm{Cl}} (V)$ for $\operatorname{\mathrm{Cl}} (V,q)$ when q is clear from the context. If $q=0$ , then $\operatorname{\mathrm{Cl}} (V) =\bigwedge \nolimits V$ , the Grassmann algebra of V. If $E \subseteq V$ is an isotropic subspace – that is, a subspace for which $q|_E=0$ – then this fact allows us to identify $\bigwedge \nolimits E$ with the subalgebra $\operatorname{\mathrm{Cl}} (E)$ of $\operatorname{\mathrm{Cl}} (V)$ .

For general q, $\operatorname{\mathrm{Cl}} (V)$ is not isomorphic as an algebra to $\bigwedge \nolimits V$ , but $\bigwedge \nolimits V$ is naturally a $\operatorname{\mathrm{Cl}} (V)$ -module as follows. For $v \in V$ , define $o(v):\bigwedge \nolimits V \to \bigwedge \nolimits V$ (the ‘outer product’) as the linear map

$$\begin{align*}o(v)\omega:=v \wedge \omega \end{align*}$$

and $\iota (v):\bigwedge \nolimits V \to \bigwedge \nolimits V$ (the ‘inner product’) as the linear map determined by

$$\begin{align*}\iota(v)w_1 \wedge \cdots \wedge w_k:=\sum_{i=1}^k (-1)^{i-1} (w_i\mid v) w_1 \wedge \cdots \wedge \widehat{w}_i \wedge \cdots \wedge w_k. \end{align*}$$

Here, and elsewhere in the paper, $\kern1.2pt\widehat {\cdot }$ indicates a factor that is left out. Now $v \mapsto \iota (v) + o(v)$ extends to an algebra homomorphism $\operatorname{\mathrm{Cl}} (V) \to \operatorname{\mathrm{End}} (\bigwedge \nolimits V)$ . To see this, it suffices to consider $v,w_1,\ldots ,w_k \in V$ and verify

$$\begin{align*}(\iota(v)+o(v))^2 w_1 \wedge \cdots \wedge w_k = (v|v) w_1 \wedge \cdots \wedge w_k. \end{align*}$$

We write $a \bullet \omega $ for the outcome of $a \in \operatorname{\mathrm{Cl}} (V)$ acting on $\omega \in \bigwedge \nolimits V$ . Using induction on the degree of a product, the linear map $\operatorname{\mathrm{Cl}} (V) \to \bigwedge \nolimits V, a \mapsto a \bullet 1$ is easily seen to be an isomorphism of vector spaces. In particular, $\operatorname{\mathrm{Cl}} (V)$ has dimension $2^{\dim V}$ .

2.3 Embedding $\mathfrak {so}(V)$ into the Clifford algebra

From now on, we assume that q is nondegenerate and write $\operatorname {\mathrm {SO}}(V)=\operatorname {\mathrm {SO}}(V,q)$ for the special orthogonal group of q. Its Lie algebra $\mathfrak {so}(V)$ consists of linear maps $V \to V$ that are skew-symmetric with respect to $(\cdot |\cdot )$ – that is, those $A \in \operatorname{\mathrm{End}} (V)$ such that $(Av|w) = -(v|Aw)$ for all $v, w \in V$ . We have a unique linear map $\psi : \bigwedge \nolimits ^2 V \to \operatorname{\mathrm{Cl}} ^+(V)$ with $\psi (u \wedge v)=uv-vu$ , and $\psi $ is injective. A straightforward computation shows that the image L of $\psi $ is closed under the commutator in $\operatorname{\mathrm{Cl}} (V)$ , and hence a Lie subalgebra. We claim that L is isomorphic to $\mathfrak {so}(V)$ . Indeed, for $u,v,w \in V$ , we have

$$\begin{align*}[\psi(u \wedge v),w] = [[u,v],w] = 4(v|w) u - 4(u|w)v. \end{align*}$$

We see, first, that $V \subseteq \operatorname{\mathrm{Cl}} (V)$ is preserved under the adjoint action of L; and second, that L acts on V via skew-symmetric linear maps, so that L maps into $\mathfrak {so}(V)$ . Since every map in $\mathfrak {so}(V)$ is a linear combination of the linear maps above, and since $\dim (L)=\dim (\mathfrak {so}(V))$ , the map $L \to \mathfrak {so}(V)$ is an isomorphism. We will identify $\mathfrak {so}(V)$ with the Lie subalgebra $L \subseteq \operatorname{\mathrm{Cl}} (V)$ via the inverse of this isomorphism, and we will identify $\bigwedge \nolimits ^2 V$ with $\mathfrak {so}(V)$ via the map $u\wedge v \mapsto ( w \mapsto (v|w) u - (u|w)v)$ . The concatenation of these identifications is the linear map $\frac {1}{4} \psi $ .

2.4 The half-spin representations

From now on, we assume that $\dim (V)=2n$ . We believe that all our results hold mutatis mutandis also in the odd-dimensional case, but we have not checked the details. A maximal isotropic subspace U of V is an isotropic subspace which is maximal with respect to inclusion. Since K is algebraically closed, q has maximal Witt index, so that every maximal isotropic subspace of V has dimension n.

The spin representation of $\mathfrak {so}(V)$ is constructed as follows. Let F be a maximal isotropic subspace of V and let $f_1, \ldots , f_n$ be a basis of F. Define $f:=f_1 \cdots f_n \in \operatorname{\mathrm{Cl}} (F)$ ; this element in $\operatorname{\mathrm{Cl}} (F)=\bigwedge \nolimits F$ is well defined up to a scalar. Then the left ideal $\operatorname{\mathrm{Cl}} (V) \cdot f$ is a left module for the associative algebra $\operatorname{\mathrm{Cl}} (V)$ , and hence for its Lie subalgebra $\mathfrak {so}(V)$ . This ideal is called the spin representation of $\mathfrak {so}(V)$ . As $\operatorname{\mathrm{Cl}} (V)$ is ${\mathbb {Z}}/2{\mathbb {Z}}$ -graded, the spin representation splits into a direct sum of two subrepresentations for $\operatorname{\mathrm{Cl}} ^+(V)$ , and hence for $\mathfrak {so}(V) \subseteq \operatorname{\mathrm{Cl}} ^+(V)$ – namely, $\operatorname{\mathrm{Cl}} ^+(V) \cdot f$ and $\operatorname{\mathrm{Cl}} ^-(V) \cdot f$ . These representations are called the half-spin representations of $\mathfrak {so}(V)$ .

2.5 Explicit formulas

We will need more explicit formulas for the action of $\mathfrak {so}(V)$ on the half-spin representations. To this end, let E be another isotropic n-dimensional subspace of V such that $V=E \oplus F$ . Then the map

$$\begin{align*}\bigwedge\nolimits E=\operatorname{\mathrm{Cl}}(E) \to \operatorname{\mathrm{Cl}}(V)f,\quad \omega \mapsto \omega f \end{align*}$$

is a linear isomorphism, and we use it to identify $\bigwedge \nolimits E$ with the spin representation. We write $\rho :\mathfrak {so}(V) \to \operatorname{\mathrm{End}} (\bigwedge \nolimits E)$ for the corresponding representation. It splits as a direct sum of the half-spin representations $\rho _+:\mathfrak {so}(V) \to \operatorname{\mathrm{End}} (\bigwedge \nolimits ^+ E)$ and $\rho _-:\mathfrak {so}(V) \to \operatorname{\mathrm{End}} (\bigwedge \nolimits ^- E)$ , where $\bigwedge \nolimits ^+E=\bigoplus _{d \text { even}} \bigwedge \nolimits ^d E$ and $\bigwedge \nolimits ^-E=\bigoplus _{d \text { odd}} \bigwedge \nolimits ^d E$ .

In this model of the spin representation, the action of $v \in E \subseteq \operatorname{\mathrm{Cl}} (V)$ on the spin representation $\bigwedge \nolimits E$ is just the outer product on $\bigwedge E: o(v):\bigwedge \nolimits E \to \bigwedge \nolimits E,\ \omega \mapsto v \wedge \omega $ , while the action of $v \in F \subseteq \operatorname{\mathrm{Cl}} (V)$ is twice the inner product on $\bigwedge E$ :

$$\begin{align*}2 \iota(v) w_1 \wedge \cdots \wedge w_k = 2 \sum_{i=1}^k (-1)^{i-1}(v|w_i) w_1 \wedge \cdots \wedge \widehat{w_i} \wedge \cdots \wedge w_k. \end{align*}$$

The factor $2$ and the alternating signs come from the following identity in $\operatorname{\mathrm{Cl}} (V)$ :

$$\begin{align*}v v_i=2 (v|v_i) - v_i v \text{ for } v \in F \text{ and } v_i \in E. \end{align*}$$

For a general $v \in V$ , we write $v = v' + v"$ with $v' \in E$ , $v" \in F$ . Then the action of V on $\bigwedge \nolimits E$ is given by

$$\begin{align*}v \mapsto o(v') + 2 \iota(v"). \end{align*}$$

We now compute the linear maps by means of which $\mathfrak {so}(V)$ acts on $\bigwedge \nolimits E$ . To this end, recall that a pair $e,f \in V$ is called hyperbolic if e, f are isotropic and $(e|f) = 1$ . Given the basis $f_1,\ldots ,f_n$ of F, there is a unique basis $e_1,\ldots ,e_n$ of E so that $(e_i|f_j)=\delta _{ij}$ ; then $e_1,\ldots ,e_n,f_1,\ldots ,f_n$ is called a hyperbolic basis of V. Now the element $e_i \wedge e_j \in \mathfrak {so}(V)$ acts on $\bigwedge \nolimits E \simeq \operatorname{\mathrm{Cl}} (V)f$ via the linear map

$$\begin{align*}\frac{1}{4}(o(e_i)o(e_j)-o(e_j)o(e_i))=\frac{1}{2} o(e_i)o(e_j); \end{align*}$$

the element $f_i \wedge f_j$ acts via the linear map

$$\begin{align*}\frac{1}{4}(4\iota(f_i) \iota(f_j)-4\iota(f_j)\iota(f_i))=2 \iota(f_i) \iota(f_j); \end{align*}$$

and the element $e_i \wedge f_j$ acts via the linear map

$$\begin{align*}\frac{1}{4}(o(e_i) 2 \iota(f_j)- 2 \iota(f_j) o(e_i))=\frac{1}{2} (o(e_i) \iota(f_j)- \iota(f_j) o(e_i)). \end{align*}$$

In particular, $\omega _0:=e_1 \wedge \cdots \wedge e_n \in \bigwedge \nolimits E$ is mapped to $0$ by all elements $e_i \wedge e_j$ and all elements $e_i \wedge f_j$ with $i \neq j$ , and it is mapped to $\frac {1}{2} \omega _0$ by all $e_i \wedge f_i$ .

2.6 Highest weights of the half-spin representations

Recall, for example, from [Reference Jacobson8, Chapter IV, pages 140–141], that in the basis $e_1,\ldots ,e_n,f_1,\ldots ,f_n$ , matrices in $\mathfrak {so}(V)$ have the form

$$\begin{align*}\begin{bmatrix} A&B \\ C&-A^T \end{bmatrix} \text{ with } B^T=-B, \text{ and } C^T=-C. \end{align*}$$

Here, the $(e_i,e_j)$ -entry of A is the coefficient of $e_i \wedge f_j$ , the $(e_i,f_j)$ -entry of B is the coefficient of $e_i \wedge e_j$ , and the $(f_i,e_j)$ -entry of C is the coefficient of $f_i \wedge f_j$ .

The diagonal matrices $e_i \wedge f_i$ span a Cartan subalgebra of $\mathfrak {so}(V)$ with standard basis consisting of $h_i:=e_i \wedge f_i - e_{i+1} \wedge f_{i+1}$ for $i=1,\ldots ,n-1$ and $h_n:=e_{n-1} \wedge f_{n-1} + e_n \wedge f_n$ (this last element is forgotten in the basis of the Cartan algebra on [Reference Jacobson8, page 141]).

Now $(e_i \wedge e_j) \omega _0=(e_i \wedge f_j) \omega _0=0$ for all $i \neq j$ . Furthermore, the elements $h_1,\ldots ,h_{n-1}$ map $\omega _0$ to $0$ , while $h_n$ maps $\omega _0$ to $\omega _0$ . Thus, the Borel subalgebra maps the line $K \omega _0$ into itself, and $\omega _0$ is a highest weight vector of the fundamental weight $\lambda _0:=(0,\ldots ,0,1)$ relative to the standard basis. Summarising, $\omega _0 \in \bigwedge \nolimits E$ generates a copy of the irreducible $\mathfrak {so}(V)$ -module $V_{\lambda _0}$ with highest weight $\lambda _0$ . Clearly, the $\mathfrak {so}(V)$ -module generated by $\omega _0$ is contained in $\bigwedge \nolimits ^+ E$ if n is even, and contained in $\bigwedge \nolimits ^- E$ when n is odd. One can also show that both half-spin representations are irreducible; hence, one of them is a copy of $V_{\lambda _0}$ . For the other half-spin representation, consider the element

$$\begin{align*}\omega_1:=e_1 \wedge \cdots \wedge e_{n-1} \in \bigwedge\nolimits E. \end{align*}$$

This element is mapped to zero by $e_i \wedge e_j$ for all $i \neq j$ and by $e_i \wedge f_j$ for all $i < j$ . It is further mapped to $0$ by $h_1,\ldots ,h_{n-2},h_n$ , and to $\omega _1$ by $h_{n-1}$ . For example, we have

$$ \begin{align*} h_n \omega_1 &= \frac{1}{2}\left(o(e_{n-1})\iota(f_{n-1})-\iota(f_{n-1})o(e_{n-1})+o(e_n) \iota(f_n)-\iota(f_n) o(e_n)\right) e_1 \wedge \cdots \wedge e_{n-1} \\ &=\frac{1}{2}(1 - 0 + 0 - 1) \omega_1 = 0, \text{ and similarly}\\ h_{n-1} \omega_1&=\frac{1}{2} (1-0-0+1) \omega_1=\omega_1. \end{align*} $$

Hence, $\omega _1$ generates a copy of $V_{\lambda _1}$ , the irreducible $\mathfrak {so}(V)$ -module of highest weight $\lambda _1 := (0,\ldots ,0,1,0)$ ; this is the other half-spin representation.

2.7 The spin group

Let $\rho : \mathfrak {so}(V) \to \operatorname{\mathrm{End}} (\bigwedge \nolimits E)$ be the spin representation. The spin group $\operatorname {\mathrm {Spin}}(V)$ can be defined as the subgroup of $\operatorname{\mathrm{GL}} (\bigwedge \nolimits E)$ generated by the one-parameter subgroups $t \mapsto \exp (t \rho (X))$ , where X runs over the root vectors $e_i \wedge e_j, f_i \wedge f_j$ and $e_i \wedge f_j$ with $i \neq j$ . Note that $\rho (X)$ is nilpotent for each of these root vectors, so that $t \mapsto \exp (t \rho (X))$ is an algebraic group homomorphism $K \to \operatorname{\mathrm{GL}} (\bigwedge \nolimits E)$ . It is a standard fact that the subgroup generated by irreducible curves through the identity in an algebraic group is itself a connected algebraic group; see [Reference Borel3, Proposition 2.2]. So $\operatorname {\mathrm {Spin}}(V)$ is a connected algebraic group, and one verifies that its Lie algebra is isomorphic to the Lie algebra generated by the root vectors X (i.e., to $\mathfrak {so}(V)$ ).

By construction, the (half-)spin representations $\bigwedge \nolimits E$ , $\bigwedge \nolimits ^+ E$ and $\bigwedge \nolimits ^- E$ are representations of $\operatorname {\mathrm {Spin}}(V)$ . We use the same notation $ \rho : \operatorname {\mathrm {Spin}}(V) \to \operatorname{\mathrm{GL}} (\bigwedge \nolimits E), \ \rho _+ \colon \operatorname {\mathrm {Spin}}(V) \to \operatorname{\mathrm{GL}} (\bigwedge \nolimits ^+ E), \text { and } \rho _- \colon \operatorname {\mathrm {Spin}}(V) \to \operatorname{\mathrm{GL}} (\bigwedge \nolimits ^- E) $ for these as for the corresponding Lie algebra representations.

Remark 2.1. The algebraic group $\operatorname {\mathrm {Spin}}(V)$ is usually constructed as a subgroup of the unit group $\operatorname{\mathrm{Cl}} ^{\times }(V)$ as follows: consider first

$$\begin{align*}\Gamma(V) = \{x\in \operatorname{\mathrm{Cl}}^{\times}(V) \mid xVx^{-1} = V\}, \end{align*}$$

sometimes called the Clifford group. Then $\operatorname {\mathrm {Spin}}(V)$ is the subgroup of $\Gamma (V)$ of elements of spinor norm $1$ ; that is, $x x^* = 1$ , where $x^*$ denotes the involution defined in Section 2.1. In this model of the spin group, it is easy to see that it admits a $2:1$ covering $\operatorname {\mathrm {Spin}}(V) \to \operatorname {\mathrm {SO}}(V)$ – namely, the restriction of the homomorphism $\Gamma (V) \to \mathrm {O}(V)$ given by associating to $x \in \Gamma (V)$ the orthogonal transformation $w \mapsto x w x^{-1}$ . For more details, see [Reference Procesi13]. Since our later computations involve the Lie algebra $\mathfrak {so}(V)$ only, the definition of $\operatorname {\mathrm {Spin}}(V)$ above suffices for our purposes.

The half-spin representations are not representations of the group $\operatorname {\mathrm {SO}}(V)$ ; this can be checked, for example, by showing that the highest weights $\lambda _0$ and $\lambda _1$ are not in the weight lattice of $\operatorname {\mathrm {SO}}(V)$ .

2.8 Two actions of $\mathfrak{gl}(E) $ on $\bigwedge\nolimits E $

The definition of the (half-)spin representation(s) of $\mathfrak {so}(V)$ and $\operatorname {\mathrm {Spin}}(V)$ as $\operatorname{\mathrm{Cl}} ^{(\pm )}(V) f$ involves only the quadratic form q and the choice of a maximal isotropic space $F \subseteq V$ . Consequently, any linear automorphism of V that preserves q and maps F into itself also acts on $\operatorname{\mathrm{Cl}} ^{(\pm )}(V) f$ . These linear automorphisms form the stabiliser of F in $\operatorname {\mathrm {SO}}(V)$ , which is the parabolic subgroup whose Lie algebra consists of the matrices in $\operatorname {\mathrm {SO}}(V)$ that are block lower triangular in the basis $e_1,\ldots ,e_n,f_1,\ldots ,f_n$ . So, while $\operatorname {\mathrm {SO}}(V)$ does not act naturally on the (half-)spin representation(s), this stabiliser does.

In particular, in our model $\bigwedge \nolimits ^{(\pm )} E$ of the (half-)spin representation(s), the group $\operatorname{\mathrm{GL}} (E)$ , embedded into $\operatorname {\mathrm {SO}}(V)$ as the subgroup of block diagonal matrices

$$\begin{align*}\begin{bmatrix} a & 0 \\ 0 & -a^{T} \end{bmatrix} \end{align*}$$

acts on $\bigwedge \nolimits E$ in the natural manner. We stress that this is not the action obtained by integrating the action of $\mathfrak {gl}(E) \subseteq \mathfrak {so}(V)$ on $\bigwedge \nolimits E$ regarded as the spin representation. Indeed, the standard action of $e_i \wedge f_j \in \mathfrak {gl}(E)$ on $\omega :=e_{i_1} \wedge \cdots \wedge e_{i_k} \in \bigwedge \nolimits ^k E$ yields

$$\begin{align*}\sum_{l=1}^k e_{i_1} \wedge \cdots \wedge e_i (f_j|e_{i_l}) \wedge \cdots \wedge e_{i_k} = \begin{cases} 0 \text{ if } j \not \in \{i_1,\ldots,i_k\}\\ (-1)^{l-1} e_i \wedge e_{i_1} \wedge \cdots \wedge \widehat{e_{i_l}} \wedge \cdots \wedge e_{i_k} \text{ if } j=i_l. \end{cases} \end{align*}$$

However, in the spin representation, the action is given by the linear map $\frac {1}{2}(o(e_i) \iota (f_j) - \iota (f_j) o(e_i))$ . If $j \neq i$ and $j \not \in \{i_1,\ldots ,i_k\}$ , then

$$\begin{align*}o(e_i) \iota(f_j) \omega = \iota(f_j) o(e_i) \omega = 0. \end{align*}$$

If $j \neq i$ and $j=i_l$ , then

$$\begin{align*}o(e_i) \iota(f_j) \omega = (-1)^{l-1} e_i \wedge e_{i_1} \wedge \cdots \wedge \widehat{e_{i_l}} \wedge \cdots \wedge e_{i_k} = - \iota(f_j) o(e_i) \omega. \end{align*}$$

We conclude that for $i \neq j$ , the action of $e_i \wedge f_j$ is the same in both representations. However, if $i=j$ , then

$$\begin{align*}\frac{1}{2}(o(e_i) \iota(f_i) - \iota(f_i)o(e_i)) \omega = \begin{cases} -\frac{1}{2} \omega \text{ if } i \not \in \{i_1,\ldots,i_k\}, \text{ and}\\ \frac{1}{2} (-1)^{l-1} e_i \wedge e_{i_1} \wedge \cdots \wedge \widehat{e_{i_l}} \wedge \cdots \wedge e_{i_k} = \frac{1}{2} \omega \text{ if } i=i_l. \end{cases} \end{align*}$$

We conclude that if $\tilde {\rho }:\mathfrak {gl}(E) \to \operatorname{\mathrm{End}} (\bigwedge \nolimits E)$ is the standard representation of $\mathfrak {gl}(E)$ , then the restriction of the spin representation $\rho :\mathfrak {so}(V) \to \operatorname{\mathrm{End}} (\bigwedge \nolimits E)$ to $\mathfrak {gl}(E)$ as a subalgebra of $\mathfrak {so}(V)$ satisfies

(2.3) $$ \begin{align} \rho(A)=\tilde{\rho}(A) - \frac{1}{2} \operatorname{tr}(A) \operatorname{\mathrm{Id}}_{\bigwedge\nolimits E}. \end{align} $$

At the group level, this is to be understood as follows. The pre-image of $\operatorname{\mathrm{GL}} (E) \subseteq \operatorname {\mathrm {SO}}(V)$ in $\operatorname {\mathrm {Spin}}(V)$ is isomorphic to the connected algebraic group

$$\begin{align*}H:=\big\{(g,t) \in \operatorname{\mathrm{GL}}(E) \times K^* \mid \det(g)=t^2\big\} \end{align*}$$

for which $(g,t) \mapsto g$ is a $2:1$ cover of $\operatorname{\mathrm{GL}} (E)$ , and the restriction of $\rho $ to H satisfies $\rho (g,t)=\tilde {\rho }(g) \cdot t^{-1}$ – a ‘twist of the standard representation by the inverse square root of the determinant’.

3.1 The isotropic Grassmannian in its spinor embedding

As before, let V be a $2n$ -dimensional vector space over K endowed with a nondegenerate quadratic form. The (maximal) isotropic Grassmannian $\operatorname{\mathrm{Gr}} _{\operatorname {\mathrm {iso}}}(V,q)$ parametrizes all maximal isotropic subspaces of V. It has two connected components, denoted $\operatorname{\mathrm{Gr}} ^+_{\operatorname {\mathrm {iso}}}(V)$ and $\operatorname{\mathrm{Gr}} ^-_{\operatorname {\mathrm {iso}}}(V)$ . The goal of this subsection is to introduce the isotropic Grassmann cone, which is an affine cone over $\operatorname{\mathrm{Gr}} _{\operatorname {\mathrm {iso}}}(V,q)$ in the spin representation.

Fix a maximal isotropic subspace $F \subseteq V$ and as before, set , where $f_1, \dots , f_n$ is any basis of F. Now let $H \subseteq V$ be another maximal isotropic space. Then we claim that the space

(3.1) $$ \begin{align} S_H := \{\omega \in \operatorname{\mathrm{Cl}}(V)f \mid v \cdot \omega = 0 \text{ for all } v \in H\} \subseteq \operatorname{\mathrm{Cl}}(V)f \end{align} $$

is $1$ -dimensional. Indeed, we may find a hyperbolic basis $e_1,\ldots ,e_n,f_1,\ldots ,f_n$ of V such that $f_1,\ldots ,f_k$ span $H \cap F$ , $f_1,\ldots ,f_n$ span F, and $e_{k+1},\ldots ,e_n,f_{1},\ldots ,f_k$ span H. We call this hyperbolic basis adapted to H and F. Then the element

$$\begin{align*}\omega_H:=e_{k+1} \cdots e_n f_1 \cdots f_k f_{k+1} \cdots f_n \in \operatorname{\mathrm{Cl}}(V)f \end{align*}$$

lies in $S_H$ since $e_i \omega _H=f_j \omega _H = 0$ for all $i> k$ and $j \leq k$ . Conversely, if $\mu \in S_H$ , then write

$$\begin{align*}\mu=\sum_{l=0}^n \sum_{i_1<\ldots<i_l} c_{\{i_1,\ldots,i_l\}} e_{i_1} \cdots e_{i_l} f. \end{align*}$$

If $c_I \neq 0$ for some I with $I \not \supseteq \{k+1,\ldots ,n\}$ , then for any $j \in \{k+1,\ldots ,n\} \setminus I$ , we find that $e_j \mu \neq 0$ . So all I with $c_I \neq 0$ contain $\{k+1,\ldots ,n\}$ . If some I with $c_I \neq 0$ further contains an $i \leq k$ , then $f_i \mu $ is nonzero. Hence, $S_H$ is spanned by $\omega _H$ , as claimed. In what follows, by slight abuse of notation, we will write $\omega _H$ for any nonzero vector in $S_H$ .

The space H can be uniquely recovered from $\omega _H$ via

$$\begin{align*}H=\{v \in V \mid v \cdot \omega_H = 0 \}. \end{align*}$$

Indeed, we have already seen $\subseteq $ . For the converse, observe that the vectors $e_i \omega _H, f_j \omega _H$ with $i \leq k$ and $j> k$ are linearly independent.

The map that sends $H \in \operatorname{\mathrm{Gr}} _{\operatorname {\mathrm {iso}}}(V,q)$ to the projective point representing it, that is,

$$\begin{align*}H \mapsto [\omega_H] \in {\mathbb{P}}(\operatorname{\mathrm{Cl}}(V)f), \end{align*}$$

is therefore injective, and it is called the spinor embedding of the isotropic Grassmannian (see [Reference Manivel11]). The isotropic Grassmann cone is defined as

where the union is taken over all maximal isotropic subspaces $H\subseteq V$ . We denote by the cones over the connected components of the isotropic Grassmannian in its spinor embedding.

3.2 Contraction with an isotropic vector

Let $e \in V$ be a nonzero isotropic vector. Then $V_e:=e^\perp / \langle e \rangle $ is equipped with a natural nondegenerate quadratic form, and there is a rational map $\operatorname{\mathrm{Gr}} _{\operatorname {\mathrm {iso}}}(V) \to \operatorname{\mathrm{Gr}} _{\operatorname {\mathrm {iso}}}(V_e)$ that maps an n-dimensional isotropic space H to the image in $V_e$ of the $(n-1)$ -dimensional isotropic space $H \cap e^\perp $ (this is defined if $e \not \in H$ , which by maximality of H is equivalent to $H \not \subseteq e^\perp $ ). This map is the restriction to $\operatorname{\mathrm{Gr}} _{\operatorname {\mathrm {iso}}}(V)$ of the rational map $\mathbb {P}(\bigwedge \nolimits ^n V) \to \mathbb {P}(\bigwedge \nolimits ^{n-1} V_e)$ induced by the linear map (‘contraction with e’):

$$\begin{align*}c_e: \bigwedge\nolimits^n V \to \bigwedge\nolimits^{n-1} V_e, \quad v_1 \wedge \cdots \wedge v_n \mapsto \sum_{i=1}^n (-1)^{i-1} (e|v_i) \overline{v_1} \wedge \cdots \wedge \widehat{v_i} \wedge \cdots \wedge \overline{v_n}, \end{align*}$$

where $\overline {v_i}$ is the image of $v_i$ in $V/\langle e \rangle $ . Note first that this map is the inner product $\iota (e)$ followed by a projection. Furthermore, a priori, the codomain of this map is the larger space $\bigwedge \nolimits ^{n-1} (V/\langle e \rangle )$ , but one may choose $v_1,\ldots ,v_n$ such that $(e|v_i)=0$ for $i>1$ , and then it is evident that the image is indeed in $\bigwedge \nolimits ^{n-1} V_e$ .

We want to construct a similar contraction map at the level of the spin representation. For reasons that will become clear in a moment, we restrict our attention first to a map between two half-spin representations, as follows. Assume that $e \notin F$ , and choose a basis $f_1,\ldots ,f_n$ of F such that $(e|f_i)=\delta _{in}$ . As usual, write $f:=f_1 \cdots f_n$ , and write $\overline {f}:=\overline {f}_1 \cdots \overline {f}_{n-1}$ , so that $\operatorname{\mathrm{Cl}} ^+(V_e)\overline {f}$ is a half-spin representation of $\mathfrak {so}(V_e)$ .

Then we define the map

$$\begin{align*}\pi_e:\operatorname{\mathrm{Cl}}^+(V)f \to \operatorname{\mathrm{Cl}}^+(V_e)\overline{f}, \quad \pi_e(a f):=\text{ the image of } \frac{1}{2}((-1)^{n-1} eaf+afe) \text{ in } \operatorname{\mathrm{Cl}}(V_e)\overline{f}, \end{align*}$$

where the implicit claim is that the expression on the right lies in $\operatorname{\mathrm{Cl}} (e^\perp )f_1\cdots f_{n-1}$ , so that its image in $\operatorname{\mathrm{Cl}} (V_e) \overline {f}$ is well defined (note that the projection $e^\perp \to V_e$ induces a homomorphism of Clifford algebras), and that this image lies in the left ideal generated by $\overline {f}$ . To verify this claim, and to derive a more explicit formula for the map above, let $e_1,\ldots ,e_{n}=e$ be a basis of an isotropic space E complementary to F. Then it suffices to consider the case where $a=e_{i_1} \cdots e_{i_k}$ for some $i_1<\ldots <i_k$ . We then have

$$ \begin{align*} eaf&=ee_{i_1} \cdots e_{i_k} f_1\cdots f_n\\ &=\begin{cases} 0 \text{ if } i_k=n, \text{ and}\\ 2 (-1)^{k+n-1}e_{i_1} \cdots e_{i_k} f_1 \cdots f_{n-1} +(-1)^{k+n}e_{i_1} \cdots e_{i_k} f_1 \cdots f_n e \text{ otherwise.} \end{cases} \end{align*} $$

Multiplying by $(-1)^{n-1}$ and using that k is even, the latter expression becomes

$$\begin{align*}2 e_{i_1} \cdots e_{i_k} f_1 \cdots f_{n-1} - a f e. \end{align*}$$

Hence, we conclude that

$$\begin{align*}\pi_e(e_{i_1} \cdots e_{i_k} f)= \begin{cases} 0 & \text{if } i_k=n, \text{and}\\ \overline{e}_{i_1} \cdots \overline{e}_{i_k} \overline{f} & \text{otherwise.} \end{cases} \end{align*}$$

In short, in our models $\bigwedge \nolimits ^+ E$ and $\bigwedge \nolimits ^+(E/\langle e \rangle )$ for the half-spin representations of $\mathfrak {so}(V)$ and $\mathfrak {so}(V_e)$ , $\pi _e$ is just the reduction-mod-e map. We leave it to the reader to check that the reduction-mod-e map $\bigwedge \nolimits ^- E \to \bigwedge \nolimits ^-(E/\langle e \rangle )$ arises in a similar fashion from the map

$$\begin{align*}\pi_e:\operatorname{\mathrm{Cl}}(V)^-f \to \operatorname{\mathrm{Cl}}(V_e)^-\overline{f}, \quad \pi_e(a f):=\text{ the image of } \frac{1}{2}((-1)^{n} eaf+afe) \text{ in } \operatorname{\mathrm{Cl}}(V_e)\overline{f}. \end{align*}$$

We will informally call the maps $\pi _e$ ‘contraction with e’. Together, they define a map on $\operatorname{\mathrm{Cl}} (V)f$ which we also denote by $\pi _e$ .

Proof. Let $v \in e^\perp $ and consider $a \in \operatorname{\mathrm{Cl}} ^-(V)$ . Then $va \in \operatorname{\mathrm{Cl}} ^+(V)$ , and hence, $\pi _e(vaf)$ is the image in $\operatorname{\mathrm{Cl}} (V_e)\overline {f}$ of

$$\begin{align*}\frac{1}{2}((-1)^{n-1} evaf + vafe) = \frac{1}{2}((-1)^{n} veaf + vafe) = v \frac{1}{2}((-1)^n eaf + afe), \end{align*}$$

where we have used $(v|e)=0$ in the first equality. The right-hand side clearly equals $\overline {v}$ times the image of $\pi _e(af)$ in $\operatorname{\mathrm{Cl}} (V_e) \overline {f}$ .

3.3 Multiplying with an isotropic vector

In a sense dual to the contraction maps, $c_e:\bigwedge \nolimits ^n V \to \bigwedge \nolimits ^{n-1} V_e$ are multiplication maps defined as follows. Let $e,h \in V$ be isotropic with $(e|h)=1$ ; such a pair is called a hyperbolic pair. We then have $V=\langle e,h \rangle \oplus \langle e,h \rangle ^\perp $ , and the map from the second summand to $V_e=e^\perp /\langle e \rangle $ is an isometry. We use this isometry to identify $V_e$ with the subspace $\langle e,h \rangle ^\perp $ of V and write $s_e$ for the corresponding inclusion map. Then we define

$$\begin{align*}m_h:\bigwedge\nolimits^{n-1} V_e \to \bigwedge\nolimits^n V, \quad \overline{v}_1 \wedge \cdots \wedge \overline{v}_{n-1} \mapsto h \wedge \overline{v}_1 \wedge \cdots \wedge \overline{v}_{n-1}, \end{align*}$$

which is just the outer product $o(h)$ . The projectivisation of this map sends $\operatorname{\mathrm{Gr}} _{\operatorname {\mathrm {iso}}}(V_e)$ isomorphically to the closed subset of $\operatorname{\mathrm{Gr}} _{\operatorname {\mathrm {iso}}}(V)$ consisting of all H containing h. We further observe that

$$\begin{align*}c_e \circ m_h=\operatorname{\mathrm{id}}_{\bigwedge\nolimits^{n-1} V_e}. \end{align*}$$

We define a corresponding multiplication map at the level of spin representations as follows: first, we assume that $h \in F$ , and choose a basis $f_1,\ldots ,f_n=h$ of F such that $(e|f_i)=\delta _{in}$ . As usual, we set $f=f_1 \cdots f_n$ and $\overline {f}=\overline {f}_1 \cdots \overline {f}_{n-1}$ . Then we define

$$\begin{align*}\tau_h:\operatorname{\mathrm{Cl}}(V_e)\overline{f} \to \operatorname{\mathrm{Cl}}(V) f,\quad \tau_h(a \overline{f}) := a \overline{f}f_n=a f. \end{align*}$$

Note that, for $a \in \operatorname{\mathrm{Cl}} (V_e)$ , we have

$$\begin{align*}\pi_e(\tau_h (a \overline{f}))=\pi_e(a f)=a \overline{f}, \end{align*}$$

where the last identity can be seen verified in the model $\bigwedge \nolimits E$ for the spin representation, where $\pi _e$ is the reduction-mod-e map, and $\tau _h$ is just the inclusion $\bigwedge \nolimits E/\langle e \rangle \to \bigwedge \nolimits E$ corresponding to the inclusion $V_e \to V$ . So $\pi _e \circ \tau _h=\operatorname{\mathrm{id}} _{\operatorname{\mathrm{Cl}} (V_e)\overline {f}}$ . We will informally call $\tau _h$ the multiplication map with h.

Proof. Let $v \in V_e$ and let $a \in \operatorname{\mathrm{Cl}} (V_e)$ . Then

$$\begin{align*}\tau_h(va\overline{f}) = va\overline{f}f_n = vaf, \end{align*}$$

as desired.

3.4 Properties of the isotropic Grassmannian

The goal of this subsection is to collect properties of the isotropic Grassmann cone that will later motivate the definition of a (half-)spin variety (see Section 5). We fix a maximal isotropic subspace $F \subseteq V$ and a hyperbolic pair $(e,h)$ with $h \in F$ and $e \not \in F$ and identify $V_e=e^\perp /\langle e \rangle $ with the subspace $\langle e,h \rangle ^\perp $ of V. We choose any basis $f_1,\ldots ,f_n$ of F with $f_n=h$ and $(e|f_i)=0$ for $i<n$ and write $f:=f_1 \cdots f_n \in \operatorname{\mathrm{Cl}} (V)$ and $\overline {f}:=\overline {f_1} \cdots \overline {f_{n-1}} \in \operatorname{\mathrm{Cl}} (V_e)$ .

Proposition 3.4. The isotropic Grassmann cone in $\operatorname{\mathrm{Cl}} (V)f$ has the following properties:

1. 1. $\widehat {\operatorname{\mathrm{Gr}} }_{\operatorname {\mathrm {iso}}}(V) \subseteq \operatorname{\mathrm{Cl}} (V)f$ is Zariski-closed and $\operatorname {\mathrm {Spin}}(V)$ -stable.

2. 2. Let $\pi _e: \operatorname{\mathrm{Cl}} (V) f \to \operatorname{\mathrm{Cl}} (V_e) \overline {f}$ be the contraction defined in §3.2. Then for every maximal isotropic subspace $H \subseteq V$ , we have

   $$\begin{align*}\pi_e(S_H) \subseteq S_{H_e}, \end{align*}$$
   where $H_e \subseteq V_e$ is the image of $e^\perp \cap H$ in $V_e$ .

3. 3. Let $\tau _h: \operatorname{\mathrm{Cl}} (V_e) \overline {f} \to \operatorname{\mathrm{Cl}} (V) f$ be the map defined in §3.3. Then for every maximal isotropic $H' \subseteq V_e$ , we have

   $$\begin{align*}\tau_h(S_{H'}) = S_{{H'} \oplus \langle h \rangle}. \end{align*}$$

In particular, the contraction and multiplication map $\pi _e$ and $\tau _h$ preserve the isotropic Grassmann cones – that is,

$$\begin{align*}\pi_e\big(\widehat{\operatorname{\mathrm{Gr}}}_{\operatorname{\mathrm{iso}}}(V)\big) \subseteq \widehat{\operatorname{\mathrm{Gr}}}_{\operatorname{\mathrm{iso}}}(V_e) \quad \text{and} \quad \tau_h\big(\widehat{\operatorname{\mathrm{Gr}}}_{\operatorname{\mathrm{iso}}}(V_e)\big) \subseteq \widehat{\operatorname{\mathrm{Gr}}}_{\operatorname{\mathrm{iso}}}(V). \end{align*}$$

Remark 3.5. If $h \in H$ , then $H=H_e \oplus \langle h \rangle $ , and since $\pi _e \circ \tau _h$ is the identity on $\operatorname{\mathrm{Cl}} (V_e) \overline {f}$ , we find that

$$\begin{align*}\pi_e(S_H)=\pi_e(\tau_h(S_{H_e}))=S_{H_e} \end{align*}$$

(i.e., equality holds in (2) of Proposition 3.4). Later, we will see that equality holds under the weaker condition that $e \not \in H$ , while $\pi _e(S_H)=\{0\}$ when $e \in H$ . These statements can also be checked by direct computations, but some care is needed since for $e,H,F$ in general position, one cannot construct a hyperbolic basis adapted to H and F that moreover contains e.

3.5 The dual of contraction

Let $e \not \in F \subseteq V$ be an isotropic vector. We want to compute the dual of the contraction map ${\pi _e: \operatorname{\mathrm{Cl}} (V)f \to \operatorname{\mathrm{Cl}} (V_e)\overline {f}}$ ; indeed, we claim that this is essentially the map

$$\begin{align*}\psi_e: \operatorname{\mathrm{Cl}}(V_e) \overline{f} \to \operatorname{\mathrm{Cl}}(V) f \end{align*}$$

defined by its restriction $\operatorname{\mathrm{Cl}} ^\pm (V_e) \overline {f} \to \operatorname{\mathrm{Cl}} ^\mp (V) f$ as

$$\begin{align*}\psi_e (\overline{b} \cdot \overline{f}_1 \cdots \overline{f_{n-1}}):= \pm e b f_1 \cdots f_n, \end{align*}$$

where the sign is $+$ on $\operatorname{\mathrm{Cl}} ^+(V_e) \overline {f}$ and $-$ on $\operatorname{\mathrm{Cl}} ^-(V_e) \overline {f}$ . The reason for the ‘flip’ of the choice of half-spin representation in the dual will become obvious below. Observe that $\psi _e$ is well defined and, given a basis $e_1,\ldots ,e_n=e$ of an isotropic space complementary to F such that $e_1,\ldots ,e_n,f_1,\ldots ,f_n$ is a hyperbolic basis, maps $\overline {e_J} \overline {f}$ to $e_{J \cup \{n\}} f$ .

Proposition 3.6. The following diagram

can be made commuting via a $\operatorname {\mathrm {Spin}}(V_e)$ -module isomorphism on the left vertical arrow and a $\operatorname {\mathrm {Spin}}(V)$ -module isomorphism on the right vertical arrow.

Remark 3.7. The statement of Proposition 3.6 holds true when replacing $\operatorname{\mathrm{Cl}} (V) f$ by either one of the two half-spin representations by considering the correct ‘flip’. For example, if $n = \dim F$ is even, and $e_1,\dots , e_n, f_1, \dots , f_n$ is a hyperbolic basis as above, then in the $\bigwedge \nolimits E$ -model, the correct grading is

To prove Proposition 3.6, we consider the bilinear form $\beta $ on the spin representation $\operatorname{\mathrm{Cl}} (V)f$ defined as in [Reference Procesi13] as follows: for $af,bf \in \operatorname{\mathrm{Cl}} (V)f$ , it turns out that $(af)^*bf=f^*a^*bf$ , where $*$ denotes the anti-automorphism from §2.1, is a scalar multiple of f. The scalar is denoted $\beta (af,bf)$ . We have the following properties:

In the proof of Proposition 3.6, we will use a hyperbolic basis $e_1,\ldots ,e_n,f_1,\ldots ,f_n$ with $e_n=e$ . For a subset $I=\{i_1<\ldots <i_k\} \subseteq [n]$ , set $e_I:=e_{i_1} \cdots e_{i_k} \in \operatorname{\mathrm{Cl}} (E) \simeq \bigwedge \nolimits E$ , where E is the span of the $e_i$ . We have seen in §2.5 that the spin representation has as a basis the elements $e_I f$ with I running through all subsets of $[n]$ .

Proof of Proposition 3.6.

Consider the bilinear forms $\beta $ on $\operatorname{\mathrm{Cl}} (V)f$ and $\beta _e$ on $\operatorname{\mathrm{Cl}} (V_e)\overline {f}$ as defined above. By Lemma 3.8, the spin representations $\operatorname{\mathrm{Cl}} (V)f$ and $\operatorname{\mathrm{Cl}} (V_e)\overline {f}$ are self-dual via $\beta $ and $\beta _e$ , respectively. Thus, it suffices to prove, for $a \in \operatorname{\mathrm{Cl}} (V)$ and $b \in \operatorname{\mathrm{Cl}} (e^\perp )$ , that

$$\begin{align*}\beta_e(\pi_e(af),\overline{b}\overline{f})= \frac{(-1)^{n-1}}{2} \beta(af,\psi_e(\overline{b}\overline{f})). \end{align*}$$

We may assume that $a = e_I$ , $b = e_J$ with $I \subseteq [n]$ , $J \subseteq [n-1]$ .

In the $\bigwedge \nolimits E$ -model, $\pi _e$ is the mod-e map, and hence, the left-hand side is zero if $n \in I$ . If $n \not \in I$ , then the left-hand side equals the coefficient of $\overline {f}$ in $\overline {f}^* \overline {e_I}^* \overline {e_J} \overline {f}$ . This is nonzero if and only if $[n-1]$ is the disjoint union of I and J, and then it is $2^{n-1}$ times a sign corresponding to the number of swaps needed to move the factors $\overline {f_i}$ of $\overline {f}^*$ to just before the corresponding factor $\overline {e_i}$ in either $\overline {e_I}^*$ or $\overline {e_J}$ .

Apart from the factor $\frac {(-1)^{n-1}}{2}$ , the right-hand side is the coefficient of f in $f^* e_I e_J e_n f$ . This is nonzero if and only if $[n]$ is the disjoint union of the sets $\{n\},J,I$ , and in that case, it is $2^n$ times a sign corresponding to the number of swaps needed to move the factors $f_i$ of $f^*$ to the corresponding factor $e_i$ in either $e_I$ or $e_J$ or (in the case of $f_n$ ) to just before the factor $e_n$ . The latter contributes $(-1)^{n-1}$ , and apart from this factor, the sign is the same as on the left-hand side.

3.6 Two infinite spin representations

Let $V_\infty $ be the countable-dimensional vector space with basis $e_1, f_1, e_2, f_2, \dots $ , and equip $V_\infty $ with the quadratic form for which this is a hyperbolic basis (i.e., $(e_i|e_j)=(f_i|f_j)=0$ and $(e_i|f_j)=\delta _{ij}$ for all $i,j$ ). We write $E_\infty $ and $F_\infty $ for the subspaces of $V_\infty $ spanned by the $e_i$ and the $f_i$ , respectively.

Let $V_n$ be the subspace of $V_\infty $ spanned by $e_1, f_1, e_2, f_2, \dots , e_n, f_n$ , with the restricted quadratic form. We further set $E_n:=V_n \cap E_\infty $ and $F_n:=V_n \cap F_\infty $ . We define the infinite spin group as $\operatorname {\mathrm {Spin}}(V_\infty ):=\varinjlim _n \operatorname {\mathrm {Spin}}(V_n)$ , where $\operatorname {\mathrm {Spin}}(V_{n-1})$ is embedded into $\operatorname {\mathrm {Spin}}(V_n)$ as the subgroup that fixes $\langle e_n,f_n \rangle $ element-wise. Similarly, we write $\operatorname{\mathrm{GL}} (E_\infty ):=\varinjlim _n \operatorname{\mathrm{GL}} (E_n)$ and H for the preimage of $\operatorname{\mathrm{GL}} (E_\infty )$ in $\operatorname {\mathrm {Spin}}(V_\infty )$ . We use the notation $\mathfrak {so}(V_\infty )$ and $\mathfrak {gl}(E_\infty )$ for the corresponding direct limits of the Lie algebras $\mathfrak {so}(V_n)$ and $\mathfrak {gl}(E_n)$ . Here, the direct limits are taken in the categories of abstract groups and Lie algebras, respectively.

The previous paragraphs give rise to various $\operatorname {\mathrm {Spin}}(V_{n-1})$ -equivariant maps between the spin representations of $\operatorname {\mathrm {Spin}}(V_{n-1})$ and $\operatorname {\mathrm {Spin}}(V_n)$ . First, contraction with $e_n$ ,

$$\begin{align*}\pi_{e_n}:\operatorname{\mathrm{Cl}}(V_n)f_1 \cdots f_n \to \operatorname{\mathrm{Cl}}(V_{n-1}) f_1 \cdots f_{n-1}, \end{align*}$$

and second, multiplication with $f_n$ ,

$$\begin{align*}\tau_{f_n}:\operatorname{\mathrm{Cl}}(V_{n-1})f_1 \cdots f_{n-1} \to \operatorname{\mathrm{Cl}}(V_n) f_1 \cdots f_n. \end{align*}$$

We have that these satisfy $\pi _{e_n} \circ \tau _{f_n} =\operatorname{\mathrm{id}} $ . Third, the map

$$\begin{align*}\psi_{e_n}:\operatorname{\mathrm{Cl}}(V_{n-1})f_1 \cdots f_{n-1} \to \operatorname{\mathrm{Cl}}(V_n) f_1 \cdots f_n \end{align*}$$

is dual to $\pi _{e_n}$ in the sense of Proposition 3.6.

Since the maps $\psi _{e_n},\pi _{e_n}$ are $\operatorname {\mathrm {Spin}}(V_{n-1})$ -equivariant, both of these spaces are $\operatorname {\mathrm {Spin}}(V_\infty )$ -modules. As the dual of a direct limit is the inverse limit of the duals, and since the maps $\psi _{e_n}$ and $\pi _{e_n}$ are dual to each other by Proposition 3.6, the inverse spin representation is the dual space of the direct spin representation.

In our model $\bigwedge \nolimits E_n$ of $\operatorname{\mathrm{Cl}} (V_n) f_1 \cdots f_n$ , the map $\psi _{e_n}$ is just the right multiplication

$$\begin{align*}\bigwedge\nolimits E_{n-1} \to \bigwedge\nolimits E_n,\; \omega \mapsto \omega \wedge e_n. \end{align*}$$

Hence, the direct spin representation has as a basis formal infinite products

$$\begin{align*}e_{i_1} \wedge e_{i_2} \wedge \ldots =: e_I, \end{align*}$$

where $I=\{i_1<i_2<\ldots \}$ is a cofinite subset of ${\mathbb {N}}$ . We will write $\bigwedge \nolimits _\infty E_\infty $ for this countable-dimensional vector space. The action of the Lie algebra $\mathfrak {so}(V_\infty )$ of $\operatorname {\mathrm {Spin}}(V_\infty )$ on this space is given via the explicit formulas from §2.5. In particular, the span of the $e_I$ with $|{\mathbb {N}} \setminus I|$ even (respectively, odd) is a $\operatorname {\mathrm {Spin}}(V_\infty )$ -submodule, and $\bigwedge \nolimits _\infty E_\infty $ is the direct sum of these (irreducible) modules.

3.7 Four infinite half-spin representations

Keeping in mind that the maps $\psi _{e_n}$ interchange the even and odd subrepresentations, we define the direct (infinite) half-spin representations $\bigwedge \nolimits _\infty ^\pm E_\infty $ to be the direct limit

$$ \begin{align*} \bigwedge\nolimits_\infty^\pm E_\infty = \varinjlim \left(\bigwedge\nolimits^\pm E_0 \to \bigwedge\nolimits^\mp E_1 \to \bigwedge\nolimits^\pm E_2 \to \bigwedge\nolimits^\mp E_3 \to \bigwedge\nolimits^\pm E_4 \to \cdots \right) \end{align*} $$

along the maps $\psi _{e_n}$ . For the sake of readability, we will abbreviate this by

(3.2) $$ \begin{align} \bigwedge\nolimits_\infty^\pm E_\infty = \varinjlim_n \bigwedge\nolimits^{\pm (-1)^n}E_n, \end{align} $$

where $\pm (-1)^n$ denotes $\pm $ if n is even and $\mp $ if n is odd. In terms of the basis $e_I$ introduced in §3.6, the half-spin representation $\bigwedge \nolimits _\infty ^+ E_\infty $ is spanned by all $e_I$ with $|{\mathbb {N}} \setminus I|$ even, and $\bigwedge \nolimits _\infty ^- E_\infty $ by those with $|{\mathbb {N}} \setminus I|$ odd. The inverse (infinite) half-spin representations are defined as the duals of the direct (infinite) half-spin representations. Using the isomorphisms from Remark 3.7, we observe

(3.3) $$ \begin{align} \left(\bigwedge\nolimits^\pm_\infty E_\infty \right)^*=\varprojlim_n \left( \bigwedge\nolimits^{\pm (-1)^n}E_n \right)^\ast \simeq \varprojlim_n \bigwedge\nolimits^\pm E_n. \end{align} $$

So the inverse (infinite) half-spin representations can be identified with the inverse limits of the half-spin representations $\bigwedge \nolimits ^\pm E_n$ along the projections $\pi _{e_n}$ .

We can enrich the inverse spin representation to an affine scheme whose coordinate ring is the symmetric algebra on $\bigwedge \nolimits _\infty E_\infty $ , recalling the following remark.

Remark 3.11. Let K be any field (not necessarily algebraically closed) and W any K-vector space (not necessarily finite dimensional). Then there are canonical identifications

$$\begin{align*}W^\ast = \mathrm{Spec}\big( \textrm{ Sym}(W)\big) (K) \subseteq \left\{\text{closed points in } \textrm{ Spec}\big( \textrm{ Sym}(W)\big)\right\}. \end{align*}$$

So $\textrm { Spec}\big ( \textrm { Sym}(W)\big )$ can be seen as an enrichment of $W^\ast $ to an affine scheme. If W is a linear representation for a group G, then G acts via K-algebra automorphisms on $\operatorname{\mathrm{Sym}} W$ and hence via K-automorphisms on the affine scheme corresponding to $W^*$ . For $W = \bigwedge \nolimits ^\pm _\infty E_\infty $ , this construction extends the natural $\operatorname {\mathrm {Spin}}(V_\infty )$ -action on the vector space $\varprojlim _n \bigwedge \nolimits ^\pm E_n \simeq W^*$ to the corresponding affine scheme.

By abuse of notation, we will write $\left (\bigwedge \nolimits _\infty E_\infty \right )^*$ also for the scheme itself, and similarly for the inverse half-spin representations $\left (\bigwedge \nolimits ^\pm _\infty E_\infty \right )^*$ . Later, we will also write $\bigwedge \nolimits ^\pm E_n$ for the affine scheme $\textrm { Spec}\left (\textrm { Sym}\left ( \bigwedge \nolimits ^{\pm (-1)^n}E_n \right ) \right )$ by identifying $\bigwedge \nolimits ^\pm E_n \cong \left (\bigwedge \nolimits ^{\pm (-1)^n}E_n\right )^*$ as in Equation(3.3).

4.1 Shifting

Let $G_n$ be the subgroup of G that fixes $e_1,\ldots ,e_n,f_1,\ldots ,f_n$ element-wise. Note that $G_n$ is isomorphic to G; at the level of the Lie algebras, the isomorphism from G to $G_n$ is given by the map

$$\begin{align*}\begin{bmatrix} A & B \\ C & -A^T \end{bmatrix} \mapsto \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & A & 0 & B \\ 0 & 0 & 0 & 0 \\ 0 & C & 0 & -A^T \end{bmatrix} \end{align*}$$

where the widths of the blocks are $n,\infty ,n,\infty $ , respectively. We write $H_n$ for $H \cap G_n$ , where $H \subseteq \operatorname {\mathrm {Spin}}(V_\infty )$ is the subgroup corresponding to the subalgebra $\mathfrak {gl}(E_\infty ) \subseteq \mathfrak {so}(V_\infty )$ . Then $H_n$ is the pre-image in $\operatorname {\mathrm {Spin}}(V_\infty )$ of the subgroup $\operatorname{\mathrm{GL}} (E_\infty )_n \subseteq \operatorname{\mathrm{GL}} (E_\infty )$ of all g that fix $e_1,\ldots ,e_n$ element-wise and maps the span of the $e_i$ with $i>n$ into itself. The Lie algebra of $H_n$ and of $\operatorname{\mathrm{GL}} (E_\infty )_n$ consists of the matrices above on the right with $B=C=0$ .

4.2 Acting with the general linear group on E

For every fixed $k \in {\mathbb {Z}}_{\geq 0}$ , the Lie algebra $\mathfrak {gl}(E_\infty ) \subseteq \mathfrak {so}(V_\infty )$ preserves the linear space

$$\begin{align*}\left(\bigwedge\nolimits_\infty E_\infty\right)_k:=\big\langle \{e_I : |{\mathbb{N}} \setminus I|=k\} \big\rangle, \end{align*}$$

and hence, so does the corresponding subgroup $H \subseteq \operatorname {\mathrm {Spin}}(V_\infty )$ . We let $R_{\leq \ell } \subseteq R$ be the subalgebra generated by the spaces $(\bigwedge \nolimits _\infty E_\infty )_k$ with $k \leq \ell $ . Crucial in the proof of Theorem 4.1 is the following result.

Proposition 4.2. For every choice of nonnegative integers $\ell $ and n, $\operatorname{\mathrm{Spec}} (R_{\leq \ell })$ is topologically $H_n$ -Noetherian; that is, every descending chain

$$\begin{align*}\operatorname{\mathrm{Spec}}(R_{\leq \ell}) \supseteq X_1 \supseteq X_2 \supseteq \ldots \end{align*}$$

of $H_n$ -stable closed and reduced subschemes stabilizes.

The key ingredient in the proof of Proposition 4.2 is the main result of [Reference Eggermont and Snowden7]. In order to apply their result, we need to do some preparatory work. We will start with the following lemma.

Proof. Equation (2.3) implies that $\mathfrak {gl}(E_\infty ) \subseteq \mathfrak {so}(V_\infty )$ acts on $\bigwedge \nolimits _\infty E_\infty $ via

$$\begin{align*}\rho(A)=\tilde{\rho}(A)-\frac{1}{2} \operatorname{tr}(A) \operatorname{\mathrm{id}}_{\bigwedge\nolimits_\infty E_\infty}, \end{align*}$$

where $\tilde {\rho }$ is the standard representation of $\mathfrak {gl}(E_\infty )$ on $\bigwedge \nolimits _\infty E_\infty $ . An $H_n$ -stable closed subscheme X of $\operatorname{\mathrm{Spec}} (R_{\leq \ell })$ is given by an $H_n$ -stable ideal I in the symmetric algebra $R_{\leq \ell }$ . Such an I is then also stable under the action of the Lie algebra $\mathfrak {gl}(E_\infty )_n$ of $H_n$ by derivations that act on variables in $\bigoplus _{k=0}^\ell (\bigwedge \nolimits _\infty E_\infty )_{k}$ via $\rho $ .

We claim that I is a homogeneous ideal. Indeed, for $f \in I$ , choose $m>n$ such that all variables in f (which are basis elements $e_I$ ) contain the basis element $e_m$ of $E_\infty $ . Let $A \in \mathfrak {gl}(E_\infty )_n$ be the diagonal matrix with $0$ ’s everywhere except a $1$ on position $(m,m)$ . Then $\rho (A)$ maps each variable in f to $\frac {1}{2}$ times itself. Hence, by the Leibniz rule, $\rho (A)$ scales the homogeneous part of degree d in f by $\frac {d}{2}$ . Since I is preserved by $\rho (A)$ , it follows that I contains all homogeneous components of f, and hence, I is a homogeneous ideal.

Now let $B \in \mathfrak {gl}(E_\infty )_n$ and $f \in I$ be arbitrary. By the previous paragraph, we can assume f to be homogeneous of degree d, and we then have

$$\begin{align*}\rho(B)f = \tilde{\rho}(B)f - \frac{d}{2}\operatorname{tr}(B)f, \end{align*}$$

and since I is $\rho (B)$ -stable, we deduce $\tilde {\rho }(B) f \in I$ . This completes the proof in one direction. The proof in the opposite direction is identical.

Following [Reference Eggermont and Snowden7], the restricted dual $(E_\infty )_*$ of $E_\infty $ is defined as the union $\bigcup _{n\geq 1}(E_n)^*.$ We will denote by $\varepsilon ^1, \varepsilon ^2, \dots $ the basis of $(E_\infty )_*$ that is dual to the canonical basis $e_1, e_2, \dots $ of $E_\infty $ given by $\varepsilon ^i(e_j)=\delta _{ij}$ .

Lemma 4.5. There is an $\operatorname{\mathrm{SL}} (E_\infty )$ -equivariant isomorphism

$$\begin{align*}\bigwedge \nolimits_\infty E_\infty \longrightarrow \bigwedge (E_\infty)_*, \end{align*}$$

which restricts to an isomorphism

$$\begin{align*}\left( \bigwedge \nolimits_\infty E_\infty\right)_k \longrightarrow \bigwedge \nolimits^k(E_\infty)_*. \end{align*}$$

We will use this isomorphism to regard $\bigwedge \nolimits _\infty E_\infty $ as the restricted dual of the Grassmann algebra $\bigwedge \nolimits E_\infty $ . We stress, though, that this isomorphism is not $GL(E_\infty )$ -equivariant.

Proof. We have a natural bilinear map

$$\begin{align*}\bigwedge\nolimits E_\infty \times \bigwedge\nolimits_\infty E_\infty \to \bigwedge\nolimits_\infty E_\infty, \quad (\omega,\omega') \mapsto \omega \wedge \omega'. \end{align*}$$

If $I \subseteq {\mathbb {N}}$ is finite and $J \subseteq {\mathbb {N}}$ is cofinite, then $e_I \wedge e_J$ is $0$ if $I \cap J \neq \emptyset $ and $\pm e_{I \cup J}$ otherwise, where the sign is determined by the permutation required to order the sequence $I,J$ . We then define a perfect pairing $\gamma $ between the two spaces by

$$\begin{align*}\gamma(\omega,\omega'):=\text{the coefficient of } e_{\mathbb{N}} \text{ in } \omega \wedge \omega'. \end{align*}$$

The map $\Phi _\gamma : \bigwedge \nolimits _\infty E_\infty \to \bigwedge \nolimits (E_\infty )_*, \; \omega ' \mapsto \gamma (\cdot , \omega ')$ induced by $\gamma $ is the isomorphism given by $e_I \mapsto \pm \varepsilon ^{I^c}$ , where $I^c \subseteq {\mathbb {N}}$ is the complement of I and for a finite set $J=\{j_1, \dots , j_k\}$ . Note that $\gamma (A\cdot \omega , A\cdot \omega ') = \det (A)\gamma (\omega , \omega ')$ for all $A \in \operatorname{\mathrm{GL}} (E_\infty )$ , and hence, $\gamma $ is $\operatorname{\mathrm{SL}} (E_\infty )$ -invariant. Therefore, the isomorphism $\Phi _\gamma $ is $\operatorname{\mathrm{SL}} (E_\infty )$ -equivariant.

Before we come to the proof of Theorem 4.1, let us recall the action of $f_i \wedge f_j \in \mathfrak {so}(V_\infty )$ on $\bigwedge \nolimits ^+_\infty E_\infty $ and its symmetric algebra $R^+$ in explicit terms. Recall from Section 3.6 that a basis for $\bigwedge \nolimits ^+_\infty E_\infty $ is given by $e_I = e_{i_1} \wedge e_{i_2} \wedge \cdots $ , where $I = \{i_1 < i_2 < \cdots \} \subseteq {\mathbb {N}}$ is cofinite and $|{\mathbb {N}} \setminus I|$ even. Then we have

$$\begin{align*}(f_i \wedge f_j) e_I = \begin{cases} (-1)^{c_{i,j}(I)}e_{I\setminus \{i, j\}}& \text{if } i,j \in I, \text{ and}\\ 0 & \text{otherwise,} \end{cases} \end{align*}$$

where $c_{i,j}(I)$ depends on the position of $i,j$ in I. (Note that there is no factor $4$ , since in our identification of $\bigwedge \nolimits ^2 V$ to the Lie subalgebra L of $\operatorname{\mathrm{Cl}} (V)$ we had a factor $\frac {1}{4}$ .) The corresponding action of $f_i \wedge f_j$ on polynomials in $R^+$ is as a derivation.

4.3 Proof of Theorem 4.1

Let $R^+ \subseteq R$ be the symmetric algebra on the direct half-spin representation $\bigwedge \nolimits ^+_\infty E_\infty $ , so that $\operatorname{\mathrm{Spec}} (R^+)$ is the inverse half-spin representation $(\bigwedge \nolimits ^+_\infty E_\infty )^*$ . We prove topological $\operatorname {\mathrm {Spin}}(V_\infty )$ -Noetherianity of $\operatorname{\mathrm{Spec}} (R^+)$ ; the corresponding statement for $\operatorname{\mathrm{Spec}} (R^-)$ is proved in exactly the same manner.

For a closed, reduced $\operatorname {\mathrm {Spin}}(V_\infty )$ -stable subscheme X of $\operatorname{\mathrm{Spec}} (R^+)$ , we denote by $\delta _X \in \{0,1,2,\ldots ,\infty \}$ the lowest degree of a nonzero polynomial in the ideal $I(X) \subseteq R^+$ of X. Here, we consider the natural grading on $R^+ = \operatorname{\mathrm{Sym}} (\bigwedge \nolimits ^+_\infty E_\infty )$ , where the elements of $\bigwedge \nolimits ^+_\infty E_\infty $ all have degree $1$ .

We proceed by induction on $\delta _X$ to show that X is topologically Noetherian; we may, therefore, assume that this is true for all Y with $\delta _Y<\delta _X$ . We have $\delta _X=\infty $ if and only if $X=\operatorname{\mathrm{Spec}} (R^+)$ . Then a chain

$$\begin{align*}\operatorname{\mathrm{Spec}}(R^+) = X \supseteq X_1 \supseteq X_2 \supseteq \ldots \end{align*}$$

of $\operatorname {\mathrm {Spin}}(V_\infty )$ -closed subsets is either constant or else there exists an i with $\delta _{X_i}<\infty $ . Hence, it suffices to prove that X is Noetherian under the additional assumption that $\delta _X<\infty $ . At the other extreme, if $\delta _X=0$ , then X is empty and there is nothing to prove. So we assume that $0<\delta _X<\infty $ and that all Y with $\delta _Y<\delta _X$ are $\operatorname {\mathrm {Spin}}(V_\infty )$ -Noetherian.

Let $p \in R^+$ be a nonzero polynomial in the ideal of X of degree $\delta _X$ . By Remark 4.4, since X is a cone, p is in fact homogeneous of degree $\delta _X$ . Let $e_I$ be a variable appearing in p such that $k := |I^c|$ is maximal among all variables in p; note that k is even. Then choose $n \geq k+2$ even such that all variables of p are contained in $\bigwedge \nolimits ^+ E_n$ (i.e., they are of the form $e_J$ with $J \supseteq \{n+1,n+2,\ldots \}$ ).

Now act on p with the element $f_{i_1} \wedge f_{i_2} \in \mathfrak {so}(V_\infty )$ with $i_1<i_2$ the two smallest elements in I. Since X is $\operatorname {\mathrm {Spin}}(V_\infty )$ -stable, the result $p_1$ is again in the ideal of X. Furthermore, $p_1$ has the form

$$\begin{align*}p_1=\pm e_{I \setminus \{i_1,i_2\}} \cdot q + r_1, \end{align*}$$

where $q=\frac {\partial p}{\partial e_{I}}$ contains only variables $e_J$ with $|J^c| \leq k$ and where $r_1$ does not contain $e_{I \setminus \{i_1,i_2\}}$ but may contain other variables $e_J$ with $|J^c|=k+2$ (namely, those with $i_1,i_2 \not \in J$ for which $e_{J \cup \{i_1,i_2\}}$ appears in p).

If $n=k+2$ , then $I \setminus \{i_1,i_2\}=\{n+1,n+2,\ldots \}$ , and since all variables $e_J$ in $p_1$ satisfy ${J \supseteq \{n+1,n+2,\ldots \}}$ , $e_{I \setminus \{i_1,i_2\}}$ is the only variable $e_J$ in $p_1$ with $|J^c|=k+2$ . If $n>k+2$ , then we continue in the same manner, now acting with $f_{i_3} \wedge f_{i_4}$ on $p_1$ , where $i_3<i_4$ are the two smallest elements in $I \setminus \{i_1,i_2\}$ . We write $p_2$ for the result, which is now of the form

$$\begin{align*}p_2=\pm e_{I \setminus \{i_1,i_2,i_3,i_4\}} \cdot q + r_2, \end{align*}$$

where q is the same polynomial as before and $r_2$ does not contain the variable $e_{I \setminus \{i_1,i_2,i_3,i_4\}}$ but may contain other variables $e_J$ with $|J^c|=k+4$ .

Iterating this construction, we find the polynomial

$$\begin{align*}p_\ell=\pm e_{\{n+1,n+2,\ldots\}} \cdot q + r_\ell \end{align*}$$

in the ideal of X, where $\ell =(n-k)/2$ , q is the same polynomial as before and $r_\ell $ only contains variables $e_J$ with $|J^c|<n$ . Let $Z:=X[1/q]$ be the open subset of X where q is nonzero.

Proof. We proceed by induction on $|J^c|=:m$ . By successively acting on $p_\ell $ with the elements ${f_n \wedge f_{n+1},f_{n+2} \wedge f_{n+3},\ldots ,f_{m-1}\wedge f_m}$ , we find the polynomial

$$\begin{align*}\pm e_{\{m+1,m+2,\ldots\}} \cdot q + r \end{align*}$$

in the ideal of X, where r contains only variables $e_L$ with $|L^c|<m$ . Now act with elements of $\mathfrak {gl}(E_\infty )$ to obtain an element

$$\begin{align*}\pm e_J \cdot q + \tilde{r}, \end{align*}$$

where $\tilde {r}$ still contains only variables $e_L$ with $|L^c|<m$ . Inverting q, this can be used to express $e_J$ in such variables $e_L$ . By the induction hypothesis, all those $e_L$ admit an expression, on Z, as a polynomial in $R^+_{\leq n-2}$ times a negative power of q. Then the same holds for $e_J$ .

Proof of Theorem 4.1.

Let

$$\begin{align*}X \supseteq X_1 \supseteq \ldots \end{align*}$$

be a chain of reduced, $\operatorname {\mathrm {Spin}}(V_{\infty })$ -stable closed subschemes. Let $Y \subseteq X$ be the reduced closed subscheme defined by the orbit $\operatorname {\mathrm {Spin}}(V_{\infty }) \cdot q$ . Since q has degree $\delta _X-1$ , we have $\delta _Y < \delta _X$ , and hence, Y is $\operatorname {\mathrm {Spin}}(V_{\infty })$ -Noetherian by the induction hypothesis. It follows that the chain

$$\begin{align*}Y \supseteq (Y \cap X_1)^{\mathrm{red}} \supseteq \ldots \end{align*}$$

is eventually stable. However, the chain

$$\begin{align*}Z \supseteq (Z \cap X_1)^{\mathrm{red}} \supseteq \ldots \end{align*}$$

consists of reduced, $H_n$ -stable closed subschemes of Z; hence, it is eventually stable by Lemma 4.8.

Now pick a (not necessarily closed) point $P \in X_i$ for $i \gg 0$ . If $P \in Y \cap X_i$ , then $P \in Y \cap X_{i-1}$ by the first stabilisation. However, if $P \not \in Y \cap X_i$ , then there exists a $g \in \operatorname {\mathrm {Spin}}(V_\infty )$ such that $gP \in Z$ . Then $gP$ lies in $X_i \cap Z$ , which by the second stabilisation equals $X_{i-1} \cap Z$ ; hence, $P=g^{-1}(gP)$ lies in $X_{i-1}$ , as well. We conclude that the chain $(X_i)_i$ of closed, reduced subschemes of X stabilises. Hence, the inverse half-spin representation $(\bigwedge \nolimits ^+_\infty E_\infty )^*$ is topologically $\operatorname {\mathrm {Spin}}(V_\infty )$ -Noetherian.

6.1 Statement

In [Reference Seynnaeve and Tairi16], the last two authors showed that in even dimension, the isotropic Grassmannian in its Plücker embedding is set-theoretically defined by pulling back equations coming from $\widehat {\operatorname{\mathrm{Gr}} }_{\operatorname {\mathrm {iso}}}(4,8)$ . Using the Cartan map, we can translate this into a statement about the isotropic Grassmannian in its spinor embedding and prove the following result.

Theorem 6.1. For all $n\geq 4$ , we have

$$\begin{align*}\widehat{\operatorname{\mathrm{Gr}}}^+_{\operatorname{\mathrm{iso}}}(V_n) = V(\operatorname{\mathrm{rad}}(\operatorname{\mathrm{Spin}}(V_n)\cdot I_4)), \end{align*}$$

where $I_4$ is the ideal of polynomials defining $\widehat {\operatorname{\mathrm{Gr}} }^+_{\operatorname {\mathrm {iso}}}(V_4) \subseteq \operatorname{\mathrm{Cl}} ^+(V_4)f.$

In other words, the bound $n_0$ from Corollary 5.8 can be taken equal to $4$ for the cone over the isotropic Grassmannian. We give the proof of Theorem 6.1 in §6.5 using properties of the Cartan map that will be established in the following sections.

6.2 Definition of the Cartan map

When we regard $e_1 \wedge \cdots \wedge e_n$ as an element of the n-th exterior power $\bigwedge \nolimits ^n V$ of the standard representation V of $\mathfrak {so}(V)$ , then it is a highest weight vector of weight $(0,\ldots ,0,2)=2\lambda _0$ , where $\lambda _0$ is the fundamental weight introduced in §2.6 and the highest weight of the half-spin representation $\operatorname{\mathrm{Cl}} ^{(-1)^n}(V) f$ . Similarly, the element $e_1 \wedge e_2 \wedge \cdots \wedge e_{n-1} \wedge f_n \in \bigwedge \nolimits ^n V$ is a highest weight vector of weight $(0,\ldots ,0,2,0)=2 \lambda _1$ , where $\lambda _1$ is the highest weight of the other half-spin representation. So $\bigwedge \nolimits ^n V$ contains copies of the irreducible representations $V_{2\lambda _0},V_{2 \lambda _1}$ of $\mathfrak {so}(V)$ ; in fact, it is well known to be the direct sum of these. To compare our results in this paper about spin representations with earlier work by the last two authors about exterior powers, we will need the following considerations.

Consider any connected, reductive algebraic group G, with maximal torus T and Borel subgroup $B \supseteq T$ . Let $\lambda $ be a dominant weight of G, let $V_{\lambda }$ be the corresponding irreducible representation, and let $v_\lambda \in V_{\lambda }$ be a nonzero highest-weight vector (which is unique up to scalar multiples). Then the symmetric square $S^2 V_\lambda $ contains a one-dimensional space of vectors of weight $2 \lambda $ , spanned by $v_{2\lambda }:=v_\lambda ^2$ . This vector is itself a highest-weight vector, and hence generates a copy of $V_{2\lambda }$ ; this is sometimes called the Cartan component of $S^2 V_\lambda $ . By semisimplicity, there is a G-equivariant linear projection $\pi : S^2 V_\lambda \to V_{2\lambda }$ that restricts to the identity on $V_{2\lambda }$ . The map

$$\begin{align*}\widehat{\nu_2}: V_\lambda \to V_{2\lambda},\quad v \mapsto \pi(v^2) \end{align*}$$

is a nonzero polynomial map, homogeneous of degree $2$ , and hence induces a rational map $\nu _2:\mathbb {P} V_{\lambda } \to \mathbb {P} V_{2\lambda }$ . Note that this is the composition of the quadratic Veronese embedding and the projection $\pi $ . We will refer to $\nu _2$ and to $\widehat {\nu _2}$ as the Cartan map.

We thank J. M. Landsberg for help with the following proof.

Proof. To show that $\nu _2$ is a morphism, we need to show that $\pi (v^2)$ is nonzero whenever v is. Now the set Q of all $[v] \in \mathbb {P} V_\lambda $ for which $\pi (v^2)$ is zero is closed and B-stable. Hence, if $Q \neq \emptyset $ , then by Borel’s fixed point theorem, Q contains a B-fixed point. But the only B-fixed point in $\mathbb {P} V_\lambda $ is $[v_\lambda ]$ , and $v_\lambda $ is mapped to the nonzero vector $v_{2\lambda }$ . Hence, $Q=\emptyset $ .

Injectivity is similar but slightly more subtle. Assume that there exist distinct $[v],[w]$ with $\nu _2([v])=\nu _2([w])$ . Then $\{[v],[w]\}$ represents a point in the Hilbert scheme H of two points in $\mathbb {P} V_\lambda $ . Now the locus Q of points S in H such that $\nu _2(S)$ is contained in a single reduced point is closed in H, as it is the projection to H of the closed subvariety

$$\begin{align*}\{(S,[u]) \mid \nu_2(S) \subseteq \{[u]\}\} \subseteq H \times \mathbb{P} V_{2 \lambda} \end{align*}$$

and since $\mathbb {P} V_{2 \lambda }$ is projective. Since H is a projective scheme, Q is projective. Hence, Q contains a B-fixed point $S_0$ . This scheme $S_0$ cannot consist of two distinct reduced points: if it did, then either both points would be B-fixed, but there is only one B-fixed point, or else they would be a B-orbit, but this is impossible since B is connected. Therefore, the reduced subscheme of $S_0$ is $\{[v_{\lambda }]\}$ , and $S_0$ represents the point $[v_\lambda ]$ together with a nonzero tangent direction in $T_{[v_\lambda ]} \mathbb {P} V_{\lambda } = V_\lambda /K v_\lambda $ , represented by $w \in V_\lambda $ . Furthermore, B-stability of $S_0$ implies that the B-module generated by w equals $\langle w,v_\lambda \rangle _K$ . That $S_0$ lies in Q means that

$$\begin{align*}\pi((v_\lambda+\epsilon w)^{2})=v_{2\lambda} \quad\mod \epsilon^2. \end{align*}$$

We find that $\pi (v_\lambda w)=0$ , so that the G-module generated by $v_\lambda w \in S^2 V$ does not contain $V_{2\lambda }$ . But since $v_\lambda $ is (up to a scalar) fixed by B, the B-module generated by $v_\lambda w$ equals $v_\lambda $ times the B-module generated by w, and hence contains $v_\lambda ^2=v_{2\lambda }$ , a contradiction.

Observe that $\nu _2$ maps the unique closed orbit $G \cdot [v_\lambda ]$ in $\mathbb {P} V_\lambda $ isomorphically to the unique closed orbit $G \cdot [v_{2\lambda }]$ – both are isomorphic to $G/P$ , where $P \supseteq B$ is the stabiliser of the line $K v_\lambda $ and of the line $K v_{2\lambda }$ . In our setting above, where $G=\operatorname {\mathrm {Spin}}(V)$ and $\lambda \in \{\lambda _0,\lambda _1\}$ , the closed orbit $G \cdot [v_{2\lambda }]$ is one of the two connected components of the Grassmannian $\operatorname{\mathrm{Gr}} _{\operatorname {\mathrm {iso}}}(V)$ of n-dimensional isotropic subspaces of V, in its Plücker embedding; and the closed orbit in the projectivised half-spin representation $\mathbb {P} V_{\lambda }$ is the same component of the isotropic Grassmannian but now in its spinor embedding.

In what follows, we will need a more explicit understanding both of the embedding of the isotropic Grassmannian in the projectivised (half-)spin representations and of the map $\widehat {\nu _2}$ . These are treated in the next two paragraphs.

6.3 The map $\widehat {\nu _2}$ from the spin representation to the exterior power

In §6.2, we argued the existence of $\operatorname {\mathrm {Spin}}(V)$ -equivariant quadratic maps from the half-spin representations to the two summands of $\bigwedge \nolimits ^n V$ . In [Reference Manivel11], these two maps are described jointly as

$$\begin{align*}\widehat{\nu_2}:\operatorname{\mathrm{Cl}}(V)f \to \bigwedge\nolimits^n V, \quad af \mapsto \text{ the component in } \bigwedge\nolimits^n V \text{ of } (afa^*) \bullet 1 \in \bigwedge\nolimits V, \end{align*}$$

where $\bullet $ stands for the $\operatorname{\mathrm{Cl}} (V)$ -module structure of $\bigwedge \nolimits V$ from §2.2 and $a^*$ refers to the anti-automorphism of the Clifford algebra from §2.1.

Lemma 6.3. The map $\hat {\nu }_2$ maps the isotropic Grassmann cone in its spinor embedding to the isotropic Grassmann cone in its Plücker embedding, that is,

$$\begin{align*}\hat{\nu}_2\big( \widehat{\operatorname{\mathrm{Gr}}}_{\operatorname{\mathrm{iso}}}(V)\big) = \widehat{\operatorname{\mathrm{Gr}}}^{\operatorname{Pl}}_{\operatorname{\mathrm{iso}}}(V), \end{align*}$$

where $\widehat {\operatorname{\mathrm{Gr}} }^{\operatorname {Pl}}_{\operatorname {\mathrm {iso}}}(V)$ is the isotropic Grassmann cone in its Plücker embedding (see [Reference Seynnaeve and Tairi16, Definition 3.7]).

Proof. Let $H \subseteq V$ be a maximal isotropic subspace that intersects F in a k-dimensional space. Choose a hyperbolic basis $e_1,\ldots ,e_n,f_1,\ldots ,f_n$ adapted to H and F, so that $H=\langle e_{k+1},\ldots ,e_n,f_1,\ldots ,f_k \rangle $ is represented by the vector $\omega _H:= e_{k+1} \cdots e_n f \in \widehat {\operatorname{\mathrm{Gr}} }_{\operatorname {\mathrm {iso}}}(V)$ where $f=f_1 \cdots f_n$ ; see §3.1. Set ${a:=e_{k+1} \cdots e_n}$ . Now

$$ \begin{align*} afa^*&=e_{k+1} \cdots e_n f_1 \cdots f_n e_n \cdots e_{k+1}\\ &=e_{k+1} \cdots e_n f_1 \cdots f_{n-1} (2-e_n f_n) e_{n-1} \cdots e_{k+1} \\ &=2 e_{k+1} \cdots e_n f_1 \cdots f_{n-1} e_{n-1} \cdots e_{k+1} \\ &=\ldots\\ &=2^{n-k} e_{k+1} \cdots e_n f_1 \cdots f_k, \end{align*} $$

where we have used the definition of $\operatorname{\mathrm{Cl}} (V)$ (in the first step), the fact that the second copy of $e_n$ is perpendicular to all elements before it and multiplies to zero with the first copy of $e_n$ (in the second step), and have repeated this another $n-k-1$ times in the last step. We now find that

$$\begin{align*}(afa^*) \bullet 1=2^{n-k} e_{k+1} \wedge \cdots \wedge e_n \wedge f_1 \wedge \cdots \wedge f_k, \end{align*}$$

so that $(afa^*) \bullet 1$ lies in one of the two summands of $\bigwedge \nolimits ^n V$ and spans the line representing the space H in the Plücker embedding. This shows that $\hat {\nu _2}$ maps the isotropic Grassmann cone in its spinor embedding to the isotropic Grassmann cone in its Plücker embedding, as desired.

6.4 Contraction and the Cartan map commute

Recall from §6.2 that we have quadratic maps $\widehat {\nu _2}$ from the half-spin representations to the two summands of $\bigwedge \nolimits ^n V$ ; together, these form a quadratic map $\widehat {\nu _2}$ which we discussed in §6.3. By abuse of terminology, we call this, too, the Cartan map. Given an isotropic vector $e \in V$ that is not in F, we write $\widehat {\nu _2}$ also for the Cartan map $\operatorname{\mathrm{Cl}} (V_e)\overline {f} \to \bigwedge \nolimits ^{n-1} V_e$ (notation as in §3.2). Recall from §3.2 the contraction map $c_e: \bigwedge \nolimits ^n V \to \bigwedge \nolimits ^{n-1} V_e$ and its spin analogue $\pi _e: \operatorname{\mathrm{Cl}} (V)f \to \operatorname{\mathrm{Cl}} (V_e)\overline {f}$ . Also, for a fixed $h=f_n \in F$ with $(e|h)=1$ , recall from §3.3 the multiplication map $m_h:\bigwedge \nolimits ^{n-1} V_e \to \bigwedge \nolimits ^n V$ and its spin analogue $\tau _h:\operatorname{\mathrm{Cl}} (V_e) \overline {f} \to \operatorname{\mathrm{Cl}} (V) f$ . The relations between these maps are as follows.

Proposition 6.5. The following diagrams essentially commute:

(6.1)

More precisely, one can rescale the restrictions of $c_e$ to the two $\mathfrak {so}(V)$ -submodules of $\bigwedge \nolimits ^n V$ each by $\pm 1$ in such a manner that the diagram commutes, and similarly for $m_h$ .

Naturally, we could have chosen the scalars in the definition of $c_e$ (or, using a square root of $-1$ , in that of $\pi _e$ ) such that the diagram literally commutes. However, we have chosen the scalars such that $c_e$ has the most natural formula and $\pi _e,\tau _h$ have the most natural formulas in our model $\bigwedge \nolimits E$ for the spin representation.

Proof. We may choose a hyperbolic basis $e_1,\ldots ,e_n,f_1,\ldots ,f_n$ of V such that $e=e_n$ and $f_1,\ldots ,f_n$ is a basis of F. We write $f:=f_1 \cdots f_n$ and $\overline {f}:=\overline {f}_1 \cdots \overline {f}_{n-1}$ .

Since the vertical maps are quadratic, it is not sufficient to show commutativity on a spanning set. We therefore consider

$$\begin{align*}a:=\sum_{I \subseteq [n]} c_I e_I, \end{align*}$$

where, for $I=\{i_1<\ldots <i_k\}$ , we write $e_I:=e_{i_1} \cdots e_{i_k}$ . We then have

$$\begin{align*}\pi_e(af)=\sum_{I: n \not \in I} c_I \overline{e_I}\overline{f}=:\overline{a}\overline{f} \end{align*}$$

and

$$\begin{align*}\widehat{\nu_2}(\overline{a}\overline{f})=\text{the component in } \bigwedge\nolimits^{n-1} V_e \text{ of } \sum_{I,J: n \not \in I \cup J} (c_I c_J \overline{e_I} \overline{f} \overline{e_J}^*) \bullet 1 \in \bigwedge\nolimits V_e. \end{align*}$$

Now note that, since $\overline {f}$ has $n-1$ factors, if $I,J$ do not have the same parity, then acting with $\overline {e_I} \overline {f} \overline {e_J}^*$ on $1$ yields a zero contribution in $\bigwedge \nolimits ^{n-1} V_e$ . Hence, the sum above may be split into two sums, one of which is

(6.2) $$ \begin{align} \text{the component in } \bigwedge\nolimits^{n-1} V_e \text{ of} \sum_{I,J: |I|,|J| \text{ even, } n \not \in I \cup J} (c_I c_J \overline{e_I} \overline{f} \overline{e_J}^*) \bullet 1. \end{align} $$

However, consider

$$\begin{align*}\widehat{\nu_2}(af)=\text{the component in } \bigwedge\nolimits^n V \text{ of} \sum_{I,J} (c_I c_J e_I f e_J^*) \bullet 1 \in \bigwedge\nolimits V. \end{align*}$$

For the same reason as above, this splits into two sums, and we want to compare the following expression to (6.2):

(6.3) $$ \begin{align} c_e(\text{the component in } \bigwedge\nolimits^n V \text{ of } \sum_{I,J:|I|,|J| \text{ even}} (c_I c_J e_I f e_J^*) \bullet 1). \end{align} $$

Now recall that the action of $e=e_n \in V \subseteq \operatorname{\mathrm{Cl}} (V)$ on $\bigwedge \nolimits V$ is via $o(e)+\iota (e)$ , while $c_e$ is $\iota _e$ followed by projection to $\bigwedge \nolimits ^{n-1} V_e$ . Hence, to compute (6.3), we may as well compute the summands of

$$\begin{align*}\text{the component in } \bigwedge\nolimits^n V \text{ of } \sum_{I,J:|I|,|J|\text{ even}} (c_I c_J \cdot e \cdot e_I f e_J^*) \bullet 1 \end{align*}$$

that do not contain a factor e. Terms with $n \in I$ do not contribute because then $e e_I=0$ . Terms with $n \not \in I$ but $n \in J$ do not contribute because when e gets contracted with $f_n$ a factor e in $e_J^*$ survives, and when e does not get contracted with $f_n$ , we use $ee_J^*=0$ . So we may restrict attention to the terms with $n \not \in I \cup J$ . Let $I,J$ correspond to such a term; that is, $|I|,|J|$ are even and $n \not \in I \cup J$ . Write $I=\{i_1<\ldots <i_k\}$ and $J=\{j_1<\ldots <j_l\}$ . Then

$$ \begin{align*} (e e_I f e_J^*) \bullet 1 &=((-1)^{n-1} e_I f_1 \cdots f_{n-1} e f_n e_J^*) \bullet 1 \\ &=((-1)^{n-1} e_I f_1 \cdots f_{n-1} e) \bullet (f_n \wedge e_{j_l} \wedge \cdots \wedge e_{j_1}) \\ &=((-1)^{n-1} e_I f_1 \cdots f_{n-1}) \bullet (e_{j_l} \wedge \cdots \wedge e_{j_1} + e \wedge f_n \wedge e_{j_l} \wedge \cdots \wedge e_{j_1}). \end{align*} $$

The second term in the last expression will contribute only terms with a factor e to the final result, and the former term contributes

$$\begin{align*}\text{the component in } \bigwedge\nolimits^{n-1} V_e \text{ of } (-1)^{n-1} (\overline{e_I} \overline{f} \overline{e_J}^*) \bullet 1. \end{align*}$$

Comparing this with (6.2), we see that the diagram commutes on terms in $\operatorname{\mathrm{Cl}} ^+(V)f$ up to the factor $(-1)^{n-1}$ . A similar computation shows that it commutes on terms in $\operatorname{\mathrm{Cl}} ^-(V)f$ up to a factor factor $(-1)^n$ .

We now consider the second diagram, where V is split as the orthogonal direct sum $V_e \oplus \langle e,h \rangle $ with $e=e_n,h=f_n$ . Consider $a \in \operatorname{\mathrm{Cl}} (\langle e_1,\ldots ,e_{n-1} \rangle )$ . By the same argument as above, it suffices to consider the case where all summands of a in the basis $e_I$ have indices I with $|I|$ of the same parity, say even. Then $\widehat \nu _2 \circ \tau _h$ in the diagram sends $a\overline {f}$ to the component in $\bigwedge \nolimits ^n V$ of $afa^* \bullet 1$ . Since the summands $e_I$ in a all have $n \not \in I$ , in $a f a^* \bullet 1$ all summands have a factor $f_n$ , and indeed,

$$\begin{align*}(a f a^*) \bullet 1 = f_n \wedge (a \overline{f} a^* \bullet 1) \end{align*}$$

(when all terms in a have $|I|$ odd, we get a minus sign). The component in $\bigwedge \nolimits ^n V$ of this expression is the same as the one obtained via $m_h \circ \widehat \nu _2$ .