2.1 Three-by-three hermitian matrices

Suppose that $B$ is a quaternion algebra over the ground field $F$ of characteristic zero. For $a\in B$ , write $a^{\ast }$ for the conjugate of $a$ , $\operatorname{tr}(a)=a+a^{\ast }$ , $n(a)=aa^{\ast }$ , the trace and norm of  $a$ . These are considered as elements of  $F$ . The symmetric pairing on $B$ is defined by

(2.1) $$\begin{eqnarray}(u,v)=\operatorname{tr}(uv^{\ast })=(v,u).\end{eqnarray}$$

We consider $3\times 3$ hermitian matrices over $B$ . If $h$ is such a matrix, then

(2.2) $$\begin{eqnarray}h=\left(\begin{array}{@{}ccc@{}}c_{1} & a_{3} & a_{2}^{\ast }\\ a_{3}^{\ast } & c_{2} & a_{1}\\ a_{2} & a_{1}^{\ast } & c_{3}\end{array}\right)\end{eqnarray}$$

where $c_{1},c_{2},c_{3}$ are elements of $F$ and $a_{1},a_{2},a_{3}$ are elements of  $B$ . The set of such matrices is denoted $H_{3}(B)$ .

For $h$ as in (2.2), define

$$\begin{eqnarray}N(h):=c_{1}c_{2}c_{3}-c_{1}n(a_{1})-c_{2}n(a_{2})-c_{3}n(a_{3})+\operatorname{tr}(a_{1}a_{2}a_{3})\end{eqnarray}$$

the norm of $h$ , and

$$\begin{eqnarray}\operatorname{tr}(h)=c_{1}+c_{2}+c_{3}\end{eqnarray}$$

the trace of $h$ . Also define

(2.3) $$\begin{eqnarray}h^{\#}:=\left(\begin{array}{@{}ccc@{}}c_{2}c_{3}-n(a_{1}) & a_{2}^{\ast }a_{1}^{\ast }-c_{3}a_{3} & a_{3}a_{1}-c_{2}a_{2}^{\ast }\\ a_{1}a_{2}-c_{3}a_{3}^{\ast } & c_{1}c_{3}-n(a_{2}) & a_{3}^{\ast }a_{2}^{\ast }-c_{1}a_{1}\\ a_{1}^{\ast }a_{3}^{\ast }-c_{2}a_{2} & a_{2}a_{3}-c_{1}a_{1}^{\ast } & c_{1}c_{2}-n(a_{3})\end{array}\right).\end{eqnarray}$$

One has that $hh^{\#}=N(h)1_{3}=h^{\#}h$ , and $(h^{\#})^{\#}=N(h)h$ .

For $x,y$ $3\times 3$ hermitian matrices over $B$ , define

$$\begin{eqnarray}x\times y:=(x+y)^{\#}-x^{\#}-y^{\#}\end{eqnarray}$$

and

$$\begin{eqnarray}\operatorname{tr}(x,y):={\textstyle \frac{1}{2}}\operatorname{tr}(xy+yx).\end{eqnarray}$$

Finally, denote by

$$\begin{eqnarray}(\;,\;,\;):H_{3}(B)\times H_{3}(B)\times H_{3}(B)\rightarrow F\end{eqnarray}$$

the unique symmetric trilinear form satisfying $(x,x,x)=6N(x)$ . Equivalently,

$$\begin{eqnarray}(x,y,z)=N(x+y+z)-N(x+y)-N(x+z)-N(y+z)+N(x)+N(y)+N(z).\end{eqnarray}$$

For $x,y,z\in H_{3}(B)$ , one has

$$\begin{eqnarray}\operatorname{tr}(x\times y,z)=(x,y,z)=\operatorname{tr}(x,y\times z).\end{eqnarray}$$

The trilinear form is nondegenerate in the sense that if $(x,y,z)=0$ for all $x,y$ , then $z=0$ . The formula

$$\begin{eqnarray}\operatorname{tr}(x,y)=\mathop{\sum }_{1\leqslant i\leqslant 3}x_{ii}y_{ii}+\mathop{\sum }_{1\leqslant i<j\leqslant 3}(x_{ij},y_{ij})\end{eqnarray}$$

shows that the bilinear form $\operatorname{tr}(\;,\;)$ is nondegenerate. Here $x_{ij}$ denotes the $i,j$ entry of $x$ , and $(\;,\;)$ is the symmetric pairing on $B$ defined in (2.1). For $x,y\in H_{3}(B)$ , one has the formula

(2.4) $$\begin{eqnarray}N(x+y)=N(x)+\operatorname{tr}(x^{\#},y)+\operatorname{tr}(x,y^{\#})+N(y).\end{eqnarray}$$

2.2 The defining representation

Fix the quaternion algebra $B$ over $F$ . We denote by $W$ the vector space $F\oplus H_{3}(B)\oplus H_{3}(B)\oplus F$ . Typical elements of $W$ are denoted $(a,b,c,d)$ with $a,d\in F$ , $b,c\in H_{3}(B)$ . The space $W$ is equipped with a symplectic and quartic form. The symplectic form $\langle \;,\;\rangle :W\times W\rightarrow F$ is defined by

$$\begin{eqnarray}\langle (a,b,c,d),(a^{\prime },b^{\prime },c^{\prime },d^{\prime })\rangle =ad^{\prime }-\operatorname{tr}(b,c^{\prime })+\operatorname{tr}(c,b^{\prime })-da^{\prime }.\end{eqnarray}$$

The quartic form $Q:W\rightarrow F$ is defined by

$$\begin{eqnarray}Q((a,b,c,d))=(ad-\operatorname{tr}(b,c))^{2}+4aN(c)+4dN(b)-4\operatorname{tr}(b^{\#},c^{\#}).\end{eqnarray}$$

We now define the (similitude) algebraic group $\text{G}$ .

Typically, the Freudenthal construction defines the subgroup of $\text{G}$ with similitude 1, but we will need this similitude version. It is clear that $\text{G}$ is a reductive group; knowledge of the generators of $\text{G}$ proves that it is connected as well [Reference BrownBro69].

We will give some explicit elements of $\text{G}$ . Most simply, we have the following maps:

1. – for $\unicode[STIX]{x1D706}\in \operatorname{GL}_{1}$ , $(a,b,c,d)\mapsto (\unicode[STIX]{x1D706}a,\unicode[STIX]{x1D706}b,\unicode[STIX]{x1D706}c,\unicode[STIX]{x1D706}d)$ , with similitude $\unicode[STIX]{x1D706}^{2}$ ;

2. – and $(a,b,c,d)\mapsto (\unicode[STIX]{x1D706}^{2}a,\unicode[STIX]{x1D706}b,c,\unicode[STIX]{x1D706}^{-1}d)$ , with similitude $\unicode[STIX]{x1D706}$ ;

3. – $(a,b,c,d)\mapsto (-d,c,-b,a)$ with similitude 1.

This last map we denote by $J_{6}$ , or $J$ , if no confusion seems likely. Under the map $\operatorname{GSp}_{6}\rightarrow \text{G}$ defined in § 5.1, $J_{6}$ is the image of the element $(\!\begin{smallmatrix} & 1_{3}\\ -1_{3} & \end{smallmatrix}\!)$ of $\operatorname{GSp}_{6}$ .

Here are some more interesting elements of $\text{G}$ . First, for $X\in H_{3}(B)$ , define $n(X):W\rightarrow W$ via the formula

(2.5) $$\begin{eqnarray}(a,b,c,d)n(X)=(a,b+aX,c+b\times X+aX^{\#},d+\operatorname{tr}(c,X)+\operatorname{tr}(b,X^{\#})+aN(X)).\end{eqnarray}$$

Similarly, for $Y\in H_{3}(B)$ , define $\bar{n}(Y):W\rightarrow W$ via the formula

$$\begin{eqnarray}(a,b,c,d)\bar{n}(Y)=(a+\operatorname{tr}(b,Y)+\operatorname{tr}(c,Y^{\#})+dN(Y),b+c\times Y+dY^{\#},c+dY,d).\end{eqnarray}$$

The following lemma says that the $n(X),\bar{n}(Y)$ are elements of $\text{G}$ with similitude 1, that $n(X_{1}+X_{2})=n(X_{1})n(X_{2})$ , and similarly for $\bar{n}$ . For a proof of the lemma, see [Reference SpringerSpr06, Lemma 2.3].

We define one more type of (fairly) explicit element of $\text{G}$ . Consider triples

$$\begin{eqnarray}(t,\tilde{t},\unicode[STIX]{x1D706})\in \operatorname{GL}(H_{3}(B))\times \operatorname{GL}(H_{3}(B))\times \operatorname{GL}_{1}\end{eqnarray}$$

that satisfy, for all $b,c\in H_{3}(B)$ :

1. – $\operatorname{tr}(t(b),\tilde{t}(c))=\operatorname{tr}(b,c)$ ;

2. – $N(t(b))=\unicode[STIX]{x1D706}N(b)$ ;

3. – $N(\tilde{t}(c))=\unicode[STIX]{x1D706}^{-1}N(c)$ ;

4. – $t(b)^{\#}=\unicode[STIX]{x1D706}\tilde{t}(b^{\#})$ and $\tilde{t}(c)^{\#}=\unicode[STIX]{x1D706}^{-1}t(c^{\#})$ .

Since the pairing $\operatorname{tr}(\;,\;)$ is nondegenerate, the first property uniquely determines $\tilde{t}$ in terms of $t$ , and then the first two properties imply the final two. If $m\in \operatorname{GL}_{3}(B)$ , and $r\in \operatorname{GL}_{1}(F)$ , then the linear map $t(b)=r^{-1}m^{\ast }bm$ satisfies $N(t(b))=r^{-3}N(m^{\ast }m)N(b)$ for all $b$ in $H_{3}(B)$ , and thus defines such an automorphism. For such a triple $(t,\tilde{t},\unicode[STIX]{x1D706})$ , one defines a map $m(t):W\rightarrow W$ via

$$\begin{eqnarray}(a,b,c,d)m(t)=(\unicode[STIX]{x1D706}a,t(b),\tilde{t}(c),\unicode[STIX]{x1D706}^{-1}d).\end{eqnarray}$$

It is immediate that $m(t)$ defines an element of $\text{G}$ with similitude 1.

We now define the rank of an element of $W$ . First, polarizing the quartic form $Q$ , there is a unique symmetric 4-linear form on $W$ that satisfies $(v,v,v,v)=Q(v)$ for all $v$ in  $W$ .

Elements of $W$ of rank one will play an important role in this paper. For example, $f=(0,0,0,1)$ is of rank one, and then applying $\overline{n}(X)$ for $X$ in $H_{3}(B)$ , one gets $(N(X),X^{\#},X,1)$ is rank one. To check that $f$ is rank one, the reader must simply observe (immediately) that the coefficient of $\unicode[STIX]{x1D707}\unicode[STIX]{x1D707}^{\prime }$ in $Q(f+\unicode[STIX]{x1D707}w+\unicode[STIX]{x1D707}^{\prime }w^{\prime })$ is zero for general $w=(a,b,c,d)$ and $w^{\prime }=(0,b^{\prime },c^{\prime },d^{\prime })$ in  $W$ .

It is easy to check that all the elements of rank one are in a single $\text{G}$ orbit. For example, one argument goes as follows. First, by applying operators $n(X)$ , $J_{6}$ , and scalar multiplication, it is easy to see that every element $v$ of $W$ can be moved to one with $d=1$ . Then applying $\overline{n}(X)$ , every element has a $\text{G}$ translate of the form $v_{0}=(z,y,0,1)$ . Now, if $v$ and thus $v_{0}$ is rank one, then $(v_{0},v_{0},w,w^{\prime })=0$ for arbitrary $w=(0,b,c,0)$ , $w^{\prime }=(0,b^{\prime },0,0)$ . The coefficient of $2\unicode[STIX]{x1D707}\unicode[STIX]{x1D707}^{\prime }$ in $Q(v_{0}+\unicode[STIX]{x1D707}w+\unicode[STIX]{x1D707}w^{\prime })$ is now easy to compute, and it is found to be $2\operatorname{tr}(y,b\times b^{\prime })-z\operatorname{tr}(b^{\prime },c)$ . Since this must be zero for all $b,b^{\prime },c$ , it follows that $y$ and $z$ are zero, by the nondegeneracy of the two pairings. Hence $v_{0}=(0,0,0,1)=f$ .

If $v$ is a rank-one element, the stabilizer in $\text{G}$ of the line $Fv$ is a parabolic subgroup of $\text{G}$ . Indeed, consider the set of flags of the form $0\subseteq \ell \subseteq W^{\prime }\subseteq W$ where $\ell$ is dimension one and $W^{\prime }$ is codimension one. The conditions $\langle v,w^{\prime }\rangle =0$ and $(v,v,w,w^{\prime })=0$ for all $v\in \ell$ , $w^{\prime }\in W^{\prime }$ and $w\in W$ clearly form closed conditions on the Grassmanian of flags of these dimensions. Since these conditions single out the rank-one lines, and since $\text{G}$ acts transitively on these lines, the stabilizer of a rank-one line is a parabolic subgroup.

The argument above also shows that if $(z,y,x,1)$ is rank one, then $y=x^{\#}$ and $z=N(x)$ . (Just apply $\overline{n}(-x)$ to $(z,y,x,1)$ .) In fact, the following is true.

A proof of the proposition may be found in [Reference Gan and SavinGS05, Proposition 11.2].

5.1 Construction of global integral

We first embed $\operatorname{GSp}_{6}$ into $\text{G}$ , preserving the similitude. Denote by $W_{6}$ the defining six-dimensional representation of $\operatorname{GSp}_{6}$ , and $\langle \;,\;\rangle$ the invariant symplectic form. We let $\operatorname{GSp}_{6}$ act on $W_{6}$ on the right. In matrices, $\operatorname{GSp}_{6}$ is the set of $g\in \operatorname{GL}_{6}$ satisfying $gJ_{6}\text{}^{t}g=\unicode[STIX]{x1D708}(g)J_{6}$ , where $J_{6}=(\!\begin{smallmatrix} & 1_{3}\\ -1_{3} & \end{smallmatrix}\!)$ and the similitude $\unicode[STIX]{x1D708}(g)\in \operatorname{GL}_{1}$ . The action of $\operatorname{GSp}_{6}$ on $W_{6}$ is then the usual right action of matrices on row vectors.

Pick a symplectic basis $e_{1},e_{2},e_{3},f_{1},f_{2},f_{3}$ of $W_{6}$ ; that is, $\langle e_{i},f_{j}\rangle =\unicode[STIX]{x1D6FF}_{ij}$ . We fix a quaternion algebra $B$ over $\mathbf{Q}$ that is ramified at infinity. For such a $B$ , denote by $W=\mathbf{Q}\,\oplus \,H_{3}(B)\oplus H_{3}(B)\oplus \mathbf{Q}$ Freudenthal’s space, as considered in § 2. Now consider the linear map $W\rightarrow (\bigwedge ^{3}W_{6}\otimes \unicode[STIX]{x1D708}^{-1})\otimes _{\mathbf{Q}}B$ defined as follows: set $e_{i}^{\ast }=e_{i+1}\wedge e_{i+2}$ and $f_{i}^{\ast }=f_{i+1}\wedge f_{i+2}$ with the indices taken modulo 3. Then we map $(a,b,c,d)$ in $W$ to

$$\begin{eqnarray}ae_{1}\wedge e_{2}\wedge e_{3}+\biggl(\mathop{\sum }_{i,j}b_{ij}e_{i}^{\ast }\wedge f_{j}\biggr)+\biggl(\mathop{\sum }_{i,j}c_{ij}f_{i}^{\ast }\wedge e_{j}\biggr)+df_{1}\wedge f_{2}\wedge f_{3}.\end{eqnarray}$$

With this definition, and the natural action of $\operatorname{GSp}_{6}$ on $\bigwedge ^{3}W_{6}\otimes \unicode[STIX]{x1D708}^{-1}$ , $\operatorname{GSp}_{6}$ preserves the image of $W$ , and preserves the symplectic and quartic form on $W$ , up to similitude. These facts are checked in the following lemma.

Lemma 5.1. Identify $W$ with its image in $(\bigwedge ^{3}W_{6}\otimes \unicode[STIX]{x1D708}^{-1})\otimes _{\mathbf{Q}}B$ . Then, the element $(\!\begin{smallmatrix}1 & X\\ & 1\end{smallmatrix}\!)$ of $\operatorname{GSp}_{6}$ acts on $W$ as $n(X)$ , and the element $(\!\begin{smallmatrix}0 & 1\\ -1 & 0\end{smallmatrix}\!)$ acts on $W$ via $(a,b,c,d)\mapsto (-d,c,-b,a)$ . If $(\!\begin{smallmatrix}m & \\ & n\end{smallmatrix}\!)\in \operatorname{GSp}_{6}$ has similitude $\unicode[STIX]{x1D708}$ , then

$$\begin{eqnarray}(a,b,c,d)(\!\begin{smallmatrix}m & \\ & n\end{smallmatrix}\!)=\unicode[STIX]{x1D708}^{-1}(\det (m)a,\det (m)m^{-1}bn,\det (n)n^{-1}cm,\det (n)\,d).\end{eqnarray}$$

It follows that the inclusion $W\rightarrow (\bigwedge ^{3}W_{6}\,\otimes \,\unicode[STIX]{x1D708}^{-1})\,\otimes _{\mathbf{Q}}B$ induces an inclusion $\operatorname{GSp}_{6}\rightarrow \text{G}$ , preserving similitudes.

Proof. The formula for the action of $(\!\begin{smallmatrix} & 1\\ -1 & \end{smallmatrix}\!)$ is immediate. We check some of the other formulas.

Consider the action of $(\!\begin{smallmatrix}m & \\ & n\end{smallmatrix}\!)$ . One first computes that under the action of $m$ , $e_{i}^{\ast }\mapsto \sum _{j}c(m)_{ij}e_{j}^{\ast }$ , where $c(m)=\det (m)\text{}^{t}m^{-1}$ is the cofactor matrix of  $m$ . Ignoring the twist by the similitude $\unicode[STIX]{x1D708}^{-1}$ in $\bigwedge ^{3}W_{6}\otimes \unicode[STIX]{x1D708}^{-1}$ , one then obtains

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{i,j}b_{ij}e_{i}^{\ast }\wedge f_{j} & \mapsto & \displaystyle \mathop{\sum }_{i,j,t,s}b_{ij}c(m)_{it}n_{js}e_{t}^{\ast }\wedge f_{s}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{t,s}(\text{}^{t}c(m)bn)_{ts}e_{t}^{\ast }\wedge f_{s}.\nonumber\end{eqnarray}$$

This proves the formula for the action of $(\!\begin{smallmatrix}m & \\ & n\end{smallmatrix}\!)$ on the $b$ values of $(a,b,c,d)$ . The action on the $c$ values is similar, and the action on the $a$ values and $b$ values are trivial.

For $X$ in $H_{3}(B)$ , denote by $L(X)$ the logarithm, inside $\text{End}(W)$ , of the element $n(X)$ of $\operatorname{GL}(W)$ . That is, set $L(X)=\sum _{m\geqslant 1}(-1)^{m-1}((n(X)-1)^{m}/m)$ . From the definition (2.5) of $n(X)$ , one sees $(n(X)-1)^{4}=0$ , and thus the sum defining $L(X)$ terminates after four terms. One gets

$$\begin{eqnarray}(a,b,c,d)L(X)=(0,aX,b\times X,\operatorname{tr}(c,X))\end{eqnarray}$$

and $n(X)=1+L(X)+L(X)^{2}/2+L(X)^{3}/6$ .

Similarly, for $(\!\begin{smallmatrix}1 & X\\ & 1\end{smallmatrix}\!)$ in $\operatorname{GSp}_{6}$ , denote by $n_{6}(X)$ its action on $W_{6}$ , $n^{\prime }(X)$ its action on $(\bigwedge ^{3}W_{6}\otimes \unicode[STIX]{x1D708}^{-1})\otimes B$ , and $L_{6}(X)$ , respectively $L^{\prime }(X)$ , the logarithms of these actions. Since $n_{6}(X)=1+L_{6}(X)$ , and the action of $n(X)$ on $\bigwedge ^{3}W_{6}$ is defined by cubic polynomials in the entries of $(\!\begin{smallmatrix}1 & X\\ & 1\end{smallmatrix}\!)$ , one obtains $n^{\prime }(X)=1+L^{\prime }(X)+L^{\prime }(X)^{2}/2+L^{\prime }(X)^{3}/6.$ Hence to check that $n^{\prime }(X)$ restricts to $n(X)$ on $W$ , it suffices to check the analogous statement for $L^{\prime }(X)$ and $L(X)$ .

So, suppose $X$ is a $3\times 3$ symmetric matrix. We consider the action $(0,b,0,0)L^{\prime }(X)$ . One has

$$\begin{eqnarray}(e_{i}^{\ast }\wedge f_{j})L^{\prime }(X)=\mathop{\sum }_{k}X_{i+1,k}f_{k}\wedge e_{i+2}\wedge f_{j}+X_{i+2,k}e_{i+1}\wedge f_{k}\wedge f_{j},\end{eqnarray}$$

where $X_{a,b}$ denotes the $a,b$ matrix entry of $X$ . Hence

(5.1) $$\begin{eqnarray}\displaystyle \biggl(\mathop{\sum }_{ij}b_{ij}e_{i}^{\ast }\wedge f_{j}\biggr)L^{\prime }(X) & = & \displaystyle \mathop{\sum }_{j,k,\ell }(b_{\ell +1,j}X_{\ell -1,k}-X_{\ell +1,k}b_{\ell -1,j})f_{j}\wedge f_{k}\wedge e_{\ell }\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD}}V_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD}}f_{\unicode[STIX]{x1D6FC}}^{\ast }\wedge e_{\unicode[STIX]{x1D6FD}}\end{eqnarray}$$

for some element $V$ of $M_{3}(B)$ . For a general element $Y$ of $M_{3}(B)$ , denote by $c_{i}(Y),a_{i}(Y),a_{i}(Y)^{\prime }$ the coordinates of  $Y$ :

$$\begin{eqnarray}Y=\left(\begin{array}{@{}ccc@{}}c_{1}(Y) & a_{3}(Y) & a_{2}(Y)^{\prime }\\ a_{3}(Y)^{\prime } & c_{2}(Y) & a_{1}(Y)\\ a_{2}(Y) & a_{1}(Y)^{\prime } & c_{3}(Y)\end{array}\right).\end{eqnarray}$$

Then (5.1) yields

(5.2) $$\begin{eqnarray}\begin{array}{@{}rcl@{}}c_{t}(V)\ & =\ & c_{t-1}(X)c_{t+1}(b)+c_{t+1}(X)c_{t-1}(b)-(a_{t}(X),a_{t}(b)),\\ a_{t}(V)\ & =\ & a_{t+1}^{\prime }(b)a_{t-1}^{\prime }(X)+a_{t+1}^{\prime }(X)a_{t-1}^{\prime }(b)-(c_{t}(X)a_{t}(b)+c_{t}(b)a_{t}(X)),\\ a_{t}^{\prime }(V)\ & =\ & a_{t+1}(b)a_{t-1}(X)+a_{t+1}(X)a_{t-1}(b)-(c_{t}(X)a_{t}^{\prime }(b)+c_{t}(b)a_{t}^{\prime }(X)).\end{array}\end{eqnarray}$$

The indices in (5.2) are taken modulo 3. Linearizing (2.3) for the map $\#$ yields exactly the formulas (5.2), and thus $V=b\times X$ . Hence $(0,b,0,0)L^{\prime }(X)=(0,0,b\times X,0)$ .

The formulas $(a,0,0,0)L^{\prime }(X)=(0,aX,0,0)$ and $(0,0,c,d)L^{\prime }(X)=(0,0,0,\operatorname{tr}(c,X))$ are much easier to establish. Hence $L^{\prime }(X)$ preserves $W$ , and restricts to the action of $L(X)$ . This completes the proof of the lemma.◻

We need to define an integral structure on $W$ . To do this, first we fix a maximal order $B_{0}$ in  $B$ . Then

(5.3) $$\begin{eqnarray}W(\mathbf{Z}):=W(\mathbf{Q})\cap \{(\wedge ^{3}W_{6}(\mathbf{Z})\otimes \unicode[STIX]{x1D708}^{-1})\otimes _{\boldsymbol{ Z}}B_{0}\},\end{eqnarray}$$

and for a finite prime $p$ , $W(\mathbf{Z}_{p}):=W(\mathbf{Z})\otimes \mathbf{Z}_{p}$ . Define $J_{0}$ to be the lattice in $H_{3}(B)$ consisting of those

(5.4) $$\begin{eqnarray}\left(\begin{array}{@{}ccc@{}}c_{1} & x_{3} & x_{2}^{\ast }\\ x_{3}^{\ast } & c_{2} & x_{1}\\ x_{2} & x_{1}^{\ast } & c_{3}\end{array}\right)\end{eqnarray}$$

with $c_{i}\in \mathbf{Z}$ and $x_{i}\in B_{0}$ . Then $(a,b,c,d)\in W(\mathbf{Q})$ is in $W(\mathbf{Z})$ if and only if $a,d\in \mathbf{Z}$ and $b,c\in J_{0}$ . It is clear from (5.3) that $\operatorname{GSp}_{6}(\mathbf{Z}_{p})$ stabilizes $W(\mathbf{Z}_{p})$ for all finite primes  $p$ .

We now define the Eisenstein series on $\text{G}$ . Recall the rank-one element $f=(0,0,0,1)$ of $W$ . Denote by $P$ the parabolic subgroup of $\text{G}$ stabilizing the line $\mathbf{Q}f$ . The Eisenstein series will be associated to a character of  $P$ . To define this Eisenstein series, we select a particular Schwartz–Bruhat function $\unicode[STIX]{x1D6F7}=\prod _{v}\unicode[STIX]{x1D6F7}_{v}$ on $W(\mathbf{A})$ . At finite primes, $\unicode[STIX]{x1D6F7}_{p}$ is the characteristic function of $W(\mathbf{Z}_{p})$ . In § 6.2 will define a Schwartz function $\unicode[STIX]{x1D6F7}_{\infty ,2r}$ on $W(\mathbf{R})$ that is appropriate for cusp forms $\unicode[STIX]{x1D719}$ that correspond to a Siegel modular form of weight $2r$ . The Schwartz function $\unicode[STIX]{x1D6F7}_{\infty ,2r}$ will be a Gaussian times a polynomial of degree $2r$ . We set

$$\begin{eqnarray}f^{\unicode[STIX]{x1D6F7}}(g,s)=|\unicode[STIX]{x1D708}(g)|^{s}\int _{\operatorname{GL}_{1}(\mathbf{A})}\unicode[STIX]{x1D6F7}(tfg)|t|^{2s}\,dt,\end{eqnarray}$$

and

$$\begin{eqnarray}E^{\unicode[STIX]{x1D6F7}}(g,s)=\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in P(\mathbf{Q})\backslash \!\text{G}(\mathbf{Q})}f^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D6FE}g,s).\end{eqnarray}$$

This is not yet the normalized Eisenstein series. Assume $\unicode[STIX]{x1D6F7}_{\infty ,2r}$ is chosen as in § 6.2, and denote by $D_{B}=\prod _{p<\infty ,\text{ramified}}p$ the product of the finite primes ramified in  $B$ . Then the normalized Eisenstein series is

$$\begin{eqnarray}\displaystyle E_{2r}^{\ast }(g,s) & = & \displaystyle \unicode[STIX]{x1D70B}^{-(s+r)}D_{B}^{s}\unicode[STIX]{x1D6E4}_{\mathbf{R}}(2s+2r-4)\unicode[STIX]{x1D6E4}_{\mathbf{R}}(2s+2r-8)\biggl(\mathop{\prod }_{p<\infty ~\text{split}}\unicode[STIX]{x1D701}_{p}(2s-2)\unicode[STIX]{x1D701}_{p}(2s-4)\biggr)\nonumber\\ \displaystyle & & \displaystyle \times \,\biggl(\mathop{\prod }_{p<\infty ~\text{ramified}}\unicode[STIX]{x1D701}_{p}(2s-4)\biggr)E^{\unicode[STIX]{x1D6F7}}(g,s).\nonumber\end{eqnarray}$$

Here, as usual, $\unicode[STIX]{x1D6E4}_{\mathbf{R}}(s)=\unicode[STIX]{x1D70B}^{-s/2}\unicode[STIX]{x1D6E4}(s/2)$ . The following theorem is proved in § 7.

For a cusp form $\unicode[STIX]{x1D719}$ associated to a level-one, holomorphic Siegel modular form of weight $2r$ , the global integral is

$$\begin{eqnarray}I_{2r}^{\ast }(\unicode[STIX]{x1D719},s)=\int _{\operatorname{ GSp}_{6}(\mathbf{Q})Z(\mathbf{A})\backslash \!\operatorname{GSp}_{6}(\mathbf{A})}\unicode[STIX]{x1D719}(g)E_{2r}^{\ast }(g,s)\,dg.\end{eqnarray}$$

By Theorem 5.2, $I_{2r}^{\ast }(\unicode[STIX]{x1D719},s)=I_{2r}^{\ast }(\unicode[STIX]{x1D719},5-s)$ .

5.2 Unfolding

To unfold this integral, we rely on the following lemma.

To do the unfolding, we need to understand the double coset space $P(\mathbf{Q})\backslash \text{G}(\mathbf{Q})/\!\operatorname{GSp}_{6}(\mathbf{Q})$ , or equivalently, the $\operatorname{GSp}_{6}(\mathbf{Q})$ orbits on the rank-one lines. Recall that $T$ denotes a half-integral symmetric $3\times 3$ matrix that corresponds to the order $B_{0}$ , and set $A(T)$ to be the rank-one matrix in $H_{3}(B)$ from Lemma 3.8. Define $f_{{\mathcal{O}}}=(0,0,A(T),0)$ in $W$ . By Proposition 2.5, $f_{{\mathcal{O}}}$ is rank one. If $v$ is a rank-one element of $W(\mathbf{Q})$ , an orbit $\mathbf{Q}v\operatorname{GSp}_{6}(\mathbf{Q})$ is said to be negligible if the stabilizer of $\mathbf{Q}v$ in $\operatorname{GSp}_{6}(\mathbf{Q})$ contains the unipotent radical of a parabolic subgroup of $\operatorname{GSp}_{6}$ .

Before giving the proof, define $J_{6}=e_{1}\bigwedge f_{1}+e_{2}\bigwedge f_{2}+e_{3}\bigwedge f_{3}$ . Note that if $a_{1},a_{2},a_{3}$ in $B$ are purely imaginary, and $c_{1},c_{2},c_{3}$ are in $\mathbf{Q}$ , then

(5.5) $$\begin{eqnarray}(a_{1}f_{1}+a_{2}f_{2}+a_{3}f_{3})\wedge J_{6}+c_{1}e_{1}\wedge f_{2}\wedge f_{3}+c_{2}f_{1}\wedge e_{2}\wedge f_{3}+c_{3}f_{1}\wedge f_{2}\wedge e_{3}\end{eqnarray}$$

is the element

$$\begin{eqnarray}\left(0,0,\left(\begin{array}{@{}ccc@{}}c_{1} & a_{3} & -a_{2}\\ -a_{3} & c_{2} & a_{1}\\ a_{2} & -a_{1} & c_{3}\end{array}\right),0\right)\end{eqnarray}$$

of $W$ .

Proof of Proposition 5.5.

Suppose we are given a rank-one element $v=(a,b,c,d)$ of $W(\mathbf{Q})$ . We will first use the action of $\operatorname{GSp}_{6}(\mathbf{Q})$ to put it into a standard form. By Lemma 5.4, we may assume $d=1$ . Then our element is of the form $(N(c),c^{\#},c,1)$ . Now use the transformations $\overline{n}(Y)$ to make $c$ purely imaginary.

Denote by $M_{P,6}$ the Levi subgroup of the Siegel parabolic on $\operatorname{GSp}_{6}$ , i.e., the elements of $\operatorname{GSp}_{6}$ of the form $(\!\begin{smallmatrix}m & \\ & n\end{smallmatrix}\!)$ . By formula (5.5), $M_{P,6}$ acts on the space of such $c$ values as $B^{\operatorname{tr}=0}\otimes _{\mathbf{Q}}\,\text{span}\{f_{1},f_{2},f_{3}\}$ . That is, $M_{P,6}$ acts on these $c$ values via its usual action on $\text{span}\{f_{1},f_{2},f_{3}\}\subseteq W_{6}$ , linearly extended to $B^{\operatorname{tr}=0}\otimes _{\mathbf{Q}}\,\text{span}\{f_{1},f_{2},f_{3}\}$ . Pick a basis $i,j,k$ of $B^{\operatorname{tr}=0}$ , and suppose $c=i\otimes v_{1}+j\otimes v_{2}+k\otimes v_{3}$ . We have four different cases, depending on whether $\text{dim}\,\text{span}\{v_{1},v_{2},v_{3}\}$ is 0, 1, 2 or 3. (Note that this dimension is not a $\operatorname{GSp}_{6}$ invariant of the orbit, just an $M_{P,6}$ invariant.) If the dimension is 0, i.e, $c=0$ , then we are in the orbit of $f$ , which is negligible. There are an infinite number of orbits with $\text{dim}\,\text{span}\{v_{1},v_{2},v_{3}\}=1$ . Given such an orbit, we can use $M_{P,6}$ to move $c$ to $h\otimes f_{1}$ with $h\in B^{\operatorname{tr}=0}$ . Then this orbit is stabilized by $U_{Q}(f_{1})$ , the unipotent radical of the parabolic subgroup of $\operatorname{GSp}_{6}$ stabilizing the line $\mathbf{Q}f_{1}\subseteq W_{6}$ . Thus, this orbit is also negligible. We will show that there is a unique orbit with $\dim =3$ , and that this orbit is represented by $f_{{\mathcal{O}}}$ . Furthermore, we will see that the orbit corresponding to $\dim =2$ is the same as the $\dim =3$ orbit.

First, for $\dim =3$ , use $M_{P,6}$ to move $c$ to $i\otimes f_{1}+j\otimes f_{2}+k\otimes f_{3}$ , proving that there is one such orbit.

Now we consider a $\dim =2$ orbit. Here we have $c=i\otimes v_{1}+j\otimes v_{2}+k\otimes v_{3}$ with one of the $v_{i}$ in the span of the other two. Thus we may use $M_{P,6}$ to move this orbit to one with $c=h_{1}\otimes f_{1}+h_{2}\otimes f_{2}$ , with $h_{1},h_{2}$ in $B^{\operatorname{tr}=0}$ . If $\dim \text{span}\{h_{1},h_{2}\}=1$ , then we are really in the $\dim =1$ case. Thus we have that $h_{1},h_{2}$ are linearly independent.

Now

$$\begin{eqnarray}c=\left(\begin{array}{@{}ccc@{}}0 & 0 & -h_{2}\\ 0 & 0 & h_{1}\\ h_{2} & -h_{1} & 0\end{array}\right),\end{eqnarray}$$

from which we obtain

$$\begin{eqnarray}c^{\sharp }=\left(\begin{array}{@{}ccc@{}}-n(h_{1}) & h_{2}h_{1} & 0\\ h_{1}h_{2} & -n(h_{2}) & 0\\ 0 & 0 & 0\end{array}\right)\end{eqnarray}$$

and $N(c)=0$ . Hence we have that $v=(N(c),c^{\#},c,1)$ is

$$\begin{eqnarray}-n(h_{1})e_{1}^{\ast }\wedge f_{1}-n(h_{2})e_{2}^{\ast }\wedge f_{2}+h_{1}h_{2}e_{2}^{\ast }\wedge f_{1}+h_{2}h_{1}e_{1}^{\ast }\wedge f_{2}+(h_{1}f_{1}+h_{2}f_{2})\wedge J_{6}+f_{1}\wedge f_{2}\wedge f_{3}.\end{eqnarray}$$

Now apply to $v$ the element of $\operatorname{GSp}_{6}$ that takes $e_{3}\mapsto f_{3}$ , $f_{3}\mapsto -e_{3}$ and is the identity on $e_{1},e_{2},f_{1},f_{2}$ . One obtains the element $v^{\prime }=(0,0,h,0)$ , with

$$\begin{eqnarray}h=\left(\begin{array}{@{}ccc@{}}-n(h_{2}) & -h_{2}h_{1} & -h_{2}\\ -h_{1}h_{2} & -n(h_{1}) & h_{1}\\ h_{2} & -h_{1} & -1\end{array}\right).\end{eqnarray}$$

Since $B$ is not split, $h_{1}h_{2}$ is not in the span of $1,h_{1},h_{2}$ . Take $h_{3}$ in $B^{\operatorname{tr}=0}$ with $h_{1}h_{2}-h_{3}$ in $\mathbf{Q}$ . Then $\operatorname{Im}(h)=(h_{1}f_{1}+h_{2}f_{2}+h_{3}f_{3})\bigwedge J_{6}$ . Since

$$\begin{eqnarray}h_{1}\otimes f_{1}+h_{2}\otimes f_{2}+h_{3}\otimes f_{3}=i\otimes v_{1}+j\otimes v_{2}+k\otimes v_{3}\end{eqnarray}$$

for some $v_{1},v_{2},v_{3}$ spanning $\langle f_{1},f_{2},f_{3}\rangle$ , the $\dim =2$ orbit is the same as the $\dim =3$ orbit.

The element $f_{{\mathcal{O}}}$ clearly represents the $\dim =3$ orbit. To compute the stabilizer, note that by (5.5) we immediately see that if $f_{{\mathcal{O}}}g=f_{{\mathcal{O}}}$ , then $f_{1}g=f_{1},f_{2}g=f_{2},f_{3}g=f_{3}$ . Hence an element of the stabilizer must be of the form $g=(\!\begin{smallmatrix}\unicode[STIX]{x1D706} & \\ & 1\end{smallmatrix}\!)(\!\begin{smallmatrix}1 & U\\ & 1\end{smallmatrix}\!)$ . Applying this element to $f_{{\mathcal{O}}}$ , we find it stabilizes $f_{{\mathcal{O}}}$ if and only if $\operatorname{tr}(TU)=0$ , completing the proof.◻

Recall the partially normalized Eisenstein series $E^{\unicode[STIX]{x1D6F7}}(g,s)$ from above, and set

$$\begin{eqnarray}I^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D719},s)=\int _{\operatorname{GSp}_{6}(\mathbf{Q})Z(\mathbf{A})\backslash \!\operatorname{GSp}_{6}(\mathbf{A})}\unicode[STIX]{x1D719}(g)E^{\unicode[STIX]{x1D6F7}}(g,s)\,dg.\end{eqnarray}$$

We will consider the integral $I^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D719},s)$ for now, and add back in the other normalizing factors later.

Write $U_{P}$ for the unipotent radical of the Siegel parabolic of $\operatorname{GSp}_{6}$ . The group $U_{P}$ consists of the matrices $(\!\begin{smallmatrix}1 & u\\ & 1\end{smallmatrix}\!)$ with $u=^{t}u$ . Define $U_{0}\subseteq U_{P}$ to be the elements with $\operatorname{tr}(Tu)=0$ . Write $\unicode[STIX]{x1D713}:\mathbf{Q}\backslash \mathbf{A}\rightarrow \mathbf{C}^{\times }$ for the usual additive character of conductor 1 satisfying $\unicode[STIX]{x1D713}_{\infty }(x)=e^{2\unicode[STIX]{x1D70B}ix}$ for $x\in \mathbf{R}$ . Define $\unicode[STIX]{x1D712}:U_{P}(\mathbf{Q})\backslash U_{P}(\mathbf{A})\rightarrow \mathbf{C}^{\times }$ via $\unicode[STIX]{x1D712}((\!\begin{smallmatrix}1 & h\\ & 1\end{smallmatrix}\!))=\unicode[STIX]{x1D713}(\operatorname{tr}(Th))$ , and set

(5.6) $$\begin{eqnarray}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(g)=\int _{U_{P}(\mathbf{Q})\backslash U_{P}(\mathbf{A})}\unicode[STIX]{x1D712}^{-1}(u)\unicode[STIX]{x1D719}(ug)\,du.\end{eqnarray}$$

Theorem 5.7. The global integrals $I^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D719},s),I_{2r}^{\ast }(\unicode[STIX]{x1D719},s)$ converge absolutely for all $s$ for which the Eisenstein series $E^{\unicode[STIX]{x1D6F7}}(g,s)$ , respectively, $E_{2r}^{\ast }(g,s)$ is finite, and thus define a meromorphic function of  $s$ . For $\text{Re}(s)$ sufficiently large, the integral $I^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D719},s)$ unfolds as

$$\begin{eqnarray}\displaystyle I^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D719},s) & = & \displaystyle \int _{U_{0}(\mathbf{A})\backslash \!\operatorname{GSp}_{6}(\mathbf{A})}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(g)|\unicode[STIX]{x1D708}(g)|^{s}\unicode[STIX]{x1D6F7}(f_{{\mathcal{O}}}g)\,dg\nonumber\\ \displaystyle & = & \displaystyle \int _{U_{P}(\mathbf{A})\backslash \!\operatorname{GSp}_{6}(\mathbf{A})}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(g)|\unicode[STIX]{x1D708}(g)|^{s}\biggl(\int _{U_{0}(\mathbf{A})\backslash U_{P}(\mathbf{A})}\unicode[STIX]{x1D712}(u)\unicode[STIX]{x1D6F7}(f_{{\mathcal{O}}}ug)\,du\biggr)\,dg.\nonumber\end{eqnarray}$$

Proof. The Eisenstein series $E^{\unicode[STIX]{x1D6F7}}(g,s)$ is a function of moderate growth wherever it is finite. Since the cusp form $\unicode[STIX]{x1D719}$ decreases rapidly on Siegel sets, the integral $I^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D719},s)$ converges absolutely for all $s$ for which the Eisenstein series is finite.

Suppose now $\text{Re}(s)\gg 0$ , and denote by ${\mathcal{F}}$ a fundamental domain for $\operatorname{GSp}_{6}(\mathbf{Q})Z(\mathbf{A})\backslash \operatorname{GSp}_{6}(\mathbf{A})$ . Similar to the above, but easier, $\sum _{\unicode[STIX]{x1D6FE}\in P(\mathbf{Q})\backslash \text{G}(\mathbf{Q})}|f^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D6FE}g,s)|$ converges to a function of moderate growth since $\text{Re}(s)\gg 0$ . Then again since $\unicode[STIX]{x1D719}$ decreases rapidly on Siegel sets, the sum

$$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in P(\mathbf{Q})\backslash \text{G}(\mathbf{Q})}\int _{{\mathcal{F}}}|f^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D6FE}g,s)|\,|\unicode[STIX]{x1D719}(g)|\,dg\end{eqnarray}$$

is finite. Hence even though there are an infinite number of $\operatorname{GSp}_{6}(\mathbf{Q})$ orbits on $P(\mathbf{Q})\backslash \text{G}(\mathbf{Q})$ , we may analyze them, and check the vanishing of the associated integrals, one by one.

By Proposition 5.5, we have

(5.7) $$\begin{eqnarray}I^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D719},s)=\int _{\operatorname{GL}_{1}(\mathbf{Q})U_{0}(\mathbf{Q})Z(\mathbf{A})\backslash \!\operatorname{GSp}_{6}(\mathbf{A})}\unicode[STIX]{x1D719}(g)f^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D6FE}_{0}g,s)\,dg\end{eqnarray}$$

where the $\operatorname{GL}_{1}(\mathbf{Q})$ is the set of elements of $\operatorname{GSp}_{6}(\mathbf{Q})$ of the form $(\!\begin{smallmatrix}\unicode[STIX]{x1D706} & \\ & 1\end{smallmatrix}\!)$ , with $\unicode[STIX]{x1D706}\in \mathbf{Q}^{\times }$ , and $\unicode[STIX]{x1D6FE}_{0}\in \text{G}(\mathbf{Q})$ satisfies $(0,0,0,1)\unicode[STIX]{x1D6FE}_{0}=f_{{\mathcal{O}}}$ . Indeed, this is the term from the open orbit, and the negligible orbits contribute zero by the cuspidality of  $\unicode[STIX]{x1D719}$ .

Now we integrate $\unicode[STIX]{x1D719}$ over $U_{0}(\mathbf{Q})\backslash U_{0}(\mathbf{A})$ and Fourier expand:

$$\begin{eqnarray}\int _{U_{0}(\mathbf{Q})\backslash U_{0}(\mathbf{A})}\unicode[STIX]{x1D719}(ug)\,du=\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in \operatorname{GL}_{1}(\mathbf{Q})}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D6FE}g).\end{eqnarray}$$

Placing this Fourier expansion into (5.7) we get

(5.8) $$\begin{eqnarray}I^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D719},s)=\int _{U_{0}(\mathbf{A})Z(\mathbf{A})\backslash \!\operatorname{GSp}_{6}(\mathbf{A})}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(g)f^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D6FE}_{0}g,s)\,dg.\end{eqnarray}$$

Integrating

$$\begin{eqnarray}\int _{U_{0}(\mathbf{A})\backslash \!\operatorname{GSp}_{6}(\mathbf{A})}|\unicode[STIX]{x1D708}(g)|^{s}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(g)\unicode[STIX]{x1D6F7}(f_{{\mathcal{O}}}g)\,dg\end{eqnarray}$$

over $Z(\mathbf{A})$ one obtains (5.8), giving the theorem.◻

6.1 The finite places

Our first task is to compute the unipotent integral $\int _{U_{0}\backslash U_{P}}\unicode[STIX]{x1D712}(u)\unicode[STIX]{x1D6F7}(f_{{\mathcal{O}}}ug)\,du$ . For an element $m\in \operatorname{GL}_{3}$ , and an element $\unicode[STIX]{x1D706}\in \operatorname{GL}_{1}$ , we set

an element of $\operatorname{GSp}_{6}$ . For $(\unicode[STIX]{x1D706},m)\in \operatorname{GL}_{1}(\mathbf{A}_{f})\times \operatorname{GL}_{3}(\mathbf{A}_{f})$ , we define

$$\begin{eqnarray}\unicode[STIX]{x1D6EF}(\unicode[STIX]{x1D706},m):=\left\{\begin{array}{@{}ll@{}}1\quad & \text{if}~\unicode[STIX]{x1D706}\in \widehat{\mathbf{Z}},m\in M_{3}(\widehat{\mathbf{Z}}),\text{and}~m^{-1}Tc(m)\in M_{3}(\widehat{\mathbf{Z}}),\\ 0\quad & \text{otherwise}.\end{array}\right.\end{eqnarray}$$

Proposition 6.1. The integral

(6.1) $$\begin{eqnarray}\int _{U_{0}(\mathbf{A}_{f})\backslash U_{P}(\mathbf{A}_{f})}\unicode[STIX]{x1D712}(u)\unicode[STIX]{x1D6F7}_{f}(f_{{\mathcal{O}}}u(\unicode[STIX]{x1D706},m))\,du=|\unicode[STIX]{x1D706}|_{f}\unicode[STIX]{x1D6EF}(\unicode[STIX]{x1D706},m).\end{eqnarray}$$

Proof. Recall the lattice $J_{0}$ in $H_{3}(B)$ defined in (5.4), and set $J_{0}(\widehat{\mathbf{Z}})=J_{0}\otimes _{\mathbf{Z}}\widehat{\mathbf{Z}}$ . Suppose $h=(\unicode[STIX]{x1D706},m)$ . One computes

(6.2) $$\begin{eqnarray}f_{{\mathcal{O}}}n(u)h=\biggl(0,0,m^{-1}A(T)c(m),\frac{\operatorname{tr}(Tu)}{\unicode[STIX]{x1D706}}\biggr).\end{eqnarray}$$

Hence for the integral (6.1) to be nonzero, one needs $m^{-1}A(T)c(m)$ to be in $J_{0}(\widehat{\mathbf{Z}})$ and the integral

(6.3) $$\begin{eqnarray}\int _{U_{0}(\mathbf{A}_{f})\backslash U_{P}(\mathbf{A}_{f})}\unicode[STIX]{x1D713}(\operatorname{tr}(Tu))\operatorname{char}\biggl(\frac{\operatorname{tr}(Tu)}{\unicode[STIX]{x1D706}}\in \widehat{\mathbf{Z}}\biggr)\,du\end{eqnarray}$$

to be nonzero. By Lemma 3.8, $m^{-1}A(T)c(m)$ is in $J_{0}(\widehat{\mathbf{Z}})$ if and only if $m\in M_{3}(\widehat{\mathbf{Z}})$ and $m^{-1}Tc(m)\in M_{3}(\widehat{\mathbf{Z}})$ . The integral (6.3) is zero when $\unicode[STIX]{x1D706}\notin \widehat{\mathbf{Z}}$ , and is $|\unicode[STIX]{x1D706}|_{f}$ when $\unicode[STIX]{x1D706}\in \widehat{\mathbf{Z}}$ . The proposition follows.◻

The global integral we must compute is

$$\begin{eqnarray}\displaystyle I^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D719},s) & = & \displaystyle \int _{U_{0}(\mathbf{R})\backslash \!\operatorname{GSp}_{6}(\mathbf{R})}|\unicode[STIX]{x1D708}(g_{\infty })|_{\infty }^{s}\unicode[STIX]{x1D6F7}_{\infty }(f_{{\mathcal{O}}}g_{\infty })\nonumber\\ \displaystyle & & \displaystyle \times \,\biggl(\int _{U_{0}(\mathbf{A}_{f})\backslash \!\operatorname{GSp}_{6}(\mathbf{A}_{f})}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(g_{f}g_{\infty })|\unicode[STIX]{x1D708}(g_{f})|_{f}^{s}\unicode[STIX]{x1D6F7}_{f}(f_{{\mathcal{O}}}g_{f})\,dg_{f}\biggr)\,dg_{\infty }\nonumber\\ \displaystyle & = & \displaystyle \int _{U_{0}(\mathbf{R})\backslash \!\operatorname{GSp}_{6}(\mathbf{R})}|\unicode[STIX]{x1D708}(g_{\infty })|_{\infty }^{s}\unicode[STIX]{x1D6F7}_{\infty }(f_{{\mathcal{O}}}g_{\infty })I_{f}(\unicode[STIX]{x1D719},s,g_{\infty })\,dg_{\infty }\nonumber\end{eqnarray}$$

where

(6.4) $$\begin{eqnarray}\displaystyle I_{f}(\unicode[STIX]{x1D719},s,g_{\infty }) & = & \displaystyle \int _{U_{0}(\mathbf{A}_{f})\backslash \!\operatorname{GSp}_{6}(\mathbf{A}_{f})}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(g_{f}g_{\infty })|\unicode[STIX]{x1D708}(g_{f})|_{f}^{s}\unicode[STIX]{x1D6F7}_{f}(f_{{\mathcal{O}}}g_{f})\,dg_{f}\nonumber\\ \displaystyle & = & \displaystyle \int _{(\operatorname{GL}_{1}\times \operatorname{GL}_{3})(\mathbf{A}_{f})}\unicode[STIX]{x1D6FF}_{P}^{-1}(h)|\text{det}(m)|_{f}^{-1}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D70B}(h)g_{\infty })|\unicode[STIX]{x1D708}(h)|_{f}^{s+1}\unicode[STIX]{x1D6EF}(\unicode[STIX]{x1D706},m)\,dh.\end{eqnarray}$$

Here $h=(\unicode[STIX]{x1D706},m)$ , $\unicode[STIX]{x1D6FF}_{P}$ denotes the modulus character of the Siegel parabolic in $\operatorname{GSp}_{6}$ , and we have applied Proposition 6.1, the Iwasawa decomposition, and the fact that $\unicode[STIX]{x1D706}=\unicode[STIX]{x1D708}(h)\det (m)^{-1}$ .

Translating between the classical and adelic language, the Dirichlet series of Corollary 4.2 exactly evaluates the integral (6.4), as we check in the following proposition.

Proposition 6.2. Assume $\unicode[STIX]{x1D719}$ is as in Definition 1.1. Then

$$\begin{eqnarray}I_{f}(\unicode[STIX]{x1D719},s,g_{\infty })=\frac{L(\unicode[STIX]{x1D70B},\text{Spin},s-2)}{\unicode[STIX]{x1D701}(2s-4)\unicode[STIX]{x1D701}_{D_{B}}(2s-2)}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(g_{\infty }).\end{eqnarray}$$

Proof. If $\unicode[STIX]{x1D708}(g_{\infty })>0$ , one has

(6.5) $$\begin{eqnarray}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(g_{\infty })=a_{f}(T)\unicode[STIX]{x1D708}(g_{\infty })^{-r}j(g_{\infty },i)^{-2r}e^{2\unicode[STIX]{x1D70B}i\operatorname{tr}(Tg_{\infty }\cdot i)},\end{eqnarray}$$

while $\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(g_{\infty })=0$ if $\unicode[STIX]{x1D708}(g_{\infty })<0$ . Suppose $h=(\unicode[STIX]{x1D706},m)\in M_{P,6}(\mathbf{A}_{f})$ . Then $h=h_{\mathbf{Q}}h_{\infty }^{-1}k$ for some $h\in M_{P,6}(\mathbf{Q})$ and $k\in M_{P,6}(\mathbf{Z}_{p})$ , where $h_{\infty }=(h_{\mathbf{Q}})_{\infty }$ is the component of $h_{\mathbf{Q}}$ at the archimedean place. Hence $\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(hg_{\infty })=\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(h_{\mathbf{Q}}h_{\infty }^{-1}g_{\infty })=\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}\cdot h_{\mathbf{Q}}}(h_{\infty }^{-1}g_{\infty })$ , where $\unicode[STIX]{x1D712}\cdot h_{\mathbf{Q}}$ is the Fourier coefficient (5.6) except with $T$ replaced by $\unicode[STIX]{x1D706}m^{-1}Tc(m)$ . Thus,

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(hg_{\infty }) & = & \displaystyle a_{f}(\unicode[STIX]{x1D706}m^{-1}Tc(m))\unicode[STIX]{x1D708}(h_{\infty }^{-1}g_{\infty })^{-r}j(h_{\infty }^{-1}g_{\infty },i)^{-2r}e^{2\unicode[STIX]{x1D70B}i\operatorname{tr}((\unicode[STIX]{x1D706}m^{-1}Tc(m))(h_{\infty }^{-1}g_{\infty })\cdot i)}\nonumber\\ \displaystyle & = & \displaystyle a_{f}(\unicode[STIX]{x1D706}m^{-1}Tc(m))\unicode[STIX]{x1D706}^{-3r}\det (m)^{-r}(\unicode[STIX]{x1D708}(g_{\infty })^{-r}j(g_{\infty },i)^{-2r}e^{2\unicode[STIX]{x1D70B}i\operatorname{tr}(Tg_{\infty }\cdot i)}).\nonumber\end{eqnarray}$$

Hence if $h\in (h_{\mathbf{Q}})_{f}M_{P,6}(\mathbf{Z}_{p})$ , with $h_{\mathbf{Q}}=(\unicode[STIX]{x1D706},m)$ , then the integrand in (6.4) is equal to

$$\begin{eqnarray}\displaystyle & & \displaystyle (\unicode[STIX]{x1D706}^{6}\det (m)^{2})\det (m)(a_{f}(\unicode[STIX]{x1D706}m^{-1}Tc(m))\unicode[STIX]{x1D706}^{-3r}\det (m)^{-r})(\unicode[STIX]{x1D706}\det (m))^{-s-1}\unicode[STIX]{x1D6EF}(\unicode[STIX]{x1D706},m)\nonumber\\ \displaystyle & & \displaystyle \qquad \times (\unicode[STIX]{x1D708}(g_{\infty })^{-r}j(g_{\infty },i)^{-2r}e^{2\unicode[STIX]{x1D70B}i\operatorname{tr}(Tg_{\infty }\cdot i)})\nonumber\\ \displaystyle & & \displaystyle \quad =\unicode[STIX]{x1D706}^{-(s+3r-5)}\det (m)^{-(s+r-2)}\unicode[STIX]{x1D6EF}(\unicode[STIX]{x1D706},m)a_{f}(\unicode[STIX]{x1D706}m^{-1}Tc(m))\times (\unicode[STIX]{x1D708}(g_{\infty })^{-r}j(g_{\infty },i)^{-2r}e^{2\unicode[STIX]{x1D70B}i\operatorname{tr}(Tg_{\infty }\cdot i)}).\nonumber\end{eqnarray}$$

Applying Corollary 4.2 and (6.5) again gives the proposition. ◻

6.2 The infinite place

Suppose $\unicode[STIX]{x1D719}$ is a cusp form corresponding to a level-one, holomorphic Siegel modular form of weight $2r$ , as in Definition 1.1. We now compute the normalized archimedean integral

$$\begin{eqnarray}I_{\infty ,2r}^{\ast }(\unicode[STIX]{x1D719},s)=\unicode[STIX]{x1D70B}^{-(s+r)}D_{B}^{s}\unicode[STIX]{x1D6E4}_{\mathbf{R}}(2s+2r-4)\unicode[STIX]{x1D6E4}_{\mathbf{R}}(2s+2r-8)I_{\infty ,2r}^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D719},s),\end{eqnarray}$$

where

(6.6) $$\begin{eqnarray}I_{\infty ,2r}^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D719},s)=\int _{U_{0}(\mathbf{R})\backslash \!\operatorname{GSp}_{6}(\mathbf{R})}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(g)|\unicode[STIX]{x1D708}(g)|^{s}\unicode[STIX]{x1D6F7}_{\infty ,2r}(f_{{\mathcal{O}}}g)\,dg.\end{eqnarray}$$

We prove the following theorem.

Theorem 6.3. Suppose $\unicode[STIX]{x1D719}$ is as in Definition 1.1, and that $\unicode[STIX]{x1D719}$ corresponds to the level-one Siegel modular form  $f$ . Then $I_{\infty ,2r}^{\ast }(\unicode[STIX]{x1D719},s)$ equals

$$\begin{eqnarray}a_{f}(T)\unicode[STIX]{x1D6E4}_{\mathbf{C}}(s+r-4)\unicode[STIX]{x1D6E4}_{\mathbf{C}}(s+r-3)\unicode[STIX]{x1D6E4}_{\mathbf{C}}(s+r-2)\unicode[STIX]{x1D6E4}_{\mathbf{C}}(s+3r-5)\end{eqnarray}$$

up to a nonzero constant.

Our first task is to define the Schwartz function $\unicode[STIX]{x1D6F7}_{\infty ,2r}$ on $W(\mathbf{R})$ . To do this, we digress slightly and discuss the Hermitian symmetric space ${\mathcal{H}}$ associated to $\text{G}(\mathbf{R})$ .

6.2.1 The Hermitian symmetric space for $\text{G}(\mathbf{R})$

We assume $B_{\mathbf{R}}=B\otimes \mathbf{R}$ is the Hamilton quaternions. Set $B_{\mathbf{C}}=B_{\mathbf{R}}\otimes _{\mathbf{R}}\mathbf{C}$ , with the involution on $B_{\mathbf{C}}$ the one from $B_{\mathbf{R}}$ , extended $\mathbf{C}$ -linearly to $B_{\mathbf{C}}$ . Associated to $B_{\mathbf{C}}$ , we have the space $H_{3}(B_{\mathbf{C}})$ of $3\times 3$ Hermitian matrices over $B_{\mathbf{C}}$ . Elements of $H_{3}(B_{\mathbf{C}})$ are formal expressions $Z=X+iY$ , with $X,Y$ in $H_{3}(B_{\mathbf{R}})$ . Note that the $i$ in this expression is not the $i$ in $B_{\mathbf{R}}$ ; this $i$ is in the center of $B_{\mathbf{C}}$ . The Hermitian symmetric space ${\mathcal{H}}$ is the set of $Z=X+iY$ in $H_{3}(B_{\mathbf{C}})$ with $Y$ positive definite.

Denote by $\text{G}^{+}(\mathbf{R})$ the subgroup of $\text{G}(\mathbf{R})$ that is the connected component of the identity. We explain the action of $\text{G}^{+}(\mathbf{R})$ on ${\mathcal{H}}$ . First, note that we have an inclusion $\text{G}(\mathbf{R})\rightarrow \text{G}(\mathbf{C})$ , and $\text{G}(\mathbf{C})$ acts on $W(\mathbf{C})=W(\mathbf{R})\otimes _{\mathbf{R}}\mathbf{C}$ . Set $e=(1,0,0,0)$ , a rank-one element of  $W$ . Identify $Z\in {\mathcal{H}}$ with the rank-one element

$$\begin{eqnarray}r(Z)=en(-Z)=(1,-Z,Z^{\sharp },-N(Z)).\end{eqnarray}$$

Consider the projective space $\mathbf{P}(W(\mathbf{C})^{rk=1})$ of rank-one $\mathbf{C}$ -lines in $W(\mathbf{C})$ . Then $\text{G}(\mathbf{R})$ acts on this space on the left via $g\cdot v=vg^{-1}$ . We will see momentarily that $\text{G}^{+}(\mathbf{R})$ preserves the image of ${\mathcal{H}}$ inside $\mathbf{P}(W(\mathbf{C})^{rk=1})$ , and acts on it transitively. In fact, the factor of automorphy

$$\begin{eqnarray}j_{\text{G}(\mathbf{R})}(g,Z):\text{G}^{+}(\mathbf{R})\times {\mathcal{H}}\rightarrow \mathbf{C}^{\times }\end{eqnarray}$$

and the action of $\text{G}^{+}(\mathbf{R})$ on ${\mathcal{H}}$ are simultaneously defined by the equality

$$\begin{eqnarray}r(Z)g^{-1}=j_{\text{G}(\mathbf{R})}(g,Z)r(gZ).\end{eqnarray}$$

We summarize what we need in a proposition. Recall the rank-one element $f=(0,0,0,1)$ .

Proposition 6.4. Suppose $Z\in {\mathcal{H}}$ , and $g\in \text{G}^{+}(\mathbf{R})$ . Then the complex number $j_{\text{G}(\mathbf{R})}(g,Z)=\langle r(Z)g^{-1},f\rangle$ is nonzero, and thus

$$\begin{eqnarray}r(Z)g^{-1}=j_{\text{G}(\mathbf{R})}(g,Z)r(gZ)\end{eqnarray}$$

for some $gZ$ in $H_{3}(B_{\mathbf{C}})$ . The element $gZ$ is in ${\mathcal{H}}$ , and thus this equality defines an action of $\text{G}^{+}(\mathbf{R})$ on ${\mathcal{H}}$ . Under the natural embedding ${\mathcal{H}}_{3}\rightarrow {\mathcal{H}}$ , the two actions of $\operatorname{GSp}_{6}^{+}(\mathbf{R})$ are the same. Furthermore, $j_{\operatorname{GSp}_{6}}(g,Z)=j_{\text{G}(\mathbf{R})}(g,Z)$ when $g\in \operatorname{GSp}_{6}^{+}(\mathbf{R})$ and $Z\in {\mathcal{H}}_{3}$ .

Proof. We sketch the proof. First, a minimal parabolic subgroup $P_{0}$ of $\text{G}$ is discussed in § 7.1. One can check that the subgroup $P_{0}^{+}(\mathbf{R})$ of $\text{G}^{+}(\mathbf{R})$ has $j_{\text{G}(\mathbf{R})}(P_{0}^{+}(\mathbf{R}),{\mathcal{H}})\neq 0$ and acts transitively on ${\mathcal{H}}$ . Now suppose $g\in \text{G}^{+}(\mathbf{R})$ , $Z\in {\mathcal{H}}$ , and $p\in P_{0}^{+}(\mathbf{R})$ satisfies $p\cdot i=Z$ . Then

$$\begin{eqnarray}\langle r(Z)g^{-1},f\rangle =j_{\text{G}(\mathbf{R})}(p,i)^{-1}\langle r(i)p^{-1}g^{-1},f\rangle =\unicode[STIX]{x1D708}(gp)^{-1}j_{\text{G}(\mathbf{R})}(p,i)^{-1}\langle r(i),fgp\rangle .\end{eqnarray}$$

This last quantity is then nonzero by Lemma 6.5 below. Hence $r(Z)g^{-1}=j_{\text{G}(\mathbf{R})}(g,Z)r(gZ)$ for some $gZ$ in $H_{3}(B_{\mathbf{C}})$ . Next, one verifies the equality $N(\operatorname{Im}(W))=(1/8i)\langle \overline{r(W)},r(W)\rangle$ for any $W\in H_{3}(B_{\mathbf{C}})$ . One deduces from this equality that $N(\operatorname{Im}(gZ))>0$ . Then that $gZ\in {\mathcal{H}}$ follows from a continuity argument.

To check that the actions and factors of automorphy agree for $g\in \operatorname{GSp}_{6}^{+}(\mathbf{R})$ and $Z\in {\mathcal{H}}_{3}$ , it suffices to check this on generators $n(X),J_{6}$ and $M_{P,6}^{+}(\mathbf{R})$ , which may be done easily.◻

Before defining $\unicode[STIX]{x1D6F7}_{\infty ,2r}$ , we need an additional lemma. Recall the element $J$ of $\text{G}(\mathbf{Q})$ that sends $(a,b,c,d)\mapsto (-d,c,-b,a)$ .

Lemma 6.5. If $v=(a,b,c,d)$ , define $\Vert v\Vert ^{2}=\langle v,vJ\rangle$ . Then $\Vert v\Vert ^{2}=a^{2}+\operatorname{tr}(b,b)+\operatorname{tr}(c,c)+d^{2}$ , and hence if $v\in W(\mathbf{R})$ , $\Vert v\Vert ^{2}\geqslant 0$ and is only $0$ when $v=0$ . If furthermore $v$ is of rank one, then $|\langle r(i),v\rangle |^{2}=\Vert v\Vert ^{2}$ .

Proof. The formula for $\Vert v\Vert ^{2}$ is trivial. Suppose $v$ is rank one. We have $\langle r(i),v\rangle =(d-\operatorname{tr}(b))+i(\operatorname{tr}(c)-a)$ . Hence

$$\begin{eqnarray}\displaystyle |\langle r(i),v\rangle |^{2} & = & \displaystyle (d-\operatorname{tr}(b))^{2}+(a-\operatorname{tr}(c))^{2}\nonumber\\ \displaystyle & = & \displaystyle d^{2}-2\,d\operatorname{tr}(b)+\operatorname{tr}(b)^{2}+a^{2}-2a\operatorname{tr}(c)+\operatorname{tr}(c)^{2}\nonumber\\ \displaystyle & = & \displaystyle a^{2}+(\operatorname{tr}(b)^{2}-2\operatorname{tr}(b^{\sharp }))+(\operatorname{tr}(c)^{2}-2\operatorname{tr}(c^{\sharp }))+d^{2},\nonumber\end{eqnarray}$$

where in the last equality, since $v$ is rank one, $db=c^{\sharp }$ and $ac=b^{\sharp }$ . For general $W$ one has the equality $\operatorname{tr}(W)^{2}-2\operatorname{tr}(W^{\sharp })=\operatorname{tr}(W,W)$ . The lemma follows.◻

6.2.2 Archimedean calculation

Define

(6.7) $$\begin{eqnarray}\unicode[STIX]{x1D6F7}_{\infty ,2r}(v)=e^{-\Vert v\Vert ^{2}}\langle r(i),v\rangle ^{2r}.\end{eqnarray}$$

This is obviously a Schwartz function on $W(\mathbf{R})$ . Note that we have $\langle r(i),vgk\rangle =j(k,i)\langle r(i),vg\rangle$ for $k$ in $K_{\infty ,\operatorname{Sp}_{6}}$ . Our assumption on $\unicode[STIX]{x1D719}$ gives

$$\begin{eqnarray}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(g)=a_{f}(T)\unicode[STIX]{x1D708}(g)^{-r}j(g,i)^{-2r}e^{2\unicode[STIX]{x1D70B}i\operatorname{tr}(Tg\cdot i)}\end{eqnarray}$$

if $g\in \operatorname{GSp}_{6}^{+}(\mathbf{R})$ , and $\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(g)=0$ if $\unicode[STIX]{x1D708}(g)<0$ . (The modular form associated to $\unicode[STIX]{x1D719}$ has negative definite Fourier coefficients when restricted to the lower half-space, ${\mathcal{H}}_{3}^{-}$ .) Furthermore, for $g\in \operatorname{GSp}_{6}^{+}(\mathbf{R})$ , one has the identity

(6.8) $$\begin{eqnarray}\unicode[STIX]{x1D708}(g)^{-1}j(g,i)^{-1}\langle r(i),f_{{\mathcal{O}}}g\rangle =\operatorname{tr}(T(g\cdot i)).\end{eqnarray}$$

Note that the left-hand side of (6.8), as a function of  $g$ , is right $Z(\mathbf{R})K_{\infty ,\operatorname{Sp}_{6}}$ -invariant. Hence to prove the equality, it suffices to take $g$ in the Siegel parabolic of $\operatorname{GSp}_{6}$ , and then (6.8) follows from (6.2).

Now, putting (6.7) into (6.6) and integrating over the center $Z(\mathbf{R})$ , we obtain

$$\begin{eqnarray}\displaystyle I_{\infty ,2r}^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D719},s) & = & \displaystyle \unicode[STIX]{x1D6E4}(s+r)\int _{U^{0}(\mathbf{R})Z(\mathbf{R})\backslash \!\operatorname{GSp}_{6}(\mathbf{R})}\frac{|\unicode[STIX]{x1D708}(g)|^{s}\langle r(i),f_{{\mathcal{O}}}g\rangle ^{2r}\unicode[STIX]{x1D719}_{\unicode[STIX]{x1D712}}(g)}{|\langle r(i),f_{{\mathcal{O}}}g\rangle |^{2s+2r}}\,dg\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D6E4}(s+r)\int _{U^{0}(\mathbf{R})Z(\mathbf{R})\backslash \!\operatorname{GSp}_{6}^{+}(\mathbf{R})}a(T)\frac{|\unicode[STIX]{x1D708}(g)|^{s+r}\operatorname{tr}(TZ)^{2r}e^{2\unicode[STIX]{x1D70B}i\operatorname{tr}(TZ)}}{|\langle r(i),f_{{\mathcal{O}}}g\rangle |^{2s+2r}}\,dg,\nonumber\end{eqnarray}$$

where $Z=g\cdot i=X+iY$ . One has

$$\begin{eqnarray}\frac{|\unicode[STIX]{x1D708}(g)|^{s+r}}{|\langle r(i),f_{{\mathcal{O}}}g\rangle |^{2s+2r}}=\frac{|\unicode[STIX]{x1D708}(g)^{-1}j(g,i)^{-2}|^{s+r}}{|\operatorname{tr}(TZ)|^{2s+2r}}=\frac{N(Y)^{s+r}}{|\operatorname{tr}(TZ)|^{2s+2r}}.\end{eqnarray}$$

Hence, $I_{\infty ,2r}^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D719},s)=a(T)\unicode[STIX]{x1D6E4}(s+r)N(T)^{-(s+r)}J(s)$ , with

(6.9) $$\begin{eqnarray}\displaystyle J(s) & = & \displaystyle \int _{U^{0}(\mathbf{R})Z(\mathbf{R})\backslash \!\operatorname{GSp}_{6}^{+}(\mathbf{R})}\frac{N(TY)^{s+r}\operatorname{tr}(TZ)^{2r}e^{2\unicode[STIX]{x1D70B}i\operatorname{tr}(TZ)}}{|\operatorname{tr}(TZ)|^{2s+2r}}\,dg\nonumber\\ \displaystyle & = & \displaystyle \int _{U_{1}^{0}(\mathbf{R})Z(\mathbf{R})\backslash \!\operatorname{GSp}_{6}^{+}(\mathbf{R})}\frac{N(Y)^{s+r}\operatorname{tr}(Z)^{2r}e^{2\unicode[STIX]{x1D70B}i\operatorname{tr}(Z)}}{|\operatorname{tr}(Z)|^{2s+2r}}\,dg,\end{eqnarray}$$

where $U_{1}^{0}=\{(\!\begin{smallmatrix}1 & X\\ & 1\end{smallmatrix}\!):\operatorname{tr}(X)=0\}$ . The change of variables implicit in equality (6.9) is $g=m_{T}^{-1}g^{\prime }$ , where $m_{T}=(\!\begin{smallmatrix}T^{1/2} & \\ & T^{-1/2}\end{smallmatrix}\!)$ and $T^{1/2}$ a positive-definite symmetric square root of  $T$ .

We now explain how to compute the integral $J(s)$ . As the global integral representation is an analogue of Garrett’s triple product [Reference GarrettGar87], we follow the archimedean calculation of [Reference GarrettGar87]. We compute up to nonzero constants. First, we have that

$$\begin{eqnarray}J(s)=\int _{Z(\mathbf{R})\backslash M_{P,6}^{+}(\mathbf{R})}N(Y)^{s+r-2}e^{-2\unicode[STIX]{x1D70B}\operatorname{tr}(Y)}\int _{\mathbf{R}}\frac{(x+i\operatorname{tr}(Y))^{2r}}{|x+i\operatorname{tr}(Y)|^{2s+2r}}e^{2\unicode[STIX]{x1D70B}ix}\,dx\,d^{\ast }Y.\end{eqnarray}$$

Here $d^{\ast }Y$ is the $M_{P,6}^{+}(\mathbf{R})$ invariant measure on $Y$ , and we have used that $\unicode[STIX]{x1D6FF}_{P}(g)=\det (Y)^{2}$ for $g\in M_{P,6}^{+}(\mathbf{R})$ . Making a variable change, the inner integral becomes

$$\begin{eqnarray}\operatorname{tr}(Y)^{1-2s}\int _{\mathbf{R}}\frac{(x+i)^{2r}}{|x+i|^{2s+2r}}e^{2\unicode[STIX]{x1D70B}i\operatorname{tr}(Y)x}\,dx,\end{eqnarray}$$

using that $\operatorname{tr}(Y)$ is positive. We now introduce a $\unicode[STIX]{x1D6E4}$ -integral, and change the order of integration to obtain

$$\begin{eqnarray}\displaystyle (2\unicode[STIX]{x1D70B})^{1-2s}\unicode[STIX]{x1D6E4}(2s-1)J(s) & = & \displaystyle \int _{\mathbf{R}}\frac{(x+i)^{2r}}{|x+i|^{2s+2r}}\int _{Y}N(Y)^{s+r-2}e^{2\unicode[STIX]{x1D70B}i(x+i)\operatorname{tr}(Y)}\nonumber\\ \displaystyle & & \displaystyle \times \,\biggl(\int _{t\geqslant 0}e^{-2\unicode[STIX]{x1D70B}t\operatorname{tr}(Y)}t^{2s-1}\,\frac{dt}{t}\biggr)\,d^{\ast }Y\,dx.\nonumber\end{eqnarray}$$

Define a logarithm on $\mathbf{C}$ by removing the nonpositive reals, and use this logarithm to define $z^{s}$ for such a  $z$ . Then switching the order of integration we obtain

$$\begin{eqnarray}\displaystyle (2\unicode[STIX]{x1D70B})^{1-2s}\unicode[STIX]{x1D6E4}(2s-1)J(s) & = & \displaystyle \int _{\mathbf{R}}\int _{t\geqslant 0}\frac{(x+i)^{2r}}{|x+i|^{2s+2r}}(1+t-ix)^{-3(s+r-2)}t^{2s-1}\nonumber\\ \displaystyle & & \displaystyle \times \,\biggl(\int _{Y}N(Y^{\prime })^{s+r-2}e^{-2\unicode[STIX]{x1D70B}\operatorname{tr}(Y^{\prime })}\,d^{\ast }(Y^{\prime })\biggr)\,\frac{dt}{t}\,dx,\nonumber\end{eqnarray}$$

where $Y^{\prime }=(1+t-ix)Y$ . The integral over $Y$ is the so-called Siegel integral, and it is a nonzero constant times

$$\begin{eqnarray}2^{-(3s+3r-6)}\unicode[STIX]{x1D6E4}_{\mathbf{R}}(2s+2r-4)\unicode[STIX]{x1D6E4}_{\mathbf{R}}(2s+2r-5)\unicode[STIX]{x1D6E4}_{\mathbf{R}}(2s+2r-6).\end{eqnarray}$$

See, for instance, [Reference Faraut and KorányiFK94, Theorem VII.1.1] for this fact.

We are thus left to evaluate the integral

$$\begin{eqnarray}K(s,r)=\int _{\mathbf{R}}\int _{t\geqslant 0}\frac{(x+i)^{2r}}{|x+i|^{2s+2r}}(1+t-ix)^{-3(s+r-2)}t^{2s-1}\frac{dt}{t}\,dx,\end{eqnarray}$$

which may be done exactly as in [Reference GarrettGar87, Proof of Proposition 5.1]. Up to a nonzero constant, one obtains

$$\begin{eqnarray}K(s,r)=2^{-(3s+3r-6)}\frac{\unicode[STIX]{x1D6E4}(s+3r-5)\unicode[STIX]{x1D6E4}(2s-1)}{\unicode[STIX]{x1D6E4}(s+r)\unicode[STIX]{x1D6E4}(2s+2r-5)}.\end{eqnarray}$$

Thus, again up to nonzero constants,

$$\begin{eqnarray}J(s)=(2\unicode[STIX]{x1D70B})^{2s-1}4^{-(3s+3r-6)}\frac{\unicode[STIX]{x1D6E4}(s+3r-5)}{\unicode[STIX]{x1D6E4}(s+r)\unicode[STIX]{x1D6E4}(2s+2r-5)}\unicode[STIX]{x1D6E4}_{\mathbf{R}}(2s+2r-4)\unicode[STIX]{x1D6E4}_{\mathbf{R}}(2s+2r-5)\unicode[STIX]{x1D6E4}_{\mathbf{R}}(2s+2r-6).\end{eqnarray}$$

Via the duplication formula $\unicode[STIX]{x1D6E4}_{\mathbf{R}}(2s+2r-5)\unicode[STIX]{x1D6E4}_{\mathbf{R}}(2s+2r-4)=\unicode[STIX]{x1D6E4}_{\mathbf{C}}(2s+2r-5)$ , $J(s)$ simplifies to

$$\begin{eqnarray}4^{-(3s+3r-6)}\frac{\unicode[STIX]{x1D6E4}(s+3r-5)}{\unicode[STIX]{x1D6E4}(s+r)}\unicode[STIX]{x1D6E4}_{\mathbf{R}}(2s+2r-6),\end{eqnarray}$$

again up to nonzero constants. Hence $I_{\infty ,2r}^{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D719},s)=a(T)D_{B}^{-s}4^{-(2s+2r-6)}\unicode[STIX]{x1D6E4}(s+3r-5)\unicode[STIX]{x1D6E4}_{\mathbf{R}}(2s+2r-6)$ , since $D_{B}=4N(T)$ by Corollary 3.4. Thus, up to a nonzero constant,

$$\begin{eqnarray}I_{\infty ,2r}^{\ast }(\unicode[STIX]{x1D719},s)=a(T)\unicode[STIX]{x1D6E4}_{\boldsymbol{ C}}(s+r-4)\unicode[STIX]{x1D6E4}_{\mathbf{C}}(s+r-3)\unicode[STIX]{x1D6E4}_{\mathbf{C}}(s+r-2)\unicode[STIX]{x1D6E4}_{\mathbf{C}}(s+3r-5),\end{eqnarray}$$

completing the proof of Theorem 6.3.

7.1 Minimal parabolic and Iwasawa decomposition

Suppose $F$ is a field of characteristic zero, $B$ is a quaternion algebra over $F$ , $J=H_{3}(B)$ , and $\text{G}$ is the reductive $F$ -group that preserves the symplectic and quartic form on $W=F\oplus J\oplus J\oplus F$ , up to similitude. We describe a parabolic subgroup $P_{0}$ of $\text{G}$ , which is minimal when $B$ is a division algebra, and the corresponding Iwasawa decomposition.