Introduction
What \(X\) is. \(X\) is the space of nondegenerate alternating forms on \(k^{2n}\), with \(G\) acting by change of basis; for \(\mathrm{SL}_{2n}\) the Pfaffian is fixed:
$$\mathrm{GL}_{2n}/\mathrm{Sp}_{2n}\xrightarrow{\ \sim\ }\{\omega\in M_{2n}:\omega^T=-\omega,\ \det\omega\ne0\},\qquad g\,\mathrm{Sp}_{2n}\longmapsto gJg^T,\qquad h\cdot\omega=h\,\omega\,h^T,$$ $$\mathrm{SL}_{2n}/\mathrm{Sp}_{2n}\xrightarrow{\ \sim\ }\{\omega:\omega^T=-\omega,\ \operatorname{Pf}\omega=1\},\qquad\operatorname{Pf}(gJg^T)=\det g.$$The map is onto because every nondegenerate alternating form has a Darboux basis, and \(gJg^T=J\) exactly when \(g\in\mathrm{Sp}_{2n}\) (\(J\) is defined in section 1).
- Cartan (1926–27). The compact form \(\mathrm{SU}(2n)/\mathrm{Sp}(n)\) is type A II of his classification; its Satake diagram has the odd nodes black (section 4).
- Symplectic periods. Heumos–Rallis (1990) introduced symplectic models of \(\mathrm{GL}_{2n}\); Jacquet–Rallis (1992) proved that the \(\mathrm{Sp}_{2n}\)-period vanishes on cusp forms of \(\mathrm{GL}_{2n}\).
- Dual group. In Sakellaridis–Venkatesh, \(\check G_X=\mathrm{GL}_n\) for \(\mathrm{GL}_{2n}/\mathrm{Sp}_{2n}\), through \(\mathrm{GL}_n\times\mathrm{SL}_2\to\mathrm{GL}_{2n}\), \((g,h)\mapsto g\otimes h\); for \(\mathrm{SL}_{2n}/\mathrm{Sp}_{2n}\) it is \(\mathrm{PGL}_n\) (Proposition III, remarks).
- Chen–Yi. Example 88 (\(\mathrm{SL}_{2n}\)) and §6.2 (\(\mathrm{GL}_{2n}\)) treat the family; this page tests it against Proposition III.
| \(n\) | \(\mathrm{GL}_{2n}/\mathrm{Sp}_{2n}\) | \(\mathrm{SL}_{2n}/\mathrm{Sp}_{2n}\) |
|---|---|---|
| \(1\) | \(\mathbb G_m\) via \(\det\), rank \(1\) | a point |
| \(2\) | alternating \(4\times4\) forms, dim \(6\), rank \(2\) | \(\{\operatorname{Pf}=1\}\subseteq\Lambda^2k^4\cong k^6\), \(=\mathrm{Spin}_6/\mathrm{Spin}_5\), rank \(1\) |
Start with \(n=1\). \(\mathrm{Sp}_2=\mathrm{SL}_2\), so \(X=\mathrm{GL}_2/\mathrm{SL}_2\cong\mathbb G_m\), and the orbits on \(LX=F^\times\) are the circles \(\operatorname{val}c=a\):
$$S=Z(\mathrm{GL}_2),\qquad\Lambda_S^+=\{(a,a):a\in\mathbb Z\},\qquad\mathcal O_{(a,a)}=t^a\mathcal O^\times.$$Verdict
Proposition III decides H1–H5 for a symmetric space by two conditions:
$$\text{(C1)}\ \ Z_{G^\theta}(S)\ \text{is connected},\qquad\text{(C2)}\ \ \varphi(\Lambda_T^+)=\Lambda_S^+\ \ \text{(every orbit label has a dominant lift)},$$ $$\mathrm{H1}\wedge\cdots\wedge\mathrm{H5}\iff\text{(C1)}\wedge\text{(C2)}.$$| \(X\) | (C1): \(Z_{G^\theta}(S)\) | (C2): a label without dominant lift | H1–H5 |
|---|---|---|---|
| \(\mathrm{GL}_2/\mathrm{Sp}_2\) | \(\mathrm{SL}_2\) ✓ | none ✓ | hold |
| \(\mathrm{SL}_4/\mathrm{Sp}_4\) | \(\mathrm{SL}_2^2\) ✓ | none ✓ | hold |
| \(\mathrm{GL}_{2n}/\mathrm{Sp}_{2n}\), \(n\ge2\) | \(\mathrm{SL}_2^n\) ✓ | \((1,1,1,1,0,\dots,0)\) | fail |
| \(\mathrm{SL}_{2n}/\mathrm{Sp}_{2n}\), \(n\ge3\) | \(\mathrm{SL}_2^n\) ✓ | \((1,1,1,1,0,\dots,0,-2,-2)\) | fail |
(C1) always holds (section 5); (C2) fails exactly at labels with two equal odd pair values (section 6), which first exist in rank two.
Smallest failing cases
The smallest labels without a dominant lift:
$$\mathrm{SL}_6/\mathrm{Sp}_6:\ (1,1,1,1,-2,-2),\ (2,2,-1,-1,-1,-1),\qquad\mathrm{GL}_4/\mathrm{Sp}_4:\ (1,1,1,1),\ (-1,-1,-1,-1).$$
Intuition on the smallest casesworked here
- What \(\varphi\) does. It adds the two entries of each pair: $$\varphi(\lambda)=(b_1,b_1,\dots,b_n,b_n),\qquad b_i=\lambda_{2i-1}+\lambda_{2i}.$$
- Why a wall label can fail. A dominant lift splits each pair value with the larger half first, so an odd value \(1\) splits as \((1,0)\); two such pairs side by side are no longer decreasing: $$\mu=(\underbrace{1,1}_{a_1},\underbrace{1,1}_{a_2},\underbrace{-2,-2}_{a_3})$$ $$\lambda=(1,\underline{0},\underline{1},0,-1,-1),\qquad\varphi(\lambda)=\mu,\quad0\lt1.$$
- Why \(n=3\) is the first. For \(\mathrm{SL}_4\) the pair values are \((k,-k)\), and \(a_1=a_2\) forces \(k=0\), which is even; for \(\mathrm{SL}_6\) the sum-zero condition allows an odd repeated value: $$\mathrm{SL}_4:\ a=(k,-k),\qquad\mathrm{SL}_6:\ a=(1,1,-2).$$
- For H5. \(Z_{G^\theta}(S)=\mathrm{SL}_2^n\) is connected, so H3 holds, and by Proposition III, step 6 no \(\lambda\) satisfies H5 at a label without a dominant lift; \(E_6/F_4\) has the same chamber and the same failing labels (its Figure 1).
1 · Model
\(G=\mathrm{SL}_{2n}\) or \(G=\mathrm{GL}_{2n}\), written with \(n\) blocks of size two. The involution twists the inverse transpose by the block matrix \(J\):
$$J_2=\begin{pmatrix}0&1\\-1&0\end{pmatrix},\qquad J=\begin{pmatrix}J_2&&\\&\ddots&\\&&J_2\end{pmatrix},\qquad\theta(g)=J\,(g^T)^{-1}J^{-1}.$$Fixed group. \(\theta(g)=g\) exactly when \(gJg^T=J\). So \(G^\theta\) is the symplectic group of \(J\), which is connected:
$$G^\theta=\{g:gJg^T=J\}=\mathrm{Sp}_{2n}.$$Torus. \(T\) is the diagonal torus. Conjugation by \(J_2\) swaps the two entries of a block, so \(\theta\) swaps each pair and inverts it:
$$\theta\bigl(\operatorname{diag}(t_1,t_2,\dots,t_{2n-1},t_{2n})\bigr)=\operatorname{diag}(t_2^{-1},t_1^{-1},\dots,t_{2n}^{-1},t_{2n-1}^{-1}).$$Split torus. The elements of \(T\) that \(\theta\) inverts have equal entries in each pair:
$$T^{-\theta}=\{t\in T:\theta(t)=t^{-1}\}=\{\operatorname{diag}(a_1,a_1,\dots,a_n,a_n)\}.$$In \(\mathrm{GL}_{2n}\) this torus is \(S\). In \(\mathrm{SL}_{2n}\) the determinant condition \((a_1\cdots a_n)^2=1\) splits it into two components, and \(S\) is the one with \(a_1\cdots a_n=1\):
$$S=\{\operatorname{diag}(a_1,a_1,\dots,a_n,a_n):a_1\cdots a_n=1\},\qquad\theta(s)=s^{-1}.$$Rank. A θ-split torus containing \(S\) lies in \(L=Z_G(S)\), so its Lie algebra lies in the \((-1)\)-eigenspace of \(\theta\) on \(\operatorname{Lie}L\). That eigenspace has dimension \(\dim L-\dim L^\theta\) (section 5), which equals \(\dim S\):
$$(4n-1)-3n=n-1=\dim S\ \ (\mathrm{SL}_{2n}),\qquad4n-3n=n=\dim S\ \ (\mathrm{GL}_{2n}).$$So \(S\) is a maximal θ-split torus, and \(\dim S\) is the rank of \(X\).
Fixed torus. \(T^\theta\) is connected and meets \(S\) in finitely many sign matrices. Since \(\dim T^\theta+\dim S=\dim T\), the two generate \(T\):
$$T^\theta=\{\operatorname{diag}(b_1,b_1^{-1},\dots,b_n,b_n^{-1})\},\qquad T=(T^\theta)^\circ\cdot S.$$Example: \(n=1\) and \(n=2\)worked here
- \(n=1\). \(J=J_2\), and \(gJ_2g^T=\det(g)J_2\), so \(\theta\) divides by the determinant and \(G^\theta=\mathrm{SL}_2\) (H1 tab): $$\theta(g)=\frac{g}{\det g},\qquad S=\{\operatorname{diag}(a,a)\}=Z(\mathrm{GL}_2),\qquad T^\theta=\{\operatorname{diag}(b,b^{-1})\}.$$
- \(n=2\). \(J\) has two blocks, and \(\theta\) swaps and inverts within each pair: $$J=\begin{pmatrix}0&1&0&0\\-1&0&0&0\\0&0&0&1\\0&0&-1&0\end{pmatrix},\qquad\theta\bigl(\operatorname{diag}(t_1,t_2,t_3,t_4)\bigr)=\operatorname{diag}(t_2^{-1},t_1^{-1},t_4^{-1},t_3^{-1}),$$ $$S=\{\operatorname{diag}(a_1,a_1,a_2,a_2)\}\ \ (\mathrm{GL}_4),\qquad S=\{\operatorname{diag}(a,a,a^{-1},a^{-1})\}\ \ (\mathrm{SL}_4).$$
- Inside \(\mathrm{Sp}_4\). \(\operatorname{diag}(g_1,g_2)\) with \(g_1,g_2\in\mathrm{SL}_2\) fixes \(J\), and so does the permutation swapping the two blocks.
2 · Lattices
Cocharacters. \(\Lambda_T=\mathbb Z^{2n}\), with \(x_1+\dots+x_{2n}=0\) for \(\mathrm{SL}_{2n}\). By section 1, \(\theta\) acts by a block-diagonal integer matrix with \(\theta^2=1\):
$$\theta=\begin{pmatrix}0&-1&&&\\-1&0&&&\\&&\ddots&&\\&&&0&-1\\&&&-1&0\end{pmatrix},\qquad\theta(x)=(-x_2,-x_1,\dots,-x_{2n},-x_{2n-1}).$$Split lattice. \(\theta x=-x\) exactly when each pair has equal entries:
$$\Lambda_S=\{(a_1,a_1,\dots,a_n,a_n)\},\qquad a_1+\dots+a_n=0\ \text{ for }\mathrm{SL}_{2n}.$$Leading-term map. \(\varphi=1-\theta\) has blocks \(\left(\begin{smallmatrix}1&1\\1&1\end{smallmatrix}\right)\). It writes the sum of each pair into both entries of the pair:
$$\varphi=\begin{pmatrix}1&1&&&\\1&1&&&\\&&\ddots&&\\&&&1&1\\&&&1&1\end{pmatrix},\qquad\varphi(x)=(b_1,b_1,\dots,b_n,b_n),\quad b_i=x_{2i-1}+x_{2i}.$$So \(\varphi(\Lambda_T)=\Lambda_S\): the point \((a_1,0,a_2,0,\dots,a_n,0)\in\Lambda_T\) maps to \((a_1,a_1,\dots,a_n,a_n)\).
3 · Orbits
Forms. On alternating forms the arc group acts by congruence, and \(\tau\) multiplies by \(J^{-1}\), since \(\theta(g)^{-1}=Jg^TJ^{-1}\):
$$LX=\{\omega\in M_{2n}(F):\omega^T=-\omega,\ \det\omega\ne0\},\qquad h\cdot\omega=h\,\omega\,h^T,\qquad\tau(\omega)=\omega J^{-1},$$with \(\operatorname{Pf}\omega=1\) for \(\mathrm{SL}_{2n}\).
Labels are elementary divisors. The label \(\mu=(a_1,a_1,\dots,a_n,a_n)\) is the block form with entries \(t^{a_i}\):
$$\tau(x_\mu)=t^\mu\iff\omega_\mu=t^\mu J=\begin{pmatrix}t^{a_1}J_2&&\\&\ddots&\\&&t^{a_n}J_2\end{pmatrix}.$$Normal form. Every \(\omega\in LX\) is \(\mathrm{GL}_{2n}(\mathcal O)\)-congruent to exactly one \(\omega_\mu\) with decreasing exponents, so these are the orbits, and dominance for the upper-triangular \(B\) (section 4) just orders the exponents:
$$\mathcal O_\mu=\{\omega:h\,\omega\,h^T=\omega_\mu\text{ for some }h\in\mathrm{GL}_{2n}(\mathcal O)\},\qquad\Lambda_S^+=\{(a_1,a_1,\dots,a_n,a_n)\in\Lambda_S:a_1\ge\dots\ge a_n\}.$$Why the normal form
- Existence. Let \(t^a\) be an entry \(\omega(e_i,e_j)\) of least valuation, rescaled by a unit. Every value \(\omega(x,y)\), \(x,y\in\mathcal O^{2n}\), lies in \(t^a\mathcal O\), so the projection onto \(\langle e_i,e_j\rangle^\perp\) is integral and splits off the block \(t^aJ_2\): $$x\longmapsto x-t^{-a}\omega(x,e_j)\,e_i+t^{-a}\omega(x,e_i)\,e_j,\qquad\mathcal O^{2n}=\langle e_i,e_j\rangle\oplus\langle e_i,e_j\rangle^\perp.$$ Induct on \(n\), then permute the blocks into decreasing order.
- Uniqueness. Congruence is a special case of \(\omega\mapsto h\omega h'\), and the Smith form of \(\omega_\mu\) is \(\operatorname{diag}(t^{a_1},t^{a_1},\dots,t^{a_n},t^{a_n})\), so the exponents are invariants.
- \(\mathrm{SL}_{2n}\). \(\operatorname{Pf}(h\omega h^T)=\det(h)\operatorname{Pf}\omega\), and both forms have Pfaffian \(1\), so the congruence has \(\det h=1\) and \(a_1+\dots+a_n=0\).
Example. For \(n=1\), \(\omega=cJ_2\) and \(\omega_{(a,a)}=t^aJ_2\): the orbits are the circles \(\operatorname{val}c=a\) of the H1 tab. For \(n=2\), \(\omega_{(1,1,0,0)}=tJ_2\oplus J_2\).
Lifts in this language. The translate of the base point by \(\lambda\in\Lambda_T\) is the form whose block exponents are the pair sums, which is \(\varphi\):
$$t^\lambda J\,t^\lambda=\begin{pmatrix}t^{\lambda_1+\lambda_2}J_2&&\\&\ddots&\\&&t^{\lambda_{2n-1}+\lambda_{2n}}J_2\end{pmatrix}=\omega_{\varphi(\lambda)}.$$4 · Positive roots
Borel. \(B\) is the upper-triangular Borel, so \(\Phi^+=\{\varepsilon_i-\varepsilon_j:i\lt j\}\). The Levi \(L=Z_G(S)\) is block diagonal, and its roots live inside the blocks:
$$L=\left\{\begin{pmatrix}g_1&&\\&\ddots&\\&&g_n\end{pmatrix}:g_i\in\mathrm{GL}_2\right\}\cap G,\qquad\Phi_L=\{\pm(\varepsilon_{2i-1}-\varepsilon_{2i})\}.$$\(\Phi_L\) is spanned by the odd simple roots \(\alpha_1,\alpha_3,\dots,\alpha_{2n-1}\), the black nodes of the Satake diagram A II.
θ on roots. \(\theta\varepsilon_k=-\varepsilon_{k'}\), where \(k'\) is the partner of \(k\) in its pair. So \(\theta\) fixes \(\Phi_L\):
$$\theta(\varepsilon_{2i-1}-\varepsilon_{2i})=-\varepsilon_{2i}+\varepsilon_{2i-1}=\varepsilon_{2i-1}-\varepsilon_{2i}.$$A positive root \(\varepsilon_i-\varepsilon_j\) outside \(\Phi_L\) has \(i\) in an earlier pair than \(j\). Partners stay in their pairs, so \(-\theta\alpha\) is positive:
$$-\theta(\varepsilon_i-\varepsilon_j)=\varepsilon_{i'}-\varepsilon_{j'}\in\Phi^+.$$5 · Centralizer
\(Z_{G^\theta}(S)=L^\theta\). On a \(2\times2\) block \(\theta\) divides by the determinant, because \(gJ_2g^T=\det(g)\,J_2\) for every \(g\in\mathrm{GL}_2\):
$$J_2\,(g^T)^{-1}J_2^{-1}=\frac{g}{\det g},\qquad\theta\begin{pmatrix}g_1&&\\&\ddots&\\&&g_n\end{pmatrix}=\begin{pmatrix}g_1/\det g_1&&\\&\ddots&\\&&g_n/\det g_n\end{pmatrix}.$$So a block-diagonal matrix is θ-fixed exactly when each block has determinant \(1\). The centralizer is connected:
$$Z_{G^\theta}(S)=L^\theta=\mathrm{SL}_2\times\dots\times\mathrm{SL}_2=\mathrm{SL}_2^n,\qquad\dim L^\theta=3n.$$This agrees with Chen–Yi, Example 88, which asserts connected \(L^+G\)-stabilizers for \(\mathrm{SL}_{2n}/\mathrm{Sp}_{2n}\). By Proposition III, step 4, this is (C1), and with it H3.
6 · Coverage
Dominant cones. Dominance orders the entries; a label is determined by its pair values \(a=(a_1,\dots,a_n)\), with \(\Lambda_S^+\) as in section 3:
$$\Lambda_T^+=\{x\in\Lambda_T:x_1\ge x_2\ge\dots\ge x_{2n}\}.$$Lemma. The label \(\mu=(a_1,a_1,\dots,a_n,a_n)\in\Lambda_S^+\) lies in \(\varphi(\Lambda_T^+)\) if and only if no \(i\) has \(a_i=a_{i+1}\) odd.
Proof. Let \(\lambda\in\Lambda_T^+\) with \(\varphi(\lambda)=\mu\), so \(\lambda_{2i-1}+\lambda_{2i}=a_i\) and \(\lambda_{2i-1}\ge\lambda_{2i}\). If \(a_i=a_{i+1}=a\) is odd, the pairs \(i\) and \(i+1\) force a strict increase, against \(\lambda_{2i}\ge\lambda_{2i+1}\):
$$\lambda_{2i}\le\frac{a-1}2\lt\frac{a+1}2\le\lambda_{2i+1}.$$Conversely, split each pair as evenly as possible. The result is dominant, because \(\lfloor a_i/2\rfloor\ge\lceil a_{i+1}/2\rceil\) whenever \(a_i\gt a_{i+1}\) or \(a_i=a_{i+1}\) is even:
$$\lambda=\bigl(\lceil a_1/2\rceil,\lfloor a_1/2\rfloor,\dots,\lceil a_n/2\rceil,\lfloor a_n/2\rfloor\bigr),\qquad\varphi(\lambda)=\mu.$$For \(\mathrm{SL}_{2n}\) the entries of \(\lambda\) sum to \(a_1+\dots+a_n=0\), so \(\lambda\in\Lambda_T^+\).
Labels without a lift. For \(\mathrm{SL}_{2n}\), \(n\ge3\), the pair values \(a=(1,1,0,\dots,0,-2)\) sum to \(0\) and have \(a_1=a_2=1\). For \(\mathrm{GL}_{2n}\), \(n\ge2\), take \(a=(1,1,0,\dots,0)\):
$$(1,1,1,1,0,\dots,0,-2,-2)\notin\varphi(\Lambda_T^+),\qquad(1,1,1,1,0,\dots,0)\notin\varphi(\Lambda_T^+).$$Both labels lift in \(\Lambda_T\), to \((1,0,1,0,0,\dots,0,-2,0)\) and \((1,0,1,0,0,\dots,0)\); for \(\mathrm{SL}_{2n}\) this is Chen–Yi, Lemma 90. No lift is dominant, because the first two pairs force a strict increase:
$$\lambda_1+\lambda_2=1,\quad\lambda_3+\lambda_4=1\quad\Longrightarrow\quad\lambda_2\le0\lt1\le\lambda_3.$$Where coverage holds. For \(\mathrm{SL}_4\) every label has \(a=(k,-k)\) with \(k\ge0\), and \(a_1=a_2\) only at \(k=0\), which is even. For \(\mathrm{GL}_2\) there is one pair, so every label lifts.
$$\varphi\bigl(\lceil k/2\rceil,\lfloor k/2\rfloor,-\lfloor k/2\rfloor,-\lceil k/2\rceil\bigr)=(k,k,-k,-k).$$A search over dominant \(\lambda\) with entries in \([-6,6]\) confirms the lemma for \(n\le4\), on every label with \(|a_i|\le4\).