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Is there a formula for the size of Symplectic group defined over a finite ring $Z/p^k Z$?

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From Christopher Perez's answer, we have $|Sp_{2n}(\mathbb{Z}/p\mathbb{Z})| = p^{n^2} \prod_{i=1}^n (p^{2i}-1)$. Following Johannes Hahn, we wish to determine the size of the kernel of the homomorphism $Sp_{2n}(\mathbb{Z}/p^k\mathbb{Z}) \to Sp_{2n}(\mathbb{Z}/p\mathbb{Z})$.

To compute this, we may induct on $k$: For $k \geq 1$, elements in the kernel of the homomorphism $Sp_{2n}(\mathbb{Z}/p^{k+1}\mathbb{Z}) \to Sp_{2n}(\mathbb{Z}/p^k\mathbb{Z})$ have the form $I + p^k A$ for a matrix $A = \left( \begin{smallmatrix} E & F \\ G & H \end{smallmatrix} \right)$ with values in $\mathbb{Z}/p\mathbb{Z}$. The symplectic condition (given in Wikipedia) is equivalent to the conditions that $F^t = F$, $G^t = G$, and $E^t + H = 0$. $F$ and $G$ are therefore symmetric matrices, while $H$ and $E$ determine each other, with no further conditions. The kernel of one-step reduction therefore has size $p^{n(n+1)/2} \cdot p^{n(n+1)/2} \cdot p^{n^2}$, or $p^{2n^2 + n}$.

The final answer is therefore: $|Sp_{2n}(\mathbb{Z}/p^k\mathbb{Z})| = p^{(2k-1)n^2 + (k-1)n} \prod_{i=1}^n (p^{2i}-1)$

Edit: To address kassabov's complaint, I'll explain the calculation in a bit more detail. The symplectic condition on $I + p^k A$ is that $(I + p^k A)^t \Omega (I + p^k A) \equiv \Omega \pmod {p^{k+1}}$, where $\Omega = \left( \begin{smallmatrix} 0 & I_n \\ -I_n & 0 \end{smallmatrix} \right)$. Because $k \geq 1$, we may eliminate the $p^{2k} A^t \Omega A$ term when expanding, to get

$$ \Omega + p^k A^t \Omega + p^k \Omega A \equiv \Omega \pmod {p^{k+1}}$$

By subtracting $\Omega$ from both sides, we get the conditions I mentioned on the blocks in $A$. As you mentioned, this coincides with the symplectic Lie algebra condition. George McNinch gave an elegant explanation in the comments, but a possibly more pedestrian reason is that $(p^k)$ is a square zero ideal in $\mathbb{Z}/p^{k+1}\mathbb{Z}$, so one has a canonical isomorphism between the kernel of reduction and the Lie algebra tensored with the quotient ring.

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    $\begingroup$ The caonditions $F^t=F$, $G^t=G$, and $E^t+H=0$ are not the same as $A$ being in the symplectic group! These conditions are for $A$ in the symplectic Lie alagebra. In this case the exponential map is a bijection between the "Lie algebra" and the "Lie group" and the counting is correct. $\endgroup$
    – kassabov
    Feb 9, 2012 at 14:04
  • $\begingroup$ Great, this is the same formula with the one derived from Joe Silverman's suggestion of using Igusa's theorem. $\endgroup$
    – hatice
    Feb 9, 2012 at 14:27
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    $\begingroup$ For an affine group scheme $G$ smooth and of finite type over $\mathbf{Z}$, the kernel of the mapping $G(\mathbf{Z}/p^k\mathbf{Z}) \to G(\mathbf{F}_p)$ has a filtration of length $k-1$ for which the successive quotients may be identified with "the $\mathbf{F}_p$-points of the Lie algebra" -- i.e. with $\operatorname{Lie}(G) \otimes_{\mathbf{Z}} \mathbf{F}_p$. This explains why the "kernel of one step reduction" as in the answer has order $p^d$ where $d$ is the dimension of the symplectic group $G_{/\mathbf{F}_p}$. $\endgroup$ Feb 9, 2012 at 19:36
  • $\begingroup$ @George: This is the right general viewpoint, I think. Is it clear where this originates in the literature? It's useful to see what is going on here based on the fact that Chevalley groups arise from $\mathbb{Z}$-schemes. By the way, there is already a similar flavor to the standard computation of units in the ring of $p$-adic integers (as in Serre's A Course in Arithmetic, II, Section 3). $\endgroup$ Feb 9, 2012 at 20:53
  • $\begingroup$ P.S. It's interesting to look at the case $n=1$ and $k=2$ in Scott's formulation, where you get the "degenerate" symplectic group of type $A_1 = C_1$ and dimension 3 whose order over the prime field is $p(p^2-1)$. As George implies, it's just the Lie algebra dimension coupled with the group order over the prime field that determine the outcome of the inductive process, rather than the specific description of symplectic groups. $\endgroup$ Feb 9, 2012 at 23:32
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From L. C. Grove's Classical Groups and Geometric Algebra, page 27: $$|\operatorname{SP}(2n,\mathbb{F}_q)|=q^{n^2}\prod_{i=1}^n(q^{2i}-1),$$ for some $q=p^k$. This works in the case of $\mathbb{F}_p\cong\mathbb{Z}/p\mathbb{Z}$, but I don't think it's helpful for the general case $\mathbb{Z}/q\mathbb{Z}$.

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  • $\begingroup$ Are you sure that you're not confusing the $q$-element field with the ring $\mathbb{Z}/q$? $\endgroup$ Feb 8, 2012 at 19:03
  • $\begingroup$ You're right, I'll fix that. Thanks! $\endgroup$ Feb 8, 2012 at 19:06
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Hi.

EDIT: As Joe Silverman pointed out, this approach doesn't work as simple as I imagined it. Sorry for that. I leave the attempted proof here in case someone has an idea how to fix it.

Yes, there is such a formula. It works for in a similar fashion all of the classical algebraic group. Consider the projection $\mathbb{Z}/p^k\to\mathbb{Z}/p$. It induces a surjection $Sp_{2n}(\mathbb{Z}/p^k)\to Sp_{2n}(\mathbb{Z}/p)$ because every transvection has a preimage and the transvections generate $Sp_{2n}(F)$ for every field $F$. The kernel of the homomorphism consists obviously of those matrices $M$ with $M-I \in (p\mathbb{Z}/p^k\mathbb{Z})^{n\times n}$. There are $p^{(k-1)n^2}$ such matrices.

Therefore $|Sp_{2n}(\mathbb{Z}/p^k)| = p^{(k-1)n^2} |Sp_{2n}(\mathbb{Z}/p)|$. The order of the symplectic group over $\mathbb{Z}/p$ is known.

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    $\begingroup$ @Johannes: Possibly I'm missing something, but why are matrices of the form $M=I+pA$ in $Sp_{2n}(\mathbb{Z}/p^k)$? I see that every matrix in the kernel must have this form, but doesn't the symplectic condition put some restriction on $A$? I assume that the definition of $Sp_{2n}(R)$ for any ring $R$ is the set of matrices $M$ satisfying $M^t\Omega M=\Omega$, where $\Omega$ is the usual anti-diagonal~$I$ and~$-I$ matrix. Then I seem to get $A^t\Omega+\Omega A+pA^t\Omega A \equiv 0 \pmod{p^{k-1}}$ as the condition on $A$ to make $M$ be in $Sp_{2n}(\mathbb{Z}/p^k)$. $\endgroup$ Feb 8, 2012 at 20:47
  • $\begingroup$ @Johannes: Adding a reference or two is desirable here, since the result is presumably known and not being written down for the first time (?) $\endgroup$ Feb 8, 2012 at 23:22
  • $\begingroup$ @Joe Silverman: Damn! You're right. I didn't think that through... @Jim Humphreys: Honestly, I don't have one. I found the analogue argument for GL on my own a few years back and I thought it could be adapted to this situation (which was premature as I now know). $\endgroup$ Feb 8, 2012 at 23:49
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    $\begingroup$ @Johannes: Right, the $GL_n$ case is easy (and well-known). In general, if $X$ is a variety defined over $\mathbb{Z}$, then the sequence of values $N_k=#X(\mathbb{Z}/p^k\mathbb{Z})$ can be complicated, and indeed it is a deep theorem of Igusa's that the generating function $\sum N_kT^k$ is a rational function. OTOH, if $X$ is smooth over $\mathbb{Z}_p$, then Hensel's lemma should allow one to lift, so then $N_{k+1}=p^{\dim X}N_k$. Presumably $Sp_{2n}$ is a (smooth) algebraic group scheme over $\mathbb{Z}$, which would give almost your answer, but the power of $p$ in front must be adjusted. $\endgroup$ Feb 9, 2012 at 0:48
  • $\begingroup$ @Joe Silverman and Johannes, Thanks for the brain storming. Then the suggested answer is $|Sp(2n,Z/p^{k}Z)|=p^{k(2n^2+n)}\prod_{j=1}^{n}(1−p^{-2j})$. hatice $\endgroup$
    – hatice
    Feb 9, 2012 at 1:51

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