This question is about the computation of $H\_2(Sp(2g, \mathbb{Z}/2), \mathbb{Z})$, where $Sp(2g, \mathbb{Z}/2)$ is the group of symplectic $2g \times 2g$ matrices over $\mathbb{Z}/2$. 
With respect to this computation I have seen quotations to the following two papers:


* Stein "Surjective stability in dimension $0$ for $K\_2$ and related functors" (<https://www.dropbox.com/s/jcy7sep2hpxy7gt/Stein.pdf?dl=0>)
* Steinberg "Générateurs, relations et revêtements de groupes algébriques"
(<https://www.dropbox.com/s/jidchwwjssnmmlm/Steinberg1.pdf?dl=0>)


According to the quotations of Stein's and Steinberg's papers the result is that
 $H\_2(Sp(2g, \mathbb{Z}/2), \mathbb{Z}) = 0$ for $g \geq 3$. 


My two concerns are:


1) The papers by Stein and Steinberg do not directly give the computation of $H\_2(Sp(2g, \mathbb{Z}/2), \mathbb{Z})$. So my first question is if it is really possible to extract this result from these papers?


2) There is a discrepancy in the case $g=3$ between the quotations of Stein's and Steinberg's papers - which say that $H\_2(Sp(6, \mathbb{Z}/2), \mathbb{Z}) =0$ - and the computation of this group done with GAP (the computational system for discrete algebra). GAP gives $H\_2(Sp(6, \mathbb{Z}/2), \mathbb{Z}) = \mathbb{Z}/2$. Checking on GAP, the process that the computer goes through to get the computation looks correct. Is there an error in Stein's and/or Steinberg's paper or is it a misinterpretation of the case $g=3$ in the quotation?



Schur multiplier of $Sp(2g, \mathbb{Z}/2)$ for $g \geq 3$
I've been corresponding via email with the OP about this (it is a paper of mine that she got these citations from), and she asked me to post an answer summarizing what I told her. I apologize for the length of this answer -- this is really quite a long story. I also apologize for sometimes butchering people's names. I am copying this answer from a long document I sent the OP, and unfortunately autocorrect screws things up without telling me (eg it wants to call Steinberg "Sternberg", though I hope caught most of those occurrences).
$\DeclareMathOperator{\Sp}{Sp} \DeclareMathOperator{\ESp}{ESp} \DeclareMathOperator{\SL}{SL} \newcommand\Z{\mathbb{Z}} \DeclareMathOperator{\HH}{H} \newcommand\tG{\widetilde{G}} \newcommand\tD{\widetilde{D}} \newcommand\Field{\mathbb{F}} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\St}{St}$ 


For $\Sp\_{2g}(\Z/2)$, the correct theorem is that $\HH\_2(\Sp\_{2g}(\Z/2)) = 0$ for $g \geq 4$. This should be attributed to Steinberg and is contained in the paper cited by the OP. More generally, Steinberg showed that a similar theorem holds for $\Sp\_{2g}(\Field\_q)$. What Stein did in the cited paper was show how to extend what Steinberg did to $\Sp\_{2g}(\Z/k)$ where $k$ is not prime. They both in fact dealt not just with the symplectic group, but also with more general finite Chevalley groups.


Another good reference for Steinberg's work is sections 6 and 7 of Steinberg's Yale lecture notes, which were never published but which are available [here](http://www.ms.unimelb.edu.au/%5C~%7B%7Dram/Resources/YaleNotes.pdf).


It is not easy to extract the above homological statement from Steinberg and Stein's papers since they are written in the language of algebraic K-theory. What they are imitating is the calculation of $\HH\_2(\SL\_n(\Field\_q))$ that is described in Milnor's book on algebraic K-theory, which I highly recommend reading.


To help you understand these papers, below I have written a guide to the calculation of $\HH\_2(\SL\_n(\Field\_q))$ from Milnor's book.




---


I will begin by recalling the theory of universal central extensions. Let $G$ be a group. A **central extension** of $G$ is a group $\tG$ together with a short exact sequence
$$1 \longrightarrow C \longrightarrow \tG \longrightarrow G \longrightarrow 1$$
such that $C$ is contained in the center of $\tG$. This central extension is a **universal central extension**
if for any other central extension
$$1 \longrightarrow C' \longrightarrow \tG' \longrightarrow G \longrightarrow 1,$$
there exists a unique homomorphism $\tG \rightarrow \tG'$ such that the diagram
$\require{AMScd}$
$$\begin{CD}
1 @>>> C @>>> \tG @>>> G @>>> 1 \\
@. @VVV @VVV @VV{=}V @. \\
1 @>>> C' @>>> \tG' @>>> G @>>> 1
\end{CD}$$
commutes. The usual argument shows that universal central extensions are unique if they exist, but they might
not exist. The following theorem summarizes their properties. A proof of it can be found in Theorem 5.7 and Corollary 5.8 of Milnor's book


**Theorem 1:** Let $G$ be a group. Then $G$ has a universal central extension
$$1 \longrightarrow C \longrightarrow \tG \longrightarrow G \longrightarrow 1$$
if and only if $\HH\_1(G;\Z) = 0$, in which case we have $C \cong \HH\_2(G;\Z)$.


For perfect groups, this reduces the computation of $\HH\_2(G;\Z)$ to the construction of the universal central
extension of $G$.




---


I now describe what happens in "infinite rank". Let $R$ be a ring. Define
$$\GL(R) = \bigcup\_{n=1}^{\infty} \GL\_n(R)$$
and let $E(R)$ be the subgroup of $\GL(R)$ generated by elementary matrices.
The low-dimensional homology groups of $\GL(R)$ and $E(R)$
are closely connected to the algebraic K-theory of $R$. In particular,
$K\_1(R)$ is by definition equal to $\HH\_1(\GL(R);\Z)$.
It also turns out that $K\_2(R) = \HH\_2(E(R);\Z)$, but this
is a theorem rather than a definition. To understand what happens in finite rank, we'll have to understand
where this comes from.


There a group $\St(R)$ called the **Steinberg group**.
It is defined via generators and relations, and informally can be described as the group generated by
elementary matrices in $\GL(R)$ with the ``obvious relations'' between elementary matrices. Its precise
definition is as follows.


1. The generators are symbols $e\_{ij}(r)$ with $i,j \geq 1$ distinct
and $r \in R$. This corresponds to the elementary matrix obtained from the identity matrix by putting an $r$ in
position $(i,j)$.
2. The relations are as follows.


	* For distinct $i,j \geq 1$ and $r,r' \in R$, we have $e\_{ij}(r)
	 \cdot e\_{ij}(r') = e\_{ij}(r+r')$.
	* For distinct $i,j,k \geq 1$ and $r,r' \in R$, we have
	$[e\_{ij}(r),e\_{jk}(r')] = e\_{ik}(r r')$.
	* For distinct $i,j \geq 1$ and distinct $k,l \geq 1$ such that $j
	 \neq k$ and $i \neq l$, we have $[e\_{ij}(r),e\_{kl}(r')] = 1$ for all
	$r,r' \in R$.


Since all these relations hold in $\GL(R)$, there is a group homomorphism $\St(R) \rightarrow \GL(R)$ whose
image is $E(R)$. By definition, $K\_2(R)$ is the kernel of this homomorphism, so we have a short
exact sequence
$$1 \longrightarrow K\_2(R) \longrightarrow \St(R) \longrightarrow E(R) \longrightarrow 1.$$
The main theorem concerning the Steinberg group is as follows (see Theorem 5.10 of Milnor's book).


**Theorem 2:** Let $R$ be a ring. Then the extension
$$1 \longrightarrow K\_2(R) \longrightarrow \St(R) \longrightarrow E(R) \longrightarrow 1$$
is the universal central extension of $E(R)$. In particular, $K\_2(R)$ is an abelian group
and $\HH\_2(E(R);\Z) = K\_2(R)$.


This allows many concrete calculations. One very simple one is as follows.


**Example:** Since $\SL(\Z)$ is generated by elementary matrices, we have $E(\Z) = \SL(\Z)$. The group
$\HH\_2(\SL(\Z);\Z)$ is then cyclic of order $2$. Identifying it with $K\_2(\Z)$, the generator
is $(e\_{12}(1) e\_{21}(1)^{-1} e\_{12}(1))^4$. Here the matrix
$$e\_{12}(1) e\_{21}(1) e\_{12}(1) = \left(\begin{matrix}
0 & 1 \\
-1 & 0 
\end{matrix}\right)$$
is the one that rotates the plane by $90$ degrees, and hence has order $4$.


Assume now that $R$ is a commutative ring and let $u,v \in R^{\ast}$ be units. Define
$$D\_u = \left(\begin{matrix}
u & 0 & 0 \\
0 & u^{-1} & 0 \\
0 & 0 & 1
\end{matrix}\right)$$
and
$$D\_v' = \left(\begin{matrix}
v & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & v^{-1}
\end{matrix}\right).$$
It is not hard to see that diagonal matrices like these can be written as products of elementary matrices, so
$D\_u, D\_v' \in E(R)$. What is more, the matrices $D\_u$ and $D\_v'$ commute. Letting $\tD\_u$ and $\tD\_v'$ be
lifts of $D\_u$ and $D\_v'$ to $\St(R)$, we obtain an element
$[\tD\_u,\tD\_v'] \in K\_2(R)$, which will be called a **symbol** and denoted $\{u,v\}$. It is not hard to see
that this does not depend on the choice of lifts.


The following theorem says that for fields, the symbols generate $K\_2(R)$; see Corollary 9.13 of Milnor's book.


**Theorem 3:** Let $\Field$ be a field. Then $K\_2(\Field)$ is generated by the set of symbols $\{u,v\}$ as $u$ and $v$
range over $\Field^{\ast}$.


For finite fields, the final piece of the puzzle is as follows; see Corollary 9.9 of Milnor's book.


**Theorem 4:** Let $\Field$ be a finite field. Then $\{u,v\} = 0$ for all $u,v \in \Field^{\ast}$.


Combining everything above with the fact that $\SL(\Field)$ is generated by elementary matrices for a field $\Field$,
we deduce the following theorem.


**Theorem 5:** Let $\Field$ be a finite field. Then $\HH\_2(\SL(\Field);\Z) = 0$.




---


Of course, what we are really interested in is $\GL\_n(R)$, not $\GL(R)$. Define $E\_n(R)$ and $\St\_n(R)$ in the obvious
way. There is still a surjection $\St\_n(R) \rightarrow E\_n(R)$; denote its kernel by $C\_n(R)$, so we have a short
exact sequence
$$1 \longrightarrow C\_n(R) \longrightarrow \St\_n(R) \longrightarrow E\_n(R) \longrightarrow 1.$$
Associated to the natural inclusion $\GL\_n(R) \hookrightarrow \GL\_{n+1}(R)$ used to define $\GL(R)$ are an inclusion
$E\_n(R) \hookrightarrow E\_{n+1}(R)$ and homomorphisms $\St\_n(R) \rightarrow \St\_{n+1}(R)$ and $C\_n(R) \rightarrow C\_{n+1}(R)$
that fit into a commutative diagram
$$\begin{CD}
1 @>>> C\_n(R) @>>> \St\_n(R) @>>> E\_n(R) @>>> 1 \\
@. @VVV @VVV @VVV @. \\
1 @>>> C\_{n+1}(R) @>>> \St\_{n+1}(R) @>>> E\_{n+1}(R) @>>> 1.
\end{CD}$$
It is clear that $\St\_n(R)$ is the limit of
$$\St\_1(R) \rightarrow \St\_2(R) \rightarrow \St\_3(R) \rightarrow \cdots$$
and that $K\_2(R)$ is the limit of
$$C\_1(R) \rightarrow C\_2(R) \rightarrow C\_3(R) \rightarrow \cdots.$$
In the ``ideal'' situation, we would have theorems of the following form.


1. For $n$ sufficiently large, the exact sequence
$$1 \longrightarrow C\_n(R) \longrightarrow \St\_n(R) \longrightarrow E\_n(R) \longrightarrow 1$$
is the universal central extension of $E\_n(R)$; and hence $C\_n(R)$ is an abelian group and $\HH\_2(E\_n(R);\Z) = C\_n(R)$.
2. For $n$ sufficiently large, the map $C\_n(R) \rightarrow C\_{n+1}(R)$ is surjective (this is called **surjective stability**).
3. For $n$ sufficiently large, the map $C\_n(R) \rightarrow C\_{n+1}(R)$ is injective (this is called **injective stability**).


If these conditions are satisfied, then we would have $\HH\_2(E\_n(R);\Z) = K\_2(R)$ for $n$ sufficiently large.


Unfortunately, one can give examples where these fail. However, they do hold for many rings; in particular, they
hold for fields. Our goal is
to talk about finite fields, so we will not try to give particularly general statements. We begin with the
bit about being a universal central extension. The proof of Theorem 2 can
be followed to deduce the following.


**Theorem 6:** Let $R$ be a ring and let $n \geq 5$. Assume that $C\_n(R)$ is contained in the center of $\St\_n(R)$. Then the extension
$$1 \longrightarrow C\_n(R) \longrightarrow \St\_n(R) \longrightarrow E\_n(R) \longrightarrow 1$$
is the universal central extension of $E\_n(R)$. In particular,
$\HH\_2(E\_n(R);\Z) = C\_n(R)$.


As the following theorem shows, the condition in this theorem is satisfied for fields; see Theorem 9.12 of Milnor's book.


**Theorem 7:** Let $\Field$ be a field. Then $C\_n(\Field)$ is contained in the center of $\St\_n(\Field)$ for $n \geq 3$, and hence
$\HH\_2(\SL\_n(\Field);\Z) = C\_n(\Field)$ for $n \geq 5$


**Remark:** In fact, it turns out that
$$1 \longrightarrow C\_n(\Field) \longrightarrow \St\_n(\Field) \longrightarrow \SL\_n(\Field) \longrightarrow 1$$
is the universal central extension of $\SL\_n(\Field)$ for $n \geq 3$ except for the following exceptions:


1. $n=3$ and $\Field = \Field\_2$, and
2. $n=3$ and $\Field = \Field\_4$, and
3. $n=4$ and $\Field = \Field\_2$.


This was proved by Steinberg.


We now turn to injective and surjective stability. The key is the notion of symbol. If $R$ is a commutative ring
and $u,v \in R^{\ast}$ are units, then we can define a symbol $\{u,v\}\_n \in C\_n(R)$ for any $n \geq 3$ in the obvious
way. We then have the following; see Theorem 9.11 of Milnor's book.


**Theorem 8:** Let $R$ be a commutative ring and let $n \geq 3$. Assume that $C\_n(R)$ is contained in the center of $\St\_n(R)$.
Then $C\_n(R)$ is generated by the set of symbols $\{u,v\}\_n$ as $u$ and $v$ range over the units of $R$.


**Remark:** Theorem 7 and Theorem 8 combine to show that if $\Field$
is a field, then $K\_2(\Field)$ is generated by symbols $\{u,v\}$ for $u,v \in \Field^{\ast}$. This is
precisely Theorem 3 above, and in fact this is how Theorem
3 is proved.


The proof of Theorem 4 above also works for the symbols $\{u,v\}\_n$ and
gives the following.


**Theorem 9:** Let $\Field$ be a finite field and $n \geq 3$. Then $\{u,v\}\_n = 0$ for all $u,v \in \Field^{\ast}$.


Combining everything above, we deduce the following.


**Theorem 10:** Let $\Field$ be a finite field. Then $\HH\_2(\SL\_n(\Field);\Z) = 0$ for $n \geq 5$.


**Remark:** In fact, as in the remark after Theorem 7, one can show that
$\HH\_2(\SL\_n(\Field);\Z) = 0$ for $\Field$ a finite field and $n \geq 3$ except for the following exceptions:


1. $n=3$ and $\Field = \Field\_2$, and
2. $n=3$ and $\Field = \Field\_4$, and
3. $n=4$ and $\Field = \Field\_2$.


