Let $\mathbb{k}$ be an algebraically closed field. The symplectic algebraic group is given by


$$
\text{Sp}(2n,\mathbb{k})=\{M\in\text{Mat}\_{2n}(\mathbb{k})\mid J=M^TJ M\}\quad\text{where}\quad J=\begin{pmatrix} 
0 & 1\_n \\
 -1\_n & 0 
\end{pmatrix}
$$
The Lie algebra of $Sp(2n,\mathbb{k})$ is given by
$$
\text{sp}(2n,\mathbb{k})=\{M \in\text{Mat}\_{2n}(\mathbb{k})\mid M^TJ+J M=0\},
$$


where for a closed subgroup $G \subseteq GL(2n,k)$ cut out by polynomials $f\_1,\dots, f\_m$, we take its Lie algebra to be the tangent space to $G$ at the identity $e$ i.e. $g=V(d\_e f\_1,\dots, d\_e f\_m)$, where $d\_e f=\sum\_{i,j=1}^{2n} \frac{\delta f}{\delta t\_{i,j}}(e)t\_{i,j}$, endowed with the commutator bracket.


One way to prove that $sp(2n,\mathbb{k})$ is the Lie algebra of $Sp(2n,\mathbb{k})$ is by writing $M$ as a block matrix
$$
M=\begin{pmatrix} 
A & B \\
 C & D 
\end{pmatrix}.
$$


Rewriting the above definition for $Sp(2n,\mathbb{k})$ in terms of these blocks gives (for example), $-AC^T+A^T C=0$, which has $(i,j)$ entry $f\_{i,j}=\sum\_{k=1}^n -a\_{i,k}c\_{j,k}+a\_{k,i}c\_{k,j}$, from which we get that $d\_e(f\_{i,j})=-c\_{j,i}+c\_{i,j}=0$, which yields $C=C^T$ in the tangent space. Continuing in this way for the other blocks, we can deduce that 


$$
sp(2n,\mathbb{k})=\Big\{\begin{pmatrix} 
A & B \\
 C & -A^T 
\end{pmatrix}\mid C=C^T, B=B^T\Big\},
$$
which is equivalent to the above.


Question 1: Is there a cleaner proof of this fact? We have $(2n)^2$ degree-two polynomials $M^T J M-J$, arranged in a matrix equation, to take the differentials of. Is there a cleaner way to compute their differentials "all at once" using the matrix structure? What got me thinking about this is that the Lie algebra in this case looks similar to the well-known formula $d\_e(fg)=d\_e(f) g(e)+d\_e(g) f(e)$ applied to $J=M^T J M$ somehow.


Question 1a: Say we have some more general situation where we have a variety cut out by a matrix equation $L M=N$, where $L, M, N$ are matrices of polynomials. What's the tangent space at a point?


Question 2: I overlooked one aspect of my computation of the tangent space-- we need to know that the polynomials $f\_1,\dots, f\_m$ (in this case, $M^T J M-J$) generate the ideal of the variety (in other words, that the ideal they generate is radical). I have blindly assumed that this is the case, but I would appreciate a proof of this fact (if it is true).



Slicker computation of the Lie algebra of the symplectic group (and computing differentials of matrix equations of polynomials)
$\newcommand{\dual}{\mathbb{k}\langle\epsilon\rangle}$Let $\dual = \mathbb{k}[\epsilon]/\epsilon^2$ be the dual numbers. A $\dual$-point of a scheme $X$ is exactly a $\mathbb{k}$-point of $X$ and a tangent vector at that point. 


1. In the case when $G$ is an algebraic group, we can thus obtain the $\mathbb{k}$-valued tangent space at the identity by the formula
$$ L(G)(\mathbb{k}) = \ker (G(\dual) \to G(\mathbb k)).$$
See Milne's book, chapter 12.
In the case of e.g. $G=Sp\_{2n}$, this provides the simplification you are looking for. A $\dual$-valued matrix is exactly a matrix of the form $A + B \epsilon$ where $A,B \in M\_n(\mathbb k)$. This is in $Sp\_{2n}(\dual)$ if and only if 
$$J = (A + B\epsilon)^tJ(A + B\epsilon) = (A^tJA) + (B^tJA + A^tJB)\epsilon, $$
and maps to the identity in $Sp\_{2n}(\mathbb k)$ if and only if $A = I\_{2n}$. Hence we recover the equation $B^tJ + JB = 0$ for elements $B$ of the Lie algebra.
2. (your 1a) Say $L,M,N$ are polynomials in $x\_1,\ldots, x\_n$. A $\dual$-valued point $p + q \epsilon = (p\_1 + \epsilon q\_1, \ldots, p\_n + \epsilon q\_n)$ at a polynomial $f$ is given by
$$ f(p + q\epsilon) = f(p) + \epsilon \sum\_{i=1}^n \frac{\partial f}{\partial x\_i}(p)q\_i = f(p) + \epsilon Df\_p(q).$$
Hence, a $\dual$-valued point of the scheme defined by $LM = N$ exactly satisfies
$$ \left(L(p) + \epsilon DL\_p(q)\right)(M(p) + \epsilon DM\_p(q)) = N(p) + \epsilon DN\_p(q),$$
so that $p$ is a $\mathbb k$-point and
$$ DL\_p(q)M(p) + L(p)DM\_p(q) = DN\_p(q).$$
3. (your 2) according to Milne, Prop 1.22, an algebraic group $G$ is smooth if and only if it is geometrically reduced if and only if its dimension equals the dimension of its tangent space. So, checking that the equations $M^tJM = J$ define a reduced scheme is essentially an inverse-function-theorem style computation of rank.




---


*References:*


Milne, James S. *Algebraic groups: the theory of group schemes of finite type over a field.* Vol. 170. Cambridge University Press, 2017.


