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I was given in class the following definitions:

Definition 1 The Lie algebra $\mathfrak{so}(n)$ is defined as the subspace of the matrices $ A \in \mathfrak{gl}(n)$ such that $AJ + JA^{t} = 0$, with $J \in \mathfrak{gl}(n)$ and

$J= \begin{pmatrix} 0 & 0 & ....& 0 & 1 \\ 0 & 0 & .... & 1 & 0\\ .. & .. & .. & ..& .. \\ 0 & 1 & ....& 0 & 0 \\ 1 & 0 & ....& 0 & 0 \\ \end{pmatrix}$

In other words, $J$ has the secondary diagonal with all entries equal to 1 and all the other entries are zero.

Definition 2 The Lie algebra $\mathfrak{sp}(2n)$ is defined as the subspace of the matrices $A \in \mathfrak{gl}(2n)$ such that $AJ + JA^{t} = 0$, with $J \in \mathfrak{gl}(2n)$ and

$J= \begin{pmatrix} 0 &...& 0 & 0 &...& 1 \\ & & . &. & &. \\ . & & . &. & &. \\ 0 &...& 0 & 1 &...& 0 \\ 0 &...& -1 & 0 &...& 0 \\ . & & . &. & &. \\ . & & . &. & &. \\ -1 &...& 0 & 0 &...& 0 \\ \end{pmatrix}$

In other words $J$ has the secondary diagonal with a sequence of n entries equal to 1 and a sequence of n entries equal to -1. All the other entries are zero.

Now, is there a standard way in which such matrices are presented? For example, I know that a matrix $A \in \mathfrak{sp}(4)$ can have the following form:

$A = \begin{pmatrix} t_1 & a & z & x \\ b & t_2 & y & z \\ u & s & -t_2 & -a \\ t & u & -b & -t_1 \\ \end{pmatrix}$

What is the analogue form for the matrices $A \in \mathfrak{so}(4)$ (or for $\mathfrak{so}(n)$)? Is the following correct, for $\mathfrak{so}(4)$?

$A = \begin{pmatrix} t_1 & a & z & x \\ b & t_2 & y & z \\ u & s & t_2 & a \\ t & u & b & t_1 \\ \end{pmatrix}$

Does it mean that in the general case for $n$, the matrix $A$ would be like the case $n=4$, but with random rows separating the two upper blocks from the two lower blocks?

cip
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    Your presentation of $\mathfrak{so}(n)$ is unusual. Typically, we say $A \in \mathfrak{so}(n)$ iff we have $A = -A^T$; I don't think that your version is equivalent. – Ben Grossmann Feb 03 '20 at 12:46
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    Your definition of $\mathfrak{so}(n)$ doesn't work for $n = 3$: The matrix $A = \begin{pmatrix} 0 & 1 & 0 \ -1 & 0 & 0 \0 & 0 & 0\end{pmatrix}$ has $A^t = -A$ but $AJ + JA^t\not = 0$. – anomaly Feb 03 '20 at 14:09
  • Oh, I see. So in the case of the definition $A^t = -A$, what would a standard presentation be? – cip Feb 03 '20 at 14:22
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    Those are standard definitions for the split forms of those Lie algebras. So e.g. over the complex numbers you can work with these, and the presentations will be helpful to see the roots etc., but e.g. over $\mathbb R$ one usually calls $\mathfrak{so}_n$ the compact form, that one is indeed given e.g. by the choice $J=Id$ (giving $A^t=-A$), and is not isomorphic to this one. – Torsten Schoeneberg Feb 03 '20 at 16:33
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    The Lie algebras as you define them, with one slight change for the case $\mathfrak{so}(2n+1)$, are spelled out and discussed in volume VIII §13 of Bourbaki's treatise on Lie Groups and Lie Algebras (types $B_n, C_n$ and $D_n$). Again, these are the split forms. Over complex numbers, this is redundant information, but over the reals, the standard notation e.g. for an even-dimensional orthogonal one would rather be $\mathfrak{so}(n/2, n/2)$. Compare also my answer to https://math.stackexchange.com/q/2953385/96384. – Torsten Schoeneberg Feb 03 '20 at 23:55

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I eventually figured out, by working with the definition, the generic matrix $A \in \mathfrak{so}(4)$:

$A=\begin{pmatrix} t_1 & a & z & 0 \\ b & t_2 & 0 & -z \\ u & 0 & -t_2 & -a \\ 0 & -u & -b & -t_1 \\ \end{pmatrix}$

cip
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  • As said in comments to the question, if the entries are supposed to be $\in \mathbb R$, this Lie algebra should not be called $\mathfrak{so}(4)$, but $\mathfrak{so}(2,2)$, as it belongs to the special indefinite orthogonal group with respect to a form of signature $(2,2)$ (https://en.wikipedia.org/wiki/Indefinite_orthogonal_group); whereas over the real numbers, $\mathfrak{so}(4)$ is virtually always understood to be the one belonging to the compact group $SO(4)$, the transformations orthogonal with respect to the standard scalar product -- and these are not isomorphic. – Torsten Schoeneberg Feb 05 '20 at 04:50
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    Yes, the entries in my case are meant to be complex (I'm working with algebraically close fields). Thanks for your clarification anyway :) – cip Feb 05 '20 at 07:31
  • @TorstenSchoeneberg Once again I came back to this problem and I encountered a new representation for $\mathfrak{sp}(4)$ which is the set of all $4x4$ matrices $A$ such that $JA = -A^{t}J$, where $J$ is a $4x4$ block matrix, where the upper-right $2x2$ block is the $Id$ (the identity matrix), the lower left block is $-Id$ and the remaining two blocks are null. How would you prove this is an equivalent representation with respect to the one written in the question? – cip Dec 31 '20 at 16:58
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    All skew-symmetric matrices over a field of characteristic $\neq 2$ are congruent, cf. https://math.stackexchange.com/q/1682385/96384, https://math.stackexchange.com/q/2719021/96384, https://math.stackexchange.com/q/388750/96384, and Bourbaki Algebra ch. 9 §5 for a generalisation to commutative rings. In your case, a relatively obvious base change matrix is $\pmatrix{1&0&0&0\0&1&0&0\0&0&0&1\0&0&1&0}$ (you flip the third and the fourth basis vector). For the matrices of the Lie algebra, you then follow the procedure in the links at the end of https://math.stackexchange.com/a/3771729/96384. – Torsten Schoeneberg Dec 31 '20 at 19:36
  • That was very helpful, thank you! – cip Dec 31 '20 at 21:54