Explicit isomorphism between the four dimensional orthogonal Lie algebra and the direct sum of special linear Lie algebras of dimension 3.

Solution 1:

I guess (edit: corrected)

$\pmatrix{0&a&b&c\\ -a&0&d&e\\ -b&-d&0&f\\ -c&-e&-f&0\\ } $

$\mapsto \pmatrix{i(c-d)&a+ie+ib-f\\ -a+ib+ie+f&i(d-c)} \oplus \pmatrix{i(c+d)&a+ie-ib+f\\ -a-ib+ie-f&i(-c-d)}$

is as explicit as you can get. This works over any field whose characteristic is $\neq 2$ and which contains a square root of $-1$, called $i$ in the formula above.

However, assuming it is correct (I leave it to you to check it's a homomorphism and write down the inverse, hoping that all my signs are correct), this shows mainly that such an explicit matrix formula is virtually useless, and one should rather understand what the theory behind it is.

And this goes like this:

Step 1: Assuming a good split form for $\mathfrak{so}_4$, construct an explicit isomorphism. Let's assume we can show that over our field we have an isomorphic representation of $\mathfrak{so}_4$ not as skew-symmetric matrices, but as matrices

$$A = \pmatrix{a&c&d&0\\e&b&0&-d\\f&0&-b&-c\\0&-f&-e&-a}.$$

Nice thing: The diagonal makes up a Cartan subalgebra. We can see two positive roots operating here, $\alpha_1$ which sends the above $A$ to $a-b$ and whose root space is

$$\pmatrix{0&*&0&0\\0&0&0&0\\0&0&0&*\\0&0&0&0},$$

and $\alpha_2$ which sends the above $A$ to $a+b$ and whose root space is

$$\pmatrix{0&0&*&0\\0&0&0&*\\0&0&0&0\\0&0&0&0}.$$

Knowing what we want and that these two roots are orthogonal to each other, we take the diagonal apart via $\pmatrix{a&0\\0&b}=\dfrac12 \pmatrix{a-b&0\\0&b-a} + \dfrac12 \pmatrix{a+b&0\\0&a+b}$ and get the isomorphism

$$ \pmatrix{a&c&d&0\\e&b&0&-d\\f&0&-b&-c\\0&-f&-e&-a} \mapsto \pmatrix{\dfrac12(a-b)&c\\e&\dfrac12(b-a)} \oplus \pmatrix{\dfrac12(a+b)&d\\f&-\dfrac12(a+b)}$$

onto $\mathfrak{sl}_2 \oplus \mathfrak{sl}_2$ almost for free. Or: Note that the triples $$H_1=\pmatrix{1&0&0&0\\0&-1&0&0\\0&0&1&0\\0&0&0&-1}, X_1=\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&-1\\0&0&0&0} , Y_1=\pmatrix{0&0&0&0\\1&0&0&0\\0&0&0&0\\0&0&-1&0}$$

resp. $$H_2=\pmatrix{1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1}, X_2=\pmatrix{0&0&1&0\\0&0&0&-1\\0&0&0&0\\0&0&0&0} , Y_2=\pmatrix{0&0&0&0\\0&0&0&0\\1&0&0&0\\0&-1&0&0}$$

satisfy the same relations as the standard basis of $\mathfrak{sl}_2$, $$H=\pmatrix{1&0\\0&-1}, X=\pmatrix{0&1\\0&0}, Y=\pmatrix{0&0\\1&0},$$ namely $[H,X]=2X, [H,Y]=-2Y, [X,Y]=H$, and are orthogonal to each other, i.e. $[\ast_1, \ast_2]=0$ for $\ast =H,X,Y$.

Step 2: Base change to that standard split form. Cf. https://math.stackexchange.com/a/3489788/96384. Remember a quadratic form (= symmetric bilinear form) is given by a symmetric $n \times n$-matrix $S$. One can in general define $\mathfrak{so}_S = \{A \in M_n(K): \, ^tA=-SAS^{-1} \}$ and check that is a Lie algebra. Now in general two matrices $S_1, S_2$ might actually describe the same bilinear form, just with respect to different coordinates, i.e. change of basis. Remember that basis change for such forms works by "congruence", i.e. existence of a basis change matrix $P$ such that

$$^tP S_1 P=S_2.$$

Now check that if such a congruence exists, then usual "equivalence" wil define an isomorphism

$$\mathfrak{so}_{S_1} \simeq \mathfrak{so}_{S_2}$$ $$A \mapsto P^{-1}AP$$

(Note: Now it's really the inverse, not the transpose!).

Now you started with the Lie algebra of skew-symmetric matrices which is the base case $S=I_n$. Turns out that written like that, one has a hard time "seeing" a Cartan subalgebra and root spaces in the matrices. So I perform a change of basis. Or rather two: First I want to get from

$S_1 = \pmatrix{1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1} \mapsto S_2 = \pmatrix{1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1}$

i.e. from the quadratic form $w^2+x^2+y^2+z^2$ to $w'^2+x'^2-y'^2-z'^2$. This works in any field with a square root of $-1$ called $i$, namely $w':=w, x':=x, y':=iy, z':=iz$ i.e.

$P_1= \pmatrix{1&0&0&0\\0&1&0&0\\0&0&i&0\\0&0&0&i}.$

Now further I want to go

$S_2 = \pmatrix{1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1} \mapsto S_3 = \frac{1}{2}\pmatrix{0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0}$

i.e. express $w'^2+x'^2-y'^2-z'^2$ as $w''z''+x''y''$. (Originally I tried to remove that factor of $1/2$, but either then it pops up elsewhere, or one has to scale with ugly numbers like $\sqrt 2$, which would not work over $\mathbb Q$, so I just left it there.) This is a standard base change for hyperbolic space, on the coefficients we have

$$w'':= (w'+z'), x'':=(x'+y'), y'':=(x'-y'), z'':=(w'-z')$$

corresponding to

$$e_1 \mapsto \frac12 (e_1+e_4), e_2 \mapsto \frac12 (e_2+e_3), e_3 \mapsto \frac12 (e_2-e_3), e_4 \mapsto \frac12 (e_1-e_4)$$

or

$P_2= \frac12\pmatrix{1&0&0&1\\0&1&1&0\\0&1&-1&0\\1&0&0&-1}$.

Putting all this together one gets

$$ \pmatrix{0&a&b&c\\ -a&0&d&e\\ -b&-d&0&f\\ -c&-e&-f&0\\ } \xrightarrow{P_1^{-1} (\cdot) P_1} \pmatrix{0&a&ib&ic\\ -a&0&id&ie\\ ib&id&0&f\\ ic&ie&-f&0\\ } \xrightarrow{P_2^{-1} (\cdot) P_2} \pmatrix{2ic&a+ie+ib-f&a+ie-ib+f&0\\ -a+ie+ib+f&2id&0&-a-ie+ib-f\\ -a+ie-ib-f&0&-2id&-a-ie-ib+f\\ 0&a-ie+ib+f&a-ie-ib-f&-2ic\\ }$$

Step 3: Combine steps 1 and 2.