# Calculating the Lie algebra of $SO(2,1)$

I am trying to calculate the Lie algebra of the group $SO(2,1)$, realized as

$$SO(2,1)=\{X\in \operatorname{Mat}_3(\mathbb{R}) \,|\, X^t\eta X=\eta, \det(X)=1\},$$ where $$\eta = \left ( \begin{array}{ccc} 1 &0&0\\0&1&0\\0&0&-1\end{array}\right ) .$$

But I am a bit unsure as how to procee:, I know that I need to take a curve in $SO(2,1)$ that passes through the identity at 0 and then differentiate at 0 but I am unsure as to what the form of curves in $SO(2,1)$ are?

So do I let $a(t)\in SO(2,1)$ be a curve with $a(0)=1$ so that:

$$a'(0)^t\eta+\eta a'(0)=\eta ?$$

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$SO(2,1)$ is isomorphic to $SL(2,\mathbb{R})$, which has a standard basis, c.f. en.wikipedia.org/wiki/Sl2-triple – Neal Jan 9 '14 at 15:59
Your equation should have $0$ on the right hand side since $\eta$ doesn't depend on $t$. Otherwise it's correct. Note that if instead of $\eta$ we would have identity matrix (group $SO(3)$), this equation would be solved by anti-symmetric matrices. – Marek Jan 9 '14 at 16:25
The equation here is then $A = - \eta A^t \eta$. It's quite simple to solve but as a baby step you can try $SO(1,1)$ with $\eta = {\rm diag}(1, -1)$ first. – Marek Jan 9 '14 at 16:30
It should be noted that @Neal's statement above is a statement about Lie algebras, not Lie groups. $\text{SL}_2(\mathbb{R})$ the Lie group is a double cover of (the identity component of) $\text{SO}(2, 1)$ the Lie group. See en.wikipedia.org/wiki/Spin_group for more. – Qiaochu Yuan Apr 23 '15 at 5:25
@QiaochuYuan this was long enough ago that I don't remember whether I was thinking of the isometries of hyperbolic space or simply forgot \mathfrak. In either case, thank you for the erratum! – Neal Apr 23 '15 at 13:17

We identify the Lie algebra $\mathfrak{so}(2, 1)$ with the tangent space $T_1 SO(2, 1)$ to $SO(2, 1)$ at the identity $1$. As hinted in the question, for any $A \in \mathfrak{so}(2, 1)$ we can pick a curve $a: J \to SO(2, 1)$ such that $a'(0) = A$. (We'll see below that the existence of such curves is all we need, i.e., we don't need to write out curves explicitly.) Then, the characterizing equation of $A$ gives $$a(t)^T \eta a(t) = \eta,$$ and differentiating with respect to $t$ gives $$a'(t)^T \eta a(t) + a(t)^T \eta a'(t) = 0$$ (note that the r.h.s. is $0$, not $\eta$). Evaluating at $t = 0$ gives $$\phantom{(\ast)} \qquad A^T \eta + \eta A = 0. \qquad (\ast)$$ (By the way, up to this point we haven't used the form of $\eta$ yet, so this characterization holds just as well for any nondegenerate bilinear form $\eta$ in any finite dimension.)

Now we can write the Lie algebra explicitly simply working out the (linear) conditions determined by the above characterization. This is just a matter of writing out $(\ast)$ in components, but observe that the form of $\eta$ suggests a natural block decomposition of the Lie algebra. (Here this doesn't save so much effort, but this technique is quite useful for computing explicitly Lie algebras $\mathfrak{so}(\eta)$ for bilinear forms $\eta$ on higher-dimensional vector spaces.) Decompose a general element $A \in \mathfrak{so}(2, 1)$ as $$A = \begin{pmatrix} W & x \\ y^T & z \end{pmatrix},$$ where $W \in M(2, \mathbb{R})$, $x, y \in \mathbb{R}^2$, $z \in \mathbb{R}$. In this block decomposition, $\eta = \begin{pmatrix} \mathbb{I}_2 & 0 \\ 0 & -1\end{pmatrix}$ and $(\ast)$ becomes $$\begin{pmatrix} W^T & y \\ x^T & z\end{pmatrix} \begin{pmatrix} \mathbb{I}_2 & 0 \\ 0 & -1\end{pmatrix} + \begin{pmatrix} \mathbb{I}_2 & 0 \\ 0 & -1\end{pmatrix} \begin{pmatrix} W & x \\ y^T & z\end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}.$$

Writing out separately the equation for each block imposes precisely the conditions $$W^T = -W, \qquad y = x, \qquad z = 0,$$ so, \begin{align} \mathfrak{so}(2, 1) &= \left\{ \begin{pmatrix} W & x \\ x^T & 0\end{pmatrix} : W^T = -W \right\} \\ &= \left\{ \begin{pmatrix} 0 & -w & x_1 \\ w & 0 & x_2 \\ x_1 & x_2 & 0 \end{pmatrix} \right\} \textrm{.} \end{align} (Of course, the condition on $W$ is just that $W \in \mathfrak{so}(2, \mathbb{R})$, where the bilinear form on $\mathbb{R}^2$ is just the standard one, i.e., the one with matrix representation $\mathbb{I}_2$.)

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I don't know if what I'm about to write is completely correct, I face the same difficulties with $SU(1,1)$, and this is what I came up with.

First of all because we are working with matrices this means we are embedding $SO(2,1)$ in $GL_{2}(\mathbb{R})$, this is the case if $SO(2,1)$ is a Lie subgroup of $GL_{2}(\mathbb{R})$, a general theorem (stated in almost every introductory texts on Lie groups) assures us that every close subgroup of $GL_{n}(\mathbb{R})$ is a matrix Lie group.

Suppose we proved that $SO(2,1)$ is a a closed subgroup of $GL_{2}(\mathbb{R})$, thus there is an embedding $i$ of $SO(2,1)$ in $i(SO(2,1))\subset GL_{2}(\mathbb{R})$.

This embedding is such that the exponential function $\exp:\mathfrak{so}(2,1)\longrightarrow SO(2,1)$ of $SO(2,1)$ is the usual exponential of a matrix in $i(SO(2,1))\subset GL_{2}(\mathbb{R})$ (this can be proved).

From this it follows that the Lie algebra $i(\mathfrak{so}(2,1))$ of $i(SO(2,1))$ is characterized by the condition:

$$A: exp(tA)\in i(SO(2,1))\;\;\;\forall t$$ therefore from the defining condition of $SO(2,1)$ we get:

$$(exp(tA))^{T}\eta \exp(tA)=\eta$$

Then:

$$\frac{d (exp(tA))^{T}\eta \exp(tA)}{dt}\left|_{t=0}\right.=\frac{d\eta}{dt}\left|_{t=0}\right.$$

Which means:

$$A^{T}\eta + A\eta=0$$

Moreover we can use the general properties of the exponential $\exp$ of a matrix:

$$det(\exp(A))=\exp(Tr(A))$$ in order to obtain:

$$Tr(A)=0$$

Thus the Lie algebra of $i(SO(2,1))$ is the set of matrices such that:

$$Tr(A)=0 \;\;\;\mbox{ and } \;\;\; A^{T}\eta + A\eta=0$$

If you want to use the curves in $SO(2,1)$ you have to find a coordinate system on $SO(2,1)$, in order to explicitely write the curve $a(t)$ and calculate its tangent vectors at the identity.

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