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

I am trying to calculate the Lie algebra of the group $SO(2,1)$ where this is defined as:

$SO(2,1=\{X\in Mat_3(\mathbb{R})|X^t\eta X=\eta, \det(X)=1\}$ where $\eta$ is the matrix defined as: $$\left ( \begin{array}{ccc} 1 &0&0\\0&1&0\\0&0&-1\end{array}\right )$$

But I am a bit unsure as how to proceed, 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 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 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 at 16:30

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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