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

Question

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:

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

Answer 1

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. Doing so 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. 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 $\mathbb{I}_2$.)

Answer 2

I'll add another answer to the very nice answers above.

First recall the definition of the Lie algebra of a matrix Lie group:

Let $G$ be a matrix Lie group. The Lie algebra of $G$, denoted $\mathfrak{g}$, is the set of all matrices $X$ such that $e^{tX}$ is in $G$ for all real numbers $t$.That is, the Lie algebra of a matrix Lie group $G$ is $ \mathfrak{g}=\{ X\in M_n(\mathbb{C}) : ~ e^{tX} \in G, ~\text{for all}~ t\in \mathbb{R} \} \subseteq M_n(\mathbb{C}). $

We're interested in finding the Lie algebra of the generalized special orthogonal group $\mathrm{SO}(n;k)$, which is defined as follows: $ \mathrm{SO}(n;k) = \{ A \in M_{n+k}(\mathbb{R}) ~|~A^{tr} g A = g ~~\text{and}~ \det(A)=1 \}, $ where $ g=\left( \begin{array}{cccc} I_n & 0 \\ 0 & -I_k \end{array} \right). $

One can prove without too much trouble that if $G$ is a subgroup of $\mathrm{GL}(n, \mathbb{R})$, then the Lie algebra of $G$ must consist entirely of real matrices. With this in mind, we'll compute the Lie algebra in question in the following proposition.

Proposition. The Lie algebra of the real symplectic group $\mathrm{SO}(n; k)$ is given by $ \{ X\in M_{n+k}(\mathbb{R}): ~g X^{tr} g = -X ~~\text{and}~\mathrm{trace}( X)=0\}.$ It is often called the special generalized orthogonal algebra and denoted $\mathfrak{so}(n;k)$.

Proof. Let $X$ be in $\mathfrak{so}(n;k)$ (as noted above, $X$ must be a real matrix). Then, $e^{tX}$ is in $\mathrm{SO}(n;k)$ for all real $t$ so that

$ \begin{array}{llllllllll} & (e^{tX})^{tr} g e^{tX} &=&g,\\ \text{which implies} & (e^{tX^{tr}}) g e^{tX} &=&g,\\ \text{which implies} & g^{-1} (e^{tX^{tr}}) g e^{tX} &=&I,\\ \text{which implies} & (e^{t g^{-1}X^{tr}g}) e^{tX} &=&I,\\ \text{which implies} & (e^{t g^{-1}X^{tr}g}) &=&e^{-tX}. \end{array} $

Differentiating the last equation at $t=0$ yields $g^{-1} X^{tr} g = -X$. This also implies $\mathrm{trace}(X)=\mathrm{trace}(-X)$, so that $\mathrm{trace}(X)=0$. Further, since $g^{-1} =g$, we have $gX^{tr} g = -X$. Hence, $\mathfrak{so}(n;k)\subseteq \{ X\in M_{n+k}(\mathbb{R}): ~g X^{tr} g = -X ~~\text{and}~\mathrm{trace}( X)=0\ \}$.

Now let $X$ be a real $(n+k) \times (n+k)$ traceless matrix such that $g X^{tr} g = -X$. Since $\mathrm{trace}(X)=0$, then, $\det(e^{tX})=e^{t\cdot\mathrm{trace}(X)}=1$. And, since $g^{-1}=g$, $g^{-1} X^{tr} g = -X$ and for all real $t$ we have

$ \begin{array}{llllllllll} g^{-1}(e^{tX} )^{tr} g e^{tX} & =&e^{tg^{-1}X^{tr} g} e^{tX} \\ &=& e^{-tX}e^{tX} \\ &=&I, %& (e^{tX})^{tr} J e^{tX} &=&J,\\ %\text{if and only if} & (e^{tX^{t}}) J e^{tX} &=&J,\\ %\text{if and only if} & J^{-1} (e^{tX^{t}}) J e^{tX} &=&I,\\ %\text{if and only if} & (e^{t J^{-1}X^{t}J}) e^{tX} &=&I,\\ %\text{if and only if} & (e^{t J X^{t}J}) &=&e^{-tX}. \end{array} $

so that $(e^{tX} )^{tr} g e^{tX} =g$, for all real $t$ so that $X\in \mathfrak{so}(n;k)$. This implies $ \{ X\in M_{n+k}(\mathbb{R}): ~g X^{tr} g = -X~~\text{and}~\mathrm{trace}( X)=0\ \}\subseteq \mathfrak{so}(n;k)$, and hence $ \{ X\in M_{n+k}(\mathbb{R}): ~g X^{tr} g = -X ~~\text{and}~\mathrm{trace}( X)=0\ \}= \mathfrak{so}(n;k)$. $\Box$

Answer 3

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.