Determine the lie algebra of the subgroup of SO(4)

Let $Gsubset SO(4)$be the subgroup given below:

a & -b & -c &-d\
b & a & -d & c\
c & d & a & -b \
d &-c & b &a
end{pmatrix} : a,b,c,din mathbb{R}, a^2+b^2+c^2+d^2=1right}$$

Find the lie algebra $mathfrak{g}$.

I know that if $Xin mathfrak{so}_4$, then $Xin mathfrak{g} iff e^{tX}in G$ $forall$ $tin mathbb{R}$.

However, I am not able to use it here, as the given group is a bit complicated. Any help is appreciated.


Mathematics Asked by Shreedhar Bhat on November 14, 2021

3 Answers

3 Answers

How does one find the Lie algebra of a Lie group? Differentiate the map $$(a,b,c,d) mapsto begin{pmatrix} a & -b & -c & -d \ b & a & -d & c \ c & -d & a & -b \ d & c & b & a end{pmatrix}quad mbox{and the equation}quad a^2+b^2+c^2+d^2=1$$at $(1,0,0,0)$ and evaluate at $(x,y,z,w)$. So $$mathfrak{g} = left{begin{pmatrix} x & -y & -z & -w \ y & x & -w & z \ z & -w & x & y \ w & z & y & x end{pmatrix} mid x,y,z,w in Bbb R mbox{ and }2x+0y+0z+0w = 0 right},$$which of course agrees with Stephen's answer $$mathfrak{g} = left{begin{pmatrix} 0 & -y & -z & -w \ y & 0 & -w & z \ z & -w & 0 & y \ w & z & y & 0 end{pmatrix} mid y,z,w in Bbb R right}.$$It is the general principle that to find the equation of a tangent space to a submanifold, you differentiate the equation defining it. Also, $G cong Bbb S^3$ is isomorphic to the group of unit quaternions, as the general expression for an element of $G$ is in the image of the composition of the maps $$Bbb H ni z+wj mapsto begin{pmatrix} z & -overline{w} \ w & zend{pmatrix} in mathfrak{gl}(2,Bbb C)quadmbox{and}quad Bbb C ni a+bi mapsto begin{pmatrix} a & -b \ b & aend{pmatrix} in mathfrak{gl}(2, Bbb R).$$

Answered by Ivo Terek on November 14, 2021

You are in the following situation. You have a Lie group $H$ (don't want to use the letter $G$ here, since you are using it) and a system of smooth equations $f_1,..., f_n: HtoBbb R$ so that $G:=bigcap_i f^{-1}_i({0})$ is subgroup of $H$. How can you determine the Lie algebra of $G$? Well: $vin T_1H$ is tangent to $G$ iff the equations $f_i$ remain unchanged when you perturb in direction $v$. This means that $vin bigcap_iker( d_1f_i)$, and you get another system of equations $d_1f_i(v)=0$ determining the elements Lie algebra.

In your example you have $13$ equations, on the one hand there are the equations $$g_{11}-g_{22}=0, quad g_{11}-g_{33}=0, quad g_{11}-g_{44}=0\ g_{21}+g_{12}=0,quad g_{21}-g_{43}=0,quad g_{43}+g_{34}=0\ g_{31}+g_{42}=0,quad g_{31}+g_{13}=0, quad g_{31}-g_{24}=0\ g_{41}-g_{32}=0,quad g_{41}+g_{23}=0,quad g_{41}+g_{14}=0$$ as well as the equation $$g_{11}^2+g_{21}^2+g_{31}^2+g_{41}^2-1=0.$$

If you take the differentials at $1$ of this the first equations remain unchanged, since they are linear, but the last equation becomes: $$2v_{11}cdot 1+2v_{21}cdot 0+2v_{31}cdot 0+2v_{41}cdot 0=0.$$

Now you can combine these equations with the additional condition that the Lie algebra elements be anti-symmetric (to get elements of $SO(4)$).

Answered by s.harp on November 14, 2021

It is infinitesimal perturbations from the identity. If $b$, $c$, and $d$ are small, then $a = 1 + O(b^2 + c^2 + d^2)$, so $a$ is constant up to first order. So $$mathfrak g=left{ begin{pmatrix} 0 & -b & -c &-d\ b & 0 & -d & c\ c & d & 0 & -b \ d &-c & b & 0 end{pmatrix} : b,c,din mathbb{R} right} .$$

Answered by Stephen Montgomery-Smith on November 14, 2021

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