3
$\begingroup$

Does a Lie group have a unique Lie algebra for each of its points? And if so, can the exponential map be defined in a similar way?

$\endgroup$
  • $\begingroup$ "Lie algebra for each of its points" is senseless... $\endgroup$ – YCor Aug 14 '16 at 21:07
3
$\begingroup$

I'm not sure I got what you meant. The Lie Algebra of a Lie group is identified with the tangent space at the identity, anyway there's a canonical way to find a basis of the tangent space in anypont exploiting the Left multiplication which is an operation that every Lie Group has. Let's make an example, let take the Heisenberg group with generic element

$$X=\left(\begin{array}{ccc} 1 & x & y\\ 0 & 1 & z\\ 0 & 0 & 1 \end{array}\right).$$ Now You have to calculate the left action which is easily found$$L_{A}X=\left(\begin{array}{ccc} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{array}\right)\left(\begin{array}{ccc} 1 & x & y\\ 0 & 1 & z\\ 0 & 0 & 1 \end{array}\right)=\left(\begin{array}{ccc} 1 & x+a & y+az+b\\ 0 & 1 & z+c\\ 0 & 0 & 1 \end{array}\right).$$ This action $L_{A}$ brings the identity $I$ into the element $A$ , i.e. $L_{A}(I)=A$ . So you can use the differential of the application $L_{A*}$ to send the tangent plane in the identity (i.e. your Lie Algebra) to the tangent plane in the point A .

First of all you have to calculate the differential of the application$$L_{\left(a,\,b,\,c\right)}\left(x,\,y,\,z\right)=\left(x+a,\,y+az+b,\,z+c\right)$$ so you just have to differentiate to obtain the differential$$L_{\left(a,\,b,\,c\right)*}=\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & a\\ 0 & 0 & 1 \end{array}\right).$$ Here I've used a shortcut to represent the element $$\left(\begin{array}{ccc} 1 & x & y\\ 0 & 1 & z\\ 0 & 0 & 1 \end{array}\right)\rightarrow\left(\begin{array}{c} x\\ y\\ z \end{array}\right)$$ otherwise You should have done the long way. Now we have the differential we take a base at the identity, i.e.

$$E_{1}=\left(\begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right),\,\,E_{2}=\left(\begin{array}{ccc} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right),\,\,E_{3}=\left(\begin{array}{ccc} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{array}\right).$$ with the ususal commutation relation$$\left[E_{1},E_{2}\right]=\left[E_{2},E_{3}\right]=0,\,\,\left[E_{1},E_{3}\right]=E_{2}.$$ Then you define a base on the point you want, namely $A$ as$$\left(E_{i}\right)_{A}=L_{A*}\left(E_{i}\right)$$ which is $$\left(E_{1}\right)_{A}=\left(\begin{array}{ccc} 0 & a & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right),\,\,\left(E_{2}\right)_{A}=\left(\begin{array}{ccc} 0 & 0 & a\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right),\,\,\left(E_{3}\right)_{A}=\left(\begin{array}{ccc} 0 & 0 & 0\\ 0 & 0 & a\\ 0 & 0 & 0 \end{array}\right).$$ If you want to define some commutation relation you'll see that$$\left[E_{1},E_{2}\right]=\left[E_{2},E_{3}\right]=0,\,\,\left[E_{1},E_{3}\right]=a^{2}E_{2}$$. I then don't know what you want to do with this base in the tangent space, anyway here you have it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.