# Is it possible to define the tangent space of a Lie group at points other than the identity?

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?

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

$$X=\left(\begin{array}{ccc} 1 & x & y\\ 0 & 1 & z\\ 0 & 0 & 1 \end{array}\right).$$ 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.