exponential map of R3

[enricfemi]

I don not know whether I was right or not, please give me a hint.

(R3,+) can be considered a Lie group. and its TG in 0 is still R3.

suppose X as a infinitesimal generater, it can give a left-invariant vector field and also an one-parameter subgroup.

but i think, this one-parameter subgroup is not exponential map r(t)=exp(Xt). it should be r(t)=Xt

must a 1-parameter subgroup whose tangent vector at 0 is X, have a unique exponential map?
is there any thing wrong?

[arkajad]

One-parameter subgroup generated by vector X is indeed tX.

The generator of this subgroup is the differential operator

\xi_X=\sum_{i=1}^3X^i\frac{\partial }{\partial x^i}

which you can get by differentiating f(x+tX) with respect to t at t=0.

Formally the exponential \exp(t\xi_x) acts on functions by power series (Taylor) expansion.

[enricfemi]

One-parameter subgroup generated by vector X is indeed tX.

The generator of this subgroup is the differential operator

\xi_X=\sum_{i=1}^3X^i\frac{\partial }{\partial x^i}

which you can get by differentiating f(x+tX) with respect to t at t=0.

Formally the exponential \exp(t\xi_x) acts on functions by power series (Taylor) expansion.

Thank you, arkajad! So you mean the 1-parameter subgroup generated by vector field can be not exponential map?

[arkajad]

Thank you, arkajad! So you mean the 1-parameter subgroup generated by vector field can be not exponential map?

It is an exponential. But the point is that what we have is not a matrix group. But, with a standard little trick, we can make it into a matrix group by representing translations as block matrices

\begin{pmatrix}I&a\\0&1\end{pmatrix}

Then it acts on R^3 via

\begin{pmatrix}I&a\\0&1\end{pmatrix}\begin{pmatrix}x\\1\end{pmatrix}=\be gin{pmatrix}x+a\\1\end{pmatrix}

Now the vector field is represented by

X=\begin{pmatrix}0&a\\0&0\end{pmatrix}

Notice that X^n=0 for n>1. Therefore

\exp(tX)=I+tX

and you get what you are looking for.

[enricfemi]

It is an exponential. But the point is that what we have is not a matrix group. But, with a standard little trick, we can make it into a matrix group by representing translations as block matrices

\begin{pmatrix}I&a\\0&1\end{pmatrix}

Then it acts on R^3 via

\begin{pmatrix}I&a\\0&1\end{pmatrix}\begin{pmatrix}x\\1\end{pmatrix}=\be gin{pmatrix}x+a\\1\end{pmatrix}

Now the vector field is represented by

X=\begin{pmatrix}0&a\\0&0\end{pmatrix}

Notice that X^n=0 for n>1. Therefore

\exp(tX)=I+tX

and you get what you are looking for.

thanks, that's a brilliant construct. i think i have finally understood it.