I am wondering why all elements could be generated by infinitesimal elements.


Here is the explanation I found from Joshi, A. W. (1997). Elements of group theory for physicists.

Lie groups definition
The dependence of the group elements $g_1,g_2$ etc. of a topological group $G$ on its $r$ continuous parameters can be written explicitly as $g_1 =g_1(a_1,a_2...,a_r),g_2 =g_2(b_1,b_2...,b_r)$,etc. Let $g_1*g_2=g3(c_1,c_2,...,C_r)$ and $g_1^{-1}=g_4(d_1,d_2,...,d_r)$. The parameters of $g_3$ and $g_4$ can be expressed as functions of the parameters of $g_1$ and $g_2$, that is, $$ c_j = c_j(a_1,a_2...,a_r;b_1,b_2...,b_r); d_j = d_j(a_1,a_2...,a_r), $$ for $1 \leqslant i \leqslant r$. A topological group is called an $r$-dimensional Lie group if there exists a neighborhood $\Omega$ of the identity element $e$ such that the continuous parameters of the product of two elements and those of the inverse of an element in $\Omega$ are continuous differentiable functions of the parameters of the elements, that is if $c_j$ and $d_j$ are analytic functions of $a_j$ and $b_j$ for elements in $\Omega$ provided that $g_3$ and $g_4$ lie in $\Omega$ when $g_1$ and $g_2$ do.

By the successive application of the product rule, we can arrive at an element of the group a finite distance away from the identity. Thus, suppose we wish to generate the element $x(0,0,...,0_j,...0)$. Let us write $a_j=N\varepsilon_j$, where N is a large positive integer so that $\varepsilon_j$ is a small quantity. Then $$ \mathbf{x(0,0,...,a_j,...,0)= [x(0, 0, . .. , \varepsilon_j, ... , 0)]^N} =[e+i \varepsilon_j I_j]^N =[e+i(a_j/N)lj]^N. $$


I think the key point is why I could use $\mathbf{x(0,0,...,a_j,...,0)= [x(0, 0, . .., \varepsilon_j,...,0)]^N}$ in Lie groups, this doesn't look like approximation at all. It seems we are assuming the differential structure of the Lie group is totally linear, like straight Euclid space.
My question is:
Is this some intrinsic property of Lie groups or is it based on some other hypothesis?

  • 1
    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Feb 20, 2022 at 6:45
  • $\begingroup$ I am not looking for rigorous proof, just wanna know how the physicist treats this issue. $\endgroup$
    – Jack
    Feb 20, 2022 at 6:56
  • 1
    $\begingroup$ You cannot generate reflection element of orthogonal group from infinitesimal elements, so the statement you are trying to prove must be modified. $\endgroup$
    – A.V.S.
    Feb 20, 2022 at 7:22

1 Answer 1


Saying that "adding $N$ infinitesimal elements gives us a finite element" is a very simplified and arguably wrong picture. What you're actually supposed to do to reach a finite element $g\in G$ is to walk along a vector field in your group that flows the identity to $g$.

What then happens is that, because of the group structure, you indeed can move in a "straight line", as it were, if you restrict yourself to left-invariant (interpret as "homogeneous") vector fields: if $X_h$ is a vector field on $G$ defined for every point $h$, and $L_g:h\mapsto gh$ then $X$ is called left-invariant if $dL_g(X_h)=X_{gh}.$ In this case, integrating the flow is trivial, and basically amounts to "adding up" a bunch of identical contributions, and is called $\gamma_X(t)=\exp(tX)$, and is basically what you did there by naively assuming you could. Your argument breaks down as soon as you start mixing coordinates, because then finding $X$ is in principle a non-trivial task.

Which brings me to my final point: the (not-so-)hard part is now showing that there is such a vector field for every point in the group. This turns out to be true at least for the identity's connected component.

  • $\begingroup$ Could you give me an example in which the group structure is not a ` straight line` but a more complex structure? $\endgroup$
    – Jack
    Feb 20, 2022 at 6:53
  • $\begingroup$ @Jack You can always move along a "straight line"- an exponential, it's just that the flow you have to follow is not obvious for most group elements. $\endgroup$ Feb 20, 2022 at 17:20
  • $\begingroup$ @Jack one way to go about this is to write an identity connected component as a product of some exponentials, $g=\exp(X_1)\exp(X_2)\dots\exp(X_n)$ and then use the BCH formula to collapse it all into a single exponential. This gives you the corresponding "straight line" $\endgroup$ Feb 20, 2022 at 17:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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