How do transformations in group G relate to curves in Lie algebras?

[vertices]

Let:

g_{j}(t) be a curve in a group G, which goes through the identity element; g_j(t=0) = identity.

and:

\xi_{j}=\frac{d}{dt}g_{j}(t)\right|_{t=0}

We know that:

\xi_{j}{\in}Lie(G)

Why can we say:

1) hg(t)h^{-1} (h is an element of the Group)

is also a curve in the group, which goes through the identity element, ie. g(t=0)=identity? [As an aside - how would you even go about doing this transformation - I mean if g(t) is a curve (for example g(t)=2t+4t^3), how can you combine this function with h and h^-1, which are, say, SU(2) matrices?]

2) g_{2}(t)\xi_{1}g_{2}(t)^{-1}{\in}Lie(G)?

I mean, these look a bit like similarity transformations - can someone clarify why these statements are true?

Thanks.

 

[George Jones]

Let:

g_{j}(t) be a curve in a group G, which goes through the identity element, ie. g(t=0)=identity

Do mean g_j(t = 0) = identity?

I mean if g(t) is a curve (for example g(t)=2t+4t^3), how can you combine this function with h and h^-1, which are, say, SU(2) matrices?]

As a function, the curve g_j in the group G maps what set to what set?

As a function, g(t) = 2t+4t^3 maps what set to what set?

 

[vertices]

Do mean g_j(t = 0) = identity?

yes (i have corrected OP)

As a function, the curve g_j in the group G maps what set to what set?

I am not sure if g_j is a map as such. In fact the definition of g_j(t) is very loose - it is any curve that simply goes through the identity of the group.

As a function, g(t) = 2t+4t^3 maps what set to what set?

Well, from parameter t to the function g(t), but the parameter is arbitrary, it doesn't matter what it is (with the exception of the constraint mentioned in the OP that d/dt og it evaluated at t=0 must be the tangent space to G at the identity)

 

[Fredrik]

A curve in a set S is a function from an interval of the real numbers into S, so if g is a curve in a Lie group G, then so is t\mapsto hg(t)h^{-1} (since g(t) and h are both members of G).

An expression like gAg^{-1} where g is a member of the Lie group and A is a member of the Lie algebra clearly makes sense if we're dealing with a group of matrices (because then g and A are both matrices). If the Lie group isn't a matrix Lie group, then we can still make sense of it, by defining that notation to mean

\lambda_g_*\rho_{g^{-1}}_*A

where \lambda_g is left multiplication by g and \rho_{g^{-1}} is right multiplication by g^{-1}. Look up "pushforward" or "push-forward" if you don't know what the * means.

 

[vertices]

Thanks for the reply.

Okay, I am convinced that:

<br /> g_{2}(t)\xi_{1}g_{2}(t)^{-1}{\in}Lie(G)<br />

May I ask a few more questions: why is it also the case that:

<br /> [g_{2}(t)\xi_{1}g_{2}(t)^{-1}-\xi]{\in}Lie(G)<br />

and can someone please explain these statements:

"by smoothness

<br /> \underbrace{lim}_{t\rightarrow0}\frac{1}{t}[g_{2}(t)\xi_{1}g_{2}(t)^{-1}-\xi]{\in}Lie(G)<br />

this implies that [\xi_{1},\xi_{2}]{\in}Lie(G)"

Thanks.

 

[Fredrik]

Because a Lie algebra is a vector space, and vector spaces are always closed under addition. If x and y are members of a vector space V, then so is x-y=x+(-y).

What does the underbrace thing mean? Is it just a limit or something else? And what does "this" refer to? Ah, is it a new sentence, so it refers to the sentence before it? An uppercase T would have helped. I think they mean that the smoothness of the function

\phi=\lambda_{g_2(t)}\circ\rho_{g_2(t)^{-1}}

implies that its pushforward is a map from T_eG into T_{\phi(e)}G=T_eG, i.e. from \mathfrak g into \mathfrak g.