# Question on Lie Algebras

1. Dec 22, 2009

### 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.

Last edited: Dec 23, 2009
2. Dec 22, 2009

### George Jones

Staff Emeritus
Do mean g_j(t = 0) = identity?
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?

3. Dec 23, 2009

### vertices

yes (i have corrected OP)

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

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)

4. Dec 23, 2009

### Fredrik

Staff Emeritus
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.

5. Dec 23, 2009

### vertices

Thanks for the reply.

Okay, I am convinced that:

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

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

$$[g_{2}(t)\xi_{1}g_{2}(t)^{-1}-\xi]{\in}Lie(G)$$

and can someone please explain these statements:

"by smoothness

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

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

Thanks.

6. Dec 23, 2009

### Fredrik

Staff Emeritus
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 $\mathfrak g$.

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