Exploring Lie Groups in $\mathbb{R}^3$

[cristo]

$\mathbb{R}^3$ has an associative multiplication $\mu:\mathbb{R}^3\times \mathbb{R}^3 \rightarrow \mathbb{R}^3$ given by $$\mu((x,y,z),(x',y',z'))=(x+x', y+y', z+z'+xy'-yx')$$

Determine an identity and inverse so that this forms a Lie group.

Well, clearly e=(0,0,0) and the inverse element is (-x,-y,-z)

Pick a basis for the Lie algebra of this Lie group and calculate their commutators, obtaining the structure constants of the Lie algebra.

This is the part I'm having trouble with. I'm not really sure where to start! Any help would be much appreciated!

 

[matt grime]

Well, the lie algebra is R^3, as a vector space. What is the definition of the commutator?

 

[cristo]

The commutator of two vector fields u,v in R^3 is [u,v]=u(v)-v(u). If the Lie algebra is R^3 as a vector space, then can we not pick the basis $$\frac{\partial}{\partial x^i}$$ i=1,2,3?

I guess not, since then the commutators will be all be zero.

 

[matt grime]

Why are the commutators zero?

 

[cristo]

Well, $$\left[\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right]=\frac{\partial}{\partial x}\frac{\partial}{\partial y}-\frac{\partial}{\partial y}\frac{\partial}{\partial x}=0$$

Is that not right?

 

[matt grime]

No. If that were right, then surely the commutator of any two elements in any lie algebra would be zero.

I said the algebra was R^3 as a vector space. I didn't say it was isomoprhic to R^3 with the trivial relations as an algebra.

 

[cristo]

Ok, so the algebra is R^3 as a vector space, with basis $$\frac{\partial}{\partial x^i}$$ i=1,2,3.

I'm really not sure about how to calculate these commutators though. I don't see how the commutator above isn't zero.

Sorry, I'm probably missing something basic here. Thanks for your patience!

 

[matt grime]

I didn't say it was or wasn't zero. I just think you should look in your notes to see how to work out the bracket.

Every (real) lie algebra is isomorphic to R^n for some n. So picking a basis we can always choose d/dx^i. But this can't just mean that the commutator is zero. The commutator depends on the Lie structure.

Find an example in your notes where they work out the commutators.

 

[cristo]

I didn't say it was or wasn't zero. I just think you should look in your notes to see how to work out the bracket.

Ok, well if we had two general vector fields, say, $$v=v^i\frac{\partial}{\partial x^i}, w=w^j\frac{\partial}{\partial x^j}$$, then the commutator would be $$[v,w]=v^i\frac{\partial}{\partial x^i}\left(w^j\frac{\partial}{\partial x^j}\right)-w^j\frac{\partial}{\partial x^j}\left(v^i\frac{\partial}{\partial x^i}\right)=\left(v^i\frac{\partial w^j}{\partial x^i}-w^i\frac{\partial v^j}{\partial x^i}\right) \frac{\partial}{\partial x^j}$$ since I was taught that d/dx^i differentiating d/dx^j is zero.

Every (real) lie algebra is isomorphic to R^n for some n. So picking a basis we can always choose d/dx^i. But this can't just mean that the commutator is zero. The commutator depends on the Lie structure.

Does this have something to do with the mapping mu?

Find an example in your notes where they work out the commutators.

There's only one example in my notes, and it's different to the one here. He calculates an effective action first, then calculates "velocity vector fields of one-parameter families of transformations." From here he manages to find a basis for the LIe Algebra.

 

[matt grime]

Try to adapt the proof to this case.

 

[HallsofIvy]

Ok, well if we had two general vector fields, say, $$v=v^i\frac{\partial}{\partial x^i}, w=w^j\frac{\partial}{\partial x^j}$$, then the commutator would be $$[v,w]=v^i\frac{\partial}{\partial x^i}\left(w^j\frac{\partial}{\partial x^j}\right)-w^j\frac{\partial}{\partial x^j}\left(v^i\frac{\partial}{\partial x^i}\right)=\left(v^i\frac{\partial w^j}{\partial x^i}-w^i\frac{\partial v^j}{\partial x^i}\right) \frac{\partial}{\partial x^j}$$ since I was taught that d/dx^i differentiating d/dx^j is zero.

You were taught what?
$$\frac{\partial}{\partial x^i}\left(\frac{\partial}{\partial x^j}= \frac{\partial^2}{\partial x^i \partial x^j}$$
Of course, that's not a derivative, it's a second derivative which is why we use the commutator as the product for a Lie Algebra.

If $\alpha= w(x,y)\frac{\partial}{\partial x}+ v(x,y)\frac{\partial}{\partial y}$ and $\beta= p(x,y)\frac{\partial}{\partial x}+ q(x,y)\frac{\partial}{\partial y}$ then $\alpha\beta$ will involve second derivatives, not just first derivatives. $\beta\alpha$ will also but since mixed second derivatives are equal, $[\alpha,\beta]= \alpha\beta- \beta\alpha$ will involve only first deriatives.

Does this have something to do with the mapping mu?



There's only one example in my notes, and it's different to the one here. He calculates an effective action first, then calculates "velocity vector fields of one-parameter families of transformations." From here he manages to find a basis for the LIe Algebra.

 

[matt grime]

The lie algebra is the space of "left invariant vector fields". The left invariance is intimately related to the product \mu.

http://planetmath.org/encyclopedia/LieGroup.html

 

[cristo]

You were taught what?

Sorry, what I said wasn't true. I remember my lecturer saying that [d/dx,d/xy]=0; not that the d/dx operating on d/dy was zero. He didn't really explain why, this is true, but of course one can still think of d/dx as partial derivative operators (as well as basis vectors) and so these obviously commute.

Thanks for pointing that out!

Try to adapt the proof to this case.

Well, here's the problem in the lecture notes. G=R x (R\{0}); $\mu$((x,y),(x',y'))=(x+yx',yy'). He finds the following effective action: $\lambda$((x,y),p)=x+yp.

However, I can't seem to find an action for my mapping \mu above; it's not obvious like in his example!