Regarding Lie algebras

  • Thread starter Kontilera
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Hello!

I thought I had a good picture of this stuff but now I suspect that I've mixed up things completely..
For su(2) we have:
[tex]f_{ab}\,^c = \epsilon_{ab}\,^c.[/tex]
Which means that, for the basis of our tangentspace we get:
[tex][\partial_a, \partial_b] = i \epsilon_{ab}\,^c\, \partial_c[/tex]
Where we can see that the space of linear approximations is closed under the Lie bracket product.

But what about the definition of a smooth manifold being locally euclidean?
For R^n we know that our orthogonal partial derivatives commute,
[tex]\partial_x\partial_y - \partial_y\partial_x = 0.[/tex]

Shouldnt we then expect the orthogonal derivatives to commute in su(2) as well?
This would mean that our structure constants are 0 both when a=b and when a is not equalt to b (for a orthonomal basis).
In other words.. it looks like the Lie algebra should be quite trivial.

This is of course not the case, but where does my mind make the mistake in this resoning?

Thanks!!
// Kontilera
 

fresh_42

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The Lie algebra elements are tangents along a path in the group. That means at its endpoint we are in a different tangent space albeit still isomorphic to ##\mathbb{R}^n##. But ##\partial_a\, , \,\partial_b## have three tangent spaces they arose from: starting point for ##a##, endpoint = starting point for ##b##, and endpoint of ##b##. The commutator measures how the endpoints of this path ##b\circ a## differs from that of ##a \circ a##. Since we have different tangent spaces at different points, we cannot expect that all this happens within one tangent space of the group. This is the difference between local and global behavior. Within a single ##\mathbb{R}^n## the paths ##b\circ a## and ##a\circ b## lead to the same endpoint, with changing tangent spaces this is no longer guaranteed.

Have a look on these insights:
 

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